text:
Iodometric titration of copper – standard electrode potentials
Upon inspection of the standard electrode potentials for the half reactions for both $\ce{Cu^2+}$/$\ce{Cu+}$ and $\ce{I2}$/$\ce{2I-}$, it becomes apparent it should be the iodide ions that act as the oxidising agent, as the standard electrode potential for the iodine half-cell is much higher.
$$\begin{alignat}{2}
\ce{Cu^2+ + e- \;&<=> Cu+}\quad &&E_\text{cell} = +0.159\ \mathrm{V}\\
\ce{I2 + 2e- \;&<=> 2I-}\quad &&E_\text{cell} = +0.54\ \mathrm{V}\\
\end{alignat}$$
To explain this, many websites that I’ve looked at suggest it’s the weak solubility of copper(I) iodide, increasing the $\ce{Cu^2+/Cu+}$ potential to around $+0.88\ \mathrm{V}$.
Such websites include:

http://www.titrations.info/iodometric-titration-copper
http://chemistry.tutorvista.com/analytical-chemistry/iodometric-titration.html

What I’d like to know, is how this value of $+0.88\ \mathrm{V}$ has been determined, and a little of the theory behind as to why it has increased from $+0.159\ \mathrm{V}$.

Let's consider the following redox couples:
$\ce{Cu^{2+} + e^- ->Cu^+\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}{ E^o_1}(\ce{Cu^{2+}/Cu^+}=0.17\, \ce{V})$
$\ce{Cu^{2+} + e^- + I^- ->CuI\,\,\,\,\,\,\,\,\,\,\,\,}{ E^o_2}(\ce{Cu^{2+}/CuI}=?)$
As the chemical species $\ce{Cu^{2+}}$, $\ce{Cu^+}$ and $\ce{CuI}$ are together in the aqueous solution, the two redox couples  $\ce{Cu^{2+}/CuI}$ and $\ce{Cu^{2+}/Cu^+}$  are in equilibrium and that means: $$ E_1(\ce{Cu^{2+}/Cu^+})=E_2(\ce{Cu^{2+}/CuI})$$ Using Nernst equation for each couple:$$ E_1(\ce{Cu^{2+}/Cu^+})=E^0_1(\ce{Cu^{2+}/Cu^+})+\frac{RT}{F}\ln{\frac{[\ce{Cu^2+}]}{[\ce{Cu^+}]}}$$
$$ E_2(\ce{Cu^{2+}/CuI})=E^0_2(\ce{Cu^{2+}/CuI})+\frac{RT}{F}\ln{[\ce{I^-}][\ce{Cu^2+}]}$$
We can write then:
$$E^0_2(\ce{Cu^{2+}/CuI})= E_1^0(\ce{Cu^{2+}/Cu^+})+\frac{RT}{F}\ln{\frac{[\ce{Cu^2+}]}{[\ce{Cu^+}][\ce{Cu^2+}][\ce{I^-}]}}$$ We rewrite the above equation after simplification:$$E^0_2(\ce{Cu^{2+}/CuI})= E_1^0(\ce{Cu^{2+}/Cu^+})+\frac{RT}{F}\ln{\frac{1}{[\ce{Cu^+}][\ce{I^-}]}}$$
$$E^0_2(\ce{Cu^{2+}/CuI})= E_1^0(\ce{Cu^{2+}/Cu^+})+\frac{RT}{F}\ln{\frac{1}{K_{sp}(\ce{CuI})}}$$
$$E^0_2(\ce{Cu^{2+}/CuI})= 0.17+\frac{8.314 \times 298}{96500}\ln{\frac{1}{10^{-12}}}=0.88 \ce{V}$$


text:
Dissociation of different compounds?
What would be the dissociation of Hydrosulfuric acid in the presence of water?
I was thinking that Hydrosulfuric acid or $\ce{H2S}_{(aq)}$ would form $\ce{2 H+}$ and Sulfur ions.
With this logic, I keep getting the wrong answer :(
The right answer of this question is it would not dissociate at all.
Shall appreciate some help on explaining this. Thanks!

$\ce{H2S}$ has a mostly covalent character. You can decide for it by the crude way textbooks give: Calculate the difference in electronegativity to decide whether it's a non-polar covalent compound, a polar one or has ionic characters.
For the dissociation part (which is usually endothermic overall) it'll need more "ionic character". And it does dissociate in water, resulting in a weak acid named  sulfhydric acid. (The spelling varies with citable sources) Your textbook likes to think about it as not dissociable, only because it's a weak acid.
Addendum:
The "ionic character" plays its role of a rule a good one as explaining acid 'strength'. However, it acts almost flawlessly in the same period, but has exceptions when it gets to groups. As Dave mentions, $\ce{HF}$ is weaker than for instance, $\ce{HCl}$ and that's because the atomic radius changes more dramatically than electronegativity difference. (See here ) But, in the $\ce{H2S}$ molecule, this exception hasn't occurred. Dave pointed out a good example of when the rule isn't working that well, but as I said, we can ignore the other factors in this very case.

so all weak acids do not dissociate?

No, but "weak acids", as the name fully suggests, dissociate partially. At times, lesser than 1% of the acid dissociates in water, so it's better to think they don't dissociate at all. It's like "Don't bother me with details".

verry helpful mannnn

@MARamezani I don't think what your are saying about ionic character and strong versus weak acids is accurate.  The series HF, HCl, HBr and HI has respective ionic charaters of 45%, 17%, 12% and 5%, and HF is a weak acid while the others are strong.

@DavePhD I'll edit my answer to cover your part.

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text:
How does HF dissolve glass?
By what mechanism does HF proceed in dissolving glass? Why is it the only acid that has this capability? Is it because of the small size and high electronegativity of fluorine?

But what exactly do you mean by "only this capability"?

@Matthew, it is  why does only it have this capability or why is only HF capable of doing so

$\ce{HF}$ reacts with glass ($\ce{SiO2}$).[1]
$$\ce{SiO2 + 4HF -> SiF4 + 2H2O}$$
$\ce{SiF4}$ is not a solid that consists of vertex-connected tetrahedra like $\ce{SiO2}$ but is a gas at room temperature. Technically, $\ce{HF}$ is not a solvent since in this case it reacts with the glass vessel.
According to Spierings:[2]

$\ce{HF2-}$ ions are adsorbed on surface silanol groups, the $\ce{HF}$ molecules on vicinal silanol groups and $\ce{H+}$ on surface bridging oxygens in siloxane units. [...] These are transformed into surface groups such as $\ce{\bond{#}Si-F}$ and $\ce{\bond{#}Si-O-SiF3}$. The adsorption of $\ce{HF}$ and $\ce{HF2-}$ increases the electronic density on the bridging oxygen in the siloxane unit. This in turn makes these oxygens more basic, so more $\ce{H+}$ ions are adsorbed, which leads to more siloxane bonds being broken per time unit, i.e. a kind of catalytic effect. [...] The catalytic action of $\ce{H+}$ ions on breaking siloxane bonds also occurs in the dissolution of glasses in acidic and weakly alkaline solutions.

Hollemann, A. F.; Wiberg, E. Hollemann–Wiberg: Lehrbuch der Anorganischen
Chemie, 34th ed.; de Gruyter: Berlin, 2007; Vol. 102.
Spierings, G. A. C. M. Wet chemical etching of silicate glasses in hydrofluoric acid based solutions. J. Mater. Sci. 1993, 28 (23), 6261–6273. DOI: 10.1007/BF01352182.

I know the reaction, but my question is about mechanism how does this happen, does fluorine go interstities of the glass or h+ or the kinetic energy is responsible for it. Technically, HF doesn't dissolve glass since a solvent does not react with your compounds.; thats why I said does it disintegrate glass then because of what: Energy or size of ions or both

What do you mean by the kinetic energy of the fluoride ion?

@Dissenter, I am reffering to maxwell velocity distribution in liquids, which is related to collision theory of reactions

I've no idea what "a solvent does not react with your compounds" means especially in this context. Hydrofluoric acid clearly does react with glass. And that is what matters since most other things do not.

Oh boy... thank you for pointing that out. I mixed things up. Now it should be correct.

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text:
Can plastic leak into water from a HDPE "2" plastic container at room temperature?
Here in USA supermarkets often sell water in plastic 1 gallon jugs. These jugs seem to all be stamped with the text "HDPE" (meaning that the material is High-density polyethylene) and a resin number of "2" on the bottom.
I had one of these jugs, containing distilled water, on a counter in the kitchen, and another one in the refrigerator. For some reason the water on the counter had a light plastic taste, whilst the water in the fridge did not.
Could this taste be because of plastic leaking into the water? Or what else could cause the plastic taste?

Polymer itself is insoluble, but additives not completely.

Nice question but do we really have an answer yet?

HDPE
has a molecular weight more than several hundreds of thousands,
"leaking" is plainly impossible.
This taste is the tase of stale water, nobody knows
what that taste is actually.
Maybe its just the taste of not cold enough water,
lacking some dissolved air, or some bacteriae
develop after contact with air and light.
(google for pseudomonas)
The only substance which ever got into the jug prior
to the water is the air (or some other gas?) used to
inflate the jug in production.
But that cause would impair the taste of the water in
the fridge as well.
BTW why do You drink such botteled water?
What about the tap water in your town/city/village?

Thank you for the explanation, Georg. I use tap water for coffee and such, the distilled water was just for other purpose.

Nah, I can tell the difference between plastic and stale.  And the water in the HDPE tasted like plastic.  It's not bacteria; it's not air; it's not temperature; it's plasticizers.  @burak-ulgut's answer is a better answer.

This should've been a comment as it only backs up Georg's answer, but it just got too long for that.
Plastic doesn't "leak", scientifically. However, it can go through the degradation process.
You don't provide enough info for anyone to decide, but if you're certain that was the "plastic" taste (I don't have much idea how you do) it's possible it may have gone through degradation, though I can tell you 95 percent of the time it isn't. And that's simply because if the plastics were too unstable to resist minor chemical or energetic "attacks" (read the wikipedia article for degradation to see what I mean), they wouldn't have produced it like that.
There is the urban hoax that doing stuff like pressurizing a mineral-water bottle by freezing the water inside releases dioxins that are carcinogenic and which cause the change of the taste of water, but this, is not true. (See here.)
Georg's answer pretty much sums it up, but I think it's better to add: It may also be possible that you (or anyone else, or mistakenly the company itself) put the "tasty" water in contact with soluble mineral salts. Thus, you drink hard water and swear for its taste!

Thank you for your comments, MARamezani. To explore your last paragraph further, could the soluble mineral salts lead to a plastic taste? Otherwise could it be that the water is not as distilled as it is marketed to be(?).

I wouldn't say it's to be blamed on the company too much. It's not only HDPE that's in contact with water. To "optimize" plastics there are many things done. (As an easy example the polymer is PE but you "HD" it) One of these things is adding material like phthalates (No phthalates, not in this case) to plastics. As mithoron points out, the minute solubility of these in the water could've caused what we know as plastic taste.

degrade? The article that is linked says 58 - 1200 years. Should we scratch that off?

I agree with the other answers, it is definitely not HDPE, however, it could be one of the chemicals known as plasticizers that are used in the process. There is always leftover plasticizer in the bottle and these are known to leach into water.
The NIH sponsored multiple studies on the levels of this, one of them published here:
http://www.ncbi.nlm.nih.gov/pubmed/21400322

Ha, why do you say you agree?  Your answer is in direct contradiction to those of the others.  It's not bacteria; it's not air; it's not temperature; it's plasticizers, like you state!  Or possibly degradation, like M.A.R. states.

@zylstra .....I will post. How do we know HDPE is degraded. The I will post again.


text:
What ratio of bicarb to vinegar do I need in order for the result to be pH neutral?
This question stems from one I just asked over at Seasoned Advice, where my recipe tasted decidedly off because the pH was wrong.
The recipe called for $\pu{16 g}$ bicarb of soda (sodium bicarbonate, $\ce{NaHCO3}$) and $\pu{15 g}$ of vinegar (mine was 5% acidity, and I assumed a density of $\pu{1 g ml-1}$). I just googled the corresponding pHs of these two, and apparently vinegar is more acidic than bicarb is basic: vinegar has a pH of 2.4–3.4 depending on concentration, and bicarb is pH 9.5.
How can I calculate whether I put too much vinegar or too much bicarb into my recipe? The ratio in the recipe is almost $1:1$.
What ratio in grams do I need of bicarb vs vinegar in order for the result to be neutral? I know there are other contributing factors to the pH in the recipe but I just want to focus on these two first. Assuming the vinegar is 5% which mine was.
Edit:
So I found the explanation for the chemical reaction between bicarb and vinegar here. Relevant bit:

$$\ce{CH3COOH + NaHCO3 → CH3COONa + CO2 + H2O}$$
$$\frac{\pu{1 g}\;\ce{NaHCO3}}{\pu{84.0068 g mol-1}} \times \frac{1}{1} \times \pu{60.0522 g mol-1} = \pu{0.715 g}\;\ce{CH3COOH}$$
What this formula tells you is that every gram of baking soda requires \pu{0.715 g} of acetic acid. But the remaining problem is that vinegar is not pure acetic acid.  In fact, most vinegars vary from 4% to 18%.
Using those numbers, you would need somewhere between $\pu{0.715 g}/0.04 = \pu{17.9 g}$ and $\pu{0.715 g}/0.18 = \pu{3.97 g}$ of vinegar, for each gram of baking soda.
If you're lucky you can get some vinegar with the percent acetic acid listed on the label, then you can substitute that number into the equation above to determine how much vinegar you'll need.

So if I plug in my 5%, I get $\pu{0.715 g}/0.05 = \pu{14.3 g}$ vinegar for $\pu{1 g}$ bicarb.
Yikes! This is way more than $1:1$. But I'm not sure if it answers my question. Is the result, $\ce{CH3COONa + CO2 + H2O}$, pH neutral? If yes, then my vinegar does hardly anything to cancel out the base, is that right?
Edit 2:
Okay so according to Wikipedia, $\ce{CH3COONa}$ is a base. So even though my reaction formula shows what happens when all the bicarb/vinegar is used up, the result is not neutral (how much does the water and carbon dioxide contribute to bringing the pH down?). I need more acid to make my sodium acetate neutral (or at least neutral enough to be palatable?). This means that the needed ratio is higher than 14:1!? So almost certainly way too much bicarb?

Your calculation is correct (14.3 grams of 5% acetic acid completely reacts with 1 g sodium bicarbonate). But you're not interested really in "completely reacts"; you want something like pH 7.
You've got it just about right with your mix; you're at the endpoint, and this is a point where the pH will drop very rapidly with even a small addition of vinegar. So it would only take a few additional drops to make the pH go down to 7... but that may not be what you really want.
Sodium acetate has a mild salty taste by itself. So I'm not sure adding extra vinegar will be worthwhile, it'll make it a bit more vinegary.
Here's what I'd do: Mix vinegar and baking soda in the ratio you've figured out, and then see how that tastes. Make adjustments at that point according to taste, rather than by "titrating" to an arbitrary pH.

Why does the ph drop very rapidly with even a small addition of vinegar?

When both acetic acid and acetate are present, the equilibrium between the two can shift when you add any more acid or base to the mix so the pH doesn't change much---this is called "buffering". When you only have one form or the other present, the mix can't buffer so any added acid or base can change the pH a lot.

I am a professional baker. A good ration in volume is for every 1/4 tsp Baking Soda, use 1 Tablespoon white vinegar.   But you can try this for yourself.
in a small bowl, measure 1/4 tsp Baking soda, pinch of salt, and a TBS sugar and 2 TBS water (because your baking formula will have similar ingredients.  At this point there will be NO reaction (bubbles). Very slowly and incrementally add the 1 TBS white vinegar while stirring.  As you add the vinegar, you will observe a reaction.  You should also notice the reaction begins to weaken toward the last drops of vinegar.  At this point you can taste the liquid.  It will taste like flat soda like sprite or 7 Up but not have any objectionable acid or base taste.  in all reality, the mix will still react with more vinegar,but this is good because you don't want to burn out all your leavener while mixing.  You may increase these amounts in your recipe as required... hope this helps

I tsp a tea spoon, or a table spoon, or is that TBS? It would be preferable if you could use units of measurement that are also know outside the English-speaking world.


text:
the particle in 1-D box momentum and total energy(P.E + K.E (x))
Employing the particle in 1-D box can momentum and total energy(P.E + K.E (x)) can be measured simultaneously?

Welcome to chemistry.SE! If you had any questions about the policies of our community, you can visit the help center or take a tour of the website.

@MARamezani The particle in the box is a classical model in quantum chemistry and found in standard textbooks on physical chemistry by Moore & Hummel or Wedler or .... From the point of view of a slightly more experienced chemist, the matter itself is definitely on-topic. Whether this could/should be closed due to the lack of personal effort is another question.

My bad then @KlausWarzecha. I just told myself that "momentum", "potential energy", "kinetic energy" are more physics, without further digging into the problem.

As this is obviously a homework problem, I'll just give you a hint:
Observables can be measured simultaneously with arbitrarily small uncertainties if their operators commute. Compute the commutator of the momentum and particle-in-a-box Hamiltonian operators; if the commutator is zero, the answer to your question is "yes".


text:
Why does potassium form peroxides but sodium does not?
As we go down the periodic table, Group 1 alkali metals hold their single outer valence electron more loosely, and so reactivity increases. Below potassium (Na, Li), we can store the metal in oil with no difficulty. Above potassium (Rb, Cs), we need to exclude all oxygen to prevent reaction in air. Potassium (K) is the unique element where we can technically store it in oil, but it is highly likely to react with any available oxygen to form explosive peroxides. I'd like an explanation of why this is; i.e. why is the reactivity favorable for the formation of peroxides. Thanks!

If you look at the standard Gibbs free energies for the alkali metal peroxides, actually, lithium should form it most easily. The standard Gibbs free energies of formation are -564.8, -430.1, -418.4, -351.5, and -359.8 kJ/mol for $\rm Li_2O_2, Na_2O_2, K_2O_2, Rb_2O_2,$ and $\rm Cs_2O_2$, respectively.
If we're looking at reaction of an alkali metal with oxygen to form explosive compounds, perhaps we should look at superoxides. The superoxides are powerful oxidizing agents, and they can explode when mixed with water, acids, organics, or powdered graphite.
At room temperature, oxygen readily reacts with K, Rb, and Cs to form superoxides ($\rm MO_2$), while Li and Na don't. There is a direct correlation between superoxide stability and electropositivity of the metal concerned (source: Cotton & Wilkinson's Advanced Inorganic Chemistry). $\rm NaO_2$ can be obtained by high pressure and temperature reaction of $\rm Na_2O_2$ with $\rm O_2$, and $\rm LiO_2$ won't form at room temperature.

Nice answer -- here's a vid of exactly how high-temperature you need to get to make $\ce{Na2O2}$: https://www.youtube.com/watch?v=ZqEWUw6sgpA

You may want to compare the quoted free energies with those of the normal oxides. For example, with lithium you have $-564.8$ kJ/mol for the peroxide versus $-562.1$ from WP for the normal oxide. These are so close that they are likely within experimental error: the peroxide may or may not be possible at room temperature. Only sodium and heavier in Group 1 have the free energies far enough apart to definitively favor the peroxide.


text:
Addition reaction of hydroiodic acid to 2-bromo-3-chloro-2-butene
How can the major product of the addition reaction of 2-bromo-3-chloro-2-butene with hydroiodic acid be predicted, since Markovnikov's rule fails to distinguish it?
$$\ce{H3C-C(Cl)=C(Br)-CH3~+~HI~->~?}$$

TL;DR - Sometimes you need to do the experiment.

Like all questions involving Markovnikov's rule, you should compare the structures of the two carbocations:

At C2 (Bromo)

$$\ce{H3C-C(Cl)=C(Br)-CH3 + H+ -> H3C-CH(Cl)-C+(Br)-CH3}$$
The two issues to consider are resonance and induction.
Resonance
You could draw a resonance structure to show that the bromo group is stabilizing the carbocation:
$$\ce{R2C+-Br <-> R2C=Br+}$$
However, since the bromine atom is much larger than the carbon atom, it cannot form as effective $\ce{p}-\pi$ overlap with carbon.
Induction
The carbocation has two stabilizing alkyl groups. What about that bromo group? Bromine is more electronegative than carbon, but not by much (2.96 vs 2.55 on the Pauling scale). Plus, bromine has a lot of electron density around it to donate, if it could. Bromine thus is a mild electron-withdrawing group by induction.

At C3 (chloro)

$$\ce{H3C-C(Cl)=C(Br)-CH3 + H+ -> H3C-C+(Cl)-CH(Br)-CH3}$$
We once again examine resonance and induction.
Resonance
$$\ce{R2C+-Cl <-> R2C=Cl+}$$
Chlorine is smaller than bromine and closer in size to carbon. Thus, chlorine is a better resonance stabilizer than bromine.
Induction
The same two alkyl groups are present, so we will discount them. Chlorine is more electronegative than bromine (3.16 on Pauling), and it has less electron density around it to begin with. Chlorine is more powerful electron-withdrawer by resonance than bromine.
Compare:
                  Chlorine              Bromine
Resonance         Stronger donor        Weaker donor
Induction         Stronger withdrawer   Weaker withdrawer

In many cases, resonance usually wins, so I might expect carbocation formation preferentially at C3 with the chloro group, but I would expect the other product to form also.
Now... what else might happen.
Regardless of the initial regioselectivity, the carbocation also can be stabilized by forming a halonium ion. Bromine is better at this since it is larger (and less electronegative so it can stabilize the positive charge better).
This halonium would be attacked at the less hindered position (C2), resulting in transfer of the bromine to C3.
So in effect, you can get three products:
$$\ce{CH3-CHCl-CIBr-CH3}$$
$$\ce{CH3-CICl-CHBr-CH3}$$
$$\ce{CH3-CClBr-CHI-CH3}$$
You would then want to actually do the experiment to determine which of these is really the major product.


text:
Why do people assume 28 oxygens in chlorite mineral formula?
I have been trying to calculate chlorite mineral formulas from microprobe data. However, all of the mineral formula spreadsheets on the internet, including those from the Open University assume 28 oxygens (http://www.open.ac.uk/earth-research/tindle/AGT/AGT_Home_2010/Microprobe-2.html)
However, all the formula I have ever seen for chlorite only have 18 oxygens? (see DHZ or http://www.mindat.org/min-1016.html)
By then calculating my formula assuming 28 oxygens, it means I have easily more cations than the mineral structure would have sites for?
Many thanks for any help

It doesn't mean that there are 28 oxygen atoms.  It means a charge equivalent of 28 oxide anions, considering that each oxide anion has a -2 charge and each hydroxide anion has a -1 charge.
This is based on the chlorite having an formula including $\ce{O20(OH)16}$
For example see $\ce{Mg10Al2(Si6Al2)O20(OH)16}$ in The effect of Cr on chlorite stability and the appearance of a new Al-rich silicate
In otherwords, as explained in OCTAHEDRAL OCCUPANCY AND THE CHEMICAL
COMPOSITION OF DIAGENETIC (LOW-TEMPERATURE) CHLORITES

All chlorite analyses were re-calculated as chlorite structural formulae on the basis of 28 oxygen equivalents (56 negative charges) i.e. assuming an ideal anion framework of $\ce{O20(OH)16}$


text:
Iodate ion and Octet rule
'm trying to figure out why the oxidation state of the iodate ion is +5. The iodine atom has 7 electrons in its outermost shell (comprised of s and p sub-shells). Two oxygen atoms receive 2 electrons from iodine to obtain full valence shells (s2 p6) and only one electron goes to the other oxygen atom. This leaves iodine with a full sub-shell s2. Is this a deviation from the Octet rule?
And how is the charge on the iodate ion negative when the third oxygen requires an electron?

First of all I strongly advise you to go through the following link as it will make my answer more clear to you .
Iodate ion

Two oxygen atoms receive 2 electrons from iodine to obtain full valence shells (s2 p6) and only one electron goes to the other oxygen atom.

The bond between the iodine atom and oxygen atom is a covalent bond . The electrons from the iodine atom are not given but shared .

This leaves iodine with a full sub-shell s2. Is this a deviation from the Octet rule?

The two double bonded oxygens ( refer the structure in my link ) share in all 4 electrons and the 3rd oxygen shares a single electron . Iodine hence has more electrons than octet rule specified 8. This is why it violates the rule .

And how is the charge on the iodate ion negative when the third oxygen requires an electron?

The third oxygen shares a single electron and hence has a negative charge on it which makes the overall molecule an anion with charge -1.


text:
Color changes of a solution of acetone and sodium hydroxide
I have put acetone and sodium hydroxide solution (approximately $\pu{1M}~ \ce{NaOH}$) in a beaker and didn't see any colour change for a day (actually, it was transparent, no colour).
After almost 2 days, I saw a dark brown colour (no precipitate, just diffused colour). Now (after 25 days) its colour seems black, but when I take a small amount and dilute it with water, it is brown. What chemical compound has formed that has this colour? What reaction has taken place? (I doubt it is an aldol condensation)

Probably your acetone has polymerized into short chain by a aldol reaction.

Acetone may have self condensed with itself. As time passed it would have got polymerized too.

Under basic condtions, acetone does condense with itself, first to form mesityl oxide, which is colorless, and then to form isophorone, which has a yellow color.
Both mesityl oxide and isophorone are still reactive carbonyl compounds likely to be attacked by deprotonated acetone carbanions.  That leads to bigger and more conjugated condensation products, which leads to darker and darker colors.
I found a paper which looks in some detail at the base-catalyzed polymerization of acetone, albeit with a different base and at high-temperature conditions in the gas phase, but I suspect a similar pathway happens, more slowly, at ambient temperatures in the liquid.  In the liquid, even larger molecules can likely form because they will stay solvated and can keep reacting (as opposed to in the gas phase where big molecules "crash" out of the phase as soot/gunk/etc.)
The color is likely in part from isophorone but much moreso from large polyconjugated molecules derived from isophorone containing many consecutive sp2 hybridized carbons. Such compounds are very often colored due to intense absorption of visible light.

So, you suggest the colour is because of isophorone?

Partially.  The color is like from large polyconjugated molecules containing many consecutive sp2 hybridized carbons. Such compounds are very often colored due to intense absorption of visible light.  I'll update my answer to reflect this.


text:
Which is more basic: aniline or pyridine?
Please explain why, thank you very much.

You can look at the pKb values.  It turns out pyridine is more basic.  This is difficult to predict from looking at the Lewis structures of the molecules.  What you should see is that the lone pair on pyridine is on an sp2 hybridized N.  This makes it less basic than an sp3 N.  The N in aniline is sp3 but the lone pair is in resonance with the pi electrons of the phenyl ring, lowering the basicity.


text:
How to Determine the Rate Law knowing time and concentrations
Consider the hypothetical reaction described by the following equation:
$$\ce{3 A + 2 B -> C}$$
Reagent B was added in excess and the following time concentration data was obtained.
\begin{array}{cr}
A\ [\mathrm{10^{-3}\ mol\cdot L^{-1}}] & \mathrm{time [s]} \\ \hline
5.0 & 10\\
2.9 & 50\\
1.9 & 80\\
1.1 & 120\\
0.7 & 150\\
\end{array}
(i) Derive the relevant integrated rate law and hence find the rate constant.
I am thinking that to solve this question you just sub the values into the integrate rate laws and see if it produces a linear relationship.
Is this correct? Also could you please suggest a faster method of solving this type of question.
Thank you.

You could plot the data to determine the order of the reaction.  I would use Microsoft Excel to do this.  Reproduce the data and then add columns that calculate the $\ln [A]$ and $\frac{1}{[A]}$.  Then plot $[A]$, $\ln [A]$ ,and $\frac{1}{[A]}$ vs. time (I would use separate plots).  The plot that gives the "best" linear fit (which you can assess using the $R^2$ value), corresponds to the appropriate integrated rate law.

If $[A]$ vs. time is linear, then the reaction is zero order.
If $\ln [A]$ vs. time is linear, then the reaction is first order.
If $\frac{1}{[A]}$ vs. time is linear, then the reaction is second
order.

Hope this helps.

In addition to Tami M.'s answer:
You might want to give Python, the IPython Notebook and matplotlib a try:

import numpy as np
import matplotlib.pyplot as plt

t = np.array([10, 50, 80, 120, 150])
a = np.array([5.0, 2.9, 1.9, 1.1, 0.7])

a_int = np.log(a)

plt.figure()
plt.subplot(121)
plt.plot(t, a, 'ro-')
plt.xlabel('t [s]')
plt.ylabel('A')

plt.subplot(122)
plt.plot(t, a_int, 'gs-')
plt.xlabel('t [s]')
plt.ylabel('ln(A)')

plt.show()

Note that the plots are not fine-tuned and that I omitted a further fitting ;-)

Least-squares regression theory can also provide estimates of the slope and intercept and their respective variance estimates. The computed slope and intercept terms are assumed to be unbiased and have a normal distribution for deviations from the true underlying value.
The regression model for the ordinary least-square case is:
Y = mX + b
with a design Matrix X:
| x1 1 |
| x2 2 |
| xn n |
where each point x is appropriately transformed depending on the presumed rate order.
The Variance-Covariance Matrix, which is central to assessing the precision of the rate order estimate, is created by taking the matrix inverse of [ matrix product of [transpose of Matrix X] times [Matrix X]]]. It is further scaled by an estimate of the variance of the regression model fit (computed from the differences between fitted and reported for each point which is further squared, summed for all points, and divided by (n-2). See Wikipedia discussion of statistical theory at https://en.wikipedia.org/wiki/Ordinary_least_squares.
Then, one could apply a hypothesis testing procedure (like the t-test) as to whether the slope is o or 1 or 2. However, this is not precisely correct, and one should examine other data or theory (like looking at the expected slow rate-determining step coefficients), and then proceed to selected just one value from (0, 1, 2) for testing.
However, in the case of getting a fractional value (as a regression model of the above data actually indicates), try a direct test for linearity (see, for example, a proposed idea of modifying the regression at http://essedunet.nsd.uib.no/cms/topics/regression/4/3.html ). This consists of introducing a quadratic term in time, and if the latter fails a t-test of being different from zero, the model is deemed to have passed a linearity test.


text:
How to understand the concept of moles?
I have trouble relating to the mole concept. For example in a question like the following:

How many moles of $\ce{K}$ are in 1 mole of $\ce{KOH}$?

How is it that in 1 mol of $\ce{KOH}$ there is 1 mol of $\ce{K}$ and x moles of $\ce{OH}$? How is it that 1 mol of $\ce{KOH}$ has 6.02×1023 atoms, and then $\ce{K}$ has 6.02×1023 atoms as well. In addition, $\ce{OH}$ will have 6.02×1023 atoms, but if we add these two it's greater than the original number of atoms in 1 mol of $\ce{KOH}$?

1 Mol of KOH has 36.0210^23 atoms, because there are three atoms in KOH.

Welcome to Chemistry.SE. Have a look at the
tour to get familiar with this site.
If necessary, mathematical expressions and equations can be formatted
using $\LaTeX$. Please note that there is a homework policy here.

hmm, ok so if I'm looking at say 1 mol of H2SO4, then the total atoms would be 36.0210^23 as well? or would it be *7 instead of 3? | why is it that people say 1 mol = 6.0210^23 if actually the KOH would contain 3 * as many total atoms in 1 mol?

I had the same problem as well, until I sketched out the actual molecules.
When you have one mole of $\ce{KOH}$, you have one mole of the collective "molecule".
If you were to break it down into $\ce{K}$ and $\ce{OH}$, you would get 1 mole of $\ce{K}$, and 1 mole of $\ce{OH}$. This is because you need 1 mole of each to make 1 mole of $\ce{KOH}$.
It's kind of like if you have one hundred keyboards, you would get one hundred 'Enter' keys, one hundred 'Up arrow key' and so on...

ah ok, an interesting analogy; so when someone says i have 1 mol of KOH, it's not actually 6.0210^23 total but 3avg constant, and each atom is 1 mol = 6.02*10^23 combining towards the total?

The concept of mole is just a number. When we say we have 1 mol of KOH, it means exactly that, there are 1 mole of KOH, not 1 mole of atoms. (100 keyboards, not 100 keys).

Ok, this is starting to make a bit more sense.
1 mol KOH = 1 mol K 1 mol OH (1 keyboard, 1 f key, 1 g key)
1 mol H2SO4 = 2 mol H 1 Mol SO4 (3 moles of ions)
(1 custom keyboard = 2 space bars, 1 escape key)
I could then take 2 mol of H and say there will be 2*(6.02*10^23) H atoms, etc. correct?
If so, thank you very much!

I think you've got it! For the H2SO4 example, you would have 2 mol of H, and 1 mol of the ion SO4, which you could count as 1 mol of S, and 4 mol of O. Just understand that the concept of mole is no different then the idea of "dozen". It is just a unit of measurement.

One mole of $\ce{KOH}$ means that there are Avogadro number of (6.023 × 10²³ molecules) $\ce{KOH}$ molecules and we need not consider the sum of number of constituent atoms.
For example,
1 mole of argon contains 6.023 × 10²³ argon atoms.
1 mole of ozone contains 6.023 × 10²³ ozone molecules (and also consists of 3 moles of oxygen atoms).

Fantastic. This is all coming together now and I have a general idea of it all; enough to stop questioning it so much.

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


text:
Concentration Inquiry
I just found a random example so I could highlight my inquiry:

A chemist wants to make .5L of a 20M solutionusing $\ce{NaOH}$. The solution will be made from solid $\ce{NaOH}$ and water. How many grams of
  $\ce{NaOH}$ are needed to make this solution?"

In this case, I am finding the moles of $\ce{NaOH}$ that would equate to the number of moles in 0.5L of a 20M solution. But, is it saying that (x grams of $\ce{NaOH}$ + x volume of $\ce{H2O}$) make up the total 0.5L ? or am I adding x grams of $\ce{NaOH}$ to .5L of $\ce{H2O}$ and not considering the changes in volume?

If the final volume of the solution is .5L then what is important here is how many moles/grams of the substance would be used to make the solution. In general when making a solution, the solution is raised to the final volume after the mixture. So whether .5L of $\ce{H2O}$ is used, or .4998L of $\ce{H2O}$ is used doesn't matter.

is it saying that (x grams of $\ce{NaOH}$ + x volume of $\ce{H2O}$) make up the total 0.5L

That would be the exact way!
Molarity is defined as $\mathrm{\frac{moles_{solute}}{\mathbf{total}\ volume_{solution}}}$


text:
HOMO and LUMOs of Carbon based Compounds
Why are the HOMO and LUMOs often $\pi$ orbitals in carbon based compounds that contain multiple double bonds or conjugation such as ethene and butadiene?

You can visit this page to find out how to make your future posts better.

Why are the HOMO and LUMOs often π  orbitals in carbon based compounds

The heat of hydrogenation of ethylene is -32.5 kcal/m

image source
In other words, ethylene is 32.5 kcal/m less stable (higher in energy) than ethane and this amount of energy (heat) is given off when we hydrogenate ethylene and convert it to ethane.
This tells us that pi bonds are, generally, less stable (higher in energy) than sigma bonds as shown in the following diagram (just forget about the delta bond orbitals, they don't play into this discussion).

image source
This diagram is a pictorial answer to your question.  Since pi bonds are generally higher in energy than sigma bonds, they will become the HOMO in most molecules.
Since bonding and antibonding orbitals are arranged symmetrically about the zero energy point, whatever type of orbital is the HOMO, then the antibonding counterpart will be the LUMO.  If a $\ce{\pi}$ orbital is the HOMO, than a $\pi^{*}%$ (pi star) orbital will be the LUMO.


text:
Etching a Screen Protector Film
I have to etch a pattern on a plastic film similar to the ones used to protect phone screens. I want a very thin etch, one that is only visible on close inspection. How can I get that done?

I don't know but I would guess laser "cutters" would probably be interesting to look at given the resolution you need.

If you don’t find a suitable chemical method, you may want to consider using an airbrush sandblaster. You can cover the area that you don’t want to be etched.

Assuming by thin, you mean a narrow line (not depth). It all depends on how thin you want to go. For a fine width line, a relatively cheap $\ce{CO2}$ laser cutter will get you down to 100–200 µm, but depending on the plastic and how thick it is, the heat produced may warp the plastic near the cut. I think most phone screen protectors are PET film. I've personally made ~200 µm features with one, and if the plastic is thin enough, you can keep the cut width pretty narrow. Things like pulsed UV lasers will give you much finer lines on plastics without thermal effects (as they tend to ablate rather than burn plastics), but they're much more expensive. If you don't have access to either, many companies will contract for this kind of work, even for small jobs.
Another option for coarser cuts that don't have too complex of shapes is a vinyl cutter—basically a plotter with a cutting blade instead of a pen. These are quite cheap, but can't cut all shapes and resolution is on the order of a hundred µm or so, though they don't cause distortions along the edges in most materials.
If you're really serious about it, it's certainly possible to coat photoresist on most plastics and make a mask for reactive ion etching, which will get you down to 1 or 2 µm easily, but it sounds like overkill for you.


text:
How to remove a stuck stopper from a flask?
How do I remove a stopper that has been stuck in the neck of a 100 mL round bottom flask since I forgot to wipe the neck of the flask. A salt was formed in the 100 mL flask and the stopper is just stuck like cement. Is there anyway to remove it without breaking the flask, breaking the stopper, or heating the flask (since I don't want to interfere with the product)?

Is it a hollow glas stopper?
Sometimes, it helps to hold the flask at the body and hit the corner of the lab bench with the stopper. If that doesn't help, heat the neck of the flask carefully with a heat gun.

No, it was a solid glass stopper. I tried the hitting method and it worked like a miracle.

@86BCP2432T I'm glad I could help! Btw, when silicone grease is not an option - and sometimes it isn't - teflon sleeves are a good alternative.

While @KlausWarzecha provided the working solution, I offer up my own adopted mantra for future reference.  "Most human problems can be solved by an appropriate charge of high explosives."

When I was younger I worked at a place which had a serious accident which could have been fatal. Someone tried to open a flask containing pet ether using a flame to heat the stuck ground glass joint holding the stopper. I tend to use the hitting method on a wooden surface, use many gentle taps and keep on turning the flask around. I never use a modern plastic type lab bench. This method deals with most stuck joints on flasks and bottles. Never use a flame on glassware to free a joint until you have removed all flammable solvent from the glassware.

Here is an alternative that worked like a charm for me.  I have a wide-mouth lab jar with a thick, ground glass stopper that stuck tight.  None of the other methods worked; oil, heat, tapping, etc. I had an aerosol spray used to blow dust from computer keyboards/screens (it is not compressed air but a liquid that becomes a gas as it is released - i.e. the can gets very cold as you use it). A quick shot of cold gas directly on the stopper and it immediately shrank enough to easily come out of the jar.

This is another reason not to use glass stoppers with separatory funnels. If one isn''t careful, pressure (from ether, CO2) can build up and the funnel may break.  With a neoprene stopper one can feel the pressure as the neoprene stopper moves slightly.

Another easy method is to get 2 other glass stoppers and tap them on both sides of the neck of the flask firmly, sometimes this will free it up immediately, other times you may need to have someone pulling the stopper while you hit the stoppers on the sides

I like the pulling while tapping. Requires more hands than the average lab worker has, tho.

Had great success with the upside down can of compressed air! I had 2 small (10 mL) volumetric flasks with stuck PTFE stoppers. I tried warming the neck slightly but that didn't help. With the can of air, I placed the flask on the counter, blasted the cap just above the flask neck for a couple seconds, the tried opening and both came open right away! Awesome!

+1 Nice to have a confirmation of a new technique.


text:
Would it be possible to create a (non-toxic) alcoholic drink that, when zapped with laser, changes color?
Would it be possible to create a (non-toxic) alcoholic drink that, when zapped with a 0.5mW laser pointer, changes color?
This would be mainly for the visual effects (bartender show) so it should not require aiming the laser at the drink for more than a few seconds. The reason for choosing a 0.5mW laser is to minimize the need for eye protection.
It would not matter what color laser is used, but would this be possible?
If it is possible, what kind of chemical(s) would be needed in the drink?

It would probably depend on the color of the pointer

thanks @Mithoron, I have added a note about that to the question.

Is it correct that you are looking for a permanent colour change?

Let me know when you find it - looks like one hell of a party trick!

I see a couple of problems here:
Beam diameter vs volume of the beverage
The volume of the sample that is actually hit by a narrow laser beam is very small, as compared to the total volume of the sample (= the drink). As a consequence, only a very small amount of the dye in drink will undergo a transformation. A wide beam of a switchable light source definitely is the better option.
Is is possible in general?
Yes, a number of compound classes undergo transformations upon iradiation and change their colour (spiropyranes, fulgides, etc.). Often, the reverse reaction can be induce thermally.
So what?
Food dyes are strictly regulated, you can't add a photochromic dye to a drink just because the colour change looks nice in the lab. If the dye isn't approved as a food dye, there's no legal way! To my knowledge, food dyes typically do not show photochromism but are photostable. Consumers usually don't like that their food changes the colour in the light.
In summary, the idea sounds funny but I rather doubt that it can be realized.
The only option to add a dramatic effect would be the use of fluorescent and food-compatible ingedients. Think in quinine (tonic water).

Do you think it might be possible to make use of a photoacid and an indicator? Only small amounts of both would be needed to change the colour of the entire mixture, if the indicator flips around neutral pH. The problem I see would be finding a photoacid safe for consumption, having to use high frequency photons, low photon intensity and the low photon absorption cross-section in a dilute mixture of the photoacid.

@NicolauSakerNeto Photoacids are a nice idea too. But as you mentioned yourself, "safe for consumption" (= approval by legal autorities) is the key.

Klaus has provided an excellent answer.  Basically your desired approach is possible, but fraught with difficulty.
Perhaps there is another way to attack the problem.  Still, I suspect considerable testing will be required.
For reasons mentioned by Klaus, it is necessary to use a food dye that is already approved.  Many food dyes are photobleachable.  For example, azo dyes exist as cis-trans isomer pairs.  Typically, the trans isomer is more stable and provides the dye color.

Often (not always, this is why you'll need to test), when the trans-dye is irradiated with laser light it will convert to the colorless (or differently colored) cis isomer. This is termed "photobleaching" - light has "bleached" the color out of the dye. Generally the cis isomer remains stable as long as you keep it cold. The same concept can be applied to non-azo dyes that can exist as isomeric pairs.

image source
Buy a number of food dyes that provide the color you desire and can exist as isomers (cis-trans or otherwise).  Dissolve each one separately (until you find one that works) in the minimum amount of one of the ingredients in your "Blue Light Special".  Irradiate the (presumably colored) solution with a general lab laser (maybe you have a chemist friend who can do this for you) -  a laser pointer probably won't have enough power.  If the experiment is successful, the color should disappear from the solution.  Keep the solution cold in your bar fridge.
When someone orders the "Blue Light Special" mix all of the other ingredients in a glass at room temperature.  Add some of the cold, colorless (photobleached) dye solution to the room temperature mix.  As the cold dye warms up it should convert back to its colored form and the drink should become colored right before the customer's eyes.

Whatever name OP'd picked out for the drink, yours is better. :-)


text:
What is the most accepted explanation of aurophilicity?
Gold has a special property called aurophilicity. Inorganic chemistry by Huheey has some explanation, but that was not enough to clear the conception.

Have a look at The aurophilicity phenomenon: A decade of experimental findings, theoretical concepts and emerging applications by Hubert Schmidbaur for some further directions and references.


text:
What is the reasoning behind the increasing pKa for organic compounds?

Arrange the following according to increasing $\mathrm{p}K_a$.

Cyclohexanol
Phenol
Cyclohexane carboxylic acid

Please tell me the logic along with the solution.
I am confused between phenol and cyclohexanol because i guess phenol is resonance stabilized-the phenoxide ion

No its not a homework question.Chemistry is my favourite subject.I am just practising organic chemistry myself before the new session begins and i got stuck with this question.

Even it is not homework for you, the policy for these kind of questions remains and it is very likely that this gets closed. Please read through the links I posted earlier, more detail is given there. For starters, why don't you look up the $\mathrm{p}K_a$ values and include them in the post, so that people willing to help you won't have to do it. Here is a table that might help.

Cyclohexanol is an alcohol. An alcohol contains an $\ce{OH}$ group connected to a tetrahedral carbon.
Phenol is slightly different; it contains an $\ce{OH}$ group connected to a trigonal planar carbon that is part of an aromatic ring.
cyclohexanecarboxylic acid contains a very different functional group, carboxylic acid.
Cyclohexanol has a $\mathrm{p}K_a$ of about 18. It is less acidic than water. It can give up a proton, but the proton is much more likely to be bound to the oxygen than disscociated. Phenol has a $\mathrm{p}K_a$ of about 9 and is considered mildly acidic. cyclohexanecarboxylic acid has a $\mathrm{p}K_a$ of about 4.9; its acidity is similar to hydrofluoric acid.
Let's compare stability of the conjugate bases of cyclohexanol, phenol and cyclohexanecarboxylic acid:

In cyclohexanol, the anion formed by loss of a proton is localized on the oxygen. There is no resonance stabilization.
In phenol, the anion formed by loss of a proton is delocalized. Resonance structures show the negative charge can be shared between the oxygen atom and three of the carbons in the benzene ring.
In cyclohexanecarboxylic  acid, the anion is also delocalized. This time the negative charge is shared between two different oxygen atoms.

Resonance delocalization plays a clear role in stabilizing the conjugate base formed after loss of a proton. Despite the similar bond polarity in cyclohexanol and phenol, the proton is much more tightly bound in cyclohexanol. The greater stability of the phenolate anion compared to the cyclohexanoxide anion makes it easier to remove a proton from phenol than cyclohexanol.
The atoms onto which the charge is delocalized still plays a major role in determining stability. Delocalization onto an additional oxygen atom may be more stabilizing than delocalization onto several carbons.
For more details, please visit: http://employees.csbsju.edu/cschaller/principles%20chem/acidity/acid%20distal.htm


text:
why colour of CuSO4 and CuCl3 is different?
These two are simple ionic compounds , not complexes [if I do not consider water of hydration] then how can I rationalise the difference in colour?

Cu3+ is still a different color than Cu2+, isn't it?

@Lighthart Honestly, I've never seen copper(III) in solution. Have you? It's not that copper(III) wouldn't exist, such as in ores, but as a naked, solvated cation?

Do you really mean $\ce{CuCl3}$ or is it just a typo and you thought in $\ce{CuCl2}$? On a second note, copper(II)sulfate itself undergoes a colour change when converting the pentahydrate to the anhydrous form.

LIGAND FIELD THEORY.  Colors in these ions are based on electron transitions between non-degenerate d-orbitals. The energy of these transitions (thus the color) is a function of the metal oxidation state, the type of ligand/counterion, and the molecular geometry. Hydration or not d-orbitals are still going to distort based on those factors.


text:
Hemoglobin equilibrium - effect of increased carbon dioxide and lactic acid
The binding of oxygen by haemoglobin giving oxy-haemoglobin is partially regulated by the concentration of $\ce{H3O+}$ and dissolved in $\ce{CO2}$ in blood
$$\ce{HbO2 + H3O+ +CO2<=>H+.Hb.CO2 + O2 + H2O}$$
If there is production of lactic acid and $\ce{CO2}$ in a muscle during excretion then is more oxygen released or is more $\ce{HbO2}$ formed?

Is this homework? What research have you done? Where exactly are you stuck?

See im a bit confused with this...i wanted to know if my guess is correct...is it oxygen because if we add more CO2 equilibrium should shift to right..i just wanted someone to verify...no its not my homework...just practising problems..


text:
How to rationalise the basicity order of phenylenediamine?
Why is the basicity of para- > meta- > aniline > ortho-
for phenylenediamine?
($\mathrm pK_\mathrm a$: ortho: 9.53, meta: 9.12, para: 7.96) I tried deducing it using resonance and inductive effects but got stuck.

related http://chemistry.stackexchange.com/questions/7683/ortho-effect-in-substituted-aromatic-acids-and-bases?rq=1

I am not being able to incorporate the ortho effect concept!Let me know please.

Isn't the basicity trend listed backwards? The para-analog has the lowest pKa for the conjugate acid (strongest acid), which means that the base is the weakest?

in phenylene-diamine shouldn't the basicity order be p>aniline>o>m since -NH2 is not much of a bulky group and thus doesn't exhibit ortho effect. SO with ortho effect out of the picture -NH2 is very strong +M group and effect of +M>-I (for -NH2). In ortho due to hydrogen bonding its basicity is reduced. Meta position only -I. so by that reasoning para is most basic followed by aniline followed by ortho and then meta....So p>aniline>o>m....Am i going wrong anywhere?

As Mithoron pointed out in a comment, we had a similar question convering the cases of 2-methylaniline (o-toluidine) and 2-methylbenzoic acid (o-toluic acid) here.
In the case of 1,2-diaminobenzene (o-phenylenediamine), it is conceivable that hydrogen bonding between the two amino groups reduces the electron density on one of the nitrogen atoms, thereby leading to a slight increase on the other. As a result, o-phenylenediamine is significantly less basic than the m- and p-isomer, but a tad more basic than aniline.

Klaus's answer is nice, but is omits one important detail:
The hydrogen bonding would necessarily require that the lone pair of electrons go 'out of conjugation' with the ring system, and hence there can be no donation of electron density via pi-effects as in the para-phenylenediamine case.


text:
Difference between continuously overlapping p orbitals and conjugated pi bonds
Is there any difference between the phrase "continuously overlapping p-orbitals" and "conjugated pi bonds" when referring to aromatic compounds? I've heard both used and I wasn't sure if they mean the same thing.

Strictly speaking p-orbitals can overlap to form a sigma-bond if the touch end-to-end

Interesting question!
To have continuously overlapping p-orbitals, the p-orbitals must be arranged in a cyclic loop.  The p-orbitals in benzene are continuously overlapping.  The p-orbitals in benzene are also conjugated pi bonds.

image source
In buta-1,3-diene the p-orbitals form conjugated pi bonds, however, they are not continuously overlapping - there are two "end" carbons that are not connected with each other, there is no continuous loop.

image source
Let's take it a step further:
Continuously overlapping p-orbitals are not always aromatic.
Example 1: Cyclobuta-1,3-diene (the square planar structure, $\ce{D_{4h}}$) has 4 continuously overlapping p-orbitals but it is not aromatic, it is antiaromatic.  For a continuously overlapping system of p-orbitals to be aromatic, it must contain 4n+2 pi electrons.

image source
Example 2: Have you heard of the Möbius strip? The mathematical (topological) concept has application in chemistry.  Imagine a large cyclic loop of p-orbitals where each p-orbital is twisted a few degrees from the preceeding p-orbital.  If the loop is large enough, then by the time we come back to the first p-orbital we have twisted the last p-orbital by 180°; there is a phase inversion (node) between the first and last p-orbital.  In a Möbius system the p-orbitals are continuous, but not continuously overlapping!  Möbius systems are aromatic if they contain 4n pi electrons.

image source
Continuous pi systems with an even number (0, 2, 4...) of nodes (cyclobutadiene, benzene, etc) are called Hückel systems in order to differentiate then from Möbius systems where there are an odd number of nodes (1, 3, 5...).
Lest you think the Möbius concept is purely theoretical take a look at these two molecules

image source
Isn't this what makes chemistry fascinating!


text:
Acidity order of acetone, methyl acetate and N,N-dimethylacetamide

Why is the acidity order true?
  $$\ce{CH3COCH3} > \ce{CH3CO2CH3} > \ce{CH3CON(CH3)2}$$

I deduced that in methyl acetate and N,N-dimethylacetamide there is a −I effect of the methoxy group, $\ce{N(CH3)2}$  and a +R effect, too so the attached keto-carbon develops a positive charge and the conjugate base becomes unstable due to the +R effect of methoxy and $\ce{N(CH3)2}$ groups.
Is my logic right?

Which protons are we considering the acidity of?

@bon Although OP should clarify, I think its the hydrogen atoms attached to the carbon atoms. Those are the only ones common in the 3 compounds.

The functional group in the second compound is an ester. I don't think its right to view it as a methoxy group attached to a keto-carbon.

The increased acidity of the protons on the alpha-carbon in these compounds, compared to the acidity of the protons in methane, is due to stabilization of the resultant negative charge through resonance with the adjacent carbonyl.  In the top line of the following figure we show a ketone where the proton has been removed.  We can draw 2 resonance structures to describe this carbanion.  In structure II the negative charge has been delocalized onto the electronegative oxygen atom - a very favorable and stabilizing situation.

To whatever degree we remove the carbonyl's ability to resonance stabilize that adjacent charge we will decrease the acidity of the alpha-protons.
In the case of both the ester (III) and amide (VI) we can draw a third resonance structure (V and VIII respectively).  To whatever extent these resonance structures contribute to the overall description of the ester and the amide, the carbonyl will be less available to stabilize the negative charge through structures IV and VII.
It turns out that resonance structures V and VIII are significant contributors, with the amide structure VIII having a positive charge on nitrogen being more important than resonance structure V which places the positive charge on the more electronegative oxygen (less favorable).  In fact, resonance structure VIII is such a significant contributor that restricted rotation occurs about the $\ce{C-N}$ bond and we can typically observe two different R group signals in the proton-nmr spectrum of amides.
So as stated earlier, as the contribution of the third resonance structure increases, the contribution of the second structure (the structure that stabilizes our carbanion) will decrease making our compound less acidic.  The ketone doesn't have a third structure so it will be the most acidic structure. Structure VIII is more important than structure V, so the ester will be more acidic than the amide.  Therefore, the overall order of decreasing acidity is
$$\ce{CH3COCH3 > CH3CO2CH3 > CH3CON(CH3)2}$$
just as you noted.


text:
Does tetramine copper dissolve cellulose?
I saw a Youtube video which showed that cotton (cellulose) dissolves in deep blue solution of tetramine copper made by adding 20% ammonia in copper carbonate salt. I tried to replicate the process using $\ce{CuSO4}$ and copper chloride. None worked. Where have I gone wrong?

As @iad22agp states, it is not sufficiently alkaline. Schweizer's reagent is tetraamminediaquacopper dihydroxide, $\ce{[Cu(NH3)4(H2O)2](OH)2}$. See Schweizer's reagent.
That is not Doktor Schweizer, I presume.

You might want to check the pH - you would have a less basic solution using copper sulfate or chloride than you would get from copper carbonate, and that could be less effective in dissolving the cotton. Also, your particular sample of cotton might behave differently from others, such as that used in the video.

There should be no anions exept OH- present.

made by adding 20% ammonia in copper carbonate salt.

is wrong, using CuSO4 is even wronger.
You have to precipitate Cu(OH)2 and wash the
precipitate thoroughly, a very tedious operation.
According to a IG Farbenindustrie Patent You
throw copper pieces in  > 25 % Ammonia
(cut from electrical wire) and percolate
with oxygen or air. When using air, you have to
maintain temperature under 8 °C. This needs days
until "ready" but has the advantage that the
reagent is pure without cleaning operations.

Is there any reason this is formatted like a cinquain?


text:
what are possible isomers of (meso-stilbenediamine)(isobutylenediamine) palladium (II)?
It can form a square planar and a tetrahedral complex but within square planar or within tetrahedral can there be isomers possible?

The use of the term "meso" in the compound name indicates that both phenyl groups are on the same side of the molecule, "cis" to one another, not "trans".
Here are the tetrahedral and square planar representations.

Note: I used $\ce{Pt}$ instead of $\ce{Pd}$ as the central metal, but it makes no difference in terms of isomers.
The tetrahedral molecule has a plane of symmetry containing the "top" 5-membered ring, bisecting the 2 methyl groups on that ring and then bisecting the "bottom" 5-membered ring with the 2 "meso" phenyl groups. Since the tetrahedral molecule contains a plane of symmetry it is achiral and has no isomers.
In the square planar form, there is no symmetry element.  Therefore the square planar molecule is chiral and exists as an enantiomeric pair of isomers.


text:
Why member ring is favoured in chelates?
why chelates which form 3 or 4 member rings or 6 or 7 member rings are as stable as 5 member ring chelates?
for example if we measure stability of chelates with oxygen donors, say oxalate, malonate, succinate etc oxalate chelated with metal shoes more stability. why malonate , succinate as chelating ligands are less stable?

Chelating increases the stability , probably due to more change in positive entropy

In its general form, your statement on the stability of these complexes is wrong!
$\ce{Be(II)}$-malonato complexes are more stable than the corresponding oxalato complexes, cf. Preparative, potentiometric and NMR studies of the interaction of beryllium(II) with oxalate and malonate.

no sir i was not trying to say it was universal, i saw in books that said 5 member ring forming chelates are relatively more stable than 6 member or 7 member or 4 member ring forming chelates.


text:
What is the explanation of the changes in stability going down a group for carbonates, bicarbonates, fluorides, and chlorides?
For carbonates and bicarbonates, I know that stability increases down the group, and for chlorides and fluorides, stability decreases down the group. Why does this happen?  Can someone explain this in detail?
(I am talking about S block alkali metals)

What do you mean by "stability"?

Well how should i explain :-P!I mean less reactive :-)!

Reactive with what?  I'm not trying to be difficult; the terms 'stable' and 'reactive' encompass a lot of different areas & answering your question well depends on exactly what you're referring to.  Can you provide more context to your question?

Ok this is a doubt that occured due to my textbook and here is the direct quotation :  "As the electropositive character
increases down the group, the stability of the
carbonates and hydorgencarbonates increases."

http://www.ncert.nic.in/NCERTS/l/kech203.pdf

Now i guess you'll get the context ! Search for "stabilty" in the pdf

Indeed, that definitely helps.  I may not be able to give a satisfactory answer; this link should assist others in answering, though.

If there is relevant information in the pdf, please include it in the question. Responders shouldn't have to search for it.

I already quoted necessary lines to explain the concept.

"...and for chlorides and fluorides, stability decreases down the group" : This is incorrect. Down the group, thermal stability decreases for fluorides, but increases for chlorides, fluorides and bromides.

Carbonates
The quote from your text:

Carbonates of alkaline earth
  metals are insoluble in water and can be
  precipitated by addition of a sodium or
  ammonium carbonate solution to a solution
  of a soluble salt of these metals. The solubility
  of carbonates in water decreases as the atomic
  number of the metal ion increases. All the
  carbonates decompose on heating to give
  carbon dioxide and the oxide. Beryllium
  carbonate is unstable and can be kept only in
  the atmosphere of CO2. The thermal stability
  increases with increasing cationic size.

So the stability that you are referring to is thermal stability. This is an important detail. So what is thermal stability? It's how resistant a molecule is to decomposition at higher temperatures.
What's happening to cause thermal instability?
So, lets look at the carbonate ion here:

This is just an illustration, and in reality the negative charge we see on the two $\ce{O}$ atoms is localized due to resonance.
Below the illustration shows where the negative charge is likely to be concentrated (colored in red).

So, when we create a carbonate complex like the example below, the negative charge will be attracted to the positive ion.

Because of this polarization, the carbon dioxide will become more stable and energetically favorable.
How does going down a group play into this?
Well as you go down the group, the charged ion becomes larger. The larger the ion, we see a lower charge density. Charge density is basically the amount of charge in a given volume. So, if a small ion has the same charge as a larger ion, the charge density will be greater for that small ion. Greater charge density, means a greater pull on that carbonate ion, and a greater pull causes the delocalized ions, and a more stable $\ce{CO2}$ molecule. So, the larger the ion, the lower the charge density, the less polarizing of an effect, and reduced stability of a $\ce{CO2}$ molecule, favoring the $\ce{CO3}$.
Chloride and fluoride stability
Stability of fluorides, chlorides, and other halogens, are likewise related to thier size. The halogens, specifically fluouride, is known for their electronegativity. Electronegativity, is the tendency to attract electrons to itself. As you move up the group, you see an increase in electronegtivity. This results in the creation of polar bonds.
Illustrated below, you see that as charge of the positive ions increase, polarizability increases (left), and as the halogen ion increases, polarizability and electronegativity decrease (right). When the ions electron cloud, is less polarized, the bond is less strong, leading to a less stable molecule.

Information and illustrations on carbonate ions were sourced from here.

As per this question and answer, it seems, the stability of alkali metal fluorides decreases down the group whereas it increases for alkali metal chlorides, bromides and iodides. In short the trend of enthalpy of formation and hence stability is reversed when comparing fluorides with rest of the halides.

"...as the halogen ion increases, polarizability and electronegativity decrease (right)" : This is incorrect. Polarizability of halides increases down the group due to increase in size and no. of electrons.

Also your answer treats fluorides and chlorides in the same way, in spite of the fact that down the group, thermal stability decreases for fluorides, but increases for chlorides, fluorides and bromides.

By Fajan's Rule you should be getting the answer and then more electropositive metal will have more ionic character and then that will increase stability.

Welcome to chemistry.SE! Could you please be a little more elaborate? Your answer might sound comment-like to some people, and I don't think it will solve the OP's problem, really.


text:
Can hydride shift and methyl shift happen one after another in a carbocation?
Can a hydride shift be followed by a methyl shift for stabilizing a carbocation?
In case both are possible, which will occur first?

Are you looking at a specific compound? In principle, the answer is yes.

Yes it is possible. But it must fulfil some basic condition for that shift.

The empty p-orbital of the carbocation must be in plane with the $\ce{C-H}$ or the $\ce{C-C}$ bond. When the alignment of a carbon–hydrogen bond with a vacant p orbital takes place allowing for hyperconjugation, a "pseudo-double-bond" develops. As illustrated in Figure 5.13, this can be envisioned as a double bond with a closely associated hydrogen ion.
If, as shown in Figure 5.13, hyperconjugation results in the formation of species possessing both double-bond character and associated hydrogen ions, equilibrium-controlled migration of the associated hydrogen ion can be expected. This transformation, shown in Scheme 5.8, is known as a 1,2-hydride shift and results in the migration of a proton
  from carbon 1 to carbon 2.
While the example illustrated in Scheme 5.8 shows equilibrium between two chemically identical carbocations, there are factors influencing the direction of these transformations when applied to more complex systems. If we consider Scheme 5.9, we notice that the positive charge migrates exclusively to the tertiary center, reflecting the increased stability of tertiary carbocations over primary carbocations. In general, where 1,2-hydride shifts are possible, rearrangement of less stable carbocations to more stable carbocations is expected.
It is important to remember that hydride shifts occur much more readily than the corresponding alkyl shifts. In fact,as a general rule, alkyl shifts will not occur unless a hydride shift cannot take place.
Alkyl migrations occur when the resulting carbocation is more stable than the starting carbocation. But for hydride shift its not such a hard and fast rule.

Source: "Arrow Pushing in Organic Chemistry, An Easy Approach to Understanding Reaction Mechanisms" by Daniel E. Levy; Copyright 2008 by John Wiley & Sons, Inc.; page 110 in chapter 6

I downvoted this answer because the quoted part does not even once mention the possibility/occurrence of multiple shifts.

Multiple rearrangements can happen if the resulting species is stabilized by it. In general hydride shift is more favourable over methyl shift because of its smaller size.
The example proposed by James was not known to me but its truly an amazing example.

James:
  It’s possible for multiple hydride/alkyl shifts to occur. One amazing example is in the biosynthesis of lanosterol.
Multiple shifts are certainly possible, and they could happen, but generally will only happen if each shift generates a successively more stable carbocation.
  For example, you probably wouldn’t see a shift if it involved turning a tertiary carbocation into a secondary carbocation.

Source
From Wikipedia:


text:
Aromaticity-Bromonium Ion

Is the bridged ion intermediate aromatic or not?
What is the way to quickly identify aromaticity?

A chemist would say that the reaction proceeds through an "aromatic transition state". The bridged, chloronium ion transition state drawn in your question involves 2 electrons.  No electrons from the electrophile were used, just 2 electrons from the pi-bond that it added to (draw the resonance structures for the bridged intermediate, each has 2 electrons in the $\ce{C-Cl}$ bond and no electrons in the carbocation p-orbital).  If a transition state contains 4n+2 electrons, then it is said to be aromatic. This case has n=0.
Here is another example. The top line in the figure below shows a suprafacial 1,5 hydrogen shift.  6 electrons are involved here, so the transition state is also referred to as aromatic (4n+2, n=1).

image source (look far down on the page)
Like the bridged halonium ion, this reaction is also very common because it too has an energetically favored (aromatic) transition state.
The second line in the figure depicts a suprafacial 1,3 hydrogen shift.  4 electrons are involved here, so the transition state does not fit the 4n+2 rule.  The transition state is said to be anti-aromatic.  Because it is disfavored energetically (anti-aromatic), the suprafacial 1,3 hydrogen shift does not occur (at least not via a low-energy concerted pathway).
Extra Credit: I don't want to jump to far ahead, but let me briefly say, that what I've written above applies to Hückel transition states, opposite rules apply to Möbius transition states.  See this earlier answer, and references therein, for an explanation of the Hückel and Möbius concepts.

Hmmm... Then what is the mechnanism by which the bromination of acetylene takes place?

@TanYongBoon  In the OP's question it is stated that the mechanisms for bromination and chlorination are similar.

I really hope I am counting that correctly; the bridged-ion transition state appears to have 4 pi electrons. It is driving me a bit mad right now. And that would contradict this whole answer, which is puzzling me even more.

Taken from Advanced Organic Chemistry: Part A: Structure and Mechanisms
By Francis A. Carey
The bridged ion may also be considered to be antiaromatic. Thus, while bromonium ions are relatively accessible intermediates for alkenes, the activation energy to reach the corresponding species is much higher for acetylenes. The bromonium ion remains the most accessible intermediate in many cases, but it is energetically relatively less favorable than for alkenes.

Your answer seems to contradict Ron's answer. Could you show how the bridged ion is in fact antiaromatic?

Read it: https://www.masterorganicchemistry.com/2017/03/27/antiaromaticity/

the concept is too simple !!

I fully understand antiaromaticity. So I suppose this is a 4 pi electron system with 2 4p electrons from Br and 2 pi electrons from the pi bond?


text:
Why is chloroethene said to be a conjugated system?
Why is chloroethene said to be a conjugated system?Is it because of p orbitals present in chlorine or something?

The conjugation is weak though because the Chlorine lone pair is in a 3p orbital but the pi bond is consists of two 2p orbitals

Conjugation can be between pi-bond and another pi-bond, lone pair of electrons and a pi-bond(as in your case) and pi-bond and a vacant orbital(p or d).

Chloroethene is said to be a conjugated system because the lone pair on the chlorine atom in in conjugation with the double bond between the 2 carbon atoms.
See this image for the resonating structures.

Here also: is called resonance and resonance structure. It has nothing to do with resonating.

Chloro ethane is a conjugated system because of the +M effect of chlorine with the double bond of ethene. Chlorine has a tendency to resonate with a double bond by donating its lone pair. What this means is that the lone pair of chlorine is in an unhybridised atomic p orbital in the same plane as the p orbital of the adjacent double bonded carbon atom, due to which, the lone pair "jumps" from the p orbital of chlorine to the adjacent carbon (forming a double bond) and displaces the original double bond to the next carbon.

resonates to

This kind of explanation simplifies the concept beyond the useful. The concept you are trying to explain is resonance, which has nothing to do with resonating. Equally bad is using jump for electrons, also when in quotation, because this simply isn't happening. There is no displacing. Resonance is a superposition of contributors.

I am sorry for the use of non technical terms, i am a little new to chemistry


text:
Why macromolecule ligands such as Crown ethers form more stable complexes than open chain ligands?
This was in J. Huheey's book and no explanation was given

The reason for crown ethers being more stable than their equally long open chain derivatives is the difference in Gibbs Free energy. (If you are not used to these terms, the Wikipedia article will help I think, though it is quite, long and thorough.)
The main point to take from the use of the Gibbs free energy in this case is this: it is made up of two parts, enthalpy ($\Delta{}H$) and entropy ($\Delta{}S$).
$$
\Delta{}G = \Delta{}H - T \cdot{} \Delta{}S
$$
If only the binding affinity of the (crown) ether chain is considered, only the enthalpy is relevant. The strength of the individual ether moieties is as strong in a crown as it is in a chain. The difference then, lies with the entropy. To have a crown ether form a structure around a ligand 'costs some entropy'. The same entropy difference should be considered for the chain ether, with added to that the cost of pulling the ends of the chain together! That difference in entropy results in the difference in $\Delta{}G$ and therefore yields a more stable complex. (The ends of the crown ether are also pulled together, this cost of entropy however, has been paid in an earlier reaction when making the crown ether, and therefore has no influence here.)
You can see this also from the relation between the Gibbs free energy and the equilibrium constant of the complex (In the imaginary reaction $\ce{A + B -> AB})$:
$$
\Delta{}G = - RT \cdot{} Ln (K)
$$
where,
$$
K = \frac{[\text{AB}]}{[\text{A}] \cdot{} [\text{B}]}
$$
A higher $\Delta{}G$ leads to a higher equilibrium constant.
Hope this helped, and I'd be happy to clarify anything further.

authors: Tʌŋɱʌƴ Jʌŋʛɩʀ


text:
Hydro-Bromination Product 1,4-Benzoquinone

What will the products on addition reaction of 1,4-benzoquinone with HBr ?

This is the mechanistic approach :

The Dissociation of HBr into H+ and Br- is implicit!

Thus final product is ortho bromo, 1,4 diphenol

Why don't you add the info in your comments to your question? And, do you have a better image, please? This one hurts eyes.

I agree, the picture is barely readable, your comments also belong into the post. Just [edit] it. The post was flagged low-quality because of these issues. For now I am not voting to delete, but these kinds of posts should not be the standard.

I apologize, the last twonsteps are tautomerism to give enol form which is aromatic, hence stable!


text:
Effect of dilution on pH for HCl
Is it necessary to consider common ion effect when suppose a strong acid like HCl is added to water?
For example in this question: 10 mL of $10^{-6 } \ce{mol/L \ HCl}$ solution is mixed with 90 mL of $\ce{H2O}$. pH will change approximately?
Normally by using $SV=constant$ I get pH will change by 1 unit (without taking common ion effect).
How will common ion effect affect the answer? Help.

Yes, you will definitely need to account for common ion effect.
However, use Kw to compute the pH. Assume that there is a complete dissociation by $\ce{HCl}$. However, you would need to solve the according quadratic for Kw.
as such, $\ce{H+= concentration~~ H+ of~~ HCl + x ~~and ~~OH- = x}$.
You should get a result around 6.79=pH.


text:
How to prevent losing solvent into the vacuum pump in rotovap solvent collection?
This is a practical question about the use of rotary evaporators ("rotovap"). So I'm asking for answers from people who actually use the things.
A rotovap typically uses a cold condenser to condense the solvent into a collection flask. But the distillation is typically done under weak vacuum as well. So the solvent sitting in the collection flask seem susceptible to re-vaporizing and getting sucked into the vacuum pump. Now I have a cold trap in front of my pump that should be cold enough (-50C) to trap most solvents under weak vacuum.
I think the answer is that if the vacuum is weak enough to vaporize the solvent under the hot water bath, but not under the cooler ambient air temperature, then the loss to the pump should be minimal. Indeed, this is what the instructions for the rotovap suggest.

Question: In practice, what are the risks to losing solvent into the pump?

We always kept the collection flask cooled with an ice bath; with volatile solvents we used a dry ice/acetone bath. The ice bath was just so easy to do, so why not.

Honestly, I never bothered to actually measure the volume of the condensed solvent in order to compare it with the original volume.
The concept described in the question, that is using a cooling trap with dry ice + acetone, is good practise and typically good enough to retain the evaporated solvent and protect the membrane pump. In rare cases, I have used a second, flat Dewar filled with dry ice + acetone on a "Laborboy" to cool the condensate-collecting flask.

Thanks--this confirms that having a trap in front of the pump, in addition to the condenser cooling, is probably a good idea, along with cooling of the receiver per ron's comment. Perhaps not vital, but why not, as ron says?

@trb456 ron's suggestion is actually a good one! The only "argument" against it is laziness. It requires an additional movement (adjusting the height of the cooling bath for the receiver) when you evaporate from flasks of different size ;-)

The idea of cooling the collection flask sounds good, but is actually useless. Makes no difference to the partial pressure of the solvent at the cooler, unless you let the rotary evaporator run for an extended time after the flask in the hot water bath is already dry.

In my experience, no further cooling other than the condensor above the collection flask is necessary. Remember that you are selecting a pressure which is low enough to boil the entire liquid at $40~\mathrm{^\circ C}$ (your flask’s temperature) but high enough so that it will recondense on cold water (I’m going to guess $15~\mathrm{^\circ C}$). Dripping down into the collection flask, the solvent will retain that temperature or at maximum reach room temperature again which should be just below a temperature for sufficient evaporation. If it does evaporate, it will be caught by the condensor.
Any solvent ‘pulled through the pump’ should be caught in a collection flask below a condensor on the high-pressure side of the pump. In $99~\%$ of my evaporations, nothing is there and the remaining percent are special cases.

Your cooling water should have around 10°C, and as long as you're not impatient and use a lower vaccum as recommended for your specific solvent, you should not loose much of it to the pump. Using an additional cold trap in front of it is not usually necessary. The manuals don't recommend or require it either.
Unless of course you (mis)use the roti to dry a product from solvent residuals. ;-)
Btw. The common membrane pumps are quite resistant to small amounts of solvent. What breaks first (if) are usually the seals, which are easy to replace. Don't soak it in chloroform, of course, and let it run dry if it got some. Put on empty flasks, leave the manual ventilation valve slightly open, and let it pump for ten minutes or so.


text:
How hydrate isomerism results difference in colours?
The standard example is $\ce{CrCl3.6H2O}$, which can have three distinctly different
crystalline compounds now known as
$\ce{[Cr(H2O)6]Cl3}$ (violet),
$\ce{[CrCl(H2O)5]Cl .H2O}$
(blue-green), and
$\ce{[CrCl2(H2O)4]Cl.2H2O}$ (dark green).
How does this difference of ligands in octahedral system affect the colour of compound? I think it may be due to the change of splitting energy but if that so then how does it actually work?

Your idea appears to be correct.
Initially, the structure of $\ce{CrCl3.6H2O}$ corresponds to the dark green compound $\ce{[CrCl3(H2O)3].3H2O}$. This complex shows hydrate isomerism; i.e. the chlorido ligands are step by step replaced by water molecules:
$$\ce{$\underset{\text{dark green}}{\ce{[CrCl3(H2O)3].3H2O}}$ <=> $\underset{\text{dark green}}{\ce{[CrCl2(H2O)4]Cl.2H2O}}$ <=> $\underset{\text{blue green}}{\ce{[CrCl(H2O)5]Cl2.H2O}}$ <=> $\underset{\text{violet}}{\ce{[Cr(H2O)6]Cl3}}$}$$
Thus, the actual complexes are:
$\ce{[CrCl3(H2O)3]}$ (dark green)
$\ce{[CrCl2(H2O)4]+}$ (dark green)
$\ce{[CrCl(H2O)5]^2+}$ (blue green)
$\ce{[Cr(H2O)6]^3+}$ (violet)
Each of these complexes has octahedral geometry. Cr(III) has a $\mathrm{d^3}$ electronic configuration. In the ground state, three electrons occupy the $\mathrm{t_{2g}}$ orbitals ($\mathrm{d}_{xy}$, $\mathrm{d}_{xz}$, and $\mathrm{d}_{yz}$). The two $\mathrm{e_g}$ orbitals ($\mathrm{d}_{x^2{-}y^2}$ and $\mathrm{d}_{z^2}$) are empty.
Three spin allowed d–d transitions are expected for Cr(III) complexes. The corresponding absorption maxima of $\ce{[Cr(H2O)6]^3+}$ are:
$$\begin{alignat}{2}
&\mathrm{^4A_{2g}\to{^4T_{2g}}}\quad&&\tilde\nu_1=17\,400\ \mathrm{cm^{-1}}\\
&\mathrm{^4A_{2g}\to{^4T_{1g}(F)}}\quad&&\tilde\nu_2=24\,500\ \mathrm{cm^{-1}}\\
&\mathrm{^4A_{2g}\to{^4T_{1g}(P)}}\quad&&\tilde\nu_3=38\,600\ \mathrm{cm^{-1}}
\end{alignat}$$
The first transition $\mathrm{^4A_{2g}\to{^4T_{2g}}}$ corresponds to promoting one electron from the $\mathrm{t_{2g}}$ level to the $\mathrm{e_g}$ level. Thus, it directly gives the value of $10\ \mathrm{Dq}$ as $17\,400\ \mathrm{cm^{-1}}$.
The energy splitting of the spectral terms depends on the ligand field strength. If the $\ce{H2O}$ ligands are replaced by the low-field ligand $\ce{Cl-}$, the value of $10\ \mathrm{Dq}$ and the wavenumber of the corresponding absorption maximum are lowered:
$$\begin{alignat}{2}
&\ce{[Cr(H2O)6]^3+}\quad&&\tilde\nu_1=17\,400\ \mathrm{cm^{-1}}\\
&\ce{[CrCl(H2O)5]^2+}\quad&&\tilde\nu_1=16\,400\ \mathrm{cm^{-1}}\\
&\ce{[CrCl2(H2O)4]+}\quad&&\tilde\nu_1=15\,600\ \mathrm{cm^{-1}}\\
&\vdots \\
&\ce{[CrCl6]^3-}\quad&&\tilde\nu_1=13\,200\ \mathrm{cm^{-1}}
\end{alignat}$$
By way of comparison, if the $\ce{H2O}$ ligands are replaced by the strong-field ligand $\ce{CN-}$, the wavenumber of the corresponding absorption maximum is increased ($\tilde\nu_1=26\,700\ \mathrm{cm^{-1}}$ for $\ce{[CrCN6]^3-}$).
The wavenumbers can be converted to the wavelength of the absorbed light:
$$\begin{alignat}{2}
&\ce{[Cr(H2O)6]^3+}\quad&&\tilde\nu_1=17\,400\ \mathrm{cm^{-1}}\quad\lambda=575\ \mathrm{nm}\ \text{(yellow)}\\
&\ce{[CrCl(H2O)5]^2+}\quad&&\tilde\nu_1=16\,400\ \mathrm{cm^{-1}}\quad\lambda=610\ \mathrm{nm}\ \text{(orange)}\\
&\ce{[CrCl2(H2O)4]+}\quad&&\tilde\nu_1=15\,600\ \mathrm{cm^{-1}}\quad\lambda=640\ \mathrm{nm}\ \text{(red)}\\
\end{alignat}$$
Since the colour of the absorbed light in this series changes from yellow to orange to red, the corresponding colour of the remaining visible light changes from violet to blue green to green.


text:
Spectral reflectance identification of types of iron rust
Based on the Wikipedia page about rust, there are red and green types of rust that occur on iron.  Red rust is the reaction with oxygen and water and the green with chlorine (in oxygen poor environments).
I understand the chemical reactions that forms these types of rust, but in the Wikipedia article, it states that:

Several forms of rust are distinguishable both visually and by spectroscopy

(bolding mine). Visually, it'd be pretty obvious (one is red, one is green); spectroscopy-wise, I would imagine it to be a similar story - but what aspect of the types of rust cause this difference - hence, the question:
Are the spectral reflectence features that distinguish between the red and green rust types based on what the iron reacts with (oxygen/water or chlorine), or is it based on the type of iron present (Fe2+ vs Fe3+)?
Full disclosure, I have a similar question on Open Data.SE here, but are asking for an explanation in terms of spectral reflectivity.

If available, the best spectroscopic technique for iron is usually Mossbauer.


text:
To what state would copper wiring revert if abandoned for centuries?
Would copper wires from power transmission or telephone lines acquire a patina but remain largely intact, or oxidize into dust after a century or more without replacement or maintenance?
Obviously environmental factors would be enormous; ideally I'd like to know what the metal's state would be in ideal (underground?), typical (temperate zone), and worst-case (near salt water?) scenarios. Clearly some circumstances allow copper objects to survive largely intact for several thousand years, since such artifacts have been found by archeologists, but does this require very exceptional circumstances?
This question is inspired by an answer on the Worldbuilding site, which claims that a post-apocalyptic society would have a difficult time developing metallurgy as the rusted remains of human civilization would be more difficult to smelt than the naturally-occurring ores which were relatively abundant during the Stone Age. Hopefully by asking about this copper wires specifically, I have an appropriate question for this site. If so, I may ask another question about iron and steel.
I did not have much luck in finding relevant tags. Any assistance there is appreciated!

Copper sheathing for roofs can last a very long time, indeed acquiring a nice patina.  It might not be a great roof after a century, but it will still be there.

True. The Clarksville Customs House is over 100 years old, and the roof has survived tornadoes. However, copper wiring is presumably much more widespread. Though I understand that most power lines are largely aluminum these days.

Let me talk first about how iron corrodes (as in rusts) and then compare that to copper. The simple answer is that when iron corrodes, the iron oxide (iron which has chemically bonded with oxygen, which it loves to do) that forms on the surface of the pure iron underneath, does not seal the surface against further intrusion of more oxygen. When copper corrodes, the copper oxide does seal the surface protecting underlying copper against further corrosion. I don't know if iron oxide is simply more porous to oxygen than copper oxide is but I do know the following. Iron can chemically combine with oxygen in multiple ratios. When you examine a highly corroded iron surface you'll notice that the first layer of rust is very red. If you brush or maybe lightly sand that red layer of rust away you'll notice a black layer underneath it. That's still rust. If you sand your way through that you'll get to the pure iron underneath which has more of a dark silver color. The red rust has more oxygen in it than the black rust but can pass some of that oxygen to the black rust turning it red. Similarly, the black rust can pass some of its oxygen on to the pure iron underneath it. That process continues until oxygen can leach all the way through the iron. Chemically, iron has more than one oxidation state. Oxidized copper atoms are much more reluctant to release their oxygen to adjacent pure copper and act as a barrier to the intrusion of more oxygen.

Right, iron is notorious that way. I heard that it's because the oxide has a greater volume, and thus the outer layer expands and separates from the inner layer, exposing it. Regarding copper, though, will the layer of oxide protect the inner part indefinitely, as long as there is no mechanical action to expose fresh metal? Will it survive for years, decades, centuries? What if it's exposed to rain, or standing water?


text:
Why should 2,6-Dimethyl-4-nitrophenol be more acidic than 3,5-Dimethyl-4-nitrophenol
I was trying to compare acidity of some organic compounds but can't understand some:
Why should 2,6-dimethyl-4-nitrophenol be more acidic than 3,5-dimethyl-4-nitrophenol?
I have been trying to figure it out using various factors like hyperconjugation and induction effect of Me group but its not working. Suppose i take H.C effect 2,6-dimethyl-4-nitrophenol be "less" acidic than 3,5-dimethyl-4-nitrophenol since it destabilizes the negative charge on conjugate base. But this logic is not working.

In the 3,5-dimethyl isomer, steric effects come into play and influence the acidity.
Resonance structures like B in the figure below, that stabilize the phenoxide anion explain why 4-nitrophenol ($\ce{R=H}$) is more acidic than phenol.  Resonance structure B requires that the nitro group be planar with the aromatic ring.  If methyl groups are placed adjacent to the nitro group ($\ce{R=CH3}$), the steric interactions between the two methyl groups and the nitro group becomes large and forces the nitro group to rotate out of the aromatic plane,  Hence resonance structure B can no longer contribute and the compound is less acidic.

So that's basically ortho effect? Thanks i got it now :-)

Yes exactly, in this case it's a double ortho effect.

In3,5 dimethyl 4 nitro phenol , the nitro group is between two ortho methyl groups . Therefore due to this steric inhibition of resonance occurs and this nitro group is pushed out of plane by two methyl groups. Such that nitro group does not take part in the resonance. And there have no extra stable structure. That is why it is less acidic than2,6 dimethyl 4 nitro phenol.


text:
Is p-bromophenol more acidic than p-chlorophenol?
p-chlorophenol is more acidic than p-fluorophenol due to presence of vacant d orbitals. But then is p-bromophenol more acidic than p-chlorophenol since it has vacant f orbitals also? But I guess that’s wrong, isn’t it? Please give your suggestions.

Disclaimer: This is not necessarily the "correct" answer as many factors play roles and it would be difficult to describe the observation so simply.
I'm not sure I can provide an easily "correct" answer, but I can try to explain some of the many factors that influence the acidity.
Most notably, the factors we should consider are inductive effects and delocalization of charge.
The inductive effects, based on electronegativity, would dismiss the experimental observation, so I'll look at delocalization of charge.
The delocalization of charge, which introduces a negative charge by the halogen is greatest by F>Cl>Br>I. This would reduce acidity, as a more negative benzene ring would not desire to gain another negative charge. Thus, this is one factor that supports the experimental observation.
Thus, we may be able to explain the observation, if only partially, by the delocalization of charge.
I adapted the answer to a similar question on http://www.quora.com/Organic-Chemistry/Which-is-more-acidic-ortho-fluorophenol-ortho-chlorophenol-ortho-bromophenol-or-ortho-iodophenol-Why . I would have preferred more sources but I could not find a very good source.

Experimental data provides the correct answer: 4-bromophenol (pKa = 9.17) is a stronger acid than 4-chlorophenol (pKa = 9.41). Scroll down to "dissociation constant". References are provided.

I looked up the pKa values of each chloro- and bromophenol and compiled them in the following little table:

pKa
chlorophenol
bromophenol

ortho-
8.49
8.45

meta-
8.81
9.03

para-
9.14
9.17

(These are just the first values that came up when I googled the name of the compound plus pKa. I did not go back and check the primary sources.)
Keep in mind that both chlorine and bromine are primarily inductively electron withdrawing (-I), and to a lesser extent mesomerically electron donating (+M). Chlorine is more electronegative than bromine, so its -I effect will be more pronounced. On the other hand, chlorine's d orbitals are smaller and overlap more efficiently with the ring, giving a stronger +M effect than bromine. This, in my opinion, nicely explains all the major trends in the table:

First, keeping the halogen constant, acidity decreases in the order ortho > meta > para, for both chloro- and bromophenols. This makes sense as the -I effect dominates and becomes weaker as the halogen atom moves further away from the hydroxyl group.
For the ortho- and para-compounds, the acidities of the chloro- and the bromophenol are practically equal. When going from chlorine to bromine, the decreased -I effect is almost completely canceled out by the similarly decreased +M effect.
However, meta-chlorophenol is significantly stronger than its bromo-analogue. Again what we would expect, as the mesomeric effect does not affect meta substituents, leaving only the inductive effect which is stronger in chlorine.


text:
Why does iodine have different colors in different media?
It is violet in trichloromethane, but brown in water. why? In the Shriver Atkins book, they tried to explain this using Molecular Orbital Theory which was not clear to me

Some interesting information can be found in this related question.

color is result of electron transfer between orbitals and anything that affect energy of these orbitals can change color of that specie, so iodine in different mediums have different interactions and because of that iodine orbitals have different energies in different medium,so iodine have different color in different mediums.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the [help]. Could you elaborate a little bit more, I cannot follow your thoughts. How do the interactions with the medium influence the energies of iodine?


text:
How does radiation facilitate the formation of tetravalent iron?
In reading the article abstract What Oxidation State of Iron Determines the Amethyst Colour?, the author states that in regards to the gemstone amethyst, from clear quartz:

The crystal was transformed into violet amethyst by gamma-irradiation. The change in colour was accompanied by changes in the Mössbauer spectrum that can be interpreted as the conversion of trivalent iron into the tetravalent state: $\ce{Fe^{3+}→Fe^{4+}}$

How does gamma radiation facilitate the conversion of oxidation state of iron, from $\ce{Fe^{3+}}$ to $\ce{Fe^{4+}}$?

The ionization energy of $\ce{Fe^3+}$ to $\ce{Fe^4+}$ is 54.8 eV.
The energy of typical gamma radiation is much higher (in the order of roughly some keV to a few MeV). Depending on the energy, the interaction of average gamma radiation with the affected electron is mainly due to Compton scattering.


text:
Organisation in Periodic table
Why do we find big spaces in the periods of the Periodic Table between $\ce{H}$ and $\ce{He}$, $\ce{Be}$ and $\ce{B}$, and $\ce{Mg}$ and $\ce{Al}$?
What is the logic of such organization of the periods?

I believe that you’re meaning groups rather than periods - unless my chemistry is rusty, the groups are the vertical arrangements of elements whereas the periods are horizontal. But I could be mistaken.

related: http://chemistry.stackexchange.com/questions/901/what-are-alternates-to-the-periodic-table-of-the-elements

I don't think that this is really a duplicate of that previous question - at least people react nervously over there that mine is not an answer. Anyway, it is an answer to this question, see: http://chemistry.stackexchange.com/questions/901/what-are-the-alternatives-to-the-periodic-table-of-the-elements/28775#28775

From a historical perspective, this was done to account for the commonalities between how the various elements behaved. In other words, the Alkali metals all exhibited a strong reaction with $H_2O$, the noble gasses all exhibited inertness in reactions, and so on.
This is explained in modern day science via the quantum description of electronic structure which indicates that elements within the same group tend to have the same number of valence electrons in the valence shell. Simply put, the reason why there are such gaps between $H$ and $He$; $Be$ and $B$, and otherwise is because there are simply different quantum arrangements of the electrons which are readily accessible to the element. That is, the ground state of $He$ has both accessible $l=0$ states occupied by electrons, which constitutes the valence shell and so is in an energetically favorable state to remain inert - becoming of the noble gasses. Hence, it’s moved far to the right from $H$ to the 8th group; consistent with the historical structure of the table.
That’s a crude description, but about what’s happening here. I agree however that the chemists may feel more at home discussing the periodic table. I’ll stick to the chart of nuclides.
Also, the Wikipedia article is fairly informative on this topic:
http://en.wikipedia.org/wiki/Periodic_table