text:
What are the products of the reaction between methanamine and nitrous acid?

What are the products of the reaction between methanamine and nitrous acid?

I'm finding this particular reaction very problematic. Here's why:
My teacher tells me that methanamine, when treated with nitrous acid (created by reacting sodium nitrite and hydrochloric acid) converts to an the unstable (at room temperature) intermediate: methane diazonium chloride, which decomposes in the (aqueous) solution to yield methanol, hydrogen chloride and dinitrogen.
A friend of mine (who happens to be a fairly trustworthy senior), tells me that the actual products are methoxymethane and dinitrogen, since methanamine is an exception and doesn't give the same kind of products the other primary aliphatic amines do (he read it up a long time ago, but doesn't remember where).
In hope of clearing up this mess, I resorted to my trusty copy of Organic Chemistry (Morrison and Boyd, 7th edition), according to which, the products of a reaction between a primary aliphatic amine and nitrous acid should be: dinitrogen and a mixture of alcohols and alkenes.

It doesn't mention anything about methanamine being an exception, and "a mixture of alcohols and alkenes" is a little ambiguous. Also I didn't discard this idea because maybe - through some method I don't even know - the methyl bits from the amines might couple and somehow result in an alkene. I wouldn't know since I'm relatively new to amine + nitrous acid reactions, so I won't take the risk.
I couldn't find any trust-worthy reference on this issue for this reaction. I need help resolving this ambiguity.

Both the teacher and the friend are partially right, but keep in mind that reaction products depend on reaction conditions, not just reactants.
According to Chemical Mutagens Environmental Effects on Biological Systems (2012):
In the presence of Cl-,

$\ce{CH3+}$,  is formed.

2a. $\ce{CH3+ + H2O -> CH3OH +H+}$
2b. $\ce{CH3+ + NO2- -> CH3NO2}$ and $\ce{CH3+ + NO2- -> CH3ONO}$
2c. $\ce{CH3+ + Cl- -> CH3Cl}$
2d. $\ce{CH3+}$ reacts with methylamine to form dimethyl amine

Methanol formed above can react with $\ce{CH3+}$ to form dimethyl ether
Methanol can also react with nitrous acid to form methyl nitrite

The above book is relying upon Austin, A. T., The Action of Nitrous Acid on Aliphatic Amines Nature 188, 1086-1088 (1960) as reference 472 to support the above information.
Austin explains that, under the reaction conditions:
0.5 mole methylamine hydrochloride
1.5 moles sodium nitrite
550 ml water
slowly acidified with 0.4 to 0.6 equivalents of 2N sulphuric acid
room temperature
products are as follows:
methanol: 6-25%
methyl nitrite: 45-35%
methyl chloride: 13%
nitromethane : 6%
methylnitrolic acid: 10-12%
Austin then states:

The multiplicity of products formed in the above reaction is rationally explained if one assumes that the electron-deficient entity $\ce{CH3+}$  is formed - a suggestion originally put forth by Whitmore [reference 7] for this type of reaction.

While Austin discusses the possibility of dimethyl ether and dimethyl amine also forming, he does not indicate that they were observed experimentally.

Nearly all chemistry that is taught in undergrad etc. was performed decades ago.

I would need to see their kinetic or mechanistic data, but I am extremely suspicious of a methyl cation in aqueous solution. This seems extremely implausible. This isn't FVP, it's at 0 degrees C.

@Zhe I think this is the primary source http://www.jstor.org/stable/43418182?seq=1#page_scan_tab_contents but I don't have full text access to that right now

I would've bought the carbocation for secondary or tertiary amines, but for methylamine, that's just a hard pill to swallow.

@Zhe "The unequivocal presence of
methyl chloride in the product gas when sodium
chloride is used in the solvent for methylamine
supports the suggestion of Austin [15] and others that
the organic portion of the amine may go through
a carbonium ion intermediate in the Van Slyke reaction." http://pubs.acs.org/doi/abs/10.1021/j150569a025

@DavePhD I'm not sure why the Okkam razaor mechanism here isn't just SN2 reaction with chloride? Not sure if I'm looking at the same Austin paper (JChemSoc, 1961, 3593)? But that definitely suggests an intramolecular displacement of dinitrogen via the carbonyl group, i.e. not via carbocation. Also, that substrate would've been a primary amine, not methyl...

According to the terminology I learnt at uni, it should be a carbenium ion (lack of a bond) not a carbonium (an additional bond).

Unfortunately I can’t access the mechanism either. But to be perfectly honest, I see no reason why the reaction must happen under $\mathrm{S_N1}$ conditions. Until proven otherwise, I will assume a mixture of $\mathrm{S_N1}$ and $\mathrm{S_N2}$ because that seems most reasonable. (CC@Zhe; I approve of bringing Occam’s razor into the discussion!)

I added information from the original A.T. Austin reference that says the mechanism is via CH3+ , but I agree it isn't proof.  The reference is very good for answering the OP's question about what the actual products are, however.

No chance to get alkenes from methanamine because there's no $\beta$-hydrogen for E2 elimination. And E1 elimination isn't going to work because you're not going to generate a methyl cation from the diazonium. The mixture of alcohols is probably because the alkyl cation could rearrange, but you're not creating the cation intermediate in this reaction.
The diazonium is a very good leaving group, so this complex is certainly going to be susceptible to SN2 reactions. The best and most available nucleophile you have in your mixture is water (the solvent), so you're going to get methanol.
Now, methanol could be a nucleophile that reacts with diazonium, but you're in aqueous solution. That means that the amount of water is going to dwarf any methanol that is produced. A negligible amount of the dimethyl ether will be produced.


text:
Is it possible to generalize whether an organic molecule is chiral or not?
Can we generalize on the basis of number of chirality centers  if an organic molecule is chiral or not?

Chirality in general is not about centers. There can be centers without chirality, as well as chirality without centers. Related: http://chemistry.stackexchange.com/questions/39830/what-kind-of-optical-activity-requires-chirality/39833#39833

There is a general statement. If the molecule has any improper axis of rotation, it is achiral.

What is an "improper" axis of rotation  by the way. Can you please elaborate on that?

Rotation + reflection through a plane perpendicular to the axis. If you can do that to the molecule and have it look exactly the same. It is not chiral.

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text:
Why is the geometry of tetracyanonickelate(II) square planar? Why is it colorless?
Why is the geometry of $\ce{[Ni(CN)4)]^2-}$ square planar given that it has a coordination number of 4?  Shouldn't its geometry be tetrahedral as it would minimise the bond angle and hence repulsion?
Also, why is it colorless? I saw that it is related to how $\ce{CN^-}$ is a relatively strong field ligand and can cause “pairing” of the electrons which would not allow the electrons to be transferred between d-orbitals, but what does it mean by pairing?

Because it has two lone pairs?

This has nothing to do with lone pairs... I think, you can start by reading my answer here: http://chemistry.stackexchange.com/a/40881/16683 If you have any questions, tell me.

that is a comparison between two different metal centres so does not answer this specific question.

@gamma1 Nowhere did I claim that it answered this specific question. However, the factors that affect the adoption of a square planar geometry vs tetrahedral geometry are the same, regardless of whether it is the metal centre or the ligand that is being compared. Hence, I felt it was related, but of course not a duplicate for the reason you mentioned. If you want to, you're more than welcome to write up an answer that actually compares the ligands.

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Smoldering Combustion
Does smoldering combustion only occur in an environment where oxygen levels are low? Will it always transition into flaming combustion (whenever there is ample supply of oxygen?
Or can it keep smoldering even when there is an ample supply of oxygen?
Some smoldering areas in forests continue to smolder for years where there is an ample supply of oxygen -> it's open air.

An adequate supply of oxygen is not necessarily the amount present in air. If there were a lot more, almost everything would burn all the time at the slightest provocation. This would be inconvenient for life. The amount we have is a balance that minimises too much burning without choking all the living things that want to breathe.

If you're smoldering, you're generally not producing enough heat to vaporize anything except in a small local environment. Only the vaporized fuel will burn.

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How do I identify lone pairs and bond pairs in beryllium(II) chloride?
How do I identify the lone and bonding pairs in $\ce{BeCl2}$?

Why is this question is off-topic???

In a simple sense, bond pairs are a pair of electrons (one from the central atom and one with the atom which is bonding) and participate in the bonding of the atom. Whereas lone pairs are the pairs of electron on an atom that do not participate in the bonding of two atoms.
To identify lone pairs in a molecule, figure out the number of valence electrons of the atom and subtract the number of electrons that have participated in the bonding. However, remember that the lone pairs are pairs and therefore if you ever find just one free electron that does not participate, it would mean that the compound has a charge.
Now for BeCl2

Usually, we only show the bond and lone pair of the central atom but if you want, just for your information, each chlorine atom in the reaction has got 3 lone pairs (the red dots.)

Be has atomic number 4 so its electronic configuration is $1s^22s^2$.
Cl has atomic number 17 so it electronic configuration is $1s^22s^22p^63s^23p^5$.
It is given that two Cl atoms bond with one Be atom so that means one $2s$ electron of Be gets excited and transitions to the empty $2p$ orbital that Be has. So now $2s$ and $2p$ of Be atom have one lone electron each. These two orbitals hybridize and form two $sp$ orbitals. These $sp$ orbitals have one lone electron each and these orbitals participate in bonding with the $p$ orbital of a Cl atom which has only one lone electron. On bonding,
the valency of the Cl atoms is satisfied.
There are only two bond pair of electrons in this case so according to VSEPR theory, these electron pairs have a tendency to be at an angle of 180 degrees to each other. By this, we can conclude that $\ce{BeCl2}$ has a linear shape.
On a side note, the valencies of Be are not fully satisfied as it does not achieve an octet. This is the reason why $\ce{BeCl2}$ acts as a Lewis acid because it has a tendency to accept more electrons in its remaining two $2p$ orbitals to complete its octet.

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Reaction between alumina and aqueous sodium hydroxide, which one's right?
My chemistry textbook has a rather annoying irregularity.
It cites two different reactions for the same set of reactants (aluminium oxide, sodium hydroxide and water) without the slightest mention of a particular temperature or concentration. Here they are:

In a chapter on the p-block elements:

$$\ce{Al2O3 (s) + 6 NaOH (aq) + 3 H2O (l) -> 2 Na3[Al(OH)6] (aq)}$$

In a chapter on metallurgy, under the topic of 'Extraction of aluminium from bauxite':

$$\ce{Al2O3 (s) + 2 NaOH (aq) + 3 H2O (l) -> 2 Na[Al(OH)4] (aq)}$$

Which one of these is correct? Or are they both correct, depending on the situation? If so, under what conditions are they correct?

Quoting from chemguide (emphasis mine):

With hot concentrated sodium hydroxide solution, aluminium oxide reacts to give a solution of sodium tetrahydroxoaluminate.
$$\ce{Al2O3(s) + 2NaOH(aq) + 3H2O(l) -> 2NaAl(OH)4}$$
Note: You may find all sorts of other formulae given for the product from this reaction. These range from $\ce{NaAlO2}$ (which is a dehydrated form of the one in the equation) to $\ce{Na3Al(OH)6}$ (which is a different product altogether).
What you actually get will depend on things like the temperature and the concentration of the sodium hydroxide solution. In any case, the truth is almost certainly a lot more complicated than any of these.
The version I am using is perfectly acceptable and is consistent with the aluminium chemistry you will find elsewhere on the site.

You can see that products of the reaction depends on the concentration of sodium hydroxide (Basic pH) but  other factors like temperature and pressure also come into play.
If you do not add water and only change the concentration of sodium hydroxide, product will also be different:
$$
\begin{align}
\ce{Al2O3 + 6 NaOH &-> 2 Al(OH)3 + 3 Na2O}\\
\ce{Al2O3 + 2NaOH &->[\pu{900 .. 1100 ^\circ C}] 2 NaAlO2 + H2O}
\end{align}
$$
Sodium aluminate is generally $\ce{NaAlO2}$ but its aqueous solution contains many other species like $\ce{[Al(OH)4]−}$, $\ce{[Al(OH)6]^3-}$ or perhaps the $\ce{[Al(H2O)2(OH)4]−}$. They are all considered to be aluminate species. So, if you are getting sodium tetrahydroxoaluminate or sodium hexahydroxoaluminate, they are no different compounds but are all hydrated form of sodium aluminate (addition of water + change in the concentration of sodium hydroxide).


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How to determine whether paper was dipped in acetone?
I was wondering if there was a solution of some sort that would cause a reaction to paper that was previously dipped into acetone much like in a counterfeit pen the iodine causes a reaction when written on standard paper .

There is iodoform test... http://www.chemguide.co.uk/organicprops/carbonyls/iodoform.html

I doubt that much acetone would be left to detect unless the test were done soon after the dipping.

If the test was done soon after the dipping what would you test it with?

I would use my nose. I think humans can detect acetone at lower concentration and with more specificity than any simple chemical test. For detecting trace amounts I would use an extraction followed by gas chromatography and mass spectrometry.

As a couple comments and an answer imply, acetone is volatile enough that there would rapidly nothing left sufficient to analyze.  You would be better off testing for residual non-volatile components in the acetone.  Hopefully it was low grade ;)

If significant amounts of acetone are left (according to my old lab manual, the analysis works with a 50 mg sample), you could use the reaction with 2-nitrobenzaldehyde in the presence of sodium hydroxide, which generates indigo dye.
See also: Baeyer–Drewson indigo synthesis


text:
Is the 5d orbital involved in triiodide ion?
Is the $5\mathrm{d}$ orbital involved in the triiodide ion, $\ce{I3-}$?
There are $5$ electron pairs around the central iodine.
(Almost) all of the hypervalent compounds involve the most electronegative elements: $\ce{F}$, $\ce{O}$, $\ce{Cl}$.
Also, the bonds involving the electronegative elements are often almost ionic:

For sulfuric acid, computational analysis (with natural bond orbitals) confirms a clear positive charge on sulfur (theoretically $+2.45$) and a low $3\mathrm{d}$ occupancy.

So, we can say that $3\mathrm{d}$ is (almost) not involved in the central sulfur atom of sulfuric acid.
Is there any experimental data to show if $5\mathrm{d}$ is involved in the triiodide ion?

related http://chemistry.stackexchange.com/questions/41952/is-the-triiodide-ion-polar

http://scitation.aip.org/content/aip/journal/jcp/19/4/10.1063/1.1748245

@DavePhD Could you make that an answer if it is?

@DHMO I meant, I'd have to read THIS http://scitation.aip.org/content/aip/journal/jcp/133/6/10.1063/1.3474571 to see if anything is majorly wrong with the older work, but usually the molecular orbitals are attributed to linear combinations of 5p orbitals

@DavePhD Then where are the two extra electrons?

@DHMO each I atom has 6 valance electron that are unshared (s, px, py) and they share the other 4 electron in two occupied molecular orbits formed by linear combinations of the 3 pz atomic orbitals.

@DavePhD But would the bond order be $1$ instead?

@DHMO the 1951 reference says "The lowest orbital [MO 1]  is
definitely bonding, while the second [MO 2] , although formally
nonbonding, does tend to place the electrons
between atoms 1 and 3 without introducing a node and
probably contributes to the stability of the bond." It also discusses that overlap of p$\pi$ orbitals is being neglected, but is small.

@DHMO see this too https://en.wikipedia.org/wiki/Three-center_four-electron_bond  and https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2276819/ but, yes, you could consider bond order 1 for the whole ion, or 1/2 between each pair of atoms

No, a four-electron-three-centre bond is all we need to explain $\ce{I3-}$

@Jan Has it been experimentally confirmed?

Professor Klüfers presents a structure of the $\ce{I3-}$ ion in his general and inorganic chemistry introductory course at the LMU Munich. The ion is linear and the $\ce{I\bond{...}I}$ distances are identical, measured experimentally to be $293~\mathrm{pm}$.
In neutral iodine, the $\ce{I-I}$ distance is measured to be $272~\mathrm{pm}$.
We see a notable increase in bond length which indicates a decrease in bond order. A reduction of bond order — say from a bond order of $1$ (single bond) to one of $0.5$ (four-electron three-centre bond) — means that we cannot add an additional bonding orbital to the mixture. Otherwise we would have something along the lines of two single bonds (bond order $1$), which is how the ‘octet expansion’ traditionally was explained.
The reduction in bond order is best explained — as mentioned — by a four-electron three-centre bond and thus without d-orbital participation.

At first, I’m fairly new to CAS calculations and to all the fancy options that one has when it comes to calculate CASSCF with Orca. Terefore I might be wrong by what I am doing here. Just keep in mind. :)

I don’t know if there are experiments and how such an experiment could show what you want to know. I also didn’t search for existing calculations. But after quite some while I could manage to calculate it with CASSCF. This is not that “easy” as the CAS needs to be very big.
Per Iodine I take the 5s, the three 5p and the five 5d orbitals into my active space. This leads to CAS(22,27), 22 electrons in 27 orbitals, which is not feasible using regular CAS. Therefor, I use the Iterative-Configuration Expansion Configuration Interaction
(ICE-CI) approximation (cistep ice) for the CI-Step in CAS, which is a new implementation in ORCA 4.
I’ve never used it before, so, as I said before, I could be wrong by, e.g., wrong input. Nonetheless, my final input with ORCA 4.0.0.2 looks like this, using the experimental geometry as Jan’s answer gives it:
! dkh ano-rcc-dzp RIJCOSX decontractaux
...
%basis
 aux "AutoAux"
end
%casscf
 nel 22
 norb 27
 mult 1
 nroots 1
 cistep ice
 etol 1e-6
 trafostep rimo
end
* xyz -1 1
I 0.0 0.0 -2.93
I 0.0 0.0  0.00
I 0.0 0.0  2.93
*

The output then gives you those molecular orbitals:

and the respective contributions to the ground state wave function:
weight   : occupation
           777777777788888888889999999 \ mol orb
           012345678901234567890123456 / index
           ssspppppppppddddddddddddddd contrib. AO types
0.86409  : 222222222220000000000000000
0.02112  : 222222222202000000000000000
0.00604  : 222222222022000000000000000
0.00276  : 222222222111001000000000000

average occupation numbers
70   71   72   73   74   75   76   77   78   79   80   81   82   83   84
1.99 1.99 1.99 1.98 1.98 1.98 1.98 1.97 1.97 1.96 1.93 0.09 0.02 0.02 0.02

85   86   87   88   89   90   91   92   93   94   95   96
0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01

There is a major CSF with 86% where all occupied orbitals are build from s- and p-type AOs, then 2% CSF where the “LUMO” is occupied instead of the “HOMO” and ignoring the next CSF, the last shown CSF hast 0.3% contribution where a single electron occupies the $\sigma$-bond from the $\mathrm d_{z^2}$ orbitals (MO 84). (With 11% remaining, there is also a large amount of other CSFs with small individual contributions. But as the average occupation numbers show, the occupation of 5d-orbitals is low.)
So from my point, I’d say that there is no real contribution of the 5d-orbitals to the ground state of $\mathrm I\,_3^-$.


text:
Is having a lone pair sufficient and necessary for a Lewis base? What about empty orbitals and Lewis acids?
Is any molecule with a lone pair and/or a pi bond considered a Lewis base? Also, is any molecule with an incomplete octet a Lewis acid?

The most common ones are the ones you described. I don't know about any specific rare cases though.

No, not every molecule that has a valence lone pair is automatically a Lewis base.
For example, tin forms divalent salts such as $\ce{SnCl2}$ wherein tin is in the $\mathrm{+II}$ oxidation state and thus its electronic configuration is $\ce{[Kr]}\ \mathrm{4d^{10}\,5s^2\,5p^0}$. Even if we say that the d-electrons are not really valence electrons, since they are not really chemically accessable, you cannot deny that tin(II) has a lone pair in its 5s orbital. However, this lone pair does not exhibit Lewis basic properties; instead, tin(II) functions as a Lewis acid. The reason for that is the spherical symmetry of the s-orbital along with its notable dispersion, which means it cannot be reasonably donated in a Lewis-basic matter.
Note that the lone pair is accessable chemically: tin(II) can be oxidised to tin(IV) e.g. by $\ce{HgCl2}$.

A somewhat similar (but much more questionable!) case can be made for electron sextets and Lewis acids. NHCs are a pretty common substance class. The most generic structure is that of imidazole, but with both nitrogens substituted so that one of them bears a positive charge (the one with a double bond to the carbon sandwiched between the two). If a base is added to remove that sandwiched carbon’s proton, a carbene is generated in which said carbon is only bonded to the nitrogen, has an $\mathrm{sp^2}$ configured lone pair and is considered to have an empty $\mathrm{p}$ orbital. However, this carbene is nucleophilic, which is why they are also called nucleophilic heterocyclic carbenes. They react as Lewis bases with metal centres; the second generation of Grubbs’ metathesis catalysts made use of NHCs.
Of course, it is slightly debatable whether the p-orbital on that NHC carbon is truly empty. One could argue that significant electron pair donation from both nitrogens should happen.
(As a side note: radicals, while not having full octets, are typically never considered Lewis acids — but then, their orbital is already occupied by a single electron, so they don’t have an empty orbital that would make them a potential Lewis acid, anyway. Thus, your question should better be worded differently to explicitly exclude radicals.

From Wikipedia:

Lewis acid is a chemical species that reacts with a Lewis base to form a Lewis adduct. A Lewis base, then, is any species that donates a pair of electrons to a Lewis acid to form a Lewis adduct.

For example, in the reaction $\ce{H3B + NH3 -> H3B-NH3}$, the $\ce{N}$ donated a pair of electrons to $\ce{B}$, so $\ce{H3B}$ is the Lewis acid and $\ce{NH3}$ is the Lewis base.

Is any molecule with a lone pair and/or a pi bond considered a Lewis base?
No. For example, $\ce{CH4}$ has no lone pair or $\pi$ bond, but it is still considered a Lewis base:
$$\ce{CH4 + H+ <<=> CH5+}$$

Is any molecule with a incomplete octet a Lewis acid?
No. For example, $\ce{NH4+}$ has a complete octet, but it is considered a Lewis acid:
$$\ce{NH4+ + :OH- <=>> NH3 + H2O}$$

NH4+ isn't Lewis acid in this example but Bronsted acid

@Mithoron but... but... wikipedia says it is... so it must be, right

@Mithoron It can be consider as both depends on how you think about it.

Well, Wikipedia isn't always right, but if you loosen the definition, you could say it's also Lewis acid.

agree with @Mith, I would argue that proton donors generate the Lewis acid $\ce{H+}$ but are not Lewis acids themselves


text:
How to identify the parent chain if two chains have the same number of skeletal atoms and the same number of double bonds?
Please could you tell me the rule to name the given hydrocarbon ... I am just confused in which chain should I take as the parent chain . The chain from A to C or from C to B. There is no rule about it anywhere on the net or in my books .

You go from carbon 'A' all the way to carbon 'C'. The 'B' chain is the substituent. Numbering the carbon chain (for your example) is pretty standard stuff, I'm surprised you couldn't find it anywhere........

@AaronAbraham : could you tell why ? please and the rule

@AaronAbraham Why not from 'B' to 'C' and name it as 3-vinylocta-2,4,6-triene?

Hint: You number it in such a way that the lowest locants are assigned to a substituent (lowest locant sum rule), all the while having chosen the longest chain......Also the 'A' end is given more priority over the 'B' end since the double bond  gets a lower locant in that case...

@DHMO The same way Loong did (albeit, without the elaborate explanation) ;) .....from 'A' to 'C'......

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

greater number of suffixes
longest chain
greater number of multiple bonds
lower locants for suffixes
lower locants for multiple bonds

The corresponding wording of the nomenclature rules (taken from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)) reads as follows:

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);
(…)

Since the compound that is given in the question does not have any principal characteristic group (a characteristic group chosen for citation at the end of a name by means of a suffix or a class name, or implied by a trivial name), the first criterion (greater number of suffixes) given in P-44.1.1 is not relevant in this case.
The second criterion given in P-44.3.2 is the longest chain. However, the  chains A–C and B–C both have eight skeletal atoms.
Also, both chains have three double bonds. Therefore, the criteria given in P-44.4.1 (a) (greater number of multiple bonds) and (b) (greater number of double bonds) cannot be used to reach a decision.
The criterion given in P-44.4.1 (h) (lower locant for the suffix) is not relevant since the compound that is given in the question does not have any suffix.
Finally, the criterion given in P-44.4.1 (j) stipulates that the principal chain shall have the lower locants for the double bonds. Therefore, the principal chain for the compound that is given in the question is the octa-1,4,6-triene from A to C rather than the octa-2,4,6-triene from B to C since the locant set ‘1,4,6’ is lower than ‘2,4,6’.
Thus, the complete name for the compound that is given in the question is (3E,4E,6E)-3-ethylideneocta-1,4,6-triene.


text:
Regioselectivity of deprotonation of 1,4-dicarbonyl

In the intramolecular aldol condensation of hexane-2,4-dione (above), why does the base abstract one of the $\ce{-CH3}$ protons instead of one of the $\ce{-CH2}$ protons? Both sets of protons are α to a carbonyl group and should therefore be acidic.
I'm confused because I've seen similar reactions where the proton is taken from the $\ce{-CH2}$ group.

Here, both the α-$\ce{CH2}$ and α-$\ce{CH3}$ groups have very similar acidities since they are both α to only one carbonyl group. [This is different compared to, for example, a 1,3-dicarbonyl such as acetylacetone, where one $\ce{CH2}$ group is α to two carbonyl groups and is therefore significantly more acidic.] In our substrate, the two protons differ very slightly in acidities, and in general, whether one deprotonates at the more or less substituted carbon can be controlled by reaction conditions.
The mechanism you are drawing is the first step of what is known as an aldol condensation. Because all the steps in this reaction are easily reversible, simple aldol condensations with $\ce{OH-}$ are typically under thermodynamic control, meaning that the most stable product is preferentially formed.
Therefore, let's consider what would happen if we deprotonated the $\ce{-CH2}$ group. The deprotonation itself is definitely a possibility, but what happens after that is quite implausible:

Once you form the enolate, the next step in an aldol condensation is the enolate attacking the other carbonyl group. In this case, it would lead to a three-membered ring, where there is a lot of angle strain and the substituents eclipse each other. The dehydration to form the α,β-unsaturated ketone is even less likely, because the cyclopropene ring thus formed would possess incredible angle strain.
On the other hand, deprotonation of the terminal $\ce{-CH3}$ group eventually leads to the formation of a five-membered ring after nucleophilic attack on the other carbonyl group. This is much more stable, and you can eliminate water via an E1cb-type mechanism to get a cyclopentenone.

So, you could certainly deprotonate the $\ce{-CH2}$ position. It is just that, after it gets deprotonated, nothing very good can come out of it. This deprotonation is said to be unproductive.

Plus, isn't a primary carbanion more stable than a secondary carbanion, in general?  It might also be worth mentioning that 1,3-dicarbonyls, in contrast to 1,4-dicarbonyls, would behave quite differently.  The acidity of the hydrogens boudn to the "di"-alpha carbon is much higher than the acidity of the alpha carbons in 1,4-dicarbonyls.

@CurtF A primary carbanion would be more stable, but in this case, the carbanion is not the predominant resonance form - the enolate is. The way I learnt it was that more substituted double bond = more stable, so deprotonation of the inner position is more thermodynamically favourable, as stated in the other answer. http://i.stack.imgur.com/i5ptd.png

You can actually take either. The other two $\alpha$ hydrogens are more acidic thermodynamically, but if you used a hindered (kinetic) base at a low temperature, you can predominatly de-protonate at the less hindered methyl (as in your diagram).
So, the answer is, it does sometimes, but it depends on the base and reaction conditions. This is why organic chemistry is interesting.


text:
Why does aqueous NaCl conduct electricity?
Could someone please explain why an aqueous solution of NaCl conducts electricity? I've Googled this up but couldn't get a satisfactory answer. This has an answer on Physics.SE but that's too complicated. My textbook doesn't answer my question.

Essentially you understand that it is because it dissociates, and your follow-up question is why it dissociates, which, as someone else pointed out, is another question.

See the answer, and linked reference therein, by @M.Farooq: https://chemistry.stackexchange.com/a/118441/79678.

Original Question:
Why an aqueous solution of $\ce{NaCl}$ conducts electricity
Because $\ce{NaCl}$ is an electrolyte. This means it yields ions in solution.
Simply put, solid $\ce{NaCl}$ consists of $\ce{Na+}$ cations and $\ce{Cl-}$  anions bound together in a rigid crystal lattice. When it melts or is dissolved in water, the crystal lattice breaks. The ions are now able to move around. Similar to charged particles in a metal conductor (in this case electrons), in liquid form or aqueous solution the ions are the charged particles that can move, allowing the solution to conduct electricity.
Follow-up question:
Why $\ce{NaCl}$ dissociates in water
The (rather oversimplified) answer is that dissociation of an ionic compound is facilitated by ion-dipole attractions between the ions of the compound and the polar water molecules.
The water molecules (dipoles) are attracted to the ions  and cause the crystal lattice to destabilize and ions to dissociate. The charged ions in solution are surrounded and stabilized by the water molecules (dipoles).
Note: Some ions migrate as ion pairs, but for a strong electrolyte most or all the ions will be dissociated and surrounded by water molecules.

As you seem dissatisfied with the answers received so far, I assume you want to fully understand the mechanism involved in dissociation. Unfortunately, this is rather complex.
There is an excellent  paper by Ballard & Dellago [1] that explains their work on the subject, but you will probably need a bit more physical chemistry / thermodynamics knowledge to understand it fully.
Reference

Ballard, AJ & Dellago, C "Toward the Mechanism of Ionic Dissociation in Water," *J. Phys. Chem. *B 2012, 116, 45, 13490–13497
Publication Date:October 19, 2012
https://doi.org/10.1021/jp309300b

NaCl is an electrolyte. When in solution it dissociates into Na+ and Cl-. When you put electrodes in the solution, the cations are drawn to the cathode and the anions to the anode. This movement produces a current and that is why NaCl solutions can conduct electricity.

Why does it dissociate?

I think this answer needs some elaboration. Does this only work with AC? If not, what happens with DC when the bulk of  the ions have already migrated?

With a DC current, you'll probably get some electrolysis of water and possible oxidation of the chloride (which helps by removing negative charge from the positive electrode).

@Zhe Agreed. I was just trying to prompt the psoter into fleshing out his answer for the OP.

$\ce{NaCl}$ dissociates into $\ce{Na+}$ and $\ce{Cl-}$ when it dissolves into water. Hydration stabilizes the ions formed. There is $\delta+$ charge on hydrogens of water and $\delta-$ charge on the atoms of oxygen. When $\ce{NaCl}$ is dissolved into water, the hydrogens are attracted to the $\ce{Cl-}$ ions and oxygen atoms to the $\ce{Na+}$ ions. Thus water molecules surround and separate the ions.

Hence, in an aqueous solution of $\ce{NaCl}$, there are positive and negative ions available to move freely (of course with some resistance due to other molecules in vicinity). When a potential difference is applied, the positive ions are attracted to the negative terminal and vice versa. Thus, a current is established.

@Abcd See this video on YouTube. link

I do not think that actual $\ce{OH-}$ ions are attracted to $\ce{Na+}$ because  the dissociation constant of water is quite low, $pK_w = 14$. So, I don't think that free $H+$ and $OH-$ ions are available. A water molecule as a whole orients itself in a particular way and surrounds the ions.

Sorry, I didn't mean that $OH-$ ions are not attracted to $Na+$ but I think that the effect wouldn't be as much.

Thankyou. So can we say that dissociation of NaCl takes place because of orientation of water molecules?

And where did u get that awesome picture from ?

Yes. I think so.

I googled it(images).

The simple answer is that water molecules have a dipole. The hydrogen atoms are slightly positively charged and the oxygen atoms slightly negatively charged. So, as you can see in the diagram above, water molecules around positively charged sodium orient themselves such that the oxygen is closer to sodium than the hydrogens. This is known as an ion-dipole interaction. The opposite is true for chlorine. @Abcd

@orthocresol and therfore they dissociate ?

@Abcd yes. In the absence of water, NaCl does not dissociate, because it is held together by ionic bonds. The dissociation of NaCl in water also requires the breakage of ionic bonds, but that is compensated for the formation of such ion-dipole interactions, which is the driving force for the dissociation.

@orthocresol sorry if i am troubling you by questioning too much .. what do u mean by driven by and how does the bond break?

@Abcd Ionic bonds are only electrostatic forces of attraction. From physics, you will know that the strength of such forces depend on the distance between the two charged species. In solid NaCl, the ions are close to each other, so a bond is said to exist. In aqueous NaCl, the ions are separated from each other, so the bond is said to be broken. A driving force simply refers to something that favours a process. So, the dissociation process is favoured by the fact that ion-dipole interactions are formed.

$\ce{NaCl}$ is strongly ionic compound. It gets completely ionized and dissociates in $\ce{Na+}$ and $\ce{Cl-}$. Both $\ce{Na+}$ and $\ce{Cl-}$ are surrounded by water molecules, $\ce{Na+}$ is surrounded with $\ce{O}$ of $\ce{H2O}$ facing towards $\ce{Na+}$, similarly $\ce{Cl-}$ is surrounded by $\ce{H2O}$ molecules with $\ce{H}$ facing towards $\ce{Cl-}$. Presence of positive and negative ions helps in conduction of electricity.
When electrodes are inserted and current is passed through these electrodes, movement of ions in opposite direction creates current.
After passing current for a long time (especially DC current), the concentration of $\ce{Cl-}$ ions will decrease. The concentration of $\ce{OH-}$ will increase gradually and there will be competition between $\ce{Cl-}$ and $\ce{OH-}$.
Electrolysis of $\ce{NaCl}$
Anode: $\ce{2 Cl- -> Cl2 + 2 e-}\quad E=-1.36\ \mathrm V$
Cathode: $\ce{H2O + 2 e- -> H2 + 2 OH-}\quad E=-0.83\ \mathrm V$
Electrolysis of $\ce{NaOH}$
Anode: $\ce{4 OH- -> O2 + 2 H2O + 2 e-}\quad E=-0.40\ \mathrm V$
Cathode: $\ce{H2O + 2 e- -> H2 + 2 OH-}\quad E=-0.83\ \mathrm V$
After Sufficient long time there will be competition between $\ce{Cl-}$ and $\ce{OH-}$ to get oxidized on Anode, whereas there is only one possible reaction on cathode. Reaction on Anode depends upon the concentration and reduction potential of $\ce{Cl-}$ and $\ce{OH-}$. Actually on cathode there is also possibility of reduction of $\ce{Na+}$ but reduction potential is $-2.7\ \mathrm V$ which is difficult than $\ce{H2O}$ so only one reaction takes place on cathode.


text:
How is it possible to increase mercury's specific weight?
Is it possible to change mercury's specific weight to make it heavier? By mixing it with anything? But not by changing its temperature.

Mix it with osmium.

Gold might be a cheaper option.

Is the mixture a liquid at room temperature? And if you could please tell me about the mixing process ? Ive heard that there's a way to "cheat"it and make it heavier.

@IvanNeretin not necessarily. At current market prices, gold is more expensive than any other precious metal.

Yeah, I know, those prices tend to oscillate wildly, hence "might be".

Use $\ce{^{203}Hg}$ instead. It's 0.5% heavier.

Even if gold is more expensive than osmium, I would still go for the gold.  Osmium generates a volatile oxide making it more hazardous to work with.  Or get a metallic asteroid with a lot of siderophiles into Earth orbit and the price problem is solved.  ;-)

Specific weight is a property of the element. It is product of the atomic mass and the volume. You can't easily alter either.
You could mix an element with another element of a different density and the density of the mixture will change. But you won't be changing the specific mass of the elements, just creating a mixture with a different density.
You can't readily change the mass of an element. But an element might be made up of multiple isotopes (mercury has 4 significant ones with masses ranging from 199 to 202 atomic mass units). If you enrich the mercury for one of those isotopes, you can change the mass, very slightly (for other elements this can be important; fissile isotopes of uranium are concentrated by doing it when making nuclear fuels).
In practice there is nothing practical you can do to alter the density (specific mass) of an element though you can create mixtures with different densities (but then you still haven't altered the properties of the element itself).

7 observationally stable isotopes with 204 being the highest according to https://en.wikipedia.org/wiki/Isotopes_of_mercury

@DavePhD Fair point. I was summarising the top 4 with >10% natural abundance. And I didn't read the full list. ;-)

Surely you can alter the density of some elements -- allotropes of the same element often have different densities. Diamond is about 1.5 times as dense as graphite; there's a similar difference between black and white phosphorus. Liquid ozone is denser than liquid oxygen.


text:
Why can copper have both four and six bonds?
I'm trying to understand the Biuret test by first understanding how copper dissolves in the aqueous solution. We first have added the $\ce{CuSO_{4}}$ to a substance than the $\ce{OH^{-}}$.
I've read the following page on the reaction, but I still confused: http://www.docbrown.info/page07/transition09Cu.htm
How is it possible to have tetraaquacopper(II) and hexaaquacopper(II) ions in the solution? How could this configuration be explained using electron configuration? $[\ce{Ar}]\mathrm{3d^{10}4s^{1}}$

http://chemistry.stackexchange.com/questions/40795/coordination-numbers-of-copper

$\ce{Copper(II)}$ complexes are $\ce{d^9}$ type complexes. This means that there are 9 d-electrons for the metal centre. When a complex forms, a number of ligands coordinate to the metal centre and the species adopts certain geometry. The five d-orbitals are no longer the same energy. They split into sets of different energies depending on the geometry and how ligands interact with them.
For an octahedral complex, orbitals split into two sets of degenerate orbitals - a lower $\ce{t_{2g}}$ ($\ce{d_{xy}},\ce{d_{yz}},\ce{d_{xz}}$) which is bonding to non-bonding and a higher $\ce{e_g}$ ($\ce{d_{z^2}}, \ce{d_{x^2-y^2}}$) which is slightly anti-bonding.
Copper is a bit funny as it experiences what is called a Jahn-Teller distortion where there is an elongation of the metal-ligand bonds along the z-axis. This is due to uneven occupancy of the $\ce{e_g}$ orbitals set. Then, the orbitals further split to remove the degeneracy and gain more stability. For the square planar complex of $\ce{tetraaquacopper(II)}$ to form you would need to go even further and remove those two ligands along the z-axis. The stabilization energy gained overall by going from octahedral to square planar configuration is what allows this switch to occur; it wouldn't happen if it were unfavorable. The gain in stability is borderline and that is why I assume that both configurations are possible.
Also, square planar fields tend to prefer stronger field ligands and water is not such ligand and this might also be considered as a reason why you get a mixture of the two types of geometry.

So,what would be the electron configuration of the tetraaquacopper(II)?

What does it means : " it wouldn't happen if it were unfavorable"? In my specific case, my copper ion is bonded with four amines. Is this creating a stronger field than water, considering that is linked with nitrogen?

The electron configuration for square planar Cu(II) would be (eg)2(a1g)2(b2g)2(b1g)1. Yes, ammonia is a stronger field ligand than water. By saying "it wouldn't happen if it were unfavourable" I mean that everything happens for a reason and as nature is lazy, no system likes having too much energy so it would utilise every possibility to lower its energy, hence the distortion I mentioned and then forming the square planar complex. However, due to the occupancy of an antibonding orbital it really isn't true square planer, two waters would still hang about at the axials.

This is at least what I think happens and how I would explain it.

@Matt The complex often described as tetraamminecopper(II) is, in fact, tetraamminediaquacopper(II) $\ce{[Cu(NH3)4(H2O)2]^2+}$. The two aqua-ligands are just further away from the copper centre.


text:
How do I determine the number and order of ligand orbitals?
I am trying to create interaction diagrams for transition metal species by reducing the representations for both $\sigma$ and $\pi$ ligand orbital sets and mixing them with the metal orbitals. The order of metal orbitals is obvious because I know I have my 5 degenerate $n\text{d}$ orbitals, then my $(n+1)\text{s}$ orbital, which is higher in energy due to its principal quantum number, then my $(n+1)\text{p}$ orbital, which is higher in energy due to penetration (it's a p orbital); and the number of metal orbitals is apparent because my character table lists the each representation's matching function, so I can tell $\mathrm{A}_{1g}$ will have s-symmetry and $\mathrm{E}_g$ will match my $\mathrm{d}_{x^2-y^2}$ and my $\mathrm{d}_{z^2}$ orbitals, for example.
I don't see exactly how I should do the same for my ligand orbitals, though. For example, I know that my ligand $\pi$ orbital irreducible representation is $\Gamma_{\pi} = \mathrm{T}_{1g} + \mathrm{T}_{2g} + \mathrm{T}_{1u} + \mathrm{T}_{2u}$, and I know from my table's matching function list that there will be three $\mathrm{T}_{1u}$ $(x,y,z)$ and three $\mathrm{T}_{2g}$ $(xy, xz, yz)$ orbitals, but I don't know how to identify that there are also three $\mathrm{T}_{1g}$ and $\mathrm{T}_{2u}$ orbitals, since my character table doesn't list the matching functions for those representations.
Also; my understanding of how to decide on the ligand orbitals' relative energy levels is dubious. Yves Jean's Molecular Orbitals of Transition Metal Complexes says "one must analyze the bonding or antibonding character" (p. 43), but I am not getting it right every time. Can anyone give me some tips?

Are you looking at the Oh group character table as it does contain the T1g and T2u representations?

yeah of course. It just doesn't have a matching function next to them, so I was unsure of how to figure out how many orbitals from my ligand set match those symmetry representations. This is also a more general question, though, but that's a good example of my problem.

T refers to triply degenerate set, E refers to doubly degenerate set, A and B are singly degenerate. The total number of orbitals should add up to the total number of atomic orbitals on your ligands.

Oh, OK that actually answers my question.

I understand that the total number of orbitals should add up to the total number of AOs, but I didn't realize the thing about T, E, A, B. For that matter, I'm also a little unsure about 1, 2, g and u.

1, 2, g, and u don't refer to degeneracy, but to symmetry of the irrep with respect to one of the operations. u/g is (anti)symmetric wrt inversion.  B/A is (anti)symmetric wrt to rotation around the principal axis. To the best of my knowledge, 1/2 differentiate between mirror planes.

Are you asking about how to find the irreps of the ligand pi-type orbitals? In general you would have to construct a reducible representation for all the relevant orbitals (by investigating how they transform under the symmetry operations of the $O_\mathrm{h}$ point group), and then reduce this into a sum of irreducible representations. The matching functions column can only help you so much here, unfortunately!

orthocresol, I already did that. My reduction yielded Γπ=T1g+T2g+T1u+T2u . I was asking how I know how many degenerate orbitals come from T1g and T2u. I know that T2g and T1u have three each and I need 12 in total (it's pi orbitals on a D4h square planar molecule), but I was unsure about the other two, but V. Vladimirov told me that T representations necessarily refer to triply degenerate sets of orbitals.

The other element of my question was how to find the energetic order of those orbital sets, but as I understand it, you need to create the interaction diagram beforehand for that, because it's based on the bonding/antibonding nature of the orbital overlaps each one contributes to.

Oh, I see what you mean. As for the energetic ordering, if your only desire is to construct a qualitative MO diagram, I would say that you can just treat them as having the same energy. This is because the overlap between the different orbitals is poor (not least because they are far away from each other) and so any bonding/antibonding effects will be small. However, if you do want to figure it out, you will have to work out which linear combinations correspond to which irrep, and look at the bonding/antibonding interactions within these linear combinations.

got it. I'm more concerned about getting the number right. The energy order is just a detail. Based on what V. Vladimirov said about T=triply degenerate, E=doubly, etc., my problem is solved.

@V.Vladimirov I'd encourage you to write that up into an answer.

T refers to triply degenerate set, E refers to doubly degenerate set, A and B are singly degenerate. The total number of orbitals should add up to the total number of atomic orbitals on your ligands. 1,2,g,u are symmetry labels (look up Mulliken symbols). One thing to remember when constructing you MO diagram is that only orbitals of similar energy and the exact same type and symmetry can mix. So, you can only mix A with A, B with B, E with E and T with T. Also, only g with g, u with u, 1 with 1 and 2 with 2.


text:
Discrepancy in Standard Hydrogen Electrode energy
From Absolute Electrode Potential on Wikipedia:
$$\ce{\frac12H2(gas) -> H+(aq) + e-(g)\quad(-(4.44\pm0.02)eV)}$$
The bond-dissociation energy of hydrogen is $\ce{4.52 eV}$:
$$\ce{\frac12H2 -> H\quad(-2.26 eV)}$$
The ionization energy of hydrogen is $\ce{13.6 eV}$:
$$\ce{H -> H+ + e- (-13.6 eV)}$$
How to address the discrepancy that the equations cannot add up in terms of energy?

Edit: of course I know that hydrogen ion in gas is different from hydrogen ion in aqueous solution... it is in my very essence. However, how come the enthalpy of formation of $\ce{H+(aq)}$ is zero?

There is a great deal of difference between $\ce{H+}$ and $\ce{H+(aq)}$.

Enthalpy of formation, unlike the absolute potential, is very much not absolute. It depends on the choice of the standard state, which in this case is... guess what?

@IvanNeretin I thought the standard for hydrogen is dihydrogen gas...

True, but there is also a standard for $\ce{H+}$.

@IvanNeretin how can one element have two standards?

Well, that's a common convention. People just thought it would be easier this way. The alternative would be to define standard enthalpy for free electron and use it everywhere, which is pretty un-chemical. See, you can easily have a 1M solution of $\ce{H+}$, but to handle a mole of electrons is an entirely different story.

@Ivan but do you have the enthalpy change of dissolution of hydrogen ion?


text:
Why does a certain number of moles of gas occupy a set volume?
In class, I was taught that 1 mole of an ideal gas occupies a set volume eg 22.4 L at STP. Why does the gas only occupy a certain volume and not just keep expanding forever, having limitless volume?
Ideal gases neglect inter-molecular forces so shouldn't they just keep expanding?

As you mentioned yourself, the relationship holds at STP (273.15 K, 100 kPa). Decrease the (outer) pressure and your gas will expand accordingly.

"At STP" implies that there is some pressure (namely 1 bar) pressing on the gas from the outside, right? If there was no such pressure, then the gas is effectively in a vacuum, and yes it would keep expanding.

Please note that the value $V_\mathrm m=22.4\ \mathrm{l/mol}$ for the molar volume of an ideal gas corresponds to the old definition of standard temperature and pressure (STP), which is a temperature of $T=273.15\ \mathrm K$ and a pressure of $1\ \mathrm{atm}$. Since 1982, $p=1\ \mathrm{bar}=100,000\ \mathrm{Pa}$ is used as the standard pressure for tabulating thermodynamic data. At this pressure, the molar volume of an ideal gas actually is $V_\mathrm m=22.710,947(13)\ \mathrm{l/mol}$.

Because the volume is set.

Say you don't wonder how restraining volume can be any different from restraining pressure. Putting into a vessel (volume) per se amounts to putting on pressure.  "Volume. you must assume". Don't put it on ...

The reason the gas doesn't expand to infinity is because it is in a vessel.
Or, to be more precise, the ideal gas equation defines the relationship between temperature, pressure and volume for a fixed amount of gas.
What is normally done to understand the relationship is to put the gas into a vessel of fixed volume and measure the pressure it exerts at a given temperature. The pressure changes with temperature according to the formula (real gases may deviate slightly).
Alternatively, we can use a more complex apparatus that allows us to measure the volume occupied by the gas as we vary the temperature while maintaining a constant pressure (say atmospheric pressure). Same equation, different constraint.
Your misunderstanding is that you have not thought through the conditions under which you are applying the rules. The thought experiment you are doing is releasing the gas in the vacuum of space where there is no external pressure. In those circumstances the gas will expand to infinity but this doesn't violate the ideal gas equation as there is no constraining volume (you could say volume is infinite).
When applying the rules you have three variables (you usually keep the amount of gas, the number of moles, fixed) and you have to constrain two of them. If you don't, are an infinite range of possible combinations of the variables (e.g. constrain temperature and there is still nothing that tells you what the volume and pressure are as they can vary together over any values that satisfy the equation).

For beginners, you rarely find such good advice:  "... you have three variables (you usually keep the amount of gas, the number of moles, fixed) and you have to constrain two of them.  ... " - The question seems to set volume on infinite, and  if the gas "keeps expanding" it seems to do exert pressure - just measure  pressure trying to stop it in the universe. Then, infinite volume seems to demand extreme temperatures to begin with.

That is based on the equation pV=nRT for ideal gases, where p is pressure, V is volume, n is amount, R is a constant, and T is the absolute temperature in kelvin.
The derivation of this equation can be found from many sources, so I am not going to include it here.

I understand that PV=nRT can be derived from Boyle's, Avagadro's, Guy-Lussac and Charle's law but I asking specifically about Avogadro's law and the volume-mole relationship.

The temperature and pressure and amount are all fixed, so the volume is fixed as well. By the way, it is assumed that they are in a closed container, for obvious purposes.

Okay I understand this but I still feel as though it is not answered fully. I am asking specifically why gas only occupies a set amount of space and does not keep expanding forever, if ideal gases neglect inter-molecular forces

This doesn't actually answer anything.


text:
Grignard reagent reaction mechanism: Which bond breaks during the decomposition of the adduct?
We just finished our Organic Chem. chapter Aldehydes, Ketones and Carboxylic acids at school, but I still have one lingering doubt.
Considering the reaction between an alkyl magnesium halide and an aldehyde or ketone: (Diagram's from Wikipedia)

Now my teacher tells me that in the adduct, it's the Mg-O bond that gets cleaved during hydrolysis and not the C-O bond, resulting in an alcohol.
He isn't willing to elaborate ("Aaron, it happens... just like that... you shouldn't worry about all the complexities behind everything...") and I can't seem to find any source on the web that deals with the particulars of bond-breakage.
I have doubts regarding that point of his, since the Mg-O bond has significant ionic character (due to the electronegativity difference between the atoms. But in comparison, the C-O bond is more covalent, and therefore must be easier to hydrolyze. Because (the way my high-schooler brain sees it) ionic bonds are established by strong electrostatic forces of attraction between the ions concerned, while this isn't the case for predominantly covalent bonds. So it seems to make sense that the weaker bond is preferably hydrolyzed, hence I'm mooting for the C-O hydrolysis.
Who's right here? Could someone cite any (reliable) sources?

Why should a covalent bond be easier to hydrolyse? If you insist on making any generalisation at all, surely it should be the other way round. Just look at NaI (ionic) vs AgI (covalent-ish). Water can only break one of those bonds.

As can be seen above, the end product is the same whether we split the $\ce{Mg-O}$ bond or the $\ce{C-O}$ bond.
The problem is, then, which pathway is more favourable?
This, I have no idea. Maybe both pathways are favoured. It doesn't matter because they give the same products.

(If I went out of my way, I would even propose a cyclic rearrangment where the $\ce{H-OH}$ is added to the $\ce{O-Mg}$ bond.)


text:
Competitive 'Wittig' vs 'Peterson' C=C bond formation

If one considers the intermediate above, there are immediately two possibilities for subsequent reaction:
1. A Wittig-type olefination which would lead to a vinyl silane: This reaction happens to be unproductive as the vinyl silane can't do any further chemistry under the reaction conditions
2. A Peterson-type olefination: Which gives a vinyl phosphine (sp.), which may further be deprotonated to give another ylid (with concomitant formation of TMSOH) which will carry on to eventually give an allene.
My question is why is the Peterson faster than the Wittig. The immediate thing that springs to mind is the difference between acid and base catalysed Peterson reactions in which the base mediated version snaps shut before the C-C bond has a chance to rotate (i.e. kinetically very fast), but this isn't an explanation.
Potentially more promising is comparing the electronegativity/positivity of the elements, and noting that silicon is significantly more electropositive and hence a better electrophile however in the Peterson, the leaving group isn't such a driving force as the phosphine oxide from the Wittig.

The Si-O bond is stronger, a stronger interaction is established faster and it is harder reverted.

Is this from a particular publication?

@Marko Are you invoking the Hammond Postulate? This is a kinetic issue, but your argument is a thermodynamic one (hard to imagine either of the two pathways being reversible).

@jerepierre, not to my understanding- just a thought experiment but I'm sure there are examples

Seems like the silicon atom is more oxophilic as compared to the phosphorus atom. This could in principal be explained by the higher strength of the $\ce{Si-O}$ bond, as indicated by the dissociation energies $D$ (see http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html):
$$\begin{array}{ccc}
\hline
\textbf{Bond} & \textbf{Dissociation energy / kJ mol}\mathbf{^{-1}} & \textbf{Bond length / pm} \\
\hline
\ce{Si-O} & 452 & 163 \\
\ce{P-O} & 335 & 163 \\ \hline
\end{array}$$
The $\ce{Si-O}$ bond is stronger thus given an explanation why the Peterson olefination is favoured.
It seems that the bond lengths $r_b$ are the same, though.
By the way, a nice discussion of oxophilicity was given recently in Inorg. Chem. 2016, 55 (18), 9461.

You're looking for the faster rate, so given that a triphenylphosphonium (ever see a model of one of these things?) is a lot bulkier than a trimethylsilyl group, I would argue that the sterics is the main driving force behind the selectivity of this reaction.

Good point : ) I guess only way to really find out if its sterics or oxophilicity would be to repeat the experiment with the $\mathrm{PMe_3}$ system.


text:
How are the atomic orbitals for multi electron atoms obtained?
The problem with Schrödinger's equation is that it isn't exactly solvable for multi-electron species. In the past, atomic orbitals were used to construct a solution for molecules (LCAO). My question is, how were the multi electron atomic orbitals (the ones used for the LCAO) discovered? . Were the atomic orbitals solutions for a one electron system? In other words, were the atomic orbitals calculated the orbital for a $\ce{He+}$, $\ce{Li^2+}$, $\ce{Be^3+}$, etc? What were some approximations that are used to obtain the solution for multi-electron species?

Already in year 1928 Hartree proposed an approximate method of solving the Schrödinger equation for a many-electron atom that became known as the Hartree method (1, 2, 3). The Hartree method is based on representing the many-electron wave function as a product of one-electron ones with the subsequent application of the variational principle. Two years later, Slater (4) and Fock (5) independently corrected the Hartree method for that did not respect the antisymmetry of a many-electron wave function by representing the wave function by a Slater determinant, rather that a simple product of orbitals. The resulting physically more accurate method became know as the Hartree-Fock method.

In the Hartree–Fock method one solve a set of one-electron equations, called the Hartree–Fock equations, of the form
$$
  \newcommand{\op}{\hat}
  \newcommand{\core}{^{\mathrm{core}}}
  \op{F} \psi_i(\vec{q}_{1}) = \varepsilon_i \psi_i(\vec{q}_{1}) \, ,
$$
where $\op{F} = \op{H}\core + \sum\nolimits_{j=1}^{n} \big(\op{J}_{j} - \op{K}_{j} \big)$ is the Fock operator and $\psi_i$ are the spin-orbitals of an $n$-electron system. It is customary to suppose that spin orbitals come in pairs: for each pair of electrons two spin orbitals corresponding to two different pure spin states are constructed out of the same spatial orbital,
$$
    \psi_{2i-1}(\vec{q}_{1}) = \phi_{i}(\vec{r}_{1}) \alpha(m_{s1}) \, ,
    \quad
    \psi_{2i}(\vec{q}_{1})   = \phi_{i}(\vec{r}_{1}) \beta(m_{s1}) \, .
$$
Substitution of such spin orbitals into the Hartree-Fock equations results in a similarly looking system of equations for the corresponding spatial orbitals,
$$
    \op{F} \phi_{i}(\vec{r}_{1}) = \epsilon_{i} \phi_{i}(\vec{r}_{1}) \, ,
    \quad
    i = 1, 2, \dotsc, n/2 \, ,
$$
although, the expression for the Fock operator is different $\op{F} = \op{H}\core + \sum_{j=1}^{n/2} (2 \op{J}_{j} - \op{K}_{j})$. Note here, that for the case of the original Hartree method there was no exchange terms $\op{K}_{j}$.

So, essentially, atomic orbitals of many-electron atoms were obtained by solving the Hartree and Hartree-Fock equations "by hands". Of course, the spherical symmetry of atomic systems greatly simplified the problem, as it can be seen already in the Hartree's original work (1), plus some additional simplification, such as the central field approximation, were often used. But still, in general, a numerical integration was required and it was usually done on some calculating machines (6). Below is the photo (courtesy of AIP) of Douglas Hartree (left) and Arthur Porter (right) viewing one such machine, the meccano differential analyzer.

And the results of numerical integrations were simply tabulated as the values of the radial part of an orbital for different values of distance from the nucleus (7).

1) Hartree, D. R. The Wave Mechanics of an Atom with a Non-Coulomb Central Field. Part I. Theory and Methods. Math. Proc. Cambridge Philos. Soc. 1928, 24 (1), 89–110. DOI: 10.1017/S0305004100011919.
2) Hartree, D. R. The Wave Mechanics of an Atom with a Non-Coulomb Central Field. Part II. Some Results and Discussion. Math. Proc. Cambridge Philos. Soc. 1928, 24 (1), 111–132. DOI: 10.1017/S0305004100011920.
3) Hartree, D. R. The Wave Mechanics of an Atom with a non-Coulomb Central Field. Part III. Term Values and Intensities in Series in Optical Spectra. Mathematical Proceedings of the Cambridge Philosophical Society, 1928, 24 (3), 426–437. DOI: 10.1017/S0305004100015954.
4) Slater, J. C. The Theory of Complex Spectra. Phys. Rev. 1929, 34, 1293. DOI: 10.1103/PhysRev.34.1293.
5) Fock V. Näherungsmethode zur Lösung des quantenmechanischen Mehrkörperproblems. Z. Physik. 1930, 61 (1), 126–148. DOI: 10.1007/BF01340294.
6) Hartree, D. R. The Differential Analyser. Nature. 1935, 135, 940-943. DOI: 10.1038/135940a0.
7) Hartree, D. R. and Hartree, W. Results of Calculations of Atomic Wave Functions. III. Results for Be, Ca and Hg. Proc. R. Soc. London, Ser. A. 1935, 149 (867), 210-231. DOI: 10.1098/rspa.1935.0058.

Wildcat's answer above had some of the early history. I'll continue a bit further.
Atomic calculations typically don't employ Hartree-Fock in the modern terminology, since due to the large amount of degeneracy of the atomic orbitals (all $2l+1$ magnetic subshells have the same energy) you actually have a problem with significant static correlation, which causes Hartree-Fock theory to fail; see e.g. my recent paper Phys. Rev. A 101, 012516 (2020).
Instead, the calculations are traditionally carried out with "multiconfigurational Hartree-Fock" - which in modern terminology is just a small complete active space calculation. The thing about atoms is that the high degree of symmetry actually allows you to figure out (with considerable effort!) what the coefficients of the various possible configurations should be in your wave function, and write down closed-form differential equations for the spatial orbitals which you can then use to solve the equations to very good accuracy. Moreover, the spatial orbitals become just one-dimensional radial functions, which are very simple to represent on a computer.
Indeed, fully numerical atomic calculations have a very long history. One of the big names in the field is Charlotte Froese-Fischer (née Charlotte Froese), who still appears to be actively engaged in research despite being over 90 years old; for instance, here is her paper on numerical solutions of the Hartree-Fock equations on an automatic computer from the early 1960s: Canadian Journal of Physics, 1963, 41(11): 1895-1910
However, the question also mentions LCAO and states that atomic orbitals used to be used to construct wave functions for molecules. However, the LCAO approach is still arguably the most commonly-used method for all-electron calculations, and is used in software packages employing Gaussian-type orbital basis sets as Gaussian, Gamess, Orca, Dirac, Dalton, Turbomole, Psi4, PySCF, Crystal, eT, MPQC, etc, as well as others using Slater-type orbital basis sets (e.g. the ADF program) as well as those that use numerical atomic orbitals (e.g. Siesta, FHI-aims).
While the numerical atomic orbitals are solutions to the gas phase atom (typically within a density functional approximation with some further constraints!), it is actually not necessary for the atomic basis functions to coincide with atomic orbitals; rather, the important thing is that the atomic basis can span the atomic orbitals in a meaningful way.


text:
Elimination of OH from spirocyclic compounds
I was wondering, what would happen if I treated the following compound i.e. trispiro[2.1.25.1.29.13]dodecane-4,8,12-triol with sulfuric acid[$\ce{H_2SO_4}$] to eliminate OH group to form double bond:

Will the double bond form? if yes then where and why? If no then what other compound will form and why?

Well, there's gonna be various rearrangements

Agree, probably a lot of 1,2-alkyl shifts.

You might get a carbonyl from a hydride shift...

Have you tried to draw the structure of a compound resulting from elimination?

@jerepierre I am really confused about what would happen. I think that there will be a ring expansion for each of those cyclopropane rings, forming a four member bridged ring. But then where would the double bonds form? Bredt's rule disallows formation of double bonds at bridgehead.

@zhe Can you explain a bit more about the formation of carbonyl be hydride shift?

Posted in chat: http://chat.stackexchange.com/rooms/47124/elimination-of-oh-from-spirocyclic-compounds       Sorry for the crappy drawing.

Ciclopropylmethanol-cyclobutanol rearrangement is  a thing.

This question has been unanswered for quite long, and meanwhile, I have found an answer, so I wanted to share it.
On treating the compound with sulfuric acid, the $\ce{OH}$ groups will be protonated to $\ce{OH_2^+}$.

The $\ce{OH_2^+}$ groups are good leaving groups, so they will leave as $\ce{H_2O}$, forming carbocation.(one by one)

As cyclopropane is under significant ring strain, it will undergo ring expansion to form cyclobutane ring, which is somewhat less strained. At the same time, $\ce{H^+}$ ions will be expelled, forming double bonds. (It should also take place step by step.)

In this case I don't think that the double bond will not cause any significant barrier to the formation of the molecules, as the double bond is here the bridging bond, not at the bridgehead[formation of double bonds at bridgehead is not possible for smaller rings(<8 carbon rings), as it causes very high strain: Bredt's rule]
I have considered only elimination in this reaction scheme. Other compounds may also be formed. Feel free to post answers if you think other compounds will be formed or if you think the above mechanism is wrong in any way.

I don’t think the steps would happen to each carbon simultaneously but other than that this sounds reasonable.

@Jan Why not simultaneously?

Carbotrications seem like a far-fetched idea to me...

@CurtF. So what is your idea?

sequential not simultaneous reaction

@CurtF. Edited answer. Sequential reaction seemed to be more favourable, but when I wrote the answer, I was thinking about the end product only.

Double bonds won't form in this scenario. It still would be a triol probably.

@Zhe http://chemistry.stackexchange.com/q/9581/7475 ;)

@Jan I'm aware of the homoallyl cation. My issue is when it's tied by with a cyclohexyl ring. Can you look in Scifinder to see if there's migration in a spiro example?

@Zhe L. Fitjer, M. Majewski, A. Kanschick, E. Egert, G. M. Sheldrick, Tetrahedron Lett. 1986, 27, 3603. DOI: 10.1016/S0040-4039(00)84860-1

I am now genuinely confused after looking at the solution of this problem. I have seen situations where in acidic medium, if the carbocation was extremely stable, then we can leave it just like that without proceeding further into the reaction. So will aromatization provide more stability, or will the sigma resonance of the cation be enough to allow the cation to exist as it is? (I mean, I thought that in aromatic compounds is due to delocalization of the electrons, so surely more energy would be released when a sigma bond breaks...?)


text:
A surface with variable roughness
As you may know, wettability is a characteristic of each surface which determines what the contact angle would be when a droplet is put on the surface (see here for more info). Surface roughness has a significant effect on wettability. For instance, if you increase the roughness of a polymer surface, the contact angle will be increased as well (i.e. it becomes more hydrophobic).
I need to create a surface such that I can control its wettability by an external stimulus (e.g. temperature, etc.). For instance, a polymer (or any other material) in which when I change its temperature, the surface roughness is also changed (due to change in the molecules chains shape or crystallization). I need to know which materials I can use to create this kind of surface.


text:
Is preference given to long alkyl group or branched cycloalkane?
Could you tell me whether it will be 1-(1-ethylcyclopropyl)hexane or 1-ethyl-2-hexylcyclopropane ? I don't know the rule or the preference order.

Related: How do I decide the parent chain of an organic compound that has a ring as well as a side chain while naming the compound?, IUPAC nomenclature Cyclic vs Acyclic, Nomenclature of cycloalkane with substituent having the same number of carbons, IUPAC Naming: 2-phenyloctane or 2-octanylbenzene, and Naming compound containing a ring (cycloalkane) side chain?

The proposed name "1-(1-ethylcyclopropyl)hexane" describes a different compound. You might want to check the locants.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), two methods are possible for naming systems composed of rings and chains: either the ring or the chain can have seniority. To construct the preferred IUPAC name (PIN), the ring is selected as the parent structure.

P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)
Two methods are recognized to name systems composed of rings and chains
  (exclusive of linear phanes).
(1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.
(2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.
(…) For selection of a preferred IUPAC name, see P-52.2.8.

P-52.2.8 Selection between a ring and a chain as parent hydride
Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).

Therefore, the preferred IUPAC name (PIN) of the compound that is given in the question is 1-ethyl-2-hexylcyclopropane.

In general nomenclature, however, the context may favour the chain, so that the parent structure has the greater number of atoms in the principal chain. The corresponding systematic name for the compound given in the question is 1-(2-ethylcyclopropyl)hexane.
Note that according to the obsolete 1979 recommendations, a hydrocarbon containing a small ring attached to a long chain was generally named as a derivative of the acyclic hydrocarbon. Many books are based on these nomenclature rules and thus may expect an answer that favours the chain.


text:
pH of buffer solution
Why does ammonium formate has less pH than ammonium acetate . Even though both the salts have the same structure..
I thought that dissociation of acetic acid and formic acid would have a role in this phenomenon, and I thought that acetic acid forms dimer but I don't know if formic acid forms a dimer or not. Do these things have a role in their pH.?

Yes, formic acids forms a dimer (nearly all carboxylic acids do), though probably not in the dilute water solution. No, this is not important. Just look up the pKa of both acids.

pKa of acetic acid is 4.8 and pKa of formic acid is 3.8 . And what factors make them different.

Well, they are different compounds, so it's only natural for them to differ in everything, including pKa. Now, which of the two is stronger? and hence which is more prone to hydrolysis?

I think formic acid is stronger than acetic acid so, it will hydrolyse more. Is that correct@ Ivan neretin

You have it backwards: salts of stronger acids hydrolyze less.

Got that salts of stronger acid hydrolyse less than salts of weaker acids, and same will be the case with bases . Thank you@ Ivan Neretin.


text:
Iodine test for Starch
The starch and iodide test is to detect amylose, a polysaccharide. We add an iodide solution to a sample and observe a color change with the starch; it turns dark blue.
While researching information online, I understood that $\ce{I_{2}}$ is not soluble in water, hence it needs to react with iodide ions present to at least create $\ce{I_{3}}$, which is soluble in water, and also forms polyiodide complex, as research shows. I also understood that this linear molecule of iodide migrates to the center of the helicoid polysaccharide structure. Then, with some electronic interactions, there is a modification in the electron configuration which produces the dark blue color.
My questions are as follows. What are the electron's movements? Which atoms are implicated in this transfer of electrons? What is the electron configuration change that results in the color change? In other words, I don't understand the electron's interactions between iodide and amylose that are creates the color change.

The amylose present in starch is responsible for the deep blue colouration.  The structure of amylose consists of long polymer chains of glucose units connected by an alpha acetal linkages. All of the monomer units are $\alpha$ -D-glucose, and all the alpha acetal links connect $C1$ of one glucose and to $C4$ of the next glucose.
As a result of the bond angles in the acetal linkage, amylose actually forms a spiral structure.
X-ray diffraction studies$^{[1]}$, demonstrate that the amylose is organized as left-handed helices, which have outer diameters of 13 Å and a pitch of 8 Å, with each turn of the helix corresponding to six 1,4-anhydroglucose units. The iodine components are linearly arranged in the 5 Å wide inner cavity of the helices with an $\ce{I–I}$ distance of approximately 3.1 Å.
Starch-iodine complexes show dichroism of flow$^{[2]}$, i.e light with its electric vector parallel to the flow lines is more strongly absorbed than light with its electric vector normal to the flow lines. The dichroism of flow is shown to require that the long axes of the iodine molecules in the
complex be parallel to the long axis of the starch iodine complex.
The exact structure of the polyiodide chain remains contentious, however, the following species are commonly implicated as key elements of the substructure:
$$\ce{I2 + I^- -> I3^-}$$
So, ignoring the structural specifics of the polyiodide chain, one notes that if an iodine molecule enters the internal channel of the amylose
helix, it may be surrounded by about six glucose residues. The following image is from reference (3):
Bear this image in mind, and note that the hydroxyl oxygen is pojected into the helix cavity.
Also, when iodine is dissolved in ether or alcohol, a new strong absorption band appears in the near ultraviolet region. In relation to the study of the benzene-iodine complex, Mulliken postulated the complex formation
between iodine and ether or alcohol, and explained this new band by the concept of intermolecular charge transfer spectra$^{[3]}$.
The "simple" idea behind a charge transfer complex is given below:
$$\ce{I2 + D -> I^{\delta -}...D^{\delta +}}$$ Reference 4 gives more details. (It also likens this bonding to a "hydrogen bond" if that helps, you understand). In this case, D is the donor atom (electronegative atoms like nitrogen, oxygen etc.)
The iodine-donor distance is usually in 2~3 Å range. Thus, based on all of this one can conclude that possible charge transfer interactions between oxygen atoms that are projecting in towards the helical cavity and the iodine molecules in the polyiiodide chain are responsible for the colouration. You can view how the oxygen molecules are pointed towards the chain of iodine here. (I recommend rotating the diagram so that the axis of the cavity is towards you, and you are essentially looking down the tube)
References
1.https://www.researchgate.net/publication/244335699_The_complex_of_amylose_and_iodine

http://pubs.acs.org/doi/abs/10.1021/ja01244a017
http://scitation.aip.org/content/aip/journal/jcp/22/3/10.1063/1.1740076 (I recommend reading this if you are not satisfied with the brief, qualitative discussion I have provided here)
Inorganic Chemistry (Nils Wiberg) (google books, relevant portion is part of the preview)

I do not have access to the article but it sounds interesting :"Also, when iodine is dissolved in ether or alcohol, a new strong absorption band appears in the near ultraviolet region." Do you know what could cause the slight color change between collagen and starch?

I can edit my answer to include some more details from the paper, and/or if you give me your personal contact in chat, I can download a copy and forward it to you.


text:
Why do spectra appear continuous despite their allowed transitions being discrete?
So in my physical chemistry class we have been looking into radiation, selection rules, etc.
Taking into account the discrete levels of energy an electron can have, I cannot understand why fluorescence and absorption spectra of compounds have continuous lines instead of discrete values.
If electrons in a system can only have a select number of energies, why is it that these graphs show a continuum of absorption and emission?
I also have this problem with blackbody radiation, since it is a continuous line for any temperature.

Related: http://chemistry.stackexchange.com/q/35658/16683 This answer may help you. I distinctly remember we had a question on this, but maybe I'm just imagining it. Or maybe I'm thinking of the one I linked you.

Related: 1, 2.

You are right that absorption and emission are quantized processes, but UV-visible peaks represent a broad range of wavelengths because there are multiple vibrational states of their representative ground and excited electronic states.

In the case on my diagram, you have six different quantized electronic transitions with slightly different ΔE values. In reality, there are more energy levels than that, plus the rotations and vibrations are constantly changing the orbital energy levels as a result of interactions with the matrix, so that helps fill in the peaks. The peak shape can actually be changed by altering the temperature. This gives rise to high-resolution UV-Visible spectrophotometric techniques, which have a few specific applications.

One reason, not mentioned in the previous comment, why this is the case is that in most cases, some of the photon energy can be accounted for by the translational energy levels, which are so close together, as to be functionally continuous, hence, various photon energies can cause the main transition plus or minus some translational energy

The difference in the energy scale between translational motion and most optical absorption/fluorescence spectroscopies means that you won't see this effect directly from translation. It would appear due to changing local environments from solute diffusion.


text:
Assigning Re and Si face for Carboxylic group
Suppose I have a compound $\ce{CH3COOH}$ and I have to assign Re and Si faces. Which of the two oxygens get a higher priority and why?

The carbonyl oxygen gets higher priority because it has a real oxygen bonded via $\sigma$ bond and also the double bond, but we count that as the oxygen bonding back to the carbon. We don't treat it like a real carbon; it's a "ghost" or "phantom" carbon with nothing attached to it. This carbon counts as a carbon but is lower priority than any other real carbon. That means that -O-(C) outranks -O-H.


text:
How to now if the OH group on the C5 carbon of a cyclic sugar is up or down in Fischer Projection
I know when using fischer projection I have to look if it's a D or L configuration. if it's D: up = right and down = left and if it's L: up = left and down = right. I think the concept is clear to me, however I dont't know if the OH group on the C5 carbon atom is pointing up or down. For example:

When I have to draw those cyclic molecules to fischer projection I never know if the OH group which will be shown in the fischer projection (not in the clyic structure because it's a hemi-acetal then) is pointing up or down. What is the rule for this?

I want to get this:

But I don't know how to know the side of the circled OH group

I'm not sure what you mean. Generally, a Fisher projection is drawn vertically with the OHs on the left and right. The carbonyl appears at the top. Up and down only make sense if there is a ring. But the C5 carbon doesn't have an OH. The C1 carbon has an OH which is clearly up, but in the Fisher projection, that's the carbonyl carbon so there is no stereochemistry in the projection. Usually, this OH is in the axial position because of the anomeric effect from the adjacent ether's lone pair. See: https://en.wikipedia.org/wiki/Fischer_projection

I updated my question hopefully it clear know @Zhe

I think it's as simple as L for left and D for right... I'm not sure what you mean by up or down still. The OH you've circled in red is the O that's in the ring. The other OH of the hemiacetal is from the carbonyl (that's the one pointing up in the ring on the right).  It could be down as well. That is still the same Fisher projection.

What I mean when you have a D Sugar the rule is in the Fischer projection of the linear form OH pointing to the left ==> up in cyclic and OH pointing to the right ==> down in cyclic. However how can you determine to which side to OH is pointing on the 5th C if you don't know if it is pointing up or down (and if you don't know if the cyclic form is a D or L) @Zhe

Your question is complicated by the fact that, in your example, the placement of  C-5 hydroxyl in the Fisher projection can't change because it is D. I think the main issue is that you need to learn how to convert a Fisher projection into the ring form and vice versa. I can't tell you explicitly up or down because it matters how you draw the ring in the L- form. To my knowledge, there is no canonical form, but I might be wrong there...

I just found my answer here: http://forums.studentdoctor.net/threads/l-or-d-sugar-from-ring.1090022/ maybe you are intersted in reading it.
For the Haworth projection:
If the -CH2OH substituent is RIGHT of the anomeric carbon, it's D.
If the -CH2OH substituent is LEFT of the anomeric carbon, it's L. @Zhe

I'm not sure right or left make sense in a Haworth projection. Honestly, I'm a bit lost on what your specific confusion is because I just think about this in terms of stereochemistry as I was trained to do as an organic chemist. With that in mind, specific mnemonics that involve relating three things aren't particularly helpful. Come to chat if you need more of a discussion.

The OH group you've circled is always on the right hand side for D-sugars and on the left for L-sugars as its position is in the very definition of D/L sugars.


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Error in calculating H+ activity
I'm starting to study thermodynamics of minerals and rocks, so sorry for the basic question. I am trying to work out some exercises of a book and I found this:

Calculate the activity of $\ce{H+}$ when the activity of $\ce{Sn^4+}$ is $10^{-10}$ for the reaction $\ce{SnO2 + 4H+ <=> Sn^4+ + 2H2O}$. Standard free energies of formations $\Delta$Fof are: SnO2 = -123.20 kcal; H2O = -56.69 kcal; Sn4+ = 0.65 kcal; H+ = 0.

I used the following equations:

$\Delta$For = ($\Sigma$$\Delta$Fof,products) - ($\Sigma$$\Delta$Fof,reactants)
$\Delta$For = -RTln(K), where R = 0.001987 Kcal/molK and T = 298.15K
K = $\frac{[Sn^{4+}]}{[H^{+}]^4}$, given that the activities of SnO2 and H2O are unit.

The result I found was $3.82 \cdot 10^{-5}$, but the book says the result is $10^{-0.40}$. I assume this is a fundamental error, but given it is also a fundamental principle I need to understand it to proceed in my study.
Thanks in advance for any help.

There got to be some constants, right? Also, your answer looks more realistic to me.

Sorry about that. Just added the standard free energies of formation

What units are the free energies of formation?  kJ / mol?  kcal / mol?  pettajoules per picomole?  picojoules per exagram?

We'll also need temperature...

Your value for $R$, above, should not be negative. It will help you to always use proper units. The Kelvin doesn't get a "degree" sign. Finally, your value for $R$ should include a per mole unit.

$$ \Delta G = \Delta G^\circ + RT \ln{\frac{a_{\ce{Sn^{4+}}}a_{\ce{H2O}}^{\;\;\;\;2}}{a_{\ce{SnO2}}a_{\ce{H+}}^{\;\;4}}} $$
At equilibrium, $\Delta G = 0$.
For $\Delta G^\circ = 2\times(-56.69) + 0.65 - (-123.20 - 4\times0) = +10.47\mathrm{\frac{kcal}{mol}}$
The question says "activity", but in the absence of any other information, let's assume that "activity" is equal to "concentration".  That is, let's assume activity coefficients are 1.
Thus, we get $$ \Delta G = 0 = +10.47  + RT \ln{\frac{a_{\ce{Sn^{4+}}}\cdot1^2}{1\cdot a_{\ce{H+}}^{\;\;4}}} $$
Thus, we get $$ {\left( \frac{-10.47}{RT}\right)}  = \ln{{10^{-10}} a_{\ce{H+}}^{\;\;-4}} = -4\ln{a_{\ce{H+}}} -10 \ln{10} $$
$$\frac{\frac{-10.47}{RT} + 10 \ln{10}}{-4} = \ln{a_{\ce{H+}}}$$

I calculated with the parameters given in the problem and your reasoning gave a third result: 0.2623.
I know why my answer and yours are different.
you used $\Delta$G = $\Delta$Go - RTlnK, whereas I just used $\Delta$G0 = - RTlnK. I guess I am now even more confused. Was I supposed to assume equillibrium ($\Delta$G = 0) even if the problem doesn't specify?

I got 0.262 without assuming $\Delta G = 0$. Perfectly workable with the equations you have above. Unfortunately, this is not the $10^-0.4$ you were looking for...

Could you reply showing me how you did it without $\Delta$G = 0?  Can't get to that using the equations I showed.

If you use the $\Delta G^\circ = - RT \ln{K}$ equation, then you are assuming that $\Delta G = 0$.  That's how you get that equation.  (It comes from $\Delta G = \Delta G^\circ + RT \ln{K})$.

Also this discussion made me realize I made a sign error above in my answer.  I'll edit to correct.


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Reaction between Sudan IV and lipids
The Sudan IV is a molecule used to detect lipids in a substance, especially the triglycerides. The Sudan IV molecule formula is  $\ce{C24H20N4O}$ and a triglyceride molecule could have this molecule formula $\ce{C55H98O6}$.
When the two solutions are mixed, there is a color change. What causes this? I assume that it is because a new compound is produced and this compound is red colored.
Is it possible to find the chemical equation of the reaction between Sudan IV and triglyceride?

The fact that it is sparingly soluble in water but v soluble in lipids means that it concentrates in lipid which is why its colour appears stronger. It does not significantly change colour on entering the lipid.

I don't think that's how Sudan IV works. Sudan IV is a highly conjugated aromatic diazo compound, so it is already highly colored. In the lipid, the Sudan IV is basically hydrophobic so it will stain any lipid droplets suspended in water. The color is due purely to the Sudan IV, not a reaction.


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Flood Control - electrolysis - limit to rate of conversion?
Flood control - is it possible to use electrolysis to remove flood waters by converting water into oxygen and hydrogan gas?   Is there a limit to the amount of water that can be converted at a time?
I know that this is a bit crazy and I respect what you are saying, BUT I am not trying to vaporize the entire Mississippi river, only enough to keep flood waters from ebbing into a house.
Maybe a different appreach - how about freezing the water place (in blocks) and then transporting it elsewhere
if only water was compressible....

Sorry, but I'm afraid you don't know what you're talking about. That's like a least viable thing. Maybe even trying to drink it could be better.

Well, it's rather late for changing ideas. You could write another question, but freezing isn't a good idea. Water is simply pumped out.

Floodwaters are immense in quantity.  Let's estimate how much energy would be required to convert that quantity of water to $\ce{H2}$ and $\ce{O2}$.

The Mississipi river has a (non-flood) flow rate of about 593,000 cubic feet per second, or 17000 cubic meters per second.  That's $1.7 \times 10^7$ liters per second, or $1.7 \times 10^{10}$ grams per second, or roughly $10^9$ moles per second.
Faraday's constant is the conversion factor between moles and Coulombs.  It's value is about 96000 ampere-seconds per mol.  Let's approximate it as $10^5$ ampere-seconds per mole.
It takes four moles of electrons per mole of water to complete the electrolysis.
Multiply the numbers together, and you get that to electrolyze all the waters of the Mississippi river, you'd need $4 \times 10^{14}$ amperes of current.  That's about the amount of current present in a billion bolts of lightning.  Yikes!
Exactly how much voltage would be required depends on your design of electrolytic cell you use for this giga-bolt scale.  Let's say it's 5 volts.  That means the power you need for the electrolysis is $2\times10^{15}$ watts.  That's 2000 terawatts.  But the world's current total energy consumption is about 12 terawatts...


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What are the factors affecting Boiling Point in Ions?
I understand that for molecular substances, intermolecular forces such as van der Waals forces and hydrogen bonding are the major factors effecting the boiling point.
However, I am not sure about what factors effect the boiling point in Ionic substances.  I assume it is to do with the strength of the bonds between ionic units, but then what effects this strength?
Thanks for your time!


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Why don't calcium chloride and sodium nitrate react?
$$\ce{CaCl2 + 2NaNO3 -> 2NaCl + Ca(NO3)2}$$
Why doesn't the above reaction occur? Does it have to do with the metal activity series or solubility rules?

It's because all the species are soluble in water and there is no driving force such as precipitation or gas evolution. You just have a mixture of ions.


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Do we cool the solution to form a precipitate because of lower melting points?
If we have two compounds in aqueous solution, in order to separate one from the other the solution is placed into an ice bath. For example the formation of $\ce{KClO3}$ from a combination of $\ce{NaClO3(aq)}$ and $\ce{KCl(aq)}$.
I am trying to understand why potassium chlorate precipitates out of solution first as the temperature is lowered. My hypothesis is that the melting points for potassium chlorate, which is $\pu{356 ^\circ C}$, is lower than that of sodium chloride, which is $\pu{801 ^\circ C}$.

Potassium chlorate precipitates because it's only slightly soluble is water compared to Sodium chlorate and Potassium chloride. The ice-bath only helps to speed up the process as solubility is temperature dependent.

The reason for cooling is related to solubility at different temperatures not melting point
The reason many precipitation reactions cool the solution to generate a precipitate (or to maximise the yield of it) is because the solubility of most substances in most solvents decrease with temperature. For those substances where the solubility is much lower at low temperatures cooling make a big difference to the amount of precipitate you get.
But some substances are so insoluble cooling won't make much difference and they precipitate immediately as soon as the reaction happens. For example: Sodium sulphate is much less soluble in cold water; temperature makes little difference for sodium chloride; and cerium sulphate is more soluble in cold water. So the general idea of cooling to precipitate a solid won't always work.
None of this has any relationship to the melting points of the solid reactants or products.


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Bond length and bond strength correlation
Is bond strength inversely proportional to bond length? As the bond length decreases, does the bond strength increase?

The decrease of the bond length is the effect of the increase in bond strength.

Decrease in bond length and increase in strength is only a trend, that is observed only between the same pairs of atoms, so its approximately true for C C bonds, single, double triple and similarly for N N bonds. But in general this is not true as can be seen in the following list of bonds with  approximately the same length e.g.
$\ce{H-Br,141 pm 366} \pu{ kJ mol^{-1}}$
$\ce{F-F, ~142 , ~~~~158}$
$\ce{C-O, 143 ,  ~~~~360}$
$\ce{O-O, 148 ,  ~~~145}$
$\ce{N-N, 145 ,  ~~~170}$


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Why do universal indicators in titration have to be swirled?

Why do universal indicators in titration have to be swirled?

Here's a specific example: I am using methyl orange along with hydrochloric acid to test the alkalinity of a set of household cleaners (titration), and I have to constantly swirl the conical flask with the cleaner inside, but why?

Diffusion isn't fast enough, so you're trying to make sure that you're making the pH of the solution as uniform as possible. You can't want a pocket of acidity or basicity to throw you off from the equivalence point.
People also use stir bars to do this if they don't want to swirl all day long.


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What is the name of the compound B2F2?
In my homework I am asked to identify the compound named B$_2$F$_2$. I tried diboron diflouride and it didn't work. Is it just boron fluoride?

It is diboron difluoride. You can even see the answer just by typing B2F2 on google.

Answer
There is a typo in your answer, as is stated in my comment.
It should be diboron difluoride instead of diboron diflouride.

Origin
"Fluoride", "fluorine", "fluorite", are all related. They all refer to the element $\ce{F}$ or its compounds.
It turns out that "fluorite" ($\ce{CaF2}$, more commonly known as fluorspar), a mineral, in the origin of the "fluor-" prefix.
The name "fluorite" comes from the Latin verb "fluō" which means "to flow", since this mineral is used "to lower the melting point of metals during smelting", hence making them easier to become liquid, i.e. to flow.
The English word "fluid" also derives from the same Latin verb.