text:
Poor aqueous solubility of chloroazodin
Below is the structure of chloroazodin:

Why is chloroazodin poorly soluble in water? It has multiple amine and imine nitrogens, all of which appear to be available for hydrogen bonding, and thus it seems it should exhibit high aqueous solubility.
Chloroazodin is mostly insoluble in water and organic solvents. I can't explain this character which is in contrast with its structure.

Welcome to Chem.SE! Thanks for posting your question here. Please take a look at my edits to your question, and feel free to edit/change anything that you don't like.  If you haven't already, please check out the tour and help pages for more information. Enjoy!

thanks.but chloroazodin is mostly insoluble in water and organic solvents.i cant explain this character which is in contrast with its structure.

It may be surprising, but seems to be standard tendency for compounds that they may be too strongly hydrogen bonded to be easily dissolved in water.

@Mithoron, interesting and maybe pretty relevant to the (now very old) question.  Can you give any examples?

Note that your link calls it a "chlorine releasing antiseptic". It doesn't give a mechanism or show the products remaining after releasing chlorine, but if this happens rapidly in aqueous solution then the compound that doesn't dissolve may be that/those product(s).  Does this seem reasonable based on what you know or have seen about this?

@airhuff Azodicarboxamide https://pubchem.ncbi.nlm.nih.gov/compound/Azodicarboxamide#section=Flash-Point

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text:
U- or RO-method for Singlet-Triplet Gap?
When I want to compare the energy between two different spin states of the same molecule, i.e. singlet-triplet-gap, do I better use unrestricted or restricted open formalism to compare the energies between both?
Further on, do I need to calculate the singlet state also with restricted open/unrestricted method? At least within Gaussian I did not recognize any energy difference between all of them. (Which sounds logic for me, as if the ground state is a singlet, also the open shell version will find that best version would be to have the same orbitals in alpha and beta.)

To expand on user1420303's answer a bit:

When I want to compare the energy between two different spin states of the same molecule, i.e. singlet-triplet-gap, do I better use unrestricted or restricted open formalism to compare the energies between both?

It depends.
The unrestricted formalism will almost always give lower absolute electronic energies (i.e., closer to the "real" value) than restricted-open will, due to the greater variational flexibility introduced by the separate spin-up and spin-down orbital sets.  So, from a purely energetic perspective, yes, unrestricted is likely preferred.
If you're interested in other properties, though, the spin contamination introduced by the unrestricted formalism may be problematic.  Of course, in systems with sufficient static correlation (viz., a degenerate or near-degenerate ground state) to result in spin contamination (radicals, bond-breaking states, transition metals, etc.), single-reference methods like Hartree-Fock and DFT may not give reliable results anyways.

Further on, do I need to calculate the singlet state also with restricted open/unrestricted method?

It depends.
For most 'non-exotic' organic systems, there is no need to calculate the singlet state using an unrestricted method since, as you note, the results will often be indistinguishable.  For systems with static correlation, though, if you still want to try to use a single-reference method, you will need to explore the unrestricted orbital space, as there may be a lower-energy solution where the alpha and beta orbital compositions appreciably differ.  The primary method I'm aware of for such exploration is the "broken-symmetry" method.  Two papers I know of that discuss the method, albeit not in extensive detail, are the following reviews by Frank Neese:

F. Neese. "Prediction of molecular properties and molecular spectroscopy with density functional theory: From fundamental theory to exchange-coupling." Coord Chem Rev 253: 526, 2009. doi:10.1016/j.ccr.2008.05.014

F. Neese. "A critical evaluation of DFT, including time-dependent DFT,
applied to bioinorganic chemistry." J Biol Inorg Chem 11: 702, 2006. doi:10.1007/s00775-006-0138-1

The method is integrated into the current version of ORCA (see Section 5.9.10 of the v3.0.3 ORCA manual), and tips for running such calculations are available at the ORCA Input Library.  Other software packages likely include this capability; search their manuals for more information.

Is an unrestricted method really a single-reference method? Different orbitals for different spins appears to me as to have one determinant (ref) per spin.

@Martin-マーチン Mathematically you're right, but I've never heard anyone call UHF 'multireference' before. It's still a mean-field, uncorrelated method... it just has greater variational flexibility as compared to RHF/ROHF.

I'd call it a dirty, dirty method (Fifty shades of spin). It's not that I would go out and say UHF is multi-reference (although in BS that is exactly what it tries to mimic), but I wouldn't call it single-reference either.

@Martin-マーチン How about a "dual coupled field" method? ;-)

1 It is better to use the unrestricted (U) formalism when the computational cost is feasible, because of course it does not impose thee extra restrictions that take place in RO. Although, RO was always suitable in my experience.
2 Normally you don't have to, e.i. in most of cases for a singlet state restricted treatment works good (giving the same results that U and RO). But you can not generalize it. Take a look at http://www.cup.uni-muenchen.de/ch/compchem/energy/stability.html

Beat me to the punch. :-)

@Brian,  because you elaborate more, +1..., after reading your answer I am curious about the relationship between static correlation and spin contamination.

I never really found a good, concise exposition of it. I would start with the relevant sections of the Neese papers, and references therein. Section 5.1.7.6 of the ORCA v3.0.3 manual actually has a nice mini-example showing how BS-UHF can treat statically correlated systems. Beyond that, a literature search for "broken-symmetry" should turn up a lot.

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text:
Which is the better leaving group: R- or H-?
I don't know how to compare them. So many factors involved. According to me, R- should be better leaving group, as it is more electronegative than H, but on the other hand, R- is very unstable due to +I effects.

R- is very vague. $\ce{-CH3}$ is a terrible leaving group, but $\ce{-CCl3}$ is a fair nucleofuge. "Leaving groups" typically means nucleofuge, which is a leaving group that takes the electrons from the broken bond with it as it leaves. $\ce{-H}$ makes for a poor nucleofuge, $\ce{H-}$ (hydride), but it is a common electrofuge, $\ce{H+}$, which is the opposite of a nucleofuge.
I just realized you might have meant $\ce{R-}$ and $\ce{H-}$ in your title question. In that case, you're still going to want to specify what "R" stands for, but $\ce{H-}$ is a very poor leaving group.

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text:
Order of priority for including functional groups in the parent chain compared to order of priority for numbering the parent chain
The following is about the IUPAC naming system.
So as far as I understand the rules for numbering a chain are as follows:
Compare the locants of functional group 1 in both cases if the are lower in one of them it wins. Compare the locants of functional group 2 in both cases if the are lower in one of them it wins. And so on for all functional groups in order of priority. And then in the end we compare the locants for all substituents in alphabetical order.
But the rules for choosing a parent chain are the same, except we firstly want to maximize its length and after that we want to maximize the number of functional group 1s, functional group 2s and so on. And after that we want to minimize the locants for functional group 1, functional group 2 and so on until any of them is better. I am not sure what we do after that though.
My main 2 questions are the following:
Is the ordering of the functional groups the same for all 3 steps (maximizing the number of functional groups in the parent chain, picking a parent chain with minimized locants of the the functional groups, and then the same but for directing it)?
Where can I find the ordering of all functional groups or at least the following functional groups: alkyl (are alkylidenes in the same category as alkyls?), halogen, OH, double bond, triple bond, and  NO2?
And I know that sometimes two or more of these are grouped together in a set (for example double and triple bonds) so an ordered list of these is not enough.
Can you please provide something like (Set of double and triple bonds)>(double bonds)>alkyls but for all of the listed? If not where can I find such a thing?
Please tell me if any of that is wrong.

IUPAC is quite complicated and produces names that wouldn't be expected by those not very familiar with it. Lots of chemicals have common names that violate IUPAC naming schemes anyway, so unless you've invented something new you'll likely not need to be too concerned with them. Sites like Chemspider might have 50 aliases for a single compound, industry names, and older names. It's one of those things I consider impossible to keep up with and ignore unless necessary for simpler compounds.

There are indeed several analogous criteria for the numbering of locants and for the choice of a principal chain; however, there are also significant differences.
The most important simplified criteria for the numbering of simple acyclic compounds are:

lower locants for suffixes
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The most important simplified criteria for the choice of a principal chain are:

greater number of suffixes
longest chain
greater number of multiple bonds
lower locants for suffixes
lower locants for multiple bonds
greater number of prefixes
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

You can find the corresponding actual wording of the nomenclature rules (taken from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)) below. However, the IUPAC recommendations have more than 1500 pages. Hence, you should be careful when applying any simplified or abridged version of the nomenclature rules.
Also note that there is no order of priority for substituents cited as prefixes (e.g. ‘bromo’, ‘chloro’, ‘ethyl’, ‘methyl’, or ‘nitro’). All such prefixes are considered together as a set. However, you need the seniority order of compound classes in order to choose the senior class that is expressed as a suffix.

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES
When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.
P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).
(…)
P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].
(…)
P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. These criteria are illustrated in P-44.4.1.1 through P-44.4.1.12.
The senior ring, ring system, or principal chain:
(a) has the greater number of multiple bonds (P-44.4.1.1);
(b) has the greater number of double bonds (P-44.4.1.2);
(…)
(h) has the lower locant for an attached group expressed as a suffix (P-44.4.1.8);
(…)
(j) has the lower locant(s) for endings or prefixes that express changes in the level of hydrogenation, i.e., for ‘ene’ and ‘yne’ endings and ‘hydro/dehydro’ prefixes (P-44.4.1.10);
(…)
P-45.2.1 The preferred IUPAC name is based on the senior parent structure that has the maximum number of substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.
P-45.2.2 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.
P-45.2.3 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes to the parent structure (other than ‘hydro/dehydro’ prefixes) in their order of citation in the name.
(…)

What's the difference between a greater number of suffixes and a greater number of double bonds? -ene is a suffixe like in hex-1,3-diene. If a molecules consists of 5 C atoms with the double bond in positions 1 and 4, and a propyl group on position 3, what would be its name?

@Marko According to current IUPAC recommendations, ‘ene’ is an ending (not a suffix). The name for your example is ‘3-ethenylhex-1-ene’ since ‘longest chain’ is senior to ‘greater number of double bonds’. (Note that this is a change from earlier recommendations. The order of seniority between unsaturation and length of chain given in previous IUPAC recommendations was reversed.)

I think I got it, I just want to ask you which things are considered suffixes if not -ene/-yne? Is -yl/-ylidene/ -ylidyne considered a suffix? Is -ol? And by that "However, you need the seniority order of compound classes in order to choose the senior class that is expressed as a suffix." do you mean that something that can be a suffixes can also be a prefix if something with higher priority is present?

@K.Smith Typical examples for functional suffixes are ‘ol’, ‘al’, ‘one’, ‘oic acid’, ‘amine’, and ‘nitrile’. Only one suffix can be placed at the end of a name to represent the principal characteristic group. Hence, the seniority order of classes ranks preferred suffixes for citation as the principal characteristic group. Characteristic groups that are not cited as principal characteristic group (i.e. suffix) are expressed as prefixes.

I see. And what do you mean by "greater number of prefixes"? Does that mean that 2-methyl-3-ethyl is better than something like 2,2,3-methyl? (I know that there is no compound that can be named by both but just as an example. I hope you got what I mean.)

@K.Smith If there is still a choice, the principal chain has the maximum number of substituents cited as prefixes. See for example here.

I forgot to ask where does the point of attachment if the thing we are naming is a substituent fit in this list. I am pretty sure that it is mandatory to be included in the parent chain but I am not sure where it is on the second list.


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Whether amino acids exist as zwitterions in the liquid and gaseous phases
I am aware that in the solid state pure amino acids are crystalline structures composed of zwitterions, however I cannot find any information about whether somehow this changes as they move into liquid and gaseous forms- are they zwitterions in these sates and if not, why not?

There's always equilibrium there - more or less of both forms.


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Can you get oxygen form silica or alumina?
If I had a lot of silica or alumina, like on the moon, would I be able to split the silica or alumina into aluminium and oxygen/ silicon and oxygen? It doesn't matter how unpractical it is. If there is a way, how can it be done?

In the Hall-Heroult process, oxygen is produced by electrolysis of $\ce{Al_2O_3}$

http://chemistry.stackexchange.com/questions/49098/how-can-i-obtain-oxygen-gas-fromplagioclase

Of course it is possible - that is how we get silicon and aluminum in the first place. On the moon, physical, rather than chemical processes might be more easily accomplished (see - isotope separation with Calutrons).

If you don't care too much about practicality, ionic compounds (i.e. salts) (which most minerals are) can be separated by melting them and then electrolysing. The only problem is that most salts have very high melting points. The needed temperature can be lowered by dissolving the salt in another lower melting point liquid, as happens in the Hall-Heroult process for aluminium production, but that only works if the dissolving liquid is more resistant to electrolysis than the stuff you're trying to electrolyse.
An even more general method of separating basically any compound is by heating it into a hot enough plasma to atomise it. The atoms can be separated using a calutron (which works similarly to a mass spectrometer). In reality such separation is only used with very small amounts of materials for analysis and historically for uranium enrichment, but there is no reason in principle why this could not be scaled up to an industrial scale.


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What type of isomerism relates an allene from a conjugated diene?
What type of isomerism exist between an isolated/conjugated diene and a cummulated diene (allene)? Are they functional isomers or position isomers?
For example, are buta-1,3-diene and buta-1,2-diene functional or positional isomers? My one chemistry teacher is saying functional and another positional!

They are just what they are. The classification into functional and positional isomers exists only in our heads.

Thinking about buta-1,2-diene and buta-1,3-diene, I'd say they are both. They are functional group isomers, because I consider an allene to be a distinct functional group from a conjugated diene. Others might not see it this way, but they have different chemistry and considerations.
They are positional isomers, because they only differ by the position of a group (in this case a hydrogen / pi-bond).
I don't see the need for it to be one or the other and not both.
However, the more important classification is that they are constitutional isomers (or structural isomers). That is, they differ in the connectivity of atoms. This is in comparison to stereoisomers, that have the same connectivity but different spatial arrangements of atoms.


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How long does it take to turn aluminum smelters on and off?
This may fit better under "engineering," but I figured I'd start here.
Aluminum is produced through an electrolysis process, and Wikipedia tells me that the electrolytic cells need constant power in order to operate - that if the aluminum solidifies, you can't just turn the thing back on.
I'm curious as to how long it takes to turn these processes on and off at a commercial aluminum smelting operation. Is it the sort of thing you can start in the morning and turn off in the evening when the plant closes, or do you really want to operate the cells 24-7?

I'll say you simply don't ever turn it off, as with blast furnaces for producing iron. If you want to stop the thing, you have to dismantle and afterwards rebuild it.


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What is the hybridization state of E (= Sn, C) in ECl₂?
What is the hybridization state of tin in tin(II) chloride, $\ce{SnCl2}$, and is the explanation similar for dichlorocarbene, $\ce{CCl2}$?
For $\ce{SnCl2}$, I read in my book that it is sp² but I am not able to explain it. I am not sure about the carbene.
I might have a misunderstanding about hybridisation, I think that the energy required for hybridization has to be compensated by the energy released in forming bond between hybridized orbitals.

Hybridisation is a mathematical concept, you never gain or lose energy due to it. Descriptions of the wave function involving hybrid orbitals or not are equally as valid. The linear combination of orbitals that result in the equilibrium geometry can however be interpreted as hybrid orbitals. Sometimes this may lead to a better understanding of the bonding situation.
From the above, it is obvious that hydbridisation is always a result from the molecular structure, never a cause for it. Hence we can deduce from the known bond angles the following:
Tin(II) chloride has a ∠Cl-Sn-Cl of about 95° (see Wikipedia). We know that p-orbitals are 90° towards each other. Therefore the tin will have more or less unhybrididised orbitals. See also this previous answer of me and this answer of Ben Norris.
Dichlorocarbene on the other hand has a ∠Cl-C-Cl of about 109° from both experimental and calculated data (Phys. Chem. Chem. Phys. 2002, 4 (14), 3282–3288). We know that the angles between sp³ hybrid orbitals are tetrahedral angles of about 109.5°. We can therefore deduce that the carbon employs sp³ hybrid orbitals in first order approximation.
The differences in the molecular structure and the utilisation of the s-orbitals for bonding stem from what is termed the inert pair effect, there is an explanation available in the this Q&A: What is the inert pair effect?

Re: "See also this answer of me and this answer of Ben Norris." Those answers talk about mainly the hydrides, while this question is about the chloride (of tin). Then, how can you say that the theory - that applies to hydride - will also apply to the chloride? Thanks for the answer!

@GaurangTandon it applies in the way that the geometry is reason for a hybridisation concept and not the other way around. It doesn't matter which atoms are involved, it only matters how the atoms are arranged. (In this kind of approximation.)

@Martin-マーチン Re: "the geometry is reason for a hybridisation concept" You mean to say that we deduce the hybridisation from the bond angle? If so, then, is it possible to predict the bond angle of $\ce{SnCl2}$ - i.e. whether it will be near $90^\circ$, $120^\circ$ or $180^\circ$ (only an approximation) - just from the atoms involved ($\ce{Sn}$ and $\ce{Cl}$)?

@GaurangTandon the geometry depends on various factors, as such it is not possible to determine it from the atoms involved a priori. You'll have to use the tools at your disposal, be it experiment or quantum chemical analysis, even VSEPR may give you a good starting point. However, don't expect to much accuracy from the latter.

1.The answer of question one is this that as Sn contain 4 electrons in its valence shell and will have to combine with two Cl atoms to form SnCl2 by sharing one electron with each Cl atom so it need two orbital which have one electron each then two free electrons will remain since there are maximum chances of these 2 free electrons to be present in the same orbital and be called as a lone pair hence Sn will undergo sp2 hybridization forming 3 sp2 hybridized orbitals and SnCl2 will have bent structure in which central atom Sn will be surrounded by 2 bond and 1 lone pair☺

I don't see why you post it after Martin's answer.

It's incorrect anyway.

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Exothermic reaction between sodium carbonate and castile soap
The other day, I tried to mix liquid castile soap and washing soda (sodium carbonate) to use in doing laundry. I mixed equal parts liquid soap and sodium carbonate by weight. To my surprise, an exothermic reation occured (my glass jar got warm), and then the substance became hard. Before it got hard, I touched the tip of my finger to it and didn't get any chemical burns.
The castile soap that I used was homemade. My soap contained: water, potassium soybeanate, and glycerin.
The soap was made of:
water 172.36g
potassium hydroxide 91.45g
soybean oil 453.59g
for a total of 717.41g before boiling down which I then dilluted to a water to soap ratio of 4:1. The soap was well cooked (5-6hrs) and cured for a week.
Could a chemist help me understand what reaction occurred?

I think the issue is that you had unreacted organic acids that were strong enough to be deprotinated by $\ce{NaCO3}$.  So then it's just an acid-base reaction between $\ce{NaCO3}$ and the leftover organic acids in the soybean oil.  Then the solids would be the  Na salts of the organic conjugate bases, which are insoluble in oil.


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What is the difference between standard chemical potential (μ standard) and the "pure" chemical potential (μ*)?
I am studying solutions for my thermodynamics course and I am a little confused about the chemical potential of a solute. I found the following from Atkins:

For a solute $\ce B$,obeying Henry's Law $p_\ce B=K_\ce B\,  x_\ce B$, where $x_\ce B$ is the molar fraction

$$\mu =\mu^*+RT\ln\left(\frac{p}{p^*}\right)=μ^*+RT\ln\left(\frac{K}{p^*}\right)+RT\ln(x)$$
It then states:
$$\mu(\textrm{standard})=\mu^*+RT\ln\left(\frac{K}{p^*}\right)$$
Therefore, the chemical potential of B in an ideal-dilute sol. is:
$$\mu=\mu(\textrm{standard})+RT\ln(x)$$
The text often states that anything with the * means it is in its pure form, so $\mu^*$ would be chemical potential of $\ce B$ not mixed in with a solvent...but how does this physically differ from $\mu(\textrm{standard})\,?$


text:
Enthalpy and Internal Energy in Constant Volume Calorimeter
It is my understanding that enthalpy differs from internal energy only in that the change in enthalpy takes into account any pressure-volume work done on the system (or by the system) in a reaction, and thus it is equal to the heat transferred to the system, at least in constant pressure.
Many sources say that in a constant pressure, "coffee cup" calorimeter, $q_p = \Delta H$. I understand that. However, they also say that the $q_v$ in a constant-volume calorimeter is equal to the change in internal energy, and not necessarily the change in enthalpy. My thought is that since $\Delta H = \Delta U + P\Delta V$, with $\Delta V$ being zero (it is by definition constant volume), is not $\Delta H = \Delta U = q_v$?

The definition of enthalpy is
$$H = U + pV$$
which gives
$$\begin{align}H_2 - H_1 &= (U_2 + p_2V_2) - (U_1 + p_1V_1) \\
\Delta H &= \Delta U + \Delta(pV)\end{align}$$
The equation you cited is only true at constant pressure: if $p_1 = p_2 = p$,
$$\begin{align}
H_2 - H_1 &= (U_2 + pV_2) - (U_1 + pV_1) \\
\Delta H &= \Delta U + p(V_2 - V_1) \\
&= \Delta U + p\Delta V
\end{align}$$
The condition for that equation is that the pressure is constant. That equation is not applicable for constant-volume (isochoric) processes unless the process is also constant-pressure (isobaric).
However, if you do have a process that is both isobaric and isochoric, you are right to say that $\Delta H = \Delta U$.

A coffee cup calorimeter is typically used for liquid phase reactions and, as such, generally involves  little or no volume change. Under these circumstances $\Delta (PV)=P\Delta V$ is very close to zero, and the change in enthalpy of the reaction is very close to the change in internal energy (particularly at low pressures like atmospheric pressure).  Also, since liquids are nearly incompressible, the heat capacity of the reaction mixture at constant pressure is virtually the same as the heat capacity of the mixture at constant volume.


text:
Electolysis of H2O to produce O2(g), in a small, portable unit
Full disclosure, I'm just a high school chem student, but I've got this question and I couldn't find the answer anywhere online.
I was reading an article about some futuristic "artificial gill", which kinda sounds like BS to me, but it got me thinking about this.
http://www.smithsonianmag.com/innovation/student-claims-have-designed-working-artificial-gills-180949465/
We were just learning in class about electrochemistry, and we just did electrolysis, so I was just thinking, instead of filtering the water, what if they used electrolysis to produce O2 gas?
There would still be hurdles, like diluting the 100% O2 to about 20% gas by volume, storing the necessary electric power to drive the reaction, and making sure that the unit wouldn't blow up, but could it theoretically work if we could overcome these other issues?
Basically, could you produce enough O2 gas to survive, through the electrolysis of H2O, in a unit small enough for a man to use?

It is a scam. Among other posts -> http://www.geek.com/news/indiegogo-backers-keep-giving-money-to-extremely-shady-artificial-gill-1651300/

Related: Wattage to produce 6L of air from Water in less than a second?

In scuba diving the main issue is pressure. To be able to breathe, you need to counteract the pressure of the column of water on your rib cage. This is why divers breathe from compressed gas tanks. Otherwise their rib cage and lungs would be crushed by the pressure. At higher pressures, air becomes too viscous to be breathed without too much effort, and the build-up of dissolved nitrogen in the tissues leads to narcosis ("being drunk"), so the oxygen has to be mixed with helium, that is much less viscous. So, with theses 'artificial gills' or with electrolysis, you still need a tank of helium, and a powerful compressor.
Overall, it is much more convenient to carry tanks of compressed mixtures of gases. Compressing gases is a form of energy storage (that could even power cars). But if you are in a nuclear-powered submarine, then you would be better off doing electrolysis.

Perhaps hydrogen produced from electrolysis is as good as helium, or even better? :)


text:
Dispersion forces - what determines which molecule will have its electron cloud repelled?
Here is an image I found which illustrates dispersion forces:

One helium has its electron cloud repelled by the other helium's electron cloud, which exposes the nucleus and gives an induced dipole which holds the atoms together weakly. But when two helium atoms approach each other, what determines which atom will have its electron cloud repelled? I would think that both electron clouds would equally repel each other. Also I am guessing that this effect will get stronger as temperature is lowered (since helium is a liquid at very low temperatures), but I am not sure why.

You are correct that the electron clouds would repel each other equally.
The induction as shown in your second picture would require another outside entity, polarizing one of the He atoms, which would then affect the other.
At an arbitrarily show distance, we might describe a molecular orbital formation between the two helium nuclei, as shown here:

(source: mhhe.com)
In this case, without the external perubation, the orbital shape (and hence the electron clouds) is (are) quite symmetrically distorted.
However, it becomes nonesense to describe a system of two atoms.  In liquid helium there are manifold other charge fields distorting these nice shapes, and you may get some asymmetry as shown in your picture above.

when you say outside entity, I'm guessing that can't be other helium atoms because they would repel each other equally, so if you had a vial of liquid helium how could the atoms be staying together? are the atoms at the walls of the vial polarised by the molecules that make up the vial, or by molecules in the air, which then has a domino effect on the rest of the atoms? or is there another factor. same thing for something like iodine? for heterogeneous molecules e.g. hexane, are the dispersion forces due to the electronegativity differences between carbons and hydrogens?

It could be other helium atoms, because they are moving and the 'chaos' of the situation could result in one being closer to anther than a third. And then a fourth makes it more 'perturbed'. And a fifth, etc. You are also correct the container walls will also matter. The field for describing this is statistical mechanics.

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Are electron orbitals in a helium atom degenerate?
In a hydrogen atom, the presence of only one electron allows various orbitals' energy states to be dependent only on the principal quantum number and not on angular momentum. Orbital degeneracy and all that fun stuff. (This has been answered many times on SE)
In helium, would this still be true? How come? I can't seem to find a satisfactory explanation.

The gist is that with two electrons the spin must either be the same or opposite. If the electrons are in the same orbit then the spins must be opposite.

The energy of orbitals for hydrogen atom is dependent only on the principal quantum number: $$E_{n}(\rm{eV})=-\frac{13.6}{n^2}$$
For atoms with more than one electron (all the atoms except hydrogen atom and hydrogenoid ions), the energy of orbitals  is dependent  on the principal quantum number and the azimuthal quantum number according to the equation: $$E_{n,l}(\rm{eV})=-\frac{13.6Z^{*2}}{n^2}$$
Where $Z^{*}$ is the  effective nuclear charge: $Z^{*}=Z-\sigma$.
where$Z$ is the atomic number and $\sigma (n,l)$ is the screen (shielding) constant and it depends on the principal quantum number and the azimuthal quantum number.
According to this, the atomic orbitals $2s$ and $2p$, that are at the same energy in the case of hydrogen atom, are no longer at the same energy for the atom of helium.

One view of the situation is that there is one 1s orbital which may house two electrons.  This, in turn means that the electron's spin is the property of the electron.
Another view of this situation is that there are two orbitals, a 1s spin up and a 1s spin down.  This implies the electron's spin is the property of the orbital.
Since orbitals don't really exist, noone may say which is a correct interpretation.  The first interpretation seems to dominate teaching.
There is a finite, greater than zero energy cost associated with aligning the electron's spins for or against the nuclear spins (indeed this is the basis for NMR).  However, the amount of energy associated with this is trivially small compared with the change in energy from 1s to 2s, that for all pragmatic purposes we make the simplifying assumption that $$\ce{\Delta E_{1s(spin down)->2s(spin down)} \approx \Delta E_{1s(spin down)->2s(spin up)}}$$
which effectively means there is no impact on the angular momentum and the energy states.
Another relevant point is that transitioning between energy levels while simultaneously 'flipping' spin is a quantum mechanically forbidden process, which has the pragmatic effect of being 'unlikely' and hence 'slow'.

I once had the same question as you and I did go to my chemistry teacher for an answer. This is his original response:
"In hydrogen, all orbitals with the same principal quantum number 'n' (1,2,3...) are degenerate, regardless of the orbital angular momentum quantum number 'l' (0,1...n-1 or s,p,d,). However, in atoms with more than one electron, orbitals with different values of l for a given value for n are not degenerate. Why is this? Surely the radial distribution functions are similar for hydrogen (in that there's still penetration of orbitals and so on). Or is it that orbitals with different values of l are degenerate for a value of n greater than would be occupied in that particular atom's ground state?
The answer is right there in your question. It's only the interaction of multiple electrons in an atom like He, Li, Be, etc. that makes the different angular momenta wave functions differ in energy.
Consider this. For the one electron system, why should a p or d orbital differ in energy from an s? What makes them differ?
In the multi-electron case, the p orbitals have different spatial extent, different angular components, so the electron density caused by an electron in that orbital will interact differently with the other electrons. In other words, you need to have more than one electron for the "shape" of the p and d and f orbitals to matter to the other electrons.
In the H atom, there's only one electron, so there's no electron-electron repulsion to differentiate the s, p, and d orbitals."
Hope this clarifies your doubt.


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Are these three molecules optical isomers?
In the structures below (2) vs (3) seem obvious optical isomers to me.
But what about (1)? Is that just a stuctural isomer (e.g. Butane vs Isobutane) or is there a more specific way to describe the relation of (1) vs (2) / (3)?

Alternative sketch of (2) & (3) based on the comment by @DGS below:

(2) and (3) look like the same molecule to me. I see the wedged and dashed bonds, but those should be up and down, and the way to flip that spot is to switch which is up and which is down.

@DGS You mean there's an error in the sketch I've used? Or you mean beta vs epi-beta are the same molecule?

I mean I don't think the wedge and dash is the appropriate way to show stereochemistry from that view. There are two different molecules, I just don't think the image properly represents them both.

@DGS Thanks! I've uploaded a different set of sketches. Do you feel this set conveys the appropriate sterochemistry?

Hmm... I don't think any of these are optical isomers of each other now that I look at them again. 2 and 3 both have three chiral centers and only one flips.

@DGS Ok. So maybe the right way to think about them all is that they are all just structual isomers of each other? Kinda like n-butanol, sec-butanol and ter-butanol?

Well, (2) and (3) are diastereomers. (1) isn't, but as Lightheart said, it may be a tautomer.

There is no exocyclic methylene in structure 1.  Is it not an optical isomer of 2 and 3.
Likely there is a pericyclic rearrangement between 1 and 2/3 which would formally render them tautomers.
2 and 3 are diastereomers.

So do you say 2 and 3 are optical isomers?

No, they are diastereomers.

Okay. Just wanted to make sure because you hadn't made it real clear. The "epi" was a good indication, but the first time I looked at it, I missed the two carbons attached to the bridging carbon.


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Can H4O 2+ form?
Like $\ce{NH4+}$ ,Is there any possibility of formation of $\ce{H4O^{2+}}$ (of tetrahedral structure)?

My theory is: it can be formed by osmosis setup where heavy acids like $\ce{H3PO4}$ or $\ce{H2S2O7}$ are passed through a semipermeable membrane like gelatinous copper ferrocyanide SPM, which allow only small ions like H+ to pass.
This then could form $\ce{H3O+}$, then its lone pair could attack the excess $\ce{H+}$ in solution to form forming $\ce{H4O^{2+}}$.

Semipermeable membranes just don't work like that. You can't separate cations from anions this way, and that's a very, very huge "can't" (as in "a mosquito can't lift the Empire State Building").
The possible (?) existence of $\ce{H4O^2+}$ in extreme conditions is another story.

see http://chemistry.stackexchange.com/questions/24342/what-is-the-ka-of-oh-and-kb-of-h3o

Osmosis can reject anions as stated in rejection of anions by osmosis  @IvanNeretin

Sure, but it would also place another ions instead of them.

The existence of $\ce{H4O^{2+}}$ has been inferred from hydrogen/deuterium isotopic exchange monitored through $\ce{^{17}O}$ NMR spectroscopy in the most extremely acidic condensed phase superacid we can make, fluoroantimonic acid ($\ce{HF:SbF5}$ or $\ce{HSbF6}$). It seems that even the slightly weaker but still very much superacidic magic acid $\ce{HSO3F:SbF5}$ is not strong enough to effect measurable protonation of $\ce{H3O+}$ into $\ce{H4O^{2+}}$. It likely also can exist in the gas phase by exposing water to bare protons, and might have some importance in astrochemistry. Relatedly, $\ce{H4S^{2+}}$ also has been inferred to exist through the same type of measurements, and is easier to make, existing in magic acid mixtures.
A nice source for your superacidic needs is books by George Olah, the father of superacidic chemistry. In particular, Onium Ions has a section on the protohydronium dication $\ce{H4O^{2+}}$ on page 435. There are many other curious cations analyzed such as $\ce{CH6^{2+}}$ and even $\ce{CH7^{3+}}$. When it comes to superacids, nothing is sacred!

Minor point.  Fluoroantimonic acid is not $\ce{HSbF6}$.  The fluoroantimonate anion is such a weak proton acceptor that it never gets protonated at all; therefore the acid forms only ionic species.  See here for a description of these ionic species.


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E / Z Isomers GC peaks for identification; Would GC-MS help?
Does a GC analysis typically seperate E / Z isomers? I'm reading a paper that confirms that the product it synthesized was indeed the (E,E) isomer by coinjecting a standard sample on the GC.
My question is, had they actually made an (Z,Z) or (E,Z) or (Z,E) would a typical GC show two different peaks?
In other words, for isomer verification of this nature is a GC co-injection a sufficient control? Or would it be more prudent to co-inject standard samples of the other samples too in other runs to demonstrate that those indeed elute differently?
The paper also claims they have verified the compound as a (E,E) by a GC-MS analysis but would (E,E) (E,Z) etc. show different patterns on a GC-MS?

These isomers would be significantly different. I don't think the diphosphate would be seen by GC, but the alcohols should separate. In the past week, I've seen several chromatograms where geraniol and nerol separated just fine.


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Is the chemical structure of an amide bond (-CONH) or (-CONH2)?
The Dictionary definition of an amide is

an organic compound obtained by replacing the −OH group in acids by the −NH 2 group.

So from this, I deduced that the bond would be CONH2.
But when I searched around, I found some images that showed the bond only had one hydrogen atom, like so:

Which is the correct bond structure? I'm looking specifically at the condensation polymerization of diamine and diacarboxylic acid to form nylon.
And also, I see online that lots of bond structures (like NH2 and COOH) are prefixed with a dash, so they'd appear like -NH2 and -COOH. Is this just a personal preference, or is there a more elaborate reason behind this?
I'm currently doing the GCSEs.
Thanks

Three types of amide - 1° 2° and 3°. -CONH2 is 1° amide, in the image it is 2° amide. If you take out that one hydrogen too you will get a 3° amide

Amides are derivative of carboxylic acids. It consists of 2 parts, and the bond between carbonyl carbon and nitrogen of ammonia is called is amide bond.
The bond structure of 1° amides is as shown in the figure.

What you have shown in the question is bond structure of 2° amide, which can be obtained by removing 1 hydrogen from 1° amide.

Now if you observe structure of nylon, you will see there are n numbers of 2° amides.

And also, I see online that lots of bond structures (like NH2 and COOH) are prefixed with a dash, so they'd appear like -NH2 and -COOH. Is this just a personal preference, or is there a more elaborate reason behind this?

The dash means it is Functional Group like -R means hydrocarbyl functional group, -COOH means carboxylic acid functional group, etc.

One more question, though: why is the coefficient of the product H2O 2, as opposed to the n for every other reagent and product?

@CommandoGeek one H from each both end of $\ce{H_2N-R'-NH_2}$


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Using Hydrogen Peroxide to clean toilet
I want to use lysol hydrogen peroxide multipurpose cleaner for cleaning my toilet. Is this a dangerous course of action?

yes i can see multiple link showing it can be use. you can check out this and this

Lysol Multipurpose Cleaner with hydrogen peroxide contains only 0.5% hydrogen peroxide. This site (using-hydrogen-peroxide.com) suggests using up to 35% hydrogen peroxide for cleaning toilets. This hydrogen peroxide website (and who knew it had one) may be biased, but that's quite a margin between the product you want to use and what the website says is safe. I would feel pretty comfortable using this Lysol product for cleaning toilets (except that it may actually not be strong enough!).


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Why isn't an Acid-base titration curve linear?
Sorry if this is a stupid question, but why is a acid-base titration curve not linear? As you add more acid solution to a basic solution, shouldn't the pH of the basic solution slowly go up in proportion to the amount of acid added? Why is there such a sudden increase?

@DGS if you just added acid to a non-reactive solvent it would be linear, but if you are titrating a base then no.


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Why is Kc not affected by change in pressure?
Why is $K_c$ not affected by change in pressure? I know the mathematical explanation, but I don't really understand the reason when only looking at $K_c$. The explanation I know is with the reference to $K_p$, but is it possible to explain without $K_p$?

Kc and Kp are.constants for a reaction. They dont get affected by changing volume or pressure. They tell you the value of Q at the equilibrium condition.

I agree. This sounds like K vs. Q confusion. When the pressure, volume, concentrations, whatever change, then Q changes. However, only one value of Q defines equilibrium. The system responds to reestablish equilibrium to that $Q=K$ once again.

Equilibrium is reached when the forward and reverse reaction rates are equal. These rates are determined by the frequency of successful collisions, which is turn is determined by concentration (seen in the rate laws), temperature, activation energy, and how often molecules collide with the correct orientation (together these last three determine the rate constant, $k$, as shown in the Arrhenius equation).
The equilibrium constant, $K_c$ is the ratio of the rate constants, so only variables that affect the rate constants can affect $K_c$. Pressure doesn't show in any of these relationships.
$$K_c=\frac{k_\text{forward}}{k_\text{reverse}}=\frac{[\text{products}]^y}{[\text{reactants}]^x}$$
$$\text{rate}_\text{forward}=k_\text{forward}[\text{reactants}]^x$$
$$\text{rate}_\text{reverse}=k_\text{reverse}[\text{products}]^y$$
$$k=A\mathrm e^{-\frac{E_\mathrm a}{RT}}$$
It may also be useful to think about different ways pressure can be changed.

The volume of the reaction can be changed. This will affect concentrations of reactants and products, and the reaction will likely have to shift left or right to reestablish equilibrium. If volume were decreased, the rates of the forward and reverse reaction will each increase due to higher concentration. If the exponents in the rate laws are different, the amount of increase for each will be different. Although the rates change, the rate constants do not, so $K_c$ doesn't change either.
The temperature can be changed. This will affect the rate constants of the forward and reverse reactions according to the Arrhenius equation. An increase in temperature will increase $k_\text{forward}$ and $k_\text{reverse}$, but by different amounts because their activation energies are rarely equal. This changes $K_c$, which is the ratio of the two rate constants.
More gas can be added without changing the volume of the container.
a. If the gas is one of the reactants or products, this would affect one of the concentrations, and the reaction will have to shift to reestablish equilibrium. Again, this changes one of the rates, but does not affect the rate constants, so $K_c$ is unaffected.
b. If the gas is inert and doesn't react with anything involved in the equilibrium, the concentration of each reactant is unchanged. Neither the rates nor the rate constants are affected, so the system should still be at equilibrium and $K_c$ is unaffected.

$K_p$ is defined in terms of free energies of the gaseous substances in their standard states at 1 atm pressure so must be independent of pressure, i.e. is a constant.  With concentrations we use $c_i=n_i/V$ for species $i$ and for example, if $K_c=c_\ce{A}^ac_\ce{B}^b/(c_\ce{C}^cc_\ce{D}^d)$ then for an ideal/perfect gas $K_c=K_p(RT)^{-\Delta n}$ where $\Delta n=a+b-c-d$ and so is independent of pressure just as is $K_p.$
So the explanation is that $K_p$ is defined that way and so $K_c$ is defined similarly as it measures the same equilibrium just expressed with different quantities, i.e. concentration instead of pressure.
What is confusing is that a typical expression could be $K_p=4a^2p/(1-a^2)$ where $p$ is total pressure and $a$ degree of dissociation. If $p$ changes $K_p$ does not but instead $a$ must to keep $K_p$ constant.

TLDR:  While $K_p$ is pressure-independent for ideal gases, it is pressure-dependent for real gases.  To get a pressure-independent constant for real gases, we need to substitute the fugacities for the pressures, giving us $K_f$.  This is not merely definitional. The pressure-dependence of $K_p$ for non-ideal reacting systems has real-world consequences for the pressure-dependence of the relative amounts of reactants and products at equilibrium.
As porphyrin and Buck Thorn nicely explained, once you establish that $K_p$ is independent of pressure then, assuming an ideal system, it necessarily follows that $K_c$ is indepedent of pressure as well.
But I'd like to take a closer look at the pressure-independence of $K_p$. The reason $K_p$ is pressure-independent is not directly because it is defined as such. Rather, the reason $K_p$ is pressure-independent is because it's assumed that the gases in the equilibrium expression behave ideally.  That's the fundamental stipulation; the pressure-independence is a consequence of this.
[I.e., an assumption of ideality is needed not only to obtain pressure-indepedence for $K_c$, but also to obtain pressure-independence for $K_p$.]
But then you might ask:  What if the gases don't behave ideally?  Will $K_p$ be independent of pressure?  Here the answer is no.  Consider the following generic reaction:
$$\ce{aA(g) + bB(g)<=>cC(g) + dD(g)}$$
Then:
$$K_p=\frac{p_C^c p_D^d}{p_A^a p_B^b}$$
Now consider that the fugacity, $f$, is related to the pressure through the fugacity coefficient, $\gamma (T,p)$: $f = \gamma (T,p)p$. Note that $\gamma$ is both pressure- and temperature-dependent*.
[*This is stated explicitly in many physical chemistry textbooks, e.g. see bottom p. 276 of Thomas Engel and Philip Reid, Thermodynamics, Statistical Thermodynamics, & Kinetics (3rd Edition), Pearson, 2012.
For more on fugacity, see my answer at:  Fugacity vs compressibility (note that there I use $\varphi$ instead of $\gamma$ as the fugacity coefficient, since there I was following Wikipedia's notation, while here I'm following that in Engel & Reid).]
Substituting $p_i = \frac{f_i}{\gamma_i (T,p)}$ into our expression for $K_p$, we have:
$$K_p=\frac{\Big(\frac{f_C}{\gamma_C (T,p)}\Big)^c     \Big(\frac{f_D}{\gamma_D (T,p)}\Big)^d }{\Big(\frac{f_A}{\gamma_A (T,p)}\Big)^a    \Big(\frac{f_B}{\gamma_B (T,p)}\Big)^b} = \frac{f_C^c f_D^d}{f_A^a f_B^b} \frac{\Big(\frac{1}{\gamma_C (T,p)}\Big)^c     \Big(\frac{1}{\gamma_D (T,p)}\Big)^d }{\Big(\frac{1}{\gamma_A (T,p)}\Big)^a    \Big(\frac{1}{\gamma_B (T,p)}\Big)^b} = \frac{f_C^c f_D^d}{f_A^a f_B^b} \frac{\gamma_A (T,p)^a   \gamma_B (T,p)^b}{\gamma_C (T,p)^c  \gamma_D (T,p)^d}$$
Substituting
$$K_f=\frac{f_C^c f_D^d}{f_A^a f_B^b}$$
We have:
$$K_p = K_f \frac{\gamma_A (T,p)^a   \gamma_B (T,p)^b}{\gamma_C (T,p)^c  \gamma_D (T,p)^d}$$
Since the $\gamma_i$'s are pressure-dependent, we can see that $K_p$ will be pressure-dependent as well, if the reaction involves at least one real gas.  [Note:  $K_f$ is pressure-independent.  Thus there is no compensating pressure-dependence from $K_f$ that cancels out the pressure-dependence of the  $\gamma_i$'s.]
To simply illustrate the real-world physical consequences of this, consider this reaction:
$$\ce{A(g)<=>B(g)}$$
We can write:
$$K_x = \frac{x_B^b}{x_A^a}=K_p \Big(\frac{p}{p^{\circ}}\Big)^{a-b}$$
Since $a=b=1$, this reduces to:
$$K_x = \frac{x_B}{x_A}=K_p$$
But we've already established that, while $K_p$ is pressure-independent for ideal gases, it is pressure-dependent for real gases.  Hence $K_x$ will be as well.
Thus, at equilibrium, the relative mole fractions of A and B will be independent of pressure if A and B are ideal.  But they will vary with pressure if A and B are non-ideal.  This directly illustrates the consequences of varying the pressure of non-ideal reacting systems.

Actually, I think you are sort of wrong (but I'm not sure, it's subtle, as your answer points out): it's true that ideality is important, but it is baked into $K_p$s definition through the expression $\Delta G^\circ=-RT\ln K_p$. Both $K_p$ and $\Delta G^\circ$ together define ideal behavior relative to an ideal standard state. Once those definitions are in place, the substances are allowed to behave nonideally, but we use fugacities (for instance) instead and continue using $K_p$.

I.e., instead of starting with $\mu(T,p) = \mu^o(T) + RT ln \frac{p}{p^o}$, we start with $\mu(T,p) = \mu^o(T) + RT ln \frac{f}{f^o}$  (see eqns. 7.13 and 7.14, p 175, Engel & Reid 3rd Ed.)

Well, I saw your point about the importance of ideality in your answer, but now your comment seems to play down that importance, so I am a little confused about how you see things. To determine a fugacity you need a measure of how far you are from ideality, so presumably you have defined an ideal state for the system (for gases, allowing the pressure to go to zero). If you did things right, you obtained $\Delta G^\circ$ by extrapolating to this ideal state.

I'll have to sit down and think some more about this, obviously. Thanks for your replies!

@BuckThorn  I edited my answer to more explicitly demonstrate why $K_p$ is pressure-dependent for real gases.

Elaborating on @porphyrin's answer, for an ideal system $K_c$ does not depend on pressure because if the temperature is constant then the concentration of a component depends only on its partial pressure $p_i$ and not on the total pressure P:
$c_i=\frac{p_i}{RT}$
In other words, at constant T a number concentration can be used as a proxy for the partial pressure.
On the other hand the mole fraction $\chi_i$ of a component depends on both the partial and the total pressure:
$\chi_i=\frac{p_i}{P}$
Therefore the equilibrium constant based on mole fraction is a function of pressure. Since % yields are based on mole fractions, varying the pressure is a potentially important way of controlling the yield of a reaction, as Le Chatelier's Principle reminds us.

You might line to add that $K_x=K_PP^{-\Delta n}$ so if $\Delta n \ne 0$ the mole fraction at equilibrium will depend on total pressure $P$ even though $K_P$ does not.

@porphyrin Thanks, I left that out as an optional exercise for the reader. Also, I thought of leaving a comment under your answer but figured there was space for another perspective.


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Is copper an element or molecule?
I am not sure is copper present in form of molecules? Can someone please describe about it? I searched on Internet and also asked many people but am not getting any clear answer.

In short, no. Copper is a metal, and metals normally don't form molecules by themselves.

related http://chemistry.stackexchange.com/questions/16261/metal-compounds-that-bond-covalently

also http://chemistry.stackexchange.com/questions/49076/do-there-exist-metals-that-are-composed-of-molecules

Metals form metallic bonds, which are a thing of its own. An easy way to rule out molecular bonding is if the boiling point is high; most metals have high boiling points, because metallic bonding is much stronger than the forces between individual molecules (such as $\ce{O_2, CH_4}$) that hold them together.

Not all, remember Hg? Be careful with generalizations.

Diamond has covalent bonding and has a very high melting point, as do many other network solids.

Copper is a metal, not a molecule. Copper's makeup is shown in the diagram below, 29 protons and 35 neutrons with one electron in the outermost ring.

I looked at https://en.wikipedia.org/wiki/Copper to see if I had any invalid or incorrect information, please let me know if I missed something. Hope this helped!

Well, if I were asking about the bonding of an element, I'd specify the temperature and pressure (ranges) I was interested in. Ground state. STP. Copper is metallic. I won't go into the discussion about whether a crystalline metal is or is not covalent. A crystalline metal by definition shares electrons broadly, and shouldn't be characterized as having covalent bonds, but that's not necessarily a clear-cut issue. At STP it (the pure element) exists as face-centered-cubic crystallites. Some natural crystals of copper are known. It (the metal) is known to exist naturally, but by far most copper on Earth is compounded in minerals (ores). I found one reference to Cu2 existing in the vapor state (there could be many more, IDK). This would be a molecule. All molecules are composed of atoms. All atoms are elements. Molecules are covalently bound atoms. Molecules are generally considered finite assemblies of atoms. Is copper an element? Yes. Can copper exist as a molecule? Yes. What is the form of the element at stp? Metal. (STP = standard temperature and pressure)


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Volume of one equivalent of a gas
How do we define an equivalent when we are not talking about any particular reaction? Like, we say that equivalent volume of Oxygen is 5.6L, so we have taken the n-factor to be 4, and in CO2 we take n-factor = 2.
What is the defining factor for an equivalent here?

In short, we don't.

So how do we define the volume occupied by one equivalent of a gas?

There is nothing to define: it is the volume occupied by one equivalent of gas. What is the equivalent, depends on the reaction.

I see, so we can't define the volume of an equivalent of a gas in absolute terms, that is without relating to a reaction? Thanks

The meaning of the word "equivalent" is relative to something. Fortunately, as a chemical term, that is still true. An equivalent is more completely called "a reaction equivalent" and is relative to the reaction. There are a few standard reactions (which are typically industry or discipline based) and it might be used usefully in limited (well constrained) situations. The problem is, especially when dealing with reactive gasses, that the words "uncontrolled" and "detonation" might also occur when explaining that you chose the wrong standard - the correct standard reaction is NOT always obvious or determinable in context. So, my best advice is that when someone uses "equivalent" ask what reaction, specifically, they are referring to. And when using the word yourself, always note the context. Examples: NCO equivalents, H(0) equivalents, H(+) equivalents may all refer to the same compound but would have different values.

You do not need to define the volume equivalent of a gas in relation to a reaction: you define it in relation to the number of molecules in the gas (or, more conveniently, the number of moles which is a convenient unit to measure the number of molecules).
Since ideal gases occupy the same volume if they have the same number of moles of their constituent molecules, volume (at constant temperature and pressure) is a proxy for the number of moles.


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Is synthesis of nitrobenzene possible without nitric and sulfuric acids?
Is nitration of benzene possible without using the mixture of concentrated nitric and sulfuric acids $(\ce{HNO3}$ and $\ce{H2SO4})$?

related http://chemistry.stackexchange.com/questions/29150/nitration-of-pyrrole

It's getting on for 3 decades since since I've done organic, so a proper organic chemist can probably give a more complete answer, but yes, there are other ways to do this. In the $\ce{HNO3}$/$\ce{H2SO4}$ system the "real" nitrating agent is nitronium, $\ce{NO2+}$, so anything that can form that will nitrate benzene. One example is simply to use a nitronium salt such as nitronium tetrafluoroborate, $\ce{NO2+ BF4-}$. As for others I would be surprised to find that something like $\ce{NO2F}$/$\ce{BF3}$ doesn't work, but I don't know for certain - it's been a looooong time!

Acetyl nitrate, AcONO2, is also used - the reagent is pre-formed by reaction of nitric acid and acetic anhydride

For some substrates, nitric acid plus nitrous acid works. In this system the generated electrophil is not $\ce{NO2^+}$, but $\ce{NO^+}$ formed from the nitrous acid accepting protons from the stronger nitric acid. If the substrate is sufficiently reactive, as with the parent benzenoid compound benzene, then it can be attacked by the $\ce{NO^+}$. Then the nitroso group is oxidized by the nitric acid (regenerating the nitrous acid, which is thus acting as a catalyst).


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Physical properties of geometric isomers
I've learnt that cis isomers have higher boiling point than the trans ones.
Why does not that rule apply on Fumaric and maleic acids?

In the case of fumaric acid, intermolecular hydrogen bonds are the reason beyond its higher melting point compared with the melting point of maleic acid ( where intramolecular hydrogen bonds dominate).

can you tell me more about this.

In the case of fumaric acid, the molecules are linked to each other via hydrogen bonds. To melt the fumaric acid, one has to provide additional  energy to break up hydrogen bonds before the solid turns into liquid. I hope it's clear now.


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Why does (3E)-3,4-dimethylhex-3-ene yield a meso dihalide when reacted with Br2/CCl4 at room temperature?

The answer is B. But why? What makes it so different than the others?
I know that it attacks from the back (anti addition). But can't it do that for all and result in a meso compound?

A meso compound is a molecule with multiple stereocenters that is superimposable on its mirror image. Meso compounds are easily identified by their internal plane of symmetry.
Using this knowledge, and given that $\ce{Br2}$ undergoes anti addition to alkenes, there are only really two answers that could possibly be meso compounds, B and D. Let us consider the products:

Notice how only B has an internal plane of symmetry, and is therefore a meso compound.


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What type of standard for HPLC capsaicin quantification?
I'm wondering which type of standard would be the most suitable for the quantitative analysis of capsaicin (8-Methyl-N-vanillyl-trans-6-nonenamide) containing products?
Sigma offers the following types:

analytical standard
United States Pharmacopeia reference standard
European Pharmacopoeia reference standard
Pharmaceutical secondary standard

While analytic standard entry gives a description (analytical standard for food analysis), assay (99.0+%) and HPLC suitability, no such data is displayed at the product pages of other type of standards.
Regarding my questions let's assume two separate situation here:
a) funds are unlimited
b) funds are limiteted


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Why is including phase info in chemical reactions important?
I would like to ask why do we have to include the phase change in a chemical reaction e.g from  solid to aqueous or liquid. Why is it important actually?

You don't have to include it, but it is often useful - for example this: $$\ce{Ag+ (aq) + Cl- (aq) -> AgCl (s)}$$

It gives a complete chemical and physics picture. Would you think that dropping mention of one of the reactants help understand all the chemistry that is going on in a given reaction? Because different phases have different Gibbs free energies, and that information is important to understanding what is going on, the phases should, in general, always be mentioned.


text:
What actually happens when citric acid is added to a highly alkaline soap solution?
A few days ago I made an experiment. I diluted a few small chunks of bar soap with 20 ml of water. I then measured the ph of the solution which was around 10. I then added a tiny amount of citric acid. I immediately noticed that the solution stopped foaming. I measured the ph again and it was around 5.
I am interested what actually happened when I added the citric acid? Since the solution stopped foaming and there were no more bubbles even with continuous stirring was the final solution still soap? How does the ph of a solution affect its detergent abilities/properties?
Thank you.

Bar soap contains sodium salts of fatty acids, e.g. sodium stearate. The anions of the fatty acids have hydrophilic groups (the carboxylate group $\ce{COO-}$) as well as a hydrophobic tail ($\ce{R}$). This amphiphilic character of these anions is responsible for them to act as a surfactant.
If citric acid is being added to a solution of soap in water the carboxylate groups are being protonated.
$$\ce{H+ + R-COO- -> R-COOH}$$
The fatty acids are poorly soluble in water. They precipitate and the solution loses its ability to act as a surfactant. The foam collapses.

Yes, free fatty acids tend to dimerize and form lipidic drops. If there are any medium-chain fatty acids in you soap, you should notice it smells like HELL. Did the same "experiment" the other day.

The difference is especially striking if you pour some acid on a solid bar of soap. All of a sudden, the thing that could wash out pretty much everything turns into a thing that can wash out pretty much nothing, and itself leaves greasy stains that need to be washed out.

Lol, this question reminds me of when I was a kid and thought it would be cool to put orange juice and drain cleaner in a plastic bag.  The mixture turned bright yellow, melted the bag, and got me in trouble for making a mess when I was in fact trying to clean one.   I was working on an answer, but it turned out to be nearly identical to this one so I scrapped my answer.

@julien Aren't most soaps these days sulfonic acids?

@SendersReagent Of course, sulfonates, sulfates or other polar groups. But plain tradtionnal Aleppo soap is nothing but carboxylates and oil (the lipid drops might as well be unsaponified laurel oil)


text:
Why do liquids vaporise spontaneously at temperatures below boiling point despite ∆G being positive?
We say that for the equilibrium
$$\ce{A(l)<=>A(g)}$$ the ∆G is zero at its boiling point (the thermodynamic definition of boiling point), and becomes positive at temperatures below the boiling point. However every liquid has a characteristic vapour pressure at all temperatures. How does this happen even though the ∆G has become positive thus the reaction has lost spontaneity?

If a reaction has a positive $\mathrm{\Delta G}$ does that mean no products whatsoever can be formed? Similarly, if vaporization below the boiling point has a positive $\mathrm{\Delta G}$, does that mean no liquid at all can vaporize? Recall the relation between $\mathrm{\Delta G}$ and the equilibrium constant.

There are two considerations about why liquids evaporate but only one of them has to do with the Gibbs free energy of vaporisation.
There is an equilibrium (heavily biased in favour of the liquid not the vapour) at temperatures well below the boiling point. In equilibrium (which is what the ΔG measures) there is enough thermal energy to put a small amount of liquid into the vapour phase but most liquid will remain liquid. This is the correct analysis in a closed vessel. The liquid doesn't continue to evaporate once the equilibrium is reached.
But, often, you don't observe liquids in closed vessels. This explains why the liquid continues to evaporate. The point is that, when the vapour can be removed from the vessel containing the liquid, the normal equilibrium is not established and the liquid can slowly evaporate (at least as long as the surrounding environment has a lower vapour pressure for the liquid than the equilibrium vapour pressure).
In open systems the simple thermodynamic numbers don't correctly describe the long term behaviour because equilibrium isn't what happens.
If you think about what is happening at the molecular level this is easier to understand. Energy is not evenly distributed among the molecules of the liquid. On average the molecules don't have enough energy to go from liquid to vapour, but some of the molecules have more energy than others. So some of them have enough energy to escape the liquid phase. And the energy will distribute among the vapour molecules leaving some with so little that they condense back to liquid.
Overall equilibrium is established when the number of low energy vapour molecules condensing is equal to the number of high energy liquid molecules evaporating. Just because there isn't enough energy to boil all the liquid doesn't mean some of the molecules can't become vapour spontaneously.

Start with a closed container with a quantity of liquid water, and no vapor. Maintain the container at room temperature. What will happen? If the temperature does not change, the overall Gibbs free energy will not change in equilibrium. Now, $G = H - TS$, and at constant temperature $\Delta G = \Delta H - T\Delta S$.
Converting liquid (take as the reference state) to vapor costs enthalpy, the $\Delta H$. But, by allowing vapor to occupy the empty space in the container you have gained a great deal of entropy, $\Delta S$. The first water molecule to go in to the vapor gains a huge amount of entropy - it has the entire volume to roam around in - overwhelming any enthalpy required. The reaction of liquid -> vapor is, in fact, highly spontaneous because of this! But, the next little bit of vapor gets a smaller entropy gain - there is already vapor occupying the head space in the closed vessel. And the next little bit gets even less entropy. Yet, it keeps costing the same amount of enthalpy to convert the liquid to vapor.
As more and more liquid goes in to the vapor phase, the additional entropy gain decreases, while the enthalpy cost remains the same. Eventually the reaction reaches an equilibrium point when $\Delta G = 0 = \Delta H - T \Delta S$ for the overall reaction. The $\Delta H$ is the total enthalpy required to make the equilibrium vapor concentration, and the $\Delta S$ is the total entropy gained by doing so. You have converted $x$ moles of liquid in to vapor, and it is now in equilibrium with the remaining $y$ moles of liquid.
Perhaps that gives you a better perspective.

$\Delta G$ is zero at all temperatures and pressures at which the vapor and liquid are at equilibrium with one another (i.e., all points along the vapor pressure curve), and not just at the atmospheric boiling point.

Then that has to mean that vaporisation can never be spontaneous because negative ∆G is the condition of spontaneity which is not being met by your arguement.

In evaporation of, say, water into a room, the partial pressure of the water vapor in the room air (away from the interface) will be lower than the equilibrium vapor pressure.  So the $\Delta G$ in going from the liquid to the partial pressure in the room will be negative.  At the interface, the partial pressure of the water vapor will be only slightly lower than the equilibrium vapor pressure, so evaporation can still occur.  The transport process involved is diffusion of the water vapor from the interface to the bulk air.

You wrote that ∆G of vaporisation will be negative in going from the liquid to the room. That is exactly my question: isnt the ∆G positive at temperatures below boiling point, and ∆G being the criteria for spontaneity, shouldnt the reaction lose its spontaneity at temperatures below the boiling point. Kindly answer my question directly, and dont take it in circles. No offence.

Inital state: liquid water at temperature T, Final state: water vapor in room air at partial pressure p < $p_{sat}(T)$ and temperature T. $ΔG=RT\ln(p/p_{sat}(T))$<0

Where does it say that $\Delta G$ is positive at temperatures below the atmospheric boiling point in going from saturated liquid to saturated vapor?  Please provide a reference.

If you don't believe that the gibbs free energy of the saturated liquid is equal to the gibbs free energy of the saturated vapor at all points along the vapor pressure curve, get yourself a set of steam tables, which gives the values of v,h, and s for both the saturated liquid and the saturated vapor along the curve, and calculate g = h - Ts for both the saturated liquid and the saturated vapor at a selection of temperatures, and compare the values you get.

Before diving deep into thermodynamics, make several thought experiments.

Open flask with liquid at a room temperature
Closed flask with a liquid at a room temperature
Ideally thermally insulated closed flask with liquid, starting at room temperature

The vapor pressure corresponds to the equilibrium between liquid and it's vapor (presumably in air). Alternatively, you can think of the chemical potential of the substance in both phases, which should be equal in equilibrium.

Please can you explain this through Gibb's free energy.

For the Gibbs energy of multiphase system equilibrium see e.g. http://pruffle.mit.edu/3.00/Lecture_25_web/node2.html


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Bond angle order of SF2, OF2, HOF

Arrange the following componds in the increasing order of bond angle.
  $$\ce{HOF,~OF2,~SF2}$$

I know that bond angle of $\ce{SF2}$ will be less than the bond angle of $\ce{OF2}$ because of larger size of sulfur atom.
How can I compare $\ce{HOF}$ with the other two? Since $\ce{H}$ is less electronegative than $\ce{O}$,  the bond pair of $\ce{O - H}$ bond will be more closer to $\ce{O}$. The opposite happens for $\ce{O - F}$ bond. In $\ce{SF2}$ both bond pairs are near $\ce{F}$ so the bonds can come closer to each other.

This answer explains why $\ce{SF2}$ will have a bond angle close to $90^\circ$ and $\ce{OF2}$ will have a bond angle closer to $109.5^\circ$. For similar reasons, $\ce{HOF}$ will have a bond angle closer to $109.5^\circ$. However, comparing $\ce{HOF}$ and $\ce{OF2}$, you need to compare both the size difference of $\ce{H}$ and $\ce{F}$ and their electron-density (not just electronegativity).

@BenNorris it given in Wikipedia that HOF bond angle is 97.2 and that in SF2 is 98.

Now, that I would not have expected. Interesting.

@SafdarFaisal I think this post might already have the answer to the question: https://chemistry.stackexchange.com/questions/119034/why-is-the-binding-angle-of-hof-lesser-than-that-of-h2o-despite-the-lone-pairs

@ShoubhikRMaiti, that answer is based on speculation and doesn't have a reference attached to it.  Note: $\ce{OF2}$ is predictable, but the trend comparing HOF and $\ce{SF2}$ is not that easily seen. Bounty will be awarded if you can  cite a reference so as to why the bond angle of HOF is unexpectedly  low..

When considering bond angles of molecules of main group ($\ce{\angle A-X-B}$) there are several different factors to consider:

The row of the the central atom $\ce{X}$ determines the energy difference between the valence s and p orbital, which determines the extent of mixing (or hybridisation if you wish)
The sizes of A and B, because steric repulsion, A and B would want to stay away from each other the bigger they get.
Electronegativities of the A and B. More electronegative atoms would prefer to have a bond with higher p-character. (This is explained in VSEPR model with the bond pair being drawn away from the central atom)
Other forces betwen A and B

The bond angles of the molecules mentioned in question are:
$$\begin{array} {c|c}\hline  & \text{bond angle} \\ \hline \ce{OF2} & 103.2^\circ \\ \hline \ce{SF2} & 98^\circ \\ \hline \ce{HOF} & 97.2^\circ \\ \hline  \end{array}$$
The difference between $\ce{OF2}$ and $\ce{SF2}$ would be usually explained with the fact that oxygen is in 2nd period so, the energy difference between $\mathrm{2s}$ and $\mathrm{2p}$ is low which means it is $\mathrm{sp^3}$ hybridised. (Or at least close to $\mathrm{sp^3}$). Whereas sulfur is in the 3rd period so, the energy difference between $\text{3s}$ and $\text{3p}$ is high which means it is almost unhybridised and the bonds are formed by almost pure $\text{3p}$ (which is why the angle is closer to $90^\circ$).
The bond angle of $\ce{HOF}$ is however, unusally small. This is difficult to explain and is usually attributed to the electrostatic attraction between $\delta+\;$ of H and $\delta-\;$ of F, which pulls the H and F close together.
From Chemistry of the Elements [1],

Spectroscopic data establish a nonlinear structure with $\ce{H-O}$ $\pu{96.4 pm}$, $\ce{O-F}$ $\pu{144.2 pm}$, and bond angle $\ce{H-O-F}$ $97.2^\circ$: this is the smallest known bond angle at an unrestricted O atom (cf. $\ce{H-O-H}$ $104.7^\circ$, $\ce{F-O-F}$ $103.2^\circ$). It has been suggested that this arises in part from eletrostatic attraction of the 2 terminal atoms, since NMR data lead to a charge of $\sim+0.5e$ on H and $\sim-0.5e$ on F.

So, the unusually small bond angle is likely from the electrostatic attraction. This type of effect is not just limited to $\ce{HOF}$, it is also found it other molecules. For instance, in hydrogen peroxide $\ce{H-O-O-H}$, the $\ce{H-O-O}$ angle is $94.8^\circ$ which is quite lower than the ideal tetrahedral angle, and has been attributed the electrostatic attraction between $\ce{O}$ and $\ce{H}$.[2]
As an aside, it should be noted that the above values of bond angles were measured in the gas phase. In solid phase, $\ce{HOF}$ has a slightly higher bond angle, at $101^\circ$.
References-

N. N. Greenwood, A Earnshaw, Chemistry of the Elements, 2nd ed., Butterworth-Heinemann, Oxford, 1998
E. Wiberg, A. F. Holleman, N. Wiberg, Inorganic Chemistry, Academic Press, 2001

I looked up the net and found the following order-
HOF(97.2°)<SF2(98°)<OF2(103°)
(Of course, the values might not be exact but this is what I found on the net.)
Now, my POV-
Yes, your analogy that bond angle of OF2> SF2 is correct. It's a direct result of Drago's Rule.
The electron cloud in OF2 will be shifted much towards the Fluorine atom. Bond pair-Bond pair repulsions are larger if the electron cloud is nearer to the central atom. But the correct reason is the distance between the electron clouds, not the "near the central atom, away from it" analogy.
In the HOF molecule, the cloud is more towards the central atom Oxygen in the H-O bond and towards F in the O-F bond. So the distance(very misleading term tho) between the two electron clouds is a bit larger than in the case of OF2. A larger distance means lesser repulsions and a smaller bond angle.
But the catch here is that the bond angle of SF2 is slightly bigger than that of HOF. This is because the effect of electron cloud repulsions ( both electron clouds in both S-F bonds shifted towards Fluorine) and the size of sulfur atom ( Drago's rule) work against each other. Also, Drago's Rule doesn't apply much when the Electronegativity of the surrounding atom exceeds 2.5 which is the case with Fluorine ( 4- on the Pauling Scale).
You should also not forget the weak but existent back bonding in the S-F bond due to the presence of vacant d orbital in the Sulphur. The double-bondish character leads to a bit stronger bond pair-bond pair repulsions. Why did I call the back bonding weak? It's because the Sulphur atom in SF2 has two lone pairs that repel the electrons flowing in due to the back bond. Also adding on to that point, the 2p-3d back bond has appreciable overlap (more than 2p-3p) but it's still weak.
Drago's Rule is explained by the unfavorability of hybridization - the energy required for hybridization is much larger than what will be obtained after hybridization. In more EN atoms(>2.5) the net charge on the central atom is more positive which reduces the size of the central atom. Thus, the sizes ( read energies) of s, p, d orbitals become similar and hybridization is favorable only when atomic orbitals involved have similar energies.

99% of the time backbonding is not required, and not valid anyway. Arguments based on s/p hybridisation should suffice. Have you come across any evidence that a 2p->3d bonding scheme is valid? Are there any studies to show whether the 3d orbital is occupied and to what extent?

Valid argument. Didn't think on those lines. Sure will look into it.

$\ce{H-O-F}$ has a lower bond angle than both $\ce{F-S-F}$ and $\ce{F-O-F}$. The reason is probably that the lower electron density around $\ce{H}$ results in significantly less steric interaction between $\ce{H}$ and the lone pairs of the $\ce{F}$ atom. In both $\ce{SF2}$ and$\ce{ OF2}$, the two $\ce{F}$ atoms with high electron densities would tend to repel each other, thus resulting in a greater bond angle.
Hence the bond angle order comes out to be-
$$\ce{F-O-F \gt F-S-F \gt H-O-F}$$
Hope this helps!


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How is it possible to get an attractive charge-dipole interaction using the modified Coulomb law?
Vqµ = − qµ / (4π ε0εr r2)
Say you had a negative charge, interacting with the positive end of a dipole then the charge would be -1.602 x 10-19 C and the dipole moment would always be a positive value. This means Vqµ would always be positive.
However a negative charge interacting with the positive end of a dipole is an attractive interaction so Vqµ should be negative. What am I missing here?

You're missing the fact that $\mu$, your quantity that measures the dipole only measures the magnitude but not the direction or alignment of the dipole. Your equation is flawed, I have no idea what it should properly be because as far as I know the interaction of a point charge with a dipole is more complex than merely a modified Coulomb law

So if you are talking about the positive end of the dipole, then μ becomes negative? My equation is from a 3rd year biophysical chem course, so I assume there is much more to charge-dipole interactions than this.

The electric dipole moment is a vector quantity: $\mathbf{p} = q\mathbf{d}$ where $\mathbf{d}$ is a vector pointing from the negative end to the positive end. You have the dipole moment as a scalar quantity: $\mu = qd$ where $d$ is the norm (aka length) of the vector $\mathbf{d}$. In simplifying from $\mathbf{p}$ to $\mu$ there is a loss of information about the orientation of the dipole - analogous to velocity/speed. The eqn doesn't capture this and you could add negative signs to make it work but 1) it is only a quick fix 2) the choice of sign is only conventional ie there is no right/wrong.

(cont) for example if you define the positive end of the dipole as having $\mu < 0$ then you need to retain the negative sign at the front of your formula. Whereas if you define the positive end of the dipole as having $\mu > 0$ and the negative end as having $\mu < 0$ then you need to get rid of the negative sign in the formula. I cannot say which approach is correct, because both give you the correct answer. Strictly speaking I would say neither are correct because $q$ is (by definition of the dipole moment) positive and $d$ being a length has to be non-negative, so $\mu$ cannot be negative.


text:
How can you explain the energy change in this phenomenon?
Tagged is the diagram of the apparatus used in this thought experiment. Pure water is taken in one side of the beaker and a solution of common salt in water in the other. After a very long time, the pure water should get transferred to the NaCl solution side because according to Raoult's law, the NaCl solution should have lesser vapor pressure than pure water, thus causing the water-vapour equilibrium in the pure water side to move forward, resulting in the condensation of the vapour in the other side (the solution side) . But how has the potential energy (=mgh) of the pure water increased during the process?

Get in a car and drive all the way to the top of hill. Your potential energy (yes, that mgh) will increase, too. How? It happened at the cost of the chemical energy. Same thing here.

Whose chemical energy is decreasing here? In the car example, fuel is being burnt causing combustion, and is being exhausted. I dont see anything being replenished or burnt here.

Energy of the salt solution and pure water. True, it is far less effective than gasoline, but still enough to do the trick. See, now that they are mixed, you can't have them separated again. Or rather, you can, but it will cost you some energy.

To disprove your logic, I can take the soultion of NaCl in water to be of infinite height. How will you then explain the infinite change in energy. Surely a finite amount of pure water cant have infinite chemical energy.

With infinite height, the setup will not work. In fact, a finite height of a few miles would be enough to stop the pure water from creeping into the solution.

How can you say that?

Like I said, chemical energy obtained from diluting a solution with pure water is quite small, while the height can be made arbitrarily big.

So, will the temperature of the pure water decrease in the process by your arguement?

That's complicated. We might have to look up the data on enthalpy and entropy of dissolution. In short, the temperature might change either way (very slightly, though) or not change at all.

Then what exactly is this "chemical energy" that you are talking about. Can you give any other example to support this?

Sorry, but I think a textbook on chemical thermodynamics would not fit in a comment.

It could, in an answer ;)

Note that in a solar still, pure water can be separated from salt water, and this process will usually leave undissolved solutes in one container and pure water in the other.    https://image.pbs.org/poster_images/assets/wgbh/ess05/ess05_vid_solarstill1/ess05_vid_solarstill1_l.jpg.resize.710x399.jpg

Some questions. Is the apparatus filled with say n-moles of room air to begin with, and sealed at a given volume.  Is the air saturated with water to begin with?

As the water level on the right rises, won't the water vapor molecules leaving need to gain less energy to get over the wall, ie they don't need as much energy to escape because they have more potential energy from the greater height?

In the specific example, with chambers on the left and right, let's say it was hung by a string.  The system would be become unbalanced and experience a torque that would cause in increase in kinetic energy of the system.  Perhaps one chamber should be in the middle and the other on the outside (a cylinder inside of a cylinder)

Initially suppose we have a mixture with mole fractions $x_1$ and $x_2$.   Let $x_1$ be the solute and $x_2$ the  water and assume for simplicity that  deviations from  Raoult’s law behaviour are insignificant.  In this case the heat of mixing is zero and entropy alone drives the mixing process.
In this case the free energy of mixing per mole at constant temperature is
$\Delta G =RTx_1\ln(x_1)+RTx_2\ln(x_2)$
and in as there are two components $x_2=1-x_1$ (and changing $x_1 \rightarrow x$  for simplicity ) gives
$$\Delta G =RTx\ln(x)+RT(1-x)\ln(1-x)$$
This energy is negative over the whole range, except at $x=0,1$ where it is zero. The minimum value at $x=1/2$ is $-RT\ln(2)$
After the pure water is added in the left hand compartment and equilibrium is established again, the initial mole fraction can only drop to half its value by mixing equal volumes of water which is a mixing energy of
$$\Delta G\left(\frac{x}{2}\right) = RT\frac{x}{2}\ln\left(\frac{x}{2}\right)+RT(1-\frac{x}{2})\ln\left(1-\frac{x}{2}\right)$$
Again this quantity is negative over the whole range, thus the solution will continue to be diluted if nothing prevents this.
Dilution of the mixture could, in principle, be prevented by the gravitational energy $mgh$ where m is the (molar) mass and h the height the liquid has to be raised to. To prevent further dilution
$mgh  - \Delta G(x_f)  \gt 0 $
where $x_f$ is the final mole fraction. If $mgh \rightarrow 0$ then $x_f=x/2$.
Using values for water gives $mgh = 0.176h$ Joule for height h whereas thermal energy $RT=2494$ J.  If, for example the height is $h=9800 $ m then $mgh \approx RTln(2)$ which is the minimum value of $\Delta G$ then no dilution can take place as $\Delta G \gt 0 $ for all x.
At smaller heights whether or not  dilution occurs depends on the mole fraction of the initial mixture but at a height of $1000$ m as long as the initial mole fraction is greater than $\approx 0.035$ then complete dilution will occur. Below $1000 $ m the final mole fraction is effectively half the initial value, $x/2$.
Of course the heights involved in effecting any change are quite ridiculous, but this is just because gravity is a weak force; you may not find it so if you ever go hill running :)
The figure shows different values of $\Delta G/RT$ per mole.

(a) shows the normal mixing free energy $\Delta G/RT$ vs. mole fraction solute x (b) the change upon no dilution and a potential of $1000$ m .(c) maximal dilution (doubling volume) at $1000$ m, thus dilution can occur for all initial values of molec fraction above $\approx 0.035$.(d) $\Delta G/RT$ plus $mgh = RT\ln(2)$ for water $h \approx 9800$ m. No transfer of water can occur in this case as $\Delta G \gt 0$

It is important to realize that you deal with a lot of particles and that you can apply the usual approach and approximations in statistical thermodynamics.
It makes more sense to ask for the potential energy of individual water molecules. Because there are so many molecules, we assume the Maxwell-Boltzmann distribution to be continuous. This means, that there will be particles with enough kinetic energy to overcome the potential energy wall (including non gravitational forces like dipole-dipole interaction).
If you had two solutions of pure water at $T > 0 K$, this would just mean that they exchange some of their water molecules over time.
If you have two different solutions as in your example there are more possible microstates for the same macrostate in the salt solution: "Putting the water into the right pot is entropically favoured, because solving salt in water is entropically favoured."
The exchange of water molecules is happening even in equilibrium.
As with any thermodynamical answer this ignores kinetic effects and other addends of the chemical potential like surface energy.


text:
Which is the better leaving group in the following Cannizzaro reaction?

In the given mechanism, is it correct to prefer hydrogen as a better leaving group than the formyl group?
The negative charge on an sp2 hybridized carbon, with the carbonyl group attached to it, seems to be more stabilized than a negative charge on hydrogen. (I have starred the step I am referring to)

Hydride is a terrible leaving group and a terrible nucleophile.

Why not just have the hydride attack the adjacent carbon directly? You need some sort of hydride acceptor.

Yes, but the mechanism is like this, even though the solution has hydroxide ions.

@Lighthart That's true, but a hydride transfer is apparently not impossible (in this reaction). I just think it needs an acceptor here.

The mechanism you have drawn is wrong. The hydride is transferred to the other carbonyl directly, it's not going to exist in solution.

Oh, okay. @orthocresol, But why is hydride leaving despite a better leaving group already present? Even the bond enthalpies of C-H is more than C-C bond!

Your mechanism is fine, just don't show the hydride going into solution. The hydride is transferred directly to the electrophilic carbon. Hydride transfer is common enough, but the hydride never just goes into solution like that. Your mechanism makes sense, just combine steps two and steps three and show an intramolecular hydride transfer so it is more representative of the real chemistry.

Okay, I understand a lot now. Just one more thing. Is it migration, @gannex, so that we prefer H over the other group for migration?

I don't know what you mean by migration. Do you mean rearrangement?
The reason you can get this hydride transfer is because the hydride has a very electrophilic place to go -- the aldehyde carbonyl.

If you'd like to read more about this mechanism, you should out this paper:
Swain, C. G.; Powell, A. L.; Sheppard, W. A.; Morgan, C. R. Mechanism of the Cinazzaro Reaction, 1979, Am. Chem. Soc. 3576
They did a kinetic study with deuterium oxide. They gave two equivalent mechanisms:

and

I think that this mechanism, which is essentially what you have shown except with intramolecular hydride shift (it cannot shift through solution) is the way I would represent it. It makes sense that this can happen if you think of the transition state. In my mind, it makes sense that this hydride transfer can occur given the very electrophilic nature of the aldehyde.
This is the transition state:

One does not 'prove' transition states 'kinetically'.

Sorry I must be conflating things. I was thinking that their kinetic study was involved because that is how they knew that the hydroxide reactant was already fully bonded before the rate determining step, so their study of the reaction's kinetics would have implications about the structure of the transition state. I guess what I'm trying to say is that this is how they know that the hydride is transferred intramolecularly from carbon to carbon, rather than through some other mechanism.

What they are doing is proposing a transition state.  The closest you can do to 'proving' a transition state is QM calcs to verify it is saddle point.  And this is not really proof, just supporting evidence that using this transition state to explain a mechanism is 'reasonable' .


text:
Why did a former emulsion of oil and water precipitate after heating and cooling when heat made it originally?
I added 100 g coconut oil, 100 g distilled water and 500 mg lecithin to an 1800 watt blender.  The motor from the blender has has a passive heating action.  I blended it at full for four minutes.  The result seemed to be a hot, uniform substance with a thin layer of foam on top.
I poured this mix into a smaller container and placed it in the fridge.  After a while it set into a solid, uniform substance with no apparent precipitation.
I then heated the substance in the microwave  for thirty seconds.  When I removed it, the components had separated and precipitated.  Cooling it in the fridge again resulted in a delicate solid sheet at the top which broke through to liquid underneath.
Why did the mix precipitate (presumably de-emulsify) after being microwaved but not occur when blended?

After cooling in the fridge the second time - (1) Is there some sort of ppt still? (2) Did you try blending it again? // Emulsions are temperature sensitive. // A ppt after second cooling in fridge would indicate that an irreversible reaction has  taken place and that probably the mixture won't be able to be emulsified again. // You do realize that heating in a microwave isn't absolutely uniform, especially when dealing with a solid.

When I get emulsions in my workups, I often hit the sep funnel with a little heat from the heat gun (assuming my product isn't terribly heat-sensitive). This seems to help things separate. The heat from the blender may not have been helping emulsify at all -- it may well have worked better making the emulsion if it was colder.

The simple explanation is that the speed of de-emulsification depends strongly on physical state and temperature.
If you have created a solid emulsion by cooling the initial blend, the de-emulsification will be very slow as there is little scope for the components to move around at all never mind separate. warming this solid up so it liquefies creates two factors that make separation easier: the mixture liquefies and the temperature rises, creating many easier ways for separation to occur. Once separation has occurred only re-blending will restore the emulsion.
Emulsions often separate slowly (milk, for example, slowly forms a layer of creamier fat on top but it doesn't happen quickly after thorough mixing). In your case the initial cooling "froze" the emulsion but the microwave heating enabled de-emulsification to occur by heating the mix. Cooling the result without blending is not going to restore the emulsion without further blending.

How long did yoi leave it to settle? Maybe it takes yet longer time to settle down than what you gave it?
Another idea is that either microwaves or heat influenced the emulsion structure and broke the oil-water complex?


text:
Is there a preferred way of naming the resonance hybrid in keto-enol tautomerism?
While answering a question about keto-enol tautomerism the question arose, how can I refer to the resonance hybrid instead of on of the resonance forms. In the case of the deprotonated butane-2-one, I can name 1 3-oxobutan-2-ide and 2 but-2-en-olate (ignoring stereochemistry). I cannot come up with a name for the hybrid 3.

While this might be quite a simple case, there are many resonance stabilised structures out there, which have contributions from more than one form. It gets especially complicated for ionic substructures. I wondered if there is either already an official recommendation (preferred IUPAC name), or any other suggestion at all handling those cases. After all, we are treating the same bonding situation with two different names, there should be something that covers both (multiple) cases.

Nomenclature of Organic Chemistry, IUPAC Recommendations and Preferred names 2013 in the section P-76 DELOCALIZED RADICALS AND IONS mentions only "totally delocalized" ions like cyclopentadiene derived cyclopentadienyl radical, cyclopentadienylium cation, cyclopentadienide anion; benzo[7]annulenylium (azulenium?) and pentadienyl radical.
However, there is a draft having this chapter expanded, allowing more systematic naming of delocalized ions.

P-76.1.2 Partial delocalization is denoted by the descriptor deloc preceded by the locants indicating the extent of delocalization. This descriptor and its locants are cited immediately before
  the appropriate suffix and enclosed in parentheses.

  cyclopenta-2,4-dien(1,2,3-deloc)ide
  ...

But I am not 100% sure how exactly would that apply to your 3 structure.. let's be brave and creative:
(2Z)-but-2-en-2-(2,3,O-deloc)olate or
(2Z)-but-2-en-2-ol(2,3,O-deloc)ate or if you don't need stereochemistry
3-oxobutan-2-(2,3,O-deloc)ide or maybe even just
butanon(2,3,O-deloc)ide ? .. or
butanon(O,2,3,-deloc)ide ?
However, that draft is from 2004, but this rule did not make it into the current official nomenclature (2013), so for now we can just speculate, or contribute to the upcoming version.

Thank you very much for digging this up. This is indeed an intriguing proposal. I guess one of the principal shortcomings in this rule is which ending takes precedence when a hetero-atom is involved.