text:
What is the reason for the blue color of an ammoniated electron?
I am quoting a line from a book regarding reactivity of alkali metals with liquid ammonia.

The blue colour of the solution is due to the ammoniated electron which absorbs energy in the visible region of light and thus imparts blue colour to the solution

Now my question is:

How does the independent electron get excited (without transitional levels)?

Why do salts of particular alkaline metal always have a fixed flame color if transitions of electrons can be from any energy level.
(I am asking this question based on my knowledge of school where we only studied transitions among hydrogen like species)

Independent (free) electron can get excited as well as any other (there are a couple of peculiarities, but that's an offtopic here). The thing is, an electron in the middle of a cavity formed by the solvent molecules is not free; hence the word "ammoniated".

@IvanNeretin-Don't you think the ammoniated electron can turn a bronze color also? I think so myself until you dilute that solution where it stays, this one is the fact.

Both of your questions could benefit from more background, so please forgive me if I understood something wrong. However, I'll try to answer as best I can:

The excess electron in liquid ammonia is not really independent, otherwise, as you stated correctly, the electron would not absorb a certain wavelength but act as a black radiator/absorber like in a plasma (for example the sun). However, how the electron is bound and how the transitional states are formed is not completely understood yet. In this paper, I found the following sentence: "The excess electron in liquid ammonia ("ammoniated electron") is commonly viewed as a cavity electron in which the s-type wave function fills the interstitial void between 6 and 9 ammonia molecules." And then they go on to present a different hypothesis...
In a flame, the spectrum is recorded in the gas phase and formed not by the salt but by the individual atoms (even if you put the salt into the flame). Thus the spectrum depends only on the elements contained in the salt, not on the type of salt. This is different, if the spectroscopy is performed in the liquid phase of alkali metal halides. There the spectrum depends on the mole ratio of the anions and cations.

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To expand on Thawn's answer to #1 a bit:
In quantum systems, a particle is bound to discrete states when it is constrained by a sufficiently deep potential well; otherwise its energy is continuous$^\dagger$.  So, the existence of solvated electrons (alt., electrides) implies that the potential field established in certain solvents in certain situations contains "deep enough" potential wells to bind an electron, but at locations "far away" from any of the nuclei in the system.
Prediction of when a solvated electron will occur remains challenging, to the best of my knowledge.  One potential method for their theoretical prediction/explanation is by a search for "non-nuclear attractors" (NNAs) in the electron density, using methods such as the QTAIM of Bader$^\ddagger$.  (QTAIM studies the spatial distribution of the electron density and its derivatives, mainly the gradient and Laplacian.)  Attractors in QTAIM are "concentrations" of electron density that most frequently occur at atoms (nuclear attractors), but sometimes are observed far from atoms (non-nuclear attractors). Thus, the presence of an NNA in an electride system provides a reasonable guess for the location of the potential well(s) holding the 'independent' electron.  However, in some electride systems no NNAs are apparent, so the absence of an NNA is not diagnostic of a non-electride system (see, e.g., the post and comment thread here).
More generally, research into solvated electrons is quite active, including work on excited states:

Electronically excited states in size selected solvated alkali metal atoms III: Depletion Spectroscopy of $\ce{Na(NH3)}_n$-Clusters
Preprint of a 1998 paper $-$ PDF link
Electron solvation dynamics in $\ce{I^- (NH3)}_n$ clusters
Faraday Discuss 115: 49 (2000) $-$ doi:10.1039/a909865h $-$ PDF link
Dynamics of Electron Solvation in Molecular Clusters
Acc Chem Res 42(6): 769 (2009) $-$ doi:10.1021/ar800263z $-$ PDF link
Time-Resolved Excited State Energetics of the Solvated Electron in Sodium-Doped Water Clusters
J Phys Chem A 118(37): 8517 (2014) $-$ doi:10.1021/jp502238c
Dynamics of electron solvation in methanol: Excited state relaxation and generation by charge-transfer-to-solvent
J Chem Phys 142: 234501 (2015) $-$ doi:10.1063/1.4922441
$-$ PDF link

Others can be easily found by searching for, e.g., 'excited states of solvated electrons' $-$ for full-text PDFs only, in Google you can add the filetype:pdf flag to the end of the search string.
$^\dagger$G. Herzberg, Atomic Spectra and Atomic Structure. 2nd Dover Edition, 2010. (Amazon)
$^\ddagger$R.W.F. Bader Chem Rev 91: 893 (1991). doi:10.1021/cr00005a012

Hmm, this "non-nuclear attractors" idea seemed rather dubious for me for quite some time now. It's just a prosthesis for limited theory imo.

@Mithoron I'm not sure what you mean. NNAs are objective features of the density distribution: local maxima not located near any nuclei. They appear in ab initio wavefunctions and DFT both. Their interpretation is certainly up for debate, but AFAICT their existence is indisputable.

Hmm, it was only me connecting it with actual physical attraction, while those are attractors from dynamical systems theory. So never mind, I guess.


text:
MgCl2 acidic or neutral in water?
I'm currently taking chemistry 12. On our test we were asked, when given a $0.1~\mathrm{M}$ solution of certain compounds whether the resulting solution when added to water would be acidic, basic, or neutral.
At first, I thought the solution would be neutral, because we learnt in class that alkali earth metals and chloride are spectator ions and therefore don't interact with water and pH.
However, I then realized that $\ce{Mg(OH)2}$ has low solubility in water, and I assume that adding $\ce{MgCl2}$, would dissociate and then precipitate $\ce{Mg(OH)2}$ from the solution, therefore the resulting solution would be acidic. My final answer was acidic.
The teacher marked my answer wrong, but when questioned could only say that $\ce{Mg}$ and $\ce{Cl}$ are both spectators, but was unable to give direct contradictory proof against my answer.
I looked elsewhere online and Wikipedia states that anhydrous $\ce{MgCl2}$ is a Lewis acid, despite being a weak one. But anhydrous seems to mean without water, so I currently have no idea which answer is correct.
I would like an explanation to why it is acidic or neutral and why the contradicting argument (both spectators vs low solubility of $\ce{Mg(OH)2}$ ) is invalid or flawed.

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. I edited your post to include MathJax for chemical formulae. You can learn more about that in the [help], this FAQ post or this meta-post.

What kind of proof would you accept? $\ce{MgCl2}$ added to pure water would not precipitate $\ce{Mg(OH)2}$, that much I can attest from firsthand experience. (This does not really prove anything, but whatever.)

@IvanNeretin Well I guess I'd like an explanation as to why either argument is invalid besides just experimental proof. But can you give me details about your experience?

As a physics teacher turned chemistry teacher, I cannot be 100% sure, but I think the answer is getting at the complex formed with water.
$$\ce{MgCl2 -> Mg^2+ + 2Cl-}$$
$\ce{Cl-}$ is pretty unreactive with its high electronegativity and completes outer shell of electrons (hence why $\ce{HCl}$ for example gives almost 100% dissoscation and does not reform).
Well, for the $\ce{Mg^2+}$ on the other hand, water molecules can now use their lone pair of electrons on the oxygen atom to form a dative bond with the metal ion (this seems to happen all the time!) forming $\ce{[Mg(H2O)6]^2+}$.
If it accepts a lone pair, it is a Lewis acid (perhaps this is what your wikipedia research showed).
ANYWAY, now the acidic bit... So you have an $\ce{Mg^2+}$ ion, surrounded by 6 water molecules. Due to the high charge density around the ion (I guess this part of my thinking means that the likes of strontium, further down the period are less acidic?), it can actually help pull the electrons in the $\ce{O-H}$ covalent bonds in the water molecule (which are already being pulled away from the H atoms due to the high electronegativity on oxygen). If the electron pair is pulled enough, it will be beyond the electrostatic force of the proton in the hydrogen atom and hence the proton will dissociate into the solution leaving behind $\ce{OH-}$ (no longer water).  This $\ce{H+}$ (no longer $\ce{H}$, because it loses its electron in its shared electron pair) goes into the solution, thus increasing the $\ce{H+}$ concentration, thus decreasing the pH.
SHORT ANSWER:

$\ce{Mg^2+}$ forms a complex with water - $\ce{[Mg(H2O)6]^2+}$
The high (positive) charge density on the $\ce{Mg}$ ion hydrolyses the surround water ligands
This releases $\ce{H+}$ (protons) into the solution, decreasing the pH slightly.

This seems like the correct answer to me, looking at the answers to https://chemistry.stackexchange.com/questions/49825/is-magnesium-sulfate-basic-neutral-or-acidic.

Long story short, the ultimate cause of non-neutral pH in the solutions of some salts is hydrolysis. If a salt is made from a weak acid and strong base, then the acid anion would get partially protonated by water, like $\ce{CH3COO- + H2O <=> CH3COOH + OH-}$, and that $\ce{OH-}$ would be responsible for the overall alkaline pH of the resulting solution. If we have a salt of a strong acid and weak base, it is the other way around.
Which acids or bases are strong and which are weak is determined by looking up their $pK_a$ or $pK_b$ in the reference books. I don't want to get deeper into this because running far ahead of your syllabus might do more harm than good. Just remember that neither $\ce{Mg^2+}$ nor $\ce{Cl-}$ would hydrolyze significantly, so the solution would remain neutral.
The precipitation is not all that important, as hydrolysis of most common salts rarely goes all the way up to metal hydroxide. For example, $\ce{AlCl3}$ would make the solution pretty acidic, but still produce no precipitate. Also, consider $\ce{NH4+}$ which is prone to hydrolysis despite having soluble hydroxide. (One might argue this does not quite qualify as hydrolysis, but anyway.) If you want an example when the precipitation actually does occur, think of $\ce{TiCl4}$.
As for the spectator ions, that's a lame argument indeed. There is no such thing as "spectator ion in general". True, certain ions are spectators in certain reactions, but there is no ion that would always remain a spectator, no matter what the situation is.
As for the Lewis acids, that's a different concept altogether, to the point that you shouldn't automatically think of "Lewis acids" as acids (though many of them are).

Okay, I know about hydrolysis, but that still doesn't seem to answer my question about MgCl2, whether or not the pH of the solution will be changed?

@user2804925 I’ll admit it’s not well written, but the answer to that question is hidden in Ivan’s answer.

@Jan okay I found the answer I accidentally skipped over it the first time. Still wondering why my argument would be invalid though. From albapa's answer, it seems that the concentration isn't high enough for precipitation to occur, but Ivan's answer doesn't seem to take the fact that Mg(OH)2 precipitates into account at all.

Technically, the magnesium ion will act as a weak acid and thus magnesium chloride solutions are going to be at least slightly acidic. However, the acidity will be less than that with most transition metal chlorides, and at least in dilute solution it's a bit of a borderline case. Calculations below indicate that the pH is estimated as 6.24, so if we regard pH between 6 and 8 as essentially neutral, then 0.1 M magnesium chloride solution would be in that range.
A compilation by IUPAC gives several values for the $\mathrm{p}K_\mathrm{a}$ of magnesium ion in water, these mostly clustering around 11.5. The measurement used based on the hydrolysis reaction which is predominantly the first stage of proton dissociation:
$\ce{Mg^{2+} + 2H2O <=> MgOH^+ + H3O^+}$
The second stage of proton dissociation is essentially inaccessible by normal techniques in aqueous solution because it equilibrates only at pH values where the solubility of magnesium hydroxide has become very low.
With the first stage of dissociation we may compute for 0.1 M magnesium chloride solution:
$[\ce{Mg^{2+}}]+[\ce{MgOH^+}]=0.1$
$2[\ce{Mg^{2+}}]+[\ce{MgOH^+}]+[\ce{H^+}]-[\ce{OH^-}]=0.2$
$[\ce{MgOH^+}][\ce{H^+}]/[\ce{Mg^{2+}}]=10^{-11.5}=3.2×10^{-12}$
$[\ce{H^+}][\ce{OH^-}]=10^{-14}$
The first equation is the magnesium ion balance, the second gives electroneutrality with 0.2 molar chloride ion, and the last two equations represent the equilibrium realtions. From the fourth equation we immediately have
$[\ce{OH^-}]=10^{-14}/[\ce{H^+}].$
Putting this into the first two equations and taking appropriate linear combinations gives
$[\ce{Mg^{2+}}]=0.1-[\ce{H^+}]+10^{-14}/[\ce{H^+}]$
$[\ce{MgOH^+}]=[\ce{H^+}]-10^{-14}/[\ce{H^+}]$
Substituting these into the third equation and simplifying gives
$a^3+3.2×10^{-5}a^2-33a-320=0$
where $a=10^7[\ce{H^+}]$ is introduced to enable better numerical calculation. This gives a root
$[\ce{H^+}]=10^{-7}a\approx5.7×10^{-7} M,$
from which the pH is $\approx6.24$.

:When substitute $a=10^7[\ce{H^+}]$ in the following equation :
$[\ce{H^+}]=10^{-7}a= 10^{-7}10^7[\ce{H^+}] \approx5.7×10^{-7} M,$? help me to clarify it

A very quick answer to this question is that $\ce{MgCl2}$ is the salt formed from $\ce{Mg(OH)2}$ and $\ce{HCl}$. Both of these are strong bases and acids respectively, so when their salt is formed, it is made up of a very weak conjugate acid ($\ce{Mg^2+}$) and a very weak conjugate base ($\ce{Cl^-}$). Since these are both very weak conjugate acids and bases, the salt is neutral and you should see a pH close to 7 (maybe slightly below 7 due to dissolved acidic $\ce{CO2}$, a small amount of hydrolysis, and whatnot).

The pH of water is defined as neutral, because it contains H$^+$ and OH$^-$ ions in equal abundance, 10$^{-7}$M. We would probably call a pH of 6.9 or 7.1 neutral. Maybe even 6.5 and 7.5 are essentially neutral.
The questions become "How wide is your neutral pH range?" and "What is the actual pH of 0.1M MgCl2?" Neither is answered; however, it is possible to make a case for either of two answers to the OP's question.
In the first case, neutral is a wide range, say from 6 to 8. This neutrality is neither very acidic nor very alkaline, perhaps a pH that registers neutral with pH paper. In this case, MgCl2 is probably neutral, because we consider Mg(OH)2 to be a strong base and HCl is considered to be a strong acid. "Spectator ions", both of them. Very approximate language, but descriptive enough for many, if not most purposes. So, with this approach, MgCl2 is neutral.
But then someone might suggest using a pH meter with very high accuracy, MgCl2 of very high purity, and distilled water freed of CO2 by boiling or repeated distillations, and setting the neutral pH at 7.00, because you want the observed pH to fall on one side or the other.
Then we discover a report that Mg(OH)2 can be considered to be a weak base (Ref 1), so we would expect a pH somewhat below 7.00, and the answer would be "0.1M MgCl2 is acidic." (But not very acidic.) An elegant way of describing the situation is that Mg(OH)2 is a strong electrolyte but a weak base (Ref 2 and 3).
None of this so far has directly addressed the issue of low solubility, which seems to be relevant, somehow. We know that Mg$^{2+}$ has a great affinity for OH$^-$ ions, because they attract each other in great numbers, precipitate, and resist the solubilizing action of water. So in solution, along with 5 water molecules, a Mg$^{2+}$ ion probably attracts one of the few (10$^{-7}$M) OH$^-$ ions tightly enough to lower the pH a bit, increase the  concentration of H$^+$ by a little.
So, the correct answer is neutral if you measure pH with paper (wide neutral range), or slightly acidic if you measure with a sensitive pH meter (sharp neutrality), because Mg(OH)2 can be considered either a strong base or a weak base.
Ref 1. https://www.numerade.com/questions/why-can-magnesium-hydroxide-be-described-as-a-strong-base-even-though-it-is-only-slightly-soluble-in/
Ref 2. https://www.toppr.com/ask/question/a-solution-of-mgcl2-in-water-has-ph/
Ref 3. https://www.sciencedirect.com/topics/chemistry/magnesium-hydroxide

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How is a side-group that contains a double bond named?
In my chemistry lecture, we are working on the naming and nomenclature of organic molecules. When we got to the naming of alkenes and alkynes, we were told that the functional group had to be in the main, name giving chain of carbons. This brought up the question: What if there are multiple functional groups arranged in such a fashion as for it to be impossible for them to all be on the main chain (see example below)? How would one go about naming such a molecule? I would much appreciate any clarification that can be provided. I have already checked the IUPAC rules, and unless I totally missed it, it isn't addressed.
Here is the Picture of the molecule in question:
$\hspace{45 mm}$
Thanks in advance for any light you can shed on this question.

Related: How do you name an alkene using IUPAC rules if the longest carbon chain in this alkene does not include the double bond?

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. Visit the [help] to learn more about it. And ask Loong for any questions on IUPAC nomenclature (;P @Loong).

When there is a choice among acyclic parent structures, a principal chain must be chosen. The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) stipulates various criteria for the selection of the principal chain.
You are right in so far as functional suffixes that express the characteristic group of a compound (e.g. the suffix ‘ol’ for alcohols) are always attached to the name of the parent structure (i.e. the principal chain). However, since the given structure does not include any characteristic group that is expressed as suffix, the order of seniority of suffixes is not relevant in this case.
The first relevant criterion in this case provides that the principal chain has the greater number of skeletal atoms; i.e., the principal chain is the longest chain (see Subsections P-44.3 and P-44.3.2 in the Blue Book).
Only if this rule does not effect a choice of a senior parent structure, the principal chain has the greater number of multiple bonds (see Subsections P-44.4.1 and P-44.4.1.1).
(Note that the order of seniority between length of chain and unsaturation given in earlier IUPAC recommendations was reversed. Thus, the first criterion to be considered in choosing the preferred chain was unsaturation; the length of the chain was the second criterion.)
In this case, the longest chain has eight carbon atoms; i.e. the unmodified parent hydride is octane. (Note that two equivalent chains with eight carbon atoms exist in the given structure.)

Then

P-31.1.1.1 The presence of one or more double or triple bonds in an otherwise saturated parent hydride (…) is denoted by changing the ending ‘ane’ of the name of a saturated parent hydride to ‘ene’ or ‘yne’. Locants as low as possible are given to multiple bonds as a set, even though this may at times give ‘yne’ endings lower locants than ‘ene’ endings. If a choice remains, preference for low locants is given to the double bonds. In names, the ending ‘ene’ always precedes ‘yne’, with elision of the final letter ‘e’ in ‘ene’. Only the lower locant for a multiple bond is cited, except when the numerical difference between the two locants is greater than one, in which case the higher locant is enclosed in parentheses.
  (…)

Therefore, the correct name of the parent hydride is oct-3-ene.
The preferred names of the substituent groups are derived from the respective parent hydrides. The presence of free valences formally derived from the loss of one, two, or three hydrogen atoms from a parent hydride is denoted by replacing the ending ‘ane’ of the parent hydride name by ‘yl’, ‘ylidene’, and ‘ylidyne’, respectively (see Subsection P-29.2).
Therefore, the preferred names for the substituent groups of the given structure are
methylidene $(\ce{=CH2})$,
ethylidene $(\ce{=CH-CH3})$, and
ethyl $(\ce{-CH2-CH3})$.
With regard to numbering of locants within the same compound class, multiple bonds have seniority over simple substituent groups (e.g. methyl, ethyl, etc.) that are used for naming of branched alkanes.

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

Note that Rule e takes precedence over Rule f.
Therefore, the structure given in the question is named as 3-ethyl-5-ethylidene-6-methylideneoct-3-ene rather than 6-ethyl-4-ethylidene-3-methylideneoct-5-ene since low locants are given first to multiple bonds (i.e. ‘ene’) and the locant ‘3’ is lower than ‘5’.

Furthermore, note that the double bond in the ethylidene group gives rise to cis-trans isomerism. As drawn, the configuration is ‘E’.

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text:
Is there an established procedure for titanium tetrachloride titration?
I’m attempting to do an aldol reaction with $\ce{TiCl4}$ as the catalyst. It worked well in the large scale ($100~\mathrm{mg}$) test system but not, so far, in my $5~\mathrm{mg}$ scale actual system. Our bottle of $\ce{TiCl4}$ is not the youngest, the seal has $\ce{TiO2}$ precipitates on the inner side, and when you finally manage to stick a needle from the argon line in, you almost get a geyser of hydrochloric acid steam — need I mention that the liquid is cloudy from precipitated $\ce{TiO2}$?
My supervisor said distillation is not an option due to unremovable residues of $\ce{TiO2}$ getting stuck in the distillation apparatus. But I would love to know what purity the $\ce{TiCl4}$ still has and thus how much I should upscale it when adding it to the reaction.
Are there any rather simple titration methods for $\ce{TiCl4}$ known or established like they are for n-butyllithium?

$\ce{TiCl4}$ is amenable to distillation if done carefully so as to exclude moisture. That doesn't mean it isn't a pain, however.
We do a lot of chemistry using $\ce{TiCl4}$ and other Ti Lewis acids (including some Ti aldol reactions on v. small scale), so end up having to distill it quite often (roughly once a quarter). Once distilled, it lasts in the fridge until it runs out so long as the flask is well sealed and everyone remembers to flush with argon before they're finished using it.
I'll quote from Purification of Laboratory chemicals for general information, but also note that we don't follow this procedure. (Mercury makes me uncomfortable)

Ref: 'Purification of Laboratory Chemicals', 6 ed.

In practice, we simply distill the $\ce{TiCl4}$ under reduced pressure (high vacuum pump), discarding the first few mL as would be common procedure. This usually yields $\ce{TiCl4}$ that is clear in colour and sufficiently pure for our chemistry. If some colour remains, a second distillation from copper turnings is occasionally done, but this is rarely needed.
The key bit of information is to make sure that you quench all of the glassware and tubing used with a weak base (sodium hydrogen carbonate is sufficient), and to ensure that the pump exhausts into a good fume cupboard incase any fumes do escape. All of the glassware used should also be heavily dried prior to use.

sufficiently dry | $\ce{TiCl4}$ may be not dry? How so?

Well we can't do Karl Fisher type work on it to identify moisture content (to my knowledge).   This kind of Aldol chemistry is very sensitive to water, especially on a small scale, so any water present would quench the enolate and our chemistry wouldn't work. This suggests to me its 'sufficiently' dry, especially as we're using it for the same purpose as Jan.

Let me expand the reason for my uncertainty. $\ce{TiCl4}$ reacts with water, producing solid $\ce{TiO2}$ and gaseous $\ce{HCl}$. I highly doubt any moisture may exist in $\ce{TiCl4}$. So, I repeat - how $\ce{TiCl4}$ may be wet?

Edited the post to reflect your comment, my initial answer was perhaps worded a little unclearly.


text:
How to find probability in different basis representations?
The matrix representation of the Hamiltonian is given in the position basis: $\{|+a\rangle,|-a\rangle\}$.
$$H = \begin{pmatrix} E_{0} & E_{-A} \\ E_{-A} & E_{0} \end{pmatrix}$$
The initial state of a system is also given in terms of the position basis:
$\mid \psi \rangle  = \frac{1}{5}[3|+a\rangle + 4|-a\rangle]$.
To find the probability of the a particular energy eigenvalue, why is it incorrect to choose the coefficient of $|-a\rangle$ so that it would be $\left(\frac{4}{5}\right)^2$?
Instead, we have to transform to the Hamiltonian basis, using the eigenvectors of the Hamiltonian. Why can't we use the position basis?

In quantum mechanics observables are represented by operators and the outcome of a given measurement can only be an eigenvalue of this operator. After the measurement, the state of the system will be projected to the eigenstate of the operator corresponding to the measured eigenvalue (consider for simplicity the case where you don't have more than one eigenstate corresponding to one eigenvalue).
Now, in you case, the probability $|4/5|^2$ corresponds to the following:
$$
P(\text{-}a) = |\langle \text{-}a|\psi\rangle|^2.
$$
This is the probability of projecting (by measurement) the state of the system into $|-a\rangle$. This corresponds to computing the probability of obtaining the eigenvalue $\text{-}a$ as a result of your measurement. If you want to know the probability of being in a give energy state, you have to compute the following probability
$$
P(E) = |\langle E|\psi\rangle|^2.
$$
where $|E\rangle$ is the state with energy $E$ (eigenstate of the Hamiltonian). Therefore, in order to compute the braket $\langle E|\psi\rangle$ you need to write $|\psi\rangle $ in the basis of Hamiltonian's eigenstates: in this basis, the solution is trivial.


text:
Why is molarity independent of temperature?
I can't find a good answer as to why molarity is independent of the temperature. Why is the concentration (in mol/l) independent of the temperature?

What are your thoughts so far on the question?

First of all, molarity is temperature-dependent.
You might have meant molality, because molality is temperature-independent.
The reason is very simple, if you see molarity definition, molarity is the amount of substance of solute per litre of solvent and that mean it depends on volume.
You know that volume depends on temperature, therefore molarity is also temperature-dependent.
Molality on the other hand is the amount of substance of the solute per mass of the solvent, both are independent of the temperature.

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text:
What is the correct way to pronounce iodine?
I have always said it like 'io-deen' because whenever 'ine' is used (eg. amine, or fluorine) it is said that way. Sometimes I'm corrected by people who think it should be pronounced 'io-dyne', which is correct?

https://www.youtube.com/watch?v=J_UB6khjYDA

Thanks @manshu but I can't use u tube with my device.

it is both of them...may be bcz of the difference in the pronunciation in Britain and USA

You might find better answers on [english.se].

Thanks @Loong but I don't want to start another site the answers here were fine.

There may not be any such thing as "correct". Customary is the best you will get and that varies from place to place. Hell, we can't get international agreement on how to pronounce or spell aluminium/aluminum even though the american preference is obviously wrong. ;-)

I feel like both pronunciations are correct.

Manshu is correct,in British English we pronounce iodine as in seen while in North American English we pronounce it as in mine.

Some countries use British English as their official language,so depends on which country you live .

I live in Australia.

So in British English do they pronounce amine and flourine differently also?

No, ine in both words are pronounced as in seen in British English.

I think it is fine for you to use British English.

Or we could just use German.  In German they do not use the suffix for halogens at all so that "iodine" is just "Jod".  Now all we have to remember is that you don't pronounce the long "i", instead using the usual "j"/"y" consonant sound.

Like the other halogens, iodine is pronounced eye - o - deen  (especially by chemists) .

Not in the USA. I’m a lifelong chemist and it is as in mine.

Who are you talking to? To non-chemists (in the US), just say i-oh-dyne.
But if you are talking with chemists, say i-oh-deen.
If you are talking to a mix of chemists and non-chemists, decide which you would like to be classified as.
Or, you could say "element 53".

Think about how other halogens are pronounced: fluor-een, chlor-een, brom-een, astat-een, and newly named tenness-een. Why would it be io-dyne and not iod-een?


text:
Why aren't the copper(II) ions in CuSO4 precipitated by hydroxide ions from water?
From my own experience and literature, I know that $\ce{CuSO4}$ is well soluble in water and dissociates into $\ce{Cu^2+}$ and $\ce{SO4^2-}$. $\ce{Cu(OH)2}$, however, is not: $K_\mathrm{sp} = 2 \cdot 10^{-19}$.
Therefore, in pure water which has a pH of $7$, $c(\ce{OH-}) = 10^{-7}$. This means that a concentration of $2 \cdot 10^{-19} / (10^{-7})^2 = 2 \cdot 10^{-5}$ of $\ce{Cu^2+}$ would be enough to cause precipitation. How come, when dissolving $\ce{CuSO4}$, the $\ce{Cu^2+}$ stays in solution, and doesn't precipitate with the $\ce{OH-}$ ions of water to form $\ce{Cu(OH)2}$?

Copper (II) sulfate forms a copper aquo complex $\ce{Cu(H2O)6^2+}$ which is slightly acidic. See: https://en.wikipedia.org/wiki/Metal_aquo_complex

Long story short, $\ce{Cu^2+}$ is slightly hydrolyzed and hence acidic. If you'd try to make it exactly neutral, a precipitate would form.

In aqueous solution, the copper(II) ion forms the aqua complex $\ce{[Cu(H2O)6]^2+}$:
$\hspace{55 mm}$
Because the central $\ce{Cu^2+}$ ion is positively charged, it polarises the $\ce{O-H}$ bonds towards oxygen, which makes the hydrogens more acidic. Therefore, the ion acts as a weak Brønsted acid, with a $\mathrm{p}K_\mathrm{a}$ of approximately $8$ (J. Phys. Chem. A, 2015, 119 (12), pp 2926–2939):
$$\ce{[Cu(H2O)6]^2+ <=> H+ + [Cu(H2O)5(OH)]^+}$$
Suppose we have a $0.1~\mathrm{mol~dm^{-3}}$ solution of $\ce{CuSO4}$. To a first approximation,
$$\begin{align}
[\ce{Cu(H2O)6^2+}] &= 0.1~\mathrm{M} \\
[\ce{H+}] &= \sqrt{(0.1~\mathrm{M})(10^{-8}~\mathrm{M})} \\
&= 3.162 \times 10^{-5}~\mathrm{M} \\
[\ce{OH-}] &= \frac{10^{-14}~\mathrm{M^2}}{[\ce{H+}]} \\
&= 3.162 \times 10^{-10}~\mathrm{M}
\end{align}$$
Therefore,
$$\begin{align}
[\ce{Cu^2+}][\ce{OH-}]^2 &= (0.1~\mathrm{M})(3.162\times 10^{-10}~\mathrm{M})^2 \\
&= 10^{-20}~\mathrm{M^3} < K_{\mathrm{sp}}
\end{align}$$
and no precipitation of $\ce{Cu(OH)2}$ occurs.
If the pH of a $\ce{Cu^2+}$ solution is neutralised by careful addition of base, then precipitation will be observed as Ivan noted in his comment. The exact pH at which precipitation will occur is of course dependent on the concentration of $\ce{Cu^2+}$. The same argument here actually applies to all of the transition metals: all of them form rather insoluble hydroxides, but they also form acidic aqua complexes, which prevent the precipitation of the hydroxide.
The calculations could have been done more accurately by introducing fewer assumptions; however, they should not affect the final conclusion. (If this was a textbook, it would probably say: this exercise is left to the reader!)


text:
Acid/Base dissociation constants relationship
From what I understand, the concentration of hydronium and hydroxide molecules is constant in pure water (and equals $10^{-7}\ \text{M}$, which is measured experimentally).
What I don't understand is why this remains true for an aqueous solution with an acid or base in it. Shouldn't it affect these concentrations?
When you write the equilibrium reactions for a weak/weak acid/base pairs (one with its conjugate), and get that $K_\mathrm a\cdot K_\mathrm b = [\ce{H+}][\ce{OH- }]$, what's the explanation for it? Why does their multiplication (either as pure water or in a acidic/basic solution) remain constant?

The concentrations do change in the presence of acid or base and are no longer $10^{-7}$M.
Instead, [H3O+][OH-] $= 10^{-14}M^2$
This comes from:
$\ce{2H2O <=> H3O+ + OH-}$
$K = \frac{a(\ce{H3O+})a(\ce{OH-})}{(a(\ce{H2O}))^2}$
Where "a(x)" is activity of species "x".
Then, approximating activity of $\ce{H2O}$ as 1, and the activity of the ions as the concentration of the ions there is a constant $K_w$, such that:
[H3O+][OH-] $= 10^{-14}M^2 = K_w$


text:
Which factors determine the keto:enol ratio for aldehydes, ketones, and 1,3-dicarbonyl compounds?

Arrange the following compounds in order of increasing enol content:

I think the order is D > C > B > A considering conjugation, hyperconjugation and hydrogen bonding. The enol formed in case D will have a substituted double bond (that is, more hyperconjugative structures) than the enol of C.
Both enols A and B lack hydrogen bonding. Also, enol of B is more stable because of more substituted alkene.

Your reasoning is along the right lines, but it is incomplete.

enol of B is more stable because of more substituted alkene.

The methyl group in acetone does stabilize the carbon-carbon double bond in the enol form for the reasons you suggest. However, the methyl group also stabilizes the carbonyl double bond in the keto form (see here for example, this is why a carbonyl in a ketone is slightly stronger than a carbonyl in an aldehyde). In fact, in simple carbonyl compounds, the methyl group stabilizes the carbonyl double bond a bit more than it stabilizes the enolic carbon-carbon double bond. As a result, simple aldehydes generally have a higher enol content than simple ketones. Hence, the enol content in acetaldehyde is greater than the enol content in acetone.
$$\ce{carbonyl <=> enol}$$
$$K_\mathrm{eq}=\frac{[\text{enol}]}{[\text{carbonyl}]}$$
$$
\begin{array}{lc}
\hline
\text{compound} & K_\mathrm{eq} \\
\hline
\text{acetaldehyde} & 6 \times 10^{-7} \\
\text{acetone} & 5 \times 10^{-9} \\
\hline
\end{array}
$$
The two β-dicarbonyl compounds have a higher enol content than the two monocarbonyl compound because hydrogen bonding and conjugation stabilize their enols. The enol content in C (a mono aldehyde) is higher than D because of the reasons outlined above.
Therefore, the overall order of increasing enol content is C > D >> A > B.
Unless you've studied this before, you might not know that a "methyl group stabilizes the carbonyl double bond a bit more than it stabilizes the enolic carbon-carbon double bond". So, a general rule that might come in handy in these situations is that the enol content increases with the acidity of the enolic hydrogen. A more acidic α-hydrogen implies a weaker $\ce{C-H}$ bond. Since the position of a keto-enol equilibrium is dependent on the relative stabilities of the keto and enol forms (the compound with the highest bond strengths overall will be more stable and will predominate at equilibrium) a weak $\ce{C-H}$ (lower $\mathrm{p}K_\text{a})$ generally implies a higher enol content.
$$
\begin{array}{lr}
\hline
\text{compound} & \mathrm{p}K_\text{a} \\
\hline
\text{acetone} & 19.3 \\
\text{acetaldehyde} & 16.7 \\
\text{2,4-pentanedione} & 13.3 \\
\hline
\end{array}
$$
Finally, this earlier answer provides a review of the key factors controlling keto-enol equilibria.

why is comparing the acidity of the hydrogen a general rule and not absolutely correct?

@YUSUFHASAN Probably because it's enol 'content' which means we are giving enough time to the enol to reach the equilibrium. Kinetic factors do not matter, only stability of enol does.


text:
Resonance in ozone and benzene
In inorganic chemistry, I learnt that stability of some atoms like ozone cannot be explained by one structure alone and hence the hybrid of possible structures is taken and assumed to be the structure and this is called resonance structure.
However, the book clearly states that it is wrong to assume that electrons move about back and forth alternating between the two structures, and that resonance is a theoretical concept only.
Here's an example for ozone.

But in organic chemistry I learnt that resonance means electrons moving around and in fact some molecules like benzene are really stable because the moving of electrons causes energy to be continuously lost.
 We also drew the electrons moving around for some structures like nitrobenzene and learnt how these cause ortho para/ meta directing nature.

The two resonances I learnt about don't seem to be the same. My question is, what is the difference? Or is there a concept I misunderstood?

some molecules like benzene are really stable because the moving of electrons causes energy to be continuously lost This is wrong. Very wrong. The first description about ozone is correct.

This is because the electrons dont actually jump between the atoms of the molecule, but rather the electron cloud density is even distributed through the resonating bonds.  If atoms had to hold formal charges this would be very unstable for the electron count. In essence all of the resonant structures exist at the same time.  This lack of formal bonding in resonace is why benzene is often drawn as a circle inside a hexagon. This explaination does involve a bit of hand waving, and ignores higher level inorganic, but I think it should provide the needed additional clairity.

Is it right to infer from your explanation that benzene is not stable because electrons move around and lose energy but because charge is more evenly distributed in the resonance structure?

Its a little more complicated than that but yes. It is worth noting though that benzene has a special kind of resonance called aromaticity which is far more stable than simple resonance.

Follow up question: could you explain why a molecule that has more number of resonance structures is more stable?


text:
How to create liquid carbon, and what kinds of properties does it have?
Carbon sublimes at about 3915K. In a high pressure environment, shouldn't it be possible to create liquid carbon at high temperatures? Has this ever been done? What does it look like and what properties does it have?

"What properties does it have?" is unfortunately too broad. Could you be more specific?

http://news.discovery.com/space/alien-life-exoplanets/diamond-oceans-jupiter-uranus1.htm

Apparently the melting point of diamond at 11,000,000 atm is only 50,000 ° F.  And under those conditions, liquid diamond is more dense than solid.

It's possible to create liquid carbon, but it's extremely hard. As you can see on the Carbon phase diagram  if you want to obtain "easely" liquid carbon you must do it at $\ce{1 GPa}$ and $\ce{5000 K}$.
And for the properties I really don't know but carbon follows the octet rule and it might be very reactive as an electrophile because it will just have four electrons around it.
I hope it can help you a bit, i'm seeking in some data-bases but there are not really things to answer your question.
Have a nice day.

It won't be monoatomic

Yes that what I'm thinking. I will add some hypothesis in my answer when i will have time. I will correct the thing which can mislead.

authors: Маноил Бакърджиев


text:
Why is ionisation energy of bismuth lower than lead?
Why is ionisation enthalpy of Bismuth less than that of Lead for it just comes after the latter in periodic table?
First ionisation energy of bismuth is 703 kg/mol while that of lead is 715 kg/mol.
I also found that the covalent radius of Bismuth was more than that of lead, though both try to show me the same cause but I couldn't figure this out.
Let me give out some reason why Bi should have higher first ionisation energy than bismuth:-
* Bi falls after Pb in the periodic table so it should be more non metallic in nature than Pb.(as from the general trend)
*  Bi have half filled p orbitals contributing to its extra stability(as from the general trend).
* Bi falling down the the group 15 elements show metallic nature but still it should be less metallic than lead as it falls just after it.
* Even if we consider d/f orbitals,it would have same effect on lead and bismuth as both falls after elements occupying 5d orbitals and much after lanthanoid series.In fact the only effect I could see would be of increase in 1 electron thus there should be decrease in size of the atom and hence increase in ionisation energy.
* I tried to find out whether there might be any relativistic corrections but I was only able to reach to the conclusion that as the nuclear charge increases for atoms the size gets shorter somewhat like in the case of radium so ionisation energy should have increased.

I could have accounted d/f orbital for bismuth but the same thing is applicable to lead so no such notable change should have been there.

I have the same doubt. https://www.quora.com/Why-is-the-first-ionisation-energy-of-bismuth-less-than-that-of-lead?share=1
Kostas Papathanasiou's answer is the closest explanation I have found so far.


text:
nomenclature of some organoselenide moieties
what are the structures of these selenide moieties : selenenylate, selenolate ?
Can anyone tell the difference between the two in the terms of if they are anions or cations , electrophiles of nucleophiles.

Based on usage:

Selenenylate is a verb. It means to attach a selenium group to an organic compound - See this passage from Comprehensive Organic Synthesis: Selectivity, Strategy, and Efficiency in Modern Organic Chemistry, Volume VII

The analogous morphinoselenamide has been shown to selenenylate the $\alpha$-ketoester...

Selenolate is a noun. It is the $\ce{RSe-}$ anion, especially in ionic or coordination compounds. See various usage at the links in this Google search


text:
Lower an electron energy level by pressure
When an atom is in extreme pressure like a little before it's Schwartzchilds radius, can the electrons in higher energy levels physically be compressed into lower energy levels?.

This question might get more traction on physics.SE.

I'm voting to close this question as off-topic because chemistry is not ordinarily concerned with black holes.

Electrons do not need to be pressed into lower energy level. If there is a vacancy in that level, the electron will go there by itself, and quite willingly. If there is no vacancy, nothing can be done. Pressure, however great, can't overrule the Pauli exclusion principle.


text:
How does the electron configuration of platinum relate to its stability?
Does platinum's electron configuration, [Xe] 4f14 5d9 6s1, influence its reactivity and stability? Is the electron configuration the primary contributing factor to platinum's relative inertness?
This seems odd to me that platinum is very inert. It seems like pulling off that 6s1 electron should not be difficult as compared to other metals.

It has to do with the relativistic effect that contracts the orbit of the 6s electron.  The relativistic effect is explained in this earlier answer.

Platinum isn't unreactive. It is resistant to some specific reactions (oxidation or tarnishing) but is a very good catalyst for others (e.g. methanol oxidation, oxidation of residual hydrocarbons in catalytic converters, ammonia oxidation...). And it has a rich chemistry.


text:
IUPAC systematic name of Carminic Acid
Every IUPAC Systematic name I have come across, I have understood up until now. This is Carminic Acid ($\ce{C22H20O13}$):

(source: chemspider.com)
This is the IUPAC Systematic Name for Carminic Acid:
(1S)-1,5-Anhydro-1-(7-carboxy-1,3,4,6-tetrahydroxy-8-methyl-9,10-dioxo-9,10-dihydro-2-anthracenyl)-D-glucitol
I have 3 main questions:

I do not understand the numbering system inside the brackets. The numbering system implies that the numbers assigned to the carbon atoms is the following:

I would have thought the numbers assigned to the carbon atoms would be the following:

Why is the numbering system the way it is? (Why does the numbering system not go around the molecule, in a normal cycle, counting off each carbon?, Why is the number 10 carbon opposite the number 9 carbon?, Why are 4 carbons [circled in the picture below] left out in the numbering system?)

I also do not understand why Carbons 9 and 10 in the anthracene group are referred to as '9,10-dioxo-9,10-dihydro'. Surely, they only need to be referred to as '9,10-dioxo'? And what is a '-hydro' group?

Doesn't '1,5-Anhydro...-D-Glucitol' imply the following: Hexan-2,3,4,6-tetrahydroxy-1-yl, a chain, and NOT a ring structure of: Oxan-3,4,5-trihydroxy-6-hydroxymethyl-2-yl, considering D-Glucitol (shown below). Also, as well as '1,5-Anhydro...-D-Glucitol' not accounting for the ring structure of that section of the molecule, it also doesn't account for the Oxygen.

What is your source for the supposed IUPAC systematic name?  A program I use (Chemdoodle) says the IUPAC preferred name of this compound is 3,5,6,8-Tetrahydroxy-1-methyl-9,10-dioxo-7-[3,4,5-trihydroxy-6-(hydroxymethyl)tetrahydro-2H-pyran-2-yl]-2-anthracenecarboxylic acid.  That is very different than what you say the systematic name is.

@Curt F. The source I used was the RSC's chemical database ChemSpider. This is the page for Carminic Acid: Carminic Acid in ChemSpider. The name you have quoted is one of the names listed but is not the IUPAC Systematic Name.

Pubchem also lists a different name than the one you give.  I guess the question is, how do you decide to name a thing as a glucitol derivative vs. a anthracene derivative if both groups are present in a molecule?

@Curt F. I fully understand the name you have quoted and I understand the other systematic names on ChemSpider, but my focus is on this name and the three questions listed above. To be honest, I would use the name you have quoted in a practical situation, I just would like to understand the name above from an intellectual standpoint.

Do you understand (6aR,9R)-N,N-diethyl-7-methyl-4,6,6a,7,8,9-hexahydroindolo[4,3-fg]quinoline-9-carboxamide? If so, please take a look at this question: https://chemistry.stackexchange.com/q/169432/108725 which for some reason has a lot of downvotes and three close votes, though it is very similar to your excellent question.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the preferred name (PIN) for the parent ring component consisting of three benzene rings fused in a linear manner is the retained name anthracene.
Subsection P-25.1.1, Table 2.7, Entry (11) shows that anthracene has a special numbering:

According to Subsection P-14.4 (a), this fixed numbering must be used in PINs and in general nomenclature.
The prefixes ‘hydro’ and ‘dehydro’ are used to indicate addition and subtraction, respectively, of hydrogen atoms (see Subsection P-31.2.1).  Hence, the PIN for the ring component of the structure given in the question is 9,10-dihydroanthracene:

The corresponding monovalent substituent groups derived from such ring systems by the removal of one hydrogen atom are named by replacing the final ‘e’ of the name of the hydrocarbon by ‘yl’ (see Section P-29). Thus, the name for the substituent group derived from 9,10-dihydroanthracene is (9,10-dihydroanthracen-2-yl):

(Note that locants are placed immediately before that part of the name to which they relate (see Subsection P-14.3.2); i.e. ‘…anthracen-2-yl’. Thus, the name ‘…-2-anthracenyl’ given in the question is not correct.)
The substitutive prefixes for the remaining groups are
‘methyl’ $(\ce{-CH3})$,
‘hydroxy’ $(\ce{-OH})$,
‘oxo’ $(\ce{=O})$, and
‘carboxy’ $(\ce{-COOH})$.
The name for the complete substituent group is (7-carboxy-1,3,4,6-tetrahydroxy-8-methyl-9,10-dioxo-9,10-dihydroanthracen-2-yl):

The bases for carbohydrate names are the structures of the parent monosaccharides in their acyclic form (see Subsection P-102.2).
Alditols are named by changing the ending ‘ose’ in the name of the corresponding aldose into ‘itol’ (see Subsection P-102.5.6.5):

The prefix ‘anhydro’ expresses the corresponding intramolecular ether, which is formally obtained by elimination of water from two hydroxy groups of a single molecule of a monosaccharide or monosaccharide derivative (hence the structure is commonly called an intramolecular anhydride). The prefix ‘anhydro’ is preceded by a pair of locants identifying the two hydroxy groups (see Subsection P-102.5.6.7.1).

Great answer (as usual) Loong.  This explains the carbon numbering mystery that the OP was wondering about.  However I'm unclear on how to decide whether the IUPAC systematic name  would regard the molecule as a glucitol-substituted anthracene or an anthracene-substituted glucitol.

@CurtF. In “normal” organic nomenclature, the carboxylic acid group has the highest seniority. Thus, the preferred name should be “3,5,6,8-tetrahydroxy-1-methyl-9,10-dioxo-7-(3,4,5,6-tetrahydroxyoxan-2-yl)-9,10-dihydroanthracene-2-carboxylic acid”. However, the section on carbohydrate nomenclature (P-102) is based on parent monosaccharides having retained names, which are modified to indicate the present characteristic groups. According to the introduction given in P-100, preferred IUPAC names (PINs) are not identified for the compounds in this chapter.

Thanks again Loong.  If I understand you correctly you are saying that there is no clear choice for the PIN for this compound?  The "systematic" name would seem to be the one that ends dihydroanthracene-2-carboxylic acid, because, the P100 chapter you allude to says When the full structure is known, a ‘systematic name’ may be generated in accordance with Rules described in Chapters 1 through 8 of these recommendations.

And it goes on to say that naming things as natural product derivatives, e.g. as a D-glucitol derivative, is a convenience and a "semisystematic" name.  I think the idea is that semisystematic names are supposed to be much shorter than the full official systematic name.  But in this case, the difference is only 30 letters saved in a full systematic name of 100 characters, so the "semisystematic" name is probably not worth it.

Thank you very much. Very decisive, and very informative answer. I now fully understand the IUPAC Systematic name for Carminic acid. You're answer is very much appreciated. I haven't heard of the Blue Book before and will use it in future.

@IUPAC struck again. +1.

Just note that 1,5-anhydro-ᴅ-glucitol would be the correct name for the carbohydrate part itself (without the anomeric C-1 OH group). However, using this name type for the whole compound is rather wrong.
This is an example of C-glycoside, or more correctly, C-glycosyl compound. The nomenclature is described in P-102.6.1.4.
Related carbohydrate (as glycosyl group is formed by removal of the anomeric hydroxy group from the cyclic monosaccharide, as stated in P-102.6.1.1.1) is β-ᴅ-glucopyranose.
The correct “natural product” IUPAC name would be
7-(β-ᴅ-glucopyranosyl)-3,5,6,8-tetrahydroxy-1-methyl-9,10-dioxo-9,10-dihydroanthracene-2-carboxylic acid.

authors: user1205901 - Слава Україні


text:
Bond lengths in hydrocarbons

It seems like the first molecule has a bond order of $1$, the second and third have a bond order of $7/6$ and the fourth molecule has a bond order of $3/2$. Why is the answer not $4 < 2 = 3 < 1$?

Well, 2 cannot be equal to three to start off, because a double bond it way shorter than a single bond.
The Answer to this is D, 2 has the shortest bond because it is a double bond, next 4 is a benzene ring, which is a resonance structure, which means that all the carbons technically have partial bonds between each other, or you can imagine the double bonds to be continuously rotating between the different carbons. Lastly 3 and 1 both have normal single bonds, so would have an equal bond length.
Therefore, since the double bond is shortest, the resonance between a single and double is a bit less short, but not as long as a single bond, and a single bond is the longest, the answer it 2(double bond)<4(between double and single bond)<3&1(which are both single bonds)

Why are 2 and 3 not also resonance structures?

@user99364 Chemical formula for 2&3 is C6H10, but benzene's is C6H6, if you draw the two molecules, you'll realize that for benzene, if a Hydrogen is placed on each Carbon, regardless of where you place the double bonds, the molecule will always satisfy the octet rule for all it's  atoms. But, if you draw 2 or 3, you will see that the double bond cannot move, but must remain where it is in the molecule to allow all its' atoms to conform to the octet rule.Therefore, benzene or (4)'s double bonds can move around freely, aka resonance, creating partial double bonds between all it's carbon atoms.

I can try to explain this by valence bond theory also:-
In first figure carbon is $sp^3$ hybridised having 25% s character and 75% p character.The same is for the  third figure where there is carbon carbon single bond.
In second figure carbon is $sp^2$ hybridised with 3 $sp^2$ hybrid orbitals which have more s character than p unlike sp3 hybridised orbital.
Now s being more electronegative makes the the double bond shorter than the single bond since it has more s character in sp2 hybridised orbital than sp3 hybridised orbital,so bond length of 1and 3 figure is shorter than that of 2 figure.
The bond in figure 4 fall somewhat intermediate to double bond and single bond and so is its bond length(intermediate).
Now the order of bond length will be:-2 less than 4 less than 1=3.


text:
Can molarity ever exceed molality
Work so far:
Molality = mol solute/kg solvent
Molarity = mol solute/L solution
So for molarity to exceed molality in the same solution, the numerical value of
kg of solvent > L of solution
From my experience this never happens. However by applying this to a very dense solvent, could the mass of solvent actually exceed the combined volume of solvent and solute. I am aware that volumes are not additive, and so if the density of a solvent is 10 g/mL the density 10 times of water and 0.5 M 2 L of HCl is added to 4 L of solvent, mass of solvent would be 40 kg solvent could the volume actually be greater then 40 L? Is there an example of a physical system where molality is lower then molarity?

Try thinking of very dilute solutions in a solvent denser than water, as you can make some very helpful simplifications. What is the relationship between molarity and molality then?

There is one possibility with glycerol With density 1.261 g/cm 3.Here the exact numerical value of mass is greater than that of volume of glycerol in kg and litres respectively.Moreover glycerol is soluble in water and vice versa.


text:
How to calculate the average quadratic velocity of carbon dioxide from given data?
I am working on this question, and I am unsure of how to proceed. The question is:

There is $31.7~\mathrm{g}$ of $\ce{CO2 (g)}$ in a container. The partial pressure of $\ce{CO2}$ is $2.62~\mathrm{atm}$ and the volume of the container is $26.8~\mathrm{L}$. What is the average quadratic velocity ($\mathrm{m/s}$) of the $\ce{CO2}$ in the container?

I think that the formula I should use is this one: $V = \sqrt{3RT/M}$, but I am unsure on how to find the temperature.

Use the Ideal Gas Law: $PV=nRT \to T = \frac{PV}{nR}$.  From the mass you can find the number of moles, $n$. $P$ and $V$ are given, and $R$ is the universal gas constant.
Even better you could rearrange the equation to $RT = \frac{PV}{n}$, substitue the value for $RT$ and do less arithmatic.


text:
Is this an exception to the rule that strong acids/bases always replace weak acids/bases?
According to the $\mathrm{p}K_\mathrm{a}$ table: hydrochloric acid is stronger than sulphuric acid. It has a $\mathrm{p}K_\mathrm{a}$ around $-6.3$ while the $\mathrm{p}K_\mathrm{a}$ of sulfuric acid is around $-3$. The second dissociation in the case of sulfuric acid is not very extensive and doesn't contribute much to its acidity.
\begin{aligned}
\ce{H2SO4 + H2O ~&<=>~ H3O+ + HSO4^{−}&&pK_{a}~=~-3}\\
\ce{HSO4- + H2O ~&<=>~ H3O+ + SO4^{−2}&&pK_{a}~=~2}\\
\ce{HCl + H2O ~&<=>~ H3O^+ + Cl^{-}&&pK_{a}~=~$-6.3$}\\
\end{aligned}
Now $\ce{BaSO4}$ reacts with  $\ce{HCl}$ to produce very less of $\ce{BaCl2}$ because of the insolubility of $\ce{BaSO4}$.
My teacher told me that in a reaction "Strong acids or bases replace weak acids or bases". This holds good for this reaction as $\ce{HCl}$ is stronger than $\ce{H2SO4}$.
However $\ce{NaCl}$ reacts with $\ce{H2SO4}$ to produce $\ce{Na2SO4}$ and $\ce{HCl}$. In this case $\ce{H2SO4}$  replaced a stronger acid $\ce{HCl}$. How do we account for this?

https://chemistry.stackexchange.com/questions/50656/how-is-volatility-useful-in-the-production-of-acids

https://chemistry.stackexchange.com/questions/59069/how-and-why-can-more-volatile-acids-be-made-with-sulfuric-acid

Every reaction is considered reversible to some extent and feasibility of a reaction is mainly dealt in the terms of favourability of either products or reactants. When we say that creating a weaker acid from a stronger base is feasible, we actually mean that the equilibrium favours the formation of the products in such a case. However some other factors can be at play and can well influence an equilibrium. One such governing principle is Le Chatalier's.
$$ \ce{H2SO4 + 2NaCl <=> 2HCl + Na2SO4 }$$
Here's a youtube video showing the production of hydrochloric acid under laboratory conditions. Notice the $\ce{HCl}$ produced is a gaseous product, and hence there's an imbalance in our equilibrium; $\ce{HCl}$ escapes out as vapors from the solution, lowering $Q_c$ of the reaction thus making the reaction go forward.


text:
Why do we say metals are monoatomic in nature?
Are metals monoatomic or polyatomic in nature? For in crystalline form they also form molecular orbitals.

Why do we say that, really? We don't. It makes no sense. It is not even wrong.

@Ivan Neretin Can we say it polyatomic.I asked this question because my book says that all group 15 elements are polyatomic and also found bismuth to be in crystal form.Now I don't know what to say about it.Though the reason I gave above should indicate it to be polyatomic.

@IvanNeretin Agree. Using the reference of monoatomic/polyatomic frame in the given situation is more confusing and meaningless then helpful.

What this question is really getting at in terms of materials science is why some metals (particularly the precious metals) appear to reveal quite distinct properties akin to different allotropes i.e. the famous "white powder of gold" when processed under extreme heat and/or pressure (as if extreme "annealing" of certain metals converts them into a semi- or nonmetallic ceramic like substance...it's monoatomic or diatomic form?)

True metals are held together by the metallic bond. Unlike ionic or covalent bonds, there is no definite ratio of ingredients: for example, electrons are shared throughout an alloy of gold and silver in any proportion.
So, one could consider any intact piece of metal a giant molecule, because electrons are shared throughout, or one could consider it monatomic, because there is no definitive chemical bond amongst the atoms. It all depends on your definition, much like the multiple Lewis, Brønsted - Lowry etc. definitions of "Acid" or "Base".

"electrons are shared throughout an alloy of gold and silver in any proportion." That is not a property of the elemental form, itself. Also, polymers (built up of covalent units) can do the same trick, so it is not truly unique for metals.

True... I should have stated it more clearly: 1. Electrons are shared throughout a metal. 2. They are shared by all elements of a pure metal or even an alloy in a "sea" of electrons. As for conductive polymers, I'm not sure that all atoms participate in the conduction band, unlike a metal.

I am not talking about conducting polymers: co-polymers do the same trick without any extended localization. What you say is neither exclusive nor relevant property of metals.

In a non-conductive polymer, electrons are not shared in a common sea, free to move amongst atoms.

I might haven't been clear: the issue of forming compounds in a fix stoichiometry or able to do it in any proportion is not much to do with metallic properties. Therefore these are rather unrelated. And yes, e.g graphene shares all the electrons, yet carbon is not a metal, neither mono-atomic..

I think metals should not be considered monoatomic for it does not made up of molecules.Look for third para and history and etymology

@ Greg I think that grapheme is a  covalent solid  and due to its delocalised electrons it shows conductivity.Metals are determined more by its electro positive nature.Carbon is not that electro positive.In grapheme carbon is covalently bonded making a broad chain of carbons.Suppose the covalent bond wasn't there I don't think carbon would have formed a sea of electrons as other metals do.Still if it would have occurred in lattice it would would have been because of Van der Waal forces.

"..electrons are shared throughout an alloy or silver in any proportion" here this alloy is an inter metallic compound so we can consider it somewhat like giant molecule of of this compound.

Not all alloys are intermetallic compounds: intermetallic compounds have defined stoichiometric ratios. See https://en.wikipedia.org/wiki/Intermetallic.


text:
What substances are typically used to produce cyan, magenta, and yellow ink?
Early in grade-school you learn blue, red, and yellow are the primary colors that can be used to produce any other color. If you're curious enough, you later learn the primary colors are actually cyan, magenta, yellow and -- for practical purposes -- key (black).
My question is, what substances are used for each of these pigments and where are they found? Are they synthesized?
To be concrete, what substances are inside the in CMYK inks found in desktop inkjet printers?

Printing ink is a pretty expensive industry, i don't think such kinds of information will be easily available.

Yes, that's probably true. Now I feel too self conscious to post my other question: "What's the difference between the Pepsi and Coke recipes?" :P

Well you could always reverse engineer your subject (inks/drinks) and find it out for yourself.

@SujithSizon reverse engineering must be done in secret, otherwise you can get a lot of troubles. But there is some info about components http://www.google.com/patents/US4818285 But it is not 20% compound A, 35% compound C...

@XuMuK damn bro,did some serious homework

If you are interested in the colour pigments, there is a lot of similar ones. And the different mixtures can be made with almost no difference.
You can find a lot of different options for CMYK (because the inkjet printers and the laser printers are completely different).
For example, this patent:

The chelating agent of preference is ethylenediaminetetraacetic acid (EDTA), and the preferred biocide is Dowicil 75 manufactured by Dow Chemical. Dowicil 75 is a powdered material composed of 1-(3-chloroallyl)-3,5,7-triaza-1-azoniaadamantane chloride, sodium bicarbonate and hexamethylenetetramine hydrochloride.
The preferred cyan dye compound is Direct Blue 86, such as Intrabond Liquid Turquoise GLL, manufactured by Crompton and Knowles.
Preferably, the magenta dye comprises a plurality of individual magenta dye compounds. The preferred magenta dye compounds are Reactive Red 40 and Acid Red 52. Exemplary magenta dyes compounds include Bay Script Brilliant Red Ink 4B, manufactured by Mobay Chemical (Reactive Red 40), and Sandolan Rhodamine E-B manufactured by Sandoz Chemical (Acid Red 52).
Yellow dye compound, preferably Direct Yellow 107. A preferred Direct Yellow 107 compound is Direct Yellow G200, manufactured by Crompton and Knowles.
A preferred black dye is SP Black Liquid, manufactured by Mobay Chemical.


text:
Gaussian09 vs home made Hartree-Fock implementation
I implemented a restricted Hartree-Fock (RHF) calculation in the STO-3G basis set, as described in Szabo and Ostlund's book [1]. I managed to reproduce the energies of all their calculations ($\ce{H_2}$, $\ce{HeH^+}$, $\ce{H_2O}$, $\ce{CO}$, $\ce{CH_4}$ and $\ce{FH}$) with precision. For example, for $\ce{N2}$ I found
$$
E_\text{tot}^\text{HM} = -107.49583784393724 \text{ a.u.}
$$
while in the book they report the following (p. 192)
$$
E_\text{tot}^\text{Szabo} = -107.496 \text{ a.u.}
$$
The situation is analogous for all other molecules I tested.
Now I am trying to implement the calculation of forces, in order to perform Born-Oppenheimer molecular dynamics. I will start to compute the gradient of the PES by finite differences (computationally costly but easy to implement). In order to check my implementation I wanted to use Gaussian09 (for the first time!) as a benchmark because we don't find many force calculations on standar books. In order to check that my input file is correct (i.e. the options can reproduce my "dumb" implementation) I performed a single point calculation for the $\ce{N2}$ dimer. Surprisingly I found
$$
E_\text{tot}^\text{G09} = -106.765838750 \text{ a.u.}
$$
which is quite different from my result. The input file I used is the following:
%NProcShared=2
%Mem=1GB
#P RHF/STO-3G SP Symmetry=None Units=Bohr Pop=None Guess=Core

RHF

0 1
N 2.074 0.000 0.000
N 0.000 0.000 0.000

I tested the same structure on the input file for $\ce{H2O}$:
%NProcShared=2
%Mem=1GB
#P RHF/STO-3G SP Symmetry=None Units=Bohr Pop=None Guess=Core

RHF

0 1
H +1.4305507125e+00 +0.0000000000e+00 +0.0000000000e+00
H -1.4305507125e+00 +0.0000000000e+00 +0.0000000000e+00
O +0.0000000000e+00 +1.1072513982e+00 +0.0000000000e+00

In this case I find
$$
E_\text{tot}^\text{G09} = -74.9629400524 \text{ a.u.}
$$
wigh Gaussian09, while with my program I have
$$
E_\text{tot}^\text{HM} = -74.962937313769473 \text{ a.u.} = E_\text{tot}^\text{Szabo}
$$
For $\ce{H2O}$ everything seems to work fine. In a similar way, I get matching results also for $\ce{CH_4}$, $\ce{FH}$ and $\ce{CO}$.
There is a problem with the $\ce{N2}$ molecule (or simply with the input file) I am not considering?
Which is the best set of options for Gaussian09 that can emulate a home-made Hartree-Fock program (written following closely Szabo's book [1])?
[1] A. Szabo and N. Ostlund, Modern Quantum Chemistry, Dover, 1996.

Diagonalization of the core Hamiltonian provides usually not the best guess for the SCF procedure to say the least, and thus, by default Gaussian uses a more sophisticated guess obtained by diagonalizing the Harris functional (Guess=Harris). With this default guess one get the same energy as OP reported:
 SCF Done:  E(RHF) =  -107.495842181     A.U. after    5 cycles

Now, Pop=None was a terrible idea, of course. It is always advisable to look at the final orbitals. Everything looks fine for the case of the default guess
 Alpha  occ. eigenvalues --  -15.51806 -15.51612  -1.44283  -0.72249  -0.57311
 Alpha  occ. eigenvalues --   -0.57311  -0.53949
 Alpha virt. eigenvalues --    0.28131   0.28131   1.12344

but results with the core Hamiltonian guess look suspicios
 Alpha  occ. eigenvalues --  -15.44896 -15.44853  -1.34440  -0.64250  -0.60670
 Alpha  occ. eigenvalues --   -0.52980  -0.16921
 Alpha virt. eigenvalues --   -0.07243   0.31047   1.20826

The energy of the first virtual orbital is negative which is odd. It is difficult to say what went wrong with the core Hamiltonian guess, but some clue can be provided if turning the symmetry back on (this is another very sensible default turning which off was a bad idea, I think).
The default guess yields the electronic state with a particular symmetry ($\sigma_{\mathrm{g}}$)
 Initial guess orbital symmetries:
       Occupied  (SGU) (SGG) (SGG) (SGU) (PIU) (PIU) (SGG)
       Virtual   (PIG) (PIG) (SGU)
 The electronic state of the initial guess is 1-SGG.

and converges within this symmetry
 Orbital symmetries:
       Occupied  (SGG) (SGU) (SGG) (SGU) (PIU) (PIU) (SGG)
       Virtual   (PIG) (PIG) (SGU)
 The electronic state is 1-SGG.

But the core Hamiltonian guess yields the electronic state with no symmetry
 Initial guess orbital symmetries:
       Occupied  (SGG) (SGU) (SGG) (SGU) (PIU) (PIU) (PIG)
       Virtual   (PIG) (SGG) (SGU)

and converges within this state with no symmetry
 Orbital symmetries:
       Occupied  (SGG) (SGU) (SGG) (SGU) (PIU) (PIU) (PIG)
       Virtual   (SGG) (PIG) (SGU)
 Unable to determine electronic state:  partially filled degenerate orbitals.

which is again suspicious.
So, there is something wrong with the core Hamiltonian guess for this system (I suspect the basis is too small for the core Hamiltonian guess to yield sensible results), thus, SCF spits out some "strange" numbers. Strictly, speaking there is nothing strange with the numbers. Remember that SCF usually converges to a local minimum, and which particular depends on which part of the wave function space the initial guess placed the system in. And besides, you're not even guaranteed to have a minimum, SCF might converge to a stationary point of any kind.

I know the options I choose are bad for an actual calculation. However, in this case the goal is to have Gaussian09 behave as close as possible to my implementation. In particular I don't use symmetry and I don't perform population analysis (but I agree that looking at least at the orbitals would have been useful!). I thought that the option Guess=Core corresponds to setting the density matrix elements to zero (approximate Fock's operator with the core Hamiltonian), which is what I do. This is not the case?

@R.M., yes, you're right, the core Hamiltonian guess ignores all the electron-electron interactions, and is thus physically terrible. For that reason it can lead to strange results.

It seemed strange to me that this guess works in my implementation (and apparently also in Szabo's one) and not in Gaussian09. In any case I will try to implement a better initial guess! Thank you.

@R.M., in your particular case the core Hamiltonian yields the electronic state that is too far away from the global minimum, so that SCF couldn't find it is way to the global minimum. Just increasing the basis set to 6-31G "fixes" the core Hamiltonian guess, so that it yields a much better starting point.

@R.M., that is a bit strange, I agree. But different implementations of the same algorithm could work differently. The devil, as usual, is in the details.

@R.M. I'm guessing that your HF uses direct  SCF, while gaussian09 by default uses an iterative (DIIS and EDIIS) procedure to find the minimum.  As previously noted, SCF is not guaranteed to find a global minimum, and different minimization procedures can end up in different minima, even starting from the same spot (Guess=Core).  I Recall this being an issue when I wrote an HF program last year in matlab and tried to benchmark results against gaussian.  You can definitely turn it off, I believe that SCF=(NoDIIS, DM) (I can never remember the exact gaussian synax) should make it more like your home written program

Direct SCF and DIIS/EDIIS are different things. Direct SCF refers to the way in which integrals are generated and stored. DIIS/EDIIS refers to how convergence of the SCF equations is achieved. They are completely independent of each other.


text:
Do volumes and pressures calculated from Boyle's Law depend on the temperature?
As the title suggests, i have a question regarding Boyle's law.
As Boyle's law states that "The volume $V$ of a given mass of a gas, is inversely proportional to the pressure $p$ applied to it when the temperature $T$ is constant."
So if we had a gas at say a pressure $p_1$, and a volume $V_1$, corresponding to this pressure at a constant temperature $T_1$.
Now if i were to increase the the temperature of this gas to a new temperature $T_2$ and make it constant. And then apply the same pressure $p_1$ on the gas would the volume still be $V_1$?
Considering the fact that volume and pressure are inversely proportional for a constant temperature, and hence they dont depend on the temperature?? Am i right?
Thanks

No,the volume would change this time  as by Gay Lussac law pressure is directly proportional to temperature  which has increased this time and can overcome the pressure you had applied earlier thus increasing the volume.
Also I think you have a misconception regarding the term constant temperature.It means that in Boyle s law pressure is inversely proportional to volume at a given temperature say 200K .But for different temperature the law will follow but with different values.For eg. at a particular temperature 200k it can be:- 200litres and 50 pascal or 100 litres and 100 pascal.For higher temperature it would be:- 200 litres and 100 pascal or 100 litres and 200 pascal.

You are wrong.As you can see in the image  for a perticular pressure volume will be greater when teperature will be higher.


text:
Will magnesium undergo a redox reaction with zinc(II) or copper(II)?
Magnesium can do a redox reaction with $\ce{Cu^2+}$ as well as with $\ce{Zn^2+}$, since Mg has the lowest reduction potential $E_0$. (I am aware that a reaction may or may not be feasible as redox, and this is according the reduction potential of the two half-equations)
\begin{array}{l|l|r}
\ce{Cu^2+ & Cu} &\pu{0.34 V} \\
\ce{Zn^2+ & Zn} &\pu{-0.76 V} \\
\ce{Mg^2+ & Mg} &\pu{-2.37V} \\
\end{array}
I'm supposing we have in one beaker, a solution of $\ce{Cu^2+}$ ions and $\ce{Zn^2+}$ ions and a piece of magnesium. So the possible reactants are: $\ce{Cu^2+}$, $\ce{Zn^2+}$, and $\ce{Mg}$.
The potential difference between $\ce{Mg}$ and $\ce{Cu^2+}$ is greater than with $\ce{Zn^2+}$.
So what's the answer?
My guess is "with $\ce{Zn^2+}$", since by comparing with another example, ethanol ($\ce{CH3CH2OH}$) reduces $\ce{MnO4-}$ to give $\ce{CH3CHO}$ instead of $\ce{CH3COOH}$, if the quantity of $\ce{MnO4-}$ was limited.
\begin{array}{l|l|r}
\ce{MnO4- & Mn^2+}     & \pu{1.51 V} \\
\ce{CH3CHO & CH3CH2OH} & \pu{0.19 V} \\
\ce{CH3COOH & CH3CH2OH}& \pu{0.05 V}
\end{array}
So $\ce{CH3CH2OH}$ reacted with $\ce{MnO4-}$ to give $\ce{CH3CHO}$, that leads me to deduce that in the case of multiple possibilities (i.e. feasible redox reactions), the reaction with the least potential difference takes place (since $1.51 − 0.19 < 1.51 − 0.05$).
From another hand, I guess it's familiar that the other option of having $\ce{Mn^2+}$ reacting with $\ce{Mg}$ is true as being held between the most oxidant which is $\ce{Mn^2+}$ and the most reductant which is solely $\ce{Mg}$ in our case.

I wonder if an answer could be in antioxidant in the body.

Let’s assume for a second that magnesium reacts only with the zinc. The reaction would occur according the following equation:
$$\ce{Mg + Zn^2+ -> Mg^2+ + Zn}$$
You have already identified, if I read your question correctly that the redox potentials allow this and that the reaction is spontaneous etc. Assume that you have equimolar amounts of magnesium, zinc and copper for the rest of this thought experiment.
Now our new solution would contain magnesium(II), copper(II) and zinc(0) (okay, the zinc would deposit itself somewhere, but you get the picture). Notice anything? Yes, zinc and copper can react according to the well-known and often-displayed electrochemical reaction (depositing copper on a blank piece of zinc metal). The equation for that one is:
$$\ce{Zn + Cu^2+ -> Zn^2+ + Cu}$$
Again, check out the standard potentials to see that this reaction is possible and spontaneous.
If we add the two reactions together, we get an effective reaction:
$$\ce{Mg + Zn^2+ + Zn + Cu^2+ -> Mg^2+ + Zn + Zn^2+ + Cu}$$
$$\ce{Mg + Cu^2+ -> Mg^2+ + Cu}$$
You notice that the zinc is little more than a spectator ion.
Now how does this differ from the ethanol case? Well, permanganate can (and will) react with ethanol to form ethanal, and it can (and will) react with ethanal to form ethanoic acid — more rapidly. However, if permanganate (and therefore: electrons) is the limiting reagent, we end up with a mixture of ethanol, maybe ethanal and ethanoic acid. By a reaction analogous to the reverse Cannizarro reaction, these can comproportionate:
$$\ce{CH3CH2OH + CH3COOH <=>> 2 CH3CHO}$$
Therefore, the overall reaction seems to be one where ethanol only reacted with permanganate to only form ethanal, while in fact ethanoic acid was formed, but in the end the equilibrium favoured ethanal.

$\ce{Mg^2+}$ will undergo a redox reaction with both the species.
Have a look at the electrode potential values at this link: http://www.physchem.co.za/data/electrode_potentials.htm via the Internet Archive
What you'll find is that the half reaction of $\ce{Mg^2+}$ has a highly negative electrode potential $-2.37$ while the $\ce{Zn^2+}$ and $\ce{Cu^2+}$  half reactions have less negative $E_0$ values [$-0.76$ and $0.34$ respectively]. A highly negative electrode potential indicates a strong tendency to reduce other species, while it gets oxidized itself. The $E_0$ values for the net reactions in both cases is positive. [Try it out: $E(\text{cathode})-E(\text{anode})$].
Consider the reverse reaction, i.e.
\begin{align}
\ce{Cu + Mg^2+ &-> Mg + Cu^2+}\\
\tag{anode} \ce{Cu &-> Cu^2+ + 2e-} & (E_0  &= +0.34)\\
\tag{cathode} \ce{Mg^2+ + 2e- &-> Mg} &(E_0 &= -2.37)
\end{align}
The $E_0$ of the net reaction is $E(\text{cathode})-E(\text{anode})$ which turns out to be negative.
The change in Gibbs free energy- given by the expression $-nFE_0$, is thus positive for this reaction, implying that it is not feasible - i.e.  $\ce{Mg^2+}$ will not be reduced by $\ce{Cu^2+}$. If you try this out with $\ce{Zn^2+}$,  you'll end up with the same conclusion.
So, in response to your question - $\ce{Mg^2+}$ will participate in a redox reaction with both the species in which $\ce{Mg}$ gets oxidized while simultaneously reducing the other species.

Yash, I'm aware that E(cathode)−E(anode) must be positive so the reaction is feasible. I have edited my question to clarify it. What I actually meant is that in one container we have Mg, and both the $\ce{Zn^2+}$ and $\ce{Cu^2+}$.


text:
Ring closure via intramolecular nucleophilic substitution or intermolecular dimerization: which is favoured?
What reaction does the following enolate undergo?

Both are nucleophilic substitution reactions. Ordinarily, the intramolecular product should be favoured, but I know that the transition state has a geometrical constraint in that the incoming nucleophile must approach the C-Br bond from the opposite side. Would that favour the intermolecular reaction instead?
I went through March's Advanced Organic Chemistry and found this:

And this from Clayden's Organic Chemistry:

That's why I am not able to decide if the reaction is intermolecular or intramolecular.

The intramolecular reaction is strongly favored entropically, but you can drive the reaction in either direction by adjusting reaction conditions such as concentration.  See this earlier answer to a similar question for more background.

@ron , I have seen some reactions in which intermolecular reactions are favoured. I have added one in the edited question. I will take one more from claydon.

It will be the first product. Generally, Intramolecular reactions happen much faster than Intermolecular reactions. Imagine you are in a room with a few people and Godzilla picks up the room and shakes it. As you go tumbling around the room are you more likely to smack your head into your own knee or someone else's knee first?

In general, intramolecular reactions are faster. Additionally, I don't know which type the metaphor is suggesting is faster.

I apologize, you are right. Thats what I meant, but due to a brain fart I swapped the terms. the answer has been corrected.

Interesting example, but what if its a fat guy and his head cant stretch upto his knees. (Is this ideal)

There are bulk molecules that cannot react between certain reaction sites or may do so with great energy input lke when making cyclobutane and cyclopropane.  Though unlike a fat man molecules can twist.

Examples are given in Marxch's advanced organic chemistry which show that it's sometimes easier to smack your head onto someone else's knees :) See the edited question for an example.

This "answer" pretty much offers no explanation, it is much more suited as a comment. If you can explain why an intramolecular reaction is faster in many cases, then this might be salvageable, but in its current state it only provides an opinion.

I'd also like to add that the intramolecular cyclisation that you proposed is a 5-exo-tet reaction according to Baldwin's classification.

The ring being formed is 5-membered.
The bond being broken (the $\ce{C-Br}$ bond) is outside the ring, or exocyclic.
The carbon being attacked is a $\mathrm{sp^3}$-hybridised carbon with tetrahedal geometry.

If you look up Baldwin's rules, you'll see that 5-exo-tet reactions are perfectly fine and allowed. Evidently the compound you drew can twist itself a little bit to accommodate the linear transition state. (I am using "linear" loosely to refer to the arrangement of the incoming nucleophile, the carbon, and the leaving group.) Therefore, in this case, the intramolecular reaction will predominate, as is usual.
The disfavoured reactions that you cited from those textbooks are 6-endo-tet reactions. In such compounds, having a linear transition state is an issue as it would lead to too much ring strain.
As an ending note: Baldwin's rules are nothing more than an empirical observation. You might ask why 5-exo-tet reactions are fine, but not 6-endo-tet: where is the threshold or boundary which makes something favoured or disfavoured? The answer is, they did the experiments and found out that that's just how it is. No more, no less.

Why is this not the accepted and most highly voted answer? It is the correct answer, i.e., stereoelectronic effects of $\mathrm{S}_{N}2$ reactions.

@Zhe I just came (very) late to the question. It takes time for votes to come in and also the accepted answer may not even change if OP doesn't frequent the site. C'est la vie - not that I need rep anyway.

@orthocrescol But someone still had to down-vote it!

I like the warning at the end! The currently accepted answer is in my opinion a comment not an answer. Apart from the guidelines you have posted, I think that we also have to look at the intermolecular reaction. There we see that in order to have an SN2 kind of attack the bromide must be close to the negative charge. Unlikely.


text:
Mass of strontium iodide contained in a 350 mL solution

What mass (in mg) of strontium iodide is contained in $350\ \mathrm{mL}$ of a strontium iodide solution that has a concentration of $1.93\ \mathrm{µM}$?

What I have done:
$0.350\ \mathrm{L}\ \ce{SrI_2}\cdot \frac{1.93\cdot 10^{-6}}{1\ \mathrm{L\ Solution}}\cdot \frac{341.42\ \mathrm{g}\ \ce{SrI2}}{1\ \mathrm{mol}\ \ce{SrI2}}=2.31\cdot 10^{-4}~\mathrm{g}~\ce{SrI2}=0.231~\mathrm{mg}~\ce{SrI2}$
What is not correct here?

Why did you ask this question? Was that your answer in a test and it was marked as incorrect? Because the numeric value in the end seems to be correct, however I as a teacher would mark it as wrong, too, for incorrect use of units.

I did write the correct units when I did the problem (just wrong on here), but it got marked completely wrong on a test, without any explanation.

The mistake is that you didn't write "mol" after $1.93 \times 10^{-6}$.
Some teachers/professors might let that go, others wouldn't, depending upon what they are really trying to teach you.
Units and dimensional analysis are important.


text:
Are graphite intercalation compounds stable in water?
I want to do an experiment with graphite sulfate, but I want to know if it will de-intercalate in a beaker of distilled water?
Thanks!

According to "Graphite sulphate – a precursor to graphene", Siegfried Eigler, Chem. Commun., 2015,51, 3162-3165:

Graphite sulphate is used as a precursor to graphene for the first
  time. The positively charged graphene layers react with water to yield
  a processable graphene derivative.

This would seem to imply that it would indeed de-intercalate in a beaker of distilled water.


text:
Differences between Fe3+ and Fe2+
I would like to know the differences between Fe3+ and Fe2+ ions. I'm guessing they must differ in reactivity but I'm not sure how. Also, is there a way that we can observed the differences

The only thing that's different about ${Fe^2}^+$ and ${Fe^3}^+$ is the number of electrons, which in turn results in different properties of these species.
${Fe^2}^+$, aka ferrous, is pale green and turns violet when added to water. ${Fe^3}^+$, aka ferric, is yellow-brown in solution. You can use several methods to distinguish these two ions. I will mention some of them.
${Fe^2}^+$ can oxidised to ${Fe^3}^+$ with the help of oxidizing agents like $KMnO_4,\space HNO_3$, etc.. $KMnO_4$ is better because its pink color decolorizes on addition of ferrous ions.
Another method to distinguish them is by using colored complexes. For eg., ${Fe^2}^+$ forms a red-orange compound with amine lignads. ${Fe^3}^+$ forms blood red solution with thiocyanate ions, which is used as fake blood in movies, dramas and so on.
Ferrous ions normally have paramagnetic properties, but can become diamagnetic due to the formation of low-spin complexes. A ferric ion is only paramagnetic due to the presence of only one lone electron.

There are plenty of high-spin ferric complexes.


text:
skeletal structure of dimethylamine? HN(CH3)2
I don't understand how to draw this skeleton.
 I know there is a NH2 in an amine structure, but this only has NH?
and dimethyl means there a 2 identical single side chains right? but where on the skeletal structure are they?
I'm totally stuck as to how the skeletal structure is supposed to look.

http://i.imgur.com/vcr20cr.png The $\ce{-NH2}$ group is only in a primary amine. This is a secondary amine.

Please look at the skeletal structure of different types of amines:

I hope it's clear now.

this helps heaps, thanks! I was almost on the right track but I thought the solution was too simple and that I had it wrong


text:
How can the ketone in a β-ketoester be protected during hydride reduction?

I am supposed to find what X is. I know that $\ce{LiAlH_4}$ will convert all carbonyl groups to alcohols, and the ester group is being reduced. The only way of ending up with a ketone that I can think of is by a pinacol reaction.
However, the answer given was that X was $\ce{HOCH2CH2OH + H+}$. How does this work?

https://en.wikipedia.org/wiki/Protecting_group Read the introduction paragraph, it is exactly the same as your question.

That was helpful.

By the way, you could technically use any alcohol you wanted to protect the ketone, although ethylene glycol is probably a reliable choice since you get the stability of a 5-membered ring. Basically, the answer isn't unique.


text:
What makes ice slippery?
I've heard and read (multiple times) two theories have have been around for a while, both of them obvious nonsense.
One states that pressure and friction make the ice melt (turn to water) where the skater's blade comes in contact with it. This is patent nonsense since a puck that weighs very little, or a pencil that weighs next to nothing, slip along the ice as well as any skater. I also find it hard to believe that the temperature of the ice (several degrees below freezing) would suddenly jump several degrees in the split second that it takes for the skate to come in and out of contact with it.
The other theory is that the ice surface itself has the same qualities as water due to some hydrogen molecules still in search of home. This is also nonsense: pour water over a concrete floor and try to skate on it; see how far you'll get.
So my question is: why is ice slippery?

‘would suddenly jump several degrees’ is a misconception. Rather, pressure alone can cause ice to melt since liquid water is more dense than ice crystals at almost any temperature.

@Jan: Instantaneously?

near-instantaneously for a very thin sheet.

Couldn't help but think of Feynman: https://www.youtube.com/watch?v=MO0r930Sn_8.

This is a question has been one of interest since 1850, and I found a paper that nicely summarizes some of the hypotheses that you yourself listed, as well as what is considered to be true. I've paraphrased it below.

Michael Faraday was the first to propose that there exists a film of water on ice that freezes when in contact with other pieces of ice, but remains liquid by itself. Though he performed several experiments that supported his claims, most of which involved sticking two pieces of ice together, his ideas were largely dismissed until C. Gurney also suggested that this film of water on play a role in ice's slipperiness.
Gurney hypothesized that molecules, inherently unstable at the surface due to the lack of other molecules above them, migrate into the bulk of the solid until the surface becomes unstable, which prompts the formation of the liquid phase. This came to be accepted, and Charles Hosler compiled very compelling evidence on the the forces required to pull ice spheres apart. He discovered that at vapor pressures below ice saturation, no adhesion occurred below $-4\:^{\circ} \mathrm{C}$. From this they inferred that the expected roughness of the surfaces was removed by the presence of a liquid film whose thickness was sufficient to provide a smooth surface of contact.
More experiments were performed trying to determine the exact thickness of the water film at different temperatures, and the phenomenon was found to occur at temperatures even below $-22\:^{\circ} \mathrm{C}$.

Here is a depiction of what the surface of ice looks like at a molecular level.

Rosenberg, R. Why Is Ice Slippery? Phys. Today Physics Today. 2005, 58, 50–54.

I recognised the figure in this article. This might also be of interest.

Could you help me understand which part precisely you mean to be "what is considered to be true" - is it everything in the paraphrase above or only some part of it? I'm not trying to be picky at all - it's just that there is so much speculation and misinformation out there, I'd love to know just what is considered to be true and what is considered to be still uncertain.

For the latest on this I suggest reading here: http://phys.org/news/2015-12-theory-ice-slippery-behavior.html
The key extract is:

"The most important result is that I have constructed a phenomenological shear stress law which is able to explain ice friction as a function of sliding speed and temperature in a wide velocity and temperature region," Persson said.

The author agrees with the premelting theory, but tries to extend it for buried surfaces which are difficult to study due to the nature they are buried.


text:
Allotropes of carbon conductivity
Give reason how can i made diamond a conductor by doping  charcoal and graphite which are allotrope of Carbon act as conductor. Whereas diamond which has a different crystal structure act as an insulator

In charcoal and graphite, conjugated pi systems (the molecule orbital configuration found in carbon-carbon double bonds) act as the conductive band, where electrons can move freely.
In diamond, all bonds are SP3 hybrid bonds (the molecule orbitals found in single bonds). Therefore, no pi systems and thus no conductive band exists.
Edit (after  Jon Custer's comment, thanks): Wikipedia says: Except for most natural blue diamonds, which are semiconductors due to substitutional boron impurities replacing carbon atoms, diamond is a good electrical insulator, having a resistivity of 100 GΩ·m to 1 EΩ·m[28] (1011 to 1018 Ω·m). Thus it is possible to dote diamonds with boron to make them more conductive.

Diamond is a wide band gap semiconductor that can be doped to make devices. Not as well as silicon, that is true.


text:
How to merge PDB-files to build trimer from monomers
Problem
Using WinCoot or SWISS PDB viewer, how would one merge monomeric structures that should be in dimer formation?
What I have done:

I have downloaded the PDB-file which shows the monomer with a bound drug (Infliximab Fab fragments), but it should be a trimetric structure as whole -although you only get the monomer when you download the file (TNFa) .
I have opened the file in both coot and SWISS-PDB viewer, and opened three copies of the monomer, but they just overlap and I am not able to combine them to the proper structure

I also know that it is possible to download the "biological assembly" PDB-file that has the entire trimetric structure, but the point here is to be able to do that on my own (which should be easy for those who know it?).
I am very new to all of this modeling, so there might be something basic I don’t understand. But I though that you can somehow "merge" the monomers by using some "layer" function? I am really unsure.

Uh, I vote to leave this open for now; I fail to see how it's a HW question.

Well, it's debatable, but OP should show some research effort and knowledge about these programs. BTW @H.T. Do you simply want to have coordinates duplicated or find "real" dimer?

I think I simply want to have the coordinates duplicated - but not just overlapped (in the sense that its a trimeric structure, but I only have one monomer). I am very new at this (modelling), so I will rephrase my question a bit and remove the homework tag (its a course assignment).

cross-posted http://biology.stackexchange.com/questions/41338/how-to-build-a-trimeric-protein-structure-from-monomeric-pdb-file

I am closing this question because the cross-posted version is more elaborate.

The PDB-file only contains the asymmetric unit, and no information on a potential biological relevant multimer. So you will need to get information about the state in solution experimentally.
That being said, one can often deduce from the size of the contact surface whether you have a potential quarternary structure. You can do this by hand, but the PISA-server is quite helpful there, and it will also output the multimer PDB-file:
http://www.ebi.ac.uk/pdbe/pisa/
If you just want to apply symmetry operators to your asymmetric unit in order to create symmetry mates, you can do so in
Pymol:
- click Action/generate/symmetry mates/within X Å
- from the command line:
        symexp prefix, selection, cutoff
        , e.g. symexp sym,1GVF,(1GVF),5
   see also http://pymolwiki.org/index.php/Symexp

Coot:
- if you just want to look at the symmetry mates, activate View/Cell&Symmetry with an appropriate radius and/or select "Symmetry by molecule"
- if you would like to actually create the symmetry mates, try Extensions/Modelling/New Molecule from Symmetry Op  - however you will have to specify the SymOp manually.

You will need to have the CRYST1-record in your PDB-file with the correct space group.


text:
Nomenclature of spiro compounds
Does the nomenclature of spiro compounds follow the same pattern as that for bicyclo compounds? That is, in absence of functional groups do we start form the largest chain to the smallest chain. What would be the case if there were functional groups?

No, according to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), numbering of spiro compounds starts in the smaller ring.

P-24.2.1 Monospiro alicyclic ring systems
Mono spiro parent hydrides consisting of two saturated cycloalkane rings are named by placing the nondetachable prefix ‘spiro’ before the name of the unbranched acyclic hydrocarbon with the same total number of skeletal atoms. The number of skeletal atoms linked to the spiro atom in each ring is indicated by arabic numbers separated by a full stop, cited in ascending order and enclosed in square brackets; this descriptor (called in these recommendations the ‘von Baeyer spiro descriptor’) is placed between the spiro prefix and the name of the acyclic alkane. Numbering starts in the smaller ring, if one is smaller, at a ring atom next to the spiro atom and proceeds first around that ring, then through the spiro atom and around the second ring.

The fixed numbering of the spiro hydrocarbon ring is not modified by the introduction of heteroatoms, double bonds, or functional groups. However, low locants must be attributed if there is a choice.

That is exactly what I needed, but is there a reason for this or is it just arbitrary?

In Spiro compounds its actually the opposite,we start from the smaller chain to the larger chain.
General Name structure:
Secondary Prefix Spiro [X,Y] Root word Primary Suffix Secondary Suffix
X,Y are the number of carbons present on both sides of bridge carbon.
X,Y should be written in ascending order.
Numbering should be started from carbon which is adjacent to fused/bridge carbon in the smaller ring and it should continue into bigger ring through the fused/bridge carbon.
Even if functional groups we will follow the same rules above (even if functional group is there in the bigger ring).
Example: spiro[3.4]octan-5-one