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What is the difference between an industrial and a laboratory production method?
I'm not sure if this is the correct SE in order to ask this, if that is true just said me that in comments and I will delete my question, there is no need of downvote or ban.
I was answering a question of World Building SE about how to make an atmosphere and I found that nitrous oxide can be used to make one. But then, I started reading how can I produce nitrous oxide (to post that information also) and I found with this on Wikipedia.
It talks about an industrial method and a laboratory method. Which is the difference?
I have the theory that a laboratory method can't be used at big scale. Is that true? But, How much is "big"? Maybe for a fabric it's too little the laboratory production (a few kilograms per minute per square meter of laboratoy???) but for me that is a lot, 1kg per minute is arround a cubic meter of air per minute (60 minutes per hour, 24 hours per day...)
P.S: Sorry for the bad tag. I don't have no idea of what to write.

Long story short, laboratory methods are too expensive for industry, and industrial methods are too hardcore for a lab.

This answer will be separate in few parts. In the first one I will just make a review of the scope between lab an industrial scale. In the other parts, you’ll find some things more deeply explained and exemplified. In any case, if something isn’t clear or the English grammar is wrong, please ask me a question and point it (the English mistake) out.

Part I - Short Overview
A laboratory method can perfectly be adapted at industrial scale however there are many things to take in consideration.
Let say you are working in a lab to perform the synthesis of a molecule. In the basics we will assume you will need a reactor (round-bottom flask), and a distillation column to separate your compound from the others (solvent, remaining reactants, possible side products).
If you do all that stuff at a laboratory scale, even if your reaction is super exothermic even involving very hazardous chemicals, you'll always have the possibility to do it safely or at least having the possibility to stop the reaction by cooling it down rapidly.
I would mention at this point that the ratio between the volume and the surface of an object isn't constant. For example, if you multiply by 10 the length of the sides of a cube, its volume rises by a factor 1000 whereas its surface by a factor 100. You can find more interesting information here.
So, if you make your reaction at laboratory scale, let say under few litre, even if it might be dangerous, you’ll control easily and readily the temperature inside your flask or small reactor.  So, at industrial scale, let say over 1 cube meter (in general I would say 5 to 50 cube meters is the average, but some reactors are bigger than 1000 cube meter), you must take in consideration safety and security details as well as the sizing of your operation. I recommend you to read here a really nice overview about most of the kinds of the reactors used in chemical plants with many details related to them.
Now I bet you start to understand the big difference. But let me go a bit deeper on that part. If you want to size a chemical operation, you must first take in consideration the hazards of all of your chemicals as well as their fundamental properties (ebullition temperature, state at a given temperature (gas, liquid, solid, supercritical fluid ,  …), the possible interactions between to compounds or more.
After that you’ll need to know about the kinetics of your reaction, is it fast or slow?, you’ll also need to know about the thermodynamics of all the chemicals to make a prediction for safety (see the next part) and to complete you’ll have to take in consideration if you have a system with only liquids, with gas and liquids or even harder with gas, liquid and solid (hydrogenation (using hydrogen gas) of a liquid organic molecule (let say an oil) catalysed by a given catalysor (a solid) (see the third part).
For years now engineers and researchers, developed methods to compute thermodynamic and kinetic properties as well as many different methods for sizing many different basic or very complex operation (distillation, crystallization, multiphase reactors, drying, precipitation, …) (and I’ll give when I’ll have time as many details as I can in a fourth part about that point). So now it exits a lot of extrapolations and models made from the experiments because of the too complex existing situations to make nice calculus involving mathematics and physics learned at school. This article gives you an idea of all the possibilities of having nightmare while sizing an operation at industrial scale.

Part II - Predicting the safety of a reaction
The Semenov diagrams
So, to give some concrete applications of what I said before, I will show you a very basic thing that safety engineers will do to know more about the danger of the reaction. They will use a Semenov-diagram (or plot).
The first idea is to consider your reaction having a zero-order kinetics, i.e. the kinetics doesn’t involve the concentration of the reactants. It’s a big assumption but it gives you really a good idea of how dangerous it can be. The second idea is, as your reactor will be quite big, to consider it as adiabatic. It’s not true at all of course because you wouldn’t be able to heat or cool it, but it allows you to know what could happen in the worst case.
Doing an energy balance on your system you’ll get:
$$Q_{acc} = k_0 \cdot C_0 \cdot \exp\left(\frac{-E_a}{RT}\right)\cdot \left(-\Delta_r H\right)-UA\cdot (T_r-T_c)$$
Where $Q$ represent the amount of heat generated by the reaction, $k_0$ is the pre-exponential factor of the Arrhenius’s law, $C$ is the concentration, $E_a$ is the activation energy, $R$ is the constant of ideal gases, $U$ is the global exchange coefficient of your reactor and $A$ the area of heat exchange.
Now if you plot this equation which tells you the amount of energy liberate by the whole process at a given temperature with the plot only $UA\cdot (T_r-T_c)$, you’ll get something like that,

reference
If you consider the blue dotted line crossing the red plot, the blue curve tells you about how much heat you can evacuate and the red one tells you about what the whole process liberates. The point A is the stable point as even if you go under in temperature your reaction will still stay in control, however the B point is the point from which you have instability. It’s clear to understand than between these two points you can control the temperature of your reactor.
Keep in mind the heat you can evacuate is a linear function of the temperature $UA\cdot (T_r-T_c)$ and the heat produced by the whole operation is exponential with the temperature as shown by Arrhenius.
Predict the dangerous nature of a reaction
Another thing you can do easily is to predict the kind of danger of your reaction (not dangerous, a little dangerous, dangerous, extremely dangerous) using the worst case (adiabatic reactor).
Then you have rises of the temperature equal to the amount of heat produced by the reaction. Then you have,
$$C_p \cdot \Delta T_{adiab} = [reactants]_0 \cdot V \cdot (- \Delta_r H)$$
Where $C_p$ is the heat capacity of your mixture, $[A]_0$ means the concentration of A at the beginning (you want to know what happens if all of it reacts), and V is the volume of the mixture.
You can then predict the variation of temperature knowing all the needed parameters. I would say if the rising of temperature is below 50 °C your reaction is quite safe; if you find something more than 100 °C you’ll have to find either another solution (a different reactor, …) or add a lot of security systems (emergency open valves, a possibility to almost instantly flush your reactor content in a inhibitor mixture under the reactor, …).

Part III - What if I have a multiphase system?
Well this is the case most of the time. It’s not common to have a system in which you only have soluble fluids (except for gases). If you are performing a treatment of your affluents or your effluents (because you can’t use row compounds and you can’t throw away your potential toxic, corrosive, … mixture in the nature) you will often have an exchange between a gas and a liquid. If you are doing a reaction involving a catalysor then you will probably have three phases if it’s a hydrogenation or a fluidized-bed reactor for a cracking operation or something else. You’ll have only one phase for vapocracking, polymerization (not in emulsion and relatives) and urea production for example.
Now in these cases you have to take in consideration of possibility of having an exchange of chemicals (the ones you want at least) between the two phases or three phases. If you have a liquid/gas system, the situation is common and even if a bit hard sometime you can get good approximation and correlation using the two films theory see that I wasn’t really confident with it some time ago.
If you use a catalysor, you have to predict the possibility of the species to go inside it so that they can react otherwise nothing will happen. Then you have to understand how your catalyst works and how to avoid the possibility of blocking it or poisoning it [ref] as in most of the cases catalysors are like sponge solids. I found two articles pdf1 and pdf2 if you want to got deeper on how complex it is.

So I hope this very long answer help you more than you need. I will also use it as a reference for some more questions in the future. You may say that all these parameters can also be taken in consideration even at a lab scale. I will anwser yes in the sense that you will never have all the data you want and you'll have to do experiment to extrapolate back to the industrial scale. But I will moderate that if you are only doing stuff in a lab scale it doesn't cost the same as when you are involving tons of chemicals and it's way more secure so even if you are wrong in something you can afford it much more easily.

I will point out that in most of the cases the chemical plants are
  sized as the lab plants a company may have meaning that the lab plants
  are made by extrapolating down the data (size of the reactors, ratios
  of length especially, ...) so that when you do your experiment in the
  lab you can then extrapolate back using your own models and then be
  more close to the reality with really tiny variations.


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Mechanism of heterocyclic ring formation
I saw a question here about the mechanism for a heterocyclic ring formation: What is a reasonable mechanism for the following reaction?. I tried solving the mechanism but I am missing something. Let me explain my reasoning:
This is the reaction:

I started protonating the nitrile N, followed by the attack of the cyclic N to the terminal alkyne carbon. Then I couldn't figure out what happens next, so I assumed the triple bond attacks a proton and forms the acrylonitrile intermediate (I am 99.99% sure this is not what happens). Then the exo N attacks the nitrile carbon and form the 6-membered ring, followed by a proton exchange between the imino nitrogen and the protonated cyclic amino group (again, I feel this is ridiculously wrong). The lone pair on the cyclic amino group form the double C=N bond, however, I still end with a protonated N that is not in the product.

Can someone explain to me what I am missing?

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Does a benzene ring continue to oscillate at absolute zero?
The carbon bonds in a benzene molecule oscillate between single and double bonds. As they do so, the distances between the carbon atoms changes (as the attached animated GIF illustrates). At absolute zero all molecular motion ceases; that is, the molecules stop bouncing into each other, but does that mean that molecules like benzene also cease their internal molecular motion?

https://chemistry.stackexchange.com/questions/51626/what-is-the-correct-way-to-verify-a-structures-geometry-for-example-for-benzen https://chemistry.stackexchange.com/questions/50005/does-benzene-structure-stand-for-a-single-resonance-form-or-the-whole-molecule https://chemistry.stackexchange.com/questions/31352/is-the-resonance-depiction-of-benzene-typically-portrayed-incorrectly

They don't oscillate like that even at room temperature. Instead, they kinda stay halfway between. The picture does not illustrate what you think it does.

The distances between the carbon atoms certainly change, but the bonds do not oscillate between single and double bonds...

This is not a duplicate, because the question conflates resonance structures with molecular motion and vibration.

"At absolute zero all molecular motion ceases; that is, the molecules stop bouncing into each other"
This statement is a common misconception about absolute zero and is totally false. At absolute zero, we are simply in the lowest possible energy state. That does not mean that molecular motion ceases. That would violate the uncertainty principle.

To be more specific, molecules still have zero-point (vibrational) energy at absolute zero. One should not confuse molecular (nuclear) motion with electronic motion, which is even less well-defined due to electron positions being probability distributions.

When I said, "...all molecular motion ceases," I tried to qualify that by adding, "...molecules stop bouncing into each other." If the last statement is false, some clarification would be appreciated. If there is a term for motion within an atom and motion as one atom relates to a neighboring atom, I am not aware of it or have forgotten it.

@BillDOe He (at 1atm) is not even a solid in the limit of absolute zero.  https://physics.stackexchange.com/questions/274910/why-doesnt-helium-freeze-at-0k It remains liquid, so the atoms are certainly bouncing into each other.

Thanks. I'm obviously a victim of that common misconception.

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What are the metamers of penta-1,2-diene?
First of all, are there any metamers for dienes? If there are any, how can I find them?

Yes. Consider metamer to be a different word for isomer, as noted by Lighthart:

The term metamer is equivalent to isomer...As an aside, I have never heard the term metamer before, I expect it is archaic.

And, according to Dictionary.com, consider a diene to be:

a hydrocarbon that contains two carbon-to-carbon double bonds in its molecules.

There are too many diene isomers to be listed here ($\ce{C5H8}$ has many atoms), but you can begin by changing the numbers in the name to generate different Lewis structures.
Penta-1,2-diene:

Penta-1,4-diene:

This is just one structure. Try cis vs. trans, position isomers, etc. You may think of many more.


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What happens to the chlorine in epichlorohydrin when a two-part epoxy is air-cured?
Where does the chlorine go? Does it evaporate as a by-product or does it remain in the epoxy? I'm using a two-part resin and hardener and curing them at room temp. The resin is bisphenol-A-epichlorohydrin, and I believe the final cured epoxy does not have chlorine as a part of its main structure. I need to know the whereabouts of Cl because I'm trying to study neutron interaction with materials in my detector, and I suspect there's some chlorine in there but not sure entirely.

There is little if any chlorine in a epichlorohydrin / bis phenol A expoxy resin, it is normally lost during production as sodium chloride into the aqueous phase while the resin forms an organic phase.
The normal way to make the resin is to react bisphenol A with sodium hydroxide and the epichlorohydrin. What happens is that a mixture of the three reagents and water is heated up together. The phenol groups on the bisphenol A are deprotonated and then they attack the epichlorohydrin to then form 2,2'-(((propane-2,2-diylbis(4,1-phenylene))bis(oxy))bis(methylene))bis(oxirane). The epoxide (oxirane) groups in this then react with phenolate anions from new bisphenol A molecules to form 4,4'-((((((propane-2,2-diylbis(4,1-phenylene))bis(oxy))bis(2-hydroxypropane-3,1-diyl))bis(oxy))bis(4,1-phenylene))bis(propane-2,2-diyl))diphenol. The unreacted phenol groups on this then are reacted with more epichlorohydrin to form new reactive groups.
This reaction continues on and on to form a polymer, it is normal to use a small excess of epichlorohydrin to give a epoxy terminiated polymer (resin). This is then reacted with a cross linking agent such as an amine to form the final epoxy. As a result the chlorine content of the epoxy should be very very low.

So to answer the OP's question, the chlorine goes into aq phase as NaCl.

I thought that if you read the answer it is clear where the chlorine goes to, but I have made it even more obvious.

@Waylander Thanks! So the aq. NaCl remains in the cured final epoxy volume right? And this amount should NOT be too low, am I correct?


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The effect of pH and solubility on coagulation in water treatment
In water treatment, a specific pH range must be met in order for the process of coagulation to occur properly. Sources have stated this is due to the pH affecting the solubility of the coagulant, but why is solubility significant?

To start, I assume you are speaking of metal salt coagulants such as aluminum sulfate (alum) and ferric chloride, as these are the most commonly used. As you have stated, their solubility is affected by pH, the simplest representation of the reaction for alum being:
$$\ce{Al^3+ + 3H2O <=> Al(OH)3,_s + 3H+}$$
Small particles in the water such as clay or bacteria are too small to settle out on their own. The purpose of coagulation is to either cause the smaller particles to aggregate and settle or to trap them during precipitation. Coagulation occurs via three main mechanisms: (1) sweep - where the metal salts precipitate (i.e., solubility) and capture particles/contaminants like a net; (2) adsorption/charge neutralization - where the positive charges on the Al and Fe cations neutralize the negatively charged e.g., clay, bacteria, etc., causing them to aggregate and settle out; (3) adsorption and bridging - where a polymer coagulant can be used to "bridge" together like charged particles, causing them to aggregate and settle out. All are affected by solubility, but sweep is the mechanism that uses precipitation of the metal salts to remove particulates.
The mechanism you will want to choose will depend on the source water and the coagulant. The mechanisms are controlled by pH and solubility. Below are the solubility diagrams for alum and ferric chloride, with solubility represented on the left y-axis, dose on the right y-axis, and pH on the x-axis. As you will see, the shaded areas are the regions for achieving optimal coagulation, depending on the mechanism you desire. For example, if your laboratory tests showed that sweep coagulation was the best for your source water and you chose alum, then you would want to be at a pH between ~7 to 8 and an alum dose of ~15 to 60 mg/L alum (the "optimal sweep" circle). This pH and dose would result in the desired solubility behavior to achieve optimum sweep conditions.
*Figure 9-11 from MWH Water Treatment: Principles and Design, 3rd Ed (Crittenden)

When aluminium sulfate (a common coagulating agent) is used for water treatment, it creates byproducts in the form of aluminium residuals.
In the process of coagulation, it is beneficial to minimise the amount of aluminium residuals produced by aluminium sulfate.
Consequently, before the coagulation process begins, the water is corrected to a certain pH dependent upon the dose of aluminium sulfate and the solubility of aluminium.
By treating the water to have a pH which exceeds the solubility of aluminium, these residuals that the process of coagulation leave behind are reduced.
Therefore, solubility is significant because it is a determining factor in the amount of byproducts that coagulation produces.


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Is there a better model than the Kinetic Theory of Gases
I'm looking for a better relationship between the Kinetic energy of a gas and it's Temperature, and also how they vary with gravity (not altitude, just gravity!). So is there a model better than the Kinetic theory of gases that I can apply?
Thanks in advance!

I'm not sure that you are wanting when you say 'better relationship'; better than what?  The Maxwell and Boltzmann distributions of velocities have been well verified by experiment. The effect of gravity between molecules is utterly minute and the effect of the earth's gravity is present as a constant force. You could re-derive the Maxwell/Boltzmann equations using, say, the van-der -Waals equation instead of using the ideal gas but I would expect the distributions to have the same shape but shifted in value this way or that depending on how the calculation pans out.

The theory is for ideal gases, And accounts for macroscopic properties.
The first assumptions that you have to eliminate to get a better result are

Spherical assumption.
Ideality assumption.

I suggest first you read through the concepts of Equations of state. Have a look at:
1.Van der Waals
2.Virial Equation of state
3.Tao Mason equation of state
Without you providing more information I can not see what do are you exactly looking for.


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Choosing Parent Chain, Complexity of Side Chains
Which parent chain should we choose for this structure? is it A or B?

I followed the rule A-2.6, but it still got me confused.
Both alternative A and B has the same number of side chains (2 side chains) - rule A-2.6(a).
Both alternative A and B has the same number of locants, which are 5 and 6 - A-2.6(b).
Both alternative A and B has the same number of carbon in each locants, which are 4 atoms on number 5 and 9 atoms on number 6 - A-2.6(c).
Both alternative A and B branched side chains are quite similar, how do we decide? - A-2.6(d)

Note that your link points to obsolete IUPAC recommendations from 1979.

Related: IUPAC naming - choosing the main branch

You have already reduced the problem to two possible parent chains:
A: 5-(2-methylpropyl)-6-(3-methyloctan-4-yl)dodecane
or
B: 5-(butan-2-yl)-6-(2-methyloctan-4-yl)dodecane
In order to identify the preferred name, we have to read further in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book). In particular, the following potentially relevant rules are actually not helpful in this case:

P-45.2.1 maximum number of substituents cited as prefixes
P-45.2.2 the lower locant set for substituent groups cited as prefixes
P-45.2.3 the lower locant set for substituent groups in order of citation

However, we finally find:

P-45.5.1 The preferred IUPAC name is the name that is earlier in alphanumerical order (see P-14.5). Alphabetic letters are considered first in the order that they appear in the name; all Roman letters are considered before any italic letters, unless the latter are used as locants or are a part of a compound or composite locant, for example, ‘N’ and ‘4a’. Then, if still there is a choice, numerical locants are considered in the order of their appearance in the name.

Therefore, the preferred name is 5-(butan-2-yl)-6-(2-methyloctan-4-yl)dodecane since ‘butan-yl’ is earlier alphabetically than ‘methylpropyl’.


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Reaction of 2,4-DNPH with aromatic aldehydes
Does 2,4-dinitrophenylhydrazine react with aromatic aldehydes, e.g. benzaldehyde (phenylmethanal), in a condensation reaction similar to 2,4-DNPH's reaction with aldehydes and ketones ?

Yes, it does react

@Waylander Just to complete your comment: Yielding a solid of m.p.=237C. It is used as derivative in the classical identification of organic compounds.

@Raoul-Kessels Thank you. I no longer have database access so cannot go looking for such data. Perhaps you should turn your comment into an answer

@Waylander I had my old book on my desk ;-)

2,4-DNPH is a derivatization reagent for the identification of ketones and aldehydes.
This was extensively used in organic analysis before instrumental methods were readily available.
You determined the boiling point of the problem, which in the case of benzaldehyde is $179\ ^{\circ}\mathrm{C}$ , then you made a solid derivative and determined it's melting point.
In the case of benzaldehyde and 2,4-DNPH that would be $237\ ^{\circ}\mathrm{C}$ .
Then you go to the tables, cross check both values and had benzaldehyde identified.
For aldehydes and ketones there are different options: oxime, semicarbazone, phenylhydrazone, p-nitrophenylhydrazone and 2,4-dinitrophenylhydrazone.
The latter one was usualy preferred because they have higher melting points and so it was easier to get a solid even if not totally pure.


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Calculating amino acid mol/mol ratios
I am trying to understand how the authors calculated the TRP/LNAA (mol/mol) ratios in Table 1 in this paper.
For example, when I calculate the TRP/LNAA (mol/mol) ratio for HPROT, I get 0.14 instead of 1.1. What am doing wrong?
I'm using the following values for the molecular weights (g/mol):

Isoleucine 131.2
Leucine 131.2
Phenylalanine 165.2
Tyrosine 181.2
Valine 117.1
Tryptophan 204.2

And here are the tables from the paper:

Table 1: Composition of the standard protein (PLC), alpha-lactalbumin whey protein (ALAC), protein hydrolase (HPROT), pure tryptophan (TRP), and synthetic dipeptide (SYN) condition
\begin{array}{llll}
\hline
\text{Source} & \text{Product}~(\mathrm{g}) & \text{TRP}~(\mathrm{g}) & \text{TRP/LNAA}~(\mathrm{mol}/\mathrm{mol}) \\
\hline
\text{PLC}   & 20~\mathrm{g}/312~\mathrm{mL}  & 0.4 & 0.04   \\
\text{ALAC}  & 15~\mathrm{g}/312~\mathrm{mL}  & 0.8 & 0.1    \\
\text{HPROT} & 312~\mathrm{mL}       & 0.8 & 1.1    \\
\text{TRP}   & 0.8~\mathrm{g}/312~\mathrm{mL} & 0.8 & \infty \\
\text{SYN}   & 1.2~\mathrm{g}/312~\mathrm{mL} & 0.8 & \infty \\
\hline
\end{array}
Table 2: Amino acid profile
\begin{array}{llll}
\hline
\text{Amino acid profile}~(\mathrm{g}) & \text{HPROT} & \text{PLC} & \text{ALAC} \\
\hline
\text{Isoleucine}    & 1.004 & 1.800 & 2.012 \\
\text{Leucine}       & 0.745 & 2.733 & 3.306 \\
\text{Phenylalanine} & 0.480 & 1.467 & 1.309 \\
\text{Tyrosine}      & 1.216 & 1.867 & 1.451 \\
\text{Valine}        & 0.554 & 2.133 & 1.322 \\
\text{Tryptophan}    & 0.8   & 0.4   & 0.8   \\
\hline
\text{Total}         & 3.999 & 10    & 9.4   \\
\hline
\end{array}

Acronyms:

LNAA: large neutral amino acids
HPROT: hydrolyzed protein

How did you calculate moles of LNAA since there are 6 of them? I'd also point out what I consider to be a flaw in the paper in general, and that is that serotonin synthesis is not directly dependent on TRP, but rather on 5-hydroxtryptophan since tryptophan hydroxylase is the rate limiting enzyme in 5-HT synthesis. Feeding tyrptophan to humans does little to raise brain 5-HT.  Giving human 5HTP however does increase brain 5-HT.

@DrJ I calculated the moles for each individual amino acid and then summed the resulting values.

By the way, the paper does not make claims regarding serotonin synthesis. Rather, they investigate links between dietary intake of TRP/LNAA, serum levels of TRP/LNAA, and mood scores

But the “mood” is ultimately supposed to be related to brain serotonin levels, and this has been fairly well documented (refer to many articles by M.A. Aprison, et.al.). But that’s off topic from your question. I didnt see where they provided the mass of each LNAA in the diet. If you direct me to those data, I can check your calculations.

Citation was meant to be M.H. Aprison. Sorry.

The masses for the amino acids are here (Table 2): https://link.springer.com/article/10.1007/s00213-008-1254-0#Tab2 It's entirely possible that I am making an elementary error (!)

Yes, the calculated ratio is 0.14. Hard to imagine how it could be > 1.0 considering the combination of 6 LNAA compared to the lonely tryptophan.  Unless there is something unique about the HPROT being used in this study, i.e. what is the TRP content of the hydrolysate? I didn't read the entire manuscript, so I don't know what else to suggest.

@DrJ Thanks for checking the calculation, much appreciated. I think there's something funny going on in this paper but I'm not sure what yet - I'll continue thinking about it. Though at this point,  I think Table 1 is just plain wrong.


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Convention for numbering of carbons
I'm trying to understand how carbons are numbered in organic molecules (i.e. as in the following image).

(source: Principles of Biochemistry by Nelson and Cox, p. 532)
I've read and heard different methods, among which:

the end of the carbon chain with the highest oxidation state is assigned the lowest number.
the end of the carbon chain with the highest total mass is assigned the lowest number. the number of bonds is used as a kind of multiplier (i.e. O=C-OH corresponds to a total mass of O*2+C+O+H). (source: What is the convention of numbering carbon atoms in organic molecules?)
the end of the carbon chain with the substituent of the highest atomic number is assigned the lowest number (source: Principles of Biochemistry by Nelson and Cox, p. 73).
the carbonyl, acetal or hemiacetal carbon is given the lowest possible number (source: Carbon numbering in carbohydrates)

These methods could come to contradictory conclusions, though. One example I could think of:

Phosphorous would be the highest atomic number constituent, while the remaining rules (carbonyl carbon, total mass and oxidation state) would assign the carbonyl carbon the lowest number.
I assume some are just simplifications. What is the correct method, or is there no single general one?

Your initial example showing derivatives of glucose and fructose might be misleading since the nomenclature of natural product parent compounds does not necessarily follow the usual nomenclature rules of general organic chemistry.
In particular,

P-102.2.2 Numbering parent structures
The carbon atoms of a monosaccharide are numbered consecutively in such a
  way that:
(1) a (potential) aldehyde group receives the locant 1 (even if a more senior
  characteristic group is present);
(…)

(This rule is probably easier to understand if you draw the monosaccharides in their acyclic form.)
Generally, however, nomenclature of organic chemistry has different rules for the numbering. The most important simplified criteria for the numbering are:

lower locants for the group that is expressed as suffix
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;
(…)

Therefore, the numbering in your own example corresponds to the name 3-phosphinopropanoic acid since the acid group is the principal characteristic group of this compound.


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Which alkyl halides can not form grignard reagent

Which compounds are not useful for preparation of grignard reagents?
a) 4-chlorobut-1-yne
b) 2-chloroethan-1-amine
c) 1-bromobutan-2-ol
d) 4-bromocyclohexan-1-ol

What are your thoughts on this?

Is this a trick question?

@Ringo yes, sort of.  The stydent is supposed to find fault with all of them.

Seems to me that there are solid reasons why NONE of the materials will form Grignards unless they are modified.
1 - The alkyne proton (pKa ~25) will quench any Grignard formed to give the Mg salt
2 - The free base is already unstable in solution wrt aziridine formation/polymerisation; any added base promotes this. Any Grignard formed will also be quenched by the NH2 protons.
3- The -OH proton will quench any Grignard forming the alkoxide which is liable to form the epoxide.
4- Similarly the -OH proton will quench any Grignard forming the alkoxide which can then form the tetrahydrofuran


text:
Electrode polarity in fuel cells
I am studying fuel cells right now and I'm confused about the polarity of the two electrodes.
In the diagram below, the anode is shown as negative. However, why do $\ce{H+}$ ions move away from the anode, when they should be attracted to it because of its negative charge?
On the other hand, if I swapped the polarity around, movement of electrons from the positive to the negative terminal wouldn't work out.
Please explain which polarity is correct, and why.

Please explain which polarity is correct, and why.

In a fuel cell, the anode does sit at a more negative potential than the cathode. This polarization is opposite that of an electrolytic cell, where the anode is positive relative to the cathode.
The chemical composition of the fuel $(\ce{H2})$ and oxidant $(\ce{O2})$ feeds at the anode and cathode, respectively, cause this.  Thermodynamically, in this system the product of the fuel cell reaction, $\ce{H2O}$, is much more stable than the reactants.  Thus, the system biases itself to move in favor of the forward reaction:
$$
\ce{2H2 + O2 <=> 2H2O}
$$
As the $\ce{H2}$ is oxidized, the electrons from this half-reaction are passed into the anode:
$$
\ce{2H2 -> 4H+ + 4e-}
$$
These electrons have to move through the circuit in order to be able to supply the electrons needed for the oxygen reduction reaction:
$$
\ce{O2 + 4H+ + 4e- -> 2H2O}
$$
Here's the key: In order for the electrons to move through the circuit in the correct direction, the potential has to be more-positive at the cathode: the negatively charged electrons are attracted toward the positively charged cathode, along the conductive pathway of the circuit.

However, why do $\ce{H+}$ ions move away from the anode, when they should by attracting because of negative charges?

For the most part, the circuit is going to be composed of (semi)conducting materials. Conductive materials in a circuit cannot sustain a very substantial charge polarization at all--as soon as they become very slightly polarized, they allow current to pass that equalizes the charge imbalance. (This is why they are used to carry current!)
As a result, the actual charging of the anode is negligible, even though it develops a finite electrical potential.  Thus, the electrostatic attraction of the $\ce{H+}$ ions to the anode can in general be ignored, as it is orders of magnitude too small to keep the protons from diffusing away, toward the cathode where they are being consumed to produce water.


text:
Maximum required order for the Boys function
I am planning to pretabulate a lookup table in order to evaluate the Boys function. We can expand the Boys function as:
$$F_n (x_t + \Delta x) = \sum _{k=0}^\infty \frac{F_{n+k}(x_t)(-\Delta x)^{k}}{k!} \hspace{16pt} \tag{9.8.12}$$
We can then use downward recursion (upward is numerically unstable) in order to evaluate the function for any value of $n$ (order):
$$F_n (x) = \frac{2xF_{n+1}(x)+\exp(-x)}{2n+1} \tag{9.8.14}$$
I would like somebody to explain how one chooses which value of $n$ the lookup table will take? My working/logic so far as:

The Hermite integral has the following recursion (McMurchie-Davidson scheme, $x$ direction):

$$R^{n}_{t+1,u,v} = tR^{n+1}_{t-1,u,v} + X_\mathrm{PC}R^{n+1}_{t,u,v} \tag{9.9.18}$$
$\hspace{16pt}$ with:
$$\hspace{16pt} R^{n}_{0,0,0}=(-2p)^{n} F_{n}(p R_\text{PC}^{2}) \tag{9.9.14}$$
$\hspace{16pt}$ and so the recursion should limit $n$ to $n\leq t+u+v$.

In 2-electron integrals, the Hermite integrals are bound by $t \leq l_\mu + l_\lambda + l_\nu + l_\sigma $. Therefore $n$ is bound by the sum of the total angular momenta of the basis functions, i.e. $n \leq L_\mu + L_\lambda + L_\nu + L_\sigma$ where $L_\mu = l_\mu + m_\mu + n_\mu$.
Therefore, considering the extreme case of four $L=3$ basis functions, $n$ should be bound by $n \leq 12$.

Does this mean that a lookup table should start at $n=12$ and always recur down from this, or should I make different lookup tables and assess which one to use depending on the largest angular momentum functions in each individual computation? Are there any flaws or holes in my above logic?

All equations taken from Molecular Electronic Structure Theory, Helgaker et al. (2000).

I think your logic seems fine.
When you make your look up tables, you want to make a table for each $n$ up to some value of $n_\text{max}$. Consider you want to calculate an integral that only requires $n=2$, then it would be very inefficient to calculate $F_{12}$ and then down recur.
In your equation (9.8.12) we can notice that in the sum we have the Boys function as $F_{n+k}$. According to "Molecular Electronic Structure Theory", we might need the first $6$ terms of the sum. I.e. when doing your tabulation you will need to calculate all $n$ up to $n_\text{max}+6$.
Also when you do your tabulation you need a maximum range of your tabulated Boys function, i.e. a maximum value for where you can start using the long range approximation afterwards (else your tables will be HUGE). The long range approximation is given as:
$$F_n(x)\leq\frac{(2n-1)!!}{2^{n+1}}\sqrt{\frac{\pi}{x^{2n+1}}}\tag{9.8.10}$$
For which values of $x$ to start using this approximation for different $n$ you can take a look at Table II in Efficient recursive computation of molecular integrals over Cartesian
Gaussian functions. I am sure newer litterature on this topic exist, but I haven't encountered it.
As a side note, you can also calculate the Boys function with out tabulation (this might be less effective). By using the following formula:
$$F_n(x)=\exp(-x)\sum_{i=0}^\infty\frac{(2n-1)!!(2x)^i}{(2n+2i+1)!!}$$
This formula makes an implementation of the Boys functions much easier. If you are making an integral code from scratch, I would suggest to implement the above formula. You can always speed up the code later. Note with the above formula you can also just calculate the Boys function for the maximum $n$ you need and then use downwards recursion. If your goal is to implement the Boys function via. tabulation, then you can also use the above formula to generate the values for your look-up tables.

Hi Erik - SlowQuant was one of my original inspirations to write my own program. I am currently calculating the Boys function with the simple implementation you mention but I read that a lookup table offers significant optimisation. Your explanation answers my question. Helgaker says that $\Delta x = 0.1$ is sufficient - should this be made lower at very small values of $x$?

I think $\Delta x = 0.1$ is sufficient for the entire interval (I am not 100 % sure).  Very glad that it could be an inspiration!
For an actual implementation of the Boys function (that is also readable) you can take a look at https://github.com/theochem/horton if you need inspiration.
They have a Python generator script https://github.com/theochem/horton/blob/master/tools/boys.py
And then the actual Boys function in C++:
https://github.com/theochem/horton/blob/master/horton/gbasis/
Look for files called boys_(something)

By the way, I think you 9.8.10 is wrong - the double factorial is $(2n-1)$. I can't edit because it's not more than 6 characters.

Yes, you are right, changed it

@obackhouse Just saw this paper, that you might find interesting, very relevant for your question https://onlinelibrary.wiley.com/doi/pdf/10.1002/jcc.23935

I know this is an old question, but does the (2x)i mean a rising factorial or just raising the value of 2x to the ith power?

It just means to the power of $i$ @HemanthHaridas

@ErikKjellgren Thanks for the clarification!

I know this question is pretty old, but for the benefit of people who would come here looking to evaluate boys function, I attach an implementation of the formula mentioned above. The reference implementation is in Fortran, and requires no additional libraries.
  SUBROUTINE BOYS(N,T,RES)
  INTEGER, INTENT(IN)           ::  N
  DOUBLE PRECISION, INTENT(IN)  ::  T
  DOUBLE PRECISION, INTENT(OUT) ::  RES
  INTEGER                       ::  CNT1
  DOUBLE PRECISION              ::  ERROR, AUX1, AUX2, AUX3, PRES

  CNT1  = 0
  ERROR = 1.0
  CALL DOUBLEFACTORIALD(2*N-1, AUX1)
  CALL DOUBLEFACTORIALD(2*N+1, AUX2)
  ERROR = (AUX1/AUX2)
  RES   = (AUX1/AUX2)
  PRES  = RES
  DO WHILE (ABS(ERROR) .GT. 1.0D-15)
    CNT1  = CNT1+1
    CALL DOUBLEFACTORIALD(2*N+2*CNT1+1, AUX2)
    AUX3  = (AUX1/AUX2)*((2*T)**CNT1)
    RES   = RES+AUX3
    ERROR = RES-PRES
    WRITE(*,*)  RES, PRES, ERROR, CNT1
    PRES  = RES
  END DO
  RES = EXP(-1*T)*RES
  ! WRITE(*,*)  ERROR
END SUBROUTINE BOYS

This code reproduced the results provide here for Electron-Nuclear Interaction integrals in water at STO-3G level. The calls to DOUBLEFACTORIALD is a subroutine that calculates the double factorial of a number and returns it as a double.


text:
acid alpha glucosidase vs glycogen phosphorylase
What’s the difference between acid alpha glucosidase and glycogen phosphorylase? Both break down the same bonds in glycogen so I thought for a long time that they are exact same enzyme until I struggled through glycogen storage disease types and found out that each one is different than the other and that deficiency of either one causes a different type of GSD.

Acid alpha glucosidase  (sometimes called α-1,4-glucosidase or acid maltase) is not among those enzymes involved the pathways of glycogen synthesis and breakdown. It however occurs in lysosomes,where it functions to hydrolyze the disaccharide maltose  and linear oligosaccharides, as well as the outer branches of glycogen,thereby yielding free glucose.
This "second pathway" of glycogen metabolism is not quantitatively important as it concerns a minor fraction of glycogen. The reason that lysosomes normally take up and degrade glycogen granules is unknown perhaps for the disposal of structurally aberrant glycogen molecules.

Glycogen phosphorylase (a dimeric enzyme)
catalyzes glycogen phosphorolysis (bond cleavage by the
substitution of a phosphate group) to yield glucose-1-
phosphate (G1P).
$\ce{Glycogen + Pi <=>  glycogen + G1P}$
This enzyme will only release a glucose unit that is at least
five units from a branch point. (Note: I am not going to describe the reaction mechanism of glycogen phosphorylase here is much more complex than it sounds)
Having briefly described the enzymes, the most important thing to note is that:

structually these two enzymes are different,
active in the different locations
the mechanism of glycogen breakdown are different (i.e phosphorylation, hydrolysis)
different regulation mechanisms

Both work on alfa 1~4 linkage but....
Phosphorylase for glycogen breakdown in muscles and liver
But 1-4 glucosidase for glycogen break down in cells of other organs especially in lysosoes which inturn help cells to git rid off xss materials like glycogen.


text:
Buffer Solution with Strong Acid?
I was thinking: Could we could make a strong acid-strong base buffer solution? Take, for instance, a mixture of 1 L $\ce{HCl}$ (0.1 M) and 1 L of aqueous salt, $\ce{Na_2SO_4}$ (also 0.1 M), note that $k_2=0.012$ for the polyprotic acid. How can the pH of the mixture be calculated? Additionally, would it be "stable" like a buffer?

Buffers can be made two ways.
Buffers
Weak Acid/Base + Salt of the Conjugate Base/Acid
Let’s take acetic acid ($\ce{HC2H3O2}$) and sodium acetate ($\ce{NaC2H3O2}$).
$$\ce{HC2H3O2 + H2O <=> H3O+ + C2H3O2-~~~~~~~~~~K_a=1.8 x 10^{-5}}$$
Acetic acid has a small $\ce{K_a}$ value, meaning that it does not dissociate much. Conversely, sodium acetate readily dissociates as sodium is an alkali metal. By having equimolar concentrations of both weak acid and conjugate base, the addition of an acid adds hydronium, which reacts with the acetate ion to shift the equilibrium to the left, and the addition of a base absorbs hydronium to shift the equilibrium to the right. Either way, the hydronium concentration, which determines the pH, remains close to the same.
Weak Acid/Base + Strong Base/Acid
This relies on the same principle as above but supplies the conjugate base in a different manner. Continuing with our acetic acid example, here, this would be added, alone. Then you would add half of the moles of acid in strong base, such as $\ce{NaOH}$. This would neutralize any available hydronium, shifting the equilbrium to the right until there was the same half-and-half of acid and conjugate base.
Your Solution
Strong Acid?
Unfortunately, strong acids do not make for good buffers as they fully dissociate -- the equilibrium is so far to the right that it is very unstable.
Salt of the Conjugate Base?
$\ce{Na2SO4}$ shares no common ion with $\ce{HCl}$, so it provides no conjugate base ($\ce{{SO4}^2-}$ is the conjugate base of the second dissociation of $\ce{H2SO4}$). Even if you used $\ce{NaCl}$, it still would not work because of the reason outlined above.
pH?
If you're still interested, the pH of the solution above is easily calculated. As $\ce{HCl}$ fully dissociates, its concentration (0.1 M, as you say above) is equal to the hydronium concentration. However, the addition of another liter of neutral solution halves the hydronium concentration to 0.05 M. Therefore,
$$\ce{-log(0.05)=pH=1.3}$$
Hope this helps!

Even if we are given the fact that $K_2=0.012$ for sulfuric acid????

@riemannium Unfortunately, yes. Due to the fact that sulfuric acid's first dissociation Ka is large, as explained in the article to which I linked, it is a strong acid. If you found a way to isolate the bisulfate ion (HSO4-), it might work, but probably not. Any other questions?

My reasoning was different...The above salt is "neutral", but as it is a solution in water, you double the amount of solvent, so I thought in the following: 1 liter HCl 0.1M equals to [H⁺]=0.1 of course, and thus 0.1 mole. Then you get another liter of aqueous salt...So a more precise answer, it seems to me, would be pH(mixture)=-log(0.05)=1.3> 1. It was obvious for me that the mixture should raise a little bit the pH, even if the salt was neutral and from a strong acid/strong base...If $K_2(H_2SO_4)=0.012$, then $K_h=K_w/k_2≃10^{-12}$. So, $\alpha=10^{-6}$ and pH=1.45 (not far from 1.3).

Nice catch on my missing the dilution's effect on the pH -- I edited my answer. However, the solution still would not buffer.

Do you have any other questions I can help with? If not, would you consider accepting my answer? I’d really appreciate it!

Read my extra answer...I only have the doubt about $K_2$...Since, I wonder if it would matter...For instance, imagine instead of $Na_2SO_4$ had been $Na_2CO_3$, and its $K_2=10^{-11}$...

In my experience, $\ce{K_2}$ has little impact on acid dissociations (relevant only to me when calculating the pH of a polyprotic acid to a number of significant figures). If you wanted to use $\ce{H2CO3}$ (carbonic acid) and $\ce{NaHCO3}$ (sodium bicarbonate), I believe that would create a reasonable buffer. Just remember: the acid/base and salt MUST share a common ion. Hope this helps!


text:
Chemical equilibrium constant problem

In a equilibrium reaction $\ce{CO {(g)} + H2O {(g)} <=> CO2 {(g)} + H2 {(g)}}$, initial concentrations of $\ce{CO}$ and H2O are equal and are 0.3 mol/dm3. What is
  the equilibrium constant of the reaction if the equilibrium
  concentrations of CO2 and H2 are equal and are 0.1 mol/dm3?

I tried to solve it this way:
$$\begin{split}
\ce{K_r} & = \ce{\frac{[CO2][H2]}{[CO][H2O]} \\
& = \frac{(0.1)(0.1)}{(0.3)(0.3)} \\
& = \frac{(0.1)^2}{(0.3)^2} \\
& = \frac{0.01}{0.09} \\
& \approx 0.11 mol/dm^3}
\end{split}$$
But the correct answer is 0.25 mol/dm3
What did I do wrong? What is the right way of solving problems like this one?

0.1 is the equilibrium concentration of $\ce{CO2}$ and $\ce{H2}$. 0.3 is the initial concentration of $\ce{CO}$ and $\ce{H2O}$. Let's plug 0.1 into an ICE-box as $x$.
\begin{array}{|c|c|c|c|c|}
\hline
\text{Initial}:& 0.3 & 0.3 & 0 & 0 \\ \hline
 & \ce{CO} & \ce{H2O} & \ce{CO2} & \ce{H2}\\ \hline
\text{Change}: & -x & -x & +x & +x \\ \hline
\text{Equilibrium}: & 0.3-0.1 & 0.3-0.1 & 0+0.1 & 0+0.1\\ \hline
\end{array}
Mass action expression:
$$K_\mathrm{c} = \frac{(0+0.1)^2}{(0.3-0.1)^2} = 0.25$$


text:
Calculating the pH of a weak acid from the Kb of its conjugate base
I'd start off by saying it's been many years since I was in chemistry lessons and have forgotten much, but recently I have had to relearn a lot of basic chemistry. The following doesn't really impede me in my work, but it irritates me that I can't figure out what is basically two sides of the same coin.
Let's imagine I want to calculate the pH of a 0.1M acetic acid ($c_\ce{CH3COOH} = \pu{0.1M}$)
Acetic acid is a weak acid and dissociates in water as follows: $\ce{CH_3COOH + H_2O <<=> CH_3COO^- + H_3O^+}$. Expressing this in terms of acid dissociation constant (and ignoring the water) yields $K_\mathrm{a} = {\large{\ce{\frac{[CH_3COO^-][H^+]}{[CH_3COOH]}}}}$
We can look up the $K_\mathrm{a}$ of acetic acid, which is $1.8 \times 10^{-5}$. If we assume that the amount of $\ce{CH_3COOH}$ change is negligible (due to it being a weak acid and only slightly dissociating), and that $\ce{CH_3COO^-}$ and $\ce{H^+}$ concentrations are equal, we come up with the following equation:
$1.8 \times 10^{-5} = \large{\frac{[\ce{H^+}]^2}{\pu{0.1M}}}$
Rearranging and solving for $\ce{[H^+]}$ gives us $\sqrt{1.8 \times 10^{-5} \times \pu{0.1M}} = \pu{1.34E-3 M}$
Since $\mathrm{pH} = -\log([\ce{H+}])$ (or $\mathrm{pH} = \log(1/[\ce{H+}])$, whichever you prefer), $\mathrm{pH} = -\log(1.34 \times 10^{-3}) \approx 2.88$
I've checked this and I know it's correct. But I don't know how to do it from the $K_\mathrm{b}$. Or rather, I get the answer, but I don't fully understand the process.
So the opposite reaction, $\ce{CH_3COO^- + H_2O <=>> CH_3COOH + OH^-}$, also has a equilibrium constant which is called $K_\mathrm{b}$, defined as follows (again ignoring the water):
${\normalsize K_\mathrm{b}} = {\large \ce{\frac{[CH_3COOH][OH^-]}{[CH_3COO^-]}}}$
We can look it up or calculate it from the relationship that $K_\mathrm{a} \times K_\mathrm{b} = 1\times10^{-14}$
$K_\mathrm{b} = {\large \frac{1\times10^{-14}}{K_\mathrm{a}}}$
In either case, $K_\mathrm{b} = 5.6 \times 10^{-10}$
Now, I don't know what to do with the $\ce{[OH^-]}$. My basic assumption is that the change in the amount of $\ce{CH_3COOH}$ is negligible, so that stay as 0.1M. But unlike previously where $\ce{CH_3COO^-}$ and $\ce{[H^+]}$ were formed in equal amounts due to the dissociation of the acid, how much of $\ce{[OH^-]}$ is there? My (wrong) logic was to assume that $\ce{[CH_3COO^-] = -[OH^-]}$, but then they cancel out, leaving me with $K_\mathrm{b} = \ce{[CH3COOH]}$, which can't be right. So yeah, I'm lost.
I solved this by substituting $\ce{[OH^-]}$ with $\large{\frac{K_\mathrm{w}}{[\ce{H+}]}}$ (because $\ce{[H^+][OH^-]} = K_\mathrm{w}$)
This yields ${\normalsize K_\mathrm{b}} = {\large \ce{\frac{[CH_3COOH] K_w}{[CH_3COO^-][H^+]}}}$. Again assuming that $\ce{[CH_3COO-]} = \ce{[H+]}$, and knowing that $K_w = 1 \times 10^{-14}$, I come up with ${5.6 \times 10^{-10}} = {\large \ce{\frac{0.1M \times 1 \times 10^{-14}}{[H^+]^2}}}$
Solving for $\ce{[H^+]}$ gives us $\sqrt{{\large \frac{\pu{0.1M} \times 1 \times 10^{-14}}{5.6 \times 10^{-10}}}} = \pu{1.33 \times 10^{-3}M}$
This is the same concentration of $\ce{[H^+]}$ that I got using the $K_\mathrm{a}$ above, making the $\mathrm{pH} = -\log(1.33 \times 10^{-3}) \approx 2.88$
So I can get to the same answer ignoring (or rather substituting) the $\ce{[OH^-]}$, but I'd really like to know how to do it using that. This is probably a really simple answer but after having spent many hours staring my many pieces of paper, my mind is coming up with a blank.

In this answer  https://chemistry.stackexchange.com/questions/60068/how-to-set-up-equation-for-buffer-reaction/95566#95566 there is a general method to solve problems of this type that might help to answer some of your queries.


text:
Rearrangement of 8a-ethyl-1,3,4,8a-tetrahydronaphthalen-4a(2H)-ol
really sorry for that previous post. I was actually half-asleep when I posted that q, hence the q was unclear. I am again reposting that question
The answer is given as A. Can anyone please explain HOW ??

The last option is methylbenzene.

Cation rearrangement should be expected here, yet the answer (C) is strange to me; I'd assume (A).

Sorry Ivan Neretin, typo error-ans is A.-but HOW ???

@Aakhyat Singh: If you are still interested, for reliable mechanism for this type of rearrangement, see my answer to this question.

Answer will be option A
First of all, name the carbon atom in the option A with $\ce{-Et}$ group attached to it as x, the $\ce{C}$ atom in the question with $\ce{-OH}$ attached to it as y, and the $\ce{C}$ atom in the Question with $\ce{-Et}$ attached to it as z
What will happen is, the $\ce{-OH}$ will get protonated forming $\ce{-OH2+}$. Concentrated acidic medium will facilitate formation of $\ce{H2O}$ by braking of bond between $\ce{C}$ and $\ce{O}$ in $\ce{C-OH2+}$ This will form a carbocation at the position y which will due to resonance reach the position x. Now, the $\ce{-Et}$ will do alkyl shift from z to x.
Now, a $\ce{C-H}$ bond at x will break and a $\ce{C=C}$ will form between the $\ce{C}$ atoms at x and z, thus forming an aromatic compound as major product i-e option A
#

Formation of $\ce{C+}$ is facilitated because of $\ce{conc. H2SO4 + heat}$ because $\ce{(conc. H2SO4 + heat)}$ acts as an oxidising agent, hence oxidising the $\ce{C}$ atom (or you may also think like, formation of $\ce{H2O}$ in a concentrate system will liberate heat thus its formation is favourable)

#

Alkyl shift from z to x is favourable as it drives the reaction towards formation of a much stable aromatic compound


text:
Can activated charcoal be used to filter bacteria out of air?
As the title says: Can activated charcoal be used to filter bacteria out of air and or gas?

It is probably a bad idea: activated carbons have most of its porosity mainly in micropores (channels smaller than 2 nm) and some have mesopores (between 2 and 40 nm). Bacteria are much larger than this.

Activated charcoal has been shown to be able adsorb certain species of bacteria, but the effectiveness varies between species [1] (which is potentially useful for medical applications), and has not been tested for many, so it is unlikely that activated charcoal could be used to adsorb all the bacteria from the air or a gas.
Activated charcoal general usage is in the adsorption of toxins.
[1] Naka, Katsumi, et al. "Adsorption effect of activated charcoal on enterohemorrhagic Escherichia coli." Journal of Veterinary Medical Science 63.3 (2001): 281-285.


text:
Why do ions form condensation centres in isopropanol vapour?
In cloud chambers the particles are detected when a particle passes through the chamber causing the vapour to condense into a "cloud". What Process causes this observed phenomenon? In other words, how exactly do the particles tracks form in a cloud chamber?

Like I said, this is because ions attract neutral molecules. In the unstable supercooled vapor, this tiny perturbation is just enough to trigger the formation of a new phase. Then again, I appreciate your effort to verify the results of an experiment (that is, your question) by repeating it, but if you are expecting a different outcome this time around, don't.

Thx Ivan, I appreciate your answer. I repeated the question because the first time I posted it, it was downvoted as “off-topic” and removed. How is it off-topic???

Anyway, I’m keeping the question up (until it’s removed again) to garner more answers. I’m not a chemist (trained in physics); do you mind elucidating your answer a little more?

Dah... As I said before, the neutral molecules are polarizable. Thus even though the whole molecules of the saturated vapor are not charged, the molecules can be attracted to ions in the vapor forming a nucleation center.

Thx Max. No need to say “dah.” I’m supposing that the molecules in the vapour polarize, and the positively charged end of each one thereupon bonds to the negative ion, created the trail, Is that about correct? As noted, I’m not a chemist.

Sounds about right. Note also that isopropanol molecules (as well as many others) are naturally polar, even if you don't polarize them.

Thx Ivan. Good to know I’m on the right track


text:
Ionization energy in Geiger counters
Geiger counters can detect radioactivity because nuclear radiation consists of particles or radiation of sufficiently high energy to eject electrons from atoms. Consequently, this type of radiation is called "ionizing radiation." What is the longest wavelength of radiation that can be detected by a Geiger counter using neon gas as the ionizing medium?
I know the ionization energy of neon is $\pu{2080 kJ/mol}$, or $\pu{3.46*10^{-18} J/atom}$ of Ne (calculated by dividing the I.E. by avogadro's number).
Since $E = \frac{hc}{\lambda}$, I can calculate $\lambda$ to be $57.3$ nm. However, the textbook's answer is $3.64$ nm. I'm not sure why.

I get your answer too. 3.64 nm is about 340 ev or 32,800 kJ/mole. I have no idea where that value is coming from. I assume that the problem answer is just wrong. // P.S. The sensitivity of a real Geiger counter depends on window thickness too. Seems like an awful thin window.

Commonly a GM tube can detect photons as a result of electrons being ejected from the steel of the tube or lead shielding around the tube as a result of the photoelectron or Compton effects. I think you should consider these effects as well as the ionization of the gas in the tube.

@NuclearChemist - I looked at various x-ray absorption edges and ionization energies assuming that the fill gas had been changed. Nothing seems to fit. // I also tried to see if there was some simple math error. //  The resolution cannot have anything to do with some hypothetical wall composition.  The textbook's answer is just wrong. I can't figure out why.

You shouldn't use the ionization energy of Ne (21.6 eV); you should use the energy that is required to form one ion pair in neon (36.2 eV). However, that still doesn't lead to the textbook result.


text:
It is possible to make radical Chlorination on the α-position of ketones?
I need examples of radical α-Chlorination of ketones. It is possible to make radical chlorination happen in the alpha position?


text:
Why aren't lasers used as light sources for atomic absorption spectroscopy?
Atomic absorption spectroscopy yields extremely narrow signals. It is difficult to separate white light with sufficient precision using monochromators, and hollow cathode lamps are often used as light sources instead. Disadvantage: a different lamp is needed for each element we wish to measure.
Why aren't lasers used as light sources? Lasers are perfectly monochromatic, and tunable lasers could be set to match absorption bands of many elements.

It could be because the lasers with the widest wavelength range have bandwidths that are too broad and the the lasers with narrow bandwidths don't have the same ability to tune over very large ranges of wavelength.

Mainly cost and ease of use. Lasers are expensive and hard to maintain with narrow bandwidths and have limited wavelength tuning. Cathode are, in comparison, dirt cheap. (In fact atomic lines are actually not that narrow and nowadays lasers can have line - widths orders of magnitude narrower than atomic lines. )


text:
Maximum possible hypochlorite concentration from electrolysis?
For reference, "electrochlorination" is just the electrolysis of a simple solution of water and table salt (NaCl) followed by reintroducing the chlorine gas produced.
But, the Wikipedia article on "electrochlorination" claims

The product of this process, sodium hypochlorite, contains 0.7% to 1% chlorine. Anything below the concentration of 1% chlorine is considered a non-hazardous chemical although still a very effective disinfectant. In addition, the sodium hypochlorite produced is in the pH range of 6-7.5. This means that the chemical is relatively neutral in regards to acidity or baseness. Also, at that pH range, the sodium hypochlorite is extremely stable and the electrochlorination extremely effective.

Is the yield really that small? It seems like we should be able to get much more! The electrolysis of NaCl in water
$$\ce{2H+ (aq) + 2Cl- (aq) -> H2(g) + Cl2(g)}$$
upon recombining the chlorine gas
$$\ce{2OH- (aq) + Cl2(g) ->  OCl- (aq) + Cl- (aq) + H2O(\ell)}$$
should lead to a net increase in pH
$$\ce{Cl- (aq) + H2O(\ell) -> OCl- (aq) + H2(g)}$$
as$\ce{\; OCl- (aq)\;}$is the conjugate base of a weak acid.
Consequently, it seems the resulting solution should have a much greater percent active chlorine, due to the hypochlorous stabilization effect of higher pH -- which would only get even higher as the electrolysis (given sufficient external voltage) continues!

About 20-21% is maximum possible and usually, its done to about 10-12%. Then some vacuum distillation if to be sold as 15% or more. Please don't ask for source, I work for an Alkali company and this is what is written on the Production records

A little bit of Wikipedia research reveals that

Sodium hypochlorite can also be obtained as a crystalline pentahydrate $\ce{NaOCl·5 H2O}$, which is not explosive and is much more stable than the anhydrous compound.5 The formula is sometimes given as $\ce{2NaOCl·10H2O}$. The transparent light greenish yellow orthorombic[10][11] crystals contain 44% NaOCl by weight and melt at 25−27°C. The compound decomposes rapidly at room temperature, so it must be kept under refrigeration. At lower temperatures, however, it is quite stable: reportedly only 1% decomposition after 360 days at 7°C.6[12]

So, it looks like 44% w/v would be an upper bound to any solution of NaOCl in water, regardless of whether it was obtained via electrochlorination or not. This corresponds to approximately 43.1% active chlorine.
Also, note that that the term "active chlorine" measures overall oxidative capability, not necessarily hypochlorite concentration — hypochlorites can easily decompose into chlorates without altering the greater solution's overall oxidative capabilities
$$\ce{3OCl- (aq) -> ClO3- (aq) + 2Cl- (aq)}$$
as demonstrated via their molar-equivalent iodometries
$$\ce{3OCl- (aq) + 6I- (aq) + 6H+ (aq) -> 3I2(s) + 3Cl- (aq) + 3H2O (\ell) \\ ClO3- (aq) + 6I- (aq) + 6H+ (aq) -> 3I2(s) + Cl- (aq) + 3H2O (\ell)}$$
so more accurate measurements of hypochlorite concentration may likely require additional techniques (such as spectroscopy, for example) as well.
Finally, due to the unstable nature of solutions with high [NaOCl(aq)] and their natural tendency to
$$\ce{2OCl- (aq) -> 2Cl- (aq) + O2 (g)}$$
maximizing [NaOCl(aq)] solely via electrolysis/electrochlorination will likely require a dedicated setup that simultaneously:

Limits atmospheric exposure around the anode to prevent loss of chlorine gas while allowing hydrogen gas to escape by the cathode (or, alternatively,
reduce atmospheric oxygen to water)
Keeps the solution at low temperatures (to limit various other decomposition reactions)
Maintains some external "tending" voltage (to preserve the Nernst equilibrium)

For practical applications, this could be a very unwieldy form of storage!

I interpret the question "Maximum possible hypochlorite concentration from electrolysis?", to imply a safe (that is, likely non-explosive) manufacture thereof. Also, as cited in this source, increased related hypochlorous acid presence (from dissolved chlorine reacting with water) promulgates accelerated chloride and chlorate formation, and can actually consume all of the hypochlorous present, thereby limiting total sodium hypochlorite creation (depending on the electrolysis cell design and operating conditions).
In respect to safety, I would comment, for the record, that the actual weak link in a high concentration chlorine electrolysis cell, with respect to safe operations, actually relates not so much to the formed aqueous sodium hypochlorite, but from dichlorine oxide, as emanating from hypochlorous acid as created from the action of electrically released chlorine interacting with water (that is, as well, increasingly being removed due to any precipitating hydrates of sodium hypochlorite).
Why my safety concerns? Some quotes from Bretherick’s Handbook of Reactive Chemical Hazards, Seventh Edition, Volume 1, available free online:

the gas [dichlorine oxide] readily explodes on rapid the end of slow thermal decomposition. Kinetic data are summarised [3]. The spontaneously explosive decomposition of the gas was studied at 42—86 C, and induction periods up to several hours were noted [4].

And further:

Several instances of containers or drums igniting or erupting on reopening after previous use have been reported, and may well have involved formation by slow hydrolysis of a relatively high concentration of dichlorine monoxide in the containers. This may have been caused to decompose from sudden exposure to light, by friction on opening the drum, or by a static spark.

I suspect the presence of sodium chloride reduces the solubility of dichlorine monoxide, further contributing to potential issues.
As such, I recommend venting, so as per the incident noted above, to lessen dangerous explosive fumes, which is likely advisable, as the electrolysis solution, itself, is caustic so that even a mild explosive event may result in injuries.
Also, based on an old text, I surmised that one should not consider working, in general, with concentrations of hypochlorous acid exceeding 20%, being described as dangerous, which I believe, would also be advisable for a safe cell operation and performance. Support is provided by US Patent 5,108,560 citing  working examples of hypochlorous solutions of 15% and 20%, and further per this already alluded to work where to quote:

The rate of decomposition of hypochlorous acid has been measured in aqueous solution in the presence of much sodium hypochlorite. The rate is nearly independent of the hypochlorite concentration, and proportional to the square of the hypochlorous acid concentration.

where the disproportionation results in an even more caustic chloric acid, and therefrom sodium chlorate.


text:
Final pressure of an ideal gas mixture, when two separate gases are put into one vessel

At constant of 200 cm³ of $\ce{N2}$ at 720 mmHg and 400 cm³ of $\ce{O2}$ at 750 mmHg are put together into one litre flask. The final pressure of the mixture is:

My question is how to use ideal gas law for mixture of two gases. For one gas, it is easy but for two it becomes difficult for me. I am assuming that these gases won't react or else it would have been given in the question. Can somebody tell me how to apply ideal gas law for two gases and what happens when they are kept together?

Imagine they are one gas. Since they don't react (which is true), this is a perfectly reasonable assumption. Then it must become easy again.

On second thought, this might be a bad idea. Look at it this way: find the resulting pressure of one gas as if the other one were not there. Then find the pressure of another, this time disregarding the first one. Then add the two pressures together.

So we have to just calculate the partial pressure of each gas

Yes, exactly.$!$

"At constant of 200 cm³ …" – Is there anything missing, like "at constant temperature a volume of 200 cm³"?

Note that since $1\ \mathrm{cm^3} = 1\ \mathrm{mL}$, I will talk in mL.
I doubt that you are supposed to use the ideal gas law for this problem since no temperature is specified, and moles are not talked about. Instead, let's look at a few instances of Boyle's Law ($P_1V_1=P_2V_2$ -- for comparing the pressures and volumes of gasses).
The gasses are, indeed, unreactive, and, according to Kinetic-Molecular Theory, have nearly identical properties where gas laws are concerned. Let's find the partial pressure of each gas with Boyle's Law and sum them with Dalton's Law of Partial Pressures ($P_\mathrm{tot} = P_1 + P_2 +\dots$).
Note that to make the units cancel, I will use $1\ \mathrm L = 1000\ \mathrm{mL}$.
$$720\ \mathrm{mmHg} \times 200\ \mathrm{mL} = x\ \mathrm{mmHg} \times 1000\ \mathrm{mL}$$
$$\frac{720\ \mathrm{mmHg} \times 200\ \mathrm{mL}}{1000\ \mathrm{mL}} = x\ \mathrm{mmHg} = 144\ \mathrm{mmHg}$$

$$750\ \mathrm{mmHg} \times 400\ \mathrm{mL} = x\ \mathrm{mmHg} \times 1000\ \mathrm{mL}$$
$$\frac{750\ \mathrm{mmHg} \times 400\ \mathrm{mL}}{1000\ \mathrm{mL}} = x\ \mathrm{mmHg} = 300\ \mathrm{mmHg}$$

$$144\ \mathrm{mmHg} + 300\ \mathrm{mmHg} = P_\mathrm{tot} = 444\ \mathrm{mmHg}$$
444 mmHg is the final pressure in the container.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. Please note that the proper term for "(number of) moles" is amount of substance. The former would be the same as referring to the mass as "(number of) kilograms".


text:
Finding equation of line of the P-T graph for a gas dissociation reaction

Gas A (1 mol) dissociates in a closed rigid container of volume 0.16 lit as per following reaction.
$$\ce{2A (g) -> 3B (g) + 2C (g)}$$
If degree of dissociation of $\ce{A}$ is 0.4 and remains constant in entire range of temperature, then correct P Vs T graph is

Slope 0.75 and passing through origin
Slope 0.8 and passing through origin
slope 1.1 and passing through origin
slope 0.4 and passing through origin

\begin{align}
\ce{2A &&-> 3B& + &2C}\\
1-2x&&3x&&2x\\
\end{align}
where $2x$ is dissociation constant. Substituting value of $x$ we get moles of A, B and C to be 0.4, 0.6 and 0.4 respectively.
I am confused about how to proceed after this. How to proceed?

The overall notion is that gases A, B and C are all behaving as an ideal gases which can be represented by the mythical ideal gas X. Then the moles of X are equal to moles of A+B+C. Then we can use the ideal gas equation:
$$pV=nRT$$
which simplifies to:
$$p = \dfrac{nR}{V}\times T$$
Given that the degree of dissociation of A is 0.4 and we started with 1 moles of A, then 0.6 moles of A remain and 0.4 moles got converted.
$$\ce{2A -> 3B + 2C}$$
or
$$\ce{A -> }\dfrac{3}{2}\ce{B + C}$$
let B and C equal mythical gas X and
$$\ce{A -> }\dfrac{5}{2}\ce{X}$$
but since A = 0.6 moles of X, the total moles are
$$n = 0.6 + 0.4\times\dfrac{5}{2} = 1.6\text{ moles}$$
We know that n = 1.6 moles and that V = 0.16 L so:
$$p =\left(\dfrac{n}{V}\right)RT = \dfrac{1.6\text{ mole}}{0.16\text{liter}}RT = 10RT\dfrac{\text{moles}}{\text{liter}}$$
Now the problem doesn't really specify what value of the ideal gas constant, $R$, to use nor what units the temperature, $T$, should be.
The temperature must be an absolute scale since the line passes through the origin. Thus of the main choices of Kelvin and Rankine, using Kelvin seems reasonable for temperature.
but there are still options for the pressure units of R

(a) $8.314\text{ L}\cdot\text{kPa}\cdot\text{K}^{−1}\cdot\text{mol}^{−1}$
(b) $8.314\times10^{−2} \text{ L}\cdot\text{bar}\cdot\text{K}^{−1}\cdot\text{mol}^{−1}$
(c) $62.36 \text{ L}\cdot\text{torr}\cdot\text{K}^{−1}\cdot\text{mol}^{−1}$
(d) $8.206\times10^{−2}\text{ L}\cdot\text{atm}\cdot\text{K}^{−1}\cdot\text{mol}^{−1}$

For $10R$ to be between 0.4 and 1.1 the only answers that fit are (b) and (d). Thus the pressure could be either in atm or bar. Presumably the OP's book would have stated a preference for R.

The units of R is indeed given in 0.08litol/K.Sorry for not including this and thanks for your answer

So even though the gases are reacting,we can consider them to be one ideal gas at equilibrium point .Can we apply ideal gases individually at the equilibrium for the gases and calculate pressure but that would be go against Dalton's partial pressure law ,right?

Also at eqiluilibrium even though  the reaction still goes on ,we can apply the gas law......is there a reason?

Can you comment on doubts pls as it will be very much helpful for me

Yes, you could calculate the individual pressures. It is just a bit more math. // You actually drew the reaction arrow wrong. You have it only going to the right, when the arrows should go both ways since there is an equilibrium (in other words both forward and reverse reactions). $$\ce{2A (g) <=> 3B (g) + 2C (g)}$$


text:
How to display ligands as ball and stick and macromolecules as cartoons (if protein) / rings (if nucleic acids) in Avogadro?
Avogadro has many good qualities but I can't seem to find a way to display ligands differently to macromolecules in the same view window. Frankly I can't seem to find a way to display any two things differently (like one as ball and stick, another as ribbons, or w/e) to one another in the same view. It seems like in a given view window everything has to be displayed the same way, there's no room for making one part displayed one way and another part displayed another way.

This was also asked on the Avogadro discussion forum
It's possible in Avogadro 1.x and discussed in the documentation

Select part of the molecule
Click on the "options" button for a particular display type (e.g. "ball and stick")
Click on the button to only show the selected atoms


text:
Oxidation of glucose

By oxidation of 9 g of glucose it is released 150 kJ of heat. Calculate the standard enthalpy of derivation of glucose (in kJ/mol), if the given enthalpies of the reactants are ΔfH (H2O(l)) = -285,8 kJ/mol; ΔfH (CO2(g)) = -393,5 kJ/mol. Mr (glucose) = 180 (The answer is -1075.8)

$$\ce{6 CO2 + 6 H2O -> C6H12O6 + 6O2}$$
$$n(\ce{C6H12O6})=\frac{\pu{9g}}{\pu{180 g mol^-1}}=\pu{0.05 mol}~\ce{C6H12O6}$$
Now, $$\frac{\pu{0.05 mol}}{\pu{150 kJ}}=\frac{\pu{1 mol}}{x}\implies x=\pu{-3000 kJ/mol}$$
$$\Delta_\mathrm rH=\Delta_\mathrm fH(\ce{H2O(l)})+\Delta_\mathrm fH(\ce{CO2(g)})-\Delta_\mathrm fH(\ce{C6H12O6(s)})$$
$$= 6 \times (\pu{-285.8 kJ/mol}) + 6 \times  (\pu{-393.5 kJ/mol}) - (\pu{-3000 kJ/mol}) = \pu{-1075.8 kJ/mol}$$
Did I take the right steps?

"Did I take the right steps?" Alex, please note that "check my work" type questions are generally off-topic. Please highlight exactly which calculation step you are suspicious of, or feel would give the wrong answer. In any case, your answer matches the one given by the answer key. So, what do you feel is the problem here?

At first it seemed like I would get the wrong answer, as I was doing this problem in my notebook. However, when I wrote it here I got the right answer. Tbh, I had some difficulties figuring out what the reaction was, ie. what are the reactants and products of the reaction. I couldn't have figure it out myself until I found the reaction on the Internet.

Note that $n(\ce{C6H12O6})=\pu{0.05 mol}$ and not $n(\ce{C6H12O6})=\pu{0.05 mol}~\ce{C6H12O6}$.

I had some difficulties figuring out what the reaction was

The question mentions "oxidation of glucose", that's a very important hint that tells you that the reaction is going to be:
$$\ce{C6H12O6 + O2 -> ?}$$
Now, note that oxidation of any carbohydrate/hydrocarbon/organic compound only yields (in general) carbon dioxide and water.
Thus, your complete reaction should be:
$$\ce{C6H12O6 + 6O2 -> 6 CO2 + 6 H2O}$$
Notice that each of the products is just the "oxide" of the elements of the original compound $\ce{(C, H)}$


text:
Why does oxygen have less negative electron gain enthalpy than sulphur?

Oxygen has less negative electron gain enthalpy than sulphur.

This statement is given in my book with a short reason:

due to compact size of oxygen

I'm unable to understand why. I would appreciate any worthy explanation for this.

https://chemistry.stackexchange.com/questions/64869/if-oxygen-has-a-lower-electron-affinity-than-sulfur-then-why-doesnt-sulfur-has https://chemistry.stackexchange.com/questions/6901/why-does-chlorine-have-a-higher-electron-affinity-than-fluorine

It is because Oxygen is first element of Oxygen family in its table and Electron gain enthaply decreasing from top to bottom in the group.
This exemption happens because of the reduced 2p orbital of Oxygen, there is enormous interelectronic r between the electrons in 2p orbitals thus Oxygen has less proclivity for an additional electron and the EGE of Oxygen is more positive than Sulphur.


text:
Why do some chemical reactions require many steps?
I was a college student of physics ten years ago, and recently started to learn biochemistry. I enjoy finding out that some familiar concepts in physics play important roles in biochemistry such as Entropy and Gibbs free energy.
For example, as a (ex-)student of physics, I am happy to know that Gibbs free energy helps to decide the directions of chemnical reactions. I feel this is a good example where a sort of fundamental law of physics determines how phenomenon looks like.
However, I still can not understand why the chemical reactions in a body need to be so complex. Many chemnical systems consists of more than several steps to acheive their purposes. According to wikipedia, glycolysis takes ten steps through its process. Why so many steps are necessary?
I tried to find out a physical law which prohibit that glycolysis process is achieved by one or two steps. But I could not find an answer.
I would like to know (or discuss) whether there is a physical law which makes the chemical system such a complex one (many steps required).
My assumption is that some physical law prohibit the existence of a enzyme which realize a one-step process of glycolysis.

...why the chemical reactions in a body need to be so complex... more than several steps to acheive their purposes....Why so many steps are necessary? What need? What purpose? Biochemical reactions do not have a mind of their own. They were not planned or designed.

Because you can find many models of biological systems in Physical Review L/E/X to explain the fundamental property of them, I think physicians are suitable for answer/discuss about this topic. But I might be wrong. I am going to transfer this question to SE.Chemistry. Due to the reputation limnitation, I can not close this. Sorry...

The number of steps required to model a reaction really depends on what time scales you are interested in looking at. Let's take the process of burning natural gas with air. If you only care about the final products of the combustion after a really long time, like let's say hours, then a single step reaction would be a perfectly good model. In fact, in situations where you only care about the global heat release and maybe the speed of the reaction at one particular condition, single-step or few-step reactions are just fine.
But let's say you are really interested in what's happening. When we look at a global reaction, let's say:
$$ O_2 + 2H_2 \rightarrow 2H_2 O $$
We'll assume this is in air, but the nitrogen doesn't really react. Nice and simple, right? Well, for the reactions to occur, you need some $O_2$ molecules to collide with $H_2$ molecules. But it's not enough that they collide. They need to collide with enough energy to break the bonds between them. And it's possible that one of the molecules breaks at lower energy than the the other. When they break, they break into 2 O's or 2 H'2. So our simple reaction really starts out as:
$$ O_2 + M \rightarrow 2 O + M $$
and
$$ H_2 + M \rightarrow 2 H + M $$
where $M$ is any other molecule. Okay, so now we've got a mixture of $H$, $O$, $H_2$, $O_2$, and $N_2$ floating around and bumping into each other. If they bump together with enough energy, we can start forming other things. Things like $HO_2$, $OH$, $H_2O_2$ and of course, $H_2O$.
Because these reactions rely on collisions, you need to form some things before you can form other things. So before an $O$ and an $H$ can collide to form $OH$, the $O_2$ and $H_2$ have to collide with other things and break apart. So there's finite time scales with all of the reactions, and they tend to cascade through a series of reactions before reaching the final product. If you are looking at time scales similar to those of the reactions, then these details really matter.
Hydrogen reactions are super simple. Imagine how complex organic molecules are. Even things like methane or propane have hundreds or thousands of steps to go from the complex molecule through all of the radicals to the final species. And the time scales the reactions occur over are all based on how frequently the collisions between them occur, and how much energy those collisions have.

Better question on Chemistry.
There is no physical law that prohibits multiple steps. Biochemical systems in particular tend to have complex mechanisms since very high specificity is required for the products.
Even for many nonbiological systems all the intermediates are often much more complex than the overall chemical reaction would indicate.


text:
Lewis structure of ClO radical
The commonly shown structure of the ClO radical described here has a single bond between the chlorine and oxygen atoms. In that structure, chlorine has a formal charge of 1, and oxygen has a formal charge of -1.
Why doesn't ClO have a double bond between the chlorine and oxygen atoms instead? That way, the radical would exist on the chlorine atom, and both atoms would have a formal charge of 0.

Your premise is false. Similar question would make sense in case of, say, SO3, but not here.

@Mithoron why is that so?

Formal charges are introduced in dative bonds, not usual single bonds!

You can calculate a formal charge for any atom. It may not be the correct approach, but there is no rule about dative versus single bonds.

"Premise is false" is not helpful. The simplest reason would be related to electronegativity; it is larger for oxygen than chlorine.


text:
Determining global minimum structures for ion-water clusters
What is the best way to find the global minimum structure(or get close to it) for an ion and N waters? (i.e. ion cluster in vacuum)

For example, consider Flouride ion cluster with N=5 waters.
I want to determine the global minimum structure (using Gaussian09 for a given level of theory and basis set).
This helped me get started (Geometry optimization: dependence on initial geometry), but initial geometry sampling is easier for molecules compared to clusters.

What is an intelligent way to generate reasonable initial configurations for this cluster (say for N=4,5,6)?
Constraint: My computational resources allows me to optimize about 10 structures at a given time for a reasonable level of DFT theory. I might be able to do a maximum of 100-200 optimizations.

@AjayMurlidharan Welcome to the site! If haven't yet, visit the [help] and take the [tour]. Not my area of expertise, but my understanding is people usually obtain reasonable starting geometries by running Molecular dynamics simulations.

@Tyberius: Thanks! Also, I thought about this. I already have an ab-initio MD trajectory (~50 ps) of the ion in bulk water (Flouride ion and 64 waters with Periodic boundary conditions).  I could in principle sample cluster configurations from here. Trying to figure out if this is the best way with my computational resource limitations.

Finding a global minimum of (generalizing here) a many-dimensional function is a great problem in numerical methods. In my opinion, a good place to start reading is about simulated annealing, which is commonly used in optimizing molecular configurations.


text:
What is the pH of this H2SO4 solution?
I'm trying to solve a simple problem which is driving me crazy. It says:

Estimate the pH of a solution that contains 1 gram of $\ce{H_2SO_4}$ dissolved in 1 liter of water.

When I solve the problem, I dissociate the $\ce{H_2SO_4}$ like this: $$\ce{H2SO4 + H2O <=> H3O+ + HSO4-}$$
And I get a pH of around 1.99
However, following my book's solution, the dissociation occurs as follows: $$\ce{H2SO4 <=> 2H+ + {SO4}^2-}$$
And they say the pH is around 1.69
Now, my question is: why is my dissociation equation wrong?

There's nothing wrong with it.

Nothing wrong with my solution (pH=1.99), or with my book's solution (pH=1.69)?

It's diluted acid so both protons are dissociated. Also this thing was done to death...

I don't understand the downvote, I'm beginning with Chemistry, I find this subject very hard, and on top of that I'm afraid of asking here because of downvotes. I don't know who else to ask to be honest.

Don't worry about the downvotes too much every new users gets a couple before they learn the ropes. Most likely it got downvoted because it's been marked as a duplicate. As for my earlier comment I meant that your dissasociation equation is correct but there will be another equation too $$\ce{HSO4- <=> H+ +  SO4^{2^-}}$$

@AvnishKabaj Thank you a lot for your kindness, now I understand the answer!

Note that this explains fully how $\ce{H2SO4}$ dissociates.

Your problem is that you only accounted for the first dissociation of $\ce{H2SO4}$, a polyprotic acid -- your book needed the extra specificity from the second dissociation. I will walk through the entire process, including the parts that you know already.
Begin by finding the molar mass of $\ce{H2SO4}$ in order to find out to how many moles one gram of it is equivalent. Then, convert to molarity (concentration) using the given volume of water.
$$\ce{MM_{H_2SO_4} = 2*1.01 g + 1*32.06 g + 4*16.00 g = 98.08 g}$$
$$\ce{\frac{1 g H2SO4}{1}\times\frac{1 mol H2SO4}{98.08 g H2SO4} = 1.0\times10^{-2} mol H2SO4}$$
$$\ce{\frac{1.0\times10^{-2} mol H2SO4}{1 L H2O} = 1.0\times10^{-2} M H2SO4}$$
Although the ICE-box is a formality for such a strong acid, it can still be shown.
\begin{array}{|c|c|c|c|c|}
\hline
\text{Initial}:& 1.0\times10^{-2} &  & 0 & 0 \\ \hline
 & \ce{H2SO4} & \ce{H2O} & \ce{H3O+} & \ce{HSO4-}\\ \hline
\text{Change}: & -x &  & +x & +x \\ \hline
\text{Equilibrium}: & 0 &  & 1.0\times10^{-2} & 1.0\times10^{-2}\\ \hline
\end{array}
The second ICE-box is a good way of organizing the second dissociation. Transfer the equilibrium concentrations from the first table. All calculations up to the line are for finding the change (using $\ce{K_{a(2)} = 1.2\times10^{-2}}$). Note that after $y$ is found, it is used again in the second ICE-box to determine the equilibrium concentrations after the second dissociation. Also note that you cannot neglect the $y$ after the second equation because of the similar magnitudes of the molarity and the $K_a$ and must use the quadratic formula.
\begin{array}{|c|c|c|c|c|}
\hline
\text{Initial}:& 1.0\times10^{-2} &  & 1.0\times10^{-2} & 0 \\ \hline
 & \ce{HSO4-} & \ce{H2O} & \ce{H3O+} & \ce{SO4^{2-}}\\ \hline
\text{Change}: & -y &  & +y & +y \\ \hline
\text{Equilibrium}: & 0.5\times10^{-2} &  & 1.5\times10^{-2} & 4.8\times10^{-3}\\ \hline
\end{array}
$$\ce{K_a = \frac{[H3O+][SO4^{2-}]}{[HSO4-]}}$$
$$\ce{1.2\times10^{-2} = \frac{(1.0\times10^{-2} + y)(y)}{1.0\times10^{-2} - y}}$$
$$\ce{1.2\times10^{-4} - (1.0\times10^{-2})y = (1.0\times10^{-2})y + y^2}$$
$$\ce{1.2\times10^{-4} = (2.0\times10^{-2})y + y^2}$$
$$\ce{0 = y^2 + (2.0\times10^{-2})y - 1.2\times10^{-4}}$$
\begin{split}
\ce{y} & = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\
& = \frac{-(2.0\times10^{-2})\pm\sqrt{(2.0\times10^{-2})^2-4(1)(-1.2\times10^{-4})}}{2(1)}\\
& = \frac{-2.0\times10^{-2}\pm\sqrt{4.0\times10^{-4}+4.8\times10^{-4}}}{2}\\
& = \frac{-2.0\times10^{-2}\pm\sqrt{8.8\times10^{-4}}}{2}\\
& \approx 4.8\times10^{-3}
\end{split}

Plug into the p function to determine the pH.
$$-\log(1.5\times10^{-2})=1.82$$
Note that $-\log(2\times10^{-2})=1.69$ so your book probably rounded to one significant figure (which would make sense given the way the problem is worded).


text:
What is the difference between X-ray photoelectron spectroscopy and X-ray absorption spectroscopy?
XPS is used for surface chemical state identification.
But other than that, is XAS just an inverse of XPS?

You could easily google this...
X-ray photoelectron spectroscopy uses low energy monochromatic x-rays to excite electrons from atoms near the surface of a sample. The kinetic energy of the electrons is then measured and then the binding energy of the electrons can be calculated.
X-ray absorption spectroscopy measure the attenuation of a x-ray beam as a function of x-ray wavelength (or energy). So what you are detecting is the absorption edges of the atoms in the sample.

@porphyrin - Yes. Thank you for the need for a correction.

But from the absorption edge, can we also calculate the binding energy?

For inner electrons you can look binding energies in a table. You really don't need to measure them again.


text:
Double surface in the pressure temperature concentration curve for a binary mixture

In the 3 dimensional diagram given, when a constant pressure plane intersects with the double surface, it produces a looped curve without a maxima or minima. What is the double surface and how is the looped curve produced? And since the line joining $T_\ce{A}$ and $C_\ce{A}$ is the temperature-pressure curve of pure A and similarly for B, I cannot understand what the curve joining the 2 critical points (of A and B) means.

The liquid phase is above the surface. The vapor phase is below the surface.
The planar cut at a constant temperature is easiest to understand. The bottom curve will show the composition of the liquid phase as a function of molar fraction, and the upper curve will show the composition of the vapor phase.
Likewise for the curves of molar fraction versus temperature at a given pressure.

Thank you for your insight.. And what about the curve linking the 2 critical points ? Does that curve mean that critical pressure of component A has exceeded and now it has become a single phase ?


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How to name long hydrocarbon chains?
It is not that I didn't google it before asking it here, I could it find it anywhere. My question must be simple but I'm really stuck upon it. How do you name long hydrocarbon chains? Like $\ce{C21H44}$, $\ce{C30H62}$, $\ce{C100H202}$ etc.
I found this when I googled but I didn't quite understand it.

Use the table mentioned in the linked article, §NT-1.1:
1   mono- or hen-*  10  deca-       100 hecta-      1000    kilia-
2   di- or do-*     20  icosa-**    200 dicta-      2000    dilia-
3   tri-            30  triaconta-  300 tricta-     3000    trilia-
4   tetra-          40  tetraconta- 400 tetracta-   4000    tetralia-
5   penta-          50  pentaconta- 500 pentacta-   5000    pentalia-
6   hexa-           60  hexaconta-  600 hexacta-    6000    hexalia-
7   hepta-          70  heptaconta- 700 heptacta-   7000    heptalia-
8   octa-           80  octaconta-  800 octacta-    8000    octalia-
9   nona-           90  nonaconta-  900 nonacta-    9000    nonalia-

just enumerate the digits of the carbon and add "-ne".
So $\ce{C21H44}$ becomes "hen-" "icosa-" "ne" = henicosane / heneicosane (the latter is just a spelling variation).
There are exceptions like propane ($\ce{C3H8}$) but these only occur for small molecules, up to $\ce{C4H10}$.


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Mathematical explanation of blank titration in Mohr method for argentometry
In my book it is written that in order to minimize systematic error in Mohr method for argentometry it is a common procedure to conduct a blank titration with $ \mathrm{Ag^+} $ against a $\mathrm{CrO_{4}^{2-}}$ solution (with the same concentration of the original solution).
The book then suggests to subtract the $ \mathrm{Ag^+} $ volume needed to observe precipitation (in the blank solution) from the volume obtained as a result of the 'standard' titration.
I don't understand why I have to subtract that number, specifically I do understand that there is a systematic error but I don't get why it is the volume of $ \mathrm{Ag^+} $ obtained by the blank titration.

$\ce{AgCl}$ ppts before $\ce{Ag2CrO4}$. But in order to visually detect the $\ce{Ag2CrO4}$ you must add a slight excess of $\ce{Ag^+}$. Hence you use a "blank" with no $\ce{Cl^-}$ to account for the amount of $\ce{Ag^+}$ needed to get enough of the $\ce{Ag2CrO4}$ ppt to form so that it can be detected visually.


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The CASSCF orbitals
I have two questions about the orbitals that we get after the CASSCF calculations.
First, what are the CASSCF orbitals that we get from these calculations? I mean after optimization, in the CASSCF wavefunction we have various configurations (CSFs) with different weights and in each of these CSFs we have MO orbitals with different MO coefficients. So my question is why CASSCF software packages (in particular Gaussian) display MO orbitals after the CASSCF calculations? These orbitals belong to which CSFs? The CSF with the highest weight/coefficient?
Second, if the orbitals after the CASSCF calculations are valid, should they approximately the same as the initial orbitals (guess orbitals)? If they are not, it means that the initial guess should be changed?

Could you post an example input to generate this? MOs are not printed by default I think for any package, just the density matrix for each state.

The MOs will not be printed unless you request them. My question is not why they are printed (upon request), but how it is possible that we have orbitals when we have many configurations in the CASSCF wavefunction. Let’s say we do a CASSCF calculations. After SCF convergence we get 20 different configurations each with different MO coefficients. So how can that be displayed?

The orbitals in CASSCF (or MCSCF in general) are optimized for a specific electronic state (= linear combination of CSFs) or a user-defined (weighted) state-average.
If they would belong just to the largest CSF, they would be equal to the HF orbitals.
Also, there is no unique set of MCSCF orbitals, but they depend on how you set up the MCSCF calculation: if you are interested in multiple electronic states, you can optimize orbitals for each state individually by separate calculations, or find one set of orbitals describing all states in a balanced way for a state-average (yielding higher energies).
The MCSCF orbitals usually look very similar to the canonical orbitals, and have similar eigenvalues. Also occupation numbers should be somewhat similar, although they are no longer restricted to being integer values. Basically, the better the HF approximation for a specific system, the larger the HF configuration in the CI vector, and the closer should the MCSCF orbitals be to the HF orbitals.
If you have large differences, then you probably have a multi-reference problem. The whole point of MCSCF calculations is to find improved orbitals for such cases.

My question is just about the ground state. Let's say we select four orbitals to be in the active space and we do a CAS(4,4) calculations. After the calculation we get 21 CSFs each with different CI coefficients. So in total I think we should have 84 orbitals (21 CSFs * 4 orbitals). However, the visualization package just show me four orbitals instead of 84. So my question is why only four orbitals are shown and what are these four orbitals? Are they just belong to one CSF?

No, please read my first sentence again. The orbitals do NOT belong to any CSF. They belong to a certain linear combination of CSFs: the electronic (ground) state. Each CSF is a certain vector in the CI space, the ground state is a vector in CI space as well, just developed in terms of CSFs. The orbitals may be optimized for any arbitrary CI vector, but you want them to be optimized for the ground state (not CSF).


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What are the differences between monotropic and enantiotropic polymorphic transformations?
I'm studying polymorphism and came across the terms "enantiotropic" and "monotropic" transformations. The first is described as reversible, whereas the latter, irreversible due to metastability. But what does "reversible" and "irreversible" means practically? What exactly is metastability? What makes these two types of polymorphism distinguishable between each other?
For instance, it's used as an example of monotropic polymorphism the transformation of graphite to diamond. However, the same source claims the reverse operation is possible. As for enantiotropic polymorphism, it used the example of silica polymorphs (cristobalite and alpha-quartz) which also can be transformed from one another. If both are reversible, why are they different?

In enantiotropic there exist an transition temperature between two crystalline state in equilibrium but reverese is true for monotropic


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Ionization energy of C2-, C2, and C2+
My textbook says that $\ce{C_2}^-$ has a lower ionization energy than both $\ce{C_2}$ and $\ce{C_2}^+$. I calculated that the bond orders of $\ce{C_2}^+$, $\ce{C_2}$, and $\ce{C_2}^-$ are 1.5, 2, and 2.5, respectively.
Why doesn't $\ce{C_2}^-$ have a higher ionization than the other two molecules? Shouldn't the molecule with the larger bond order have the larger ionization energy, because it would be more difficult to remove an electron from the stable bond?

Bond order has to do nothing with ionization energy, it is the bond dissociation energy which has to be dealt with bond order. For minimum ionization energy, the electron has to be lost from the highest energetic orbital (outer).
In $\ce{C2^-}$, the electron is lost from high energetic $\sigma(\ce{2p_x})$ (if you take the bonding along x-axis) orbital, whereas, in the other two molecules electrons are lost from low energetic  $\pi(\ce{2p_y})$ or $\pi (\ce{2p_z})$ orbitals. So $E_{\infty} - E_\text{initial}$ is higher for the latter case, which makes $\ce {C2^-}$ having lower ionization energy.


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How does association within vapor or dissociation within liquid affect the ideal nature of a binary solution?
As I understand, when there is positive deviation of a binary solution from Raoult's Law, the equilibrium partial pressure curve approach ideality (i.e a straight line dependence of equilibrium partial pressure and pure component vapor pressure) when the mole fraction approaches unity for any one the components. This is true except when there is association within vapor or dissociation within liquid. What is this association and dissociation and how does this affect the ideal nature?