text:
Is lithium L-threonate a potential brain medicine?
Lithium ions have known neuroprotective qualities. That would explain its therapeutic benefit in some mental disorders. It is also well known that lithium carbonate must be taken in toxic doses to get even small amounts into the brain. This makes lithium carbonate a very non-optimal medicine. On the other hand, lithium orotate promises to reach the brain much more efficiently and therefore require smaller doses. Unfortunately it contains orotic acid, which is a known carcinogen.
Magnesium L-threonate is a form of magnesium that easily penetrates the blood-brain barrier and therefore confers therapeutic effects on the brain. Is it then conceivable that lithium L-threonate is both safe and effective as a medicine for brain disorders?

Maybe yes? Perhaps the synthesis is easy, I believe that the metabolic route of l-threonate should be taken into account too.

Do you know why lithium ion has a good neuroprotective qualities? Is it because is small and has high redox potential? I never searched for this.

I’m pretty sure that elemental lithium is bad for the brain as it will immediately react with the water around to create $\ce{LiOH}$ which will cause even more damage … ;)

Jan, elemental lithium is good for the brain in trace amounts. Cities that have higher levels of lithium in the water supply have lower suicide rates, at least suggesting a potential connection. The fact that lithium carbonate is administered to mentally ill patients means it must have tonic effects on the brain. As stated in the original post, lithium carbonate is so inefficient at getting to the brain that toxic amounts must be used to get therapeutic results.

Orr22, this might be of some use to you: http://www.psychiatrictimes.com/bipolar-disorder/lithium-neuroprotective-effects

Of course, it's the psychiatric times and in the same article they praise the infamous anti-psychotic meds. But take it with a grain of (ahem) salt.

Consider also the fact that the positive charge is less polarizable on the small lithium compared to potassium. I am no pharmacologist but it may play a role for the permeation of the blood-brain barrier.

Lithium has a diagonal relathionship with magnesium, so in effect, it makes sense to compare it to magnesium compounds. It may also substitute for sodium ions in transporters. I doubt the counter ions is relevant, since most Li coumpounds are quantitatively dissociated. Stable concentrations of lithium are only achieved after several week of treatment. Lithium carbonate, citrate and sulfate are used to my knowledge. Plasma concentrations are monitored so as to avoid reaching toxic levels (more than 1 milli molar). Some physicians also test for the concentration inside red blood cells to get an idea of the concentration in the brain, but this has not been confirmed to be relevant. What is important is not the amount of it, but the clinical effect : is the patient responding to treatment ? does he/she support it ?
Lithium has not only neuroprotective effects, but also mood-stabilizing effects. It is mostly used in case of bipolar disorder (mood swings), but also as an antidepressant. The mechanism of action is still elusive. It may enhance serotonin pathway, and, by competing with magnesium ions, inhibit some magnesium-dependent enzymes (typically enzymes handling MgATP). One of these enzymes is called glycogen-synthase-kinase (GSK).


text:
What acid can remove Calcium in Ca2SiO4 to leave SiO2 (mostly)
What acid can remove calcium in $\ce{Ca2SiO4}$ to leave $\ce{SiO2}$ (mostly)?
It seems that calcium silicate is stable in most acids, but I need to etch it and leave only silica (mostly).

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
Compounds that release carbon dioxide
What are some compounds that release carbon dioxide when heated, hydrated, or reacted with sodium, aluminum, silicon, or iron?

Mix baking soda with dry citric acid; that would work pretty well once it gets in contact with water.

You see, the problem with these questions is that many answers can be correct and you can't rightfully judge which single answer is the best. SE model is not good for these kind of questions and we close them as "too broad", which is what's going to happen to this question, sadly.

A solvent such as water would be required and carbonate or bicarbonate salts will release CO2(g) when reacted with most acids.
Dry ice will turn to CO2(g) when exposed to room temperature.
These are only a couple of examples as others have stated there could be many other methods.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Does adding ice to sparkling mineral water extend the 'fizziness' of the solution before the drink goes flat?
To my understanding the carbonated beverage loses its $\ce{CO2}$ content as gas bubbles which gives the drink its 'fizziness'. The loss of $\ce{CO2}$  occurs due to the low solubility of carbonic acid formed during carbonation resulting in an excess of unreacted $\ce{CO2(g)}$ which is trapped in the container when bottled and escapes when opened:
$$\ce{H2O(aq) + CO2(g) -> H2CO3(aq) + CO2(g)}$$
Temperature and volume of the solvent play a critical role in the solubility of matter.
Would the decrease in temperature and slight increase in solvent caused by the ice cube being added to a carbonated beverage exposed to air cause the solution to remain 'fizzy' for longer than the same solution at RT°C with no solvent increase?

Let me lightly gloss over the issue of nucleation sites and ice. While one observes bubbles forming on ice in the glass, they continue to form on the glass as well.
Instead, one could well ask the impact of temperature on release of the carbon dioxide, and thermodynamically that does come down to the solubility of carbon dioxide in water. Since this is of great importance to the soda industry, the study goes back some ways. R. Wieb and V.L. Gaddy, "The solubility of carbon dioxide in water at various temperatures from 12 to 40 degress and at pressures to 500 atmospheres - Critical phenomena', Journal of the American Chemical Society 62 815-817 (1940) shows measured solubility curves close to the temperatures of interest. One should note the increase in solubility as the temperature is dropped from room temperature down towards zero C.
While the carbon dioxide is supersaturated in the soda water, the driving force relates to the difference of the supersaturation to the solubility limit. As the temperature decreases, the thermodynamics favor retention of more carbon dioxide, which will then come out in your (warmer) mouth, giving you that 'fizz'.
Further, ignoring nucleation sites on the ice being much better than on the glass, as the temperature is decreased, the kinetics of nucleation will be at least partially suppressed as well.
Plus, I like my soda (and beer) cold, not warm!

Lots of nucleation sites on a glass indicates a dirty glass. http://www.craftbeer.com/craft-beer-muses/coming-clean-about-dirty-beer-glassware

Yes, but I'm not quite ready to use a piranha clean at home!

I worked in a bar. Having clean glassware was a matter of pride (and kept health inspector happy...).

I guess there is a difference between the bubbles that stick to the side of the glass, and that one place where a steady stream of bubbles comes from - it is the second one that is a nucleation site. But even with a nucleation site, there is still the thermodynamic driving force as part of the kinetics... And, sometimes that site is an actual flaw in the glass (or that is what I tell myself).

It absolutely does if you are using dry ice. I love dry ice in my drinks.

Your analysis of the situation is wrong. Most of the carbon dioxide in the solution is present as dissolved gas, not carbonic acid. Remember that the beverage was under pressure when packaged. When the liquid is poured into a cup there is no longer any great carbon dioxide pressure and the drink (solution) becomes supersaturated with carbon dioxide. Adding ice cubes creates nucleation sites which promote the supersaturated dissolved gas to form bubbles and release carbon dioxide into the atmosphere. So adding ice causes the drink to go flat faster.

On the other hand, the decrease in temperature should really slow down the release of $\ce{CO2}$. It is not immediately clear to me what would be the overall effect.

@Ivan Neretin , that's exactly what I meant about the temperature variable. I understand it is always going to release the CO2 likewise if the ice is added or not but does the  drop in temp slow the rate it is released. If the sparkling mineral water was heated would the CO2 be released faster?

@MaxW , I appreciate your input,  but the ice isn't cooling the drink to the point of freezing so the water in the drink wouldn't be forming nucleation sites . The gas should become more dense at cooler temperatures and expand on heating.  I'm only talking about slight changes.

@Joel - The nucleation sites are on the surface of the ice cube, not into the solution. Drop an ice cube from the freezer into a carbonated drink and watch it fizz. You can reduce the fizz by pretreating the ice cube. Put some ice cubes from the freezer into a glass. Fill glass with water to cover the cubes. Stir for a couple of seconds and let glass sit for about 30 seconds. Pour off water and use those cubes to cool the drink. You have less ice, but the surface is now much smoother and the ice is at 0 C not freezer temperature. This greatly reduces fizzing of drink.

The other thing to reduce fizz is to start with soda that is already cold from the fridge. So pouring cold soda over pretreated ice cubes works much better than warm soda over cubes straight from the freezer.

A pre chilled glass would be good also ...


text:
What is the chemical test for identification of benzophenone?

How to identify that a unknown compound is benzophenone?

My working:-
First I added 2,4- DNP(2,4-Dinitrophenyl hydrazine) to the compound. Yellow-orange crystals were formed which indicated presence of carbonyl group. Then, I added Tollen's reagent to the compound. It did not gave silver mirror which means aldehyde group is absent. So, it must be a ketone. So, I added m-Dinitrobenzene to the compound. It did not gave violet colour. Alternatively, I added Sodium nitroprusside to the compound. It did not gave a red color. Both the test is for confirmation of ketone compound.
I asked my teacher what is the unknown compound. He said that it is benzophenone. Also, in the practical book where all the chemical test were mentioned, it was mentioned  as a side-note that benzophenone does not give the above mentioned test. So, what is the chemical test for identification of benzophenone?

You can use iodoform test after Tollens test.

@AdityaDev, benzophenone will not give iodoform test because it does not contain ketomethyl group       ($\ce{-CH3CO}$).

oops. i thought benzophenone was CH3-CO-ph. my mistake.

After identification of ketone, to identify benzophenone you need to find the range of melting point using melting point apparatus which you should get in range of $48.3\ \mathrm{^\circ C}$ to $50.6\ \mathrm{^\circ C}$.
Then you can find IR of unknown(benzophenone).
Source: Prezi

experiment
To 50 mg of benzophenone taken in a dry test tube add 1 gm of naphthalene mix and heat until melts. Add a small piece of sodium metal and heat again.
observation
The mixture turns green due to the free radical formation which is in delocalisation  with two phenyl rings

Could you provide a source from which you found the experiment?

@koteswararao D  How do you say that the radical will be formed in the benzophenone? A radical anion formation in naphthalene with sodium is known.


text:
Is conversion of L-glutamate to 6-phospho-D-gluconate a feasible reaction?
I want to know whether a reaction is feasible or not. Can L-Glutamate ($\ce{C5H9NO4}$) be converted to 6-Phospho-D-gluconate ($\ce{C6H13O10P}$) or vice versa in one step?

Take a look at the structures of the two molecules you're interested in (below). There are many chemical changes required if one were to theoretically carry out the transformation.
In a synthetic sense this would require a very long sequence- a quick glance on SciFinder doesn't reveal anyone having done this (probably because it would be incredibly lengthy and has no real purpose).
Even biochemically, where the cells have highly specialised enzymes capable of effecting all manor of changes, I find it unlikely that the transformation can be achieved by a single enzyme acting upon the glutamate, and indeed a cursory flick through the biochemical pathways atlas shows no direct transformation between these two molecules.

Structure of L-Glutamate

Structure of 6-phospho-D-gluconate

... and the conversion would probably not be stereo-specific enough.


text:
Why does dry boiling discolor steel cookware?
I have a stainless steel kettle that I let boil dry. That is, I left it on a hot coil burner with nothing in it, and noticed a few minutes later that the bottom was red hot and smoking. To make matters worse, I panicked and quenched it with cool water.
Both the inner and outer surface were left discolored with various shades of yellow and purple. This seems to be a known consequence of overheating, but I can't find an explanation of what actually happened.

I understand why heat (especially combined with quenching) would warp the metal, but why would it cause discoloration (that persists after the metal has cooled)?
After hunting around, I saw several suggestions that boiling white vinegar could remove the discoloration, and to my suprise, it worked pretty well. Why does this work?

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center.

It is all about thin oxide layer on the surface. Looks like stainless steel is not absolutely stainless, after all. Strange enough, the Wiki page about it does not exist in English.

Steel (even stainless) will oxidize at high temperature. Vinegar, which is acidic, will etch the oxide, leaving a slightly thinner and probably warped kettle.

Stainless steel have many different compositions. If you heat it up the surface undergoes oxidation. Diffusion at higher temperatures plays also its roll - shifts surface composition of the kettle. The various shades of yellow and purple are metal oxides. The vinegar removes the oxides, thus removes the coloration. Vinegar should remove the oxides even without boiling. It will just take more time.


text:
Addition of HOBr across C=C

Source: Carey Advanced Organic Chemistry B, problem 4.1 b

Looking for interesting problems about addition to C=C, I came across this in the back of Carey. It looks like a simple addition across the alkene, but using HOBr which is something I've never came across in the lab (and perhaps not seen used as an example since I was an undergrad...).
The most obvious way to dissociate HOBr is to lose a proton, leaving the OBr anion behind, Wikipedia confirms this:

[...] hypobromous acid partially dissociates into the hypobromite anion OBr− and the cation H+.

So far so good. And based on this fact, you might expect the product to be as drawn below. However the original literature points towards a different product, also shown below.

Obviously, the product I've drawn as expected looks a little odd, with the unstable Br-O bond present. However, I cannot rationalise how the actual product has formed.

Ref.: JOC, 1968, 33, 3953.
  The original literature from which the problem was taken

Three possibilities:

The HOBr doesn't dissociate as I'd imagined. There is the possibility of the Br forming the bromonium ion which is then opened by hydroxide, however to get the product as isolated, you have to attack in the anti-Markovnikov sense (i.e. the product would look like the one formed if you'd attacked a primary carbocation with hydroxide).
The product forms as I drew it, but rearranges to give the product they isolate. This is what I consider to be the most likely explanation, but I'm unable to see how this happens.
The secondary carbocation forms upon protonation, but this rearranges. Again, struggling to see how this would work as the secondary cation is the most stable one that can form in this system, as there is no possibility of a tertiary carbocation.

Any suggestions to point in the right direction? Even the name of a rearrangement would be helpful.

Are you sure about that "observed product"? I heard that halohydrin reaction follows markovnikov rule. Same reaction will happen if you use Br2 in H2O. First bromonium ion is formed followed by nucleophilic attack by water molecule.

I can't guarantee that the product in the paper is correct, obviously in 1968, characterisation wasn't what it is today. But the majority of my question holds up even supposing it was wrong.

Look halfway down the page here at the hypohalous acid examples. $\ce{Br^{+}}$ is the electrophile.  One of their examples involves addition of $\ce{HOBr}$ to methylenecyclopentane.  The product is different than what the JOC reference suggests.

So the product follows markovnikov rule?

@AdityaDev Yes, it follows Markovnikov's rule.

Markovnikov's rule applies to the addition of $\ce{H+X-}$ across a double bond.  Considering $\ce{HOBr}$ does not add via $\ce{H+OBr-}$ I don't know whether it's a useful "rule" here.

The 'expected' product comes from analogy of addition of strong acids like HBr or acid (often sulfuric acid) catalyzed hydration. However, The pKa of HOBr is 8.7 (ref), much lower than those conditions, so it's not reasonable to make that analogy. Put another way, the base in those reactions is an alkene, which is very weak. Only a very strong acid will protonate an alkene.
In some quick SciFinder searching, I haven't found any overwhelmingly convincing evidence for the regioselectivity observed in the product, but it does seem reasonable to me by a mechanism similar to what Aditya Dev suggests.

I look at HOBr as an oxygen with a leaving group (analogous to Br-Br). Attack by the alkene on the oxygen releases bromide and gives the tertiary carbocation. Kick back by the oxygen closes a protonated epoxide, which is analogous to the bromonium. In the bromonium case, these are considered resonance structures. I don't know if that's definitely the case here (oxygen is much smaller than bromine), but I've written it that way to continue the analogy. In any case, protonated epoxides are typically attacked at the site best able to stabilize positive charge, which is the tertiary carbon. The resulting bromohydrin is the 'observed' product.

Why do you think that the double bond would attack the electron rich oxygen atom instead of the more positively charged Br? The electrophile gets attacked first and the table here says that in HOBr, Br will be the electrophile.

@pH13 Either is reasonable... Bromide is a much better leaving group than hydroxide. That makes the process similar to reaction with percarboxylic acids (peracetic or mCPBA, e.g.) where an alkene attacks an oxygen with a leaving group. In those cases, the leaving group is a carboxylate. I would believe it either way, and I would like to see experimental evidence for one over the other. In my brief searching, I have not seen anything convincing. I saw that table as well, but unfortunately there are no references so I don't know how much stock to put into it.

Traynham, J. G.; Pascual, O. S., Tetrahedron 1959, 7 (3-4), 165-172, 2) Carey, F. A.; Sundberg, R. J.; Electrophilic Addition to Alkenes.  Advanced Organic Synthesis, Part B: Reactions and Synthesis, 5th Edition; Springer: New York, 2007; 302 f. which cites Masuda H.; Takase, K.; Nishio, M.; Hasegawa, A.; Nishiyama, Y.; Ishii, Y., J. Org. Chem. 1994, 59 (19), 5550–5555

Traynham and Pascual1 were first to investigate the reaction of hypohalous acids with methylenecyclohexane 1 and that of the epoxide 3 of this olefin with HBr. While both regioisomeric chlorohydrins were reported, only one bromohydrin was obtained, regioisomer 2 (mp.82o-83o). The structural assignments were made without the aid of nmr spectroscopy. Sisti2 reinvestigated these reactions and provided convincing evidence with the aid of nmr that the structure 2 is the other regioisomer 8 (mp.82o-83o and undepressed mixture melting point). Predictable oxymercuration of olefin 3 led to mercurial 6 which was readily converted to bromohydrin 8 and also reduced with NaBH4 to form tertiary cyclohexanol 7. Preparation of bromohydrin 2 was achieved in part by reduction of bromoester 4 with LiAlH4. The non-ring methylene groups were distinguished by nmr: 2 (CH2OH, δ 3.70); 8 (CH2Br, δ 3.37).
To answer this inquiry properly, it is important to have the correct structure of the product before providing a mechanism for its formation.

1) Traynham, J. G.; Pascual, O. S. Tetrahedron, 1959, 7, 165.
2) Sisti, A. J. J. Org. Chem., 1968, 33, 3953.

Sorry to dig up an old question but I think some important points are missing:
The experimental result is easily explained by steric hindrance and polarization of a bromonium ion formed due to addition of $\ce{Br+}$ to the alkene (three membered ring). SN2 attack of water occurs (hydroxide is not very common in aqueous acidic media!) at the bromonium ion. Nucleophilic attack at the cyclohexyl ring is more difficult (slow) than at the exposed site (fast). Additionally, the cyclohexyl ring delivers a +I effect, deactivating the cyclohexyl ring carbon for nucleophilic attack.

Also, I wanted to add it's not actually possible to obtain pure HOBr. These species exist in complicated equilibria in aqueous solution. They are known to disproportionate to more stable oxidation states (including bromide) and they exist in an equilibrium with bromine and water. Similar to other electrophilic halogenating agents such as N-bromosuccinimide they supply a small concentration of the molecular halogen to the solution. You can treat this as the reactive entity in your reaction.


text:
How to distinguish between carbon-carbon double bond lengths?

Which of the following compounds should have the longest carbon-carbon double bond?

$\ce{CH2=C=CH2}$
$\ce{H3C-CH=CH2}$
$\ce{(H3C)3-C-CH=CH2}$
$\ce{(H3C)2C=CH2}$

I know that single bond > double > triple length-wise, but how do you distinguish between double bonds?

You can draw hyperconjugation structures to compare bond length. The more α-hydrogens the alkene has, the more hyperconjugation structures it has, and the higher is the degree of delocalization of the π-bond and therefore the longer is the bond. When double bond character decreases, bond length increases. Therefore the answer should be (d).

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


text:
Detect a baked diamond?
Diamond suppliers often "bake" stones. This causes them to expel yellow-causing nitrogen and become white, but also makes them more brittle. A baked diamond can often be recognized by its unnatural pasty white color.
Is there any scientific or chemical way to detect that a diamond has been heat treated?

Quote from link "Baked diamonds, or annealed diamonds as some people refer to them, though they are undetectable to labs or independent appraisers have one major flaw, they are brittle!" http://diamondcuttersintl.com/monarch-diamond/ // So it doesn't seem like the standard gemological equipment can detect such a treated diamond. The website refers to such treated diamonds as causing an industry crisis.

This really intrigues me. // A "pasty white color" would be due to internal cracks which would reflect/refract light in odd ways. // I'd expect that such a treated diamond wouldn't break by cleaving along a single plane in the crystal, but shatter. // It wouldn't seem that the somewhat exotic technique of thermoluminescence would help. It would be one shot, but you could reenergize the diamond using UV light or x-rays.

A change in properties means a change in structure, therefore it seems that X-Ray diffractography should be able to tell the difference.

@A.K. Maybe it's just not accurate enough to detect the difference though.


text:
Aren't unit cell and the relevant crystal similar?
"A crystal is formed by a large number of repetitions of basic pattern of particles in space.
The basic structural unit which when repeated in three spatial directions generates the crystal structure is called unit cell."
Problem:
Assume I take some same perfect geometrical cubes and arrange them in all three dimensional directions, a larger box similar to the smaller ones will be introduced.
From this assumption and the information above it, I deduce that a unit cell should be similar to the relevant crystal. That means, the only difference would be lying between their size.
If this is the case, why consider the geometry of unit cell and not directly the respective crystal? We know in the assumption above; the fundamental cubical boxes would be determined by same geometrical information as would the derived cubical box.
That is; if the lengths of three axes of cube are represented by letters $a,b$ and $c$ and the angles between axes are represented by Greek letters $\alpha$, $\beta$ and $\gamma$ then for a cube,
$$a=b=c$$$$\alpha=\beta=\gamma=\pi^c$$
This information is general. It is for all size of cubes. Then, why to define a unit cell and think it simpler to handle with than the crystal?

The unit cell doesn't really define the crystal shape. The 3D crystal shape depends on a multitude of factors. I think 2500 odd shapes have been detected for the 3D crystals of the mineral calcite. http://www.mnh.si.edu/earth/text/2_2_1_4.html

But the unit cell for all those crystals is same. Right @MaxW?

There three calcium carbonate minerals. Calcite, aragonite and vaterite.

Forget that (though that's true, too). Yes, unit cell is the same, so what?

The Bravais lattice and the unit cell combine to make a space group. See here: http://chemistry.stackexchange.com/questions/28507/are-all-lattices-bravais-lattices/28510#28510 for more details.

@IvanNeretin, the information I provided at the very beginning of my question is from my textbook. It is suggesting that unit cell is similar to the the relevant crystal. What about that? Do you have any explanation for it? Please tell me.

@Volunteers, the basic structural pattern which is being repeated throughout the crystal is the unit cell. This suggests that unit cell is similar to that crystal.

Well if I agree upon all the calcite crystals having same unit cells then it must be some other thing that is a unit cell. Because, in that case, all calcite crystals would become same (after combining their unit cells).

A crystal is not similar to its unit cell, nor does the textbook say so. See, you are made out of a large number of atoms; do you look like an atom? Certainly not!

I think you are ignoring the CONTENT of the unit cells

The symmetry limitations (rotations, reflections and mirror images) of the unit cell also limit the crystals. So calcite won't ever form a perfect cube like pyrite can. But as you can see from the number of different calcite crystals the symmetry limitations don't prevent a great deal of variety in the 3D crystal shape.

@IvanNeretin Draw a small square. Draw 3 more same boxes. Put them in all 2 dimensional directions over your paper. A larger box similar to the smaller box would appear. This example is still criticizing me. As far as a human composition is concerned, so there is no basic structural unit or a unit cell of human shape. You are considering an atom as a unit cell. However, unit cell is the least volume consuming repeating structure of particles in a crystal. An atom can not describe the repeating pattern of a human composition unless it is a sphere.

@Man_Of_Wisdom OK, a crystal may grow like that. But it may do otherwise as well. Draw a small square. Draw 4 more similar squares, and attach them to all sides of the initial square. What you see is not similar to the smaller box. Continue by adding more repeating unit cells to all sides of the new figure. This is still a crystal with the same unit cell, but it does not look quite like yours.

According to my textbook, placing them in all three spatial directions would introduce the crystal structure. For a moment if I ignore it and do it as you just told me, how I am gonna know that in what way I should place them so as to get the crystal structure? If I do it my way, I am going to have similar larger unit cell (crystal structure) and if I do it yours way, I am going to have an unspecified geometrical shape maybe.

Think of unit cells as Lego blocks. As long as you attach new unit cells to similar faces of existing unit cells, you are doing it right. But the overall shape of the crystal can be just about anything.

Along; I can see in my book that the crystals are drawn just beside their respective unit cells and they look exactly similar

Then your book does a poor job at explaining the crystal shape. Sure, sometimes a crystal may look like its unit cell. But more often than not, it will be more complicated.

Then why does the crystals are specified by their unit cells? Unit cell fall into seven well known categories (Isometric-Triclinic).  Following your example of 5 squares, the combinations of same number of unit cell may introduce several crystal structures/crystals. Unit cell are of 7 kinds and one kind may further introduce several crystal structure. Then, actually why define a unit cell?

Let us continue this discussion in chat.

The statement from your book is correct, though the devil lies in the details. The point here is that the book is treating a theoretical crystal (i.e. one that is infinitely large). Let me elaborate and then answer your question.
A unit cell is simply a representation of a periodic structure. It is a means of conveying translational symmetry (and perhaps other types of symmetry as well, but let us focus on translational symmetry). If you have for instance a cubic unit cell, than a 3x3x3 grid of these unit cells is also a unit cell of that same lattice. If you extrapolate this idea, then the reasoning you pose in your question makes a lot of sense!
Let us now discuss the details. Assume that we have a theoretical crystal of infinite dimensions. What is then the unit cell? Can we define a unit cell which is also the infinite crystal? Every finite unit cell would obviously not be that crystal. This explains why we even use a unit cell in the first place: By defining the translational symmetry by means of the unit cell, we can define an infinite crystal.
In the field of solid state physics, it is attempted to keep things simple and well-defined and therefore a so-called primitive unit cell is introduced. This unit cell only contains one lattice point and you could see this as the smallest unit cell possible that is still able to convey the translational symmetry of the infinite crystal.
Your last question relates to why we want to use, for instance, the primitive unit cell or some other unit cell rather than the complete crystal? Let me answer this question by giving an example: If you cut a crystal, you will create a surface facet. Such surfaces also have translational symmetry and you can define these using Miller indices. Your textbook will probably treat these as well. Those Miller indices are defined by the reciprocal lattice vectors of your unit cell. You already touched upon on the fact that there exists an infinite set of unit cells that define that infinite crystal, yet the question arises which unit cell to use if we, for instance, want to use these Miller indices. In principle, any unit cell in the set would be valid. To avoid any kind of ambiguity, we tend to use the smallest unit cell possible.
A nice book treating these concepts is "Introduction to Solid State Physics" from Charles Kittel.


text:
Why is the Friedel-Crafts alkylation of nitrobenzene disfavoured?

Suggest a synthetic route to m-nitrotoluene, starting from benzene.

The conversion can be achieved in two ways:
$\hspace{10 mm}$
or:

However, the first route is not possible as the Friedel-Crafts reaction does not yield a meta-substituted product. Why is that so?

Hmm, I don't like second reaction either, you have proof it would work?

Also, I don't believe Na will leave that nitro group intact.

The use of a methyl halide does not seem feasible to me, as it suggests that the electrophile would be a methyl carbocation, which does not form in practicality. I would imagine otherwise that

The nitro group is so deactivating that yields are really poor for Friedel-Crafts reactions with nitrobenzene; not even the meta-substituted product is found in good yield. In fact, nitrobenzene is actually used as a solvent in Friedel-Crafts reactions.

Agreed. Most undergraduate organic chemistry texts come with a big disclaimer that Friedel-Crafts reactions work poorly on deactivated rings. I would love to see a link or reference for the use of nitrobenzene as a solvent for FC reactions. It would really solidify this answer.

@BenNorris http://onlinelibrary.wiley.com/doi/10.1002/bscb.19660750903/abstract


text:
Why was the Deville process of Alumina production given up in favor of the Bayer process?
I guess this is a sort of historical question. The DeVille process seems fairly simple. Does anybody know the history or practical reasons for the switch?

The reason has to be cost. I have no specific knowledge of either process but looking at the DeVille process https://en.wikipedia.org/wiki/Deville_process and the Bayer process https://en.wikipedia.org/wiki/Bayer_process I can see that the Deville process uses an extra calcination step and bubbles carbon dioxide. Neither process seems to adequately deal with the process waste.

I think  the DeVille would produce waste with a lower pH.

This might have been a better fit on [hsm.se].

Related: When is Serpeck's process preferred over Bayer's process?

This looks like a promising source. I think you may have to buy it though if you want to read it.

@bon I should have been searching in french It is so obvious. Thank you so much.

I suspect the problem is that the differential step between the two processes involve calcination at high temperature for Deville vs. heating with sodium hydroxide at relatively lower temperature for Bayer. That's a huge energy and cost savings.

@GeoffHutchison Apparently both processes originally used CO2 bubbling and reclamation of the  Na2CO3 the modern Bayer process as I understand  it skips this step.

The product of the Bayer process is purer than that of the Deville process.
Trevor I Williams, Aluminium: latecomer to the metal industry, Endeavour, Volume 17, Issue 2, 1993, Pages 89-93


text:
Stability of isomers of N4O
I have googled and found that the structure of $\ce{N4O}$ is linear with a positive and negative charge on 2 nitrogen atoms. But, the following structure seems more stable because it fulfils octet for all elements, the structure has no charge and it is resonance stabilised:

Why is this not the actual structure of $\ce{N4O}$? Are there any destabilizing effects that outweigh the stabilizing effects in this structure? The structure doesn't seem to have much angle strain. Where am I wrong?

The real structure, too, has a complete octet for all atoms and is resonance stabilized. As for having charge, that's not a big deal.

https://en.wikipedia.org/wiki/Nitrosylazide, looks sth good about oxatetrazene is harder to find, but they both decompose easily.

This is the kind of molecule you’ve drawn, where I’m scared to touch the monitor because it might explode …

@Jan, could you please explain why it is so unstable?

I consider every molecule that has more nitrogen atoms than insert-other-atom-type-here to be prone to explode, that’s why ;)

But isn't this structure more stable than the linear structure?

You're wrong in thinking that everything can be explained using few rules of thumb.

@Mithoron I am not an expert in inorganic chemistry. I only know the basics of inorganic chemistry taught at my school. My teacher was unable to explain as to why this structure was unstable because I followed the basic rules to draw the lewis dot structure. That is why I approached you guys. Many of you have great knowledge in chemistry and I am hoping I can get a clear answer. SE has never let me down.

Cyclic is not inherently more stable than linear.

@Jan no it is not. My question was whether this particular molecule is more stable in the cyclic form than in the linear form.

@ShankRam Ah, if that is your question then the answer is ‘we cannot say a priori’.

@Jan with the help of various stabilizing and destabilizing effects we have in our hands, is it possible to tell which one is expected to be more stable?

Not as far as I know.

Let us continue this discussion in chat.

From The Nitrosyl Azide Potential Energy Hypersurface: A
High-Energy-Density Boom or Bust? J. Am. Chem. Soc. 1996, 118, 4860-4870 :

[Abstract] ...While the ring isomer is predicted to be the most stable structure on the
  hypersurface, the barrier to dissociation is most likely between 1 and 2 kcal mol-1 (including zero-point vibrational
  energy [ZPVE], the existence of any barrier becomes questionable) making isolation theoretically possible but
  experimentally difficult. This small barrier also detracts from the attractiveness of the N4O ring structure as a high energy-density material. The trans-chain isomer, however, lies in an energy valley with higher sides, consistent
  with its previous experimental observation.
[full text]...In an ideal five-membered ring
  with 6 $\pi$ electrons, the $\pi$ electrons would be distributed evenly
  among all the bonds. In the present case, the highly electronegative
  O atom prefers to keep electrons around itself, leading
  to a partial negative charge on the oxygen. Energetically, there
  is a certain degree of stability associated with the ring isomer,
  although not on the order of common aromatic systems. The
  ring isomer is predicted to be at most (DZP CISD) 20.9 kcal
  mol-1 more stable than the trans-chain isomer; however, this
  value decreases to 13.2 kcal mol-1 with TZ2P CCSD.
  The ring isomer TS to dissociation into N2 and N2O is shown
  in Figure 5. ... Energetically, the barrier to dissociation
  is at most 15.3 kcal mol-1 (DZP CISD) and drops lower
  with improvements in both basis set and correlation scheme.
  In going from a DZP to a TZ2P basis set, for example, this
  barrier drops by 5.2 kcal mol-1 for CISD and 4.3 kcal mol-1
  for CCSD. Assuming a similar trend in moving from DZP
  CCSD(T) to TZ2P CCSD(T), the ring dissociation barrier is
  expected to drop to 1-2 kcal mol-1 with the addition of f-type
  functions possibly making it even lower. A barrier of this size
  lies below the ZPVE, throwing doubt on the existence of the
  N4O ring isomer.

So thermodynamically, yes, the ring form is the lowest energy isomer.  However, all the isomers are unstable to decomposition and the linear isomer is in a deeper potential energy well than the ring isomer. Therefore, the linear isomer is easier to observe experimentally for kinetic reasons.
There may be additional information in Theoretical study on structures and stabilities of N4X (X = O, S, Se, Te) series International Journal of Quantum Chemistry Volume 109, pages 226–235.


text:
Absolute configuration in stereochemistry: R/S nomenclature
I have all the rules of Fischer projections figured out.
But, when the problem asks for finding the configuration of a molecule with 1 chiral carbon and in a Fischer projection form, do we need first to make the manovrations that are allowed ( like fliping 180 degrees, or changing the places of 2-2 substituents )  or do we directly find the priorities ?

Just to remember the rules to assign an R/S configuration:
In order to assign the configuration as R or S:

Identify each of the chirality centers (most commonly an sp3 C with 4 different groups attached) Then at each chirality center....
Assign the priority (high = 1 to low = 4) to each group attached to the chirality center based on atomic number.
Reposition the molecule so that the lowest priority group is away from you as if you were looking along the C-(4) σ bond. If you are using a model, grasp the lowest priority group in your fist.
Determine the relative direction of the priority order of the three higher priority groups (1 to 2 to 3)
If this is clockwise then it is the R-stereoisomer (Latin; rectus = right handed)
If this is counter-clockwise then it is the S-stereoisomer (Latin; sinister = left handed)
If there is more than one stereocenter, then the location needs to be included with the locant, e.g. (2R)-

You can identify these previous rules on this example:

Source

You can assign the priorities directly. When assigning rotation, keep in mind that the atoms on the horizontal are coming out of the plane of the paper. Since there is usually a hydrogen on the horizontal, usually you will be looking at the stereocenter backwards. If it appears R, it is actually S, and vice versa.


text:
About how much carbon monoxide is produced in these two different reations?
The U.S. Department of Agriculture produces a gas cartridge to poison subterranean rodents.  The product is provided as a cardboard tube sealed at one end and then loaded with a mixture of 144.6 grams of the following ingredients:
53.0% Sodium Nitrate
28.0% Charcoal
19.0% Inert Ingredients
Once filled, the cardboard tube is sealed at the other end with a thin cardboard cap which can be punctured and a fuse inserted when ready to be used.  The gas cartridge's advertised function once lit is to generate a large volume of carbon monoxide that kills the rodent within a few minutes in its burrow.  My experience has been it is very effective when used according to the instructions.
The USDA gas cartridge is getting harder to obtain anymore and a commercial producer sells a similar product which is utilized in the same applications.  But it is smaller and uses a somewhat different formulation.  It does not advertise how the rodent is poisoned, so I don't know if the resulting gas from combustion is primarily CO or something else.  My experience has been that it doesn't seem to be as effective as the USDA gas cartridge, but I don't know if it is because of its different formulation or just its smaller size.  The 56.7 gram mixture of ingredients in this commercial gas cartridge are as follows:
50.0% Sodium Nitrate
38.0% Sulfur
9.0% Charcoal
3.0% Inert Ingredients
The question is how much carbon monoxide is generated by each of these cartridges when their ingredients are ignited and what are the quantities of other resulting gases?  Just straight theory here is good enough for me.  I know there are probably variables such as how well the ingredients are ground and mixed and what are used as the inert ingredients in each.  But lets just leave those variables out of the equation.

Straight theory would not tell you much, as long as you are leaving out the crucial variables. That being said, the latter cartridge seems to be acting primarily via $\ce{SO2}$, not $\ce{CO}$. Both are deadly; which death is better, I'm not sure.

I think the CO is probably more humane although it looks like it takes much less Sulfur Dioxide than Carbon Monoxide to result in an effect that is Immediately Dangerous to Life or Health at 100 ppm versus 1200 ppm.  See my comment below.

About the best that can done is simply to look at the ratio of carbon in the two cartridges.

$\dfrac{56.7*0.09}{144.6*0.28} = 12.6$%

The ratio of carbon dioxide to carbon monoxide formed on burning would depend on particle size, how well mixed an so on. The second formulation with sulfur is akin to black gun powder. The sulfur burns first at a lower temperature to ignite the charcoal. The sulfur dioxide would be toxic as well, but not as fast acting as carbon monoxide.

It looks like the IDLH for carbon monoxide is 1200 ppm where for sulfur dioxide it is 100 ppm.  IDLH is an acronym for Immediately Dangerous to Life or Health, and is defined by the US National Institute for Occupational Safety and Health (NIOSH)  Assuming rodents may react similar to humans, it looks like sulfur dioxide would be toxic in much lower concentrations than CO.  This may explain why the commercial manufacturer uses a smaller quantity of ingredients.  Maybe the better results I see with the U.S.D.A gas cartridge does have something to do with the inert ingredients.

Well if the "inert" ingredients increase the lethality, then they really aren't inert are they?

You would think that would be true.  I read somewhere that the inert ingredients are usually the binder, whatever that is.  But with 19% inert ingredients in the U.S.D.A cartridge, you kind of have to question why that is so high.  Is it something that allows for more complete combustion of the primary ingredients.  And if that is the case, could you really call it inert?  For example, if water was one of the ingredients, could that be classified as inert on the label when it may actually contribute to more efficient combustion?  Do you have any other theories?

I would really believe that the scientists creating such a mix understood what they were doing. So I'd assume that inert really meant inert relative to the purpose.

Maybe the inert materials slow down the combustion process as opposed to participating in it so you get a longer burn time as opposed to a "flash in the pan" so to speak that might burst the cartridge and blow some of the material away before it has a chance to ignite.


text:
Sulfur atomic absorption
Would the atomic absorption from FAA for Sulfur be from 3s2 -> 3p4 or 3p4 -> 4s0?
I think according to the electronic partition function, there would be zero electrons in the 4s0 state at the temperatures for ICP
N = pe^(-E/kT), where p is the # of electrons in the state.
Perhaps I'm thinking about this wrong.

It is for the transition $\ce{3s^2 3p^3 4s -> 3s^2 3p^4 }$ of neutral sulfur. http://articles.adsabs.harvard.edu//full/1994A%26A...286..344R/0000344.000.html

How is that for neutral sulfur? Neutral sulfur starts with 3s23p4

I whittled it down. The gist is that the notion now in my answer isn't sufficient to really define the electronic state since it doesn't give details for the spin coupling.

I think that's showing the emission. So I guess if it emits from 4s it must absorb from 3p to 4s, right? Why would 3s to 3p absorption not occur?

Not necessarily. The excitation could be from the ground state to some other yet higher excited state. The higher state then decays first to state with 4s electron then decays to ground state from there.

dah... we're talking AAS here. So yes the tube would emit that line and the free atom of sulfur would absorb it at that same wavelength.

Ok but in simple terms it's essentially absorbing from 3p to 4s, right?

Yes. In coarse terms it is a transition (emission and absorption) between 3p and 4s. To completely define the transition though you need to consider all the spin coupling states.


text:
Experimenting with electrochromic prussian blue
I am attempting to preform this experiment. Chemistry isn't really my cup of tea, so I have a couple questions.
The instructions say to provide 40µA per cm2, where does that number come from?
The TIO material I'm using has a different resistance than the glass listed, will that change anything?
All the pt wire I could find was very pricy so I though I would use a graphite electrode instead, but all I can find is carbon electrodes. Is that close enough?
Here is the list of materials I have come up with, let me know if any of them are wrong.

HCI
Potassium Ferricyanide
Iron III
KCI
ITO Plastic
Carbon Flat Electrode This was much larger than I thought. Will the size be fine, or should I find a smaller electrode?

Thank you! I'm really excited about attempting this, hopefully I can pull it off.

You can't really make an electrode out of just graphite. It would fall apart. To make a carbon paste electrode (i.e. a carbon electrode) you mix graphite with a binder. https://en.wikipedia.org/wiki/Carbon_paste_electrode

Ah. So when it says you can substitute a graphite electrode it really means carbon.

A fine point here is that there are a lot of allotropes of carbon. Diamond and graphite the best known. Any sort of "carbon electrode" would conduct electricity and that is all you care about. An "electrode" made from diamond and a binder would still be carbon but it would be useless since it wouldn't conduct electricity. https://en.wikipedia.org/wiki/Allotropes_of_carbon

Carbon and Graphite are the same thing in respect to electrodes, though I understand the frustration of their ambiguous language.

You pose a series of questions. Let me attempt to answer them all:

The instructions say to provide 40µA per cm2, where does that number
  come from?

This comes from an article of Itaya and coworkers where they studied the electrochemical synthesis of Prussian Blue. (http://pubs.acs.org/doi/abs/10.1021/ja00382a006) In fact; the website in your question refers to an article in the Journal of Chemical Education (http://pubs.acs.org/doi/abs/10.1021/ed075p881) which in turn refers to the JACS article above.
From my understanding, this number should be seen as an experimental condition. It is given so that the experiment is reproducible.

The TIO material I'm using has a different resistance than the glass
  listed, will that change anything?

The material you are using has roughly twice the resistivity as compared to the one used in the experiment. In principle, this would not pose too much of a problem. You can tune your electric system using the variable resistor. Just make sure you stick to 40µA per cm2.

All the pt wire I could find was very pricy so I though I would use a
  graphite electrode instead, but all I can find is carbon electrodes.
  Is that close enough?

This has already been answered by A.K. The carbon flat electrode should work fine. Though I am not sure whether you can use it repetitively (in multiple experiments).
As a boost of confidence. The procedure you are using is both reported in JACS as well as in J. Chem. Ed.; it should work. If you are conducting this experiment at your university or high school, perhaps someone can give you a copy of the articles to which the website refers. The experimental procedures in these articles can be really helpful in preparing your experiment. Best of luck!


text:
Ostwald Law of dilution makes no sense
According to wikipedia
the Ostwald's dilution law says that for a weak acid
 $K_{d}={\cfrac {[A^{+}][B^{-}]}{[AB]}}={\frac {\alpha ^{2}}{1-\alpha }}\cdot c_{0}$
when I read the derivation of it ,I knew something was wrong
specially this line :

If α is the fraction of dissociated electrolyte, then αc0  is the concentration of each ionic species

here is how i see it :
$AB\rightleftharpoons A^{+}+B^{-}$
each Molecule of AB gets separated into A and B when put into water
so we can make a small table
|                     |     AB    |   A    |   B     |
| Before dissociation |   1 Mol   | 0 Mol  | 0 Mol   |
| Concentration Before| 1/V Mol/L = C acid Mol/L | 0 Mol/L| 0 Mol/L |

| After dissociation  | (1-X) Mol | X Mol | X Mol |
| Concentration After | (1-X)/V Mol/L | X/V Mol/L | X/V Mol/L |

V = Volume of the Acid AND Water,
After dissociation is an equilibrium between AB and (A,B), so we can apply the law of mass action
$$
K_{dilute}=\frac{\frac{\alpha}{v}\cdot \frac{\alpha}{v}}{\frac{\left(1-\alpha\right)}{v}}
$$
and then
$$
K_{dilute}=\frac{x^2}{v\cdot \left(1-x\right)}
$$
and as we referenced the AB Concentration Before dissociation with $c_{acid}$ = 1/V , then
$$
K_{dilute}=\frac{x^2\cdot c_{acid}}{\left(1-x\right)}
$$
and from here the non-sense starts,
the law says that X is a very tiny value so (1-x = 1) making x = 0 but forgetting that this way $x^2$ is even a tinier value and would also would also = 0,then $k_b$ = 0 , but in my opinion the law should be
$$
\alpha ^2+\frac{K_{dilute}}{C_{acid}}\cdot \alpha -\frac{K_{dilute}}{C_{acid}}=0
$$
to get $\alpha$, and to calculate the concentration:
$$
\left[A^+\right]=\left[B^-\right]=\frac{K_{dilute}\ \cdot \left(1-\alpha \right)}{\alpha }
$$
Any explanation would be appreciated

You made $K_b$ = 0. That is where you were wrong. Taking approximation of Neglecting $\alpha$ with respect to 1 only gives us advantage of not solving the quadratic equation.
From there you can write $\alpha = √(K_b/C)$ here you can see $\alpha$ comes inversely proportional to $√C_o$ or we can say directly proportional to $√V$. From here you can deduce if $V$ approaches infinity $\alpha$ approaches 1. And that's the law.

my problem with this is that this is not mathematically accurate, setting a value to 0 means deleting a value, if 1 -  x = 1 because x was too small with respect to 1,then ka = 0 because x^2 is too small with respect to ka

That's why its just an approximation for small values of $\alpha$. The value of $\alpha$ is that much small that we neglect it in front of 1 since it is much much less than 1. But we cannot say value of K equal to zero. It could be of order of -5 or less than that but it is not zero. That's How this approximation works. If you are still not satisfied yet take a weak acid pH calculation problem and do by two methods. First by neglecting $\alpha$ with respect to one and other by solving quadratic. You will see the results are almost same in both cases

I'd appreciate it if you don't add stuff like "hi", "thanks", "hope this helped" etc. to your posts. They don't add anything to the answer or the question.

authors: محمد ابو العز


text:
Need to chemically heat water
I have a holding tank on rv, non potable water that I need to heat up to drain . I live in northern Alberta, so the average temperatures are -20. I don't want to wreck anything or gas myself out. I am looking for over the counter chemicals.

Ideas (1) Overnight in heated garage (2) Can you run RV long enough to get inside warm? (won't help if tank outside cabin space) (3) space heater inside RV like a kerosene heater. (again won't help if tank is on "outside" of cabin space.) (4) Since this is non-potable water might be able to dump car windshield washer fluid into system if there is any headspace in tank left. Might have to put warm fluid in, and siphon out since -20 is really cold. (too low for salt). (One gallon windshield washer fluid won't melt 20 gallons of ice...) // I'd worry that the freezing broke/cracked something.

Possible duplicate of Can you heat water with additives?

I'm assuming you only need to liquefy ice to drain a septic/grey water tank. Calcium chloride (sold as damp rid) will melt ice down to $-56^\circ C$.  Just add about 10 Pounds to the tank wait a few hours (perhaps even drive a little to slosh it) and drain.

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


text:
Why we have mass of adsorbent in Freundlich equation?
In Freundlich equation why we have $m$ the mass of adsorbent. I mean it is surface chemistry and on the surface of adsorbent the adsorption takes place so why we have mass of adsorbent in that equation.
Let us understand my problem by an example.
In my book it is given that when we add charcoal to a solution of acetic acid in water then the acidity of the solution is decreased because charcoal has adsorbed some amount of acid.
Now let us have two beakers with acetic acid in water with same concentration and let us have $m$ gram of charcoal. In the first beaker you add the charcoal in shape of a cube and in the second beaker you add the charcoal in the powdered form.
Since the surface area of charcoal available for adsorption in second case is more and so there will be more adsorption and this means the final concentration of the second beaker will be less as compared to the first beaker.
Please tell me why we have mass of adsorbent in Freundlich equation rather than surface area? In the example above we were having same mass of adsorbent but the extend of adsorption was different and this is the reason why I think that mass of adsorbent in useless. If there is some use of mass of adsorbent in the Freundlich equation then please tell me.
EDIT
In the above example of acetic acid we have considered two cases in one we adding charcoal in the shape of cube and in the second one we are adding charcoal in powdered form.
In both the situation adsorbate and adsorbent are same and are at same temperature so the value of $K$ and $n$ is same for both situations. Also we have taken concentration to be same in both the situation so $c$ is also same for both the situations. Mass of the adsorbate that is mass of the acetic acid is also same in both the situations.
Also take the same mass of charcoal. Now according to the Freundlich equation the concentration of acetic acid in water after adsorption has to be same in both the situations because in both the case we have same value of $K,n,c,x,m$. But this is not true since we know that more the surface area more will be the adsorption therefore in the second case when we have added the charcoal in the powdered form then it will adsorb more acetic acid as compared to the first case where we have added the charcoal in the shape of a cube.
Sir I don't have equipments to perform this experiment but I have some logic.
Statement 1 Freundlich equation is true.
Statement 2 More the surface area more will be the adsorption.
Now here I can say without any doubt that one of the above statement is clearly wrong.
I request the readers to please see the example and please try to understand  what I mean to say.

It seems from your example that you are considering charcoal to be non porous? In any case, physical adsorption depends on the physical nature of the adsorbent, and so one does not expect a powdered adsorbate to behave the same as a monolith of the same material. The Frueundlich model is expressed as mass adsorbed per mass adsorbate because that's how Freundlich decided to express it. It's not clear to me what you are asking really, as the accessible surface area can be directly related to $K$ in each system, which are physically different as you point out

@repurposer Sir thank you for you time, I have edited the question and I hope that it will be clear this time. So please have a look.

$K$ and $n$ are not the same for both situations. You may be using the same material, but the surface area matters. Thus they are not the same adsorbent


text:
boiling point and distillation relation
i need help with the fractional distillation method of a couple of substances at least more than two substances , just like air
actually what am i looking for is that why do substances in air with a low boiling point like nitrogen reach higher the fractionating column even if it looks like that substances with higher boiling point will absorb more energy and thus eventually result in greater kinetic energy (at least to me). please help . heres how i came to know that nitrogen rises higher the fractionating column
!(http://www.hazemagazine.co.uk/wp-content/uploads/2014/07/fractional-distillation-of-air.jpg)

Please attempt to use proper capitalisation and punctuation. 2) You seem to have to accounts. If you wish to have them merged, ping a mod such as @Mart.

Suppose you have a mixture of two compounds, one with a low boiling point and another with a higher boiling point in a flask.Prepare a distillation setup connected to a collector. As you start the distillation process, the lower boiling compound would start condensing (through the condensor to the collector). If you had only two compounds, you are left with the compound with the high boiling point in the flask.
In case you have a mixture left, you could distill out the another compound on increasing the temperature.

The boiling point is in relation to the strength of the intermolecular forces.  If the forces between the molecules are strong, the molecules require a greater average kinetic energy to overcome those forces sufficiently well to separate and become a gas.
The boiling points, in Kelvin, for the these gasses are:
  Gas      K
Nitrogen  77.4
Argon     84.2
Oxygen    90.2

In the distillation column the temperature increases and eventually nitrogen boils off at 77.4K, argon and oxygen are still liquid.  Then as it gets hotter the nitrogen molecules get more energy and the gas becomes less dense and rises up the column. By the time it gets to 84.2K at the bottom (where the heat is), argon boils but oxygen is still liquid.
Imagine it is 84.2K for all the gases, they all have the same average kinetic energy.  The argon atoms only just have enough kinetic energy to become a gas, the forces between them are present pulling them together.  If it loses a small amount of energy it will condense, and as a gas it is relatively dense.  The nitrogen has enough kinetic energy to easily overcome these forces and expands, rising above the argon.  The oxygen molecules have stronger intermolecular forces so even at this temperature, they do not have enough kinetic energy to overcome the forces and become a gas, they are still a liquid in the bottom.
Usually a fractional distillation column is hot at the bottom and gets cooler as you go up away from the heat source.  Things become a gas and rise until they get to the place in the column where the temperature makes them condense into a liquid again onto some collectors, but in this case you would probably pump the nitrogen out as a gas out the top and drain the liquid oxygen off the bottom.


text:
Is FeS2 both an ionic and covalent compound?
Iron (II) disulfide.  Iron pyrite.  Fool's gold to some.  But what bond does it fall under?  It could possibly be used for both?
This would give a deeper understanding into other like examples, where it is a fine line between the determination of an ionic compound or a covalent compound.

Related: http://chemistry.stackexchange.com/questions/5227/is-there-a-word-for-a-compound-that-has-both-ionic-and-covalent-bonds

http://chemistry.stackexchange.com/questions/17064/why-is-kf-the-most-ionic-compound http://chemistry.stackexchange.com/questions/32306/can-100-covalent-bonds-exist

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

Thank you! :) This clears certain things up.  You've been a great help!

Why do you feel a need to classify it under one or another? Chemistry is not about having a list of ionic compounds and another list of covalent compounds.
There are "ionic" bonds (see below) between $\ce{Fe^2+}$ and $\ce{S2^2-}$, and there is a $\ce{S-S}$ covalent bond in $\ce{S2^2-}$. That's all there is to it.
If I absolutely had to pick one, I would say it is an ionic compound. $\ce{S2^2-}$ is a covalently bonded species; $\ce{FeS2}$ is an ionic crystalline solid. The fact that it melts at $1193~^\circ\mathrm{C}$ supports this "classification".

As @Mith notes in the comments, all ionic bonds are not 100% ionic, and a lot of ionic bonds are in fact covalent to a large degree. The fact that $\ce{Fe^2+}$ is a transition metal ion (meaning it has empty 3d and 4s orbitals available for overlap) and also that sulfur is a rather polarisable ion (because of its larger size) means that there is likely to be significant covalency in the $\ce{Fe-S}$ bonds.
What does this mean? It just shows that trying to classify something as "ionic" or "covalent" is futile. $\ce{FeS2}$ is likely somewhere right in the middle of the spectrum.

Are there not more ways to go about categorizing FeS2 though?  Per se, could iron (II) disulfide instead be an iron (VI) disulfide, and then the bonds are singularly covalent?  Would it still be classified at FeS2, for it technically would be, granted, or would it acquire a different name, with the alternate properties of the transition metal in the equation?

Actually S and Fe are bonded covalently, only with slight polarisation. It's popular misunderstanding that all so called salts are ionic; most of them have stronger covalent component.

@Mith I assumed that it was mostly ionic - however given that $\ce{Fe^2+}$ is a transition metal ion and sulfur being rather polarisable I can see why there would be Fe-S covalency. Well, for the purposes of this question, I can probably call it "ionic", but I'll make a note.

@Navysubgirl the Roman numeral indicates oxidation state and not the valency of the compound - for iron(VI) disulfide to be electrically neutral, you would need two $\ce{S^3−}$ ions - I don't think that is happening. The truth is that if you have a molecular covalent compound with the formula $\ce{FeS2}$, it would extremely quickly rearrange to give you the pyrite structure, just like how if you tried to prepare an ionic lattice of $\ce{H+}$ and $\ce{O^2−}$ ions it would say no thank you and go back to being ordinary water.

@Mith (and orthocresol) Thank you.  I somewhat understand the ideas of preferential bonding, as in the case of H2O and suchlike.  Thus, as a clarification, compounds following similar characteristics and properties to FeS2 would prefer covalency to being to alternate ions, or in this case, an ion and covalent bonding between two sulfurs, yes?


text:
overall energy and the boiling point
i need help with the fractional distillation method of a couple of substances at least more than two substances , just like air actually what am i looking for is that why do substances in air with a low boiling point like nitrogen reach higher the fractionating column even if it looks like that substances with higher boiling point will absorb more energy and thus eventually result in greater kinetic energy (at least to me). please help . heres how i came to know that nitrogen rises higher the fractionating column

Distillation relies on the fact that, at the same stage (tray), all the substances are approximately at the same temperature. Then, at the same temperature, the vapour will be composed preferably by substances with a lower boiling point and the liquid composition will be dominated by higher boiling point compounds.
Very quickly, distillation works as follows. The vapour flow from one stage (i) is then richer in the light compound (with lower boiling point). That vapour is enters the immediately higher stage (i-1) making the composition richer in the light compound. The vapour from stage i-1 is even richer in the light compound. This process is repeated until you reach the top stage where the light compound can be almost pure. Likewise, the liquid of stage i is rich in the heavy compound and goes down to stage i+1. The liquid leaving stage i+1 is even richer in the heavy compound. At the bottom, you can reach and almost pure heavy compound.
Formally, the fractionation of a mixture is dictated by the relative volatility of the compounds (which for ideal solutions does not vary along the column). At low/moderate pressures and for an ideal solution, the relative volatility is simply
$$
\alpha =\dfrac{\dfrac{y_1}{x_1}}{\dfrac{y_2}{x_2}} = \dfrac{P^s_1}{P^s_2}
$$
We usually take the heaviest compound as the bottom line for comparison. In your example, at the pressure and temperature conditions of the column, the relative volatility of nitrogen/oxygen is higher than argon/oxygen. Therefore, nitrogen leaves the column at the top, argon at an intermediate stage and oxygen at the bottom.

that hopefully explains why at the same stage(tray) the substance with lesser boiling point dominates . but why do the higher melting points condense at a particular tray and the lower melting points rise higher ?
thanks in advance

I have edited my answer to respond your query

thanks , it explains everything now
is this your own opinion or have you taken it from a source
please mention if there is any
i am asking you this because i had a similar post and someone posted some diffrent explanation. i would had provided th link but the post is gone now

No, this is not my own opinion. You can take any textbook on distillation and it will say the same. I would recomend "Separation Process Principles" by Seader and Henley; "Separation Process Engineering" by Wankat

one second
i pondered over your explanation and it suggests that if we achieve the boiling points of any two of three liquids in diffrent containers , then in a specific interval of time all should have achieved the same height ? but we know that this is not the case ! please explain

I don't understand your remark. Achieve boiling points? Achieved the same height?

actually what i am saying is that liquids with diffrent boiling points if kept in diffrent containers and then  beginned to heat till they convert into gases will then the gas of the liquid which had higher boiling point travel farther or the gas with the liquid having lower boiling point ?

The gases in the distillation column boil off according to their boiling point.
   Gas          Boiling point
   Nitrogen     77.355 K (−195.795 °C)
   Argon        87.302 K (−185.848 °C)
   Oxygen       90.188 K (−182.962 °C)

The distillation column shown is setup for continuous distillation. But imagine that a pot distillation was being done. Nitrogen would boil off first, then argon, then oxygen. So the various trays in the distillation column represent different temperatures. Nitrogen boils off at the top since it has the lowest boiling point, and oxygen collects in the "pot" since it has the highest boiling point.

that answered my whole question , thanks !
actually i did not knew that the trays in distillation flask are not of the same trempreature throughout .
But what about a fractionatig column with marbles in it ? it has no trempreature diffrences

The marbles would act as the trays. There has to be a temperature difference from the bottom to the top of the column or it isn't a "fractioning column."

but the laboratory setup for the marble distillation column does not show any trempreature difference !

In a typical pot chemistry experiment you heat the liquid in the pot and the vapor carries the heat up the column.

thanks !
for a second i thought that in the marble fractionating column
you need to collect the pure gas from below the container!
hopefully i was wrong , right ?


text:
Stereoisomerism in 1,2-cyclobutanediol
For 1,2-cyclobutanediol, it's given in my book that it has 3 stereo isomers.
Here's the passage:

My questions:

How can it have stereo-isomers?
What will the meso compounds look like?

These two isomers are enantiomers, i.e. mirror images, of each other:
$\hspace{50 mm}$
This one has a plane of symmetry through the molecule and is therefore meso:
$\hspace{60 mm}$


text:
Cat litter crystals causing toxic smell upon rubbing and pouring out of the package, safe?
I am under-educated in terms of chemistry hence the question and assumptions.
Basically, for our cats, we are using what seems to be the latest scientific achievement in litter sand - the litter crystals. Might be about 2 years now, since we switched from clumping sand to crystals.
As of late, I have started to pick up a somewhat toxic, disturbing smell, that feels to me as if those could be dust particles or something. Now, since I read that the litter contains something of silica family:

Most cat litter crystals are composed of silica gel which is a combination of water, oxygen, and silica dioxide sand.

And also that the silica dust is somewhat toxic - this kind of matches, though, I may just be getting paranoid.
But yeah, when the cat rubs the litterbox after he has done his business, the smell picks up and it's heavily noticeable, really unpleasant for my and wifes nose.
Obviously, when the packages are being transported - the crystals rub/grind against each other, making them crumble and when I pour the crystals out - the crumbled parts get up in the air making the smell again.
Yeah, I'm starting to become a little concerned about the possible health hazards that the crystals could cause, therefore I'm looking for some clarification from you guys.
Are these things safe?

Not a complete answer to your question, but with respect to the silica gel, it's only toxic when its a fine powder of the type used for chromatography. The toxicity is not due to the compound itself, but in the size of the particles.  In its bulk form, silica  (aka. silica dioxide) is even taken as a homeopathic treatment (http://dailymed.nlm.nih.gov/dailymed/fda/fdaDrugXsl.cfm?setid=97903a05-fd97-420c-836f-b8d73277b683&type=display), no idea what health benefit it supposedly gives but its certified as not being toxic in any event.

(1) It is safe (2) Yes cat boxes stink so the litter generally has a perfume which the cat releases when it digs in the litter. (3) The litter needs to be a deep enough layer so that the cat urine doesn't reach the bottom of the box. (4) The whole litter box needs to be dumped occasionally rather than just replenishing it.

If the smell of scented cat litter bothers you, I would recommend switching to an unscented cat litter product. It still controls the odor quite well since it dehydrates the urine and feces. Either way, you are better off not putting off cleaning the litter box too long. The scented product masks but does not eliminate the natural odor of used cat litter. Why introduce extra chemicals into the environment that we do not actually need?

Oh, no, it's not the scent of the urine, it's clearly the sent of the crystals themselves. The box can be freshly filled with litter, I can shuffle the grains around with shovel and the toxic smell comes up. Same as when the grains are poured into a freshly cleaned box.

Not to nitpick , but a smell cannot be "toxic".  Cyanide apparently smells like almonds, or to put it differently, almonds smell like cyanide.  Trying to interpret toxicity using only smell is a fool's errand.

@NotNicolaou Since I came across potassium dichromate as a homeopath(et)ic drug I no longer consider any of them to be safe by default — except if it is merely due to the concentration. (e.g. I would gladly swallow as many C10 $\ce{K2Cr2O7}$ pills as someone wants to give me for any amount of money.)


text:
Three general questions in silanization of a glass surface
For silanization, I put a glass slide horizontally and pour the silan solution on it to cover the glass surface. Then I leave it for a specific time and then wash with water and acetone.
I have three questions:

Is this method correct or I should necessarily deep coat the glass plate into a container of silan solution (because in the 2nd method, there are larger number of available agents than can react with glass or have better accessibility to the glass surface)?
How durable is the silanized glass? i.e. after silanization, when I want to clean the glass surface from reaction products, can I rub something onto the surface to help clean it better or the treated layer of glass might be damaged? (rubbing with cleanex, etc)
For getting maximum hydrophobicity, can I use Octadecyltrichlorosilane>90% directly on glass surface (for how much time?) or I should dilute it?

Is this method correct or I should necessarily deep coat the glass plate into a container of silan solution (because in the 2nd method, there are larger number of available agents than can react with glass or have better accessibility to the glass surface)?

Your method is fine, anything method that covers the surface with solution will work.

How durable is the silanized glass? i.e. after silanization, when I want to clean the glass surface from reaction products, can I rub something onto the surface to help clean it better or the treated layer of glass might be damaged? (rubbing with cleanex, etc)

In my experience hydrophobic coatings are durable, but will wear due the the small surface thickness.  My recommendation is to use rinsing only, if you can and use abrasive cleaning when you have to.

For getting maximum hydrophobicity, can I use Octadecyltrichlorosilane>90% directly on glass surface (for how much time?) or I should dilute it?

I find that hydrophobic solutions typically make  very effective coatings nearly instantaneously.  You can use it directly which should only require a few seconds maybe a whole minute to work.  If you want to conserve octadecyltrichlorosilane you can dilute it. I have had 5% solutions also work nearly instantaneously, but if you want the most hydrophobic surface you should wait 1-5 minutes.

I understood that even a shallow silan is enough AND maximum 5 min is required. 1) If I spend more time (60 min) for silanization, will I receive a higher hydrophobicity than spending just 5 min ( I need to know the APPROX. MINIMUM time that I should wait to make SURE that the solution has affected the surface COMPLETELY? 2) If I change the concentration to decrease the level of hydrophobicity (e.g. using a mixture of 1 to 10 silan in kerosene), then should I still spend the same amount of time OR a dilute solution requires more time to react even though it creates a LOWER hydrophobic surface?

The marginal gain in hydrophobic nature decreases exponentially with time. For a concentrated solution, anything past a minute won't hurt but only gives immeasurable gains.  For a dilute solution even a minute is fine, you may get better results with longer time but past five minutes, the difference in properties will be a number indistinguishable from zero.

Thanks for helpful info. First, I used only Octadecyltrichlorosilane>90% and coat one glass, and then  I mixed Octa with kerosene (1 Octa+9 kerosene) too see how different hydrophobicity I will get. However, both surfaces have almost similar contact angles. Do you think that I should dilute the Octa more (e.g. using a solution of 1 to 20) or better to decrease the time (BTW, I'm trying to do both but I prefer to know what you think as well). Thanks again!

Now I get a contact angle of 108 with OctaDecylTrichloroSilan. If I deep the glass surface into H2SO4 for a few hours and then perform silanization, is it likely to get a higher contact angle?

I think you will get more consistent results.


text:
Is there such thing as too much fermentation
When you make a wine the more you age it the better the taste. But is there such a thing as too much fermentation where it makes the wine taste worse? If not is there a time where fermentation doesn't really affect the taste anymore?

Ageing is usually done after fermentation has finished, though this doesn't mean the question of whether you can ferment too much is invalid.

Not every wine gets better with age. Even those that do would eventually start getting worse. And of course there is such thing as too much fermentation, for a variety of reasons.

Aging wine is a science.  You Can' just leave a bottle in your cellar to be opened on the 18nth birthday of your firstborn. The cork will dry, opening its pores for air and the wine will oxidize. The popular method to avoid this is to turn the bottles at least once a year (or more frequently, I won't pretend to be an expert), to moisture the cork to prevent oxygen influx.

Continued fermentation of wine produces vinegar, which is not necessarily a bad thing. However, further fermentation may encourage growth of "vinegar eels" (harmless worms, actually, but not very appetizing).
Vinegar itself can be metabolized by microorganisms. According to the Compendium of the Microbiological Spoilage of Foods and Beverages, "several molds, yeast and bacteria are capable of spoiling vinegar... [such as the appropriately named] Lactobacillus acetotolerans."


text:
Is this structure make-able?
I was wondering if it would be possible to make this chemical?

Any structure, is, in theory, make able. The real question is whether anyone would want to waste a sufficient amount of time to make it.
The structure you've drawn doesn't seem so unreasonable. With enough time planning a synthesis, and enough hours of graduate student time I see no reason to object

Well, many many molecules wouldn't  be possible even according to current theories.

And since you’re KC, you know all about Maitotoxin … wait, whoops, got your username wrong there ^^'

I do hope you're not putting maitotoxin into the category of wasting time Jan. It's a Nobel pursuit ;) plus it has so many rings to shade with clashing colours


text:
Why is diene more reactive than alkene?
The chemical reduction of 1,3-butadiene with sodium in alcohol gives mainly the 1,4-addition product that is but-2-ene.
Under these condition the isolated double bond are not reduced. This suggest that dienes are much more reactive than simple alkenes.
Why is diene more reactive than alkene even though the conjugated double bond of diene (in the example of excerpts given above) is stabilized with resonance?
How is 1,4-addition product the major product?

It is not only the diene itself which is stabilised by resonance, but on electrophilic addition the carbocation formed is also stabilised by resonance if the diene is conjugated. Thus, forming the intermediate for a conjugated diene is easier than forming the intermediate for the alkene, even though the starting compound itself is more stable in case of conjugated dienes and therefore conjugated dienes react faster.

Also can you tell how 1,4-addition product is the major product?

The 1,4 addition product is the major product at high temperature, as the actual stability of the product comes into picture rather than the kinetics, which dominates at lower temperature.The 1,4 adduct is thermodynamically stabilised by hyperconjugation.

Can you elaborate it comparing how it is major product than 1,2 addition product?

The formed carbocation tells you only about kinetics.Thermodynamic formation of 1,4 adduct is easier.Draw the structures of the 1,2 and 1,4 addition products and you will see that there are 2 alpha C-H bonds for the 1,2 product and there are 6 for 1,4 addit ion product.See http://chemwiki.ucdavis.edu/Organic_Chemistry/Conjugation/Electrophilic_Attack_on_Conjugated_Dienes for more details

There is a long and elaborate answer by Philipp which will tell you a lot about why — possibly more then you need to know. It boils down to reactiviy meaning easier accessable orbitals meaning a lower LUMO or a higher HOMO in energy terms. Well, going from ethene to butadiene, you go from two π to four and thus you can imagine two additional orbitals to be sandwiched in the middle between the (all)bonding and (all)antibonding ones. This is also shown including an energy scale in Philipp’s answer:

If you liked this answer, please also go and upvote Philipp’s.

Conjugated dienes have enhanced stability due to resonance. So they are less reactive when compared to alkenes in general. But many reactions proceed through high-energy cation or free radical intermediates; in these cases the resonance stabilization of the intermediate allyl species makes conjugated dienes much more reactive than alkenes.

As to your other question, a Michael addition is a 1-4 addition, where a nucleophile attacks the β carbon, and produces the thermodynamically favored product, which happens at high temperatures. On the other hand, a 1-2 reaction (on the carbonyl) gives the kinetic product, and is obtained at low temperatures. At higher temperatures, it is not only important that the product forms, but it is also important as to whether the formed product will be stable under the conditions.


text:
What are the dangers of melting HDPE and silicone? Is it even possible?
I've searched the internet and couldn't find a specific answer.
There doesn't seem to be a single recycling company that takes empty silicone tubes (also know as mastic tubes) and recycles them. As a result, here in the UK it goes to landfill. Such a shame and doesn't seem right!
The tube is made from HDPE but is contaminated with the silicone inside it, usually used for sealing windows or bathrooms.
There is a chemical solution available that will strip out the silicone by dissolving it but that doesn't make commercial sense for two reasons;

The cost and time involved
You are left with a chemical to dispose of

Nobody wants these tubes and it has begun to annoy me!
I'm wondering if anyone could advise of the possibility / dangers of simply melting the empty tubes which result in a plastic mix of HDPE and silicone?
Would it work? Could it work?

It's not really possible to answer your question since there are many kinds of silicone. For instance, some contain biocides and some others don't. And without even taking into account the commercial stickers or the inks on it. Concerning specifically HDPE plastic, there is a question and an answer about it.


text:
Do electron donating groups make a molecule nucleophilic?
Do electron donating groups make a molecule nucleophilic?
One group, the alkyl group in this example, is electron donating in $\ce{CH3CH2O-}$. Will the molecule be nucleophlic?

The molecule is already nucleophilic by virtue of the negative charge and lone pair on oxygen.

Do electron donating groups make a molecule nucleophilic?

Sometimes.
Sure, electron donating groups can make a molecule more nucleophilic by virtue of increasing the negative charge density on the nucleophilic atom.
However, nucleophilicity is a kinetic property, and having a lot of electron donating groups attached can significantly affect the ability of the nucleophile to approach the electrophile. This reduces nucleophilicity. An example would be tert-butoxide. Yes, oxygen with a negative charge is generally nucleophilic, and yes, the tert-butyl group is an inductive donor, but tert-butoxide is not very nucleophilic because of steric hinderance.

One group, the alkyl group in this example, is electron donating in $\ce{CH3CH2O-}$. Will the molecule be nucleophlic?

Maybe. Nucleophilicity also depends on the solvent. A charged ion such as ethoxide ion would be well-solvated in a polar solvent, such as water, and less well-solvated in a non-polar solvent, such as hexane.
Solvation impacts reaction kinetics, and in turn, impacts nucleophilicity. Consider the halides for example. In polar protic solvents, iodide ion is the strongest nucleophile. Protic solvents are not able to solvate iodide ion as well as they can solvate the fluoride ion, which has a much higher negative charge density. In polar aprotic solvents, however, fluoride ion is generally a stronger nucleophile than iodide ion because fluoride ion is not as strongly solvated and it also has a great negative charge density.
In sum, nucleophilicity is not an inherent property. It depends on a variety of factors - both external and internal to whatever molecule or ion that you are considering.

Yes electron donating group increases nucleophillic nature either by $+I$ (Inductive effect) or  by $+M$ (Mesomeric Effect). They do so by increasing the electron-density over nucleoplillic atom. In $CH_3CH_2O^-$ the alkyl group has its $+I$ effect which Increases nucleophillic nature.
Hyperconjugation also increases nucleophillic nature by $+H$ effect. According to Baker Nathan Hyperconjugation dominates Inductive effect. Therefore $CH_3-$ will Increase more nucleophicity than $CH_3CH_2-$ because of presence of 3$\alpha H$ rather than 2$\alpha$.
The general order of effects is :-
$+M > +H > +I$
Hope it Helps...!


text:
Different possible Lewis structures of peroxymonosulfate anion
I was trying to draw the Lewis structure of $\ce{HSO5-}$ (peroxymonosulfate), so I came up the structure on the left that satisfies the octet rule. Then, I saw that the correct structure is the one on the right. Why is the structure on the right favored over the one on the left? They both have the same formal charges. So shouldn't the one that satisfies the octet rule be favored over the one with an extended octet?
$\hspace{15 mm}$

You can modify the structure on the right to obey the octet rule creating the one which is closest to the real picture …

The structure on the left is less favored because of the presence of too many peroxide bonds. You must have come across a few peroxides like $\ce{H_2O_2}$. Have you ever noticed why we handle these compounds with great care, like storing them away from sunlight in a dark bottle? It is because the peroxide bond is quite weak when compared to other bonds. So, more the number of less energetic bonds, less stable is the compund
$\ce{S=O}$ has a bond energy of around $\pu{128 kcal/mol}$, but $\ce{O-O}$ has a measly bond energy of $\pu{35 kcal/mol}$. So, in future, if you are asked to draw Lewis structure for compounds, avoid using multiple peroxide linkages. There can be a maximum of one* in your structure.

Some compounds with many $\ce{O-O}$ bonds exist. But in general, such compounds are highly reactive in nature. There could be exceptions as well, since it is not possible to give a single set of rules to determine stability for all compounds.

The second part of your great answer could benefit if you gave the name of this compound or a link to it, so interested readers could further read about it.

So shouldn't the one that satisfies the octet rule be favored over the one with an extended octet?

That's only true when you are comparing resonance structures. Resonance structures only involve movement of electrons and not atoms.
It's not true for different Lewis structures, which are the ones you drew. There are lots of compounds which are perfectly stable but do not (in principle) satisfy the octet rule.
I would have mentioned the extreme instability of your multiple $\ce{O-O}$ single bonds in your alternative proposed structure, but the other answer beat me to it.


text:
Thermodynamics entropy and temperature
In thermodynamics, whenever heat is released from the system to the surroundings, temperature of surroundings remains unchanged but entropy increases. But entropy is a function of temperature. How is it possible?

can you mention the source which says that says "whenever heat is released from the system to the sorrounding , temperature of sorrounding remains unchanged but entropy increases"
you have got it wrong whenever trempreature passes from someting more hotter to something less hotter there is a increasein trempreature of the less hotter substance as well as the trempreature of the less hotter substance

In scenarios like the one you describe, the surroundings are typically modeled as an infinite reservoir, with the capacity to supply or absorb heat with negligible change in temperature.  If you treated it as a large finite reservoir instead, and evaluated its entropy change based on $\Delta S=mc\ln ((T_0+\Delta T)/T_0)$ and its heat absorbed by $Q=mc\Delta T$ (where m is the mass of the reservoir and c is its heat capacity), you would find that, in the mathematical limit of "mc" becoming infinite (at constant Q), the entropy change would approach $\Delta S=Q/T_0$.

The entropy function is of the form $S(T,p,n) = n\, s(T,p)$, where $s(T,P)$ is the entropy per mole.
The surrounding is always much, much larger than the system. Hence, for the surrounding, $n$ is an extremely large number. Then, $S(T,p,n)$ can change by a finite amount with little change in $s(T,p)$, in which case $T$ and $p$ also stay more or less the same.

Practically speaking, although the external temperature may not appear to increase (e.g. the room doesn't get noticeably hotter) the energy is transferred to the kinetic energy of the surroundings and the average energy of the surroundings increases fractionally.  This increases the entropy of the surroundings.  Depending on the the circumstances that energy can be used to do work, i.e. increasing the volume of the universe which also corresponds to an increase in entropy.

Start with entropy function $S=S(T,V,N)$. Then
$$dS = \frac{\partial S}{\partial T}dT + \frac{\partial S}{\partial V}dV + \frac{\partial S}{\partial N}dN$$
Since entropy is an extensive quantity and temperature is an intensive quantity, it follows that the coefficient $(\partial S/\partial T)$ has to be extensive as well. The bigger is the system (e.g. the environment) the bigger is this coefficient and then smaller is the variation of temperature. If volume and composition are held constant and if the change in entropy is of order $dS \sim 10$ in arbitrary units and $(\partial S/\partial T) \sim 10^3$, then $dT \sim 10^{-2}$.


text:
Stability of alkenes by hyper conjugation
I have learnt that for carbocations, electrons are delocalized from the $\ce{C-H}$ $\sigma$-bond to the empty p-orbital of carbon.
Moreover, alkenes are stabilized by the delocalization of electron from $\ce{C-H}$ $\sigma$-bond to the empty anti bonding $\pi$ molecular orbital.
But, don't electrons in anti bonding molecular orbitals lead to instability? So, how does hyper conjugation lead to stability of alkenes?
I believe this question has already been asked before, but I couldn't find any satisfactory and/or relevant answers.

You are confusing the mixing of orbitals with the promotion of electrons from one orbital to another. You're thinking of this:
$\hspace{50 mm}$
I don't blame you, because the phrasing "delocalisation into an antibonding orbital" implies that, but that is not actually what happens.
In what actually happens, the two original orbitals are said to mix. They cease to exist. They overlap to give two new MOs, one which is even lower in energy and one which is higher in energy, like in this diagram:
$\hspace{45 mm}$
The "new MO 1" has characteristics of both the original $\ce{C-H}$ $\sigma$ as well as the original $\ce{C=C}$ $\pi^*$, which is why we loosely say "donation into the $\pi^*$ orbital". However, don't take it too literally. The phrase means that two orbitals are mixed. It does not mean that electrons are promoted from one orbital to another.

But still, we are DELOCALIZING electron density into an anti bonding orbital.So, shouldn't that be unstable?

Wait, which orbital is "lower in energy" ?


text:
Abnormal deviation of CO2 from ideal gas behaviour
Why does carbon dioxide show larger abnormal deviation from ideal gas behaviour than other gases. Why does carbon dioxide first show negative deviation and then positive?

What does showing negative deviation of positive deviation mean (pressure, temperature, volume, z factor??)?

When the compressibility factor z for CO2 is expressed in terms of reduced pressure and reduced temperature (i.e., using the law of corresponding states), the PVT behavior of CO2 is not very different from that of other gases.  The reduced temperature and reduced pressure are defined as the actual pressure and temperature normalized in terms of the critical pressure and critical temperature, respectively.  The real question is, "what causes the critical temperature and critical pressure of CO2 to be what they are?"

An ideal gas is the assumption that:

you have a large number of particles in the gas with a Boltzmann distribution
the size of the particles is negligible, and
there are no forces acting between the particles except for elastic collisions

In a gas with very weak forces, such as hydrogen which has only van der Waals forces, these assumptions are quite reasonable.  In a gas with strong intermolecular forces, such as ammonia which can form hydrogen-bonds as well as reglar dipole-dipole interactions, these assumptions no longer make as much sense and the gas deviates from 'ideal' behaviour.
Molecules of CO$_{2}$ do not have an overall dipole so it only has van der Waals forces acting on it, but it is much larger and has more electrons than hydrogen molecules so the van der Waals forces are stronger.  When you begin to compress the gas the molecules are compressed within range where these forces attract the molecules closer together and the volume is smaller than expected (your negative deviation from ideal gas).
Eventually there comes a point where this extra attraction is overcome by repulsive forces, because the particles of all gases do have size and can't be compressed infinitely. At the point the gas occupies a larger volume than you would expect for the pressure (the positive deviation from the ideal gas shown by all gases at high pressure).