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Clarifying the concept of the dative covalent bond
Part 1: I was just reviewing concept of the dative covalent bond in some simple molecules. One of which was carbon monoxide.
Some structures I saw featured the use of formal charges, in which the carbon atom had a negative formal charge while the oxygen atom had positive formal charge. These did not show that a dative bond was involved.
Some I saw claimed the molecule remained neutral and showed a dative bond between the two atoms.
I understand that the formal charge is a theoretical construct and does not refer to actual charges on atoms. Rather, they indicate to the chemists where are the regions of electron-richness and electron-deficiency.
So these two structures do not contradict each other and each are limited in some way.
However, dative bonding in complex metal ions always show dative bonding which is the latter structure, never the former structure, with formal charges. Why is this so?
Part 2: Secondly, what are the differences between a dative covalent bond, where one species donating both electrons, and a typical covalent bond, where each species donates one electron. Other than the process of bond making, is every other characteristic of the two bonds the same?
If you think about it, a lot of covalent bonds are actually "dative". Consider all the bonds between nucleophiles and electrophiles. Isn't it always the case where the nucleophile donates a pair of electrons?
In my opinion, the only difference appears to only be the difference in the bond making process and all else is exactly the same.

Possible duplicate of The bond in coordination complexes

Second part may be a dupe of https://chemistry.stackexchange.com/questions/39293/how-to-distinguish-between-a-coordinate-bond-and-a-covalent-bond

I agree with Mith. This is a dupe of two different questions. Since it is also to not-so-related questions in one, I’m going to vote to close it as too broad.

@Mithoron I agree that my question is a duplicate, sorry for the mistake on my part as I did not see those posts when I searched the site. However, I would like to point out the accepted answers to the two questions on those two posts are somewhat lacking and do not give complete and coherent responses to my questions.

Philipp has just recently revealed in his reply to my comment claiming that based on his knowledge of MO, there is not much of a difference between the coordinate covalent bonds and non-coordinate covalent bonds. Jan has also clarified in his replies to my comments that the difference lies in whether the bond cleaves homo- or hetero- lytically upon addition of bond dissociation energy. Both answers are not substantiated by any sources regarding the views they hold and I find Jan's answer very dubious, based on my knowledge of chemistry.

Dupes happen all the time and earlier answers usually aren't satisfying, that's how life goes ;) You can get to chat if you wanna clarification or simply to chat.

However, the common understanding that both answers seem to share is that they both describe coordinate bonds as being polar covalent bonds, especially Philipps' response.


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Why is the enthalpy of combustion higher for n-butane as compared to 2-methylpropane?
N-butane has an enthalpy of formation of $\pu{-125.6 kJ/mol}$, whereas it is $\pu{-134.2kJ/mol}$ for 2-methylpropane. When comparing the heat set free at the combustion process, why is it higher for n-butane? Isn't the enthalpy of formation the negative of enthalpy of combustion and therefore 2-methylpropane should be higher?

"Isn't the enthalpy of formation the negative of enthalpy of combustion" – No.

Take 4 moles of carbon and 5 moles of H2. Synthesize a mole of n-butane and measure the heat evolved (125.6 kJ). Then repeat the process, but make 2-methyl propane (you'll measure 134.2 kJ). Then, with sufficient oxygen,  prepare 4 moles of CO2 and 5 moles of H2O from each of the previous two procedures and measure the heat of combustion (more for n-butane).
The total heat evolved to form the final oxidation products from carbon and H2 will be the same, but since the paths are different, the heats evolved in the two steps of each procedure will be different. n-Butane evolves less heat upon formation, therefore can (must) evolve more heat in the next step when it is oxidized to final products.

Isn't the enthalpy of formation the negative of enthalpy of combustion

No! That certainly isn't true.
Consider this equation: $\ce{CH4 + O2 -> CO2 + H2O}$. If you reverse it, you'll get: $\ce{CO2 + H2O -> CH4 + O2}$ which is clearly not an enthalpy of formation because:

the left hand side of the reaction does not have elements instead compounds
the right hand side of the reaction should have had a single product, but instead has two

I cannot comment on the second part of your question because I could not find standard data for heats of combustion of both the compounds.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

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Oxidizing agent, reducing agent, acid or base
In concrete reactions, how can you define whether Ammonia reacts as an oxidizing agent, reducing agent, an acid or a base? I'm confused because in the examples given it's not an acid/ base when an $\ce {H+}$ is donated/ accepted.
$$\ce {2NH3 + 3Cl2 -> N2 + 6HCl}$$ (oxidizing agent)
$$\ce {Na + NH3 -> NaNH2 + \frac{1}{2}H2}$$ (reducing agent)
$$\ce{NH3 + H2O + CO2 -> NH4HCO3}$$ (base)
$$\ce{NH3 + LiCH3 -> LiNH2 + CH4}$$ (acid)

In the example reactions, ammonia is incorrectly stated to be an oxidizing agent in the reaction with chlorine (here it acts as a reducing agent); when reacting with sodium metal, ammonia is an oxidizing agent and not a reducing agent as shown. Not sure that helps address your question.

The use of the terms "oxidizing agent" and "reducing agent" in this example apparently refer to the overall reaction. Ammonia certainly can act as an acid in the process of reacting with sodium metal, and as a (Lewis) base during the reaction with chlorine. However, the net effect of the reaction is a reduction or oxidation, rather than a simple proton transfer.

I'm basing this answer from translation of the original question (in German) which I believe is supposed to read:
Ammonia can exist in oxidised form, reduced form, acid or base. Name the form in which ammonia exist in the following: (correct me if I am wrong)
Focusing on the acid/base forms (3) and (4):

ammonia acts as an base by accepting a proton $\ce{H+}$ from $\ce{H2O}$.

Recall the Brønsted-Lowry theory of acids and bases which states: An acid is a proton (hydrogen ion) donor, and
a base is a proton (hydrogen ion) acceptor.

similarly on the last equation $\ce{NH3}$ donates a proton to $\ce{LiCH3}$ to form lithium amide $\ce{Li^+ NH2^-}$ ($\ce{LiNH2}$) and $\ce{CH4}$

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How do you calculate the concentrations of multiple reactants in a certain point in time?
For $\ce{A + B -> C + D}$, given that $v = k [\ce{A}] [\ce{B}]$. I know the derivation for $\ce{A->B + C}$ , being $\ln[\ce{A}]=\ln[\ce{A}]_0−νkt$, but I'd like to know how you can, given a point in time, calculate the concentrations of A and B in the first equation, given the initial concentrations.

You need to perform integration for that. Relate both $\ce{[A]}$ and $\ce{[B]}$ with a single variable.

Assuming $[\ce{A}]_0 =  [\ce{B}]_0$, it all boils down to a second order kinetics with $v = k[\ce{A}]^2$.

$$\frac{dA}{dt}=-kAB$$
$$\frac{dB}{dt}=-kAB$$subject to initial conditions $A=A_0$ and $B=B_0$ at time t = 0.  Let x = number of moles of A that react = number of moles of B that react.  Then,$$\frac{d(A_0-x)}{dt}=-k(A_0-x)(B_0-x)$$and$$\frac{d(B_0-x)}{dt}=-k(A_0-x)(B_0-x)$$Therefore, from both equations  $$\frac{dx}{dt}=k(A_0-x)(B_0-x)$$ subject to the initial condition x = 0 at t = 0.  This differential equation can readily be integrated using separation of variables and partial fractions.

Given dX/dt = k[A][B] and A0 and B0 where you failed to define what X is, along with no value for k? It can not be done. see https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Kinetics/Reaction_Rates/Second-Order_Reactions (and probably hundreds of other sites) for the integration over time. In short, a parameter x is inserted so that [A]= [A]0-x & [B]= [B]0-x, etc. and we integrate dx/dt which requires two cases: [A]0 = [B]0 and [A]0 ≠ [B]0 but I won't repeat the rest here. If they're equal [A] = [A]0 ÷ ([A]0kt + 1) and if they're not equal then generally we consider the ratio of [A]:[B] but determining either [A] or [B] is simple ; i.e. don't replace [A]0 - x, etc. with [A], etc. at equation 1.22 in the above link and then solve for x. Once you have x, you know [A]0 - x, etc.

Situation 2a on that site was exactly what I was looking for -- thank you!


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Acid-base reaction: Al(OH)₃ + NaOH
Why is the product of $\ce{Al(OH)3 + NaOH -> Na[Al(OH)4]}$ and not $\ce{NaAlO3H2 + H2O}$? Is this an acid-base reaction?

Think about it, $\ce{Al(OH)3}$ has a vacant orbital, desperately looking for anything that would generously give it a pair of electrons. With $\ce{NaOH}$ around, possessing free $\ce{OH-}$, why not?

Note that aluminum hydroxide is not fully acidic. It is actually amphoteric in nature i.e.  It dissolves in acid, forming $\ce{[Al(H2O)6]^3+}$ (hexaaquaaluminium) or its hydrolysis products and dissolves in strong alkaline solutions, forming $\ce{[Al(OH)4]-}$ (tetrahydroxidoaluminate).


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Ground state oxidation potential and Ionization potential from density functional theory (DFT)
Can someone please tell me how to calculate the ground state oxidation potential (GSOP)?
I understand that according to Koopmans' theorem, the negative of HOMO can be considered as the GSOP. Is there any other way to calculate GSOP?
Also, are GSOP and ionization potential same?

Bear in mind that Koopmans' theorem, as you stated, applies specifically to Hartree-Fock theory.

Short answer - compute the relative energies of the neutral and cation species
Typically, when people discuss the "oxidation potential," they're considering electrochemistry, so there's solvent around. Keep this in mind if you're trying to compare a calculation done in gas phase - the best comparison would be with an appropriate solvent calculation, although calibrating between a calculation and an experimental electrochemical potential requires work.
On the other hand, ionization potential is typically used for a gas phase measurement - for example, with XPS or UPS spectroscopy, where an electron is removed from a species.
Finally, there's a difference between the vertical ionization potential when an electron is ripped out of a species, but the cation hasn't had time to relax (change geometry) and the adiabatic ionization potential, which is the difference of energy between the neutral and the relaxed cation.

That's a long way of saying that if you want to compute an accurate ionization potential, you need to:
1) Optimize the geometry of the neutral species using an accurate method.
2) Take the optimized geometry, and run a geometry optimization with one fewer electron. (You will get both the vertical ionization energy as the first SCF energy, and the adiabatic ionization energy of the optimized cation.)
Subtract the total energies and convert to your unit of choice.

Note that I don't say anything about orbital eigenvalues because there's no guarantee that for DFT/DFA methods, that these will correlate with ionization potentials. Koopmans' theorem is not guaranteed to apply.


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Why does Le Chatelier's principle not apply to pure solids and liquids?
If we had a homogeneous reaction of pure liquids only, would Le Chatelier's principle still apply ?
Also, wouldn't increasing the amount of solid in a heterogeneous reaction mixture at equillibrium result in more product, therefore indicating a shift in equilibrium?

Le Châtelier does apply to pure solids and liquids - what happens is that pure solids and liquids experience almost no change in concentration[*] in chemical equilibria, so for all intents and purposes they are not affected by Le Châtelier considerations.
Heterogeneous systems
Imagine a system with two phases; for instance, a gas phase and a solid phase, and we have some sort of equilibrium between them; for instance if we put water in a closed container (remember that Le Châtelier only properly applies to closed systems), it reaches an equilibrium with its own vapour:
$\ce{H2O(l) <=> H2O(g)}$
Here we have two phases: a liquid phase and a gas phase. The molecules in both are the same but in different aggregation states, which crucially differ in a fundamental behaviour regarding volume: in the liquid, molecules occupy a fixed volume each in certain conditions. If you have twice the number of molecules, you get twice the volume, and the density remains constant. So, if the equilibrium above is displaced to the left, the concentration of water molecules in the liquid phase doesn't change: the size of the liquid phase does. You simply get more volume of liquid. The same is true about solids.
For the gas it's different: the gas phase will always occupy the whole free volume in the container, and what changes is its pressure (which is to say, its concentration). If you double the number of gas molecules, the volume doesn't change - it still fits the whole container - but their concentration doubles.
This is a simplified picture; formally, the concentration of real solids and liquids is not exactly constant. However, it is much, much less sensitive to conditions than gases or solutes, so for all practical purposes it's a constant. See the next section for a numerical example that also applies here.
Dilute homogeneous systems
When we deal with dilute homogeneous systems (only one phase), the concentration of the solvent can be safely treated as constant because it's so much larger than anything else that changes simply don't impact it.
Imagine that you are applying Le Châtelier to the self-ionisation equilibrium of water; if you have a sample of pure water, a very small fraction of the molecules dissociate per the following reaction:
$\ce{2H2O<=>H3O+ + OH-}$
If you have a pure water in equilibrium and you add $\ce{OH-}$ ions, Le Châtelier predicts that this will displace the equilibrium to the left, and some of the previously dissociated water molecules will associate again to form $\ce{H2O}$.
However, water itself is the medium of the reaction, so its concentration is pretty much constant. Let's calculate some actual figures to get an idea. The self-ionisation constant is $\mathrm{K_w=10^{-14}}$, so for a chemically pure sample of water we'd get these concentrations:
$[\ce{H3O+}] \approx 10^{-7} \mathrm{M}$
$[\ce{OH-}] \approx 10^{-7} \mathrm{M}$
$[\ce{H2O}] \approx 55.55 \mathrm{M}$
Now, imagine we add a lot of $\ce{OH-}$ ions - and we make their concentration a thousand times bigger than before - that will have a big impact on the concentrations of both $\ce{H3O+}$ and $\ce{OH-}$. When we reach equilibrium we'd have
$[\ce{H3O+}] \approx 10^{-10} \mathrm{M}$
$[\ce{OH-}] \approx 10^{-4} \mathrm{M}$
Note that this means that 99.9% of the dissociated molecules have been displaced to the left and formed water. However, this number of new water molecules (around 0.0000000999 M) is simply irrelevant compared with the previous concentration of water (55.55 + 0.0000000999 is still 55.55 for all practical purposes). Even further, as with heterogeneous systems, even if the addition of more water were relevant, its impact would be increasing the volume of the phase, not changing its concentration.
That's why the concentration of water isn't even included in the self-ionisation constant - yes, formally the equilibrium depends on $[\ce{H2O}]$, but since that's pretty much always the same, it's easier to factor it into the constant and write the equilibrium as $\mathrm{K_w=[\ce{H3O+}][\ce{OH-}]}$.
[*] More precisely, activity

But wouldn't adding more solid to a reaction mixture lead to more product , which means a shift in equilibrium

No, because the concentration is the same, and that's what matters, thermodynamically speaking.

But when we increase the amount of liquid H2O in a closed system, doesn't this increase the pressure? Why this pressure is not increasing the concentration of gas H2O?


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What metal is under the stainless layer when it wears of?
I have this belt for several years now and the belt buckle is naturally subject to wear and tear.
On the most exposed spots the outer layer is now completely worn of and I wonder

Is this an example of passivation of the inner material?
What is the outer layer made of? Stainless steel?
What is the metal under the outer layer? It looks (and smells) like copper but why would they make it out of copper? Copper is – as far as I know – more expensive (by volume comparison, because the belt buckle can't be made smaller) than stainless steel?
Wouldn't it be cheaper to make it only from stainless steel (assuming the outer layer is from stainless steel)? It would also be better from health perspective because I suspect the inner layer causes skin irritation.

I intentionally scrubbed the bottom edge of the buckle before taking the photo to reveal the real color of the metal, on the top edge it is presumably covered by its natural oxide layer (copper has it green, right? But it's just a thin layer so it can look different I suppose).
I am kind of ashamed answers to these questions aren't already known to me because this effect happens on a lot of daily used metal objects...

I believe it may be bronze. That was a bit cheaper considering the fact that the Bronze Age came before the Iron Age.

Looks like nickel or chrome plated copper, maybe. If it's old enough, it may come from an era when 18-8 stainless was a lot more expensive than it is now. Almost certainly it would make more sense to make a modern buckle out of 18-8 SS.

1 =no , 2= no, 3 Likely brass , 4= no. The buckle is likely a brass although the bottom edge where you polished it is copper color . That could result if you used some cleaning agent. It could be made from copper but modern mass production would use brass. It was plated with nickle and / or chromium. Nickel plating is usually put down first , under chromium plating.

Thank you for the answer! I haven't used any cleaning agent, just a metal scourer.


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Intermolecular forces between C and Na2O
I was trying to do an experiment called Pharaoh's Serpent and I read that the intermolecular forces between the $\ce{C}$ and the $\ce{Na2O}$ (formed in the thermal decomposition of the $\ce{NaHCO3}$) produced the "serpent" in the experiment, but I can't find what intermolecular force (if there's really one) that is.
What is the structure of this compound?
Here's a link to the experiment: https://www.youtube.com/watch?v=vQdK7gaZS0k
edited:
Here are the reactions that I can think of and I researched :

Sodium bicarbonate decomposition ( with temperature ) :

$$\ce{2NaHCO3 -> Na2CO3  +  CO2 +  H2O}$$
$$\ce{Na2CO3 -> Na2O + CO2}$$
2.Fructose complete combustion:
$$\ce{C12H22O11 + 12O2 -> 12CO2 + 11H2O}$$

Fructose incomplete combustion:

$$\ce{C12H22O11 -> 12C + 11H2O}$$
$$\ce{C12H22O11 +6O2 -> 11H2O + 12CO}$$
So , I found that the $\ce{C}$ ( produced in the fructose incomplete combustion) reacts with the $\ce{Na2O}$ ( produced in the sodium bicarbonate decomposition), composing the "body" of the "snake".

And where do you have Na2O molecules there, I wonder, cause not in solid...

https://en.wikipedia.org/wiki/Mercury(II)_thiocyanate#Pharaoh.27s_serpent There's no Na2O or elemental carbon or bicarbonate in this reaction so I'm afraid you deeply confused here and question should be closed as unclear.

You can do the experiment using NaHCO3 or Mercury(II) thiocyanate. But since Mercury(II) thiocyanate is toxic  , i decided to do using NaHCO3. Heres a link showing how to do the experiment https://melscience.com/en/experiments/sugar-snake/

Bicarbonates decompose to CO2 not C and still no idea what "intermolecular forces" you want.

It's sucrose not fructose and combusts completly. I have a feeling that editing won't help here :( I appreciate your responsiveness and additional effort, though.


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Law of Mass action VS Rate law
The Law of mass action has 2 parts - the equilibrium rate part & the reaction rate part.
The "reaction rate part" states that rate of a reaction is directly proportional to the product of active masses of the reactants with each reactant term raised to it's stoichiometric coefficient. The rate law states that rate of a reaction is directly proportional to the product of active masses of the reactants with each  reactant term raised to some power which may or may not be equal to stoichiometric coefficient of the reactant.
Why are the two laws different when in fact we are talking about the powers to which active masses of reactants should be raised in both the cases ? And which law is correct?

As far as my knowledge is concerned reactant can be one or more. In case when there is one reactant first part is applied and when there are two or more reactants second part is applied.

This question has been answered in a separate question:https://chemistry.stackexchange.com/questions/68195/law-of-mass-action-collision-theory

First of all, you need to understand the basic difference between the two.
Consider the reaction $$\ce{A + B -> C}$$ Law of mass action states that in a rate equation the powers that the active masses of components raised are always stoichiometric factors.
For a reaction they are always fixed. However, the rate of  reaction differs with the physical conditions. This is applicable to both equilibrium reactions and moderate reactions that are unidirectional.
This makes law of mass action kind of theoretical in nature. In the rate law, the rate equation consists of actual factors that changes with change in experimental conditions. These may or may not be equal to stoichiometric factors. These are generally responsible for determining the order of reaction.
Both the laws are correct because both have a wide area of application. Rate law is although quite experimental in nature and is generally considered to be more appropriate.

In which cases do the law of mass action give the exactly correct relations?

Your statement of the Law of Mass Action is wrong. From Wikipedia:"The law is a statement about equilibrium and gives an expression for the equilibrium constant, a quantity characterizing chemical equilibrium." You apparently confuse the dynamic equilibrium where concentrations of reactants and products do not change with the rate of chemical reactions not at equilibrium (generally at constant temperature, and always with changes in concentrations). Keq is the equilibrium rate constant, the Reaction Quotient is Q. Q changes as the reaction proceeds and reaches Keq as dynamic equilibrium is attained.
It is NOT true that the rate of a (forward) chemical reaction is proportional to the concentration of the active masses of the reactants raised to some empirical power. Some reactions proceed at constant rates. These are known as zeroth order reactions - catalyzed reactions can be examples of these. (While you can argue that if R → P and if d[R]/dt = -k then d[R]/dt = -k[R]° (since concentration taken to the zeroth power is 1) that the proportionality holds, but this is trivial and not useful.)
In the historical development of the Law of Mass Action, there have been many instances where the powers for the Reaction Quotient at equilibrium have matched (more or less) the empirically determined powers of the rate expression. This is quite different than CONFUSING a situation at equilibrium with the situation which must, by definition, be non-equilibrium (since concentrations are changing).
Which is correct?
Logically invoking a dynamic equilibrium requires concentrations to be unchanging. Say R → P. With [R] and [P] constant, then taking the ratio of [R]α/[P]ß for some arbitrary constants α & ß is also going to produce a constant. Choosing the stoichiometric coefficients for α & ß doesn't change that.
For the rate of a reaction, the claim is that the rate of (forward) reaction is proportional to [R1]α[R2]ß...[Rn]n. This often isn't even an acceptable approximation. (although over a small enough time interval, any continuous (non-pathological) data can be approximated by a straight line). The more difficult question to answer here is does the rate of reaction for simple single-step reactions follow this rule? I'd say in dilute gas phase reactions (without significant effects from products) (and of course at constant temperature and pressure) it generally will. The problem with this, is determination of which are and which are not "simple single step" reactions isn't trivial. You could start there and confirm the proportionality. You'd have to confirm it, which makes it mostly unnecessary to assume it to begin with, but you have to start somewhere.

The argument that you could choose any powers to raise the concentrations to (when at equilibrium as the concentrations don’t change, so neither will the quotient of those concentrations) is not a justification for always choosing to raise to the power of the coefficients. Raising to other powers will still give a constant, but it won’t generally have the same value as the actual equilibrium constant.


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Why methanol is more acidic than water
All alcohols are less acidic than water except methanol. Methyl group exhibits positive inductive effect. So water should be more acidic than water. But according to experimental evidence it is not so. Can someone explain the reason for this.

Your question has been asked previously, and has an answer here.  For future reference, you're able to search chem.SE using the 'search Q&A' bar at the top of the page, or equally the linked question appears on google if you search for "why is methanol more acidic than water".

@NotEvans. the answer there is wrong. Water has a pKa of 14.0 (not 15.7 like Evans pKa table shows. https://chem.libretexts.org/Core/Organic_Chemistry/Fundamentals/What_is_the_pKa_of_water%3F
http://pubs.acs.org/doi/abs/10.1021/acs.jchemed.6b00623

@DSVA. Irrespective, the question is a direct duplicate and should be closed as such, you could always answer the original question, correcting the 'wrong' data

I am sorry. I didn't knew that the question was asked before. Please provide me with the answer.


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Concentration of acid when titration has reached equivalence point?
I'm currently working with a lab report concerning titration. One of the tasks is to calculate the concentration of the weak acid ($\ce{HOAc}$) and the strong acid ($\ce{HCl}$) at the equivalence point during titration (through $\ce{NaOH}$).
But shouldn't the concentration of $\ce{HOAc}$ and $\ce{HCl}$ in the new solutions per definition be 0. Hasn't all acid at that point been neutralized and converted to its conjugate base at the equivalence point?


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How can the trityl tetrafluoroborate function as a Lewis-acid catalyst

Trityl tetrafluoroborate is a reagent sometimes used in synthesis as a very mild Lewis acid catalyst, and recently I've been (unsuccessfully) using it in some protecting group chemistry.  Several other uses are mentioned in eROS:

...for various transformations, e.g. the Mukaiyama-type aldol reaction using a dithioacetal and silyl enol ether. It has also been used as the catalyst for the formation of glycosides from alcohols and sugar dimethylthiophosphinates and for the formation of disaccharides from a protected α-cyanoacetal of glucose and a 6-O-trityl hexose. Michael additions of various silyl nucleophiles to conjugated dithiolenium cations also proceed well. Finally, the [4 + 2] cycloaddition of cyclic dienes and oxygenated allyl cations has been effected with trityl fluoroborate.
Ref: Encyclopaedia of Reagents for Organic Synthesis

Clearly, the trityl cation could act as a Lewis acid catalyst (or at least its reactivity suggests that it does), and indeed it has a vacant p-orbital allowing it to accept an electron pair (the IUPAC gold book definition being a molecular entity that is an electron-pair acceptor).
What makes less sense to me is the 'mechanism' this catalysis occurs by. The cation itself is pseudo-stable (relative to a standard planar carbocation intermediate such as that observed in an SN2 reaction)  since the aromatic rings are able to stabilise the positive charge, but even if it did react, surely one would expect this to be irreversible, such that a nucleophile would attack the vacant p-orbital to form a stable compound, killing off the reaction.
To give an example of this, trityl tetrafluoroborate may be used to protect an alcohol with a DMB protecting group. What isn't observed is the alcohol attacking the cation to form an O-trityl bond (the trityl group itself is sometimes seen as a protecting group).

Doesn't it work via SET mechanism? You get intermediate radical that is oxidised later?

One point is that you're assuming that the "p" orbital in the center is the Lewis acidic site. This may not be the case, you can draw a resonance structure that breaks aromaticity but puts the cation at the para position of the phenyl ring. For example, the ortho and para positions are known hydride acceptors (link).
Another issues to consider is that the phenyls are bulky enough that they can't even remain coplanar.
So we can return to your issue of reversibility by looking at trityl via these two properties:

If the Lewis acidic site is via one of the phenyl rings, then the resultant adduct is destabilized by breaking aromaticity. It may be kinetically accessible and allow the trityl ion to serve as a Lewis acid, but the adduct is not stable.
If you do create an adduct with the central carbon, this adduct is greatly destabilized by steric repulsion. Enough so that fragmentation becomes likely.

In both cases, there is a kinetically accessible Lewis acid/Lewis base adduct, but this adduct has some kind of feature that makes it thermodynamically less stable. Therefore, the trityl ion can serve a catalytic role as a Lewis acid.
The exact mechanism of action (1) or (2) is likely dependent on the exact reaction you're looking at.

I'm a little unconvinced by (2). Trityl protected alcohols are readily formed and require moderately forcing conditions to remove. I will however look for some literature on the phenyl group being the LA centre (intuitively this makes me uncomfortable but I can maybe buy into that argument)

I think it depends on how hindered the alcohol is. I gave away my copy of Wuts/Greene, but you might want to check there for the types of alcohols that can be protected. My guess is that tertiary and maybe secondary alcohols are not effectively protected with trityl.


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Is turpentine (as a solvent in plant-based paints) toxic after application? How long?
Turpentine is often used as a solvent in plant-based paints. It is believed that
turpentine is equally if not more toxic than white spirit (see Green Building Handbook. Volume 1 by Tom Woolley,Sam Kimmins, Paul Harrison and Rob Harrison. p.229). Is it toxic after paint application and for how long?
Updated:
I believe that question can be divided on two simpler subproblems (it's worth to consider them together):

what is the speed of turpentine level reduction in regular environmental conditions? As function of time in analytical form or better graphical curve.
determination of the safe level of turpentine in room according to some authoritative standard. This level likely is disputable, so knowing the speed will help - maybe waiting some additional time (a couple of months) before starting to settle in the room will eliminate the risk completely.

Yes it's toxic, but the "how long" part depends on ambient conditions - temperature, humidity, ventilation, and so on. Eventually it goes away: standing in front of a van Gogh is hardly a danger. You might want to narrow the scope of your question - for instance, if you are inquiring about a certain amount within a confined space or the like.

Thank you. humidity 40%; 25 C  with opened windows. Painted walls of the room. 1) Is period from van Gogh was alive till present time sufficient for 100% elimination of Turpentine (when it's impossible to detect Turpentine) or for level when amount of  Turpentine is negligible (recognized by some strict standard as safe for humans)? 2) What is the shortest period when the level of Turpentine in the paint of the walls under specified conditions is very close (say difference is <0.1% from the level in the very first days after paint application) to level in 1).i.e. when  it becomes safe to live?


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Stabilisation of carbocations through hyperconjugation
Our teachers told us that greater the number of alpha H, greater is the stability of carbocation.
But consider this:
$\ce{CH3CH2+}$      and.      $\ce{CH3CH2CH2+}$
The first one has 3 alpha H while the second one has 2. So first one is more stable, as our teachers say. But even C-C sigma bond can participate in hyperconjugation, isn't it?
Going with that, second one has equal no bond resonating structures and even higher inductive effect of ethyl group. So second one should be more stable (without overruling the dominance of hyperconjugation).

See this earlier Q&A, Which carbocation is more stable, the ethyl- or 1-propyl-carbocation?

Hyperconjugation
Hyperconjugation, as you point out, is able to stabilise carbocation intermediates, IUPAC define this as:

the interaction is ... between σ-bonds and an unfilled or partially filled π- or p-orbital
Ref: IUPAC Gold Book

In a simple sense, we can visualise this interaction by drawing the carbocation and considering the adjacent σ bonds able to interact.

This can also be visualised in a more 'quantitative' way by considering a simple mixing of orbitals, in which we can see that the σC-H -> p+ lowers the energy relative to the vacant p-orbital.

Ref: Molecular orbitals and organic chemical reactions, Fleming
Stability of carbocations
If we now consider the various types of cation, we should be able to rationalise the order of stability.

The methyl group is completely unstabilised through hyperconjugation, due to the stereoelectronic requirement that the σ bond be located on the adjacant carbon (of which the methyl cation has none).
Primary carbocations have a maximum of 3 σ-bonds capable of hyperconjugation, secondary a maximum of 6, tertiary a maximum of 9– hence why the tertiary is the most stable cation out of the simply alkyl groups (the allyl and benzyl systems in the diagram above are stabilised through delocalisation of the charge onto the pi system).
Going back to your question of ethyl vs propyl carbocations, you can hopefully now see that they have the same number of σ-bonds in the correct position to participate in hyperconjugation.
An additional complication
Primary carbocations are sufficiently unstable that if they form, they tend to rearrange into more stable carbocations, such that the primary propyl cation will quickly undergo a 1,2-hydride shift to give the secondary propyl cation. A more complex example is shown below:

And leaving that secondary carbocation at the very bottom lying around for too long should result in another Wagner-Meerwein this time with a methyl group migrating to give a tertiary carbocation


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Folin–Ciocalteu formula
I am searching a long time now. I have a question about the Folin–Ciocalteu formula for quantification of polyphenol. How is the $62.5$ in the equation explained? Where does it come from?
This is the protocol [1, p. 38]:

$\pu{5.0 mL}$ of extract were placed in calibrated flask and filled up to $\pu{25 mL}$ by distilled water. $\pu{1.0 mL}$ of Folin-Ciocalteu reagent, $\pu{10.0 mL}$ of distilled water were added to $\pu{2.0 mL}$ of diluted extract (from $\pu{25 mL}$ calibrated flask) and filled up to $\pu{25 mL}$ by sodium carbonate solution ($\pu{290 g/L}$). An absorbance of prepared sample was measured by means of UV spectrophotometer at $\pu{760 nm}$ after $\pu{30 min}$ of incubation in darkness. The same liquid (with pure water instead of plant extract) was used as a blind test. All determinations were performed in triplicate. The content of polyphenols was calculated from the formula:
$$X = 62.5 \cdot \frac{A_1 \cdot m_2}{A_2 \cdot m_1}$$
where $A_1$ - absorbance of pyrogallol solution, $A_2$ - absorbance of investigated samples, $m_1$ - mass of investigated sample [$\pu{g}$], $m_2$ - mass of pyrogallol [$\pu{g}$].

Reference

Modnicki, D.; Balcerek, M. Herba Polonica 2009, 55 (1), 35–42.

Authors mention that pyrogallol (PG), among other standards, has been purchased from Sigma Aldrich US and used as is without further dilution. The amounts of polyphenols were calculated as PG equivalent by external standard method, and utilizing additivity of absorption coefficients the proportion has been established.
$62.5$ is just a dilution factor for the herb extract probe. From the protocol, first time the extract was diluted $5$ times (from $\pu{5 mL}$ to $\pu{25 mL}$), second time – $12.5$ times (from $\pu{2 mL}$ to $\pu{25 mL}$), resulting in total dilution factor of $5 \cdot 12.5 = 62.5$.


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Reaction of weak alkali and strong acid
How the alkali molecules can dissociate completely in the reaction of weak alkali and strong acid?

Welcome to Chem SE. I'm not exactly what you're trying to ask. Could you clarify a bit more?


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Functionalization of nanoparticles
I want to functionalize silver nanoparticles with thiol group.
My question is "how can I find that the thiolated compound is attached to the surface of nanoparticles or not?"

If the thiolate bounds to the Au nanoparticle, both the electronic, as well as the mechanical properties of the thiolate and the Au nanoparticle change.  So you are going to monitor these changes with a spectroscopy susceptible to this, comparing the spectra of the pure starting materials with the one(s) of the decorated NP.
One of the easier available techniques is to rely on FT-IR spectroscopy, as by the established bond between the two partners, some absorption bands newly appear, and others change their position.  The former are the ones more easily taken as a quantitative reference, too (because yes, IR spectroscopy allows both qualitative, as well as quantitative characterisations, too).
In the instance of Au NP, to check for a bonding of the NPs to the mercapto decoration is possible both by IR spectroscopy, as well as by (surface enhanced) Raman spectroscopy.  In a work deploying $\omega$-mercaptoundecanoic acid (MUA) by Tripathy and  Yu, the later was the preferred check to monitor the S-Ag interaction:

(source, doi 10.1016/j.saa.2008.12.004, Spectrochimica A, 72, 2009, 841-844).
Similar as in the case of the Au NP (vide infra), the key feature is the disappearance of the S-H vibration.  To quote from the section about their characterisation by Raman spectroscopy:

The most interesting feature was the disappearance of S–H stretching
  ($\pu{2547 cm^{−1}}$) peak. The disappearance of S–H stretching indicated the covalent interaction of sulphur and silver, which is in good agreement with the FT-IR data [16]. However some of the expected peaks such as C–Sgauche stretching could not be identified which may be due to their low intensity values.

the earlier answer, erronusly about Au NP
As an example, were gold nanoparticles were decorated by dodecanthiol (DT) chains:

(source, Sharma et. al. in Indian Journal of Pure & Applied Physics
52, 2014, 93-100.)
Other than the absorption band at $694$ and $\pu{728 cm^{-1}}$, the authors equally draw attention that

the "S-H vibration band at 2576 cm−1 for pure DT is observed while it is almost absent for AuNPs. This suggests that as the thiol gets bound onto the  surface of gold, the S-H bond of thiol breaks in order to form the S-Au bond." (loc. cit., p. 97).

It may be helpful to mention that short wavelengths of light ($\pu{< 400 nm}$) may break the $\ce{Au-S}$ bond between nanoparticle, and thiol, hence if you doubt the presence of thiolate decoration on the NP, select an absorption band and monitor its attenuation while shining the short wavelength light on the sample.
And complementary to IR spectroscopy, is Raman spectroscopy.

Nice answer. My only comment is that the OP specifically refers to Ag nanoparticles, whereas your answer exclusively discusses Au nanoparticles. How do you think the FTIR spectra in particular would compare between and Ag and Au analogs?

I find it kind of ironic that if there is a question explicitly asking about silver nanoparticles, the answer that exclusively talks about gold, gets upvoted, while the one that tells silver is different from gold, gets the downvotes. Nice answer by the way.

@airhuff Sorry, this was an error by mine.  I found a work including spectra of the two worlds (IR and Raman) of Ag NPs and mercaptans and added this ahead of the section of Au NPs.  Of course the best were to have an example "same mercaptan, once to decorate AgNP, the other on AuNP" that however I did not find (yet).  On the other hand, attenuation of the SH-stretch appears to be (again) the indicator.

@Greg The error to display Au instead of Ag is on my side; by an edit I rectify for the situation. The later surfacing answer by #Vic Lineal offers a sound additional insight, appreciated by a tip to the hat (+1), too.

It's complicated, because there is no general pattern of reactivity for thiol compounds and silver nanoparticles (unlike, say, gold nanoparticles, for which the reaction with thiols is widely studied and understood). The chemistry and local structural changes are varied; cysteine for instance is known to form well-defined layers around silver nanoparticles, but unfortunately it also tends to aggregate them. Different thiols vary in the local surface chemistry they produce - some generate $\ce{Ag2S}$-like layers, and that can compromise the stability of the nanoparticles.
I'd suggest reducing the list of possibilities to the reagents you can use and then looking for literature to see how they characterise the reaction.


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What can be done to increase the yield of a product in a chemical reaction that has reached equilibrium?
Let's take the following reaction that has reached equilibrium
$$\ce{CaCO3 (s) <=> CaO (s) + CO2 (g)}$$
What can be done to increase the yield of calcium oxide ?
My initial thought was to remove the carbon dioxide produced in the reaction mixture. I understand that this is correct since, according to Le Chatelier's principle, the reaction would shift to increase the yield of the products.
Why wouldn't adding more calcium carbonate lead to a shift to the products and hence lead to a greater yield of the products? Doesn't more reactant mean more product?

Is that reaction a true equilibrium? The loss of carbon dioxide makes it somewhat irreversible...

If you ran the reaction in a closed system, you could reach a state of true equilibrium under certain conditions of temperature and pressure. A complicating factor is that two of the components of the system are solids, which may  be slow to react, especially if not in the form of very small particles. You can "shift the equilibrium" by removing or adding a component, but you no longer have a closed system, nor do you have a true equilibrium. The modified system can of course be closed again and brought to a new state of equilibrium, perhaps as a way to increase the yield of the desired produc

You could also leave the system closed, and get a higher yield of product by shifting the conditions (T, P)  and letting the system re-equilibrate.

But why would adding more calcium carbonate not cause a shift , my textbook just states that's it's because it's  a solid , so no shift means no greater yield of product by why is that so, shouldn't the yield be greater  because of the greater amount of reactant

Generally only the surface of a solid reacts so it matters whether there is, say, one large crystal or a fine powder. Also a product formed on the surface may inhibit any more reaction.

@porphyrin Could you please explain in more detail, because wouldn't more surface area due to added reactant mean more molecules available to react , hence this could lead to an increase in product therefore signifying a shift in equilibrium

it was just a general comment that it is the surface area that might be limiting rather than mass of material.

Note that the question about adding more solid reagent was already asked here: https://chemistry.stackexchange.com/questions/76387/why-does-le-chateliers-principle-not-apply-to-pure-solids-and-liquids/76390

While it can be frustrating sometimes when systems don't act in the way we intuitively expect them to, asking again won't change that.

I agree with your initial starting point, that the thermal decomposition of calcium carbonate into calcium oxide and carbon dioxide may be treated as an example of an heterogeneous equilibrium
$$\ce{CaCO3 (s) <=> CaO (s) + CO2 (g)}$$
that still is of relevance today

(source, Hughes et al. in Ind. Eng. Chem. Res. 2004, 43, 5529-5539).
and actually at around $\ce{900 ^\circ{}C}$, the equilibrium pressure of $\ce{CO2}$ over $\ce{CaCO3}$ and $\ce{CaO}$ would equal to $\pu{1 atm}$.
Your first reasoning, while keeping the other parameters of this equilibrium fixed ($T, p$), a removal of $\ce{CO2}$ may increase the yield of $\ce{CaO}$ since this eventually (macroscopically) leads to an exhaustive decomposition of $\ce{CaCO3}$ is correct.

For the second part, you have to balance the general idea about chemical equilibria (which in this case indeed means, keeping other parameters fixed, adding more $\ce{CaCO3}$ as starting material yields more $\ce{CO2}$) with the fact that this particular reaction is heterogeneous.  Looking up the data for molar mass and density, (1, 2, and 3) allows you to estimate the volume one mole of each pure compound under standard conditions occupies:
| compound | molar mass |  density | molar volume |
|          |    [g/mol] | [g/cm^3] |   [cm^3/mol] |
|----------+------------+----------+--------------|
| CaCO3    |     100.09 |    2.711 |        36.92 |
| CaO      |      56.08 |     3.37 |        16.64 |
| CO2      |      44.01 | 0.001977 |     22261.00 |
|----------+------------+----------+--------------|

The value of the molar volume of $\ce{CO2}$ may remind you to the value of the volume occupied by the Ideal Gas at standard conditions ($\pu{22.7 L/mol}$) and demonstrates that the influence of $\ce{CO2}$ on the position of the equilibrium is much more important than the two other of the solids with their -- in comparison -- often negligible molar volumes. Hence, to underline that this equilibrium is this much pressure dependent, it is described by $K_{\text{p}}$, rather than by a normal $K$ value.
The general idea of influencing a chemical equilibrium along "an increase of  partial volume (or partial pressure) of starting material(s) will increase the yield of the product(s)" better full-filled in the case of homogeneous reactions, like the synthesis of ammonia
$$\ce{N2(g) + 3 H2(g) <=> 2 NH3(g)}. $$

Could we use Le Chatelier principle even for endergonic reactions to increase the yield?

@Anton Good question if global change of the volume (e.g., ammonia from nitrogen and hydrogen) may simply overrule an endergonic energetic balance.  Right now, I have no example which would support this.  Note, the generation of ammonia from nitrogen and hydrogen is exergonic.

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Why is the force constant for double and triple bonds approximately double and triple of that of a single bonds?
I am new to spectroscopy, and I am referring to Donald Pavia's Introduction to Spectroscopy. I came across this statement in the book, under the chapter Infrared Spectroscopy:

$K$ is a constant that varies from one bond to another. As a first approximation, the force constants for triple bonds are three times those of single bonds, whereas the force constants for double bonds are twice of those of single bonds.

As far as I know, double bonds are not considered as two single bonds. Double bonds are rather one $\sigma$ bond, and one $\pi$ bond. Neither the bond energies of double bonds are twice the energy of corresponding single bonds, nor is the bond length of double bonds half of the single ones. Then how is the force constant doubles as the bond order doubles?

IIRC from my undergrad, the 1:2:3 ratio is an approximation based on the two-mass model, which estimates 500 N/m, 1000 N/m, 1500 N/m respectively for single, double, triple C=C bonds .

@NotEvans. Um whats the two-mass model?

Force constant is not the same as bond strength. Remember that an approximate measure of bond energy from the bond as a simple harmonic oscillator would be $\frac{1}{2}kx^{2}$ where $x$ is the bond length.

@Zhe Ahh, I haven't thought of that. Thanks for your input. Do you want to transform it into an answer so I can accept it?

Force constant is not the same as bond strength. Remember that an approximate measure of bond energy from the bond as a simple harmonic oscillator would be $$E=\frac{1}{2}kx^{2}$$ where $x$ is the bond length. In this case, you would also need to account for differing lengths of single, double, and triple bonds to recover the bond energy trend.
Note that this is not how a quantum mechanical harmonic oscillator is normally described. There, you would have quantized energy levels of $(n+\frac{1}{2})\hbar\omega$, where $\omega = \sqrt{\frac{k}{m}}$. But since the higher order wave functions "concentrate" at the classical turning points, you might still be able to use the classical harmonic oscillator as a model for what is going on, with strong caveats.
For reference, consult: https://en.wikipedia.org/wiki/Quantum_harmonic_oscillator


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Will Sodium Silicate dissolve in acetone?
I recently discovered (through Wikipedia)  that water glass / sodium Silicate dissolves in water (22 g / 100 ml @ 25 degrees and 160 g / 100 ml @ 80 degrees).
Doesn't that mean it'll dissolve in acetone?

Welcome to SE. The statement you gave and what you're asking aren't really related. Could you clarify a bit more?

I would like to infuse wood with water glass.  Acetone is used to dissolve epoxies to a very thin consistency.  I was hoping that since water glass is soluble in water, it can dissolve in acetone, as well.  Or at least it won't react if I added acetone to the water glass solution (of water and sodium silicate).

Sodium silicate partially dissolve in acetone solution forming a homogeneous mixture. Hence, it is miscible in acetone solution. From pubchem:

GLASS FORM SOL IN STEAM UNDER PRESSURE; PARTIALLY MISCIBLE WITH
  PRIMARY ALC & KETONES; MISCIBLE WITH SOME POLYHYDRIC ALC
Sax, N.I. and R.J. Lewis, Sr. (eds.). Hawley's Condensed Chemical Dictionary. 11th ed. New York: Van Nostrand Reinhold Co., 1987.,
  p. 1072

Well acetone is ketone, isn't it? You're contradicting yourself.

sigh Dissolving means forming homogeneous solution....


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In atomic diffusion in Au/Cu metal contact, where do the atoms come from ? (need melting temperature to break bonding in lattice ?)
It is known that Au metal is diffused when it is contacted with copper metal. And Au metal have lattice which consist of atoms. If so, may I think that atom is diassociated from lattice by heating ? If so, it needs to break bond between atoms before diffuion ?
I think that it needs melting temperature to break bonding in lattice.

Yes, it takes energy to get an atom off a lattice site. However, as an activated process with an energy barrier, Arrhenius' equation shows you that even with that requirement, some number of atoms will be off their lattice sites in thermal equilibrium, making vacancies (empty sites) and interstitial atoms.
This was realized some time ago by Simmons and Balluffi. They investigated many metals, using precision measurements of the change in length of a bar of material in a furnace, coupled with x-ray diffraction to determine the lattice constant at the furnace temperature. The result is you can back out how many lattice sites are vacant from the difference in the change of length actually measured vs what you would expect from the change in lattice spacing.
The linked article focuses on gold, but has some data on other fcc metals as well. Just at the melting point of gold (not yet melted), the concentration of vacancies is about $7 \times 10^{-4}$, or 0.07% of lattice sites are vacant. The activation energy is given as 0.94 eV.
Bottom line - there are native populations of point defects even at room temperature, and those drive diffusivities. Certain point defects or defect complexes may be more mobile than others, and the concentration of a given defect will vary with temperature.


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Why boiling can't take place in closed containers
I am being told that boiling can't take place in closed containers. Can someone please explain why this is so.

Can you reproduce the reason you've been given for why boiling can't occur in a closed container?

Recall the fact that boiling point depends on external pressure.

If 'closed container' means 'closed system', boiling of course can take place. But I'm not sure whether both stands for the same thing or not...

To me closed container very much means closed system. Container as in a tupperware, or a pot with a lid.

I am being told that beyond the ocean there is an island inhabited by two-headed people. Boiling surely can and does happen in closed containers; it is just that the pressure increases in the process, and so does the boiling point.

You can understand this well by thinking about why a liquid boils in the first place.  Imagine you have a pot of water, and it's just sitting there in your kitchen.  Intermolecular forces (specifically hydrogen bonding for H20 molecules) keep the water in liquid state.  The air molecules zipping around the pot and above the water's surface represent a certain pressure.  Let's say your kitchen has standard conditions, so 1 atm.  Then, you put the pot on a stove, and heat the water up.  This heat breaks the intermolecular bonds between the liquid molecules.  Now, the only reason the molecules rise up and become water vapor is because the kinetic energy of the water molecules matches that of the surrounding air.  Pressure moves from high to low, and as the water boils, certain molecules acquire the necessary speed (and therefore pressure) to enter the gaseous state.  Keep in mind that heating the pot negligibly changed the kitchen pressure.  The pressure in your kitchen should still be about 1 atm, since the vast air space should distribute the heat produced by the stove.
In a closed container, however, the story is different.  Let's say you put a lid over your pot.  Now, there is a relatively small amount of air trapped between the lid and the water's surface.  Heating the pot heats the air as well, and pressure does not remain at 1 atm.  That is because that small amount of air has to handle all that heat, whereas in the situation without the lid, a bigger volume (the kitchen) allowed for more heat dissipation.  So now the small amount of air molecules are moving really fast.  The heat is still breaking the intermolecular forces of the water, but this time, there is no (entropic) reason to move up into the gaseous state.  The air molecules are moving just as fast, and therefore have just as much pressure, as water molecules that have broken bonds.
The key to your question is rooted in why the water molecule (without a lid) moves up out of the surface of the water and joins the air molecules, effectively becoming water vapor.  Simply put, it is because the water molecule (because of the heat) has enough speed to zip out of the water.  Pressure is intrinsically based on speed of molecules, and so you can think about this situation as: the water molecule(s) has a higher pressure than the air molecules, and thus they move up to join the air molecules, in order to balance out the pressure.  A lid prevents this from happening because the water molecule(s) pressure will never be higher than the air molecules, since the air takes the same amount of heat as the water.  (The preceding statement is true only in usual cases.  There are caveats to this - see comments.)
Therefore, water does not (usually) boil in a closed container.  A couple of things: There are many more explanations/videos online about this topic (boiling water, pressure cookers, more altitude = lower pressure, higher boiling point etc..)in case you want to strengthen your understanding further and explore the core of this question from different angles.  Also, the intuition in this Khan academy video https://www.youtube.com/watch?v=hA5jddDYcyg is pretty good.

Mostly OK as an explanation except that water can and does boil in closed containers. But the pressure and temperature can get very high so the point where boiling happens is very different. And can be very dangerous.

Apologies for leaving out the caveats to my explanation.  You are correct, water CAN boil in a closed container, for example, if you add heat really quickly, or like you said, high temperatures.  I was only trying to confine the problem to the context of the question, and I chose not to venture outside of what the asker was getting at.  Thanks for the comment.

I would clarify that water molecules are always being exchanged between the gaseous and liquid state. Boiling is the phenomenon where there is a bulk movement of water molecules, in the characteristic form of bubbles, into the gaseous phase.

@oceandevil24 Hey,I have a doubt. When the container is closed does the pressure of air trapped inside the container is now 'atmospheric pressure ' or are we still talking about the pressure outside the container? And so the water can boil in closed container when the vapour pressure becomes equal to this 'atmospheric pressure ' of the container?

I am being told that boiling can't take place in closed containers. Can someone please explain why this is so.

Source: https://upload.wikimedia.org/wikipedia/commons/thumb/9/9c/Simple_distillation_apparatus.svg/2000px-Simple_distillation_apparatus.svg.png
A liquid can boil in a closed container under certain conditions. For example, you can close off a distillation apparatus by putting a balloon on the only opening with the surrounding (position 9 in the picture). As long as you condense the vapors created by the boiling liquid using a cooler, you can boil off the entire liquid. (You should allow for some expansion, otherwise the glassware will be under high pressure - a dangerous condition.)
Also, if the container is large enough (like a house), you could boil water in it on a stove top with an open pot. You can also put a lid on the pot, and it will boil a little bit faster (because the lid keeps cold air from coming in). The boiling point might rise a little over the normal boiling point because the pressure in the house will increase a little bit as you boil off the water.
Heating liquid in a closed constant volume container
If the container is closed, all vapor generated by boiling will stay in the container. If it is at constant volume, the pressure will rise in it. That is what happens in a pressure cooker at the beginning of the cooking process. The pressure rises, and so the water in it will not boil at the normal boiling point. Instead, it will start boiling at a higher temperature. In a pressure cooker, there is a valve to release gas when the pressure reaches a certain value. Without the valve, the pressure would continue to rise until it stops boiling again, or the pressure cooker will explode.

boiling point is the temperature at which vapour pressure of the liquid becomes equal to the external pressure. But in case of closed vessel on boiling volume remains the same so pressure increases due to increase in temperature,which also cause the pressure of the air molecules in the container to rise, along with that of the liquid present so in that case vapour pressure of the liquid remains lower than that of the air molecules and therefore boiling does not occur.

Boiling water in a closed container essentially requires you to add heat to a constant volume. As noted above, this increases pressure, and ways to get around this basically open the system.
However, heat pipes are small closed systems (really closed, and really small), and liquids are boiled in them to transfer heat from one end to the other. Pressure increase is nominal (i.e., within design specs). https://en.wikipedia.org/wiki/Heat_pipe
The key is that you can't keep adding something (neither material nor energy) to a closed system without exceeding some limitation. Heat pipes boil a liquid at one end and condense it at the other. So the system can remain closed because the contents don't increase: you put in heat, you take out heat.

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Are there any poisons with an activation threshold?
Are there any poisons that lay dormant unless a significant quantity of one has been added, i.e. poison X might as well be water in concentrations less than 5M, but at 6M, all of a sudden a threshold has been reached and the poison has activated.

Anything that that the body has a mechanism to filter out fit into this description you pose. If the does was acute and high volume, the poison would overwhelm the body's ability to process and excrete the poison and produce the negative effect. I think you can find more detailed discussion and an actual answer by searching this site.

How about alcohol then? Seems to me it also has some activation threshold until negative effects kick in ; )

Poisons based on inactivating an enzyme might work like that: as long as there is some critical amount of a vital enzyme left inactivated, the organism could be comparatively healthy. It seems you're writing a murder mystery...

This is pretty much how anything that enters your body works.. even water.

Yes, there are certainly some chemicals that are harmless in low quantities but really dangerous in higher doses.
The common painkiller paracetamol (or acetaminophen) is one such candidate. In low doses it is a mild analgesic but in higher does it is deadly. The reason is that there is a metabolic side-reaction in the liver that produces a small amount of a toxic compound. Under normal circumstances this is mopped up by other pathways in the liver and causes no damage. But when large amounts are taken the liver's capacity to mop up the toxin is exhausted, the toxin builds up to dangerous levels and it causes severe liver damage often resulting in death.

Close, but it's not exactly what I'm looking for, since that would be how the body deals with it rather than the mechanism of the poison itself.

@Tzacol You might struggle to find a poison with those requirements: all poisons depend on how the body interacts with them; you can't separate the effects.

Well the question was about the existence of them, if such a thing doesn't exist, then it doesn't exist and I have the answer.

@Tzacol You don't have a useful answer since no poison "exists" except through its interaction with something. So the question is meaningless unless you specify what is being poisoned and how (and, once you specify that context, threshold poisons exist).

Seems pretty useful to me. I've got my answer, what I'm asking doesn't exist.


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How to determine concentration limits of a buffer that's limited by solubility?
I need to maintain a certain range of % of ionisation from a buffer.
For e.g. Histidine buffer. When making high concentration of this buffer at (say) pH 7.2, 25'C, [concentration] is limited to 75mM solution (250mL), due to the solubility. Adding HCl increases solubility just as much, but it also decreases pH of the solution.
I want to know if there's a calculation or a formula to figure out the concentration limits (at least the highest concentration) of a particular buffer one can plan to make instead of trial and error in the lab?

@Martin - マーチン even giving a bounty, I did not get an answer or response to my previous questions. I just wonder if there's anyway to receive community attention and support?

bon, you can only ping someone in a comment if they're the author or an editor of that particular post. You would need to ping Mart in chat, for him to get a notification. As to not getting an answer, the site's seen a significant uptick in the number of questions posted lately, and a lot of questions are getting overlooked. Asking specifically for assistance, in chat or in a comment as you did, is probably your best bet.

@hBy2Py oh yeah that's good to know. well do you have any advice for finding a solution to my question? it doesn't have to be Mart.

I've been thinking about this on and off -- I can sort of piece together a general approach, but solubility of mixed solutions, especially ~concentrated ones, can get kind of complicated and it's always system-dependent.


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Can I use a digital multimeter as substitute for a zero resistance ammeter?
I am testing the galvanic corrosion between two metals, following ASTM G71. In this method, a zero resistance ammeter was the material presented to determine the galvanic current measurements. Unfortunately, we don't have access with it. So can I use a digital multimeter instead?

There are no truly zero-resistance ammeters - just very low. You should look at the specs for your digital multimeter to see what the resistance is in ammeter mode. It may be higher than (or lower than if lucky) what you need for your experiment. There certainly are very low resistance digital ammeters.

Yes, using Ohm's law: I=V/R.
The only true "zero" resistance ammeter are superconductive. In practical use, the voltage drop across a small resistance is measured. If your DVM can measure down to a millivolt (mV), then a 1 mV drop across a 0.001 ohm resistor shows a flow of 1 ampere current. You'll need to consult the guide (~$45 online) for the maximum permissible voltage drop, and use an appropriate resistance.
Recently, I used the voltage drop across a piece of #16 copper wire to measure currents in the 1 to 5 A range... a switch in the circuit had greater resistance!

Not at all, Zero Resistance Ammeters are dedicated testing equipment provided with internal feedback to automatically compensate for voltage losses caused by the measurement. They actually have "infinite" input resistance, so don't load the source of current in any way, but are backed by an op-amp based circuit which acts like a zero impedance current source.
Without a solid theoretical evaluation (impedence matching), you cannot substitute such instrument with a common digital ammeter.


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Restricted open-shell versus unrestricted methods for open shell systems
Open-Shell systems (radicals and so on) can be modeled by single-determinant quantum chemical methods using unrestricted (u) or restricted open-shell (ro) methods (I am aware that single-determinant approaches are not always reasonable for open-shell systems, but let's ignore that for a second and suppose we have to use a single-determinant method).
Unrestricted calculations may suffer from spin-contamination. Restricted open-shell calculations remedy this, they are unaffected by spin-contamination. However, restricted open-shell methods are much more computationally expansive. Also, Koopmans' theorem cannot be rigorously applied in the case of ro calculations.
My question is:
Apart from the before mentioned issues, are there any disadvantages of restricted-open-shell computations in comparison to unrestricted calculations? Are there any situations where an unrestricted calculation would be more appropriate than a restricted-open-shell calculation?
[Methods of interest are - for example - MP2, DFT and HF.]
There is already a somewhat related discussion: U- or RO-method for Singlet-Triplet Gap?. However, my question is much broader, it includes for instance: Are there cases when it is absolutely not ok to use ro methods, and you have to use unrestricted methods?

One rumored disadvantage of RO is less reliable and possibly slower convergence. This may also be a result of the implementation at hand.

To be more precise, it's not so much that restricted open-shell calculations are unaffected by spin-contamination as that you are forcing them to give back the multiplicity you want.
Broadly speaking, closed-shell methods, restricted open-shell methods and constrained unrestricted open-shell methods all give correct spin multiplicities, but that's because all of them assume from the beginning which is the correct multiplicity and force it into the calculation - there is always a restriction in place. For closed-shell methods, it is that the orbitals occupied by both spin populations are the same; for restricted open-shell methods, the same applies to all but the excess spin electrons from the larger spin population. But more recent developments point to restricting multiplicity after the fact - by adjusting orbital population to minimise spin contamination. There are a couple of methods to do that, but they start from an unrestricted calculation (much faster, as you mention), and then force the population in a way that minimises spin contamination - either by forcing orthogonality constraints between the $\alpha$ and $\beta$ determinants, or by parametrising spin contamination in a way that allows it to be fitted to the expected total multiplicity by constraining occupation. This retains the computational advantage and simpler mathematical implementation of unrestricted methods but makes sure that the multiplicity of the model system reflects the multiplicity of the modelled system with generally very small additional computational effort.
I'd suggest that this is going to be an active area of research that will see increased use in the next few years.

You don't even have to consider open-shell systems to see differences between RHF and UHF. A very popular example is the dissociation of $\ce{H2}$. RHF is simply not capable of describing the dissociated fragments and will not converge with respect to the internuclear distance $R$. UHF will be spin contaminated, but you get the right energies (apart from correlation effects, which are negligible for 2 separated hydrogen atoms.)
You should be able to find this easily in standard quantum chemistry text books.

Thanks for reminding me of this simple example system. Is there a way to generalize the learning from this model to larger systems? Would you - based on your answer to my broad question - argue, that unrestricted methods are always superior to ro methods?

To cite from your question: "Unrestricted calculations may suffer from spin-contamination." So no, you cannot argue that one or the other will always be superior.
In UHF you sacrifice a well defined spin quantum number to get the degree of freedom of splitting your spatial orbitals into separate spin orbitals.
If such a result is not good enough, you will need to go to more sophisticated methods.

The real question is:

What do I gain from spin contamination?

Spin contamination is usually presented as a bad thing. However, the extra (sometimes large) spin polarization that can occur when relaxing the spatial constraint on the wavefunction is what gives UHF a little bit of electron correlation beyond RHF and ROHF. This is why you should be wary of people who blindly state "there is no correlation in Hartree-Fock". They probably mean "when compared to wavefunction-based methods", but without context, it isn't true!
I will steal from Frank Jensen's book, section 4.4:

By including electron correlation in the wave function, the UHF method introduces more biradical character into the wave function than RHF. The spin contamination part is also purely biradical in nature, i.e. a UHF treatment in general will overestimate the biradical character of the wave function. Most singlet states are well described by a closed shell wave function near the equilibrium geometry and, in those cases, it is not possible to generate a UHF solution that has a lower energy than the RHF. There are systems, however, for which this does not hold. An example is the ozone molecule, where two types of resonance structures can be drawn.

The biradical resonance structure for ozone requires two singly occupied MOs, and it is clear that an RHF type wave function, which requires all orbitals to be doubly occupied, cannot describe this. A UHF type wave function, however, allows the $\alpha$ and $\beta$ orbitals to be spatially different, and can to a certain extent incorporate both resonance structures. Systems with biradical character will in general have a (singlet) UHF wave function different from an RHF.

If you look for the singlet with RHF, there will be almost no biradical character in the resulting wavefunction, unless there is a biradical-looking orbital delocalized over the whole molecule. I don't know if there is, just guessing. With the biradical you can also see how the triplet state will enter the picture; I am leaving the details to Jensen's chapter. There is an analogy between UHF including too much biradical character with MP2 overestimating the dispersion interaction; it is the first level of theory where the effect is being considered qualitatively, and should not be expected to be quantitatively accurate. Though, UHF is probably worse at describing biradical character than MP2 is at quantifying the dispersion energy.
I haven't thought much about if this is considered to be dynamic or static correlation if we perform that artificial division. For example, this singlet biradical can be described by two determinants composing a single configuration state function (so TCSCF would work, and some flavors of ROHF), but you can capture the same effect after the fact with EOM-CCSD. It might be considered a static effect, since the presence of higher spin states in the resulting wavefunction results in some configurational mixing.

HF describes exchange correlation exactly, so the statement "there is no correlation in Hartree-Fock" should always be considered wrong.

One particular application where it is a bad idea to use the assumptions of RO, is the simulation of NMR or EPR spectra.
If you include the coupling of nuclear and electronic spin (which you have to) you have to allow different energies and population for the inner s-orbitals.

So what you really want to say is that spin polarization of the core orbitals is needed, not just the valence orbitals, especially for something like the Fermi contact interaction. One would not want to perform ROHF indirect spin-spin coupling constant calculations.