text:
Schrödinger equation for multi-electron atoms
Does the Schrödinger equation alter its form when we apply it for an atom having a configuration other than hydrogen-like? I think it should happen because of difference in nuclear attraction and shielding effects.
If there is a change, how significant is that?

May be useful: https://chem.libretexts.org/Textbook_Maps/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Quantum_States_of_Atoms_and_Molecules_(Zielinksi_et_al.)/09._The_Electronic_States_of_the_Multielectron_Atoms/9.1%3A_The_Schr%C3%B6dinger_Equation_For_Multi-Electron_Atoms

The Schroedinger equation doesn't change. The components of the Hamiltonian do.

@ToddMinehardt That's much better than what I said... :)

That probably depends on what you consider to be the "form" of the Schrödinger equation.
As Todd succinctly said, the Schrödinger equation[s]
$$\mathrm i\hbar \frac{\mathrm d|\psi\rangle}{\mathrm dt} = \hat{H}|\psi\rangle \quad \text{(time-dependent)} \tag{1}$$
or
$$\hat{H}|\psi\rangle = E|\psi\rangle \quad\text{(time-independent)} \tag{2}$$
don't change. The Hamiltonian will, of course, be different, but the question is how different? The general form of the (non-relativistic) Hamiltonian for an atom has three components
$$\hat{H} = \color{red}{\underbrace{-\frac{\hbar^2}{2m_N}\nabla_N^2}_{\text{KE of nucleus}}} \,\,\,\, \color{blue}{\underbrace{- \sum_i \frac{\hbar^2}{2m_e}\nabla_i^2}_{\text{KE of electrons}}} \,\,\,\, \color{green}{\underbrace{-\sum_i\frac{Ze^2}{4\pi\varepsilon_0r_i}}_{\text{nucleus-electron PE}}}\,\,\,\, \color{purple}{\underbrace{+ \sum_i\sum_{j>i} \frac{e^2}{4\pi\varepsilon_0r_{ij}}}_{\text{electron-electron PE}}} \tag{3}$$
where the nucleus has mass $m_N$ and charge $Ze$; and $r_i$ denotes the distance between electron $i$ and the nucleus, and $r_{ij}$ denotes the distance between electrons $i$ and $j$.
Whether you are dealing with a hydrogen atom or a silver atom, this form remains valid. The only special thing about the hydrogen atom (or in a hydrogenic ion) is that there is only one electron, so the first two sums only have one term, and the last sum has no terms. So, the Hamiltonian for the hydrogen atom can be simplified to
$$\hat{H}_\text{H atom} = \color{red}{-\frac{\hbar^2}{2m_p}\nabla_p^2} \color{blue}{- \frac{\hbar^2}{2m_e}\nabla_e^2} \color{green}{- \frac{e^2}{4\pi\varepsilon_0r}} \tag{4}$$
but I'm of the opinion that the form of the Hamiltonian
$$\hat{H} = \color{red}{\hat{T}_n} + \color{blue}{\hat{T}_e} + \color{green}{\hat{V}_{en}} + \color{purple}{\hat{V}_{ee}} \tag{5}$$
doesn't change, it's just that there are fewer terms in each sum. You might disagree, but I don't really care, as long as you write the correct Hamiltonian out.

There are many terms for a given problem as well as a system of equations for all of the interactions. The answers for even 3 body gravitational problems are chaotic, and gravity is less complication than schrodinger.
However, using numerical methods, symmetry, and observation we can model how many-electron atoms behave.
Here is an example of how the mechanics of a helium atom is
determined.
Quantum numbers arise as discrete parameters for solutions to the schrodinger equation as well as being observable. The Pauli exclusion principle is empirically found. We combine concepts together to model a larger atom, and then further to model molecules (Molecular orbital theory).
The Schrodinger equation works well, but it isn't the whole story.


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Which complex is the most stable?
Between $\ce{[Rh(Ph2PCH2PPh2)3]^{3+}}$ and $\ce{[Rh(Ph2PCH2CH2PPh2)3]^{3+}}$, which one is the most stable thermodynamically? I think it's about the size of the chelate ring effecting their stability, but i'm not sure.

I'm not sure that $\ce{Rh}$ can handle either of these  three bulky bis(diphenylphosphino)alkane ligands. Bis-rhodium seems more plausible to me.

I've just found in Science of Synthesis: Houben-Weyl Methods of Molecular Transformations, Vol. 1, Georg Thieme Verlag, 2014 on p. 964 that $\ce{[Ru(\eta^2-H2C=CHPh)(dppm)2)]}$ is way more stable (towards light and heat) that $\ce{[Ru(\eta^2-H2C=CHPh)(dppe)2)]}$. Whether better stability of bis(diphenylphosphino)methane-complex can be extended to you case, I'm not sure.

It is risky to predict just by looking at the structure, of course there is the ring size effect and one could easily conclude that $\ce{[Rh(Ph2PCH2CH2PPh2)3]3+}$ is the most thermodynamically stable compound since you would have a 5-membered ring instead of a 4-membered as in $\ce{Rh(Ph2PCH2PPh2)3]3+}$. However, there is also the contribution of the electronic effect, which is the increase or decrease in electron density of the metal ion caused by the distribution of ligands around the metal center. The combination of electronic and steric effects was reviewed by C.A Tolman in 1977 for monodentate ligands and he proposed a way of measuring these effects experimentally and quantitatively, they are the so-called Tolman's parameters (Tolman, C. A. (1977). "Steric effects of phosphorus ligands in organometallic chemistry and homogeneous catalysis". Chem. Rev. 77 (3): 313–348.)
For bidentate ligands such as DPPE (2 $\ce{-CH2 -}$ groups between phosphorus atoms) or DPPM (1 $\ce{-CH2 -}$  group between phosphorus atoms) another important parameter is the bite angle "β" (the circular sector angle between the metal and the two phosphorus atoms). For example, in square-planar and octahedral complexes the ideal bite angle is 90º , deviations from this value will electronically destabilize the complex (KUMAR, M.  et al. Importance of Long-Range Noncovalent Interactions in the Regioselectivity of Rhodium-Xantphos-Catalyzed Hydroformylation. Organometallics, v. 34, n. 6, p. 1062-1073, Mar 2015.).
Bidentate phosphines with a natural bite angle of 120° preferentially occupy diequatorial sites in a trigonal bipyramidal complex whereas a bidentate phosphine with a natural bite angle of 90° preferentially occupy apical-equatorial positions (Casey, C. P.; Whiteker, G. T.; Melville, M. G.; Petrovich, L. M.; Gavney, J. A.; Powell, D. R. (1992). "Diphosphines with natural bite angles near 120° increase selectivity for n-aldehyde formation in rhodium-catalyzed hydroformylation". J. Am. Chem. Soc. 114 (2): 5535–5543.).
DPPM has a bite angle of 78º while DPPE has a bite angle of 85º (closer to 90º), so considering only the angular aspect, the $\ce{Rh-DPPE}$ complex would be more stable (Ligand Bite Angle Effects in Metal-catalyzed $\ce{C−C}$ Bond Formation
Piet W. N. M. van Leeuwen,, Paul C. J. Kamer,Joost N. H. Reek, and, and Peter Dierkes, Chemical Reviews 2000 100 (8), 2741-2770)*. Electronically, I assume that the 2  $\ce{-CH2 -}$ groups of DPPE could donate more electronic density to the P atoms than DPPM, which would somehow strengthen the $\ce{Rh-P}$ bonds.
However, it is the combination of sterical + electronic effects that can unequivocally decide which of the two complexes are more stable, and this would only be possible using computational chemistry or experimentally, though the above can be used when discussing about it on a statement or paper.

As @andselisk stated, the bis-Rh species is expected to be more thermodynamically stable. You also need to consider the metal itself; Rh. Rhodium is famous for often forming $\mathrm{d^8}$ square planar complexes (a lot more than octahedral complexes) which results in a four coordination number complex instead of six - this would strongly suggest that the bis-Rh complex involving bidentate ligands would be thermodynamically more stable based primarily on the chemistry of the metal itself. Whether this is is reflected experimentally I have no idea.

This is d6 Rh(III), not d8 Rh(I), though.

That's true, I missed that. My bad.

source: Stack Exchange

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

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Help finding a thixotropic suspension with specific properties for 3d printing
I currently doing research on 3d printing on circuit boards. The technique we are using is to put a conductive material, I.e ceramics/metal in a gel/polymer, to print the suspension and then when is complete to sinter away the gel/polymer. The problem I am having is the suspension needs to be thixotropic. Furthermore it must have a high viscosity i.e 500 cpa under low shear and to quickly drop to low viscosity as the shear increases
So far the closest thing I have been able to find is in this paper but if you look at the graphs none of them have a high enough viscosity and more importantly the drop to low viscosity under increasing shear is too shallow.
I have tried to use models like the ones in this paper but I am finding it quite difficult.
Does anyone have any suggestions to find more materials like the one I require? Thanks

"500 cpa" -- what is cpa? I'm aware of $\pu{Pa \cdot s}$ and $\pu{cP}$, but cpa is something new to me.

my apologies I meant Pa s

@IntentFilters there's an SE site specifically for 3d printing you know. Do check it out, maybe you will get better answers: https://3dprinting.stackexchange.com

I have done so thanks. Unfortunately no answers so here I came.

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


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Synthesis of 2,3-dihydrobenzo[b][1,4]dioxin-2-yl)(piperazin-1-yl)methanone
My interest regarding Doxazosin made me curious, what's an easy way to synthesis Doxazosin. The first part is 2-chloro-6,7-dimethoxyquinazolin-4-amine. However, I would like to find a simple synthesis 2,3-dihydrobenzo[b][1,4]dioxin-2-yl)(piperazin-1-yl)methanone in order to get Doxazosin. I tried to form a synthesis, however it looks not very effective (too many steps, etc.).
My questions:
Is there an easier method?
Are there critical errors in my synthesis?

methyl acrylate bromation
formation the 1,4-dioxane via pyrocatechol
cleavage of the methyl group with LiOH
forming an acetylchloride
piperidine

I edited/ improved the steps with the help of Zhe.

The biggest issue I see here is that step 3 probably results in a Baeyer-Villager oxidation, which is not what you want. Probably easier is to start from methyl acrylate and then remove the methyl group with LiOH after cyclization. Steps 5 and 6 seem unnecessarily complicated. Just add the acyl chloride to an excess of piperidine (it's pretty cheap), and you should get very little of the double addition.


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Can a reference electrode be inserted into an electrochemical cell horizontally?
If I have an opening on the side of my electrochemical cell, I'm wondering if I can use this to insert a reference electrode. It is normally used to insert an L-shaped tube where you place the reference electrode, and is connected to luggin capillaries. I'm wondering if we can take out the entire connection with the capillaries and replace it with only the reference electrode (sans capillaries). This is in a 3-electrode system, so if I could insert the counter electrode in this way, that would work too. In my setup, I cannot do this with the working electrode.
From what I've read, it's good practice to not have the reference electrode vertical because of bubbles on the bottom of the frit (so most tilt it slightly), but I haven't read anything about a sideways electrode.

As long as there's nothing about the design of the reference electrode (RE) that will make it malfunction when oriented horizontally, it should be fine.  Obviously if there's free liquid inside the RE that will slosh away from the frit or the electrode internals (or spill out onto the lab bench!) when you tip it, then that won't work well at all. The concern about bubbles collecting on the bottom surface is very real, especially if you're not stirring very vigorously.
One practical downside to inserting the CE directly like this is that the frit of the RE will be liable to clogging/degrading much more rapidly than if you use a Luggin. Also, the composition of the internal electrolyte of the RE will change more rapidly if directly exposed to your working solution without a Luggin.
Of course, depending on your setup, this will allow you to get your reference much closer to your working electrode, reducing the $iR$ drop in your measurement.


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How do repulsive solute interactions affect the van't Hoff factor?
Context. Consider an ideal solution, one in which the intermolecular solvent-solvent, solute-solute, and solute-solvent interactions are all equal. The presence of a solute therefore produces (solely) an increase in entropy due to the increase in configurational microstates. This increase in entropy leads to the well-known colligative properties, characterized by the van't Hoff factor, the number of effective solute particles per molecule or salt of solute.
Now consider a solution in which the solute-solute intermolecular interaction is changed by $+\epsilon$, so that solutes are more attracted to each other. This decreases the van't Hoff factor due to clumping of the solutes, which we can rationalize from two different perspectives:

probabilistic: solute particles are more likely to be found close to each other (due to the favorable attraction between them), so we effectively have less solute particles as a whole, resulting in a lower van't Hoff factor
statistical-mechanical: microstates in which solute particles are close to each other are weighted more heavily; this perturbation from the uniform distribution of microstates decreases the entropy of the system, effectively decreasing the "strength" of the colligative effect and hence resulting in a lower van't Hoff factor

Both factors agree, so that's good.
Now consider a solution in which the solute-solute intermolecular interaction is changed by $-\epsilon$, so that solutes are less attracted to each other. Consider the two perspectives again:

probabilistic: solute particles are less likely to be found close to each other (due to the unfavorable attraction between them), so we effectively have more solute particles as a whole, resulting in a higher van't Hoff factor
statistical-mechanical: microstates in which solute particles are close to each other are weighted less heavily; this perturbation from the uniform distribution of microstates decreases the entropy of the system, effectively decreasing the "strength" of the colligative effect and hence resulting in a lower van't Hoff factor

Questions.

Why don't the probabilistic and statistical-mechanical viewpoints agree in the latter case?
How can the incorrect viewpoint(s) be modified to make them correct?

I am inclined to believe the statistical-mechanical viewpoint is correct, since it's more fundamental. I'm also somewhat leery of the implication that solute particles being less likely to be close to each other leads to effectively more solute particles.

If this implication is indeed wrong, why is it wrong? How about the converse, that because solute particles are more likely to be found close to each other, we effectively have less solute particles as a whole?
If this implication is actually right, why is it right?

I think that your first probabalistic argument is correct, i.e that as the solute interaction increases the effective number of them is reduced. In the second case, as the interaction become repulsive, the effective number of them cannot increase over the maximum number there. They are restricted in their configurational entropy because the repulsion  means that the chance of finding one next to another is greatly reduced. Thus the probabilistic aligns with your stat-mech one.
You can calculate the entropy of mixing in an ideal solution then in a 'regular' one in which the components interact with one another.
In the ideal solution a statistical argument based on number of ways of placing particles on a grid,  leads to the entropy being given by
$$\Delta S_{mix}= -R[n_1\ln(x_1)+n_2\ln(x_2)]$$
for $n_1,\,n_2$ moles and $x_1,\,x_2$ mole fractions of species 1 and 2.
In a solution in which there is interaction between (equal sized) molecules the free energy is
$$\Delta G_{mix} = R[n_1\ln(x_1)+n_2\ln(x_2)] +(n_1+n_2)x_1x_2E$$
where E is is an energy of mixing. This can be calculated by considering that one solute molecule interacts with energy $\epsilon _{aa}$, the solvent with energy $\epsilon _{bs}$ and between solute and solvent with $\epsilon _{ab}$. Again a statistical argument based on molecules being on lattice sites means that nearest neighbour interactions of the type a-a  and b-b occur with probability $x_a^2$, and $x_b^2$ respectively and between a and b as $2x_ax_b$ and the 2 arise because a_b and b-a interactions are counted.
The total mixed energy is then $E_{tot} \approx (\epsilon_{aa}N_a^2+ 2\epsilon_{aa}N_aN_b+\epsilon_{bb}N_b^2 )$$
but the change on mixing is this energy less that due to the two components unmixed which is $\approx (N_a\epsilon_{aa}+N_b\epsilon_{bb})$
which means that the extra mixing energy is $E = zN_0(2\epsilon_{ab}-\epsilon_{aa}-\epsilon_{bb})$
where $z$ is related to the coordination number on the lattice.
The entropy on mixing is then
$$\Delta S_{mix}= -R[n_1\ln(x_1)+n_2\ln(x_2)] -(n_1+n_2)x_1x_2\frac{\partial w}{\partial T}$$
where the second term $\displaystyle \Delta S^E =-(n_1+n_2)x_1x_2\frac{\partial w}{\partial T}$ is a positive number about $0.2R$ J /mol/K for benzene/cyclohexane and $0.01R$ for benzene /$\ce{CCl4}$
Changing $\epsilon_{bb}$ as you suggest will still increase or decrease $\Delta S^E$ acting through the derivative.


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General rules to predict solubility in water
We were taught these rules to predict the order of solubility in general cases.

This is the theory we were taught:
Case1) When number of cations is greater than number of anions,
Solubility increases down the group
Case2) When number of anions is greater than number of cations,
Solubility decreases down the group
**
Case3**) When number of cations = number of anions,

          Case3.1) Radius of cation> radius of anion (for anions like                        >.                                Chloride, oxide, fluoride, hydroxide, sulphide)

                           Solubility increases down the group.

           Case3.2) Radius of anion> radius of cation (for anions like        >.                                   Iodide, bromide and polyatomic anions**)

                           Solubility decreases down the group(but for                              >.                                  hydroxides, solubility increases down the group)

Case4) For heavier metal compounds, we always use the concept of
polarization which in turn is directly proportional to "intensity of
color" and inversely to "solubility" .

I haven't seen these rules in any book or elsewhere so I am assuming my teacher came up with these rules on his own after analyzing solubility order in many compounds.
Here are things I don't understand-
A) Why does solubility depend on the number of cations and number of anions?
B) Ionic compounds are soluble in water. For a general compound which doesn't show hydrogen bonding, its solubility should depend upon its percentage of ionic character. Comparison of percentage ionic and covalent character can be simply done using Fajan's rule. So is comparing solubility order by comparing their percentage ionic character wrong?
C) In case3, how does the radius of cation and anion affect solubility?

Solubility in what? Water, benzene? If you're talking about water then number 2 isn't right for group-2 elements. Eg. Solubility of $\ce{CaCl2, MgCl2}$ is in 100 mL 74.5 g & 54.3 g respectively.

According to Fajan's rule even your case 3 is wrong.


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Mechanism of Dehydrogenation Reaction
In the hydrogenation of alkenes over a metal catalyst, the two hydrogen atoms are added with syn stereochemistry.
In the metal-catalysed dehydrogenation of alkanes, is the removal of hydrogen likewise concerted with syn stereochemistry? And if so, why?

The answer you seek can be found here uwyo.edu/roddick/site/alkane_dehydrogenation.html , the catalytic cycle of dehydrogenation of alkanes by Iridium complexes does not follow a simple syn addition such as in the hydrogenation of alkenes. In the hydrogenation of alkenes, the $\ce {H2}$ dissociates and is added syn to the substrate in a well understood mechanism.
In the dehydrogenation though, the coordination of the alkane to the Ir metal center occur first, then a migratory insertion forming a 5-coordinated trigonal bipyramidal complex, then a β-elimination occur forming an unstable 6-coordinated complex which readily loses the alkene (final product). The catalyst is then regenerated by a hydrogen acceptor which can be a bulky alkene (so its product can't participate in the cycle due to steric hindrance). If no regio inducing group is present in the catalyst, the formation of the Saytzeff product (most substituted alkene) is favored.


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How can I perform calculations including the relativistic approximation SC-ZORA and solvent approximation COSMO in Gaussian 09?
I am working on transition metal centered oganometallic complexes.
How can I apply or run the relativistic and non relativistic zero-order regular approximation (SC-ZORA) using Gaussian 09 in a DFT calculation?
I would also like to know about the process of performing COSMO analysis in Gaussian.
I have only available Gaussian 09 package for this purpose.

Gaussian can’t effectively treat relativistic effects (look up integral=dkh [1] and try its limited use. AFAIK you can only have 50 parameters in internal coordinates which is not very realistic for common molecules today.) and has no ZORA implemented. It is also not able to do COSMO PCM calculations. Instead one would use other PCM schemes, e.g. IEFPCM which is the default in Gaussian.
So can you perform your calculations with Gaussian 09? Unfortunately not.
Instead, you should change to other programs that fit your needs better. As it is free for academic use [2], you can have a look at ORCA and if it fits to your needs.

[1] http://gaussian.com/integral/
[2] read the licence carefully, it is quite strict

Thank you very much Mr. Martin for more clarification in questions and thanks a lot to pH13 - Yet another Pjilipp and paracetamol for answer and further modifications. I just like to know another thing as I am a novice in the sector. What is the difference between the relativistic and non-relativistic approximation, I mean, if I use Gaussian 09 program with integral =dkh0 and cpcm continuum polarization field instead of COSMO? I am expecting a comment. Of course I shall check for ORCA whether I can use it suitably or not now. Thank you again for your kind attention and suggestions.

pH13 Yet another Philipp, Thank you. But your Ref [2] is not linked properly. Anyway I am checking and will inform later. Thank you very much for your kind suggestion and help with valuable comments.

@KeshabK.Adhikary The second is only a comment but no link.

Regarding your two questions; I think you can either search on the site for solutions to that or ask it as two new questions. First about the difference between relativistic and non relativistic treatment (of your system, give an example on at least a very similar system ... if a similar question does not exist) and another question on the difference between COSMO and CPCM.

More of a side-note, but the most recent version of ORCA no longer supports COSMO by the way.


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Number of atoms per unit cell
I don't understand the concept of unit cell and the number of atoms per unit cell in a cubic lattice also the calculations for the number of atoms. For example in the $\ce{fcc}$ lattice, number of atoms per unit cell is:
$$8\cdot\frac{1}{8} + 6\cdot\frac{1}{2}=4$$

what does the 2 and 8 in the denominator stand for?
also 4?

The denominator signifies the number of cubes that are needed to completely encompass the whole point. For example, a corner point can be thought of as a center of 8 whole cubes, while a face centre is enclosed by 2 cubes and an edge center by 4. Hence, only 1/8 of a corner atom is in a specific unit cell and so on and so forth.
Consequently, the total number of atoms in a unit cell (say a FCC) would be equal to -
(no of corners)(fraction of corner in the unit cell) = 8(1/8)
plus
(no of face centers)(fraction of face center in the unit cell) = 6(1/2)
which equals to 4

It should be noted that this answer (and the question!) only make sense for a unit cell which contains an atom in the corner and one on each symmetry-equivalent face. You might then translate the unit cell (or the atoms) to show how one can arrive at the end result of four in a different way.


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How can I lower the flash point of stearic acid (Palm Based Fatty Acids)
I am trying to build a environmentally neutral firelighter that does not use petroleum based accelerant. I have found that the flash point is way to high to be viable in Stearic Acid (Hexadecanoic acid / Octadecanoic acid). Is it possible to have a "safe", "environmentally friendly" and biodegradable additive to lower the flash point? I am open to suggestions and alternatives. Any assistance in this is appreciated.

Why do you need a low flash point? Candles are made of things with a high flash point and burn well because they have a wick which is easy to ignite and raises the temperature high enough for the wax to burn. Structure the lighter correctly and you can exploit this effect.

could you have a small amount of a more volatile liquid with the hexadodecanoic acid - such as white spirit or kerosene? This would support a flame and raise the temperature to the point at which the hexadodecanoic acid acid will burn.

Thank you for the response on this. Unfortunately as part of the Organic and biodegradable target i need to make it as non-toxic as humanly possible. If i use your suggestions then the product will require being sealed to prevent evaporation/leakage. thus the use of plastic and moving away from the environmentally neutral goal. I am experimenting with natural oils to force the Flash point down and increase the volatility. Once again thank you for your input, really appreciated. Comments always welcome.

What about a plant based terpene,  maybe limonene?  It has a flash point of 50 degree C and is very cheap (it's sometimes considered as a potential "green" solvent).

Do you mean it is "too high"? ESL student? Setting aside your use of the word "safe", sure: use ethanol. It's conceivable that you do not know that "flash point" is a measure of a materials' safety with respect to fire. By definition, a material with a lower FP will be less safe than one with a higher FP (ceteris paribus). I'd suggest you study the history of the match, if you haven't already. You need to explain why a butane lighter isn't sufficiently "safe", "environmentally neutral" and "organic" (and dare we say "natural") for you. Same with safety and kitchen matches.
In broad strokes, I can think of nothing which will be both exothermically reactive enough to work and be safe enough; adding a volatile to a non-volatile to make it more volatile seems to (obviously) fail the safety test. To initiate a flame, you need an (intense) energy source and a fuel. This reaction must (with 99.999% certainty) occur in the gas phase. Vapors are one possibility (the most efficient) the other is very, very finely divided powders. I can imagine a "firestarter" which uses a flint and a puff of powder to initiate the flame.
More specifically, it is only the rare azeotrope which will lower the minimum boiling point of the mixture below that of the lower boiling component. I doubt the difference will be enough to help you. Based on the chemical functional groups present, the slope of the vapor pressure vs temperature curves are indeed different. You want a steep curve, but you'd have to look into that correlation yourself. (You don't explain why you can't just enclose your "firestarter" to prevent evaporative losses...) One of the oldest "rules" in chemistry is "like dissolves like". In other words acetic or propionic acid might have significant solubility in stearic acid (but would be acid (corrosive to metals) and stinky, although mostly - imho - "non-toxic"). This type of mixture might be a way to lower FP. (or how about salts like ammonium acetate? (or other nitrogen bases) IDK...) Actually, flash point is a lousy way to quantify a compounds' fire properties. It is used because it is simple. Fire safety experts (which I am not) find measures of energy required to create a flame (not to sustain it) at a given temperature and (gas phase) composition much better. Because fuels provide the source of energy to sustain the flame reaction, it turns out that flame initiation it the process which requires the most energy input per molecule. Chemicals differ in the ease at which they start burning, at a given VP (of course, VP and Temperature are highly correlated in or near equilibrium, but that's not necessarily what you have when starting a fire). I'd look at it like this: 1. Find the most acceptable chemical with the lowest energy requirement at the temperatures the device "sparks" at. 2. find the most acceptable fuel to sustain that fire. (Note that here I mean sustaining for a second or two, long enough to start the "real" fire (or candle, etc.)). This implies that the first thing you need to do is locate a "environmentally neutral" source for your spark. (Do batteries meet your criteria? They wouldn't mine, how about ferrocerium or flint and steel? Or, what seems to be the "most" environmentally neutral energy source, a magnifying lens (only works when sun is shining, however.))

FFS could you write in paragraphs! And don't ramble so much on topics barely related to the actual question.


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Single-molecule magnet with electrically-controlled permeability: How does the Titan Shield from Deus Ex work?
In a recent installment of Deus Ex game series there is an augmentation called "Titan Shield" (it has nothing to do with $\ce{Ti}$ element):

A neodymium skin underlay matrix built of nano-meshed rare earth magnets and powered with hook-ins to the Biocell electrical system, the TITAN skin augmentation can be activated at will and then dissipated instantaneously.
When activated, dimorphic magnetorheological fluid is ejected from tiny nozzles installed throughout the skin. Simultaneously, the neodymium underlay electrifies, causing the fluid to solidify and seal the user inside an iron shell, effectively protecting them from all physical damage. The shield is resistant to gunfire and all forms of conventional explosives, but is vulnerable to EMP-based weaponry.

Titan Shield in action. From the official trailer of Deus Ex: Mankind Divided.
The second part regarding dimorphic magnetorheological fluid (MRF) looks quite plausible: such smart fluids, combining micro-sized spherical iron particles with nano-sized iron wires in order to prevent sedimentation and increase its responsiveness to external magnetic field are well known and can be prepared relatively easy [1].
Also, MRFs are used in vehicle suspension systems for more a decade now and they can withstand physical impact and dampen incoming momentum [2]. Typical design suggests putting a piston rod with MRF inside a coil which is upon electrifying controls solidification [3].
But the first part regarding "nano-meshed rare earth magnets" that can be triggered via sending electrical impulses caught me off guard. I thought that as long as conventional lanthanide-based magnets are permanent and have constant permeability in absence of external magnetic field, nanoscaled/single-molecule magnets are also permanent and cannot be used to dinamically change rheological properties of magnetorheological fluid.
Are there any existing examples of single-molecule magnets with permeability that can be controlled via altering current flow just like in case of electromagnets? If not, is there an explanation why it is not possible?
And yes, I do ask for this:)
References

Ngatu, G. T.; Wereley, N. M.; Karli, J. O.; Bell, R. C. Smart Mater. Struct. 2008, 17 (4), 045022. DOI: 10.1088/0964-1726/17/4/045022.
Gołdasz, J.; Sapiński, B. Insight into Magnetorheological Shock Absorbers; Springer, 2015. DOI: 10.1007/978-3-319-13233-4.
Liu, Q.; Jing, T.; Mo, A.; Xu, X.; Zhang, W. IEEE, 2015,; pp 2495–2500. DOI: 10.1109/ROBIO.2015.7419714.

Your question is much better then this in-universe description :)

Thinking of adding a clarification for magnetorheological fluid in the lines of "(when this kind of smart fluid is under a magnetic field, its viscosity increases to the point of becoming a viscoelastic solid)" but can't find where to do it without breaking the text flow, so I just suggested an edit to include a link to wikipedia instead.

@agaitaarino Thank you for the suggestion and grammar corrections, gratefully accepted. Note that you don't have to add a comment when suggesting an edit, this is an arbitrary choice. OP receives notification about pending edit, and your comment can be placed in Edit Summary field.

Is there any existing examples of single-molecule magnets with permeability that can be controlled via altering current flow just like in case of electromagnets? If not, is there an explanation why it is not possible?

Short answers: no, and not yet.
Longer answer, and a warning that I'm using recent results from my group as references:

Yes, single molecule magnets exist, but... The field of single molecule magnets is hot, but the working temperatures of single molecule magnets are still below liquid nitrogen. Things seem to be improving, but they do not seem to be improving a lot. See the plot below for how the maximum temperature at which long-term magnetic memory has been reported has improved over the years (for single-ion magnets based on lanthanides, which are the best single molecule magnets right now). While there is no consensus that room-temperature single molecule magnetism is impossible, it has not happened yet, and there is no explanation why it could not happen in the future.

Maximum magnetic hysteresis temperature vs year when the Single Molecule Magnet was reported. Plot generated dynamically by the SIMDAVIS (AGPL-3.0) web app: https://rosaleny.shinyapps.io/simdavis_dashboard/ [1]

Yes, magnetism in molecules can be controlled via electricity, but... In this work, and in several related works, Wernsdorfer and coworkers show electric control (and electric read-out) of the magnetic state of a single molecule magnet. [2] However, this happens at temperatures below liquid helium, and the machinery needed is only able to do this for a single molecule at a time. Furthermore, the device is so delicate that hundreds of attempts are needed to trap the molecule in the right configuration. If the device heats up by accident, the molecule is lost forever and one needs to start from scratch. Again, there is no solid reason why this could not be achieved at some point in the far future. As promised, an alternate way of coupling the magnetic (spin) state of molecules to electric current, reported recently by my group.[3] In this approach, molecular helicity can be used to generate magnetic current (like a molecular electromagnet), and in principle this could eventually be employed to control the magnetic state of a single molecule magnet (again, at some hypothetical point in the far future).

To complete the answer, while neodymium-based alloys behave very well, neodymium-based single molecule magnets are, so far, greatly outperformed by dysprosium, terbium and even erbium (see figure below). Since there are fundamental reasons for this, it is more likely that any future technology resembling the one described in the question will be based on dysprosium (as of may 2021 holding the record for highest temperature single molecule magnet) or terbium (the ion employed in the cited examples about electrical control of magnetism in molecules).

Maximum magnetic hysteresis temperature vs lanthanide ion. Plot generated dynamically by the SIMDAVIS (AGPL-3.0) web app: https://rosaleny.shinyapps.io/simdavis_dashboard/ [1]
References

Duan, Y. et al. Data-driven design of molecular nanomagnets. Nat. Commun. 13, 7626 (2022). https://www.nature.com/articles/s41467-022-35336-9
Godfrin et al., Operating Quantum States in Single Magnetic Molecules: Implementation of Grover’s Quantum Algorithm,  Phys. Rev. Lett. (2017) 119, 187702, https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.119.187702
Torres-Cavanillas et al., Reinforced Room-Temperature Spin Filtering in Chiral Paramagnetic Metallopeptides, J. Am. Chem. Soc. (2020), 142, 41, 17572–17580, https://pubs.acs.org/doi/abs/10.1021/jacs.0c07531]

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text:
Decarboxylation of Salicylic acid
I would like to know how and under what conditions can decarboxylation of salicylic acid (o-hydroxybenzoic acid) occur. I couldn't find any sources for it online. Here are some reasons I think it would happen:

The $\ce{COOH}$ group is out of plane of the benzene ring due to the ortho $\ce{OH}$ group, making it easier to decarboxylate.
The keto form is a beta keto-acid, which makes it easier to undergo decarboxylation.

From (1):

Salicylic acid decarboxylated to give phenol and carbon dioxide, at a convenient rate for kinetic measurements, at $200-230^\circ$, in a homogeneous solution of benzoic acid and certain other high-boiling solvents.
... there was a considerable amount of interest in the possibility of a bimolecular decarboxylation mechanism involving an attack by a proton on the free acid or its anions. A variety of compounds such as anthracene-9-carboxylic acid, substituted cinnamic mesitoic acid, aminosalicylic acid, and salicylic acid, appeared to represent examples of this type of decarboxylation mechanism. In most cases, pseudo-first-order rate constants were
  observed experimentally.
... mechanism is also proposed for the decarboxylation of salicylic acid in benzoic acid solution. A competition between an oxygen atom and the terminal ring carbon atom for the proton to give either the undissociated acid or a $\sigma$-complex leading to products is shown by eq. 3.

The effect of substituents on the rate of decarboxylation of salicylic acid has been used to support a bimolecular mechanism. Electron-donating groups in the ortho or para positions have led to an increase in rate. The increased electron density on the ring carbon atom bonded to the carboxyl group would presumably facilitate an attack by a proton at this position.
... An attack by a proton on the terminal ring carbon atom of the salicylate ion would appear, at first glance, to be extremely unfavorable because of the close proximity of a formal negative charge. Any interaction should merely lead to undissociated acid. The hydroxyl group in the ortho position appears to influence the distribution of the charge because of the steric configuration.
  Resonance and tautomeric forms, eq. 5, which do not involve an extensive shift of atoms, can be drawn which would distribute the charge in a manner favorable
  to the proposed mechanism, eq. 3.

This unique steric arrangement is not possible with m- or p-hydroxybenzoic acid, the effect of which is borne out by the fact that these two acids did not decarboxylate at a measurable rate in benzoic acid solution at $212^\circ$. Since inductive and resonance effects would be expected to be similar for the ortho and para isomers, the steric effect appears to exert a critical influence.

(1) Kaeding, W. W. Oxidation of Aromatic Acids. IV. Decarboxylation of Salicylic Acids J. Org. Chem. 1964, 29 (9), 2556–2559. DOI: 10.1021/jo01032a016


text:
How does non-electrolytic tinning work?
Having made diy pcb 's at home I have used things like tinnit which is a clear liquid . When the pcb is dropped into them , a layer of tin forms on the exposed copper regions on the pcb . I was wondering if someone could explain what is happening or if someone could give an equation .
Here is a video of the process on youtube https://www.youtube.com/watch?v=wOqkI-S-wdI

It works due to Redox Flipping
https://youtu.be/Hsw3lOnHaas

This is a good question: Tin is above copper in the galvanic series, so logically, copper should displace tin, not the other way around.
Electroless (autolytic) tin plating can be accomplished by disproportionation of a tin compound. $\ce{2Sn(OH)3- -> Sn + Sn(OH)6^{2-}}$, where the less stable Sn(II) yields Sn(IV) + Sn(0), or metallic tin. The divalent tin is usually $\ce{SnCl2}$.
Electroless tin can also be deposited by reduction of a tin compound, e.g.  thiourea reducing $\ce{SnCl2}$ to metallic tin. In a similar fashion, silver nitrate can be reduced to metallic silver to make a mirror on a glass surface. Note that the silver is deposited without  any effect on the glass (other than a nice, shiny coating).

So the metallic tin precipitated deposits on the surface . But in silvering of glass we see that whatever surface is exposed to the solution gets silvered but in this case only the copper surface gets tinned . Why is that?

Another good question... but I am not sure of an answer. In the case of thiourea reduction, I assume copper is a catalyst, similar to the way a hot copper wire can catalyze methanol oxidation. Hopefully, someone more knowledgeable will comment on this and the disproportionation

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text:
What is the chemical formula of KIT-6?
I have seen KIT-6 used in literature as a catalyst support, but cannot find the chemical formula for it.

According to original paper (1), KIT-6 is a highly ordered cubic $Ia3d$ silica ($\ce{SiO2}$) templated by Pluronic P123 ($\ce{[EO20PO70EO20]}$) block-copolymer surfactant. As a source of $\ce{Si}$, tetraethoxysilane (TEOS) or sodium silicate are primarily used.
(1) Kleitz, F.; Choi, S. H.; Ryoo, R.; Chem. Commun. 2003, 0 (17), 2136–2137. DOI: 10.1039/B306504A

KIT-6 is a mesoporous silica, the chemical formula is the same for all silica which is $\ce{(SiO2)_n}$. The tridimensional structure though is usually represented as

According to this data sheet (LINK) is SiO2.


text:
Are there any non-Jahn-Teller type distortions?
My question is essentially, if ANY gradual type of deviation from a higher possible symmetric arrangement of the nuclei in a molecule is covered by the Jahn-Teller model.
More specifically I have in mind some lets say "ionic complexes" between a very small cation and some larger (possibly weakly coordinating) anions.
Very specifically, I remember I have seen for example a calculated equilibrium structure of a beryllium bis-2-fluorobenzoate $\ce{(Be[o-C6H4FCOO]2}$, isomer with $\ce{F\bond{...}Be-O}$ coordination), which showed a distinct deviation from $\ce{C2}$ symmetry. The only type of JT-distortion this would qualify for is 2nd order one. But then since its largely an ionic species I suppose the HOMO-LUMO gap should be so large that 2nd order JT is very unlikely (I am aware that $\ce{C2}$ is not Jahn-Teller active but through vibronic coupling of close-lying $\ce{^2A}$ and $\ce{^2B}$ also JT-like distortions can occure).
I imagine that if the cation is very small so that it "dangles" in the gap between the anions it might be energetically more favourable if the cation decides to stick on one than to balance in the centre. Just a naive picture. But would such a thing - if it exists - be subsumed under JT as well?
So, is distortion possible for ionic aggregates and would such a thing also be described by JT?

Are sterically driven distortions in scope?

@OscarLanzi: Yes.

Apparently there is some theorem e.g. to be found in an article from Isaak B Bersuker with title "SPONTANEOUS SYMMETRY BREAKING IN MATTER INDUCED BY DEGENERACIES AND PSEUDODEGENERACIES" in Advances in Chemical Physics, Volume 160, First Edition. Stuart A. Rice and Aaron R. Dinner. © 2016 John Wiley & Sons, Inc. Published 2016 by John Wiley & Sons, Inc; p 187-
The theorem reads like:
"DEGENERACY AND PSEUDODEGENERACY ARE THE ONLY SOURCE OF SPONTANEOUS SYMMETRY BREAKING (SSB) IN ATOMIC SYSTEMS"
And can be reformulated by:
" [...] can we get structural instability of high‐symmetry configurations in any electronic state without involving interaction with other states [...]? The answer to these questions is no: any instability and SSB in polyatomic systems leading to their formation, deformation, and transformation involves at least two electronic states, degenerate or pseudodegenerate. This is the two‐state paradigm in structural instability."
So in essence that means all types of molecular distortions can be traced back to some JT effect.


text:
What is the causing my solution to precipitate?
I have a solution that contains M9 Minimal Media 5X, Glucose, Magnesium Sulfate, Calcium Chloride and L-Phenylalanine. When the solution is pH'd to 8.5 with 4M Sodium Hydroxide, it starts to precipitate. What is the causing the solution to precipitate?

It is probably precipitating Calcium Hydroxide and Magnesium Hydroxide since you are adding a very concentrated NaOH solution to it. Not sure if this is what you've asked though.


text:
What chemical should I use to thin UV cure epoxy?
The epoxy that I am working with is UV curable with a cycloaliphatic epoxy resin base. I am trying to reduce the viscosity. Whatever I add to thin the epoxy, it needs to be able to evaporate off so that when I cure it, it's just the epoxy remaining. I have used solvents like acetone to thin 2 part thermal cure epoxies, but I am not sure if this is appropriate for a UV cure epoxy or which solvent would be best to use.

Adding solvents that need to evaporate, to your epoxy, can cause pinholing. You need a integratable solvent. Styrene monomer works well. You can get it from a fibreglass supply or by pyrolizing poly-styrene.
The monomer will combine with the epoxy so it won't need to evaporate out of the epoxy.

You can heat up the epoxy resin to reduce the viscosity.
To make it thicker, you could add any kind of inert powder filler,
like talcum, silica variants, whatever.
I understand what you want to do with the solvent - but it seems not to be in common use.
But it is worth a try with a small batch. The curing by UV makes it easier, because you have much time to get the solvent removed before curing. And you van even heat it up without shortening the workable time by it.
It is important in what shape it will be used: Is it a thin coating, or casting a complex object into a block. I assume a coating, and with low viscosity, it will be a very thin coating.
When the epoxy can wet the surface, it will find just about any pore to go into - independent of viscosity. It just takes longer if it is thick.
As for the solvents to try, Xylol and Toluol could work - evaporating quickly, but not very quickly. Try a common "paint thinner" - some are a mixture of these


text:
Help understanding this passage on optical activity

Other than mirror images, compounds having opposite configurations do not necessarily cause opposite directions of optical rotation any more than do compounds of the same configuration necessarily cause the same direction of optical rotation.

I know what optical rotation is and I know what "molecular configuration" means but I still can't understand what I just wrote above.

The meaning's pretty clear (at least when I read it). I would paraphrase it as: "Compounds having opposite configurations do not necessarily cause opposite directions of optical rotation... just as compounds of the same configuration do not necessarily cause the same direction of optical rotation; with mirror images of a given molecule being an obvious exception." Give this a re-read ;)

@paracetamol  but if two molecules have opposite configurations, doesnt that mean they are mirror images of each other ?

The main confusion is with the word "compound." We are not talking about a property that holds true in all compounds. We are saying no relationship of the type described exists when comparing compound A with compound B, unless compound B is the mirror image of compound A (thus implying opposite configurations). This is just a terribly worded statement.

The issue is that knowing the absolute configuration around a chiral centre doesn't help you work out which way the molecule will rotate polarised light. So two different molecules with a similar chiral centre might rotate light the opposite way even though the immediate configuration around the chiral centre is the same.

@Max Example (If you're familiar with Fischer's D-L notation): D-(+)-glucose is dexterotary, while its mirror image, L-(-)-glucose is levorotary. But, D-(-)-fructose, which has the same "conformation" as D-(+)-glucose, is  levorotary. The mirror image of D-(-)-fructose is L-(+)-fructose and has the same "conformation" as L-(-)-glucose, but is instead, dexterotary.

@paracetamol  sorry i am very late to reply but i hope that i can still get an answer from you. Anyway, what you are trying to say is that if molecule A has an opposite configuration to molecule B which is of a different compound, they dont necessarily cause opposite effects on plane polarised light. And even if these molecules had same configurations , they wouldnt necessarily cause same effect on plane polarized light. They are just putting a line between the D/L system and +/- system, right ?

@Max Yup, sounds like you got it right ;). The D/L system (No, they don't mean "dextro/levo"...that's another thing) is assigned to a molecule based on its stereochemistry (the relative positioning of its atoms/ groups in space); whereas the +/- system (This one is "dextro/levo") is based on the optical activity of the molecule (which is experimentally obtained). Just because some molecules have, say, the (D) form doesn't mean they're all going to be dextrorotary (+) or levorotary(-)... each molecule polarizes light in whatever way it "wants" to; it can only be experimentally found.

Once again, I suggest you compare optical activities (+/-) of the D/L forms of glucose and fructose; it might help to have a look at the molecular structures for these 4 molecules). D-glucose and D-fructose have the same configuration (obviously, since they both carry the  prefix "D"), but they have opposite optical behavior (D-glucose is dextro, D-fructose is levo). Similarly, L-glucose and L-fructose have the same configuration, but different optical activities (L-glucose is levo, L-fructose is dextro).

Another useful comparison of optical activities would be between the D/L forms of glucose and glyceraldehyde. D-glucose and D-glyceraldehyde have the same configuration, and they happen to have the same optical behavior (both are dextro). Similarly, L-glucose and L-glyceraldehyde have the same configuration and they happen to have the same optical behavior (both are levo).

@paracetamol  but my book was saying that D/L is the same thing as Dextro/levo, it is only the short form of the words. If not then what is the difference between these two ?

So this means that levo signifies one direction of optical rotation (and dextro signifies the opposite direction) , right ? And the D/L system is only about the arrangement of the molecules in space ?

Forgot to mention you @paracetamol

@Max Nope. When I was in middle-school, we used "d/l" for "dextro/levo" (not D/L). When you start taking high-school or undergrad Chem, you'll realize that +/- is the preferred notation for "dextro/levo". The D/L system was proposed by Fischer to help him with sugars/sugar-like molecule book-keeping. Gimme, minute....I'll get to what the D/L system's about.

This might help: [ https://chemistry.stackexchange.com/questions/44260/what-is-the-difference-between-d-and-l-configuration-and-and-%E2%88%92 ] Also, read the comments under the answer to that question.

Yeah i got what you said. Thanks a lot man i finally understood this. @paracetamol

Glad to hear that. @Mods If I get the time (don't think it'll be anytime soon), I'll try to compress all these comments into an answer and post it here. It should make it easier for the OP and the others to refer to later ;)

Oops, I missed this comment: "So this means that levo signifies one direction of optical rotation (and dextro signifies the opposite direction) , right ? And the D/L system is only about the arrangement of the molecules in space ?" Yes, that's right. I went and said "Nope" to D/L = dextro/levo. Also, "levo" is specifically LEFT-polarized, while dextro is specifically RIGHT polarized. And obviously, they are opposite. O:)

I believe what is causing you confusion is the difference between optical activity and absolute configuration.
A molecule with a single stereocenter such as limonene and carvone have two stereoisomers, one rotates polarized light to the left (-) and the other to the right (+). Additionally, each one has its own absolute configuration in their chiral center, which can be R or S. The thing is, R and S configurations are not attached or are, anyhow, synonyms with (+) and (-) configurations. Carvone and limonene are good examples:

In carvone, the S-enantiomer rotates polarized light to the left and so it is named S-(+)-carvone. The S-enantiomer in limonene rotates polarized light to the right and so it is named S-(-)-limonene. The two nomenclature systems are different since they specify distinct properties.
I hope this answers your question. If not, I would be glad to help in anyway I can.

Sorry but i dont know what the terms "absolute configuration" and "stereocenter" mean. I have tried looking them up and i will continue trying to understand what they mean after i post this comment.

do you think the guy in the first comment (paracetamol) paraphrased it correctly ? because i thought that "opposite configuration" means mirror image. If so then why wouldn't they necessarily cause opposite directions of optical rotations ?

I am sorry, I didn't know how much you knew about this topic so I assumed you already knew what those terms mean. A stereocenter in organic chemistry is a carbon atom that can originate chirality or is chiral (i.e has 4 different substituents attached to it). If the molecule has only 1 stereocenter (or chiral carbon), it will have 2 optical isomers, one is called R, the other S (a nice introductory video by Khan Academy about this can be found at https://www.khanacademy.org/science/organic-chemistry/stereochemistry-topic/chirality-r-s-system/v/introduction-to-chirality )

What i am not understanding about stereocenters is how every website is saying that exchanging two groups connected to the stereocenter leads to the formation of a stereoisomer but i can see the molecule looking the same if we just rotated it.

Chirality is a broad concept that is present not only in organic chemistry, but in everything, from inorganic compounds to everyday structures such as your hands! Something chiral does not have apparent symmetry.

If you happen to have access to molecular modeling kits (those with sticks and balls used in introductory organic chemistry courses) you can create your own chiral molecule and see for yourself that they are not the same thing just inverted. If you don't, then watch some videos suchs as this one from Khan Academy, they will eventually convince you that those are different compounds :)

@Max white:" other than mirror images" reads as "except for mirror images". Then explanation is as in comment of Zhe.


text:
What is the liquid coming out of these Duracell batteries?
I picked up a remote control to a DVD player that I had not used in a couple months.  The remote control was stored in a clean plastic bag.
The bottom of the remote control is covered with a liquid.  Upon further inspection, I discovered that the liquid is coming out of the Duracell Coppertop (non-rechargeable) batteries in the remote.
The amount of liquid is substantial.  It almost completely covers the bottom of the remote control.
Typically, when batteries have problems, you see crystals forming.  But not so much in this case.  In this case, it is almost all liquid that has been discharged.
I did not initially notice the liquid, and so I picked up the remote, and then set it down on the furniture.
What is this liquid?  What chemical process took place that resulted in the liquid being discharged from the batteries?
(And, as a bonus, how to I get the liquid off the furniture?)

Well, it depends on which type of battery you have. Alkaline batteries such as Duracell Coppertop usually leaks KOH (potassium hydroxide) under conditions of misuse as a white foam. However, KOH is hygroscopic and will easily form a oily like liquid which is what you might be seeing in your batteries. Handle it with proper care since KOH is caustic and can cause tissue damage.

Thanks.  I got it all over my hands, as I picked up the remote control in the dark.  Any idea what to do?

Just wash it with water, is not a big deal... Just don't let it in contact with your eyes and you're fine.

Thanks.  Will it bleach the furniture it was sitting on?  If so, best way to clean it off?

Clean it with a humid cloth, but I really don't know if KOH bleaches the furniture :(

Thanks.  By "humid cloth", do you mean slightly dampened with water?


text:
Potential energy versus reaction coordinate (graph of Gibbs free energy)
Does the potential energy scale in the above mentioned graph measure the Gibbs free energy of the system?
As the system always tries to achieve a minimum of free energy and chemical reactions are taking place can be considered a non-mechanical work (I'm not sure of it, as the system isn't in a state of chemical equilibrium if a reaction is taking place), potential energy should represent Gibbs free energy.
If not, how should I understand the potential energy of a reaction mixture?

There is a long explanation for that, however in short they are not the same thing. Free energy relates to thermodynamic systems with many particles and is governed by Boltzmann's distribution, a sum of all possible energy configurations, where F is the free energy of the system.
$Z = e^{- \frac{F}{kT}} = \sum_i e^{- \frac{E_i}{kT}}$
Potential energy is non-statistical and well-defined for one particle. Thermodynamic free energy (one type being Gibbs) is a subset of the total potential energy. Even though they may seem similar, they should not be confused and should not be interchanged.


text:
Must lactose-free milk be ultra-pasteurized?
In this question I asked why lactose-free milk lasted such a long time. The answer was because it is ultra-pasteurized. This leads naturally to another question: does lactose-free milk have to be ultra-pasteurized?My initial guess is that it does because:Since the lactase added to make the product lactose free breaks down lactose into twice as many sugar molecules, any living bacteria have twice as much nutrients to thrive in.Since the disaccharide lactose has been broken down into the simple monosaccharides glucose and galactose, the bacteria do not need to produce their own lactase to use the sugar; thus any non-lactase producing bacteria can thrive in the milk.And finally, I have been adding lactase to regularly pasteurized milk to produce my own lactose-free milk. However, this milk almost always goes bad a couple of days before the Best-By date stamped on the container.This all leads me to think that lactose-free milk must be ultra-pasteurized in order for it to have any reasonable shelf life.

Regular pasteurised milk always goes bad within a few days after opening the bottle, which contaminates it with a million of funny organisms from the air we breathe.

Hmm...good point. So your view would be that it would probably go bad before the best-by date with or without the added lactase? Although...the bacteria that live in the air wouldn't find much sugar to consume in regular milk, since they would probably not be able to use the lactose in the milk.

I'll say the answer is no, you don'T have to, but should probably  drink the lactose-free stuff a bit faster. What you say sounds reasonable, but the sugar contents might not be the important factor.

Consider the question of how long you'd leave some other random meat product out in the refrigerator.

You can now buy (at least in Switzerland where I live) lactose-free milk which is not ultra high-temperature pasteurized. They are kept chilled in the supermarket alongside "normal" milk. So I would say based on this, no lactose free milk does not have to be ultra pasteurized. Though I will note, they hardly tasted sweet at all and they are quite expensive, nearly double the price of ordinary milk.

I can't answer why they ultra pasteurize it with absolute certainty but my guess would be because there is less demand for lactose free milk so the extended shelf life will prevent waste if it doesn't sell quickly.
Sugar, like salt, retards bacterial growth, it doesn't increase it. That's why candy and sugary baked goods aren't refrigerated.  The 2 types of milk contain the same amount of sugar.
Regarding your milk going bad after you've added the enzyme, you could unwittingly be introducing more bacteria into your milk during that process.
I know this is an old post, but I came across it because of having the same question about lactose free milk and this was the conclusion I've come to.


text:
I'm looking for something as strong as Acetone
We use Acetone to remove nail polish and gel polish at our salon.  Is there another  alternative substance out there with less irritated smell like Acetone?

Most nail polish remover now days is actually made from ethyl acetate. It does the same job but with lower toxicity

Hello and welcome to Chemistry.SE. Feel free to take the short [tour] or visit the [help] if you have any questions or to learn more about how this site works. Best of luck with your question!

Reviewers please note, this question isn't asking for the best or anything. He/She simply wants an alternative to acetone, with a clear definition that it should be less volatile (and less irritating to the nose) than acetone.

Hmm...I like the smell of acetone better than that of EtOAc, but well, that's a personal opinion...

Mineral oil is often used to remove nail polish. Give it a try (J&J has a great one for babies with a soft smell).


text:
Identification: was this gallium or mercury from broken thermometer?
I had accidentally broken a thermometer against my computer screen, resulting in the splash of the silvery substance seen below:

I cannot remember whether the thermometer is a mercury-free thermometer, and was wondering if you could help with identification of the substance given the following information:

This picture was taken about 1-2 minutes after the initial spill
The temperature of the room was around 70 degrees F.. at most 80 degrees
The silvery substance clung to the screen as shown.
Small bits of the silvery substance seeped into the computer. However, it was fully functional in the few instances I have used it

Any suggestions or additional follow up questions would be much appreciated. Thanks!

How can you accidentally break a thermometer on the screen of your computer? Next time hang is on a wall.

It was the type where you put it in under your arm to measure your temperature. I was shaking it, lost grip, and had it splash against the screen.

I've never seen gallistan, the alloy used in gallium thermometers, but that just doesn't look like mercury to me. If I were really certain though, I'd have make this a formal answer ;)

By the way, you have probably destroyed the casing of your computer. Gallium (not sure about the exact alloy used in thermometers, but it contains a lot of gallium) will destroy aluminium.You might not notice it immediately, but the structural integrity of the case will eventually cease to exist. Unless you got it off the case very quickly.

Gallium has melting point of 29,76 °C. If the alternatives are Ga or Galinstan, than surely Galinstan.

Keep Gallium off your aluminium objects. It amalgamates with it and it's even able to penetrate the oxide film. The result is an alloy of gallium and aluminum.

Gallium, and Galinstan have the ability to wet glass, while mercury does not.
So it's most probably gallium (or Galinstan).

Ah - I definitely encountered the wetting glass issue when cleaning. Thanks for this

@daspianist Either way I would get rid of the substance from the body of your MBP. Anodized layer ideally should protect aluminium, but theoretically gallium readily dissolved aluminium.


text:
How can we tell the particle size of solutes in solutions?
This is a general question in regard to solutes dissolving into solvents. I am interested in the reality of the particle size of the solute during dissolution.
For example, take $\ce{NaCl}$ and dissolve it into water. Are the salt particles, which are now invisible to the eye, truly dissolved? Meaning have they turned into single atom particles of $\ce{Na+}$ and $\ce{Cl-}$ clinging to $\ce{H2O}$ water structures?
Chemical equations seem to imply this. However, when you look at colloids, then particle size matters and can vary. Do ionic salts still have large clusters of their cations/anions upon dissolution?
I am really interested in this topic and would love someone to give me a good explanation.

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Colloidal solutions are not solutions in the sense of the definition, as they are not a homogeneous phase. They can be separated by conventional means, like centrifugation. Apart from this, I think your question is too broad as it seeks information about not only solutions, but also the process of dissolution and is not a good fit for this format. If you limit the scope to only be about NaCl, then that would be more appropriate.

Martin, the question is about ionic compounds. Thanks!

I assure you there are many reasons to believe that the ions really are solvated individually. The first thing to look at is the relative strength of interactions between the ions in a lattice and the ions when in a solution. For water, these interactions must be very strong because a number of hydrogen bonds must be broken, the interaction between ion and water must overcome this initial deficit and the loss of some entropy. So, the interaction is surely quite strong.
It is of course possible for two ions which dissolve to get back together and interact with each other. They are oppositely charged ions after all. By looking at the relative size of equilibrium constants for these two processes, however, I think you'll be convinced it is far preferable for the ions to be solvated. For instance, a quick google search will tell you that the $K_{sp}$ of $\ce{NaCl}$ is around $36$. That's an astronomically large solubility product. This means we dump sodium chloride into water until the concentration of both ions is around $6\ M$.
Thus, the easiest experiment to do is to just saturate a solution of water with salt and write the mass of salt added in molarity units. This doesn't really answer your question though because we assumed that the salt existed as two ions and then made a correspondence between that concentration of ions and the concentration of water which each ion is occupying. Plus, we would generally be interested in more dilute solutions than this.
This is why people have used methods which probe the structure of ions in a solution directly. Namely, using x-ray diffraction and neutron scattering experiments, it is possible to determine the local environment around water (or something else solvating an ion) and calculate the hydration number. This is the number of ions, of a specific ion, solvated per molecule of solvent.
Here is a review which discusses x-ray diffraction and neutron scattering experiments and one which discusses coordination numbers of ions solvated in water.
Using these structural methods, which show it is extremely unlikely there are chunks floating around in solution, the following data can be found for hydration numbers of ions in water:
$$n_{\ce{Li+}}=4$$
$$n_{\ce{Na+}}=5$$
$$n_{\ce{K+}}=6$$
Here is also an old review which describes many simple but outdated methods for determining the solvation/hydration numbers of ions in a solvent.
Hopefully some of this is helpful and convincing. The experiments are quite definitive, but if you believe the equilibrium constants, then it is very hard to believe these lattices would remain intact in solution.

Using your example of $\ce{NaCl}$, yes, upon dissolution into water the solid is broken down into individual $\ce{Na+}$ and $\ce{Cl-}$ ions in a process called solvation. Each ion is surrounded by a solvation sphere, or solvation complex, of solvent molecules creating a more stable configuration than undissolved $\ce{NaCl}$ in the presence of water (up to the limit of solubility, about $\pu{260}$ g/L at room temperature).
The following image was taken from the Wikipedia article for solvation:

Here you can see that the positively charged sodium ion is surrounded by the electron-rich end of water molecules, creating a solvation spere (of course this is happening in 3 dimensions, thus a sphere). The same thing happens for the negatively charged chloride ions, except that the more electron-poor end of the water molecules are attracted to the ion.

So, if you took silver metal and dissolved it into nitric acid, based upon this information you would have single atoms of silver surrounded by a solvation sphere. But, when you reduce it again, it turns into colossal metal particles, not a perfect colloid.

I guess I am asking, how do we know that what is at the core of this solvation sphere is truly a single atom and not multiple? As we know colloids that contain many thousands of atoms can appear dissolved in solution.

The definition of a colloid, in terms of size, is a bit fuzzy. Typically they can be considered to be a few nm up to a micron in size. While this seems small compared to bulk material, a colloid still contains thousands to billions of of atoms in size. They can be perceived by light scattering, they can be seen in optical or electrical microscopes. Electrochemical experiments also show that individual ions are at play, and not colloids.

Yes, I don't doubt that ions are indeed ionized. I have played a bit with electrolysis myself and it does follow that the positive cations are drawn to the cathode and the negative anions are drawn to the anode. Does this necessarily prove that these are single ions? Or could there be ion spheres with superficial charges that behave as single ions, but really they are large particles?

Another example is ion chromatography (IC). In IC, individual ions interact with a column of material containing charged sites. If the solution being run through the column contained large "superficially" charged colloids rather than individual ions, the interaction would be negligible. I'm not sure that's a great example, but I'm also not sure I understand what you mean by "ion spheres". Do you mean a colloidal particle with a charge to it? Let me think about this a bit and see if I can come up with a better example.

Well, it may not be pure colloid elements like pure sodium colloid with superficially charged +1 sodium. It may in reality be more of a salt matrix of Na/Cl/Na/Cl (in a crystal lattice) and the surface of this salt matrix, which was attacked by the solvent is the "dissolved" part. So, the solvent being water would "float" this large particle in the solution by interacting with the outer part of the salt lattice. Hope I was clear.

In regard to IC, with this salt lattice in mind (not a colloid surrounded by ions), the ions being large particles may still get caught in the column, right, because they are ionic? To be honest I am not too familiar with the process and will need to further research it.

Another thing that would be affected is the change in melting point and boiling point of having the ions in solution. If the dissolved material were clumped the effect would be much smaller and we would observe it only in proximity to the clumps.

Perhaps, this clumped material is very well dispersed still, according to the human macro standards. In mining, 600 mesh is a good grind, pure talc. Now, imagine a particle 1nm in diameter with hundreds of atoms - that is seriously attenuated. This may have an appreciable effect on melting/boiling points. However, they aren't single atoms.

Colligative properties, electrical properties are the first two that come to my mind to experimentally access the solute at the level required. Uhmm, pH can be used too, in a more general case.


text:
Is a temperature decrease required for hydrogen bonds to form in water?
As I understand it, increasing the temperature of water causes hydrogen bonds to break.  Does this mean that a decrease in temperature is required for the formation of those hydrogen bonds?

You are right, in a sense.
A more accurate statement would be that as the temperature decreases, the average lifetime of a hydrogen bond increases. Thus, on average the liquid/solid/molecule will be more stable. This is because, as you point out, hydrogen bonds are weak enough to be broken by collisions due to thermal motions (translation and rotation). They are almost always immediately formed again (in liquid), but the more frequently they are broken, the less hydrogen bonding there will be on average.
One indicator of the degree to which hydrogen bonds stay formed is the average number of hydrogen bonds over time per water. In water, this number comes out to be around 3.4 hydrogen bonds, where it could have a maximum of 4.0. Two donors and two acceptor molecules.
Thus, you can rationalize the transition from liquid water to ice in this way by considering that in ice you come quite close to the number of 4.0 hydrogen bonds per water. As you say, then, when temperature decreases, more hydrogen bonds are formed in general. I would hesitate to call it a requirement though because hydrogen bonding can take place at any temperature, but is just rarer when the temperature increases.


text:
Synthesis of Haloperidol with 4-chlorobutanoyl chloride and piperidin-4-one
I searched for a easy synthesis on Haloperidol.

Step 1 and 2 are taken from https://www.slideshare.net/SagarJoshi2/antipsychotics-33785528 (Slide 23). I just tried to work out a way to prepare 4-(4-chlorophenyl)piperidin-4-ol via Grignard.
Does the 4-chloro group has any negative influence on the reaction.

The reaction mechanism for the step in question simply involves initial formation of a Grignard reagent followed by nucleophilic attack into the ketone.

When forming Grignard reagents from halides, there is a general trend that the rates of formation are in the order  I > Br >> Cl, making it possible to get selectivity where multiple halides exist.


text:
Haloperidol synthesis - 1-chloro-4-(prop-1-en-2-yl)benzene with NH4Cl and formaldehyde
However I have no idea how the first two steps work. Especially the first step makes me wonder how the amine and the aldehyde form the 1,3-oxazinane.

Suggestions:
Ammonia and formaldehyde form via condensation the imine.
The imine and formaldehyde form the 1,3-oxazinane, however I looks kinda wrong to me. Moreover, if this is mechanism would be correct, why does the formaldehyde binding to prop-1-en and the imine don't?

Step 2 probably starts with protonating the oxygen and i think the 1,3-oxazinane cleavage follows after. However how does the methyl group bonds with the other methylene group?

Related: https://chemistry.stackexchange.com/q/79050/4945 And again it might be better to ask step 2 as a separate question. Link back to this one for context.


text:
Rayleigh–Jeans Law
In McQuarrie's Quantum Chemistry, the Rayleigh–Jeans law is given as:
$$\operatorname{d}\!\rho(\nu,T)=\rho_v(T)\operatorname{d}\!\nu=\frac{8\pi k_bT\nu^2\operatorname{d}\!\nu}{c^3}$$
where $\rho_v(T)\operatorname{d}\!\nu$ is the radiant energy density between $\nu$ and $\nu+\operatorname{d}\!\nu$.

Why does the author consider the range of the energy density, and why does he take the differential of the function $\rho(\nu,T)$?
Also, what is the difference between the two functions $\rho_v(T)$ and $\rho(\nu,T)$?

Why would one be interested in knowing radiation between certain frequencies?

In this context, radiation is understand as a continuous function of wavelength or, equivalently, frequency.
At an exact frequency $\nu$, according to Rayleigh-Jeans law, radiation is
$$\frac{8\pi k_bT\nu^2}{c^3}\text{.}$$
In practice, it is hard to measure a single frequency.
We are thus interested in frequency intervals.
An exact frequency is the limit of a sequence of smaller and smaller intervals, so there is no problem here.

If we make the assumption that, for a sufficiently small interval, $\rho(\nu, T)$ doesn't vary, we get your definition for the differential $\operatorname{d}\!\rho(\nu,T)$:
$$\operatorname{d}\!\rho(\nu,T) = \frac{8\pi k_bT\nu^2}{c^3}\operatorname{d}\!\nu\text{.}$$
The assumption is fair due to the continuity of $\rho(\nu, T)$.
This is the approximation of an integral on a very small interval $\operatorname{d}\!\nu$ by the height of a point inside this interval ($\frac{8\pi k_bT\nu^2}{c^3}$) times its length ($\operatorname{d}\!\nu$).
No magic here.

So, if we sum an infinite amount of small intervals like the one above we get an integral.
The total radiation between $\nu_1$ and $\nu_2$ will be:
$$\int_{\nu_1}^{\nu_2}\operatorname{d}\!\rho(\nu,T) = \int_{\nu_1}^{\nu_2}\rho(\nu, T)\operatorname{d}\!\nu = \int_{\nu_1}^{\nu_2}\frac{8\pi k_bT\nu^2}{c^3}\operatorname{d}\!\nu = 8 \pi k_b T \frac{v_2^3 - v_1^3}{3 c^3}\text{.}$$
Observe that $\rho(\nu, T)$ is quadratic in $\nu$.

Also what is the difference between the two functions $$\rho_v(T)\text{ and }\rho(\nu,T)\text{?}$$

None.
Compare the definitions.
I changed the nomenclature because, in my opinion, there is no reason not to treat $\nu$ and $T$ on equal footing.


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Reaction mechanism of redox reaction
So on the data, it said that we can reduce potassium permanganate by utilizing oxalic acid, thus producing carbon dioxide.
But can anyone propose a reaction mechanism of how the whole reaction happens, and how did oxalic acid break up into $\ce{CO2}$?
Thanks in advance!

I think it would go through a manganese-oxalate complex (oxalate is a good ligand for many metals), and that would fragment via a reaction such as:

The Mn(V) species would likely disproportionate to manganese dioxide and permanganate.

Couldn't understand can you do a more formal explanation ?Thanks a lot!

More formally, the complex shown results from adding oxalic acid to permanganate ion with elimination of a water molecule. The three curly arrows each represent a pair of electrons moving to shift the pattern of bonding from reactants to products. The result is a change of oxidation state (REDUCTION) of manganese from VII to V. Hope this helps!

But wouldn't the result became Mn (II) ion?

Either that or manganese (IV) dioxide, depending on pH and other factors--as a final result. But other intermediate oxidation states may be initially formed, but then disproportionate in a manner like what I gave in my original answer. If you want to go deeper on this, I am not your man. I urge you to post another question about this specific issue and let a bona fide inorganic chemist respond.


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Is protein folding thermodynamically favourable? If yes.. why?
The second law of thermodynamics dictates that the sum of entropy of the universe is always increasing. Is the process of protein folding a spontaneous process which is increasing the entropy? Is this why it can be called thermodynamically favourable? Or is there something else I am forgetting..

https://chemistry.stackexchange.com/questions/28589/truly-understanding-the-second-law-of-thermodynamics-and-entropy

What determines whether a process is spontaneous or not, is the change in free energy, not the increase of entropy. Entropy is one important factor, as is enthalpy. Entropy in this case is not increasing since the system is becoming more organized, however,enthalpy is greatly favoured since protein folding creates many Hydrogen bonds between its polar aminoacid fragments and surrounding water and also many hydrophobic interactions between their non-polar aminoacids constituents. Take a look at this page to have a better understanding about protein folding thermodynamics (introductory text, though it´s a good starting point).
http://www.chembio.uoguelph.ca/educmat/phy456/456lec02.htm

You're ignoring the hydrophobic effect, which is usually considered a major driver of protein folding. Packing the hydrophobic amino acids inside the protein core during folding does increase the entropy of the water around the protein.

To be precise, for a process to be spontaneous, the entropy of the system can increase or decrease, but the entropy of the universe must increase. To say that a process decreases the free energy of a system is exactly equivalent to saying that it increases the entropy of the entire universe.

@NicolauSakerNeto you are absolutely correct! When I said entropy decreased I meant the system entropy.


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How does binding energy "manifest" itself?
The concept of binding energy is, that it is equal to the energy that is needed to separate atoms in a covalent bond. I think I have understood that. The electrostatic powers between the atoms create a bond which needs to be "broken" to free the atoms.
My question is: If for example an exothermic covalent bond is created how does this bond energy get transferred to the surroundings? Atoms witch want to fulfill the octet rule create an electrostatic bond. HOW/WHERE is energy released and where did it come from?

related https://chemistry.stackexchange.com/questions/710/fundamental-forces-behind-covalent-bonding

Any way and any source possible.

https://chemistry.stackexchange.com/questions/44546/what-decides-whether-a-reaction-releases-light-or-heat

Possible duplicate of What decides whether a reaction releases light or heat

If you want to break a bond you first need to excited the right degrees of freedom so the molecule can move along the reaction coordinate of interest. In other words, the correct vibrational modes need to be excited. At equilibrium distance this will be reflected in the kinetic energy of the atoms. As you move towards the dissociation limit this kinetic energy gets transformed to potential energy, much like you give a ball a push to move up a hill. Any excess energy (if you excite more than binding energy is required) will result in kinetic energy after the dissociation.
Forming bonds happens in reverse: Two fragment come closer and convert their potential energy into kinetic energy.
After such an reaction (bond forming or breaking) several options can happen:

the excited states go into lower states by emitting photons (e.g. fire, chemoluminescence)
the energy dissipates into other degrees of freedom (e.g. by momentum transfer in collisions with other molecules)
the reaction may happen backwards and you get your initial educts again
(there are probably further options ...)


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What is the difference between statistical mechanics and quantum mechanics in terms of accuracy?
It is my impression that everything relevant to 'quantum' is the truth and others is just a simplification. Is this the case in those two kinds of mechanics?

Quantum mechanics is about the physics of very small things, molecules and smaller.
Classical mechanics is about macroscopic things. Quantum mechanics covers the whole of classical mechanics as well, but in the macroscopic limit both become equivalent. For example discretized energy states become so close, that you can thing of them as a continuum of states.
Statistical mechanics can be employed to consider many, quantum or classical, like systems and how they evolve. One application is to get thermodynamic quantities from quantum mechanical energy states, by considering how these states are statistically populated.

From plain Wikipedia,

Statistical mechanics is a branch of theoretical physics that uses probability theory to study the average behaviour of a mechanical system, where the state of the system is uncertain.

(Thus, very useful for things like thermodynamics, where everything we measure is an average in some sense.)
Statistical mechanics per se doesn't rely on any specific set of laws for the interactions of the "small parts" that compose a mechanical system.
This set of laws could come from either quantum or classical mechanics.
So, in a way, the comparison posed is not valid.
But, by comparing the statistical treatment that uses classical mechanics with the one that relies on quantum mechanics, the later gives us a broader agreement with experiment.
Again, from plain Wikipedia,

The term "statistical mechanics" was coined by the American mathematical physicist J. Willard Gibbs in 1884. (...) Gibbs' methods were initially derived in the framework classical mechanics, however they were of such generality that they were found to adapt easily to the later quantum mechanics, and still form the foundation of statistical mechanics to this day.

They operate in different ways, quantum mechanics allows us to derive, for example, the energy levels of molecules and statistical mechanics allows us to study the effects produced when these energy levels are populated, usually thermally via the Boltzmann distribution.
Neither theory has been shown to be false under the conditions where they might be expected to apply. However, in quantum mechanics there is no satisfactory derivation of the Schrodinger equation, he just wrote it down. Therefore, even though QM works for all molecules and explains very many phenomena  there remains the possibility that it will eventually be shown to need modifying.
(note : While Qm is often thought to only apply to small things, the quintessentially quantum Double-Slit experiment, often performed with photons and electrons, has now been performed on molecules as large as $\ce{C60}$. So there is no clear distinction between 'small' and 'large'. Similarly Bose-Einstein condensates can contain many thousands of atoms. Remarkably, White dwarf stars are thought to be supported by 'electron degeneracy pressure', a consequence of the Pauli exclusion principle.)


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Why are peroxides unstable but disulfide bridges considered stable? Why are esters stable but thiolesters are unstable?
I can not understand why a peroxide $\ce{R-O-O-R}$ is considered reactive and unstable.
Going down one row on the periodic table, a disulfide bridge ($\ce{R-S-S-R}$) is apparently super stable and super important to proteins 3d structure.
At the same time, a thioester is considered to be as unstable/reactive as ATP and the ester is the stable one?
What am I conceptually missing here to explain this contradiction?

Why do you think the trend should be the same, though?

You're trying to compare trends between -O-O- and -S-S- bonds, and -C-O-C- and -C-S-C- bonds.

https://chemistry.stackexchange.com/questions/431/why-does-sulfur-but-not-oxygen-catenate

From @Mithoron's link, I also found this https://chemistry.stackexchange.com/questions/47056/why-is-an-s-s-bond-stronger-than-an-o-o-bond   which explains why s-s bonds are stronger than o-o bonds.

The whole point is that you're comparing apples and oranges. The O-O bond is weaker than the S-S bond, so peroxides are less stable than the corresponding disulfides. But there aren't any O-O or S-S bonds in esters/thioesters, so there's no reason why the same argument should apply, and there's no "contradiction" in the fact that thioesters are more reactive than esters (which is due to poor overlap of sulfur's 3p orbital with the 2p orbitals of C=O).

As you can see at the top of this comment chain, I asked you a question when you first posted this, as I wanted to see what you were thinking - that way it would be easy to identify a mistake, and maybe you learn something extra along the way. But if you don't engage with people who would like to help you, then it simply makes it harder for them to help you. What you are conceptually missing has already been pointed out - it is the fact that you are comparing two very different types of bonds.

The confusion arose because in my non-majors biochemistry class they said thiolesters are easily broken because the longer bonds between the carbonyl carbon and the sulfur, make the partially positive end of the carbon atom more susceptible to nucleophillic attack, and in gen chem we learned that longer bonds tend to be weaker, this makes sense... However, I do not understand why a longer bond in a disulfide bridge would somehow be MORE stable than a peroxide. I probably have a poor understanding of orbital overlap?

Like I get how esters are stable bc of how much the minor resonance structure contributes to the hybrid vs thiolesters, I guess I should word the question differently: why are peroxides unstable?

Some hard data: bond enthalpies (in $\pu{kJ mol-1}$)
$$\begin{array}{c|c|c|c} \text{Bond} & \text{Enthalpy} &
     \text{Bond} & \text{Enthalpy}  \\ \hline
\ce{C-C} & 350 & \ce{Si-Si} & 226\\
\ce{N-N} & 163 & \ce{P-P} & 201\\
\ce{O-O} & 146 & \ce{S-S} & 226\\
\ce{F-F} & 155 & \ce{Cl-Cl} & 240 \\
\end{array}$$
We observe a decrease in bond energy from carbon to nitrogen and from silicon to phosphorus. It is clear and probably have same reason: appearance of lone electron pair of the atom. There is a repulsion between lone pairs, that is in partially compensated by stronger bonds formed by smaller atoms (the atomic radii falls down to the end of the row). When moving from second to third row, this repulsion is reduced thanks to larger size of atoms of the elements of the third row.
The general decrease in energy from carbon to silicon is thanks to increased size of the orbitals of valence level, making them more diffuse with less efficient overlap and longer bond length.
The stability of esters vs thioesters is ruled by different reasons. Sulfur is not as electronegative as oxygen, but when you consider an anion, sulfur anions have larger radius and are less prone to grabbing first positive charge they find. So sulfur anions are more stable intermediates (kinetically), and thus polar dissociation of thioester bonds is easier.

Regarding the last paragraph; it is correct if you are considering a dissociative mechanism of the form $\ce{RC(=O)SR' -> RC#O+ + ^-SR'}$. I'd suspect, though, that the usual mechanism for thioester cleavage is associative. So I think it would be to do with the electrophilicity of C=O, which is in this case dependent on resonance with the lone pair on sulfur (poor overlap = less donation into π* orbital = more electrophilic).

@orthocresol The C-S bond is still easier to break than C-O bond in this case as well.

It has to do with electronegativity (EN). Remember the concept of "formal charge"?
In a bond between molecule 1 and 2, if molecule 1 is more electronegative than molecule 2, then the electrons will be drawn closer to molecule one.
That is general, but to get to your specific question, Oxygen's EN is 3.5. Sulfur's EN is aprox 2.5. Carbon's EN is also 2.5.
So in a Carbon oxygen bond, the electrons will shift to oxygen.
In a carbon-sulfur bond, the electrons will be relatively in the middle, thus one can expect a carbon-sulfur, and sulfur-sulfur bonds to be relatively equal in strenght.
On the other hand, the peroxides (when bonded to carbons) will always pull the electrons towards them more than the carbons wil, and at some point, the bond will eventually snap.

Electronegativity is only one factor, and rarely the deciding factor, in bond strength. I highly doubt that C-C, C-S, and S-S bonds are all of the same strength. Your last paragraph is also not exactly on topic. The issue is not the C-O bond breaking, it is the O-O bond. The argument is also incorrect, as C-O bonds are generally not easy to break at all.

And another factor is....


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Is there any empirical upper limit for order of reaction?
I learned from one of my teachers that empirically reaction with order greater that 3 are never found. But, I am suspicious of the truth value of his claim.
To add to my suspicion there is this paper from 1932 referring to a fourth-order reaction
It seems to me theoretically there should be no upper limit for a  reaction as it is possible for 4,5,6... chemical species to come together with the right activation energy in the right direction.
However, empirically making 4 or 5 chemical species to form a transition species is very hard. So, there should be a empirical limit to order of reaction in the light of chemistry we know today.
What is this limit?

Are you asking why he didn't provide a reference for something that is empirically not observed?

@Zhe Ha ha,  made a blunder. But forget it. Answer my question. Is there an empirical upper limit to order of reaction?

If you can't find an example of a fourth order reaction, then the empirical limit is 3...

Also, consider what you are asking. You want four separate species to come together in a transition state?

@Zhe read this. http://pubs.acs.org/doi/abs/10.1021/ja01350a021

@Zhe Yeah! Why not? Is it 'impossible'?

Your question does not ask about impossible. As far as I can tell, you asked about empirically found. And you just found one, so I'm not sure what you are asking any more...

Can't you simply edit my question? That would be better. Better than pointing mini errors.

I don't know what to edit your question to. Impossible and empirically not found are not the same thing.

Do you mean overall reaction order or for a reaction step?

@Eashan Godbole I don't wanna complicate things. So, I am talking about reaction steps.

The order of a reaction is an experimentally determined quantity and can be  positive, negative or fractional. The order need not be related to the stoichiometric coefficients, although sometimes it is. In some reactions it is not possible to define an order. There seems to be no reason therefore why the order cannot be greater than three; the iodate-iodine reduction (Dushman) reaction $\ce{IO3^- + 5I^- + 6H^+ -> 3I2 + 3H2O}$ has the rate expression $r \approx \ce{[I^-][IO_3^-][H^+]^2  }$.
If the reaction is $aA+bB \rightarrow cC+dD$ then the rate is
$$ r= - \frac{1}{a}\frac{d[A]}{dt} = \frac{1}{c}\frac{d[C]}{dt}= \cdots =k[A]^\alpha[B]^\beta[C]^\gamma \cdots$$
where the order n is $n=\alpha+\beta+\gamma$ and these need not be the same as a, b and c.
If the rate expression is not of this form then it is hard to define an order; one familiar example is the chain reaction $\ce{H2 + Br2}$ where the experimentally determined rate law for the production of HBr has the form $\displaystyle r=\frac{k_2[\ce{H_2}][\ce{Br_2}]^{1/2}}{1+k_2'[\ce{HBr}]/[\ce{Br}]}$ where an overall order cannot be defined.
The order should not be confused with molecularity which describes the number of species reacting in a  postulated elementary reaction and which is always a small positive number. Effectively this means 1 or 2. A molecularity of 3 is not impossible but the chance of three  reactive species colliding at the same time is vanishingly small. In practice when a molecularity of 3 is suspected it is often found that in  subsequent experiments an intermediate is formed between of the two species which then reacts with the third at a slightly later time.

In an order 4 reaction, the transition state for the rate determining step still requires that four species are coming together. Otherwise, I don't see how you would get order 4. It doesn't mean that they're all combining in a single step, but you do need to have 4.

@Zhe, I'm afraid not, the order is an experimentally determined quantity, and is not related at all to what actually happens at a molecular level.  This is given by the molecularity.  The order is obtained by observing the dependence of rate on the  concentration of various species, hence the complicated and unexpected rate expression for $\ce{H2 + Br2}$

For a reaction with a well-defined mechanism, I am forced to disagree because the rate law of the reaction reflects the difference between the states that define the activation barrier. I agree that this is not in general the case because some reactions are not so easy to describe.

@Zhe, its just the difference between order and molecularity;  certainly a molecularity of 4 in not possible and probably not 3 as I mention in my text. The order tells us nothing about  what happens at the molecular level and this is why chemical kinetics is difficult and reaction mechanisms hard to determine, e.g. an example is the reaction $\ce{H2 + Br2}$ mentioned above.

OK. I don't see us disagreeing on anything. There are times when order reflects the molecularity of a well-defined mechanism.

Molecularity can itself become ambiguous. In this answer the rate determining step for nitric oxide reduction by hydrogen is discussed. Two nitric oxide molecules first form a temporary dimer, which reacts with a hydrogen molecule. If you regard dimer formation as separate from the interaction with the hydrogen, molecularity is 2; if you assume the hydrogen arrives during a relatively long-lasting collision, molecularity is 3. Either way gives the same experimental predictions.