text:
Why did my glass pot explode when I boiled sugar in it and added cold milk?
I decided to try some cooking with sugar and put a glass bowl on the flame with half a cup of sugar in it and some water. After bubbles appeared I stirred the sugar for about a minute and then I decided to add milk, and as soon as I poured the milk, everything exploded and the glass pot shattered into little pieces rapidly.
What just happened? Why did my glass pot explode when I boiled sugar in it and added cold milk? And does this also happen with laboratory equipment?

You have discoverd that cheap glassware does not like thermal stress. Congratulations. Good thing you did this with milk and sugar.

@Karl Yes, but an explosion ? and so rapid ?

Of course, immediately. And that was not an explosion. What on earth were you expecting from this experiment?

@Karl Nougat,Honestly.

I'm sure there is a SE for cooking and bakery. What happend is this: The boiling point of your thick sugar solution is well above that of the water in the milk. So the cold milk not only broke the beaker, but also quasi instantly evaporated. 18 gramms of water are >25 liters of gas. I'm sure this looked a bit like an explosion.

Yes, i know about the Se for cooking, i just wanted a chemists explanation for this, Would this happen with any other solution where with such difference between the boiling points and with cheap glass?  and i don't know if it's related or not, but i decided to try milk and sugar in the microwave in a plastic bowl i use in the microwave, and it melted the plastic. so i thought it has to do with thermal absorption of sugar

You said that you used a "glass bowl." You didn't specify if the bowl was Pyrex. But even Pyrex is susceptible to breaking with "great" thermal stress, especially if the glass has been weaken by scratches. I had a Pyrex beaker break when  making dilute sulfuric acid. // You get a lot of thermal stress when only a part of the glass container is made much hotter or colder than the rest of the glass. Such as pouring a relatively cold liquid into a hot glass container.

@soundslikefiziks Look, if you cannot judge that by yourself, now that you have the info given, you should not be doing such experiments at all. You are endangering your health, especially you eyesight and fingers, and that of all people around you. Sorry. Stick to the procedures given your cookbook.

@Karl I couldn't agree more, it was dangerous and the glass pieces could've easily harmed me, but i did not predict this outcome, and to me it was not taken as an experiment, but a genuine attempt to prepare nougat. i am definitely going to stick to metal cookware from now on.

@MaxW Out of curiosity, why did your Pyrex beaker break ? was it a temperature different accident as well ? and my glass was not Pyrex, it was an old glass.

@soundslikefiziks - When you dilute sulfuric acid the reaction is very exothermic. The sulfuric acid is also denser than water so it tends to sink. thus the bottom of the beaker can get very hot. Glass doesn't conduct heat well, so it is very easy to have spots of glass much hotter than other spots. This sets up immense stresses.

Assuming initially your glassware contained a cup of sugar and a little of water, you basically prepared caramel.  Depending on the sugar employed, temperatures needed to trigger caramelization vary, but $\pu{160 ^\circ{}C}$ ($\pu{320 F}$) mentioned succrose, as example (which is the normal household sugar) is quite high. Most materials, including glass, dilate upon heating, which is fine as long as this is evenly done.
Now if you add cold milk, say of $\pu{20 ^\circ{}C}$, the glass aims to contract back to its initial state. Because of the quantity of milk added, and the large heat capacity of water as a major constituent of milk, milk served momentarily as considerable heat sink.  Now taking into consideration that the glass ware's walls are unevenly heated -- at the outside still heated by gas or stove to more than $\pu{160 ^\circ{}C}$, and the inner of $\pu{20 ^\circ{}C}$, this suddenly generates a lot of mechanical strain and stress on the material.  The more the walls of your glass ware are thick, the more easy these may then crack and shatter into pieces just by the sudden temperature change.  (For the same token, you place hot glass ware on a plank of wood to allow slow cooling to room temperature.)
In addition, if the glass ware were closed tightly just after addition of the cold milk, the hot ($\pu{160 ^\circ{}C}$) is able to boil off the water in the milk, too; generating steam that likes to expand, or -- if confined in volume -- will build up pressure.  Under normal circumstances, for each litre of (liquid) water, up to $\pu{1.7 m^3}$ of steam may be generated (at normal pressure). This represents an additional stress for the material, and standard kitchen glass ware is not designed to withstand such pressures.

What if i had boiled the milk separately before i added it ? would that be an example of "evenly done" ?

@soundslikefiziks Your right, gentle addition of milk already boiling to the caramel (which often need not be at $\pu{160 ^\circ{}C}$ for mixing) lowers the temperature change, hence lowers the mechanical stress the glass ware will experience and consequently will lower the chance that the glass ware will shatter (and getting the content spilled all over ...). Beside looking glass ware dedicated for cooking, why not simply switching altogether to a sauce pan, made of metal? (So far, I still did not see a sauce pan crack...)

I will definitely follow that advice and switch to metal made cookware. i was just surprised by this explosion.

" if the glass ware were closed tightly just after addition " That is impossible.

While certainly not as tight as a pressure cooker, we all experience once upon a time a lid accidentally crammed to the pan, for example. Then, the comment above, "but also quasi instantly evaporated. 18 gramms of water are >25 liters of gas. I'm sure this looked a bit like an explosion." has "Karl" as signature tag, too.

Mind that caramelization can be exothermic, which doesn't help even heating :)

I did a calculation of how much thermal stress can develop merely from pouring boiling water into a glass teapot in an answer on Physics The pure size of the numbers involved is impressive.

@Buttonwood I don't get it. How do you pour water (or milk) in an overheated pot (160°C!) and keep it from boiling over long enough so you can cram a lid on it?

Whilst it's almost certainly down to thermal stress and/or the milk effectively "exploding", there's another possibility that is interesting enough to note: "glass cancer".
When glass is made there are often impurities in it, including nickel sulphide crystals. These can exist in two forms, alpha and beta, depending on temperature. When glass is toughened it is heated up and then cooled rapidly - the heating required is coincidentally high enough to change the NiS inclusions into the more compact form, and when the glass is chilled the NiS is trapped in this form. Over time they will revert to the larger form, potentially shattering the glass in the process. Even a UV photon can trigger the process, I've seen a pint glass sat on a table in a beer garden spontaneously explode, it's quite impressive.
The heat shock could be enough to trigger this even if it's not enough to mechanically shatter the glass. The fact that you say it shattered into "small pieces" (relatively regular, quasi-cubical pieces?) is typical of toughened glass failing in this manner.

Yes, it is interesting and well worth noting. i also wonder how much of these impurities are needed in order for this to happen.

This does not happen with pyrex glass, only standard safety glass. The former does not contain these impurities, and the latter is unsuitable for heating anyway. And the story about the UV photon is, well, less than scientific.

I have already experienced the sudden explosion of two glasses. Temperature unchanged. In one case I did spoon an aspirine tablet, glass went into lots of small cubes (I do not know how to translate, perhaps safety glass but this could be a glass with metal grid or polymer films. In the second case the glass exploded in two tre pieces and was even noiser. I think these case are related to annealing induced tensions, isn't like that?

Older Pyrex is true borosilicate.  to save money, some of the newer pyrex is tempered glass.  I think the true borosilicate glass has pyrex in lower case and the tempered glass has PYREX in upper case.


text:
Why lone pair occupies more space than bond pair?
I'm not a chemistry expert and merely a high school student. In my book there is a slight introduction to VSEPR theory and therein they mention that lone pair-lone pair repulsions are maximum, followed by lone pair-bond pair and then bond pair-bond pair repulsions.
Now the reasoning they provide to this is that as lone pairs are localized on the central atom, and each bond pair is shared, the lone pair occupies more space and hence the repulsions are greater.
Now I don't understand how localisation of lone pair leads to its large size and secondly what is meant by"size"here, because it's hard to imagine size of a lone pair for me.
It's a request not to mention about advanced theories like molecular orbital theory, because I'm not familiar with aaj that right now.
EDIT
I would say that the other is also in a way similar but I could not find an answer of my question there. That's why I posted it.
Any of your help is appreciated.
Thanks.

@Ayushmaan I referred to this but could not get an answer to my question from here. That's why I posted it. If you know the reason , please help.

At this stage we are thinking in this box, so we view, electrons clouds and bonds as concrete blocks sticking at one place. But they float...move rapidly pushing the surrounding with repulsion and what not...to spread as far apart as possible. Given that lone pair has a clotted electrons cloud and near to nucleus than to bonding pairs, wouldn't it exerts the most repellence asking for biggest space?

This is one of the worst parts of the theory. It would be better not to try and understand it, because any explanation of this effect is only an attempt to make the theory fit the observation. This part has no footing in actual physical phenomena.

@bon codigo Sorry, but I still didn't get an answer. I tried to understand yours in this comment section, but it's beyond my understanding. I've edited to explain why I posted this though there is a similar question put there.May be Martin is right , that the answwr is above my level now. Thanks and would be happy to get a correct and simple answer, if possible.

Abhinav, I second what @Martin-マーチン mentioned. I am too a naaive victim of trying to fit theory to observation (has no footing in actual physical phenomena). What I learnt here in CE was, that in order to grasp the basic theory well, I stay in the box at that stage. As there's more theory and application comes along, I attempt to question and expand my knowledge a little. It is a slow process for me.

It is very unfortunate that this theory is taught in such a universal fashion as  a truth. (And it has many flaws.) You therefore learn something, that later, when you advance to a more complex understanding, have to unlearn. And that usually is more difficult. The very questions you ask are proof that you would be ready for a deeper understanding and a more fundamental theory. My personal opinion is that molecular geometry can be taught in a different way, not resorting back to oversimplified theories. Stay curious and a bit patient, over time, things will get clearer.

The basic premise of this statement is that the repulsion exerted by an electron system depends upon the area it occupies. More the area, more is the repulsion.
As the bond pair is formed from the sharing of electrons, it experiences attraction from not one, but two nucleii. On the other hand, the lone pair is attracted by only one nucleus.
So, the bond pair is more localised while the lone pair is spread out and occupies a larger area.
The 'size' here refers not to the size of electrons but to the size and nature of the orbitals these electrons occupy.

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Is it safe to evaporate hydrogen peroxide in test tube?
Is it safe to evaporate 3% hydrogen peroxide in test tube? What will be left in the test tube if the solution totally evaporated?

Don't do chemical experiments if you cannot predict what will happen! You are bound to injure yourself and others!

Possible duplicate of Is it safe to immersed metal in hydrogen peroxide in a closed container? In hydrochloric acid solution?

@andselisk This definitely isn't a duplicate.

At what concentration and is it aqueous solution?

I just used the one that can be bought from pharmacies, with 3% concentration only.

NO, do not do this.  It is too dangerous.
Pure $\ce{H2O2}$ is known to be an explosive material, especially if heated it may undergo thermal decomposition.  This is one reason why this chemical is sold as an aqueous solution, most often at a maximal concentration of up to about $\pu{30 m\%{}}$.
If $\ce{H2O2}$ is distilled, the process is performed under reduced pressure with specialized equipment.  A test tube, on the other hand, is not designed for a vacuum distillation; so do not do this.
If the distillation mentioned by you (which, for safety reasons, you should NOT perform) aimed to obtain a higher concentrations of $\ce{H2O2}$ as a reagent, consider alternative oxidizers; like the adduct of urea and $\ce{H2O2}$; or oxone.  Both solids are powerful oxidizers.

What if it is only 3% concentrated?

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MO Theory: Why do hydrogen and lithium bond, but hydrogen and helium don't?
I've been cracking my head trying to make sense of Molecular Orbital Theory, but all sources I find are simplistic {If anyone has an online course or a book to recommend, I'd really appreciate}. Therefore, my questions are quite simplistic:
1 - Why should atoms in the first row {Hydrogen} interact with atoms in other rows  {Lithium, Beryllium, Boron, etc}? I mean, as a way to answer why two Hydrogen atoms form a Molecular Orbital in the first place, I have the postulate that it is due to Pauli Exclusion Principle: as the 1s Orbitals - and, by extension, its electron - overlap, a new orbital is formed to avoid* having two electrons with the same quantum numbers. But, when Lithium approaches Hydrogen, there is only a significant overlap between 1s and 2s Orbitals. In this sense, there is no threat* to Pauli's principle, given that the approaching electrons do NOT have the same quantum numbers. Which yields the question: how do Lithium and Hydrogen bond?
. * I understand that the universe doesn't care about what we, smart monkeys, think of it. It doesn't see things as "threats". Still, interpreting things this way makes everything easier.
2 - Why won't Helium bond with Hydrogen? The approach of He and H should form bonding and anti-bonding Orbitals, which will be filled with three electrons. This should place two electrons in the Sigma Bond, but only one in the Sigma* Bond. Therefore, the bond order is 0.5! So why HeH isn't observed?
Another way to phrase it: If He2+ can exist as a molecule, why can't HeH?
Thank you very much for your time.
Sorry if I messed something up.

For LiH, I understand that the two nuclei create a combined potential that results in a redefinition of all orbitals. We then have two electrons in what could be called the common 1s orbital, but which is almost the same as the only-Li 1s orbital, just minimally deformed, and therefore it makes sense to still call this the 1s orbital of the Li. And  next we have two electrons in a next hiher common orbital, which is not so much owned by either of the nuclei alone because 2/3 of the Li charge are mostly shielded off by the 1s electrons

You've asked seemingly simple questions that get into quirky considerations. A molecule of HeH would very unstable:
$\ce{2HeH -> 2He + H2}$
As Oscar points out HeH probably can be made. In all of deep space where molecules could go days before colliding with another particle there must be at least one molecule of HeH. So it can be "made" and it does "exist." But how would you detect it?
Typically when we are talking about "making" something we are talking about making it in a earth bound laboratory. You make enough to at least observe it somehow. For instance a lot of radioisotopes have half-lives less than a second which really doesn't give an experimenter time to do any really "chemistry" with them. So in any practical sense the HeH molecule doesn't exist.

https://en.wikipedia.org/wiki/Helium_hydride_ion#Neutral_molecule

Probably $\ce{HeH}$ can be made, but, like the dihelium cation, it would not be stable in low-energy environments such as a typical condensed medium. And its nonionic character means we can't see $\ce{HeH}$ in a mass spectrometer.

"it would not be stable in low-
energy environments such as a typical
condensed medium" why?

HeH doesn't lay in minimum on PES - even if you'd make it, molecule would break down instantly.

@Mithoron I don't understand your comment.

@Mockingbird it would be unstable in a condensed medium because the H could form a stronger bond with virtually anything else except He. I'm not sure Mithoron's comment "HeH doesn't lay in minimum on Potential Energy Surface" is true, it may well be that the forming of a He-H bond does release energy. But so little energy (bond so weak) that it a temperature only slightly above absolute zero would be all it took to dissociate into He and H. Getting something that cold when the hydrogen is desperate to react with anything else is probably only achievable in the hard vacuum of interstellar space

What drove you to this kind of considerations is the use of that postulate, which is not true as your second question made clear.
First combine the relevant AOs, then fill in the electrons following the Aufbau, considering the exclusion principle by Pauli, and applyibg Hund rule. Find out the energy of the molecule, the bond order, and compare to the energy of the atoms.
Alternatively, you can still regard MOs as forming by virtue of Pauli exclusion, nevertheless this alone does not tell you anything about bond formation (only in H H it happens to be the case).
If you draw a qualitative MO diagram for hydrogen helide (HHe), taking into account the stability of H and He (in other words He as being noble has a much higher electronegativity than H, so following the common rule its AO lies lower down in energy), you see that overall you end in an antibonding situation, different than the isolectronic - but homonuclear - case of dihelium+ that you've mentioned as example.

Hmm... I see. So, basically, when electrons approach they can reinforce each other, hence increasing the resulting wave function and, due to this, creating a lower energy state. We interpret this as the fusion of AOs, creating MOs.

@Andre Liston. Somehow yes. Not really the electrons reinforce themselves, nor that the wave function gets increased. But if the amplitude of the wavefunction gets increased along the axe of the atoms (at least with sigma bonds) and upon filling it you find a energetic minimum as compared to the isolated atoms, then the latter two are bound.

Bond order 0.5 means that it CAN exist, but it's just not very stable. At the end of the day, it probably exists in deep space or something, but would immediately decompose if we synthesised it.
Also, one must note that the orbital interaction is non-degenerate, meaning that your diagram is very much inaccurate. Due to the immense energy gap, there would be little-to-no interaction whatsoever.

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What are the byproducts in a Suzuki reaction?
I am looking precisely at a coupling of a trifluoroborate with an aryl bromide. I know the mechanism of the Suzuki reaction, yet I cannot find any mention of the side products or degradation products that one could obtain if the coupling fails. Maybe those products are obvious for those who know well how palladium reacts, but this is not my case. I tried to find a book that would give me a feel for what might happen and could find anything close enough.
I'm expecting a protodeboration as a side reaction, yet I can't find any source nor mechanism. Could anyone shed some light on the subject?

I believe homocoupling of the aryl bromide is a minor byproduct.  There is enough literature on the reaction, but the information may be tucked away in the supporting info

It is postulated that the trifluroborates are not the transmetallating specis, rather than to be the source of slow hydrolysis to yield boronic acids:

(source 1, slide 41; citing Molander and Biolatto, 2003 (Ref.1))
which then enter the catalytic cycle:

(preview in Kürti & Czakó, page 448 bottom (Ref.2)
Organic Reactions (vide infra) explicitly states that complete hydrolysis of the trifluoroborates is necessary to allow successful transmetallation.
In addition, the fluoride anion is said to interact in the following way with the palladium catalyst:

(source 2)

References:

Gary A. Molander, Betina Biolatto, "Palladium-Catalyzed Suzuki−Miyaura Cross-Coupling Reactions of Potassium Aryl- and Heteroaryltrifluoroborates," J. Org. Chem. 2003, 68(11), 4302-4314 (https://doi.org/10.1021/jo0342368).
László Kürti, Barbara Czakó, In Strategic Applications of Named Reactions in Organic Synthesis: Background and Detailed Mechanisms; Forwarded by E. J. Corey, Introduction by K. C. Nicolaou; Elsevier Academic Press: Burlington, MA, 2005 (ISBN-13: 978-0-12-429785-2).

Thanks, but I had already found these mechanisms. I'm wondering more precisely about the mechanisms that don't give the desired product and the factors influencing them.

Oh. Sorry, I misunderstood "byproduct" as "what is naturally formed beside the product of normal coupling" rather than "product(s) obtained if the intended coupling reaction fails".  Let us see suggestions by others.

I clarified the question. Thanks anyways, this is an excellent summary of the general mechanism with sources.

Buttonwood has given an excellent answer based on mechanism and inorganic byproducts. Since OP has indicated that the anticipation is about unintended products, I have done a little search and found some help in Ref.1, Its abstract states that:

The outcome of the Suzuki–Miyaura cross‐coupling for the direct competition reaction between two boronic acids was evaluated under routine synthesis conditions. The reaction selectivity was found to depend on the amount of the base used, with fewer bases favoring the reactivity of the boronic acid with lower $\mathrm{p}K_\mathrm{a}$ (stronger acid). The dependence of the reaction selectivity on base stoichiometry was found to increase with the increase in the difference in the $\mathrm{p}K_\mathrm{a}$ values of the competing boronic acids. These results confirm a relationship between acid–base chemistry and the Suzuki–Miyaura reaction catalytic cycle. Moreover, the results indicate that under these specific conditions, the most reactive organoboron species toward transmetalation is the borate anion $\ce{R-B(OH)3−}$ instead of the neutral boronic acid $\ce{R-B(OH)2}$. Hence, the main role of the base in the reaction mechanism is to increase the reactivity of the boronic acid toward the $\ce{Pd}$–halide complex by converting it into the respective organoborate. In addition, boric acid, an important reaction byproduct, affects the selectivity in the Suzuki reaction because its gradual formation in the reaction medium disturbs the acid–base equilibrium.

Following scheme would show the outcome:

Reference:

Carlos F. R. A. C. Lima, Ana S. M. C. Rodrigues, Vera L. M. Silva, Artur M. S. Silva, Luís M. N. B. F. Santos, "Role of the Base and Control of Selectivity in the Suzuki–Miyaura Cross‐Coupling Reaction," ChemCatChem 2014, 6(5), 1291-1302 (https://doi.org/10.1002/cctc.201301080).


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Potassium sulfate as a byproduct of manganese heptoxide synthesis?
Manganese heptoxide may be  formed by the action of concentrated sulfuric acid on potassium permanganate (ref):
$$ \ce{2KMnO4 + 2H2SO4 -> Mn2O7 + H2O + 2KHSO4} $$
Could the amount of potassium permaganate as a reactant in the reaction be increased to yield potassium sulfate as a product instead of potassium bisulfate as shown?

Can you propose a chemical equation for that reaction?

The reaction you mentioned shows the formation of manganese heptoxide and potassium bisulfate. But if you alter the stoichiometry of the reaction, i.e. change the moles of the reactant, you will obviously get different products like manganese(II or III)sulfate, manganese oxide, potasium sulfate etc. For this, the reaction conditions should be changed.

$$\ce{2KMnO4 + 2H2SO4 ⟶ Mn2O7 + H2O + 2KHSO4}$$

Potassium permanganate react with sulfuric acid to produce
  manganese(VII) oxide, potassium bisulfate and water. Potassium
  permanganate - solid. Sulfuric acid - 98% solution. The reaction
  proceeds at room temperature. Impurities: manganese(IV) oxide
  $\ce{MnO2}$(source)

$$\ce{2KMnO4 + H2SO4 → Mn2O7 + K2SO4 + H2O}$$

Potassium permanganate react with sulfuric acid to produce
  manganese(VII) oxide, potassium sulfate and water.(source)

$$\ce{4KMnO4 + 6H2SO4 → 4MnSO4 + 2K2SO4 + 5O2 + 6H2O}$$

Potassium permanganate react with sulfuric acid to produce
  manganese(II) sulfate, potassium sulfate, oxygen and water. Sulfuric
  acid - 60% solution. Impurities: ozone $\ce{O3}$.(source)

$$\ce{2KMnO4 + 4H2SO4 ->[70-75°C] Mn2(SO4)3 + K2SO4 + 4H2O + 2O2}$$

Potassium permanganate react with sulfuric acid to produce
  manganese(III) sulfate, potassium sulfate, water and oxygen. Sulfuric
  acid - 98% solution. This reaction takes place at a temperature of
  70-75°C.(source)


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Space filling model for nitrogen molecule
I want to make space filling model for nitrogen molecule.
I've learned that van der Waals diameter is to be taken for atom radius.
Since van der Waals radius for nitrogen is $\pu{155 pm}$ and bond length is $\pu{110 pm}$, does this mean that nitrogen molecule space filling model is two blue spheres of diameter $\pu{155 pm}$ and center to center distance $\pu{110 pm}$?
One of the reasons for this question is that samples of space filling model for this molecule (see for example Wikipedia) do not agree with these data; there's about a 11% difference between the length/width ratio of the drawn molecule, and what it should be based on the van der Waals radius and bond length.

\begin{array}{cccccc}
  \displaystyle\frac{d}{2r}
    &\sim& \displaystyle\frac{\pu{1000 pix}-\pu{716pix}}{\pu{716 pix}}
    &\sim& 0.397\\
  \displaystyle\frac{d}{2r}
    &\sim& \displaystyle\frac{\pu{110 pm}}{2\times\pu{155 pm}}
    &\sim& 0.355
\end{array}

Yes, two spheres with a radius of 155 pm and the distance between the center is 110 pm.
The axis in that picture is not aligned, that's why you get an error. Here's a picture with aligned axis and van der vaals spheres produced in Avogadro:

The corresponding *.xyz:
2

N   -0.55   0.0 0.0
N   0.55    0.0 0.0

I'm also unable to follow your math here. height of the molecule should be $2\times 155=310$ and length should be $155\times 2+110=420$ so height to length ratio should be $0.73$, you got $0.72$ which is pretty close.

I've edited your answer to make it match the question and removed the reference to my (now) deleted comment. (I also change \cdot to \times, because sometimes that looks like a decimal delimiter on some devices.) [Flag this comment obsolete when read ;)]

Thanks. Could you please compare d/2r to make direct connection to the question?


text:
How to know which ionic compound have the highest lattice energy without knowing the values of the radii?
Given two ionic compounds, for instance:
$$\ce{CaS} \quad \mathrm{or} \quad \ce{KCl}$$
What is the procedure to predict which of the two have the highest lattice energy (in absolute value)?
(Suppose that I do not know the values of radii of the atoms but I can only look at the periodic table).

I know that I must look at the ratio
$$\frac{Z_1 Z_2}{r_1 + r_2} \propto E_\text{lattice},$$
where $r_1 + r_2$ is the distance between the two atoms in each molecule of the compound, $Z_1 Z_2$ is the product of atomic numbers of each element in the molecule.
But, for example, in the case above I have both $r_1 + r_2$ and $Z_1 Z_2$ bigger for $\ce{KCl}$ then $\ce{CaS}$, so I cannot say (without knowing the numerical values) if one of the two ratios would be higher.
Am I missing something?

The lattice energy is proportional to the charges of the atoms in the formula, not their atomic number! (There are no molecules in such simple ionic substances, btw.)

The Lattice Energy of Ionic Compounds is directly proportion to the Charge Density. As Ca, S have 2 units of charge while K, Cl have one and there is not much of a difference in size ( atleast not enough to overcome the doubling of the charge), the charge density of Ca and S ions would be more. Consequently, the lattice strength of CaS would be more.

So it depends mainly on how many electrons are exchanged! Thanks a lot! But, for istance, if I take $MgO$ and $MgS$, there are two electron exchanged in both, nevertheless the same problem showed in question remains, so is there a "secondary" criterium to select one of the two?

@Sørën As the no. of e- exchanged are the same, we now have to focus on the size of the receiving atom (which will give us an idea about the strength of the electrostatic forces).


text:
Can phenol be detected with a ceric ammonium nitrate test?
Today, I came across a question which asked name three  reagents can be used for identification of phenol. The answer given was:

Neutral $\ce{FeCl3}$
$\ce{ NaNO2 +HCl}$
$\ce {(NH4)2[Ce(NO3)6]}$.

I understand (1) and (2), but I am doubtful about CAN test (3). As far as I have studied in various books, it is used for testing presence of alcohol, and phenol is not alcohol.
Can phenol be detected with a ceric ammonium nitrate test?

With unsubstituted phenol ($\ce{PhOH}$) dark-brown precipitation is formed:
$$\ce{(NH4)2[Ce(NO3)6] + 2 PhOH -> [Ce(NO3)4(PhOH)2] + 2 NH4NO3}$$
Other phenols should produce wine-red solution as expected [1, p. 478]. Also, this and other phenol tests are discussed in [2].

Pavia, D. L. A small-scale approach to organic laboratory techniques, 3rd Ed.; Brooks/Cole, Cengage Learning: Belmont, CA, 2011. ISBN 978-1-4390-4932-7.
Atul, S. The Pearson Guide to Objective Chemistry for the AIEEE; Pearson Education India, 2010. ISBN 978-81-317-3381-3.

Could you please give some examples of the "other phenols"? Would phluoroglucinol work? Or are only monohydroxy derivatives are acceptable?

I have mentioned CAN test of alcohol and phenol in a previous answer of mine, so I am going to directly quote that:

Genereally, CAN test is done for alcohols which give pink or red
  colour. But for phenols and phenolic compounds, it gives brown or
  black color. So, this test helps to differentiate phenols from
  alcohols.

Unfortunately, the source link is broken. :(


text:
What is the product of the reaction of 2-bromo-1-fluoro-4-nitrobenzene with sodium methanethiolate?
In the reaction depicted below shouldn't $\ce{Br}$ be replaced by $\ce{SCH3-}$ ion as $\ce{Br-}$ is a good leaving group and the $\ce{C-F}$ bond is quite strong due to back bonding of 2pπ-2pπ orbitals?

What is the rate limiting step in this reaction? What is the structure of the intermediate? This should help you think about what happens.

I'd suggest looking up nucleophilic aromatic substitution in a textbook; the issues you are asking about should be described in any decent text. In particular, any decent text should be able to explain why (1) the 2- and 4-positions relative to the electron-withdrawing group are activated towards substitution and (2) the order of reactivity is F > Cl > Br > I.


text:
Ring expansion in the reaction of (hydroxymethyl)cyclobutane and HBr
I saw a question on how to predict the product of a reaction between (hydroxymethyl)cyclobutane and HBr. The correct answer was given as 1-bromocyclopentane, however, I do not understand why a ring expansion would occur in the reaction.
My attempt: I would think that the final product would by (bromomethyl)cyclobutane. This is due to the fact that HBr would first protonate the alcohol, causing it to become a good leaving group. Afterwards, an $\ce{S_N2}$ reaction between this protonated alcohol and $\ce{Br-}$ would take place, as the substrate is a primary alcohol, where no ring expansion/carbocation formation could occur as the steps are concerted, resulting in (bromomethyl)cyclobutane.
Why is this mechanism wrong, and why does a ring expansion occur instead?

Reduce ring strain. Also secondary carbocation is more stable than primary, so there's a double driving force for rearrangement via ring expansion. There might be a duplicate somewhere, I'm not sure.

@orthocresol Would the reaction not occur in a concerted SN2 mechanism, in where there is no chance of C+ rearrangement? The rearrangement you suggested would involve a SN1 mechanism, which would not occur due to the fact that the alcohol is primary, right? Or am I misunderstanding something here?

True. The rearrangement can still occur in the same step as loss of H2O.

Plus intramolecular is faster than intermolecular with bromide

https://chemistry.stackexchange.com/questions/29559/ring-expansion-from-a-given-cyclic-carbocation

Does this answer your question? Why and how does ring expansion occur in the dehydration of (cyclobut-3-ene-1,2-diyl)dimethanol?


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Why the boiling and melting points of n-alkanes are higher than those of iso- or neo-?
Starting with an example, n-butane has boiling point of $\pu{55^\circ C}$, whereas that of iso-butane is $\pu{-10.2^\circ C}$.
Why there is so much difference in the boiling points?
*EDIT : On the contrary, branched alkanes are more stable. For example, 2-methylpropane is more stable than n-butane. Why?*

@Abdul Ahad. Do not equate low boiling point and stability because the two things are not related, at least not in general.

@Alchimista No! Obviously not.What I'm saying is that how come a compound (n-alkane) having a higher boiling point is less stable ?

Is the same. They, stability and high or low boiling point, are two different things and in general they are unrelated!

Well! Thanks for clearing the fact that they are unrelated.If you could provide me with an explanation on the second part of the post,it'll be much better.So the question can be isolated as "Why n-alkanes are less stable than iso- ?"

The point is that myself I cannot recall an easy answer I started to constructing one. I proceed by considering linear and branched carbocations and or radicals. But it did not seemed to me a proper way to go. With my surprise I was somehow on the track. I just googled your question and found Protobranching. Still I have to read. It is indeed an electronic effect getting stronger as the number of Cs in germinal position - respect a given C - is higher. I forgot what I learned / I was told about this years ago. Probably some attempt to rationalise the situation via steric effects, now dismissed


text:
How to evaluate enthalpy variation of combustion of sucrose in a constant volume calorimeter?
Consider combustion reaction (with the formation of $H_2O$ and $CO_2$) of a mass $m$ of  sucrose in a constant volume calorimeter with thermal capacity $C$. This reaction increases the temperature of $\Delta T$.
How do I find $\Delta H$ of this reaction?

Since $$\Delta U= C \Delta T= \Delta H+\Delta n_{gas} R T$$
I would use this formula, but what is $\Delta n_{gas}$ in this case? It should be zero (from textbook) but I do not undersand why it is zero. Furthermore, can I really use this formula even if $T$ is not a constant and there is a variation in temeperature in the reaction?

$\ce{C6H12O6 (s) + 6 O2 (g) -> 6 CO2 (g) + 6 H2O (l)}$. 6 moles of gas on the left and 6 on the right, so $\Delta n_g = 0$.


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Can corrosion from hydrogen peroxide solution model the corrosion protection activities of Al coating?
Immersion in hydrogen peroxide can make steel corrode very fast. However, the coated steel resisted the corrosion attack for a span of time, and as time passes, the coating becomes thinner. Some got removed from the subtrate. Can this mechanism be a model for the corrosion protection of the Al coating for research studies? Does this reflect the behavior of metal coating in immersion with other corrosive media like NaCl?

Still on the same matter. The question is not clear. If you have understood the answer(s) to your previous question, and it is convincing that the Al coated sample got mass due to Al oxide formation, ie the underlying steel was protected by passivation, now you must clarify what you model? If you simply replace NaCl of the previous exp. with H2O2, obviously you will see the same but in a (much) shorter time. Obviously the microscopical and mechanical properties of the Al oxide layer (which in turn depends on those of the initial Al layer) do not guarantee a long standing protection....

... and after some time you should get two samples of the same steel with the same properties immersed in hydrogen peroxide solution. At a given time, their relative weight loss will be different due to the fact that one was protected for a while. From a moment, assuming the same surface to volume ratio, they will loose weight at the same ratio.


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What are the parameters of protonation of a molecule?

Adding an acid replaces the $\ce{-MgBr+}$ with an $\ce{H+}$. Why does this happen? Is $\ce{H+}$ a better electrophile than $\ce{MgBr+}$?
Or, to put it another way, why doesn't $\ce{MgBr}$ stays attached to $\ce{O}$?

Well, it's symbolic scheme of reaction not telling what exactly Mg(OH)Br in H2O/t-butanol solution "looks like". t-butanolate isn't monomer and hydroxide is solvated, but those are kinda details.

Isn't that a Grignard reagent? If so, it's a strong base and a nucleophile, isn't it? Making the intermediate carbocation less stable, more reactive with acidic hydrogens?

We are taking about electrophilicity of ions.
It basically depends upon:
1. charge on the ion(+)
2. density of plus charge on it
So comparing $\ce{H+}$ and $\ce{CH3+}$, the former has a higher plus charge density so that's why it will be more attracted towards 'O'.


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What source of error in an acid-base titration could lead to a higher reading of the acid concentration?

Methanoic acid ($\ce{HCOOH}$) is found in stinging nettles. In an analysis of the concentration of methanoic acid in a solution extracted from nettles, the solution was titrated with a standard solution of potassium hydroxide. Aliquots of the methanoic acid were transferred to a conical flask and titrated against the potassium hydroxide solution in a burette.
Which one of the following would cause the calculated concentration of methanoic acid to be higher than it actually is?
A Rinsing the conical flask with deionised water
B Rinsing the burette with deionised water
C Rinsing the pipette with deionised water
D Rinsing the conical flask with potassium hydroxide solution

For this question, im not sure if A and B are correct answers. The book only has B.
For A: if you rinse the conical flask, would this mean that there would be a greater volume in the flask after the titration than expected? Therefore when you have a lower concentration?
For B: if you rinse the burette with deionised water, this would dilute the $\ce{KOH}$, therefore causing a greater titre required. This means a greater mole of $\ce{KOH}$ and hence acid. Would this then lead to a greater concentration?
So B is the only correct answer, is my reasoning correct?

Option A implies that the rinse residual would dilute the methanoic acid that would be added to the flask after rinsing and thus your titration would indicate that the acid is a LOWER concentration than the calculated value.
You are correct in your assessment of Option B.

That is wrong! Excess water in the conical flask will have no influence whatsoever. Volumetric analysis is the quantitative determination of the amount of substance. The volume you read is from when you take the aliquot with the pipette. The water you add will not change that reading.

A Excess water in the conical flask will have no influence whatsoever* (in first order approximation). Since you are determining the the amount of substance of methanoic acid with the titration, the volume is irrelevant. The volume you use to determine the concentration is from when you take the aliquot with the pipette.
This option is therefore incorrect.
B You are correct, the excess water left in the burette will dilute you potassium hydroxide solution, causing you to use more of it, and therefore also reading a higher amount of substance of methanoic acid than would be correct. Obviously this leads to a higher concentration.
This is a correct option.
C Rinsing the pipette with water will cause you analysis solution to be diluted. Therefore the calculated concentration would be lower than the actual concentration.
This option is therefore incorrect.
D Rinsing the conical flask with potassium hydroxide solution will lead to having some of the methanioc acid already neutralised. You will use less of your analysis solution, therefore calculating a lower concentration.
This option is also incorrect.

In order to achieve the highest accuracy, you should

rinse the burette with the potassium hydroxide solution you use to analyse.
rinse the conical flask with plenty of deionised water (to remove any residues).
rinse the pipette with the solution which is to be analysed.

* If it is too much water, then the volume significantly changes. This will cause the indicator substance to react ever slightly differently, which might lead to you not seeing the equivalence point early enough. This could lead to reading a higher concentration.
However, you eyes are a larger source of error if it is a few drops of water.


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How do I run MP2 and/or CCSD-level hyperpolarizability calculations with GAMESS?
I have used GAMESS before for TDHF (RPA) / TDDFT level first hyperpolarizability ($\beta$) calculations. However, I cannot figure out how to calculate $\beta$ with MP2 and CCSD for comparative purposes. Do I have to use the FFIELD keyword for the finite field approach? If that is the case, what is the general workflow necessary to calculate the time-varying hyperpolarizability?

Welcome to ChemSE. Please feel free to take a [tour] of this site to get a brief idea of who we are, and what topics are suitable here. Good luck, and hope to see you satisfied soon.

As you guessed, to calculate $\beta$ within GAMESS for MP2 and CCSD, you would need to perform first- and second-order finite difference, respectively, as there is no coupled cluster gradient needed, so no true CCSD dipole moment.
However, there is a more fundamental problem: it is impossible to calculate time-dependent response using a finite field approach. Take the general form of 1-dimensional 2nd-order central difference,
$$
f''(x) \approx \lim_{h \to 0} \frac{f(x+h)-2f(x)+f(x-h)}{h^2},
$$
which can directly translate to a diagonal element of the polarizability tensor formed from energies at finite electric fields:
$$
\alpha_{zz}(x) \approx \lim_{h \to 0} \frac{E(x+h)-2E(x)+E(x-h)}{h^2}
$$
where $h$ is a finite electric dipole field of some strength applied along the $z$-direction. What is $x$? Is it as simple as converting the desired frequency $\omega$ to units of electric field strength? No; that still corresponds to a static field rather than a time-varying field. How are the different perturbation directions handled? $x$ must be set to 0.
From a molecular properties review by Jurgen Gauss, page 5:

(...) disadvantages of the numerical differentiation scheme are
a) that there is no straightforward extension to the computation of frequency-dependent properties (...)

Page 37:

While analytic derivative theory is sufficient for the theoretical treatment of time-independent (static) properties, the underlying theory needs to be extended for the calculation of time-dependent (dynamical) properties. In particular, the fact that there is -- unlike for the static case -- in the time-dependent case no well-defined energy explains why the simple derivative theory discussed so far is not applicable.

Page 39 (with some manipulation by me):

It can be shown that the linear, quadratic, etc. response functions $\left<\left<A;B\right>\right>_{w_{b}}$, $\left<\left<A;B,C\right>\right>_{w_{b},w_{c}}$, etc. and thus the frequency-dependent properties of interest can be determined as derivatives of the so-called time-averaged quasi energy
$$
Q(t) = \left<\tilde{\psi}\left|\left(H - i\frac{\partial}{\partial t}\right)\right|\tilde{\psi}\right>
$$
with the phase-isolated wavefunction
$$
\tilde{\psi}(t) = e^{iF(t)} \psi(t)
$$

where $F$ is a time-varying external field, and $\psi(t)$ is the usual wavefunction where MO coefficients and thus the density are explicitly time-dependent.
The only numerical (rather than analytic) method for calculating dynamic properties I am aware of is direct time-propagation of the wavefunction. If you want to calculate frequency-dependent CCSD hyperpolarizabilities, your best (free) bet is DALTON. Note that their coupled cluster module is closed-shell only.

Thank you for your response; it was very informative and I am grateful for the resource you have provided.  I believe analytic gradients for CCSD and MP2 for the first hyperpolarizability have previously been published, is it just the case  that they have not been implemented into GAMESS?  Or am I mistaken about their derivation?

The point is that they are not implemented in GAMESS. Analytic gradients for MP2 and CCSD are in most packages. Very few programs (DALTON, Gaussian) can calculate correlated hyperpolarizabilities.

@ComputationalNovice If my answer was useful, please consider accepting it.

NWChem has had CCSD quadratic response hyperpolarizabilities with non-zero frequencies since 2008.  You can do RHF, UHF or ROHF, but I strongly recommend ROHF for open shells, for reasons discussed in the paper linked last.
https://nwchemgit.github.io/TCE.html#input
https://pubs.aip.org/aip/jcp/article-abstract/130/19/194108/296267/Parallel-computation-of-coupled-cluster
https://pubs.aip.org/aip/jcp/article-abstract/128/22/224102/924805/Coupled-cluster-dynamic-polarizabilities-including

https://github.com/nwchemgit/nwchem/tree/45332c485cdc3143c13b382848b3acdbd967f4f3/QA/tests/tce_hyperpolar_ccsd_small is an example input file, since it seems the user manual is not complete (my fault)


text:
Why does sunlight cause plastics to become a faded blue color?
Due to the UV absorption, I knew that the plastics can be easily affected, especially PP. But is it normal that the change is always into faded blue from white color? and Why?
Because I made an experiments, but the results made me surprised. One sample of opaque white PP material was put on sunlight, one sample of same material having same conditions was put in lightproof black color bags. Then measurements was conducted every hour till 8 hours by using spectrophotometry which works in CIE Lab color space.
According to the results, b-value of sample exposed to sunlight had changed into the minus value, describing blue color in L a b color space, while L- and a-values are relatively stable. The b-value of another sample in lightproof bags had changed into the plus value, describing the yellow color in L a b color space, while L- and a-values are again relatively stable. So we can see that the effect of sunlight is only on b- value, which describes the values of blueness and yellowness of the sample.
My question is that what is the phenomena behind this?
What do you mean by reflected and transmitted lights? Even looked with naked eyes, easily seen that it becomes slightly faded blue color. The object is opaque white color PP material.

Think about what happens to blue light that hits something "blue", and to blue light that hits something "red". I wish I could find a photo of an old Pepsi can that's been in the sun a decade. It's a great illustration of the phenomenon.

Look more faded blue in transmittance? Or in reflection? Undefined?

What color is the plastic to start with? Dark blue? Transparent? Any color? The question likely needs additional context to give a good answer.

Please don't re-post your question, you can edit it to bump it

It is very hard to generalise about things like this since plastics often contain a variety of minor additives (for example colourants or stabilisers). In addition, plastic degradation in light is complex. So the phenomenon could be scattering-related or due to some differential reaction of some other additive (e.g. red compounds tend to fade quickly in sunlight as the light they absorb has higher energy and loss of a small amount of red will lead to a bluer colour).

I don't think its actually blue, I think the blue is a result of Rayleigh scattering.
Over time the dyes are destroyed by light removing any color. simultaneously the plastic is degraded and becomes rough on the surface and acquires an inhomogeneous refractive index within the structure due to the destruction of bonds and oxidation and the end of chains. These two structural changes cause scattering, and since blue is scattered more than the other colors due to the relation of scattering with wavelength to the fourth power, the material appears blue when looking at the reflected light.  If you look at transmitted light going through the plastic it should appear yellow.

Yes, it is why I asked the author to specify in which way he/she looks at the plastic. I really think as you do - scattering. But in principle you can have absorbers in the blue region, that once destroyed let more blue passing through your object.

More context was added to the question


text:
Why are heterogeneous catalysts calcined?
Most of the literature I read says to calcine heterogeneous catalysts when preparing them. When synthesizing heterogeneous catalysts, what is the reason for calcining them?

I guess this is only true for heterogeneous catalysis, when you need a mechanically-stable porous matrix with developed surface area. In homogeneous catalysis you want the opposite, so that your catalyst remains dissolved all the time and doesn't precipitate, so the calcined compositions are probably the worst choice here. Maybe you need to clarify the question a little.

For heterogeneous catalysis calcining of the catalyst allows for the following:

mechanically and thermodynamically stable macrostructure (matrix) with defined geometry, surface and pore size;
predictable activity based on calcining regime (generally, there is a temperature peak for maximum activity, and further extreme temperature treatment often lowers catalyst's activity);
pure composition without traces of organics and majority of other undesired impurities;
production convenience and scalability (for example, metal oxide catalyst manufacturing process can be easily implemented within mining and processing plants);
when distributed over support, more uniform distribution of metal oxide is achieved, and upon slow treatment crystallinity of the catalyst can also be tweaked.

One of the most important reasons to calcine a catalyst preparation is to decompose the impregnated metal salts.  This includes decomposition of nitrates, chlorides, carbonates and organic chelates, leaving a metal or metal oxide on the support surface.


text:
Will bonding molecular orbital be formed?

Which of the following statements is not correct regarding bonding molecular orbitals?
A Bonding molecular orbitals possess low energy than the atomic orbitals from which they are formed;
B Bonding molecular orbitals have low electron density between the two nuclei;
C Every electron in bonding molecular orbitals contributes to the attraction between the atoms;
D They are formed when the lobes of the combining atomic orbitals have the same sign.

I am confused about option D, I think this should be right as same sign means lobes are in the same phase, so they should form BMO. But the answer key states that the statement is incorrect.
Can anybody please explain me this and is my reasoning correct?

Statement (D) doesn't quite make sense. Same sign? Same sign where?

Consider the canonical bonding molecular orbitals for butadiene. One of the bonding molecular orbitals has a node, so the signs are not the same.

@Gem. Perhaps because D) is intentionally unclear? A statement like ".... have the same sign in the interatomic region" which is more commonly said, as you did, "lobes overlap with the same sign". It is like that D) means that both lobes of two p orbitals have a + and a - lobe, the same signs.

So if 2 - lobe combine would they form a BMO

The question (of the exercise, not that by the OP) Is not well formulated. Zhe is correct with the example of butadiene. However the question seems to refers to biatomic molecules (choice B) "... between the two nuclei". Anyway the general answer is A and one has not to run in distinctions, nor to pi-sigma separation, etc.

I have studied MOT only for diatomic molecules and very few heteroatomic such as CO NO

This question shouldn't be closed. The author has included enough of their thoughts to give an unambiguous and useful answer.

I also don't think this should be closed, at least not according to the HW policy. To me it is fairly obvious that the exercise is misleading at best and relies heavily on a certain form of definition. I guess the whole thing needs to be dissected. Please add the source (citation) of the question for context. If it is a textbook you can also add [tag:erratum].

I thought the question was which statement is correct, while it asks for the one (it) which is not. So the answer is B). I think is a mess of an exercise :) If one has to find more, than D) is also wrong - in the sense I've tried to explain, or better as Zhe said "same sign where?".


text:
Why does mmHg persist in organic chemistry?
In most areas of chemistry, standard units are adopted - we never would think to use ounces (even American chemists resist) BUT even in modern journals or textbooks, we commonly still see the unit mmHg used. Why is this the case and why have people not moved to adopt SI units in this situation?
To give an example, consider the "Purification of Laboratory Chemicals" book that is often referenced here. Many distillations are given using pressures in mmHg.

In comparison with inHg and psi, mmHg looks pretty decent. Also, habits and arguably convenience when interpreting information from the resources as dusty as the first Merck's Index.

Predominantly habit, I suppose.  Most people of a certain age remember ACTUAL manometers that used millimetres of mercury.

Actual millimeters of actual mercury, I would stress. I remember them myself, and I'm not that old. Once in a while they would break and spill mercury all around, which was a peculiar kind of fun.

Unfortunately the same reason why the modern HPLC I use registers pressure in Megapascals on the hardware (which is fine that is SI) and psi on the software (pisses me off:). Convention, and stubbornness.

I just graduated from grad school last year and I exclusively used Schlenk lines with actual mercury bubblers with (metric) yard sticks attached for distillations. It was very convenient to measure the pressure in mmHg.

Possibly it's academia.  No modern chemist in Industry would use that unit. I saw inchHg from a CMO the other week and it rang alarm bells. It meant they were using an analogue guage and hence, an old possibly uncalibrated one.


text:
Is the Arrhenius Equation a solution to a differential equation?
I just learned about the Arrhenius Equation,
$$k(T)=A\exp{\left(\frac{-E_\mathrm a}{RT}\right)}$$
and noticed that it appears to be in the same form as the solution to a linear homogeneous differential equation, so I attempted to find the associated differential equation.
Since $E_\mathrm a$ and $R$ are both constant parameters, you can treat it as a single constant. Defining $$\beta = -E_\mathrm a/R$$
we have $$\ln(k(T)) = \ln[\exp(\beta/T + \ln A)]$$
Take the derivative of each side and solve for $\mathrm dk/\mathrm dT$ ($A$ is treated as a constant of integration and goes away), and we get:
$$\frac{\mathrm dk}{\mathrm dT} = -\frac{\beta k}{T^2}$$
Here is the slope field for the differential equation when $β=-1$

and here is the slope field for $β=1$
However, since $k$ is a rate constant, I'm not sure if this method is actually applicable. Is this, or a similar differential equation actually used on this topic?

The last equation looks suspiciously similar to Van't Hoff equation.

The limitations for your plots are that $E_\mathrm a,; T$ and k are always positive. The expression 'rate constant' is misleading as this clearly varies with temperature; 'rate coefficient' would be better but is not commonly used.

To answer the question of the title. In Adam-Gibbs theory the activation enegy is defined as [1]:
$$E_\mathrm a=-R\left[ \frac{\partial \ln k}{\partial (1/T)} \right]$$
Now clearly this is a first order linear differential equation, as you were looking for. This can easily be solved by seperation of variable:
$$\int E_\mathrm a\, \mathrm d(1/T) = -R\int \mathrm d\ln k $$
$$\frac{E_\mathrm a}{T}=-R\ln k+A'$$
$$k(T)=A\exp\left( \frac{-E_\mathrm a}{RT} \right)$$
Here $A=\exp(-A'/R)$, now it can be seen that the Arrhenius equation has been regained.
[1] S. Petrucci; Ionic Interactions: From Dilute Solution to Fused Salts; equation 125


text:
Are the canonical orbitals of Hartree-Fock also the natural orbitals?
My question stems from the comments on my answer to After a unitary transformation, is Koopmans' theorem still valid?. There was some confusion relating to differing terminology referring to different, mostly unrelated density matrices. As of now, I'm uncertain as to what density matrix one diagonalizes to obtain the natural orbitals. Is it the density matrix that appears in the HF calculation or some different density matrix?

There is a bit of a terminology problem in the field that makes things very confusing and I will try to clarify some of this here. Part of the problem arises from the fact that sometimes only one kind of density matrix is relevant and so people drop the adjectives when talking about specific approximations.
The Density Matrix
We're talking electronic structure here (density matrices are used in other areas of quantum and physical chemistry. Honestly, it's the same density matrix but people use different language and representations and normalizations to discuss them and everyone gets confused when non-EST and EST people talk to each other about density matrices) so I will restrict myself to that topic.
The ($n$-particle) density matrix is given by:
$$\rho=|\Psi(x_1,x_2,\cdots x_n)\rangle\langle\Psi(x_1,x_2,\cdots x_n)|$$
where $x_i$ is the coordinates (space and spin) of the $i$th electron. That is, it gives the probability that the system is in a state with a given set of coordinates (space and spin) for all electrons in the system. The density matrix has many great properties, including being Hermitian and having the property that $\text{tr}{\rho A}=\langle A\rangle$ for any operator $A$. In other words, the $n$-particle density specifies everything about the wavefunction of your system. If you have the full wavefunction, you have the full $n$-particle density and vice versa.
Now, most operators are not $n$-electron operators. For example, the kinetic energy operator is a 1-electron operator (only the coordinates of a single electron are needed to calculate a kinetic energy), the standard electronic Hamiltonian is a 2-electron operator (because the Coulomb potential is only dependent on the positions of two electrons at a time). Hence, we don't need the full $n$-particle density matrix to get energies and properties.
The 1(or 2 or 3...)-particle (reduced) density matrix
Since we don't need all of those other coordinates we can integrate them out. Indeed, the trace operation that we described above would do just that. Say we want the expectation value of a 1-particle operator (like kinetic energy). We can define a 1-particle reduced density matrix (sometimes the word "particle" or "reduced" is omitted), often abbreviated as 1PDM, 1RDM, $P^{(1)}$, or $\gamma^{(1)}$ (also others).
$$P(x_1)=N\int \rho(x_1,x_2,\cdots x_n)dx_2\cdots dx_n$$
where $N$ is some normalization (commonly the number of electrons in the system ($n$) because, typically, people want the trace of the 1PDM to be the number of electrons in the system).
This is the matrix which is diagonalized to form the natural orbitals.
Relatedly, we can also define a 2-particle reduced density matrix (aka 2PDM, 2RDM, $\Gamma$, $P^{(2)}$, others) as
$$\Gamma(x_1,x_2)=N\int \rho(x_1,x_2,\cdots x_n)dx_3\cdots dx_n$$
where $N$ is a different normalization (usually, $n(n-1)$). Most operators that we care about (like the energy) are 2-particle operators so the problem of electronic structure is determining the 1 and 2PDMs.
The Hartree-Fock density
There are many ways to think about the Hartree-Fock approximation, but one way that is related to density matrices is that the HF approximation is that the 2-particle, 3-particle, etc. density matrices can be expressed as direct products of the 1PDM. Intuitively, since HF ignores electron correlation, it makes sense that the probability of finding electron 1 at position $x_1$ and electron 2 at position $x_2$ is related to just the product of the those individual probabilities (with appropriate fermionic antisymmetrization). In this way, the HF 1PDM specifies all density matrices in the HF approximation (which is why in HF you can compute $J$ and $K$ which are two-electron operators by contracting twice with the HF density).
$$P(x,x') = \sum_i^\text{occ} \phi_i(x)\phi_i^*(x')$$
where $\phi_i$ are the spin-orbitals. Note that this means for HF, the canonical orbitals are the natural orbitals. For a correlated method though, the 1PDM will generally be different. For example, CI methods like CAS express the 1PDM as the weighted sum of several HF-like 1PDMs, with coefficients determined by the CAS-CI optimization. Also, note that for Kohn-Sham DFT, the noninteracting reference system also looks like a system without electron correlation and so the 1PDM specifies terms that look like higher order density matrices. However, one does not obtain the 2PDM with KS DFT methods. One obtains the energy from the 1-particle density and an exchange-correlation functional although $i$-particle operators value can be extracted from this non-interacting $i$PDM.

That was a really long answer, but I sense there is a lot of confusion about these different density matrices, so I figured I would take the opportunity to give a very clear answer. In short, the answer to your question is that the HF (1-particle) density specifies all of the HF $i$-particle densities. Generally though, the $i$-particle densities are determined somewhat separately, as each are impacted by electron correlation differently. The orbitals that diagonalize the 1-particle density give the natural orbitals.

Thanks for expanding this into an answer! I think you may have left a thread hanging in the 2nd paragraph of "The Density Matrix," btw.


text:
How fast does the diamond lose the atoms at sunlight?
recently I've read that the diamonds lose their atoms at sunlight on the surface. Would you be able to explain me how fast is the rate of this process? Does the diamond lose the atoms each time when is exposured at sunlgiht? Thank you. :)

https://chemistry.stackexchange.com/questions/76709/will-a-solid-object-lose-or-gain-atoms-on-standing-over-time-without-being-acte

Thats true, but even an UV lamp would take 10 bilion years to remove a microgram (one-million part of a gram). It is caused by carbon atoms oxidizing, forming CO2. This process is accelerated by the presence of UV light.

Does that mean that Graphene will disappear in UV light too? :)


text:
Why wouldn't you consider the [H+] produce from the weaker acid in a mixture of two acid?

Calculate the pH of a solution that contains $\pu{2.0 M}$ $\ce{HF}$ and $\pu{2.0 M}$ $\ce{C6H5OH}$. Also calculate the concentration of $\ce{C6H5O-}$ in this solution at equilibrium.

In these situation, you are told to first find which is the strongest acid. $\ce{HF}$ is.
Knowing this, you can use the ICE table and the $K_a$ of $\ce{HF}$ ($7.2 \times 10^{-4}$) to find $[\ce{H+}]$ in the dissociation of $\ce{HF}$. This $[\ce{H+}]$ is then use to find the $\mathrm{pH}$ of the mixture, but my question is why wouldn't we add the $[\ce{H+}]$ produced from the dissociation of $\ce{C6H5OH}$? Also the $[\ce{H+}]$ that could be made from $\ce{H2O}$ reacting with itself? Is it because the $[\ce{H+}]$ from $\ce{C6H5OH}$ (weaker acid) and $\ce{H2O}$ is so low that it won't really affect the $\ce{pH}$ number?
This question pours over to when you calculate $[\ce{C6H5O-}]$. Why would we just substitute the $[\ce{H+}]$ we found from $\ce{HF}$ for the $[\ce{H+}]$ in the $K_a$ formula of the dissociation of $\ce{C6H5OH}$:
$$K_a = \frac{[\ce{H+}][\ce{C6H5O-}]}{2.0 - x}?$$
Is it because at equilibrium of the mixture we know $[\ce{H+}]$ must be close to the $[\ce{H+}]$ that would be produce at the equilibrium of a $\ce{HF}$ reaction? So it will somehow affect which way the $\ce{C6H5OH}$ reaction will go (Le Chatelier's principle). So because of the addition of HF $[\ce{H+}]$, the reaction will go left and less $\ce{C6H5O-}$ should be made compare if it was just a solution of $\ce{C6H5OH}$?

You don't have to consider the system's sub-components separately, but it makes the algebra much easier.
If pKa of $\ce{HF}$ is 3.14 and pKa of $\ce{PhOH}$ is 9.95 and the problem gives you concentrations to ONE SIGNIFICANT figure, it isn't clear to me why you think the difference between .00072 and .00072000011 is worth worrying about...It can be ignored, clearly. It's not a question of the weaker acid, its a question of whether the values are close enough so that the smaller is significant with regard to the larger. In this case, clearly not. $\ce{PhOH -> PhO- + H+}$ is the chemical equation for the 2nd part, so
$$
\ce{\frac{[PhO][H]}{[PhOH]}} = K_a
$$
is the equation you derive from it (ignoring difference between concentration and activity, and ignoring water).
The fact that $K_a$ for this is $10^{-10}$ and that $\ce{[H+]}$  is on the order of 0.0007 means you get $\frac{X(0.0007)}{(2-X)}=10^{-10}$ which if you have any done a minimal amount of algebra practice/drill immediately suggests that X is going to be on the order of 3E-7 and again is there really a difference between 2 and 2-0.0000003? You'll note I'm not asking the slightly different question about whether the difference between 2.0000000 and 2.0000000-0.0000003 = 1.9999997; 2 is not the same as 2.0000000 when significant figures are in play.

Alphonse, please format your answers so they are more readable.

For example, see the edit made here: https://chemistry.stackexchange.com/a/76519/402

You've practically answered it yourself... The $\ce{H^+}$ released from $\ce{HF}$ will be so much that it will cause an effect called the common ion effect, which is basically Le Chatelier's rule applied to ions in equilibrium. This effect will cause a reduction in the total amount of $\ce{H^+}$ that dissociates out of the weaker acid, $\ce{C6H5OH}$, due to which the amount of $\ce{H^+}$ that dissociates out of $\ce{C6H5OH}$ will be negligibly small compared to that produced from $\ce{HF}$. The same thing happens to $\ce{H2O}$. Thus, in the pH calculation, we only substitute the concentration of $\ce{HF}$, so that we can get the answer with least trouble. In reality, even the smallest quantity that dissociates out should be considered, but to make our calculations simple, we use the most significant amount, leaving out the others.
In the equilibrium equation, the same rule applies, giving us an $\ce{[H^+]}$ value same as that of $\ce{[HF]}$. Also, the $x$ in the denominator is negligibly small, so even that is neglected in the calculations.


text:
How do delocalized π electrons satisfactorily fill π orbitals?

The above example is benzene.
Only six electrons contribute to the 6 π  bonds and pπ  orbitals. Normally, an orbital requires two electrons to satisfactorily "fill" it. I don't understand how this rule is not abided yet bezene is still considered a stable molecule.
Am I to understand that a single electron can be shared by 6 orbitals simultaneously? Do all 6 π orbitals completely spread out amongst each other to each this?

Well, it seems you still don't get resonance and try to get to MO theory... 6 electrons are still in three orbitals which are spread out on molecule.

@Mithoron Are you saying that each pair of π orbitals form a single orbital? Do you have any good resources to understand resonance?

That's for MO ^ and about resonance - https://chemistry.stackexchange.com/questions/31352/is-the-resonance-depiction-of-benzene-typically-portrayed-incorrectly


text:
How to delete everything but a specific ligand from a PDB file using Linux-compatible free software?
I would like to delete everything from a PDB file except a specific ligand that I know the PDB code for. Is there a simple way to using free Linux-compatible software? If you want a specific example for your answer 1YAT and its ligand FK5 (tacrolimus) is an example I'm presently working on. I think PyMOL might have this capability using the delete command but running:
delete /1YAT/A

which should delete the macromolecule (chain A) at least but I saw no visually-obvious change. I have viewed the PyMOL Wiki article on the delete command but it was not helpful as what name meant was not obvious to me.
By free software I mean free of charge not free as in freedom. One piece of software that I used to do this with on Windows was Accelrys (or BIOVIA) Discovery Studio Visualizer, tried to run it on Arch Linux, however, as it has a Linux binary, but it failed to run. This is presumably because it is for Red Hat Enterprise Linux/SUSE Linux Enterprise which has too old of system libraries for Arch Linux compatibility.

The PDB is human readable, so the quickest way is just to use a text editor and delete all the bits you don't want; its messy but if you have only to do it once then probably ok. Alternatively if you have many files use python (with, say, Atom(apple mac) or notepad++(PC)  code editors) to read line by line and print same for only the header parts (those not starting with word 'ATOM' ) and then  print the 'HETATM' parts you want.  Python,  Atom and Notepad++ are all free and there is lots of online help to use python.

Sed and/or vim.

I'm having trouble envisioning why you want to get the conformation of the ligand using this method.  There is a plethora of open-source, linux-compatible chemoinformatic software to do this.  Here's an example using rdkit: http://nbviewer.jupyter.org/github/tentrillion/ipython_notebooks/blob/master/visualize_tacrolimus_conformation.ipynb .  (Note that you should be able to pan/zoom with the mouse on "Out[6]" to look at the conformation in detail).  There are similar capabilities, I'm sure, in cdk and other software.

I usually use Avogadro, Jmol, Mercury and PyMOL only and I was unaware of ways to remove everything but the ligand with any of these pieces of software. Guessing that was what you were asking since your rdkit example was along the line of suggesting another method of doing this. If you're asking why this method of determining the 3D structure as opposed to optimization and other methods well I want to draw structures for Wikipedia (e.g. here is one of mine https://commons.wikimedia.org/wiki/File:Tianeptine-xtal-2012.png) and crystal structures are preferred there.

Avogadro

Open the PDB file. Then choose Select → Select by Residue:

Pick Select → Invert Selection followed by Edit → Clear.

Done. Well, save the new file… :-)

I'll note that there must be some way to do this with PyMol, but it's pretty easy to do in Avogadro.

Is it possible to do this with Avogadro2? I don't see a "Select by Residue" option in is Select menu, only All, None and Invert Selection.

@BH2017 - Avogadro2 at the moment is in beta and has no residue/biomolecule support yet.

grep
Substitute AAA with the name of the residue you want:
grep -w " AAA " file.pdb > ligand.pdb

While you don't get visual confirmation of the ligand match, I think this is a nice hack. :-)

Well, usually before redirecting the output to a file you try few things. grep -w is not immediate, one learns it step by step. First you grep only and then you refine the command, maybe even piping to sed or head or whatever. Or you redirect using tee.
Or maybe I misunderstood and you meant "visual" as in "actually visualizing the structure"...but I don't think it's really necessary!

I meant "actually visualizing the structure" but agree that all the points you mention are important in building up such hacks.

PyMOL

Create a new object for your residue:
create myobj, r. fk5

Delete the original object with the entire protein:
delete 1yat

I have written some simple code in python to do what you ask, keeping just FK5.
# remove rows from pdb file

import numpy as np

dataname = '1yat.pdb'
keyword = 'FK5' # keep row with this in col 4

isatom = False
with open(dataname) as f:
    for line in f:
        vals = new_str.split(' ')

        if vals[3]==keyword:
            print(line, end= ' ')

For me this script returns the error (and yes 1yat.pdb does exist in the folder where I am running the script) shown here https://gist.github.com/fusion809/11278bea129cee1adfa495e0767c8439. The script is also shown at that gist.

Does indenting the whole thing by four spaces, to enable code formatting, not make it render correctly?

I used tabs but I did use it for each separate level. So, for example, for the if vals[0] line I used two tabs as it is level 2. While for line in f: is level 1 so one tab.

Oh, sorry, @BH2017, I wasn't referring to the error you observed; I was asking porphyrin about his troubles getting the code to render correctly in his post here.

@BH2017 It looks like you're running this as a shells script instead of a python script. To fix this, add the following line as the first line of the script: #!/usr/bin/python3

Adding the shebang did get the script to work. But the script is not what I wanted. The title of this question is how to delete everything but the ligand. Not how to delete just the ligand. See I want a PDB file just for the ligand. This script deletes just the ligand.

sorry, yes I was running inside 'atom' . To get just the lines you want junk everything after vals = new_str.split(' ') and add as next line if vals[3]==keyword : print(line, end= ' ')

@hBy2Py thanks for tip on formatting code, I was unaware of it.

Jmol/JSmol

Load your file:
load "=1YAT"

Select the residue:
select FK5

Write all the selected coordinates. The file name is arbitrary except for the extension .pdb:
write ligandFK5.pdb


text:
Vinegar reacting with metal pan?
I was following a recipe for a seasoning containing rice vinegar. The directions noted that a metal saucepan should NOT be to used to heat the seasoning because it would react with the vinegar.
What would happen, chemically, if I used a metal (stainless steel) pan?
My thought is that the acetic acid and Iron could react to form Iron(III) acetate. Would this complex be in a concentration high enough to warrant the, "do not put in a metal pan," warning or would the reaction be something else?

Vinegar was used as one of the main household cleaning and odor removal agents for a good reason. I think in general you can get numerous iron (III) and chromium (III) (which are less pleasant) salts and complexes aside from acetate, e.g. chloride/iodide if you use iodised salt or hypochlorite from chlorinated water.
Diluted acetic acid (vinegar) etches majority of steels quicker than the glacial one, dissolving its passivating $\ce{Cr2O3}$ layer. Aside from concentration, reactivity also greatly depends on the steel type and acetic acid purity [1]:

Carbon and Alloy Steels. Steel is attacked quite rapidly by all concentrations of acetic acid, even at room temperature. Glacial $\ce{CH3COOH}$ at room temperature is less aggressive than aqueous solutions of the acid, but still gives a rate of attack of $\pu{0.8 to 1.3 mm/yr}$ ... Therefore, steel is normally unacceptable for use in $\ce{CH3COOH}$ service.
Stainless Steels. The chromium stainless steels of the 400 series occasionally exhibit low corrosion rates in laboratory tests in dilute acetic
acid. However, because field experience with these materials indicates high corrosion rates and pitting attack, they are rarely used for $\ce{CH3COOH}$ production equipment.
Impurities present in the manufacture of acetic acid, such as acetaldehyde, formic acid, chlorides, and propionic acid, are expected to increase the attack of stainless steels. Contamination with chloride can cause pitting,rapid stress-corrosion cracking, and accelerated corrosion of type 316 stainless steel. Up to $\pu{20 ppm}$ of chloride can be tolerated, but higher concentrations are likely to cause rapid equipment failure.
Transferring heat through a metal wall, as in heat exchangers, can drastically alter the corrosion characteristics of the metal.

If you know the exact steel type used, you can check its stability against various acidic media in several corrosion handbooks, e.g. [1] contains cumulative tables for corrosion behavior of manufactured steels tested under various conditions.

Craig, B. D.; Anderson, D. S. Handbook of Corrosion Data; ASM International, 1994. ISBN 978-0-87170-518-1.


text:
Mixture of water and acid 175 litre, in that 88 % acid, how many litres of water you should add to do 23% of water
Mixture of water and acid 175 litre, in that 88 % acid, how many litres of water you should add to do 23% of water


text:
Ring opening of epoxides

For the above question, here's what I'd thought:
1.  H+ ion from HCl protonates the oxygen
2. Chloride ion attacks the electrophilic carbon, opening the epoxide ring
3. So the final product formed is a chlorohydrin (which it really isn't)
Also, I didn't know what to do with methanol (guessing it just acts as the solvent, HOWEVER, the product contains a methoxy group)
Could someone please explain the mechanism to get to the actual product, and also highlight the mistake here?

Methanol attacks, not chloride. It's present in large excess (as it's the solvent)

Is it that major product is formed due to attack by methanol and minor is formed as a result of attack by chloride?

If you think about the quantities involved in this reaction, you'll realize why does the product contain the methoxy group and not the halide. Methanol is present in large excess since it is the solvent of this reaction. Cl- ions will also be present and a very small amount of the chlorohydrin may be formed as well. The reason why methanol is a reagent and not only act as a solvent in this reaction is because its oxygen atom acts as a nucleophile, just like halides or any other molecule containing electronegative atoms (such as O, N, S etc).
I found the following mechanism on Google and it is draw as if the reaction was being performed in water, however you can replace the blue water molecule for methanol or any other alcohol since the mechanism is the same for both.

Whenever you look at a reaction schemes like the one you've posted, you should always address the reactivity of solvents and other reaction components, sometimes they end up reacting with your substrate rather than your desired reagent, this is a very good example of it. So if you intend to synthesize a chlorohydrin, you cannot use a nucleophilic solvent, otherwise, it will react with the epoxide and form the correspondent adduct.
I hope this answers your question.


text:
Reaction of aluminium trichloride with water
Since $\ce{Al^3+}$ is an acidic cation, shouldn't it react with water like this?
\begin{align}
  \ce{AlCl3 + 6H2O &<=> [Al(H2O)6]^3+ + 3Cl^-} \tag1\\
  \ce{[Al(H2O)6]^3+ +H2O &<=> [Al(H2O)5(OH)]^2+ + H3O^+} \tag2\\
  \ce{[Al(H2O)5(OH)]^2+ + H2O &<=> [Al(H2O)4(OH)2]^+ + H3O^+} \tag3\\
  \ce{[Al(H2O)4(OH)2]^+ +H2O &<=> [Al(H2O)3(OH)3](s) + H3O^+} \tag4\\
\end{align}
The 3rd step of the equilibrium rarely happens in water. But if we add a base, strong or not ($\ce{NaOH}$, or $\ce{Na2S}$) the equilibrium weighs to the right and $\ce{Al(OH)3}$ is formed.
But I've seen people write it as this.
$$ \ce{AlCl3 + 3H2O -> Al(OH)3 + 3HCl}$$
Is this correct? I need a certain answer.

It's not incorrect, but certainly as it stands there, it is missing a few key details. The aluminium water complexes are very sensitive to pH change, it really depends on all conditions and on the context whether this can be seen as correct or not.

In 25 degrees celcius when the pH in water is 7, if  we add anhydrous AlCl3 into the water. Can we see any precipitation? And if we heat the system all equlibrium constants move right? And then will Al(OH)3 be formed? We have very limited lab recources here. I have no way of knowing it practically. Thats why im asking for your help. Thanks so much for everyone who contributed thus far. More help is appreciated.

I doubt you'll see precipitation. Aluminium is a very delicate system and you need to get the pH just right to get it. Also keep in mind that the  equilibria are necessarily linked and happening at the same time. If I have the time I'll add some authoritative resource. Just give it a little time, it's a good question and I'm positive you'll find a good answer.

Aluminum chloride is very soluble in water, according to wikipedia 44-49g/100ml water depending on the temperature, so you shouldn't expect any preciptation. The reaction is very aggresive and exothermic.
Assuming you dont have any buffers or bases in solution when you add $AlCl_3$, the reaction will lower the pH. At a pH lower than 4-ish, $Al^{3+}$ is the most stable cation. Thus, the reaction
$$\ce{AlCl3 + 3H2O -> Al(OH)3 + 3HCl}$$
is pretty accurate. Aluminum hydroxides are decreasingly stable at low pH, as you can see below. Since HCl is formed upon dissolution of $AlCl_3$ you can safely assume that $Al^{3+}$ is the only Al-species present.
!1

You didn't read what OP wrote? such "reaction" is terrible.

Wow. All you guys are awesome. Thanks for devoting your precious time towards me. Its apperciated.  Mithoron. Would you please elaborate? Youre saying writing that last reaction is wrong?

The displayed reaction equation is just plain wrong. And please cite the source of the graph you use.

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


text:
Heat capacity of mixtures is higher than their pure contituents?
I'm currently studying the effects of how changing mixed-refrigerant compositions effects their heat capacity. Surprisingly, my data suggests that some combinations of mixed-refrigerants have heat capacities larger than their pure constituents. This seems to violate Kopp's Law which states:

The molecular heat capacity of a solid compound is the sum of the atomic heat capacities of the elements composing it; the elements having atomic heat capacities lower than those required by the Dulong–Petit law retain these lower values in their compounds.

The mixed refrigerant compositions include: methane, ethane, propane and nitrogen. The graph below shows average, minumum and maximum heat capacity values over the temperature range from $\pu{130 K}$ to $\pu{280 K}$ and pressure range from $\pu{1 bar}$ to $\pu{100 bar}$.
Is there another possible explanation as for why a combination of the above components would have a higher capacity than their pure constituents? I was thinking along the lines of molecular interactions and deviations from the ideal gas law.
Here is a graph of my findings:

Note: I am using a gas physical property software for these calculations.
link: https://www.dnvgl.com/services/gasvle-8331

This is probably not so relevant here, but ideally you might want to provide a reference or a link to the github repo for the software you are using.

Are these mixtures of liquids or of vapors?

Liquid, vapour and liquid/vapour over the specified temperature and pressure range. Boiling points differ between each component quite drastically.

So you are measuring "apparent" heat capacity values over phase changes, correct?

Yes, this is totally theoretical. Though, I am assuming the software I am using is fairly accurate for these predictions. I am calcualting the heat capacity at each temperature over the full range of pressures (1bar-100bar in 1 bar steps).

Well, this includes the latent heats of vaporization, so you shouldn't be too surprised.

Could you expand upon this? Are you saying that the molecular interactions between the individual components are increasing the latent heat of the mixture (compared with the individual components? ).

Can you please give a specific example.

If you look at the average heat capacity of each componenet C1,C2,C3 and N2 compared with the average heat capacity of MR1, MR1 is a lot higher, despite being a mixture of the said individual components. Is this purely down to molecular interaction causing the latent heat to rise as the molecules are more strongly interacting?

By the way, I am not measuring the latent heat of vaporisation. Heat capacity and latent heat are two separate quantities. One being the energy absorbed as a function of temperature and the other the amount of energy required to induce a phase change.


text:
What is the IUPAC name of borax?
Please don't say sodium tetraborate decahydrate since that is not the real composition of borax.
Borax contains tetranuclear coordination complexes $\ce{[B4O5(OH)4]^-^2}$. I'm asking about the name of this complex in borax.

Sodium tetraborate decahydrate is a legitimate IUPAC name. If you want to impress a girl, use sodium bicyclo[3.3.1]tetraboroxane-3,7-diolate hydrate (2:1:10), or pick further German/French names from here.

This might help you-https://chemistry.stackexchange.com/questions/59273/structure-of-borax

The name suggested here (via the question comments), the anion-corresponding part, ‘bicyclo[3.3.1]tetraboroxane-3,7-diolate’ ($\ce{[B4O5(O)2]^2-}$) obviously does not match with the tetraborate formula $\ce{[B4O5(OH)4]^2-}$ present in borax, whose structure, confirmed by X-ray crystallography, including hydrogens[1], is

However, the ‘bicyclo[3.3.1]tetraboroxane’ part is correct, for the parent hydride:

This type of names is known e.g. for the ‘cyclotriboroxane’ name for boroxine, or ‘cyclotetrasiloxane’ (in octamethylcyclotetrasiloxane).
So the name would be rather 1,3,5,7-tetrahydroxybicyclo[3.3.1]tetraboroxane, or bicyclo[3.3.1]tetraboroxane-1,3,5,7-tetrol, however it still does not explicitly tell that it's an anion.
Interestingly, this boroxane, siloxane type names (interchanging element/oxygen pattern) is rather marginally mentioned in the inorganic nomenclature, notably more detailed in the organic nomenclature.[2]
Inorganic additive nomenclature is able to constructs names for similar skeletons,[3] but somewhat differently than organic nomenclature.
E.g. bicyclo[x.y.z]Nane (where N is the numerical prefix root, with the exception for 1..4, ‘meth’..‘but’, for the organic names) would be named: Ncarby-[0a.z1,b]dicycle, where a = x+y+2 and b = x+2. If the structure is an ion, it's ‘…cyclate’ followed by the charge in parentheses. One advantage is that the first numeral is the size of the (largest) cycle, not just a branch.
Some examples from the book:

1,7-diazyundecasulfy-[012.11,7]dicycle
(IR-1.5.3.4 Example 6)

cyclo-triphosporic acid
2,4,6-trihydroxido-2,4,6-trioxido-1,3,5-trioxy-2,4,6-triphosphy-[6]cycle
(IR-8.3 Example 5)

tetraaluminy-1-lithy-[05.01,301,402,5]tetracyclate(1−)
(IR-7.4.3 Example 11)

(The leading zero in the first numeral in the square brackets is intentional for the cycles, however in the latter example, it's not used, despite IR-7.4.2 Nodal descriptor, example 3, that states that the nodal descriptor for hexacycle is ‘06’.)
With this nomenclature we can get very descriptive (but almost unusable) names
1,3,5,7-tetrahydroxido-1,3,5,7-tetrabory-2,4,6,8,9-pentaoxy-[08.11,5]dicyclate(2−),
1,3,5,7-tetrahydroxidotetrabory-2,4,6,8,9-pentaoxy-[08.11,5]dicyclate(2−), or
1,3,5,7-tetrahydroxido-1,3,5,7-tetraborypentaoxy-[08.11,5]dicyclate(2−).
References:

Levy H.A., Lisensky G.C. Crystal structures of sodium sulfate decahydrate (Glauber's salt) and sodium tetraborate decahydrate (borax). Redetermination by neutron diffraction Acta Crystallographica Section B 34 (1978): p. 3507
IUPAC Blue Book 2013, P-52.1.6.2
IUPAC Red Book 2005, IR-7.4 Additive nomenclature: Inorganic chains and rings

snort dicyclate! :D Yep, that almost unusable, indeed...


text:
Aluminum to prevent Iron corrosion
I have heard of how zinc is used to prevent corrosion of iron. So why not use aluminium coat instead? After all Aluminum oxide forms a stable coat over the surface.

Both are used, even together (Zincalume).

Oh I didn't know. Everybody I asked this question to in the past thought I was crazy

Aluminum, just like zinc, is already widely used to coat iron surfaces to prevent corrosion (in the elemental form, or in combination with other metals, like zinc). You can deposit aluminum on practically any surface (including plastics). On metals it is done by electrochemical deposition.
There are several companies working with this type of coating nowadays, including AlumiPlate (http://www.alumiplate.com/).

Not all aluminum on metals is deposited electrolytically.  Hot dip aluminizing is a thing in steel strip manufacturing.  Also aluminum is pasivating rather than cathodicat ambient temperatures but becomes truly protective at high temperature.

No one mentioned aluminum anodes; So many tons of anodes are used on an oil platform that the weight must be considered in the design. Some Al is dipped, some is sprayed , some is vapor deposited: some  coatings are diffused into the steel at high temperature.


text:
Valence Bond theory, VSEPR theory and predicting the shapes of the molecules
I read about the above mentioned theories then used them to predict the molecular geometry. However, I found that the shape/geometry of a molecule obtained from both the theories was same. For example, water has bent/angular shape according to VSEPR theory as well as Valence Bond Theory. So I wonder if there's any difference between the two theories in predicting molecular geometry.
Do these two theories ever yield different results?

I'm not the downvoter. But, you were probably downvoted because this is not a very good question. Why would you expect the two theories to differ? What research have you done on your own to try to solve your own question?

Yes they do yield different results. What do you predict the shape of $\ce{PH3}$ to be?

@bon as they do yield different results, on what occasions we employ each separately?

VSEPR theory is a purely observational model, based on empirical knowledge. Many of the "rules" were added to fit the theory to what has been observed. It has quite a lot limitations and may only provide an initial guess for the structure of a molecule.
According to VSEPR, the structure of water is bent. While we would expect $\angle^\mathrm{theo}_\mathrm{bond}(\ce{HOH})=109.5^\circ$, it is indeed smaller $\angle^\mathrm{expt}_\mathrm{bond}(\ce{HOH})=104.5^\circ$. This discrepancy is explained with the reasoning that lone pairs require more space. This is wrong, and a caveat added to make the theory fit.
It would also predict the lone pairs of water to be equivalent. A statement which has experimentally been disproved. Quoting myself:

One popular example is water. In many texts the central oxygen is described as having (approximately) four sp3 orbitals, which essentially makes the lone pairs equivalent. That view, however, is not in agreement with the photoelectron spectrum, which clearly shows that the lone pairs are not equivalent. At this point I'd like to refer you to Michael Laing's article: "No rabbit ears on water. The structure of the water molecule: What should we tell the students?" (J. Chem. Educ. 1987, 64 (2), 124.).

The theory sometimes predicts correct structures, but that has no theoretical foundation. It is basically educated guessing.
When it comes to more complicated molecules, for example hyper-coordinated compounds like $\ce{IF7}$, most explanations (the involvement of d orbitals) break completely down and have been disproved many times. However, just by minimising the inter-ligand distances, it yield approximately the correct shape. This is a nice coincidence.
Valence Bond theory is on the other hand a highly sophisticated and complex approach, deeply rooted in theory. If applied at a modern, computational level, it will always yield the correct molecular geometry for the correct reasons (within its other approximations).
While we can approximately use a piece of paper and a pencil to do these calculations for simple molecules, it really is an approach at formulating a wave function and approximately solving the electronic Schrödinger equation. That requires a lot of maths and integral solving. It is not so much a theory that can be done on-the-go.
As it is such an extensive and accurate theory, it is complementary to molecular orbital theory. That means that they will yield the same results at their respective accurate (infinite) expansions.
For the example of water, VB theory would predict a bent structure. It would also predict non-equivalent lone-pairs, which agrees nicely with the photoelectron spectrum.
In conclusion; A sad fact is that both theories, VSEPR and VB, are usually (more than just often) taught completely wrong at below university levels (sometimes even there) and will introduce many artefacts into young scientists brains. Your statement, that you "used" valence bond theory is a reminder of that.
These theories could not be more different from each other. In most cases they predict structure on completely different grounds. It is basically a coincidence if they predict the same thing.
Take home message: VSEPR theory is a rule of thumb, VB theory is an accurate description of physics (within its approximations).

@Karsten I'm not sure I can agree with this edit. I really personally do not like to include these kind of images if I have not verified them myself. As they currently stand, without the necessary details of how they came about, I find them distracting. The left image is definitely incomplete, the right image is just trickery. In any case: images like these must be cited accordingly; a link hidden behind the word source is the worst possible way of attribution.

Not a problem, I will revert it. I liked the web site that had the interactive Jmol scenes for different molecules, from this collection: https://chem4.cns.uaf.edu/orbitals/index.htm


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Why is a racemic mixture formed in the Diels-Alder cycloaddition?
Why is a racemic mixture formed in the Diels-Alder cycloaddition? I note that for both endo and exo products, either enantiomer can be formed. Why is this so?

In the absence of any other external effects, why would we expect anything but a racemic mixture? You have totally achiral starting materials.

Hi Zhe, I'm not exactly sure how the mechanism of this reaction would directly lead to a racemic mixture.

Both enantiomers are formed because all intermediate states including the end products are energetically identical.

If you take the diene as a reference plane, the dienophile can "attack" from "above" or "below".

This is a great time for me to point out that I didn't need to even look at what the reaction was or its mechanism. >99.999% of the time, if you start with achiral reactants, you end up with either a racemic mixture or achiral products. In the remaining <0.001% of the time, you might have some amplification like in the Soai reaction, where a small imbalance in the amount of one enantiomer is magnified because it aids in catalysis. This is not one of those cases.

The concerted Diels-Alder reaction has in principle eight different possible transition states. Depending on the substrates some of them will lead to the same products.
Below you can see an overview of the (non-optimised) transition state arrangements for the reaction of isoprene and propene.

From these images you should be able to determine that the following behave like enantiomers:

(a) and (g)
(b) and (h)
(c) and (e)
(d) and (f)

When you think a little bit longer about it, you will also see that (a) and (h) will form the same product. This is also true for (b) and (g), (c) and (f), as well as (d) and (e).
In total, you will therefore obtain four products two different regio-isomers and their corresponding enantiomer each.