text:
Froth floatation process - why CuSO4 is activating floating character of ZnS?
I read the following :

"In a mixture of $\ce{PbS}$, $\ce{ZnS}$ and $\ce{FeS2}$, each component is separated from each other in froth floatation process by using the reagents potassium ethyl xanthate, $\ce{KCN}$, $\ce{NaOH}$, copper sulphate and acid. Here potassium ethyl xanthate acts as a collector. $\ce{KCN}$ and $\ce{NaOH}$ depress the floatation property of $\ce{ZnS}$ and $\ce{FeS2}$ particles. Thus, only $\ce{PbS}$ particles go into the froth. Now copper sulphate is added to the mixture which activates floating character of $\ce{ZnS}$ and this time only $\ce{ZnS}$ comes along with froth. The remaining slurry is acidified and $\ce{FeS2}$ floats along with froth."

Can someone help me in exactly understanding the role of copper sulphate clearly along with its specific property associated?

There is obviously some interesting chemistry in this. I'd guess that the   $\ce{Cu^{2+}}$ ions react with the cyanide to form complexes of some sort, probably of copper (I). Adding acid to a solution of plain cyanide anions would of course release HCN fumes which would be very dangerous otherwise.

Ullmann’s Encyclopedia mentions “…, sphalerite ($\ce{ZnS}$) does not readily become hydrophobic in the presence of xanthates unless copper (II) ions are introduced into the aqueous pulp. In this case $\ce{CuSO4}$, used to supply copper (II) ions, is an activator.” and also includes a plot of this effect, but does not give further chemical explanation.

Can you add a link or reference to the quoted text please?  It would make your very nice question even better.

I navigated the net and came across the following details.....
The most common collectors for sulfide minerals are the sulfhydryl collectors, such as the various xanthates and dithiophosphates. Xanthates are most commonly used. Xanthates are highly selective collectors for sulfide minerals, as they chemically react with the sulfide surfaces and do not have any affinity for the
common non-sulfide gangue minerals.

Sulfhydryl collectors such as xanthate ions compete with $\ce{OH-}$ ions to adsorb on mineral surfaces, and so adsorption is a function of pH. This makes it possible for sulfhydryl collectors to be used to progressively separate specific minerals. The pH where the xanthate ion wins the competition with $\ce{OH-}$
ions depends both on the concentration of xanthate in solution, and on the specific sulfide mineral present.For the same purpose, alkali like $\ce{NaOH}$ may be used and not $\ce{Ca(OH)2}$ because the sodium cation
generally does not have any significant effect on the particle surface chemistries unlike the calcium ion.
Activators are specific compounds that make it possible for collectors to adsorb onto surfaces that they could not normally attach to. A classic example of an activator is copper sulfate as an
activator for sphalerite ($\ce{ZnS}$) flotation with xanthate collectors. When
untreated, xanthate cannot attach to the sphalerite surface because it forms a zinc-xanthate compound that quickly dissolves:
$$\ce{ZnS(s) + Xanthate -> S(s) + ZnXanthate (aq)}$$
The surface of the sphalerite can be activated by reacting it with a metal ion that does not form a soluble xanthate, such as soluble copper from dissolved copper sulfate:
$$\ce{ZnS(s) + CuSO4(aq) -> CuS(s) + ZnSO4(aq)}$$
This forms a thin film of copper sulfide on the sphalerite surface, which allows for stable attachment of the xanthate, rendering the sphalerite particle hydrophobic and floatable. Other metals such as silver and lead can also be used to activate zinc, but copper is cheaper than silver and less toxic than lead.
I read the following in a website which mentions the research as follows :

The role of copper sulphate in the flotation of pyrite has been
  studied. A survey of a number of pyrite flotation plants using copper
  sulphate showed that in correct dosages it was able to increase both
  grades and recoveries. This was confirmed in batch tests. Copper
  sulphate had no significant effect on the rate of adsorption of
  potassium n-butyl xanthate (PNBX) onto pyrite. Copper adsorbed
  irreversibly onto the pyrite when the latter was slurried in a copper
  sulphate solution. Copper sulphate had a significant effect on the
  maximum froth height obtained in static three-phase froth tests.
  Proposals to explain these observation are made.

If the above-mentioned thing is taken into consideration, the role of $\ce{CuSO4}$ is to provide $\ce{Cu}^{2+}$ which will act as an activator in the manner explained above.
Also, the role of cyanide ions may be:
Cyanide ($\ce{CN-}$) is a particularly useful depressant in sufide mineral flotation. Its activity is believed to be due to its ability to complex with, and in some cases dissolve, a number of metal ions, preventing them from attaching to the xanthate molecules. In particular, it is a strong
depressant for pyrite ($\ce{FeS2}$), and can be used to “deactivate” sphalerite that has been activated by copper ions in solution. It is interesting to note that flotation of galena ($\ce{PbS}$) is unaffected by the presence of cyanide.
References

https://www.researchgate.net/file.PostFileLoader.html?id=567e6a16614325ed178b4589&assetKey=AS%3A310856070434816%401451125270248

I'm going to consult the literature later, but for me it seems to be as follows.
In the compounds discussed, there are following cations:

$\ce{Fe^2+}$. It is a transition metal with a moderate affinity to oxygen, a slightly larger affinity to sulfur and a high affinity to cyanide.
$\ce{Zn^2+}$. It is a small cation with a high affinity to oxygen, a slightly larger affinity to sulfur and a moderate affinity to cyanide.
$\ce{Pb^2+}$. It is a large cation with a moderate affinity to oxygen and a slightly larger affinity to sulfur.

Xanthate bind to the surface via the sulfur atoms. So, in a basic cyanide solution the binding sites on the surfaces of pyrite and sphalerite are both completely filled with cyanide and hydroxide respectively (please, note that sulfur in xanthate is a lot less nucleophilic than in sulfide form). Copper has a high affinity to sulfur, a lot larger than zinc or iron. So it binds to xanthate and to the sulfur in either pyrite or sphalerite. However, the surface of pyrite is still occupied with cyanide, which binds really strongly there. And in an acidic environment, cyanide is transformed into $\ce{HCN}$ and thus the pyrite surface is open to binding with xanthate.

Re; "And in acidic environment cyanide is bound into $\ce{HCN}$" - chemically this is true, but I'd doubt that a flotation process would be setup to form HCN. That would seem to be very dangerous to the operators.

@MaxW Given that quicksearch on HCN+flotation returned quite a bit of links, I wouldn't bet on it.

I guess something is fishy as my navigation through the net yielding me with details which contrast your answer. But anyway, I am not sure of it. Thanks for your approach.


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Why o-Nitrophenol is more volatile than p-Nitrophenol?
Why o-Nitrophenol is more steam volatile than p-Nitrophenol?

Related, possibly duplicate: Physical properties of phenols, Intermolecular Hydrogen Bonding, and Strength of types of hydrogen bonds?

o-Nitrophenol forms intramolecular H bond  whereas molecules of p-Nitrophenol get associated through intermolecular H bond. During boiling, the strong intermolecular H bonding increases the boiling point but intramolecular H bonding cannot do so.
Therefore, o-Nitrophenol is more steam volatile than p-Nitrophenol.

I understand that there is intramolecular hydrogen bonding in o-Nitrophenol. But, how does that prevent it from interacting with surrounding o-Nitrophenol? Consider 3 charges in space. Say Charge 1 interacts with Charge 2 (let's name it 1-2 interaction). The '1-2 interaction' is not going to affect the '1-3 interaction or 3-2 interaction'. Does it?

There are two types of hydrogen bond.One is intermolecular hydrogen bond and other is intramolecular hydrogen bond.Intermolecular H-Bond increase the melting point,boiling point of compounds having them while the intramolecular H-Bond have mostly opposite effects to that of intermolecular H-Bond(it decreases melting and boiling point).Since o-nitrophenol forms intramolecular H bond it have low Melting boiling point than p-nitrophenol(which form intermolecular H-Bond).That's why o-Nitrophenol is more steam volatile than p-Nitrophenol.

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text:
What Is The Solution Of This Reaction?
I am confusing in between these two br groups.
Which is the best Leaving group I am not deciding
help me please

And answer is C

welcome to chemistry SE.  Be sure you are not infringing on anyone's copyright? ...  http://meta.chemistry.stackexchange.com/questions/3268/on-the-issue-of-copyright ... also also unclear ...  please read through the tour http://chemistry.stackexchange.com/tour

No I am not Pasting Any copyright material.I respect DMCA ,This image is available on google.If any problem i will remove this picture right now.Thanks For the suggestion

@AkashGoel Just a random fact: That an image is available via Google does not imply, that no copyright law has been violated.

It isn't best leaving group but which is the most reactive centre.
You have a choice of two reactions between the cyanide anion and (i) an alkyl bromide or (ii) a benzyl bromide.  The benzyl bromide is more reactive than the alkyl bromide which leaves you with C or D.
The problem with D is that the bromine atom has an inverted stereochemistry which doesn't make sense.
C has the correct regiochemistry (the position in the molecule) and the correct stereochemistry.  As to the stereochemistry: the reaction appears to be SN2 meaning that the cyanide anion attacks the carbon on the opposite face of the bromine atom of the starting material.

The bromine closer to the aromatic ring system is replaced because it is easier to replace.
The reason is obviously not sterics, but it is the stability of the transition state for the SN2 reaction.
The transition state is stabilised by the pi electron density from the aromatic system, and hence, that is the preferred product.

The answer is C because the nucleophilic attack of the C - Br adjacent to the benzene ring produces a carbanion transition state that can be stabilized by the pi system in the benzene ring. The geometry is obvious as the cyanide ion must come in from the side opposite to the Br.  Try it using a model kit.

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Thermal stability of interhalogen compounds
I am striving over the question that what would be the stability order of the following interhalogen compounds : $\ce{ClF3, BrF3}$ and $\ce{IF3}$.

One probable answer I got is that stability of interhalogen compounds decreases as the size of central atom decreases due to electronegativity difference. Hence, the stability order should be $\ce{IF3} > \ce{BrF3} > \ce{ClF3}$.

Is the above mentioned logic correct? Can someone please elaborate?

Stable with respect to what?

Thermal stability

More electronegativity difference generally gives you stronger bonds,  therefore higher thermal stability (i.e. you need to put in more energy to break them). Also, there are 2 lone pairs on the central atom and 3 on each F. Thus, interhalogen compounds of this type with larger central atom would experience less electron pair repulsion. So, I believe that the correct order is $\ce{IF3>BrF3>ClF3}$ as you suggested.


text:
Electric Attraction Vs Resonance Energy in Electrophilic Aromatic Substitution Reaction

Image Credits: MHHE
In the electrophilic aromatic substitution attack of Benzene by the nucleophile, as shown in the figure, nucleophile attacks hydrogen of the intermediate, instead of attacking the positive charge distributed ring; leading to substitution reaction.
Bezene's closed shell of six pi electrons is said to give it an extra stability (resonance energy)$_1$, but it is not clear and precise in the explanation of how the electric attraction is overcome by this another tendency which is described. So, the question is, how is this electrostatic attraction towards positive ring overcome by other tendency of stability?
$_1$ Organic Chemistry 11e by Solomons, Fryhle, Snyder, pg.671.

The positive charge, instead of being localized, is actually delocalized throughout the ring because of resonance, making the charge shared between three carbons, lessening the ability for the nucleophile to attack the positive charge.

But, it is of high probability for the nucleophile to get attracted towards a less positively charged ring than the neutral hydrogen.

@ImmortalPlayer The problem is that addition breaks the $\pi$-system created by the benzene ring. The $\pi$-system stabilizes the benzene.

@ImmortalPlayer I think I'll leave it to someone who knows more than me.

It seems that, I need to add more description, I will be back soon. Thank you for the time.

I might need to understand QM first, before any trial of thinking into Chem.


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Stability of ortho and para hydrogen
I have two books on inorganic chemistry, one says that ortho hydrogen is more stable while other says para is more stable. Even on Internet, there are websites some supporting the former fact and some the latter. So, which one is more stable?
I think that para hydrogen is more stable because opposite spins cancel each other's magnetic effect, hence leading to lower energy (it's purely a guess, I'm not at all familiar with spin or magnetism) and if my answer is incorrect then what explains the stability of correct spin isomer?

What do you know about nuclear spin statistics? How do they affect the symmetry of the other components of the nuclear wavefunction, most notably rotational?

@orthocresol Nothing. I have not even heard about nuclear spin statistics and so looking for a bit simplified explanation.

@DavePhD I have already read that question but that was a little off the scope, is there any simpler answer? Or explanation of the same thing but in a more intuitive way?

A bit meh on the duplicate. That answer doesn't answer this. @user375072 I don't think there is an intuitive way. The explanation is not about the relative orientation of the nuclear spin and whether they attract/repel each other. It is about the symmetry of the wavefunction upon exchange of identical particles (the two H nuclei) which is constrained by the spin statistics theorem (fermions, like H nuclei, have to have an antisymmetric wavefunction). para-H2 has a lowest rotational state of J = 0 and ortho-H2 has J = 1, which has a higher energy.

@user375072 the oversimplified explanation is that orthohydrogen cannot exist without rotating, while parahydrogen can; therefore, parahydrogen is lower energy (because ortho must have the energy of rotation).

@DavePhD Thank you very much, this explanation is enough to explain me the answer, thanks again:)


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What exactly is basicity?
In my textbook, it’s given that the trend of the basicity of the elements of group 15 is $$\ce{NH3 > PH3 > AsH3 > SbH3 \ge BiH3}$$
Also, it’s said that the reducing character decreases down the group. (Which is the ability to lose $\ce{e-}$)
It’s also give (on a different page) that $\ce{H3PO3}$ and $\ce{H3PO4}$ are di- and tri-basic respectively.
Phosphoric acid (presumably, a Brønsted-Lowry base) has a basicity of 3. This would imply that basicity is a property of acids. However, Ammonia (which is a Lewis base) has the highest basicity. Further, quickly Google search of the definition of basicity states: “Basicity is the number of hydrogen atoms replaceable by a base in a particular acid.”
My question is, how does Ammonia (a Lewis base) have a basicity (i.e. acid character)... shouldn’t it be a weak base? Or is my understanding of the definition of basicity wrong?

Equivocation at its best.

related http://chemistry.stackexchange.com/questions/16804/what-is-the-word-for-how-polyprotic-an-acid-is

The issue is that basicity has a couple of meanings and you have to be able to tell from the context what it means.
The first is logical, and is used to describe how basic a base is. That is the trend that you mentioned about the Group 15 hydrides. (In my experience, this is by far the more common usage of the word.)
The second is somewhat confusing, and it is used to describe how many protons of an acid can be removed by neutralisation with base. For example, $\ce{HCl}$ has one acidic proton, so it is monobasic; $\ce{H2SO4}$ has two acidic protons, so it is dibasic.

So, basicity can mean both the strength of the acid as well as the base? Does the meaning change when talking about either a Lewis acid or Bromsted base?
My bad.. the error is on my part and not my textbook. I'll edit it now.

It is not about the strength of the acid. It is simply a measure of how many protons can come off; it does not say anything about how difficult it is to get those protons off. Acetic acid and HCl are both monobasic, but acetic acid is a weak acid and HCl a strong acid.

Okay. My question came off wrong. Let me rephrase.

So, basicity can mean both how acidic an acid is as well as how basic a base is? Does the meaning change when talking about acids or bases in the Lewis acid-base theory or the Bromstead-Lowry acid base theory?

Basicity meaning number one applies to all kinds of bases, although it is mostly applied to Bronsted bases. The meaning does not change depending on the type of base. Basicity meaning number two only applies to Bronsted acids.

Thank you. Though this hasn't fully made sense to me, it has helped a tonne.


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Why is potassium permanganate coloured?
$\ce{Mn}$ in  $\ce{KMnO4}$ is in $+7$ oxidation state, has no d electrons so  d-d transition is not possible.
Then, how is $\ce{KMnO4}$ coloured?

It is not ionic bond; there is still $\mathrm d$ electrons.

In $\ce{MnO4-}$, there are no electrons in the $e$ or $t_2$ orbitals which arise from the metal 3d orbitals and therefore what are conventionally known as d-d transitions are not possible. Whether it is ionic or covalent has no impact.


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Charge and spin multiplicity change in IRC search
I have successfully isolated the TS for a radical cation catalysed C-C bond formation using Gaussian with input
1 2

for charge and spin multiplicity.
But when I do an IRC search with the correct input for charge and spin multiplicity (1 2 as above), the IRC log files terminate normally but when seen on GaussView, it says the molecule is neutral with singlet spin.
Is this a problem with GaussView? Will it affect the interpretation of the IRC in any way? In other words, does the IRC search still occur on the doublet PES or has it (somehow) gone to the singlet PES?

In other words, does the IRC search still occur on the doublet PES or has it (somehow) gone to the singlet PES?

You could check that by looking at the output. Right at the end of each and every SCF optimization Gaussian prints a block that starts with SCF Done and contains information about energy of the state to which SCF converged and its total spin. For instance, below is one such block in the results of test170 calculations from Gaussian 09 test set:
 SCF Done:  E(UHF) =  -40.1568456632     A.U. after   10 cycles
             Convg  =    0.6432D-08             -V/T =  1.9978
 <Sx>= 0.0000 <Sy>= 0.0000 <Sz>= 0.5000 <S**2>= 0.7832 S= 0.5165

Now, regarding GaussView, interestingly, it also wrongly shows that the state is singlet and not a doublet for that test job. Besides, GaussView reports the following when opening the output:

Unresolvable inconsistency between charge and multiplicity and orbital occupancies. Orbital data will be ignored.

Honestly speaking, I suspect that is a bug, because while reading the output I could not see any inconsistencies. I've tried few other IRC jobs on radicals form the Gaussian test set (test171, test452, test453) and GaussView always reports that a state is singlet when in fact it is a doublet. Moreover, in some cases GaussView even failed to read all the steps of a normally terminated IRC calculation. So, the only inconsistency I've noticed so far is that GaussView could not always correctly read Gaussian outputs of IRC calculations on states that are not singlets.

Hi Wildcat, thanks for your reply. Yes, indeed when i open my normally terminated log file, i get the same warning as you have got. I did not find any inconsistency in the gaussian output too. I think you must be right that it is a problem with GaussView rather than the calculations.


text:
Kinetic isotope effect - why transition states taken as isoenergetic

I am working on the derivation of the equation below for the difference in activation energies for the two reactions (1) and (2).
I've done the derivation, but I am unsure on some of the approximations I have used. When considering the activation energies I have just used the zero point energy for the C-H and C-D bonds - I am not sure why I can do that, but I assume its because the rest of the zero point energies cancel out.
Secondly I am confused why I can consider the transition states as isoenergetic, as to me they look like they would different energies.
Thank you for your help.

When you deal with energies, absolute energies never matter. Energy only makes sense when you consider it relative to some reference point. For example, for standard values, we consider standard states. You're perfectly fine declaring that both TS's have the same energy because you're not comparing them to each other; you're comparing them to two ground states.

H and D each have 1 electron so the electronic energy can be considered to be the same. In calculating energy of the vibrational quanta then the difference in mass becomes important, as you indicate, via $E = \frac{h}{2\pi}\sqrt{k/\mu} \pu{~~(J/molecule)}$ where k is the force constant, equal for C-H and C-D bonds, and $\mu $ is the reduced mass $m_1m_2/(m_1+m_2)$.


text:
How can an α,β-unsaturated ketone tautomerise?
I was taught that for aldehydes and ketone to exhibit keto-enol tautomerism, it is essential that there must be at least one α-hydrogen.
For example, acetophenone and butan-2-one show keto-enol tautomerism, but benzaldehyde and benzophenone don't show keto-enol tautomerism, since they don't have any α-hydrogens.

However, I was subsequently told that the cyclic enone 6,6-dimethylcyclohex-2-en-1-one also exhibits tautomerism:

How is this possible? It does not have any α-hydrogens (the vinylic proton cannot be enolised as it is orthogonal to the C=O π* orbital).

If you have an α,β-double bond next to your carbonyl, a γ hydrogen can also take part in tautomerism. Take crotonaldehyde $\ce{H3C-CH=CH-CHO}$:
$$\ce{O=CH-CH=CH-CH3 <<=> HO-CH=CH-CH=CH2}$$
In general, a double bond is able to ‘extend’ the mesomeric properties of a carbonyl compound across two additional atoms. This is known as vinylogous; crotonaldehyde can form a vinylogous enol.
In the specific case of 6,6-dimethylcyclohex-2-en-1-one, these are the two tautomers:

There should be possibility of an intramolecular acid-base reaction. Acid-base pair are usually Lewis acid-base but there may be others. Such molecules show tautomerism. There are a large number of tautomerisms that can be explained by this way.
Such as keto-enol (your problem too), oxime-nitroso tautomerism and many more.

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text:
Is it really a reduction?
In my chemistry book they write $\ce{MnO4^- -> Mn^{2+}}$ is a reduction but is this correct? Electrons are lost aren't they because it becomes less negative so is it not an oxidation?

A simple analogy to see whether oxidation or reduction has occur is as following.
If oxidation number of element decreases,there will be reduction and if oxidation number increases there will be oxidation.In your case,Mn had +7 oxidation state initially  and finally it has a oxidation state of +2 so it is a reduction.

So how does your technique work with hydrogen peroxide?

in peroxides,oxygen atom always have -1 oxidation state.

You have to write out the complete half-cell reaction, and then it is obvious that $\ce{MnO4^−}$ is reduced.
$\ce{MnO4^− + 8 H^+ + 5 e^− <=>[reduction][oxidation] Mn^{2+} + 4H2O\quad\quad\quad\rm{E}^0 = +1.51~\rm{V}}$


text:
Is there a T-V or P-V chart relating LiBr temperatures for recovery?
I'm working on a project and would like to consider a LiBr solution to be used as a desiccant. I don't know however the temperature of air vs pressure to recover it. I was thinking of using a nozzle/diffuser and room temperature air to recover it's desiccant properties, but I'm not sure that this is possible.

Without some fancy equipment to measure the salt concentration in the solution,  a simple saturated solution of lithium bromide would have some relative humidity which would be a function of temperature.  I looked quickly for some relative humidity table of lithium bromide and found the following reference which has an equation that can be used for calculating the relative humidity ($RH_{sat}$) over a saturated solution of lithium bromide as a function of the temperature in Celsius ($T_C$).

"Humidity Fixed Points of Binary Saturated Aqueous Solutions" JOURNAL OF RESEARCH of the National Bureau of Standards, Vol. 81A, No.1, January-February 1977, Lewis Greenspan (pdf link)

$RH_{sat} = 7.75437 - 0.0654994T_C + 0.420737\times10^{-3}T_C^2 $
Some values for $RH_{sat}$ are:

7.75 % at 0 C
5.41 % at 100 C

So as long as some solid lithium bromide remained then the solution would be saturated. If all the lithium bromide dissolves, then you'd simply boil the solution to regenerate it. Lithium bromide is about twice as soluble in a boiling solution (somewhat over 100 C) as at room temp (20 C). So boil down to some starting volume of solution or until you see solid lithium bromide in the boiling solution. When the solution cools to room temperature about half the lithium bromide will crystallize out regenerating the saturated desiccant solution.
If you do want to boil this to regenerate then be sure and use some boiling stones. The solution would seem likely to bump.


text:
Best reagents state for air free chemistry. Choosing halide for grignard synthesis
I am planning to synthesize a grignard and I have a choice between the iodo or chloro aryl halide. The iodo is a solid and the chloro is borderline. For my previous reaction I melted the chloro and then tranfered it using a flushed sringe. As an alternative I could just weigh out the iodo and then pull vacuum on it a few times before running my reaction. Is there a better option between the two or they are just the same?

The chloride can be a bit slow to react, particularly if the aryl halide is hindered. You might need a small amount of iodine to get the reaction started. (Maybe you ran into this already?)
Once you form the Grignard, they should be identical for most applications.
Personally, I would just go with iodide because the chloride is probably a bit less reactive, and frankly, weighing out substances that are almost solid is extremely annoying. Strongly prefer the actual solid.
The main reason I would elect the chloride over the iodide in this case is if the cost of the chloride were much lower, and even then only if the reactivity weren't seriously impeded by using the chloride.

Also, I should point on that the air-free part doesn't matter in terms of the choice. You'll going to pull vacuum on the substrate, and if you really need to avoid oxygen/water vapor, you'd backfill with an nitrogen/argon and then transfer into your reaction vessel with a cannula.

While I haven’t done this with halocarbons to create Grignards, I have performed quite a large number of reactions where I either had a very highly viscous liquid or a solid.
If you have a choice, my advice is to go for the solid for these reasons:

If a compound is ‘borderline’, it’ll typically be rather viscous as a liquid just above its melting point.
Viscous liquids are a pain to deal with. Taking them into a syringe is hard.
Getting exact amounts of liquids is harder than getting exact amounts of solids. Typically, the scale on a syringe is too broad to adequately determine how much substance you have added. You would have to weigh the syringe with its contents. Adding or removing a few milligrams is much harder than just picking a few crystals/needles.
Solids, having relatively high melting points, are unlikely to evaporate notably when being subjected to HV conditions. (But this applies to viscous liquids, too.) This makes them easy to handle as you can simply add them to a flask and then evacuate.


text:
An acid and Sodium carbonate
I have to explain why it 'fuzzes' when an acid is added to Sodium carbonate? And what are the relevant Chemical equation(s)..
I'm stuck with this question. I'd appreciate some help here.
Thanks a lot.

Acids in the presence of carbonates produce carbonic acid which then decomposes into carbon dioxide gas (the fizz) and water. Sodium is a 'spectator ion'; i.e., does not react. Reaction equation is a metathesis double replacement:
$$\ce{HA + 2Na^+ + CO3^2- <=> A^- + 2Na^+ + HCO3^-}$$
$$\ce{HA + HCO3^- <=> A^- + H2CO3}$$
$$\ce{H2CO3 <=>  H2O + CO2(g)}$$


text:
Why do we titrate from a colourless solution and why should we remove unused sodium hydroxide from the burette as soon as possible?
In my lab we performed an acid base titration, where NaOH was used to titrate a standard solution, potassium hydrogen phthalate (KHP, or $\ce{K+C6H4(CO2H)(CO2^-)}$), using phenolphthalein indicator.
Why is it preferred to titrate from colourless to a pink solution rather than pink to colourless?
Also, why is it advisable to remove NaOH from the burette as soon as possible after we are done titrating?

Why is it preferred to titrate from colourless to a pink solution rather than pink to colourless?

It is much easier to detect the appearance of a color rather than the disappearance of one.
I really liked the large white ceramic ringstands for phenolphthalein titrations. The bright white background really made it easy to monitor the titration. At first you can just let the base (sodium hydroxide) pour into the solution as your are swirling the solution. You just get a pink cylinder where the base is entering the solution. As you get nearer to the end point then the color will start to go out more into the solution. Finally you do drops at a time, and then a partial drop.

Also, why is it advisable to remove NaOH from the burette as soon as possible after we are done titrating?

Sodium hydroxide is a strong base and it will etch (dissolve) glass given enough time. If you're using a burette with a glass stopcock the sodium hydroxide can also weld the stopcock joint.
I'll add that you never should pour the unused sodium hydroxide back into the stock solution. Any that you don't use is waste. You want to avoid any possibility of contamination.


text:
Does trans-decalin really have a plane of symmetry?
My teacher told me that trans-decalin (see below) is achiral due to the presence of both, a centre of symmetry and a plane of symmetry. But I could not spot the plane of symmetry untill now. Can someone point out the plane of symmetry in a diagram?

It passes through the two central carbons, perpendicular to the molecule.

It's not easy to see from a diagram, because it distorts bonds and angles. I recommend building it with a balls-and-sticks model set. You can also use a molecular viewer to model it; there are a couple of open-source (or at least free) ones out there.
I have calculated the molecule on the DF-BP86/def2-SVP level of theory. The point group of the molecule is C2h. In the following model I have highlighted the plane of symmetry, and the rotational C2-axis. At their intersection is an inversion centre Ci.

(I needed to downscale this a lot, click on the image to get to a high resolution still. Images created with ChemCraft, assembled with GIMP.)

Is it possible for one ring to have a boat conformation at room temperature? In this case the plane disappears.

@porphyrin I found another structure, where one ring is in twist-boat conformation. On the GFN2-xTB level of theory (semi-empirics) it is about 23 kJ/mol higher in energy. This is at room temperature negligible.

The question has already been answered nicely by Martin - マーチン♦ who pointed out that there is a mirror plane (i.e. a $\sigma = S_1$ element) and a center of inversion (i.e. an $S_2$ element), which both are already enough for a molecule to become achiral (existence of any $S_n$ element suffices for that). There is one important caveat though that often bugged me during stereochemistry lectures:  Even if the ground state of a substance might not be achiral (for example if you placed some substituents on your decalin), then it can still be possible for the chemical to be achiral due to quick racemization when an achiral transition state with low energy is available (e.g. inversion of $\mathrm{NR_3},\mathrm{PR_3}$ systems or the $^1C_4\mapsto ^4C_1$ inversion of cyclohexanes). I think that is a little trap that is often missed in stereochemistry lectures.


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Is there a chemical which fluoresces only after coming in contact with water?
I require an easily available chemical compound which in its pure form, does not  fluoresce under UV light. However upon coming in contact with water (at room temperature) it fluoresces under UV light.

Many fluorescent dyes are not fluorescent in their solid states. Fluorescein and similar dyes (rhodamines, eosines) for example. They need to be dissolved in order to be fluorescent (so just wetting them with a bit of water won't work either).

Riboflavin (Vitamin B2) fluoresces under UV light only when in solution, i.e. in contact with water.


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Why is the term not squared in the antigen-antibody interaction formula?
From Antigen-antibody interaction (Wikipedia):

We've basically multiplied two similar terms, because inverted Kd is Ka - why isn't there any square signs (power of two)?

Why would you expect there to be squared terms?

@CurtF. - we're multiplying two similar terms, as my last sentence says. [AbAg]*[AbAg] =  [AbAg]^2

Oh, I see what you mean.  Wikipedia is just wrong.

I edited Wikipedia to correct the problem.

@CurtF. - thank you! I also wonder why we multiply [Ab]*[Ag] and not, say sum them up, but I should try to read up on this first.


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Soaps come in different colors, but why is soap lather always white?
Soap comes in different colors, but why is soap lather always white?

This is a physical phenomenon...

Though @DHMO's answer is quite interesting, you shouldn't take it without a pinch of salt. I've to disagree with it in some aspects.
Soap colorants range from various kinds of Mica to dyes found naturally in plants. Now typically when you dye anything the quantity of dye used in comparison to the quantity of the substance to be dyed is very small. Too much dye, and it could possibly stain your hands/clothes, so soap manufacturers use as little dye as possible, but just enough to give the soap a good color.
Now for the rest of this answer, I will assume (reasonably enough) that by 'colored' soap you are referring to soap that goes by colors other than white.
Now when you lather using soap, only a really teeny-tiny bit actually goes into the water. As @DHMO correctly points out, the dye/pigment used to color the soap is greatly diluted. Why don't you carve off a piece of colored soap (having more or less the same size as as an almond) and dissolve it a mug of water by gently stirring it. I emphasize on "gently" so that you don't stir up a lather.
Have a look at the water. You'll see that the soap, apart from making it more cloudy, has not visibly imparted any particular color to the water. This is not unexpected, as I've already mentioned, the quantity of dye used is very small.
Now if you go ahead and agitate that soap solution, it will give rise to white lather, and this shouldn't be surprising anymore.
Now @DHMO's goes on to mention that Total Internal Reflection (TIR) is what imparts the white color to the foam. But this is grossly incorrect. While TIR can result 'white light', it is not the dominant phenomenon acting here (I'm not saying it's completely absent here either).
What is largely giving foam it's white appearance is another phenomenon called Scattering.
Now you might ask, 'Then why isn't soap water white?'. Well, since the foam is made up of lots of teeny tiny bubbles, light passing through it will have to encounter several surfaces, and it's these surfaces that scatter the light in so many directions.
So to say it straight DHMO's answer is incorrect. (No offence DHMO)
Remember I said you can't see any visible coloration in the water because of the dye is present in really small quantities? Well here's a way to validate that claim: Simply combine a teaspoon of red food-color with a bit of hand-wash (now that's really concentrated). Now you can lather the soap and lo behold! You have red foam.
[Credit to @ACuriousMind from Physics.SE, for that bit on why soap water isn't white. Also he confirmed that the white color of foam is due to scattering]
Edit- More kudos to @ACuriousMind for providing this amazing link on scattering.

And what causes scattering?

The only bits of DHMO's answer to disagree with are the "total" and to a lesser extent "internal" reflection.  If you ground up metal to a fine  enough powder  it would approach whiteness with only external reflection

You might have observed that large soap bubbles are transparent.
This is because the color has been diluted to a great extent that it basically became transparent.

Now, the foam is just a multitude of tiny bubbles that are transparent.
You would observe that the unorganized mass of transparent particles (the bubble is quite organized) would appear white.
For example, $\ce{NaCl}$ crystal (organized) is transparent while as a powder (unorganized) it is white.
Another example; a small water column when you open the water tap (organized) is transparent, but when you open the water tap to a greater extent, the water no longer falls together, and it appears white.

This is because of total internal reflection.
When the particle is organized, the light is uniformly bent, so the image is not very distorted and it appears transparent.
However, when there are many transparent particles, a small change in the angle of incidence can cause a large change in both the position and the angle of the resultant light.
Therefore, the colors are mixed together, and you see white.

Edit: After a short debate with Aaron Abraham, I figured that I only mentioned total internal reflection, but forgot refraction. Therefore, I now clarify that refraction instead of total internal reflection plays a dominant role in bending the light.

I do not think that this explanation is correct. Bubbles do not exhibit total internal reflexion (TIR) to any great extend as they are full of air , and so any ray is hardly deviated on passing through. The walls of a bubble are very thin thus refraction causes effectively no deviation before air is met inside the bubble. TIR conditions can hardly be met. Water droplets do show TIR as the refractive index is approx 1.33 throughout. Here TIR occurs & causes rainbows. The effect of bubbles is due to light scattering. If you used  white light white scatter, red light red scatter.

It is due to multiple scattering events of the incident light off the surface of the many small bubbles in the foam? Large bubbles (of the order of centimetres) do not scatter but show the typical colours of thin film interference fringes.
Common table salt or pure sugar crystals look white for the same reason. Even though the polished crystal is optically transparent, any roughened crystal surface causes scattering.  Scattering occurs due to the difference in refractive index of the air, approx 1, and the material in the foam or crystal, which is in the range 1.3 to 1.5 for many  substances.  Scattering is sensitive to the size of the imperfections on a crystal surface or bubble sizes in foam. It increases with more imperfections or denser bubbles.
You can remove the scattering by placing crystals into a solvent of the same (or very similar) refractive index as the crystal: harder to do with bubbles. However, you can make  surface roughened glass beads 'disappear' in the right solvent; water is quite effective.


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Is there a melting analogy for evaporative cooling
Evaporative cooling can occur in an air-liquid water system initially at the same temperature. If the air is not saturated with water vapor, the liquid water will evaporate until the partial pressure reaches its vapor pressure. In this process, the temperature of the liquid water decreases. This is because there is a distribution of speeds among the water molecules, and only the water molecules moving fast enough with enough kinetic energy to break their hydrogen bonds will escape into the gas phase. This selection based on kinetic energy leaves the slower molecules in the liquid phase, thereby lowering the temperature of the liquid.
My question concerns the parallel process of an ice cube melting. If the ice cube is in thermal equilibrium with a super-cooled salty brine, say -20 degrees C, there will still be a distribution of speeds among the water molecules in the ice lattice. Since only the fastest moving water molecules in the ice lattice will have enough energy to break the hydrogen bonds, would the melting of the ice cube cause a similar cut-off of the speed distribution curve and lower the ice cube temperature? If the volume of the brine solution was large enough so that the brine temperature doesn't rise much, could there be a situation where the ice cube both melts and also get cooler?

I'd call your first paragraph a simplistic explanation (don't worry - it is used a lot). The real, thermodynamic answer is that the entropy gain of the system by making some liquid in to vapor compensates for the enthalpy required to do so.  Just how the system does that is a kinetics question. A comparison of the liquid/gas situation to the solid/liquid phase balance is left as an exercise for the reader.

Thanks, I'm a teacher and trying to help students grasp some thermodynamic phenomena without invoking energy and state functions.


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Effects of atmospheric gases on colour of aurora
Recently I moved to the north of Norway, so I've seen quite a few auroras. Because the colour is mainly green, I was sure this must be due to electron transitions in $\ce{N2}$ (because the atmosphere is like $75\%$ $\ce{N2}$)
Looking into this, however, I found out that the colours of the aurora are mainly due to electron transitions in oxygen gas. Why is it that the electron transitions of oxygen are most prevalent in the colours of the aurora, even though it is far less abundant than nitrogen?
I also read that oxygen gas gives one colour up to $150$ miles altitude(green), and another above $150$ miles (red). Can someone explain the reason for the difference in colour? Does oxygen exist in different oxidation states at different altitudes, leading to different electron configurations => different available electron transitions?

Note: Follow up question to this can be found here - https://chemistry.stackexchange.com/questions/64414/aurora-borealis-less-favored-energy-transitions-higher-in-atmosphere

There is good information at Glowing Gases - Aurorae
There are many factors that need to be considered.
Once an atom or molecule is excited, it can lose the energy by collision or by emission of light.
The longer the lifetime of the excited state, the higher the altitude is required to make radiation vs. collision the way energy is lost.
The atomic oxygen transition with green light has a lifetime of about 1 second, while the transition for red light is 110 seconds, so a higher altitude is required for the red transition to become significant.
For molecular nitrogen, radiative lifetime is only $4 \times 10^{-8}$ seconds according to Electronically excited molecular nitrogen and molecular oxygen in
the high-latitude upper atmosphere Ann. Geophys., 26, 1159–1169, 2008
Also, human eyes are more sensitive to green light than red light.

This article is very useful, providing a good explaination for the variation of colour from Oxygen with heigth. However, I still remain unsatisfied With the explaination of why Nitrogen does not play a larger role in determining the colour of the aurora. Given that the atmostphere is 78% N2, but only 21% O2, how could this be? The article simply states "[...] nitrogen N2, is exceptionally stable and there are not many nitrogen atoms below 400 km to make aurorae."

@Adroit the composition of the atmosphere is much different at high altitude https://malagabay.files.wordpress.com/2014/07/daytime-ionospheric-and-atmospheric-composition.gif

@Adroit - Nitrogen is really hard to ionize, even with energetic charged particles. Then you've got the non-radiative energy loss mechanisms thrown in. It is not simple.

@DavePhD When your source says "The transition is O 1S to 1D" about the green emission line from atomic oxygen, does this mean that the transition is from S to D orbitals both in the first Principal energy level(Bohr Shell n=1)? From my knowledge of quantum numbers the first Principal energy level only has one S-orbital. Is this what it means by "The singlet S to singlet D state transition is not allowed by the quantum selection rules for electric and magnetic dipole transitions"?

@Adroit No, it doesn't mean first principal energy level (n=1).  This about the n=2 electrons, particularly the 2p electrons.  See Fig. 1 here: https://three.jsc.nasa.gov/articles/RadChemO2Sidebar.pdf

@DavePhD "The Laporte rule is a selection rule formally stated as follows: In a centrosymmetric environment, transitions between like atomic orbitals such as s-s,p-p, d-d, or f-f, transitions are forbidden." I've understood is so that this is the reason the green line transition is forbidden. Are all single atoms considered "centrosymmetric environments"?

Yes, a free atom is a centrosymmetric environment.

This phenomenon is already explained in this website, but I will paraphrase it.
All images here are from that website.
The atomic oxygen causes the green and the red color, while diatomic nitrogen causes blue and red:

The effect of altitude on the color is not explained.

However, an explanation is found on Wikipedia:
The highest altitudes have fewer particles, hence fewer collisions. In this state, the oxygen emits the lower-energy $630.0 \mathrm{nm}$ (red).
At lower altitudes, there are more collisions, and the nitrogen molecule also helps exciting atomic oxygen, so the higher-energy $557.7 \mathrm{nm}$ emission (green) dominates.

the cartoon has $\ce{N2}$ transmuting into $\ce{O}$

The anwer lies in understanding forbidden transitions. These are transitions which are "forbidden" by quantum mechanical electric dipole selection rules, have long radiative lifetimes and only occur radiatively if the density falls below some quenching threshold. Above this density the dexcitation takes place collisionally.
So let me advance a plausible explanation. The green northern lights are formed high up ($\geq 100$ km) in the Earth's atmosphere, largely by photons at 557.7 nm emitted from excited oxygen atoms. This is an example of a  forbidden transition with a long radiative lifetime (a second or so).
Forbidden lines get "quenched" by collisions if the density is high enough - i.e. the atoms are de-excited by a collision rather than by emitting a "forbidden line" photon. At lower heights in the atmosphere the densities increase and the forbidden emission is quenched, thus we may not expect to see any green light emitted lower down in the atmosphere. This also accounts for the stripes of different colours that are sometime seen, caused by transitions with different radiative lifetimes and different collisional cross-sections that quench at different densities and hence heights.
A quantitative answer to the question would require a detailed model of the atmosphere (since the quenching densities are also temperature dependent), a model for the energy spectrum and flux (as a function of height) of the incident exciting particles as well as the detailed quantum mechanical radiative lifetimes and collisional cross-sections for each of the relevant transitions. I think that the relative abundances of oxygen and nitrogen play a very minor role compared with these factors.


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Why is Fmoc base-labile and Moz acid-labile?
I'm learning about the Zuckermann approach to combinatorial chemistry in my MedChem lecture. My textbook claims that the Fmoc protecting group is base-labile while the Moz protecting groups is acid-labile. Please see image below.

Both protecting groups have the same exact structure that is bonded to the amine (methoxycarbonyl) so should be equally reactive to either an acid or a base. How come this isn't true?
In other words, why  does hydroxide succesfully attack the carbonyl to liberate the amine in Fmoc, but not in Moz. Conversely, why does hydronium protonate the amine to create a good leaving group for the resultant water molecule to attack Moz and kick it (the amine) off, but Fmoc is not open to this pathway? (NOTE: These are my assumptions for how the mechanisms work. They could be (and most likely are) wrong.)

Are you able to draw the mechanisms?

The mechanisms are really straightfoward. Just a simple acid or base catalyzed ester hydrolysis

You are wrong about the mechanism of FMOC deprotection. It proceeds by deprotonation of the doubly benzylic methine, followed by CO2 loss.


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Way to establish the acid/base properties of a compound from its chemical formula?
Is there a way to figure out the acid/base properties of a compound just by looking at the chemical formula?

$\hspace{4.3cm}$
Consider the two compounds above. Both have the formula $\ce{C5H10O2}$, yet one is about 20 orders of magnitude stronger an acid than the other. Without knowing the general structure of a molecule (what functional groups it is comprised of), suffice it to say no, you cannot know how acidic the molecule is.

It depends on the type of formula you are talking about. You need to know the structure before you can start assessing acid/base behavior, and not all formulas are equal when doing so.
A structural formula can be useful, as they are textural representations of the chemical structure. It is possible to determine the acidity difference between $\ce{CH3CH2CH2OH}$ and $\ce{(CH3)2CHOH}$, even when both have a molecular formula of $\ce{C3H8O}$.
An empirical or molecular formula, however, tells you very little, unless you can identify the compound based on it, or work out the molecular structure from the formula. It is generally easier to do so for smaller molecules. An empirical is even less useful as it tells you even less about the molecule itself; you cannot even determine the size of the molecule with certainty, which makes determining the structure nearly impossible.
You might be able to make educated guesses about the compound given its empirical formula, but they will be just that, guesses. For example, seeing sodium or potassium can imply a base (or something more basic, at least), while halogens are equally associated with acids.


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Why is absolute zero unattainable?
We were dealing with the Third Law of Thermodynamics in class, and my teacher mentioned  something that we found quite fascinating:

It is physically impossible to attain a temperature of zero kelvin (absolute zero).

When we pressed him for the rationale behind that, he asked us to take a look at the graph for Charles' Law for gases:

His argument is, that when we extrapolate the graph to -273.15 degrees Celsius (i.e. zero kelvin), the volume drops down all the way to zero; and "since no piece of matter can occupy zero volume ('matter' being something that has mass and occupies space), from the graph for Charles' Law, it is very clear that it is not possible to attain the temperature of zero kelvin".
However, someone else gave me a different explanation: "To reduce the temperature of a body down to zero kelvin, would mean removing all the energy associated with the body. Now, since energy is always associated with mass, if all the energy is removed there won't be any mass left. Hence it isn't possible to attain absolute zero."
Who, if anybody, is correct?

Edit 1: A note-worthy point made by @Loong a while back:

(From the engineer's perspective) To cool something to zero kelvin, first you'll need something that is cooler than zero kelvin.

Edit 2: I've got an issue with the 'no molecular motion' notion that I seem to find everywhere (including @Ivan's fantastic answer) but I can't seem to get cleared.
The notion:

At absolute zero, all molecular motion stops. There's no longer any kinetic energy asscoiated with molecules/atoms.

The problem? I quote Feynman:

As we decrease the temperature, the vibration decreases and decreases until, at absolute zero, there is a minimum amount of motion that atoms can have, but not zero.

He goes on to justify this by bringing in Heisenberg's Uncertainity Principle:

Remember that when a crystal is cooled to absolute zero, the atoms do not stop moving, they still 'jiggle'. Why? If they stopped moving, we would know were they were and that they had they have zero motion, and that is against the Uncertainity Principle. We cannot know where they are and how fast they are moving, so they must be continually wiggling in there!

So, can anyone account for Feynman's claim as well? To the not-so-hardcore student of physics that I am (high-schooler here), his argument seems quite convincing.
So to make it clear; I'm asking for two things in this question:
1) Which argument is correct? My teacher's or the other guy's?
2) At absolute zero, do we have zero molecular motion as most sources state, or do atoms go on "wiggling" in there as Feynman claims?

Both are utter nonsense. Absolute zero is physically possible (not that we can attain it, though).

Since systems can only approach it logarithmically, one would have to seriously question if it is physically possible.

Both explanations are wrong. (a) It is quite possible nowadays to obtain temperatures in the milli-kelvin range. This is done to remove thermal noise on very sensitive instruments. (b) very briefly, temperature is a measure of motional energy, molecule still have what is called zero-point vibrational energy at zero K, and crystals have lattice motion (phonons) that also have zero point energies, so it is not true that the total energy is zero at zero K. Atoms still have the same electronic energy as at room temperature.

@porphyrin Temperature is not a measure of motional energy. All it is is a ratio between dU and dS at constant V and N. That's how it can be negative in many exotic systems and (effectively) in some that have become quite familiar, for example. You can perfectly have motion at $T=0$. A harmonic oscillator does.

I think that what I wrote is  correct in the context of the question but not complete in a general sense. The average energy, translation, vibration, rotation is $=k_BT/2$ for each 'squared' energy term with $k_B$ being Boltzmann's constant. This assumes that a Boltzmann distribution applies, i.e. that each energy level has a smaller population than the one immediately below it in energy and that the number of levels is infinite. A population inversion as in the laser or in nuclear spin pulsed NMR experiments is then described as having a negative temperature but just for those levels.

There was a story in my days about a physical chemist who was asked to explain some effect, illustrated by a poster on the wall. He did that, after which someone noticed that the poster was hanging upside down, so the effect appeared reversed in sign. Undaunted, the guy immediately explained it the other way around, just as convincingly as he did the first time.
Cooking up explanations on the spot is a respectable sport, but your teacher went a bit too far. What's with that Charles' law? See, it is a gas law; it is about gases. And even then it is but an approximation. To make it exact, you have to make your gas ideal, which can't be done. As you lower the temperature, all gases become less and less ideal. And then they condense, and we're left to deal with liquids and solids, to which the said law never applied, not even as a very poor approximation. Appealing to this law when we are near the absolute zero is about as sensible as ruling out certain reaction mechanism on the grounds that it requires atoms to move faster than allowed by the road speed limit in the state of Hawaii.
The energy argument is even more ridiculous. We don't have to remove all energy, but only the kinetic energy. The $E=mc^2$ part remains there, so the mass is never going anywhere.
All that being said, there is no physical law forbidding the existence of matter at absolute zero. It's not like its existence will cause the world to go down with error 500. It's just that the closer you get to it, the more effort it takes, like with other ideal things (ideal vacuum, ideally pure compound, crystal without defects, etc). If anything, we're doing a pretty decent job at it. Using sophisticated techniques like laser cooling or magnetic evaporative cooling, we've long surpassed the nature's record in coldness.

Comments are not for extended discussion; this conversation has been moved to chat. If anyone needs any content from them to be incorporated into the answer, I can undelete those as needed, or you can copy them back from the chatroom.

Um, Ivan? I seem to be having a little issue with the 'no kinetic energy' bit. Could you have a look at the question again?  (I just edited it)

Of course I did simplify things a lot; so did Feynman. True, at absolute zero the atoms still have some uncertainty of position, which can't be further diminished because of Heisenberg's principle. Whether or not it constitutes a motion is a philosophical (that is, pointless) question.

@IvanNeretin Six months later, an article was published in "Nature Communications" by people from UCL who claim that (acc to this site) "you can’t actually cool a system to absolute zero with a finite amount of resources and we went a step further…. it is impossible to cool a system to absolute zero in a finite time..." This appears to me to be in direct contrast to your answer, but I may be wrong. So could you please clarify a bit? Thanks!

I just said that a matter at absolute zero is not forbidden by physical laws; I never said we can reach it. On the contrary, see "the closer you get to it, the more effort it takes" part?

@IvanNeretin Can you separate never being able to do something from a physical law? How does one know something can never be done unless one argues from other physical laws, or in essence, discovers a new law?

@LinearChristmas Why, of course we can. An ideal vacuum is probably the simplest example. Or imagine a 100% pure crystal of NaCl, without any impurities or defects. For all we know, it would persist just fine. Good luck trying to obtain it in real life, though. Same thing with absolute zero.

@IvanNeretin You talked about uncertainty in position and asked whether this can interpreted as motion, but of actual relevance is the uncertainty in momentum. This is undoubtedly a problem for even defining a motionless substance. You cannot get down to zero kelvin as having no motion is not a thing that exists by Heisenberg. And this is not addressed by your answer at all.

A nice explanation (+1). At temperatures close to absolute zero, the laws of quantum mechanics become important for the study of a system of atoms, and the effects of Heisenberg's uncertainty principle become more and more apparent. But this problem would be much important if the size of system is comparable with the de broglie wavelength of the atoms.

Absolute zero is a tricky concept, particularly once you start getting precise about it.  Thermodynamics and quantum mechanics is a tricky business!  I'll try to avoid the precise parts, and see if I can give you an answer which is more intuitive than a pile of equations.
The first question is what does it mean to "attain a temperature of absolute zero."  Typically when we phrase things like this, we are thinking in terms of equilibrium thermodynamics.  In this context, we are interested in bulk objects that have a uniform temperature.  We can quickly see that if there is any heat transfer between an object "at absolute zero" and any object not at absolute zero, then the first object will be warmed as thermal energy from the warmer object flows into it.  This mean our object at absolute zero can only remain there if it is in thermal isolation.  There is no known way to do this (especially when it comes to radiative heating), unless your object at absolute zero is completely surrounded by other objects at absolute zero.  This forms a sort of tower of babel that eventually falls when some outside objects must be subjected to the 3K background radiation.  Empty space is "warmer" than absolute zero.
What if we consider the world of non-equilibrium thermodynamics.  This is the study of systems that are not currently at equilibrium.  This is a strange place where some things can occur which don't make sense at first sight.  One of them is negative temperatures.  Negative temperatures occur because of how physicists define temperature: $\frac{1}{T} = \frac{\text{change in entropy}}{\text{change in energy}}$.  It's easy to show that in equilibrium situations (the ones we are used to) it is impossible to have a negative temperature (it also points out that if you ever set T=0, you would have an undefined value in your equation).  However, in non-equilibrium thermodynamics, we can consider strange compounds that are metastable.  You can think of them like a ball perfectly at the top of a smooth hill. If the ball is tapped in any direction, it will roll down the hill to the bottom.  However, at the top, it can theoretically stay motionless (temporarily).
We have corralled atoms into traps, and cooled them until they were very very cold (a few billionths of a kelvin).  Then, we flipped a switch which turned the trap around.  Suddenly a position that was very stable became an unstable equilibrium.  If you run the math on this weird state, it turns out that this implies a negative temperature!
Now this would suggest that, since a temperature went from positive to negative, it must have crossed through 0K, proving that we created something at absolute zero.  However, this is not the case.  What actually happens is that the temperature rushes towards positive infinity, reaches a discontinuity, and then wraps around to negative infinity.  It then approaches its negative temperature from negative infinity.  So even in this case, we can't reach absolute zero.
Quantum mechanics also poses an issue in that you could never prove you attained absolute zero if you tried.  Thermal energy is kinetic energy, which is related to momentum.  Let's say you found a hypothetical approach to reach absolute zero.  When you go to prove your findings, you must prove the momentum is also 0.  However, by proving that to be true, with no error, the uncertainty principle states that you can know nothing about the position of those particles.  They might be anywhere in the universe!

Nice answer! Thanks for posting, and welcome to Chem.SE!

I'd prefer to state the uncertainty argument as: observing, in any way, that the object is in a particular place, is an interaction that imparts some momentum to it, and after that interaction there's a 0% chance that the object is at absolute zero. It goes together with the first paragraph. Even if we theoretically have an object at absolute zero, it only exists there as long as it's never observed or interacted with.

@hobbs That is true.  I was being a bit cagey about that way of wording it.  Given that I'm starting with a hypothetical approach to reach absolute zero, I wanted to stay as close to the raw mathematical limits as I could manage.  I wanted to avoid the natural loophole of developing a hypothetical measurement approach which could do the job.

@CortAmmon fair enough. I just feel that the "if you know momentum perfectly, then position could be anywhere!" argument is one of those that causes more confusion than it's worth. People don't understand it as a mathematical bound on the possible information gained from all possible measurements, they imagine all kinds of mystical things going on :)

@hobbs Yeah but quantum physics is based on the assumption that the uncertainty relation is a physical concept and not just a limit of our measurements. In reality a particle really can't have a definite momentum. We can't measure the momentum more exactly because our measurements are flawed or affecting the particle in some way, we can't measure it because it just doesn't have definite momentum. To use an analogy, even God wouldn't know the exact momentum of the particle.  Even Einstein had problems with this notion and famously said: "God doesn't play die".

I think you misused the term "metastable", what you describe is highly unstable. It has zero resistance to any perturbation. A metastable state is a local minimum in potential energy, what you describe is a local maximum, and I believe it's called a saddle point.

@luk32 I suppose you could draw that distinction.  When getting near absolute zero, there's enough limits being taken that the difference becomes less important.  One could easily define a saddle point as the limit of metastable regions as the region of stability approaches zero.  In reality, of course, there will be imperfections which either make the system unstable or metastable in the sense you mention, where an imperfection creates a local minimum energy with a small region of stability around it.

Those are terms with precise mathematical definitions and physical interpretation. I don't think the application matters. I don't know how the potential functions of gases look like in near-0K. Maybe the distinction is not important, but those terms do not lose their meaning. "Suddenly a position that was very stable became very metastable." what you describe here is changing the sing of potential function, so the minima become maxima, therefore from a metastable state, you create an unstable equilibrium (that's the name for a local maximum, i.e. the saddle point).

@luk32 I see. I'll change the term to match the physics terminology.  In electrical engineering, that point is called a metastable point, and its considered metastable in philosophy (see Buridan's Ass)

I know it from quantum chemistry, and it's in agreement with the 1st picutre here https://en.wikipedia.org/wiki/Metastability . However, after your comment I've checked Metastability in electronics, and the very 1st sentence puts both states into "one bag", so I clearly see where you come from. Though, personally I don't think any real system can be in a state of unstable equlibrium, it's simply impossible to keep it from any perturbation. I am curious why electronic engineers care about it. Gotta read on that.

@luk32 We care about it because, for us, metastability means the state lasts for an indeterminate amount of time before the perturbations pull the voltage to one side or the other, and we have lots of circuits that misbehave if left in that half way state.  For example, transistors can overheat if they're left in an amplifying region rather than a switching region.  In chemistry, having to wait 1000 times as long for a system to eventually fall off its unstable equilibrium may not be a big deal, but when you're dependent on the timings to keep your circuit working, it can be nasty.

Ok, so it takes so long to fall out of the equilibrium into an actual stable state, so even if it's unstable, for the practical reasons it's as good as metastable. Thank you very much for the explanation. Makes much sense. Cheers.

The horribly named principle of uncertainty does not say that you can't know where  in the universe a particle is located at momentum zero. I says that the particle is spread all over the universe in that state until it eventually condenses anywhere in the universe. So, practically speaking, bringing a particle REALY close down to a momentum of zero would just make it go far away. Of course, it won't let you squeeze it past its ground state, as you cannot actually squeeze it but only kindly ask it to cool down a little bit further through friendly photon emissaries created by your laser :)

You have many good points here, but you might think about tying the argument back to heat again.  A system with negative temperature via the entropy definition will actually transfer heat to something at 0K.

Leaving quantum mechanics aside (it gives me a headache) the second law of thermodynamics prevents absolute zero from being reached in practice.  To cool something down, its heat must be transferred to something cooler than it. Since nothing can be cooler than absolute zero, one cannot cool something to absolute zero.
One can sidle right up close to $0\ \mathrm{K}$ and be amazed by its quantum awesomeness, but as Cort explained, at absolute zero, quantum effects make the concept of temperature rather awkward.
Your teacher's explanation is, as Ivan points out, based on the ideal gas law and there is no such thing as an ideal gas, especially not close to absolute zero.
And let's not forget about the physicist who fell into a vat of liquid helium.  He's 0K now.

"To cool something down, its heat must be transferred to something cooler than it" is plain wrong. E.g. the freezer is  by far the coolest object in my apartment, yet it has no trouble working.

@DmitryGrigoryev On a macro level, yes. However, at each physical stage within it, heat is moving from hotter to colder. What is hotter and what is colder is manipulated by pressure. There can be no colder than absolute zero, no matter how much you pressurize, depressurize, or otherwise mess with a substance; by definition the closest you can get is the same temperature, so there is never a heat sink available to reach abs 0.

@WilliamKappler Heat doesn't have to move to something colder. If you expand a gas it cools. An example of another effect is a common method for reaching  ~0.3 mK. You draw a vacuum on liquid He3 causing some to evaporate and cool down the remaining liquid He3. It obviously isn't the case that the only way to get colder is to have heat move to something colder, otherwise a refrigerator would need some sort of "cold source". But that isn't the case, they use a compressor to cool down the refrigerant without the need for something colder.

@Matt No matter how much you decompress something, it will retain the exact same thermal energy. Absolute zero has no thermal energy. You might get closer to absolute zero, but you cannot reach it by playing with pressure. And that's ignoring that you're not going to have gas phase (or indeed any of the 4 common phases) near absolute 0.

@Matt The refrigerator's cold source is the room, because the temperature of the coolant can be increased above room temperature by raising the pressure. You can never create a relative temperature difference that will cool something to absolute zero, which is required for heat transfer, unless you have something below absolute zero. Which does not meaningfully exist.

@WilliamKappler I'm not proposing getting to absolute zero using any of the methods I outlined. I'm just disagreeing with your assertion that the only way to cool something is by letting heat flow from that object to something colder. I did so by giving two practical examples of ways to cool something without needing something colder: 1) gas expansion, 2) phase change. These are two commonly used methods for cooling things. Again, not saying they get you to 0 K, just that they are cooling methods that don't rely on heat transfer from the object-to-cool to something colder.

@Matt I think you are missing the whole point of the first comment. See: "On a macro level, yes. However, at each physical stage within it, heat is moving from hotter to colder." That is to say, you do need hotter and cooler but can generate those within a system such that it appears you are moving heat from colder to hotter. That breaks down with absolute 0.

The coldest object in the universe is the 3K background radiation.  Your argument implies that we can never cool something below 3K, which is clearly false.

But what about a perfectly spherical, frictionless gas?

The coldest temperature recorded so far is 0.006 K and this in a cubic metre of copper which is quite something.  http://www.interactions.org/cms/?pid=1034217   I am quite tickled by the fact that they're using ancient Roman lead in this experiment to reduce levels of radioactivity.

The usual answer is that it actually is possible because even though the ground state of spacetime itself has non-zero energy, it is not kinetic energy, which is the very definition of temperature.
Due to the discrete nature of bound states, it is possible and has been achieved to bring single particles or even ensembles to their ground states or common ground state. If you define your system as just that and ignore the fact that very soon it'll get hit by something and suddenly no longer been at zero, goal achieved.
Macroscopic systems or timespans, though: no dice. As a single kinda random example, you have lots of neutrinos everywhere, which you need to count, and which you can't really shield yourself from. A massive galaxy-sized sphere of gold might do that to some degree, but unfortunately you can't build such a thing because of general relativity (i.e. it would become a black hole long before).
Numbers extremely close to zero are just as hard to get to as their inverse, i.e. things like massive galaxy-sized golden spheres, so even without the fundamental reason I gave first, it would still be impossible because our universe is full of stuff.

Comments are not for extended discussion; this conversation has been moved to chat.

Absolute zero can definitely exists (see the later edit), and there is at least one theory, that says that absolute zero will kind of be the norm in the universe at one point.
Absolute zero cannot be observed. Observation always implies interaction. Absolute zero implies no movement whatsoever. Observation implies that you somehow either receive a particle from the observed object or you send some particles that somehow get back to you, or you have a device on the other side and you measure the interaction of your particles with the other particles.
If a position in space is $0\ \mathrm{K}$, it means that nothing moves, so nothing from there can reach you, unless:
You send a particle to a position in space, which you expect to have $0\ \mathrm{K}$, but by the time your particle is there, the space will not have $0\ \mathrm{K}$ anymore, because it will have a moving particle that you just sent and in the end you will observe the temperature that you just produced.
Later Edit: I suppose it depends on how you define temperature. If you consider that something has to exist for it to have temperature, then $0\ \mathrm{K}$ cannot exist, because any fluctuation in any field means that something exists and that something has a temperature higher than $0\ \mathrm{K}$. But if you extend the definition of temperature to non-existing objects, then such an object should have $0\ \mathrm{K}$

Like the uncertainty principle, this is not about observation but way more fundamental.

@yeoman how is observation not fundamental and why is this not about observation?

In quantum mechanics, a particle is not a dot flying through space. Instead, it's a blob in space-time, described by the Schrödinger / Dirac equation. The blob can either be very focused around a specific position, then it implicitly is very fuzzy about its momentum. Or the other way round. or something in between. It's a fundamental property of its state, thus a fundamental property of the particle itself that exists with or without measurement.

To put it simply, when something isn't measurable it can't exist. When you can't measure the exact position without interfering with the momentum then the particle just doesn't have definite momentum. There are no hidden variables that just can't be measured. To put it in an analogy, even God wouldn't know the exact momentum.

@JannikPitt then the whole concept of observation for me is completely blown off. How can you observe something without any interaction? In my opinion, observation involves a transfer of information. A transfer means that something which used to be somewhere it is not there anymore, but at a position, from which the information can be interpreted for humans to read through one of their senses. Even if the observation does not actively change the object, the object has to change itself for others to be able to observe it. The object though, imo, still exist, even in the absence of observation.

Absolute zero is unattainable. You could do it in theory according to classical laws of physics, but it's quantum mechanics (including quantum electrodynamics) which prevents absolute zero to be reached. According to these laws, there is a possibility to have a fluctuation in energy even at zero level, which means that temperature will also fluctuate above zero. And the lower the temperature, the bigger the impact of quantum effects is. Special relativity (things like $E=mc^2$) doesn't matter here.

If anything, it is classic theory that prevents us from reaching absolute zero, and quantum theory provides a faint hope that it can actually be done, if only in some limited sense. See, if the energy is infinitely divisible, then you may suck out half of the remaining heat, then half of the rest, then again half of the rest, and so on. But if the energy goes in tiny discrete portions, that's another story.

That is true for any temperature, not only absolute zero. Any given temperature value is unattainable exactly due to fluctuations.

Please don't call the ground state "zero level" when the whole point is that it has non-zero energy :D

This argument makes no sense. The zero point energy of any QM system is certainly finite, while we can cool any object down infinitesimally close to zero K.

Two days ago a publication (Ref. 1) appeared proving mathematically that it is impossible to reach Absolute Zero.
This is kinda hard to read but well this is a serious proof.
Have a good read !
Reference

Masanes, L., Oppenheim, J. A general derivation and quantification of the third law of thermodynamics. Nat Commun 8, 14538 (2017). https://doi.org/10.1038/ncomms14538

Feynman is correct. Stop and think a moment. What does it mean to be "at absolute zero" if we accept the proposition that absolute zero is unobtainable?  Assume that absolute zero is unobtainable. Which of these statements is more false:1. At absolute zero, there is no motion. or 2. At absolute zero, there is no mass. Both are equally false; since the predicate is false, the statements are, logically, meaningless. (Which means both statements are equally true, too. But while #1 has often been used, I sure wouldn't use #2 unless I wanted to be pelted with rotten fruit. #1 is "close enough" to be a "useful" (if false) explanation while #2 has no physical justification.).  Classically, the Laws of Thermodynamics are empirical. They is what they is because they is. No "why" there. However, we can derive them from classical statistical mechanics (which was invented many years after the acceptance of the 2nd Law) or even better after the invention of quantum mechanics (1925 on) and its application to statistical mechanics. Temperature is a property of large populations of quantum particles, an atom does not have a (classically defined) temperature. So, in order to understand why this population can't have a temperature of 0 K, we need to understand how temperature is derived from the quantum mechanical properties of the particles that the system is composed of. That is a semester course in college, and not an Introductory one! If you understand Feynman's explanation, accept it (although it's not perfect - it was, after all, targeted at college freshmen with no previous exposure to quantum mechanics)

I'm not sure what Feynman means here. The lowest quantum state of a finite system is definitely reachable. At that point, we cannot remove any more energy from the system, and are at zero thermodynamic temperature. Of course any macroscopic object is practically infinite, in terms of an exact quantummechanical description.


text:
Energy levels database
Does anybody know where to find a database for atomic and molecular energy levels, with the corresponding energy values for each level? It would be needed just for small molecules. I tried the NIST website but it only has the ground state energies.

It does not get you all orbitals, but the Pitt Quantum Repository contains at least calculated HOMO and LUMO values for over 100,000 molecules.

Hi Ben, I could not find the energy levels on that website.

Go to the site and search for a compound, like formaldehyde. Then, in the properties pane on the left, click the link to "Detailed". HOMO and LUMO energies will show up in the list.

The Pitt Quantum Repository can get you what you want, at least for molecular species.

Go to the PQR at http://pqr.pitt.edu.
Search for a molecule, like benzaldehyde.
Click "Detailed" on the properties pane.
You can see HOMO and LUMO in the list.
Scroll down to the bottom of the page and click on the link to "Get JSON data".
The JSON file will open in your browser or you will download. You should be able to open it with a text editor. There will be a list of orbital energies for the set of molecular orbitals computed from the valence atomic orbitals. For example, for benzaldhyde.
"homo": -10.168,
"lumo": -0.712,
"orbitals": [
"-31.27648",
"-30.67710",
"-26.81219",
"-26.23865",
"-23.01445",
"-21.46816",
"-20.14594",
"-18.05465",
"-16.39792",
"-15.70143",
"-14.95762",
"-14.67105",
"-14.63486",
"-14.38581",
"-13.35746",
"-12.66484",
"-12.26937",
"-10.28746",
"-10.23037",
"-10.16786",
"-0.71215",
"-0.10196",
"1.01720",
"2.30473",
"3.35595",
"3.89337",
"4.34241",
"4.54954",
"4.71066",
"4.82113",
"4.99402",
"5.26380",
"5.29598",
"5.54486",
"5.70599",
"5.83103",
"5.89240",
"6.03437"
],

Though it seems that the database doesn't have all the molecules. For example I cannot find simple molecules as LiH or BeH.

They are there. You need to search by their compound name: Lithium hydride, beryllium hydride


text:
Why do (did) they make trans fat instead of using saturated fat?
When making chocolate unsaturated fat is used for a good melting point (and perhaps health reasons.)
To make chocolate edible it has to be a bit more solid, so other fats were required. Therefore they use trans fats, which are unsaturated fats made saturated in cis or trans form.
Why do they do that instead of just mix it with normal saturated fat? Why use trans fats?

https://cooking.stackexchange.com/questions/63747/why-most-chocolate-bars-have-trans-fat

they don't deliberately make trans fats. The reactions that partially hydrogenate polyunsaturated fats cause isomerisation leading to unnatural trans fats. This creates the desired properties of solid rather than liquid fats and nobody worried about it until trans fats were found to be much less healthy. Many places now ban trans fats forcing food firms to use more careful reactions.

What do you mean with  'not deliberately' ? Isn't (wasn't) it the purpose of processes to create them? Isn't there any control whether hydrogenate induce a cis or a trans form? But I always wonder why using trans and not just saturated fats? Are trans easier to make/get than saturated? And to get the right solidity and meltingpoints etc. are trans better equiped for that than saturated fats? Or did they believe that trans were healthier than saturated fats with almost equal properties?

Keep it simple... No, it is not the purpose. They are trying to make saturated fats, via the process that is hydrogenation. Trans unsaturated fats are simply an unpleasant byproduct of that. That's all.

Catalytic hydrogenation tends to make trans fats... unfortunately catalytic hydrogenation is the easiest (read lowest cost) way of hydrogenating fats so it became the industry preferred way of hardening fats before anyone realized there was a health risk to trans fats. Cheap plant oils can be hardened by hydrogenation... and hydrogenated plant oils are generally cheaper than naturally saturated fats.
Because they are unnatural, trans fats also don't go rancid so quickly... so goods made with trans fats can have longer shelf lives. Which cuts costs. Given so much of mass food production is severely cost conscious, it was inevitable that trans fats became widely used and are now having to be removed as ingredients.

Consider olive oil spread. Olive oil is composed largely of unsaturated fats which do not  "stack" easily onto each other at the molecular level, and thus are a liquid at room temperature. If you want to use it as a spread or in baking it is usually needed to be a solid. This is done in a process called Hydrogenation where the unsaturated molecules become saturated with hydrogen, thus as you stated making trans fat. This process is conducted in order to make often healthier vegetable oils (usually unsaturated) into solids at room temperature.


text:
Is it true that todays crude oil is getting more 'dirty'?
I heard that crude oil, which is used to make other products like gasoline, is getting contaminated more and more by atoms like nitrogen and metals, so that the catalysts have more difficulty making it clear for gasoline. If so what causes these contaminations?

It's not like it becomes more contaminated. It's like 'sweet' light and nice oils are not available in sufficient qualities and market demands deeper refinement, including heavy fractions, that were dumped in the past. The graphs of oil demand (readily available in the internetz) show steady growth for many years, and economical shocks are only a minor setback.

The other significant point is that modern cars have a catalytic converter to control emissions. The fuel has to be produced in such a way that it won't poison the  catalytic converter.

We've already extracted most of the easy-to-get, easy-to-use, relatively pure, oil, and are now digging up oil from other sources, like the tar sands of Alberta, which contain a lot more contaminants than most other sources we've drawn from in the past.

We've already extracted most of the easy-to-get, easy-to-use, relatively pure, oil,  || I request a proof on this. As far as I'm aware, that's not true. There is a plenty of light oils on the market, (WTI, Brent); and arabian oil that is offered in plenties, is still very easy to process.  The offering is by no mean sufficient (and tails are gathered from many sources) but it is still significant, and there is no need to move to tar sands yet. (Afaik, tar sands are calculated to be profitable to use at oil prices on sweet oils above $80-120/barrel) .


text:
What is the hybridisation in dimolybdenum?
Dimolybdenum's $\mathrm{4s}$ and $\mathrm{4p}$ orbitals are full. It is going to use its $\mathrm{4d}$ and $\mathrm{5s}$ orbitals to form a sextuple bond (2 $\sigma$, 2 $\pi$, 2 $\delta$). I am wondering if the $\mathrm{4s}$ and $\mathrm{4p}$ orbitals are going to hybridize with the $\mathrm{4d}_{z^2}$ orbital (the one forming the first $\sigma$ bond) and the $\mathrm{5s}$, given the differences in energy. The hybridisation would be $\mathrm{ds^2p^3}$, which is certainly uncommon.
If so, do the $\mathrm{s}$ and $\mathrm{p}$ lone pairs always hybridize, even if they don't form any bonds?

The electronic configuration of $\ce{Mo2}$ is given in Bursten, B.E.; Cotton, F.A.; Hall, M.B. Dimolybdenum: nature of the sextuple bond. J. Am. Chem. Soc. 1980, 102 (20), 6348–6349:

$$\ldots\mathrm{
9\sigma_g^{1.88}
5\pi_u^{3.78}
2\delta_g^{3.42}
10\sigma_g^{1.92}
9\sigma_u^{0.08}
2\delta_u^{0.58}
5\pi_g^{0.22}
10\sigma_u^{0.12}
}$$

Which can be simplified as:
$$\ldots\mathrm{
(9\sigma_g)^2
(5\pi_u)^4
(2\delta_g)^4
(10\sigma_g)^2
(9\sigma_u)^0
(2\delta_u)^0
(5\pi_g)^0
(10\sigma_u)^0
}$$

$9\sigma_{\mathrm{g}}$ is not degenerate; it is $\mathrm{4d_{z^2}+4d_{z^2}}$.
$5\pi_{\mathrm{u}}$ is doubly degenerate; it is $\mathrm{4d_{xz}+4d_{xz}}$ and $\mathrm{4d_{yz}+4d_{yz}}$.
$2\delta_{\mathrm{g}}$ is doubly degenerate; it is $\mathrm{4d_{xy}+d_{xy}}$ and $\mathrm{d_{x^2-y^2}+d_{x^2-y^2}}$.
$10\sigma_{\mathrm{g}}$ is not degenerate; it is $\mathrm{5s+5s}$.

Now, let us have a look at the beautiful 4d orbitals, kindly provided by our user Philipp:

From here, you can see that no hybridization is necessary, because the $\mathrm{4d}$ and the $\mathrm{5s}$ orbitals can all form bonds within themselves.

In $\ce{Mo2}$ the six bonding and six anti-bonding $\sigma(s), \sigma(d_{z^2}), \pi(d_{xz,yz})$ and $\delta(d_{xy})$ molecular orbitals with occupation numbers of less than 2 and more than 0, the active space, are only formed by $\ce{5s}$ and $\ce{4d}$ atomic orbitals (see DHMO’s answer for the $\ce{4d}$ atomic orbitals).

So, to answer your questions:

[Does] Dimolybdenum [...] use its $\ce{4d}$ and $\ce{5s}$ orbitals to form a sextuple bond (2 σσ, 2 ππ, 2 δδ)[?]
Yes, but the MOs are as described above.
[Are] the $\ce{4s}$ and $\ce{4p}$ orbitals [...] going to hybridize with the $\ce{4d_{z^2}}$ orbital [...] and the $\ce{5s}$[?]
No, they don’t hybridize the way you imagine it.
[Do] the s and p lone pairs always hybridize, even if they don't form any bonds?
That question is de facto answered, as there is no hybridization.

(The shown molecular orbitals and the occupation numbers result from a fully relativistic (DKH2) CASSCF(12,12)/def2-QZVP calculation with ORCA 3.0.3; RI was used. The calculation is based on A. C. Borin, J. P. Gobbo, B. O. Roos, Chem. Phys. 2008, 343, 210-216.)


text:
Why does a dough with margarine absorb more oil than one with butter?
My family makes deep-fried cakes. Normally, we use butter in the dough and fry it in canola oil. Recently though, we tried using margarine instead of butter and the cakes absorbed much more of the cooking oil than ever before.
What could be the reason for this?
First I thought that, by having a relatively similar, more unsaturated, molecular structure, the margarine had a greater intermolecular attraction with the canola oil than the butter with it's more saturated structure. I tried to search the web to find something to prove or disprove this, but, unfortunately, I found nothing of the sort.

I'd wonder if the margarine had water in it, and if it also had some sort of emulsifying agent.


text:
Is equilibrium potential for K+ always ~-90 mV?
Equilibrium potential is the difference in charge relative to the inside of the cell. -90 mV means that the inside of a cell is more negative than the outside. However when you talk about equilibrium potentials, I thought you only consider an individual ion like potassium? Potassium usually has 150 nM inside the cell and 5 nM out. If you allow potassium to freely pass through the membrane, wouldn't it eventually balance out so that the net charge on both sides is 0 (concentration would also be equal)? So why do people say that the equilibrium potential for K+ is -90?

Most ions don't pass freely through the membrane. This is prevented by the hydrophobic (water "fearing") fatty acid tails of the phosholipids that provide the basis of the membrane. The regions outside and inside the membrane are aqueous and attract ions much more than does the hydrophobic fatty region in the interior of the membrane bilayer.
Under usual physilogical conditions, the only way ions can cross the membrane is through protein channels. But these channels are gated and only open in response to certain stimuli. Stimuli can include a chemical substance binding to the gate and opening it, this is how neurotransmitters work. Many gates respond to changes in the overall potential across the membrane. These are termed voltage regulated gates and, among other finctions, explain how action potentials (impulses) travel down neurons and muscle cells.
Regardless of how these gates open, the effect is temporary. Neurotransmitters break down, the membrane potential continues to change in a manner that voltage regulated gates once again close.
During these events only small amounts of ions cross the membrane compared to the vast supplies on whichever side has the surplus. That would be the interior of the cell in the case of potassium. Yes, this might eventually deplete the concentration if allowed to continue without some mediation.
That mediation is the sodium-potassium pump. The Na-K pump, using the energy in ATP, pumps Na+ out and K+ into the cell. But it pumps out 3 Na+ for every 2 K+ it brings in which makes the overall resting membrane potential -70 mV. This pump is busy and important. It is responsible for using about 20% of the ATP your body produces. Even more if you're inactive.
So, after some of the K+ ions are allowed out through gated channels, the potential changes very locally. Then other K+ ions migrate into this area, down their concentration gradient, restoring mmost (but not all) of the normal K potential. Operation of the Na-K pump provides the additional K+ ions to restore it completely


text:
Polyatomic Cations
While learning about poly-atomic ions, our instructor stated that ammonium is one of the few poly-atomic cations. Are there other poly-atomic cations? And why are positively-charged poly-atomic ions "rarer" than negatively-charged poly-atomic ions?

https://en.wikipedia.org/wiki/Polyatomic_ion#Examples_of_common_polyatomic_ions

There is an infinite number of polyatomic cations, much like there is an infinite number of polyatomic anions. So your instructor is wrong there. But there is still some reason in the statement. Why? Polyatomic simple cations are rare (if you discount metal-ligand complexes).
Generating a simple polyatomic anion is easy: Take an element that is not fluorine, bond it to as many hydroxy groups as its valency is and deprotonate the hydrogens — voilà. You can then play around with oxidation states (i.e. add more oxygen) to generate even more anions, e.g. with $\ce{PO3^3-}$ and $\ce{PO4^3-}$.
This works because oxygen is a very electronegative element that, when bonded to a hydrogen, is strong enough to displace said hydrogen. For a polyatomic cation to form, we would need to reverse the idea: Take a stable, neutral molecule that has a lone pair and then add a hydrogen (one of the few rather electropositive elements that still has a tendency to form bonds).
And here are the problems. Most elements simply aren’t basic enough to cope with the additional hydrogen. In fact, basically any element that isn’t nitrogen will be too acidic in its protonated state. Also, you can’t really play with additional oxygens to create further oxidation states since increasing the oxidation state would destabilise the cation. Thus, simple polyatomic cations are few while simple polyatomic anions are plenty.

That was a great exploration, I didn't know that electropositivity was a thing. Thanks!

I think it could phrased better as inorganic polyatomic cations are rare. Every organic amine could form a cation of course.

@MaxW Hence simple. Taking my arbitrary definition of simple, it would only include two elements, but any organic amine requires at least three elements (carbon, nitrogen and hydrogen).

Jan - understand. My point was just that "inorganic" is a fairly well defined term but "simple" is more fuzzy.


text:
How to order the reactivity of t-butyl iodide and neopentyl bromide for SN2
The relative leaving group ability of iodide is better than that of bromide but steric crowding is more important as the neucleophile will attack the back lobe of carbon atom.
So how should I order the steric hinderance produced by t-butyl and neopentyl groups?

The order will be $\text{neopentyl bromide} > \text{t-butyl bromide}$. This is just an addition to @user377386's answer. From Solomons and Fryhle's Organic Chemistry, the relative rates of reaction (SN2) of different kinds of halide are (for the same halogen):
$$ \begin{array}{c|c}
 \textbf{Type of halide} & \textbf{Relative Rate} \\ \hline
 \text{Primary Halide} & 1 \\
 \text{Secondary Halide} & 0.03 \\
 \text{Neopentyl Halide} & 0.00001 \\
 \text{Tertiary Halide} & \approx 0 \\
\end{array} $$
You can easily see that neopentyl halides and tertiary halides are almost unreactive towards SN2 reaction, so it makes little sense to compare them. Neither of them is going to react by SN2 ever, they will instead react through SN1 mechanism. The slightly increased leaving group ability of iodine does not matter much. If you insist on comparing though, neopentyl halide will be better at SN2.

The order would be neo-pentyl bromide > tert-butyl iodide . This is because SN2 has concerted mechanism. Although, better is the leaving group, higher is the reaction rate for SN2, but, if the substrate is bulky, it has more tendency to follow SN1 mechanism because of bulkiness of the substrate which hinders the backside attack of the nucleophile. In case of tert-butyl iodide, the carbocation formed is very stable, so SN1 tendency is much more than SN2.


text:
Does the Clemmensen reduction reduce the C=C bond of α,β-unsaturated ketones?
The Clemmensen reduction is a chemical reaction which reduce aldehydes and ketones to alkanes.
In my book, it is written that under such conditions, double and triple bonds are not reduced, but my teacher told me of one exception:

α,β-unsaturated carbonyl compounds undergo reduction of both the olefinic and carbonyl group.

But according to my book, when this compound undergoes Clemmensen reduction, the product is pent-2-ene.

Who is correct?

Both are not quite correct. A quick literature search doesn't turn up any hits for the Clemmensen reduction of this particular aldehyde, but March's Advanced Organic Chemistry, 7th ed. (p 1541) writes about the Clemmensen and Wolff-Kishner reductions:

Neither method is suitable for α,β-unsaturated ketones [...] Under Clemmensen conditions, both groups of these molecules may be reduced or if only one group is reduced, it is the C=C bond.

Large structural rearrangements are common under the vigorous conditions: see, for example, Helv. Chim. Acta 1976, 59, 962. "In all cases but one the formation of cyclopropyl acetates was observed."
It seems that there are some situations in which the ketone is selectively reduced, but this is not general: see Tetrahedron 1986, 42, 6615.