text:
Counting body centre as lattice point when the basis consists of 2 atoms
An article in my book gives the following informatiom :-

I understood part a) and b) but faced problems with c).
It says that here the basis consists of 2 atoms which doesnt seem to be very convincing. It would have been clear if it said something like 'a molecule' or 'an ion'. How can it contain 2 particles at once?
Then it says that the smaller particle is at the body centre but we won't count it as a lattice point thus making it a simple cubic rather than a body centered cubic. Can't a particle be only placed at a lattice point ? If yes, why we hadn't we counted it here?

Please cite the chapters, helps others as well. It's NCERT, no?

It's a private institution book ,coaching/tution book to be precise. Citing the source won't do any good since it is only available for the students part of it.

The book is rather sloppily written, I'd say. To begin with, a basis can't consist of two, or one, or any number of atoms. Basis is a mathematical concept. It doesn't consist of atoms; instead, it consists of vectors. Then again, what's that with particles placed outside of lattice points? Of course it may happen; you saw an example as early as (b), where big atoms are placed in lattice points, but the small ones are not.

In maths, yes, a basis is a set of vectors. However in crystallography a basis usually means the set of atoms that are associated with each lattice point and their positions. See e.g. https://ocw.mit.edu/courses/earth-atmospheric-and-planetary-sciences/12-108-structure-of-earth-materials-fall-2004/lecture-notes/lec7.pdf

I'm not an expert here, hopefully somebody who really does know their stuff can answer this definitively. But anyway here goes ...
From http://www.physics-in-a-nutshell.com/article/4/lattice-basis-and-crystal "A lattice is in general defined as a discrete but infinite regular arrangement of points (lattice sites) in a vector space". Notice how this says nothing about atoms, molecules or ions; it is purely a regular layout of points in space. We define the atoms and their positions via the basis . This is simply a list of species types and their positions relative to an arbitrary origin. We then associate the basis with the lattice points, that is we use the lattice points as the origin for the basis, and as we have a crystal (and so translational symmetry) every lattice point is associated with the same basis. Thus to define a crystal structure

We first define a lattice - a set of regularly laid out points
We then define the basis - a list of atom types and positions and offsets
We then associate an identical basis with every lattice point

Thus for a given lattice we can have any number of bases, and each basis can contain any number of atoms.
Now we can simplify things a little bit by noting that we can always shift the origin of the basis so that one of the atoms is at the origin, and thus in the crystal a lattice point coincides with one of the atoms in the basis. This is what has happened in your case c). Now note that for us here atoms are always point particles, so rather than large or small I'm going to use blue and red - I think giving the atom a size may be confusing here as here atoms only have a position and a type. So

We have a simple cubic lattice, so the lattice points (which have nothing to do with atoms) are on a cubic layout throughout all space
For the basis (i.e. the atoms) we have chosen to place a blue one at the origin, and a red one offset along the diagonal
We then associate the basis with the lattice to make the crystal, the result being a blue atom at every lattice point (because the basis has a large atom at its origin), and a red atom offset diagonally from the lattice point, coincidentally in this case half way to the next lattice point due solely to the choice of basis, not the lattice

Also note that the basis may indeed be the positions and types of atoms in a molecule. We would then form a crystal of that molecule. But bonding is not important in this context, all we are interested in is the layout of the atoms in space, so for the moment we can forget all about chemical bonding and think purely in terms of atoms.

I understood your answer and I think it's very much correct but what you said raises another doubt in my mind. Could you please check out https://chemistry.stackexchange.com/questions/94456/body-centered-cubic-lattice-vs-primitive-cubic-unit-cell

How can it contain 2 particles at once?

[ we are discussing crystal structures here - - of course crystals can contain two different atom - e.g. sodium and chlorine - common salt ]

Then it says that the smaller particle is at the body centre but we won't count it as a lattice point thus making it a simple cubic rather than a body centered cubic. Can't a particle be only placed at a lattice point ? If yes, why we hadn't we counted it here?

[- a particle can be placed anywhere in space - and there can be more than one sort of particle-it is up to us to categorize it mathematically and topologically ... and the text is trying to explain the three different catagories
what they are saying is that in a crystal of two different atoms , if
the second atom is  at the centre of the cube being repeated in the
crystal - then you can ignore it in terms of geometrical symmetry
and crystal lattice maths categorization . ]

What I feel is that the basis consists of the two particles(blue and red) and they are placed in such a manner that the blue one is present on the lattice point(corner one) and the orange point happens to be on the body centre position. And then just like case b) where we took lattice point on the center of the blue one, we will do the same here therefore its still a simple cubic but the way the basis was placed at the lattice point, it makes it appear like a body centered cubic.


text:
Isothermal compression at critical temperature
Suppose I am compressing a gas isothermally at its critical temperature. Initially, it has a large volume. I start compressing it and reach the critical point. At this point, the gas will start to form liquid and both of these phases cannot be distinguished, right? Now if I further compress it isothermally at its critical temperature so that the pressure reaches beyond critical pressure, what will my gas look like? Will it be a liquid or a gas or something in between, I don't know. Is there a video where I can see something of this sort? Just curious

https://chemistry.stackexchange.com/questions/37088/why-must-both-the-critical-temperature-and-pressure-be-exceeded-to-achieve-the-s

https://youtu.be/RmaJVxafesU?t=242
carbon dioxide has a critical point at 31 degrees celsius.

source: Stack Exchange

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

sort: votes


text:
Chirality in coordination octahedral compounds

Isn't this molecule achiral? I mean if I flip the mirror image on right by 180 degrees(like we do to determine chirality in tetrahedral carbons) it would be superimposeable? But apparently it is chiral. How am I flipping wrong?

EDIT
I finally got the hang of it (I think). Are the images below correct?

This is a $\pu {3D}$ structure. So use of model would be very helpful to see the difference.

http://www.chemtube3d.com/TM-Ru(en)3enantiomers.html it shows here but I still cannot understand :(

@yashasvigrover can't you attempt to rotate them in the link and see that they can't align.

Check edit please

I know it is optically inactive I just needed to know where I was wrong?Also it does not address the superimposeable mirror age aspect of the question which essentially the crux of the question? @mithoron

"I mean if I flip the mirror image on right by 180 degrees" when checking chirality, you don't flip the mirror image! You just check the superimposition of the mirror image and the original molecule. The mirror image was already obtained by flipping the original molecule, a second flip will always make it identical.

No it won't.You can rotate the mirror image as you want because if it is a separate molecule,it won't superimpose.Take your hand for example,it won't superimpose no matter what you do

This interactive image of $\ce{[Fe(en)3]^{3+}}$ might help you understand: https://chemapps.stolaf.edu/jmol/jmol.php?model=N1CCN[Fe3+]123(NCCN2)NCCN3

See following image:
Look at the purple loop (en) on upper left hand side structure. Now, if you rotate the whole molecule ${90^o}$ clockwise around an axis going through top $\ce {NH2}$- $\ce {Ni}$-bottom  $\ce {NH2}$, you will get that loop overlapping on similar loop on the mirror image. But, now look what happens to the other two loops? They are not overlapping on other two loops on mirror image. For example, the purple loop on bottom of left hand side structure will move to back side of the same side (it would be ${180^o}$ far from similar loop in the mirror image). Thus this molecule would not superimpose with its mirror image.

When you say 180° you mean 180°in perpendicular plane right?

See edit please

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


text:
Role of solvents in ozonolysis and oz0nolysis of alkynes with water
My text book says that when an ozonide (formed from an alkene) is treated with $\ce{Zn/CH3COOH}$, it cleaves to give carbonyl Compounds.
1) Does $\ce{ Zn/CH3COOH}$ mean Zinc in acetic acid (or) zinc or acetic acid. I at first thought that it should be zinc in acetic acid, but then I found a separate mechanism for acetic acid alone, (given below) which made me confused. So, which one is it?

2) If the above mechanism is correct, then why doesn't the oxygen-oxygen bond (being unstable and weak) cleave in the step I've encircled?

Acetic acid may facilitate the collapse of the ozonide to a carbonyl cmpd. and a zwitterion. Reduction of the zwitterion with acetate would also form peroxyacetic acid that would be have to be reduced by Zn. Zinc can effect the same reduction of the ozonide directly. You may wish to see: http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/ozonolysis/Oz.html

You want to edit your title as teh question is about alkenes not alkynes

@user55119 So, reduction by zinc directly would follow the same mechanism as  dimethyl sulphide? This would mean that Zn would donate its electrons to the ozonide forming Zn+ and then be stabilized by the O- released from the ozonide, resulting in the formation of ZnO, correct?

You bet! The ozonide is at the same oxidation level as the sum of the zwitterion and the carbonyl cmpd. You need to reduce the ozonide (or the methanol addition product of the zwitterion if methanol is the solvent) by two electrons to obtain two carbonyl cmpds. E.g., the ozonide of 2-butene is at the oxidation level of latent acetaldehyde and acetic acid. The role of the reductant is two get "acetic acid" to acetaldehyde.

Ozonolysis:
Ozone, an allotrope of oxygen, is highly reactive molecule that can undergo a 1,3-dipolar addition across alkenes, forming a 1,2,3-trioxolanes. This compound is unstable and undergoes a cycloreversion to form a carbonyl compound and a carbonyl oxide. These resultant compounds (the carbonyl compound and the carbonyl oxide) then proceed through a second 1,3-dipolar addition to form a 1,2,4-trioxolane, commonly known as an ozonide. To complete the cleavage of the alkene, the ozonolysis (implies that ozone causes to break (-lysis)) is followed by a work-up designed to break apart the ozonide.  The work-up, which can be oxidative or reductive, determines the oxidation state of the products. For example, cis- or trans-alkenes gives two corresponding aldehydes with reductive work-up while same alkene gives two corresponding carboxylic with oxidative work-up.
The most common reductive work up procedure uses dimethyl sulphide (DMS) in the
presence of methanol. Methanol reacts with the carbonyl oxide to form a hydroperoxy acetal, and this is in turn reduced by the dimethyl sulphide to give relevant carbonyl compound and dimethyl sulfoxide. It is also possible for dimethyl sulphide to reduce the ozonide without the presence of methanol as well, although reation would proceed slowly (A new and convenient method for converting olefins to aldehydes: Tetrahedron Letters, 1966, 7(36), 4273–4278). Other reductive methods: $\ce {Zn/AcOH}$, triphenylphosphine ($\ce {PPh3}$), or thiourea ($\ce {NH2C(=S)NH2}$).
Note: Recently, there is a  publications claiming an efficient and convenient method for reductive workup of the ozonolysis reaction using sodium hydrosulfite. However, the final product could be either aldehyde or acid, thus the method is not selective.
Oxidation: The direct conversion of terminal and 1,2-disubstituted alkenes to carboxylic acids is described as oxidative ozonolysis. This can be easily achieved through treatment of initial ozonolysis products with hydrogen peroxide ($\ce {H2O2}$) or other oxidants.
You may also find answers to your Q3 and Q4 from the reference given above.


text:
Jahn Teller Effect for d4 low spin configurations
Why does Z-in take place in $d^{4}$ low spin configuration whereas in Z-out one can get all the 4 electrons in $d_{xz}, d_{yz}$ which would lead to lower energy

Can anyone tell me why it has been downvoted?

Dunno, but elaborating should do the post some good.

First off, I don't think all 4 electrons will be occupying the $d_{xz}$ and $d_{yz}$ orbitals in z-out distortion. If you do that, you'll be changing spin multiplicity on distortion. That doesn't happen. Rather, all the (originally) $t_{2g}$ levels would fill up, leaving you with a paired up electronic configuration in either $d_{xz}$ or $d_{yz}$ for z-out and a spin paired $d_{xy}$ for z-in.
Now, calculate the relative stabilization such that:
a. z-out:  $d_{xz}$ and $d_{yz}$ are stabilized by 1/3$\delta$ and $d_{xy}$ is destabilized by 2/3$\delta$.
b. z-in:  $d_{xz}$ and $d_{yz}$ are destabilized by 1/3$\delta$ and $d_{xy}$ is stabilized by 2/3$\delta$.
Your electronic configurations would then read ($e_g)^3$($b_{2g})^1$ for z-out, ($b_{2g})^2$($e_g)^2$ for z-in. Now calculate the relative energies. You will find that z-in is favoured by a factor of 1/3$\delta$.


text:
Does burning graphite give off carbon monoxide?
Put $\pu{3.7V}$ through a graphite rod, I know graphite doesn't burn but it did smoke. So I automatically assumed it to have gave off $\ce{CO}$, so I've aired out the house $\ce{CO}$ Detector, but it did not go off. I know they only detect above $\pu{30PPM}$, but does burning heating graphite rods give off $\ce{CO}$?

It could do - but I doubt you had enough carbon in the rod for any significant risk . Hence no detector signal.
( carbon monoxide is normally formed where oxygen is restricted in the combustion process )


text:
Why does formaldehyde exist primarily as the gem-diol in aqueous solution?
From Wikipedia,

Methanediol is the product of the hydration of formaldehyde $\ce{H2C=O}$, and predominates in water solution: the equilibrium constant being about $10^3$ and in a 5% by weight solution of formaldehyde in water, 80% is in the methanediol form.

As far as I know, carbonyl compounds usually don't exist in their gem-diol form, unless there's a stabilizing effect (like H-bonding, release of angle strain, inductive effect etc.), which we observe in ninhydrin, chloral, hexafluoroacetone, etc. (relevant image) But such effects don't seem to be the case for formaldehyde.
My question: What is the stabilizing factor in the case of hydration of formaldehyde?

Possible duplicate of What makes C=O more stable that C(OH)₂

In acetone the carbonyl form is more stable than the gem-diol, whereas with formaldehyde the situation is reversed and the gem-diol form is preferred. Overall, the equilibrium shifts by 6 powers of 10 between these two compounds.

It takes $\pu{\approx1.4 kcal/mol}$ to shift an equilibrium by 1 power of 10. Therefore in the acetone-formaldehyde comparison there is a stabilization change of $\pu{\approx8.4 kcal/mol}$.
It is difficult to see how changing the substituents in the gem-diol could have such a large effect, after all the gem-diol carbon is a saturated  $\mathrm{sp^3}$ carbon, no p-orbitals are avaliable for resonance interaction. Considering steric effects as proposed in the other answer, methyl groups and hydroxyl groups are not particularly large, nor is the gem-diol carbon especially congested. Hence it is difficult to see how steric factors could provide the necessary $\pu{8 kcal/mol}$ change in stabilization. For reference, Cis-di-t-butylethylene only has $\pu{9.3 kcal/mol}$ of strain energy.(1)
On the other hand we might expect the substituents on the carbonyl carbon (an unsaturated $\mathrm{sp^2}$ carbon with a p-orbital available for interaction) to have a significant effect on stability. Just as alkyl substituents stabilize a carbon-carbon double bond through both an electron-releasing inductive effect (2) and a hyperconjugative effect, we would expect the same effects to operate with a carbon-oxygen double bond.  In fact, we might anticipate that the alkyl stabilization effect will be even greater with the carbonyl double bond since the carbonyl carbon is more electron deficient than the carbon in a carbon-carbon double bond.
If we compare the heats of combustion for the isomers acetone ($\pu{-427.9 kcal/mol}$) and propanal ($\pu{-434.1 kcal/mol}$), we see that attaching a second alkyl group to the carbonyl carbon stabilizes the system by $\pu{6.2 kcal/mol}$.(3) Of course, adding the "first" alkyl group will also provide significant stabilization. Hence, it is clear that adding two alkyl groups to formaldehyde will provide enough electronic stabilization (inductive and hyperconjugative) to shift the equilibrium by $10^6$ from the enol to the keto form.

see here
the alkyl group has an $\mathrm{sp^3}$ carbon while the double bond carbon is $\mathrm{sp^2}$ hybridized; since the latter is more electronegative, electron density will flow inductively (through sigma bonds) from the alkyl group to the $\mathrm{sp^2}$ carbon.
data from K. Peter C. Vollhardt, Neil E. Schore, "Organic Chemistry, Fourth Edition: Structure and Function" p. 125; see here

The reason is mainly a lack of steric hindrance due to the hydrogen atoms. The substituents on the carbonyl in formaldehyde - hydrogen, offers negligibly small steric hindrance in shifting from a $\ce{sp^2}$ hybridization on the carbonyl to a $\ce{sp^3}$ hybridization of the gem-diol, the $\ce{C-H-C}$ angle decreasing from 120 degrees to ~109.5 degrees.
The other examples you mentioned are because of electronic effects (chloral hydrate, hexafluoroacetone), hydrogen bonding (ninhydrin) or angle strain release (cyclopropanone). Read the relevant portion from Clayden/Warren's textbook for a sound comprehension!

And how "negligibly small" is supposed to be non-negligible? You're negating yourself.

Where did I even mention non-negligible? o_o

Your saying reason is steric and then that the steric effect is negligible (which is true). Talking about inductive is enough here.

Ah, okay. Well, most books classify anything that has to do with size effects of substituents as steric effects. Using the word steric does NOT automatically imply that I am talking about steric hindrance. It can just as well be a lack of steric hindrance, as is the case here. Hope this clarifies your doubt. :)

Oh, yeah, that makes sense. Edit it too make it more clear and I'll be able to undownvote.

But alkyl groups stabilize double bonds, how do you know the effect is not caused predominantly by electronic factors?

You think that the carbonyl C=O is a normal double bond? Why on earth would alkyl groups stabilize a carbonyl? The only sort of electronic factor that is at play when you are discussing hydration of a carbonyl is something that makes the carbonyl carbon more electrophilic (hence, more susceptible to a nucleophilic attack by the lone pairs based on oxygen in the HOMO of water). Case in point: chloral, hexafluroacetone.

The carbonyl carbon is electron deficient and will therefore be stabilized (destabilized) by electron releasing (withdrawing) groups.


text:
What is the difference between chiral molecule and chiral carbon?
I read in my book, that substituted allenes do not have chiral carbons, but actually the molecule is chiral, so they show optical isomerism. But what is the difference between them( chiral carbon and chiral molecule)?

The difference? They are about as different as two things can be. Chiral carbon is an sp3 carbon with four different substituents, and chiral molecule is a molecule that can be left or right, that is, does not coincide with its mirror image. Indeed, it is possible to have chiral molecules without chiral carbon or even without any carbon whatsoever.

It is even possible to have a molecule containing several chiral carbon atoms, but which is not chiral itself. Such a molecule is called a meso compound.

@Masrat : the post linked by Mithoron is a must read. Please do read it carefully, it will clarify many things. Very briefly, though, remember that chirality is a property of the whole molecule. And if your book really uses the term chiral 'carbon', rather than 'stereocenter' or 'stereogenic carbon', then it's probably not a very good book. Allenes can have stereogenic 'axes', but the main concept doesn't change: the whole allene molecule is chiral if it is not superimposable to its own mirror image. BTW, optical isomerism is a possible (not always observable/ed) consequence of chirality.

Thanks for explaining this, I will read the Mithoron link for sure. Thanks again! @user6376297

In addition to the excellent Mithoron link, heed Lord Kelvin.


text:
Where to get electrodes for Microbial fuel cell
I am currently working on a soil-based MFC. So far I have built myself some carbon electrodes by putting glue on a stainless steel mesh as a current collector and sprinkling activated carbon power over it. I have yet to try this but after some research people with far more advanced materials have only gained the amount of energy I am looking for.
Could anyone point me to a source of carbon electrodes, preferably flat ones, since I only found brush electrodes and those won't help my cause. Pricetag does not matter too much for now, I an curious to see what options I have.

I'm voting to close this question as off-topic because it's likely to attract spam. Sorry.

Proton Exchange Membrane (PEM) is designed for polymer electrolyte membrane fuel cells (PEMFC). Read this article first to decide: PEM fuel cell electrodes (review): Journal of Power Sources, 2004, 130(1-2), 61–76 (https://doi.org/10.1016/j.jpowsour.2003.12.055). The review is available online, courtesy of Univ. of Connecticut: http://www.engr.uconn.edu/~jmfent/electrode%20review.pdf
PEM is available to buy online (albeit expensive): http://www.fuelcellstore.com/fuel-cell-components/membranes
http://www.fuelcellstore.com/index.php?route=product/search&search=Proton%20Exchange%20Membrane
Also available following carbon cloth electrodes, if you want cheap replace ment. One example is:
0.03 mg/cm² 20% Platinum on Vulcan - Carbon Cloth Electrode (Product Code: 1610000): A low-cost electrode utilizing a woven carbon cloth substrate and a 20% Platinum on Vulcan Carbon support catalyst.  This low loading is designed for cost sensitive applications and is suitable for electrochemical systems needing low amounts of Platinum to support the reaction.  This electrode is perfect for Hydrogen/Air Fuel Cell applications where the extra cost of the higher loading GDE may not be justified.
You can order on line,
http://www.fuelcellstore.com/003-ptc-cloth-electrode

My pleasure! Good luck on your project.

The growing interest in organic photovoltaics and the potential for a future mass production urges to find alternatives to the presently employed materials that are well performing but not convenient from the point of view of large area fabrication. Electrodes based on non abundant elements, or that constitute an issue for devices (i) long term stability, (ii) mechanical robustness and (iii) continuous fabrication process, shall be possibly soon replaced by earth abundant, easy processable and sustainable materials. Many groups have recently started to devote their research work on materials not containing metals or metal oxides, and the time has come to summarise the progress that has been reached so far.
https://www.sciencedirect.com/science/article/pii/S0927024811007185>

https://www.intechopen.com/books/carbon-nanotubes-current-progress-of-their-polymer-composites/carbon-nanotubes-supported-conducting-polymer-electrode-for-supercapacitor
Carbon Nanotubes Supported Conducting Polymer Electrode
highly stable chemically


text:
Why is the number 2 needed for 2-methylpropanal?
I think there is just one possible position for the methyl group to 'fit in' to propanal, however, my textbook and wiki suggests the correct names should be '2-methylpropanal'. Why is the '2' needed?

Related (concerning 2-methylpropanoic acid): IUPAC rules on omission of locants in carboxylic acid

Well I suppose you are correct in a way- but it does help us see the molecule quicker in our heads...
it should be written methyl-2-propan-1-al or methyl-2-propan-2-al
otherwise it would be a butan-1-al  or butan-2-al
let me explain further>
the new IUPAC nomenclature  advises the use of numbers instead of the iso epithet- in organic chemistryIn old naming -as with propanol and iso-propanol - so with propanal
Isopropyl alcohol
Preferred IUPAC name
Propan-2-ol
-
Isopropyl alcohol (IUPAC name propan-2-ol; commonly called isopropanol) is a compound with the chemical formula C3H8O. -It is a structural isomer of 1-propanol.
wiki  Chemical Names: ISOBUTYRALDEHYDE; 2-Methylpropanal; 78-84-2; Isobutanal; Isobutylaldehyde; Propanal, 2-methyl- More... Molecular Formula: C4H8O or (CH3)2CHCHO.


text:
Can halogens exist in their elemental state in nature?
Question 7 from the 2015 Australian Chemistry Olympiad was:

Which list below gives non-metals that can be found in their elemental form in nature?
(a) neon, phosphorus, fluorine
(b) helium, hydrogen, iodine
(c) helium, carbon, fluorine
(d) nitrogen, chlorine, sulfur
(e) oxygen, chlorine, phosphorus

I was somewhat confused when all the options included reactive elements such as sulfur, phosphorus and the halogens. I know neon is a noble gas and is present in minute quantities in the atmosphere, and nitrogen and oxygen gases are present as diatomics.
However, where are hydrogen, phosphorus, sulfur, fluorine, chlorine and iodine found in their elemental forms in nature? (Answer is B)

It is not a particularly easy question. But some examples should be more obvious than others. Native sulphur is found in some volcanos; a large part of the solar system is made of hydrogen. And the least reactive halogen is iodine. I'd have guessed 2 by elimination.

I agree with Matt Black's comment on the question. The question is rather tricky and the answer can only be determined by the process of elimination.

Have to assume we're talking about STP.
"In nature" is ambiguous, so assume the question means "on earth."
It is also somewhat ambiguous what "elemental form" means. Assume that it means that element is only combined with itself since only the noble gases exist as monoatomic species.

The noble gases (helium and neon) and exist as monoatomic gasses at STP and can be found in the atmosphere.
Oxygen and nitrogen are diatomic gases and both exist in earth's atmosphere.
Sulfur is found in volcanic vents.
Carbon is found both as graphite and diamonds on the earth.
So, the list becomes:

neon, phosphorus, fluorine
helium, hydrogen, iodine
helium, carbon, fluorine
nitrogen, chlorine, sulfur
oxygen, chlorine, phosphorus

Oxygen is ubiquitous on earth and phosphorus is so reactive with oxygen that elemental phosphorous just won't be found on earth. However, elemental phosphorous might be found on the moon for example.

neon, phosphorus, fluorine
helium, hydrogen, iodine
helium, carbon, fluorine
nitrogen, chlorine, sulfur
oxygen, chlorine, phosphorus

So now we are left with the halogens -- fluorine, chlorine, and iodine. Iodine is the least reactive, so "elemental iodine" is the most probable to be found in nature.
Thus the answer must be #2. However this is really an "educated guess," not a "fact" since the question is so ambiguous.

As @MattBlack and MaxW state, iodine is the only halogen likely to be in elemental form in any quantity; see https://agupubs.onlinelibrary.wiley.com/doi/pdf/10.1029/JZ068i013p03989 . That said, the amount is minuscule.

According to Volcanic emissions of molecular chlorine Geophysical Research Abstracts Vol. 14, EGU2012-12386:

When eruption ceases, the cooling volcanic cone can work as a giant
  chemical reactor to convert residual HCl into
  molecular chlorine (Cl2). Up to 60 ppmv (180 mg/m3) of Cl2 together with 30-77 ppmv HCl were measured in
  gas emissions from the Tolbachik scoria cones, Kamchatka, which are still hot after the 1975-1976 eruption.

...

The estimated total discharge of Cl2 from the Tolbachik cones is ca. 12-16 t/yr.

There is also a corresponding full paper in Geochimica et Cosmochimica Acta Volume 87, 15 June 2012, Pages 210-226.
So, although the Olympiad wanted you to answer "b", the above necessitates that "d" is a correct answer.
Note also that there was a highly similar, but more sensible, question (#7), in the 2014 U.S. NATIONAL CHEMISTRY OLYMPIAD:

Which list gives nonmetals that are found in their
  elemental forms in nature?
(A) neon, phosphorus, fluorine
  (B) helium, hydrogen, iodine
  (C) nitrogen, oxygen, sulfur
  (D) oxygen, chlorine, phosphorus

with "(C)" being correct.
Here, the American Chemical Society is on record indicating "(B) helium, hydrogen, iodine" is incorrect, directly contradicting the Australians.

Halogenes usually do not occur in their elemental state in nature.
However, under certain circumstances there are exceptions from this rule.
E.g.

Fluorine ($\ce{F2}$) exists in calcium fluorite (antozonite) that has been exposed to radioactive radiation.
Bromine ($\ce{Br2}$) can occur in atmospheric plumes of degassing volcanoes.
Iodine ($\ce{I2}$) is produced by some kelp in response to oxidative stress.

Chlorine ($\ce{Cl2}$) is also said to occur in volcanic gases, but I could not find a reference that supports this claim.

Iodine is volatilized from the sea as methyliodide, elemental iodine,
  and possibly other inorganic compounds...

Reference
Thus iodine is naturally occurring as the element.  That settles the original question.

hydrogen, phosphorus, sulphur, fluorine, chlorine and iodine will exist in their elemental states, as this simply means that they aren't bonded to any other element; their structure is only formed from that particular element. this means that they ill exist as H2,P4, S8, F2, Cl2 and I2.
but the answer is two as we typically won't find fluorine, chlorine and phosphorus as their elemental molecules in nature as they are too reactive.


text:
Derivation of the Temperature Dependence of the Boltzmann Distribution
In lecture notes, Atkin's Physical Chemistry, etc. the Boltzmann Distribution is derived by first claiming that the system takes on properties of its weightiest configuration. The weight is therefore maximised, utilising Lagrange multipliers.
\begin{align}
\nabla \ln{W} - \alpha \ \nabla \sum_i n_i - \beta \ \nabla \sum_i n_i\epsilon_i &= 0
\\
\to \frac{\partial{\ln{W}}}{\partial{n_i}} - \alpha - \beta \epsilon_i &= 0
\end{align}
The Stirling Approximation transforms the derivative to something we can work with, and the Boltzmann distribution is eventually derived.
\begin{align}
\frac{\partial \ln{W}}{\partial n_i} &= -\ln{\frac{n_i}{N}} \\
&\vdots \\
\frac{n_i}{N} &= \frac{e^{-\beta\epsilon_i}}{\displaystyle \sum_i e^{-\beta\epsilon_i}} \hfill
\end{align}
The derivation is left there with no mention of the value of $\beta$, though it is later claimed $\beta = \frac{1}{kT}$. How is the value of $\beta$ ascertained?

Isn't that the definition of temperature?

Hardly an expert on this, but afaik you can derive $\beta = 1/kT$ if you define $S = k\ln W$, but I've also seen some cases where one defines $\beta = 1/kT$ and derives $S = k\ln W$ from it. I guess either way it is a postulate.

The relationship between $\beta$ and $T$ comes from experimental observations like $E=\frac{3}{2}RT$, i.e. if you add $E$ amount of energy to a mole of monatomic gas, then you raise the temperature by $\frac{3}{2}R$ degrees.
$E = N_A\langle \varepsilon \rangle = N_A \sum_i p_i \varepsilon_i =  \frac{N_A}{q} \sum_i \varepsilon_i e^{-\beta\varepsilon_i}$
$\varepsilon_i e^{-\beta \varepsilon_i}  = - \frac{d }{d \beta} e^{-\beta \varepsilon_i} \to \langle \varepsilon \rangle = - \frac{1}{q} \frac{\partial q}{\partial\beta}$
$q = \left( \frac{2\pi m}{h^2\beta} \right)^\frac{3}{2}V \to E = \frac{3N_A}{2\beta} \to \beta = \frac{N_A}{RT} $

What OP cited is a derivation of the Boltzmann distribution starting from a micro-canonical ensemble (which has a fixed energy). Here, $\beta$ arises as a Lagrange multiplier, and one needs to relate it to the thermodynamic tenperature $T = 1/(\partial S / \partial E)$. On the other hand, you started from a partition function, which means you are working with a canonical ensemble that has a fixed temperature. In this case, you have $\beta=1/kT$ from the very beginning, and you don't need to compare against the internal energy of the monatomic ideal gas to derive what is already there.

What OP cited is the derivation of the Boltzmann distribution starting from the micro-canonical ensemble, characterized by a fixed energy and a fixed number of constituent particles. In mathematical terms, the problem statement is as follows:

Maximize the function $\ln W(\{n_i\})$ under the constraints $\sum_i n_i \epsilon_i = E$ and $\sum_i n_i = N$.

The solution is a function of $E$ and $N$, so let's define

$\ln \Omega(E, N)$ := (constrained maximum of $\ln W$ as a function of $E$ and $N$),
$\tilde{n}_i (E, N)$ := (value of $n_i$ at the constrained maximum).

Notice that $\ln\Omega = \ln W(\{\tilde{n}_i\})$. Also, this quantity is what appears in Boltzmann's entropy formula $S=k\ln\Omega$ because $\Omega$ is the number of available states for given values of $E$ and $N$. This observation enables us to relate $\ln\Omega$ to the thermodynamic temperature:
\begin{equation}
\frac{\partial \ln \Omega}{\partial E} = \frac{1}{k}\frac{\partial S}{\partial E} = \frac{1}{kT}.
\end{equation}
Now, OP's question can be rephrased as follows:

When deriving the Boltzmann distribution starting from the micro-canonical ensemble, $\beta$ is introduced as a Lagrange multiplier associated with the constraint $\sum_i n_i \epsilon_i = E$. How can we relate $\beta$ to the thermodynamic temperature, i.e., establish that $\beta = \frac{\partial \ln \Omega}{\partial E}$?

The underlying principle is really the chain rule in multivariate calculus:
\begin{equation}
\begin{split}
\frac{\partial \ln \Omega}{\partial E} &= \sum_i \frac{\partial \ln W}{\partial n_i}\bigg|_{n_i = \tilde{n}_i} \frac{\partial \tilde{n}_i}{\partial E}\\
&= \sum_i (\alpha + \beta\epsilon_i)\frac{\partial \tilde{n}_i}{\partial E}\\
&=  \alpha\frac{\partial}{\partial E}\Big(\sum_i \tilde{n}_{i}\Big) + \beta\frac{\partial}{\partial E}\Big(\sum_i \tilde{n}_{i}\epsilon_i\Big)\\
&= \alpha \frac{\partial N}{\partial E} + \beta \frac{\partial E}{\partial E}\\
&= \beta.
\end{split}
\end{equation}

I'm not sure I follow where $\beta = \frac{\partial \ln{\Omega}}{\partial E}$ is coming from. Could you clarify that?

If you are talking about where I rephrased your question, the reason is that $\Omega (E, N)$ is what enters Boltzmann's relation $S = k \ln \Omega$, where $S$ is a function of $E$ and $N$. Then, one should have $\partial \ln\Omega /\partial E = k^{-1}\partial S / \partial E = 1/kT$.


text:
Number of hyperconjugating structures due to a C=N system

How many hyperconjugating strutures are there for the following compound?

I thought the answer to be 8 (as there are eight alpha hydrogens) but the correct answer given is zero. Why isn't the answer eight here, as expected from a $\ce{C=C}$ alkene? Does the explanation invoke the very definition of hyperconjugation?

this question came in my test today. Answer given is 8

a chemistry faculty told me it is  0.(that too an experienced one)


text:
Chemical potential of mixtures of ideal gases and the Gibbs-Duhem equation
As is well known for a constant composition system, $\mu$ (symbolizing chemical potential) is equal to the molar Gibbs energy. Using
$$
\mathrm{d}G = V\mathrm{d}p-S\mathrm{d}T,
$$
we can write
$$
\mu = \mu^{0} + RT\ln P.
$$
But for mixtures,
$$
\mathrm{d}G = V\mathrm{d}p - S\mathrm{d}t + \mu_{\ce{A}} \mathrm{d}n_{\ce{A}} + \mu_{\ce{B}} \mathrm{d}n_{\ce{B}}
$$
Can the original formula for chemical potential in the case of mixtures?
Moreover, using the Gibbs-Duhem equation,
$$
n_{\ce{A}} \mathrm{d}\mu_{\ce{A}} + n_{\ce{B}} \mathrm{d}\mu_{\ce{B}} = 0,
$$
which means that $\mu_{\ce{A}}$ changes upon adding B to the mixture, even when the partial pressure of A is constant. This contradicts the formula.

Who says that the partial pressure of A is constant when you add B to the mixture at constant pressure?

I meant adding b at constant volume of the mixture

The chemical potential of a the two species in your ideal gas mixture are $$\mu_A=\mu^0_A(T)+RT\ln\left(P\frac{n_A}{(n_A+n_B)}\right)$$
$$\mu_B=\mu^0_B(T)+RT\ln\left(P\frac{n_B}{(n_A+n_B)}\right)$$
and the total free energy of the mixture is:
$$G=n_A\mu_A+n_B\mu_B$$So, the chemical potential of A depends on the number of moles of B and the chemical potential of B depends on the number of moles of A.  Do these equations satisfy the Gibbs-Duhem equation?  Try them out and see.

I also had one more aspect in my question as to how can we arrive at this equation for mixtures. because deriving it is straightforward for one component systems but not for mixtures.

I don't quite understand what you are asking here.  Are you asking about your third equation, and where it comes from?

no i mean the logarithmic dependence of chemical potential... how do you reach to that in the case of mixtures..

I still need a little more clarification.  Are you asking how I know that the chemical potential of a species in an ideal gas mixture is determined by the log of its partial pressure in the mixture?  Are you familiar with the change in entropy between an ideal gas mixture at T and P, and the same amounts of pure species, each at P and T?

I just need a simple derivation of the explicit expression of u if possible from the expression of G=VdP-SdT+u(a)dn(a)+u(b)dn(b)..

if youre asking about the gibbs energy of mixing and the associated entropy change then yes..

You're familiar with Gibbs' theorem that the contribution of each species in an ideal gas mixture to an overall extensive property of a mixture is equal to that of the pure species as the same temperature and at its partial pressure in the mixture?

Suppose I asked you to derive the equation for the entropy change between the following two ideal gas states:  (1)  na moles of pure A at T and P, and nb moles of pure B at T and P.  (2) (na+nb) moles of a mixture of A and B at T and P.  Could you do it?  What would be your answer?

thanks a lot for your help @chester miller. But it would be great if you could provide me a link to the proof or intuition behind the Gibbs Theorem.because it explains quite a lot of things...

I don't have an online link, but I can provide a book reference:  Introduction to Chemical Engineering Thermodynamics by Smith and McCabe.  Here is a question for you:  If I have an ideal gas mixture containing species A and an ideal semi-permeable membrane that is permeable only to A, what pressure of pure A do I have to apply on the other side of the membrane so that the mixture is in equilibrium with the pure A?

that is precisely the point i was hitting on. even semipermeable membranes should maintain same potential rather than same pressure on both sides. but in light of the gibbs theorem it all makes sense as same pressure implies same potential. anyways thanks for your help sorry for the late reply.


text:
Determining the depression in freezing point experimentally
I was reading about Beckmann's method for determining the depression in freezing point:

First, the solvent is placed inside the cooling chamber directly and the approximate freezing point is determined. Next, the freezing tube is again kept directly in the cooling chamber and then placed inside the air jacket.
I have the following questions:

Why not directly place the freezing tube into the air jacket and determine the freezing point? Why do we have to determine the approximate freezing point?
How is the latent heat released ?
Lastly, I think that the last line should say, "The temperature of the cooling bath should be 4° to 5°....." for all of this to make sense.

I'm not expert on this apparatus. But I'll try to give you logical reasons for your questions:
1) Assume you do not know the freezing point of your solvent. Even if you know it, that reference value is under different conditions such as atmospheric pressure (suppose you are in Denver of Mexico City, for instance). Thus, you need to determine the approximate freezing point of your solvent before you do careful reading. It's exactly like you are trying to determine the melting point of given compound. First time you set up MelTemp in high heating range to get an idea of melting range. Then you do careful heating with second capillary containing your solid to get melting point within $\mathrm {\pm 1 ^\circ C}$. The same approach applies here: First, direct heating to get approximate freezing point, then warm it to melt back, cool again directly to close to that temperature, wipe the outside of the tube, and this time, air jacketed it before do slow heating and get accurate reading. Whole this process is just to  save your time and energy. That air jacket make long time to make your freezing tube with solvent cool (prime example: Dewar flask, with vacuum though). And, imagine if you have to stir that whole time with one stroke per second?
2) The latent heat of fusion release when a liquid undergoes liquid to solid transformation at its freezing point. If cooling is not uniform, the releasing energy may cause some solid near by to melt again (remember, at freezing or melting point, $\ce{liquid <=> solid}$). That's why frequent stirring is important, to give uniform cooling for this case.
3) You are correct about changing the instruction (c), but it should have been, "The temperature of the cooling bath should not be more than 4° to 5° below the freezing point of the liquid" to prevent the supercooling. But it's also worth noting that the set point of cooling bath temperature is subjected to change by each addition of the solute, because of the freezing point depression of the solvent.


text:
what keeps an identical atoms of an element together?
I have learned that the atoms of different elements are held together by covalent bonding and ionic bonding but in the case same element which force keeps the atoms together i.e in  sodium which is the force that keep sodium atoms in position?which is the force acting between the atoms of sodium?

Types of bonding are not limited to covalent and ionic.

You'll really want to wait till you're taught metallic bonding.

Possible duplicate of Metallic bonding and covalent boding

well atoms of the same element will typically either, forma metallic lattice, a giant covalent structure ( a macro-molecule), or from a simple molecular structure. but all the bonds will either be metallic or covalent bonds. if it forms a simple covalent structure, there will be van der Waal's forces acting between the molecules to keep it as a solid or liquid.


text:
Body centered cubic lattice vs Primitive cubic unit cell
I was reading about Bravias lattices(cubic crystal system) when got confused about an article in my book.
Please refer to Counting body center as lattice point when basis consists of 2 atoms.
The answer there makes me think about these two cases:-

In figure 1, the crystal is produced by conbining the body centered lattice frame with a single atom motif.
In figure 2, the crystal lattice is produced by combining a primitive cubic lattice frame with a motif made of 2 atoms.
Both things lead to that same result so would we calk the final crystal formed a a body centered cubic lattice or a primitive cubic lattice?
P.S : The book I am using(a private institution book, tution book to be precise) has unclear definitions creating a lot of problem so pardon me if you had found me saying weird things.

This is about the difference between the conventional unit cell and the asymmetric cell. Have a look at https://www.crystallographiccourseware.com/unitcells/UnitCellMain.html for a very quick and colourful summary, and if I get time I'll try to write an answer


text:
Is there a relationship between boiling point and van der Waals constants
Is there a relation of the boiling point of a gas with its van der Waals constants, $a$ & $b$?
I know if constant $a$ is greater, then the boiling point will also be greater, because of more attraction between the molecules, but does the boiling point depend on $b$ too? If two gases have the same $a$ but different $b$, can I tell which of the two gases will have a higher boiling point or nearly same boiling point?

Boiling point is a fact of nature; van der Waals constants aren't. That said, there surely is a relation, albeit a complicated one (it involves the roots of a cubic equation and stuff).

Without information on the liquid side of the phase transition, this is not answerable.

There will be a relationship, but as the comments on the question indicate, the relationship will not be simple. There are two complications:

The van der Waals equation is an empirical model built to fit experimental pressure/volume/temperature data for real gasses.
The boiling point of a liquid is not a single number but a line on the pressure/temperature phase diagram.

Supposed the van der Waals equation can predict phase behavior (source: 1) However, I made some graphs. The van der Waals constant data are from Wikipedia and the boiling point data looked up from the Wikipedia articles.

Constant a

The blue line is a cubic polynomial fit, while the grey shaded area is the error. With the exception of the metal outlier (mercury), this fit seems pretty good.

Constant b

The blue line is a logarithmic fit, while the grey shaded area is the error. Again, with the exception of the metal outlier (mercury), this fit seems pretty good.

Finally, I decided to plot the van der Waals constants against each other.

The van der Waals constants are fairly co-linear, so any clear relationship between one and boiling points implies a clear relationship between the other and boiling points.

Source:

Terrell L. Hill, 2012 [1960], "An Introduction to Statistical Thermodynamics" [Dover Books on Physics], Chicago, IL, USA:R.R. Donnelly (Courier/Dover), ISBN 0486130908, referenced on the Wikipedia article, but I do not have access.


text:
Significant figures conflict
So I am asked to solve this question from Silberberg:
Use Appendix C to calculate $[\mathrm{H_2S}]$, $[\mathrm{HS}^-]$, $[\mathrm{S^{2-}}]$, $[\mathrm{H_30}^+]$, $\mathrm{pH}$, $[\mathrm{OH}^-]$, $\mathrm{pOH}$, in a $0.10 \mathrm{M}$ solution of the diprotic acid hydrosulfuric acid.
In Appendix C (Equilibrium Constants for Selected Substances at $298 \mathrm{K}$),
$\mathrm{H_2S}$ has a $K_a$ of $9 \times 10^{-8}$ (which has one sig.fig.)
I answered it with one sig.fig. because $K_a$ has one sig.fig. but my teacher told me that $K_a$ is a constant (I did not understand why and he told me that the burden of proof is up to me) so it should be two sig.fig.
I checked the Answers appendix of the book, and all of the answers were in one sig.fig except $[\mathrm{H_2S}]$, which had $0.10 \mathrm{M}$.
What is the correct answer?

An exact constant has infinite significant figures.For example you can say 2 eggs,2.0eggs2.0000000eggs or 2.00 eggs and it will all mean exactly the same.
Therefore you are limited by expression 0.10 molar since it has 2 Sig. Digits,hence you should report it in 2 digits.


text:
Participation of Pi-bonds and lone pairs in resonance
Below is a quote from Klein's Organic Chemistry 3rd edition:

Here is the bottom line: Whenever an atom possesses both a π bond and a lone pair, they will not both participate in resonance

However, I am confused as a few pages prior it was shown how allylic lone pairs and pi-bonds participated in resonance:

The first curved arrow goes from the lone pair to form a π
  bond, while the second curved arrow goes from the π bond to form a lone pair:

What have I understood wrongly?

In your example there is no atom that possesses both a π bond and a lone pair at once.

In the 2-D resonance structure shown, no distinction is made between the pair of electrons in the pi-framework and the two pairs of electrons in the sp2 non-bonding orbitals. The choice of the electron pair on the left of the oxygen is arbitrary. One pair of electrons on oxygen and the pair of electrons in the double bond are in the pi-framework of the canonical resonance structure that is orthogonal the plane of the structure. A 3-D resonance representation of the enolate would have clarified this point.

You have probably misunderstood what the sentence actually means.
It means that only one, either $\pi$ bond or lone pair will participate in resonance if the atom has both.

For example, in pyridine, the nitrogen has lone pair as well as is attached with a $\pi$ bond.
While resonance, its $\pi$ bond participates in resonance because it (and not the lone pair) is in the plane perpendicular to the molecule.

Observe that the lone pair remains at the place it is meaning it doesn't participate.

Conclusion: The pair of electrons present in the orbital perpendicular to the plane of molecule participates in resonance if it is present.

What's meant is that at a time for an atom only one lone pair will participate in resonance.
Oxygen over here has three lone pairs  but only one of them can participate in resonance since the other other two will be perpendicular to the $\pi$ orbitals of the carbons.


text:
Dimensional Analysis
In a dimensional analysis problem they take 6.20L x (1ft^3/28.3L) x (12in^3/1ft^3) to get 378 in^3. Why do they express the 12in^3 as 12x12x12? To me it would be expressed as 6.20/28.3*12 to get 2.6289.... ? I may be missing an easy concept here?
In addition, This relates to volume in chemistry as is why I am asking this; How does one understand or express cubic units? For example, is 12cm^3 essentially 12x12x12? But.. A rectangle could be 3x4x1 which is 12cu. But then what is confusing do you then take the 12x12x12?

The source of your confusion would be eliminated if you format your braces properly. 12cm^3=12cm*cm*cm. But, (12cm)^3=12*12*12 cm*cm*cm.

@GaurangTandon So 12 cubic centimeters is simply 12cm12cm12cm?

12 cubic centimetres is not that. Take a cube of side 1cm1cm1cm, that's one cubic centimetre. Multiply it by 12, that's 12 cubic centimeters.

12 cubic metre means $$\pu{12 \times 1 m \times 1 m \times 1 m}$$So if you wish to convert it into centimetres, it becomes $$\pu{12 \times 100 cm \times 100 cm \times 100 cm}$$ as $\pu{1 m = 100 cm}$

So what would 12 cubic centimeters be ? 12x1cmx1cmx1cm?

Yeah $\mathstrut$

How would this be represented by a shape? Is the shape irrelevant here?


text:
What is chemically different about tea for hard water?
At the supermarket they sold tea for hard water. (Hard water is water with limestone/calcium carbonate in it).

There appears to be a chemical modification for foods using this type of water.
My question is: What is chemically different about tea for hard water?
Edit: here is the list of ingredients: tea (only)

Edit 2:
- it appears hard water is more basic so a tea mix that is more acidic is required (perhaps)

A quick Google search seems to implicate that hard water tea is made by using a different blend.
"Hard water is the opposite. Tea flavour is dissolved slowly and inefficiently. Therefore, in the most general terms, a hard water tea needs to be better than a soft water tea."
"A hard water blend needs lots of quality Assams and East African teas, with plenty of “guts” and strength."
https://allabouttea.co.uk/tea-news/hard-water-tea/

@Bdrs can you brew this into an answer?

Hard water tea
By definition hard water is defined as water that has more than 0,6mg/l of dissolved CaCO3. Therefore soft water that doesn't have that much minerals has a greater dissolubility and can dissolve more tea. It can also dissolve the tea quicker. So the tea companies need to find what tea blend dissolves quickly and efficiently in hard water.

A hard water blend needs lots of quality Assams and East African teas, with plenty of “guts” and strength. It needs flavour too, of course, but this flavour will be the rich, malty type that punches through the hard water. -https://allabouttea.co.uk/tea-news/hard-water-tea/ (as Bdrs commented)

Thanks this is helpful. Is there a chemistry dimension in your answer I’ve missed?

Well, its all about solubility so there isn't much to it. Some say that theres something about acidity (so the pH is constantly 7?)


text:
Calcium carbonate and hydrochloric acid
I am trying to solve an exercise, where a block of 605.5g calcium carbonate should be completely dissolved in 30% hydrochloric acid (w/w) and the concentration of the acid should be 3% in the end. The question is: How many g of 30% HCl-acid is needed in the beginning.
$$\ce{CaCO3 + 2 HCl -> CaCl2 + H2O + CO2}$$
So for each dissolved molecule of $\ce{CaCO3}$:

2 molecules of $\ce{HCl}$ are used
1 molecule of $\ce{H2O}$ is produced
1 molecule of $\ce{CaCl2}$ is produced
1 molecule of $\ce{CO2}$ is produced, but leaves the solution

I got a result (about $\pu{1671.524g}$ 30% hydrochloric acid), which I verified this way:
Let $p$ be the number of dissolved $\ce{CaCO3}$ moles, which is about $\mathrm{605.5/100.0869 = 6.05}$
$\ce{HCl = 1671.524g * 0.3}$
$\ce{H_2O = 1671.524g * 0.7}$
$\ce{HCl_{end} = HCl - 2 * (p * M(HCl)}$)
$\ce{H2O_{end} = H2O + p * M(H2O)}$
$\ce{CaCl_{2end} = p * M(CaCl2)}$
$c_{end}$ = $\ce{\frac{HCl_{end}}{HCl_{end} + H2O_{end} + CaCl_{2end}} = 0.03}$
But my solution is claimed to be incorrect. Am I missing something here?

You need to tell us what % means in your question. Is it $\mathrm (w/w)$ or $\mathrm (w/v)$ since the solvent is $\ce {H2O}$? That make a difference as well.

@MathewMahindaratne it is w/w, I added that in the question

Okay, now what is your objective? Is it to find how many $\mathrm {g}$ of $\ce {HCl}$ you have to use? I think that is the most obvious.

@MathewMahindaratne Yes, the g of hcl

Is the correct answer $\pu {1526 g}$?

@MathewMahindaratne How did you calculate it? I get 1671.52 as result for the 30% HCl (that is, mass of acid+water). I cannot see the correct answer, I only can verify it and we need two decimals precision

Your approach is i n correct track, but your calculations probably need attention. I'll post  an answer in few minutes.

Look  at my answer below so you can see. Let us know if you still have problems. Good Luck. :-)

We'll assume the reaction loses only $\ce {CO2}$ from the system (although it is a exothermic reaction, we'll assume $\ce{H2O}$ produced would not be lost as vapors):
$$\ce{CaCO3 + 2 HCl -> CaCl2 + H2O + CO2}$$
Mass of $30\%~\ce{HCl}$ needed to react with $\pu {605.5 g}$ of $\ce{CaCO3}$ $\pu {= 605.5 g \ce{CaCO3} \times \frac{\pu{1mol}~\ce{CaCO3}}{\pu{100.09g}~ \ce{CaCO3}}\times \frac{\pu{2mol}~\ce{HCl}}{\pu{1 mol}~\ce{CaCO3}}\times \frac{\pu{36.45g}~\ce{HCl}}{\pu{1mol}~\ce{HCl}}\times \frac{\pu{100g}~\text{solution of 30%}~\ce{HCl}}{\pu{30 g}~\ce{HCl}}= 1470.04 g of 30\% \ce{HCl} solution}$.
Similarly, mass of $\ce{CO2}$ released when $\ce{HCl}$ reacted with $\pu{605.5 g}$ of $\ce{CaCO3}$ $\pu{= 605.5 g \ce{CaCO3} \times \frac{\pu{1mol}~\ce{CaCO3}}{\pu{100.09g}~\ce{CaCO3}}\times \frac{\pu{1mol}~\ce{CO2}}{\pu{1mol}~\ce{CaCO3}}\times \frac{\pu{44.0g}~\ce{CO2}}{\pu{1mol}~\ce{CO2}} = 266.18 g \ce{CO2}}$.
According to the law of conservation of mass,
$\pu {mass of reactant = mass of products}$
Assuming no solvent left the system,
$\pu {mass of reactant + solvent = mass of products + solvent}$
$\pu{mass of reactant + solvent = 1470.04 g + 605.5 g = 2075.54 g}$
$\pu{mass of products + solvent remained in the flask = mass of reactant + solvent - mass of \ce{CO2} released = 2075.54 g - 266.18 g = 1809.41 g}$
Now, suppose extra $\pu{A g}$ of $30\%~\ce{HCl}$ added to the solution, so that final concentration is $3\%~\ce {HCl}$.
$\pu {mass of \ce{HCl} in A g of 30\% solution = A g of 30\% \ce{HCl} solution \times \frac {\pu{30g}~\ce {HCl}}{\pu{100g}~ \text{of 30% $\ce{HCl}$ solution}} = 0.3A g of 30\% \ce{HCl}}$
Thus, since final concentration of solution is $\mathrm {3\%}$,
$$\pu {\frac {(0.3 A) g}{(1809.41 + A) g} = 0.03}$$
$$\pu {0.3 A = 0.03 \times 1809.41 + 0.03A}$$
Thus, $\mathrm {A} = \frac {0.03 \times 1809.41}{0.3-0.03} = \pu{201.04 g}$
Finally, the mass of $30\%~\ce{HCl}$ initially added is $\pu {(1470.04 + 201.04) g = 1671.1 g}$.

Hello,  @Mathew Mahindaratne - can you elaborate on what does "your reaction loses only CO2 from the system" means? That is gasifys and exits the solution?

@Stipe Galić: Yes, therefore, mass of $\ce{CO2}$ should be eliminated from calculations.

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


text:
Maximum (Z) to (E) ratio in the photoisomerization of azobenzene?
Considering the photoswitching of azobenzene, I was told the following argument: Considering that the photoexcited state of azobenzene has a about 50% to 50% chance of yielding either the (E)- or (Z)-isomer, the maximum (Z)/(E) ratio achievable should be 1:1.
However, I have seen UV-VIS spectra where the (Z)-isomer seems to be in excess. So I figured that the above argument must be wrong. I would analyze the situation in the following way:

In this picture, the fact that the photoexcited state has about equal probability to yield (E) or (Z), i.e. $k_{-1} = k_2$.
So wouldn't the overall (Z) to (E) ratio be given by the expression
$\frac{k_1\cdot k_2}{k_{-1}\cdot k_{-2}}$?
I would then argue that in many cases, the cis isomer exhibits lower absorption at the wavelength of excitation, meaning that $k_{-2} << k_{1}$, which would explain that in many cases, an excess of the cis isomer is obtained.
By this reasoning, the initial argument would be completely wrong, as in principle an infinite excess of the cis isomer could be obtained if $k_{-2}$ is sufficiently smaller than $k_{1}$.
Is this analysis correct or am I overlooking something?

Your question is not completely clear to me, so I am answering based on thinking that you are asking about why, after photoisomerization, the system can relax with roughly equal probability to either the E or Z minimum.
Strangely I know something about the photochemistry of azobenzene. One overarching theme is that azobenzene molecules undergo very complicated isomerization processes which are sensitive both to how you substitute them and to the solvent in which the experiment is done. Additionally, the photoisomerization and thermal isomerization processes don't necessarily share anything in common. There is a pretty in-depth interesting literature on different types of azobenzene molecules discussing the relative roles of inversion and rotation relaxation pathways, solvent effects, and so on... I just bring this up in case you are interested to read more.
The case you describe, I think has a relatively straightforward explanation actually. In ref. [1], a computational study is done on azobenzene to find which pathway, rotation or inversion, is dominanant upon photoisomerization of azobenzene. To clarify, rotation means rotation about the CNNC dihedral, and inversion means moving one of the rings along the CNC angle.
The paper shows that there is a conical intersection along the rotation pathway which is only 2 kcal/mol above the $\ce{S1}$ excited state minimum. Thus, the molecule will follow this conical intersection upon relaxation. Because the molecule is already above both states, and the molecule can essentially end up going toward either the E or Z minimum from this conical intersection, to first order, the two should end up equally populated on relaxation. Of course, from there, one would expect to move back to a Boltzmann distribution of the two states thermally.

[1]: Cembran, A., Bernardi, F., Garavelli, M., Gagliardi, L., & Orlandi, G. (2004). On the mechanism of the cis− trans isomerization in the lowest electronic states of azobenzene: S0, S1, and T1. Journal of the American Chemical Society, 126(10), 3234-3243.

So, is the maximal cis:trans ratio really 1:1? Because my argument is that the trans state exhibits a higher absorption at the excitation wavelengths compared to the cis state.

I am not saying the maximum ratio is necessarily 1:1. It is possible it could be skewed either way based on the most likely geometry when the system follows the conical intersection back to the ground state. Also, the diagram you have seems to show a higher excitation wavelength,  but the same intensities for the two states. The intensities are what can be correlated to population, not the wavelength. Are you confusing these things or am I misunderstanding?

Btw, thanks for the references and the well-thought-out text : )  Well the intensity (aka absorption) of the cis isomer is also lower compared to the trans, which isn't really well represented in the figure. Anyway, as the excitation light is monochromatic (365nm), the wavelength of maximal absorption is also important concerning the photoisomerization: If the cis state doesn't absorb at the excitation wavelength, it will also not be excited.


text:
Why is caustic soda (NaOH) used to clean beer bottles?
I see in many blogs in the internet that cleaning bottles with $\ce{NaOH}$ is very effective. I also realize that industries use caustic soda to clean bottles. Why is caustic soda very effective at cleaning bottles? What makes it so special from other solvents which made it preferable for cleaning?

Most organic chemistry lab use $\ce{KOH + i-PrOH}$ to clean dirty glassware.

Caustics such as $\ce{NaOH}$ and $\ce{KOH}$ are effective in converting insoluble fatty acids in organic fats and oils into soluble soaps in a proccess called saponification. The caustics also react with cell membranes, DNA and RNA, destroying undesired yeasts, bacteria and other microorganisms.
That said, caustics react with the glass itself, removing the surface and etching the glass. Use will weaken and damage containers.
Use caution! Caustics also destroy skin and clothing, and are particularly damaging to the eye.

Did you mean "Caustics such as $\ce{NaOH}$ and $\ce{KOH}$ are effective"  instead of "Caustics such as $\ce{NaOH}$ and $\ce{NaOH}$ are effective" ?

@ Avnish Kabaj , Thanks! So much for copy-&-paste.

Slightly etching the glass is actually a benefit if you really, really want to clean the surface of a vessel. That's why chemists often clean their lab glassware this way.

@matt_black, true,  caustics are effective cleaners. However, getting rid of the last bit of alkali (or sodium silicate that forms) can be a problem for some uses. Also, I've seen caustic reagent destroy glass bottles when left too long.

The caustics also react with  glass, cell membranes, DNA and RNA, destroying undesired yeasts, bacteria and other microorganisms.

Hi Baraka, could you elaborate on your answer?

during the cleanness of bottle  in bottle washer in industry it need strong and effective disinfectant to ensure all contamination are removed when deal with contamination its about stains, bacteria, cell membranes and others undesired material, so that caustic is strong and more effective cause it react with glass effective and also removed simple and also have the ability to destroy all sources of microorganism that its  my cause to select caustic to best plus coast


text:
How do the pressure acting on the reactants and the initial temperature of the reactants affect the enthalpy of reaction?
In an exothermic reaction that occurs under constant pressure, does the pressure acting on the reactants affect the enthalpy of reaction? Does the initial temperature of the reactants affect the enthalpy of reaction? If so, how and why?
The idea of "standard enthalpy of reaction" seems to imply that the pressure acting on the reactants and the initial temperature of the reactants do affect the enthalpy of reaction.
According to Wikipedia,

The standard enthalpy of reaction (denoted $\Delta H_r^{⊖}$) is the enthalpy change that occurs in a system when matter is transformed by a given chemical reaction, when all reactants and products are in their standard states.

"Standard states" means that the reactants are at $\pu{10^5 Pa}$ and $\pu{298.15 K}$, implying that if the reactants are not at their standard states, the enthalpy of reaction will be different.
However, I don't understand why the enthalpy of reaction would change. After all, if the same number of moles of reactants form/break bonds in the same way, shouldn't the heat released be the same, regardless of the pressure and the initial temperature?
EDIT: I recently learned about Le Chatelier's principle, which describes how changes in concentration, temperature, volume, and pressure affect a system in dynamic equilibrium. Is this principle related to my original question? Or am I heading in the wrong direction?

Comments are not for extended discussion; this conversation has been moved to chat.


text:
Is Using HSAB to Explain Michael Donors and Michael Acceptors Valid?
Basically, I've recently been thinking about using HSAB theory to rationalize the concept of michael donors and michael acceptors. Essentially, if we have a deprotonated diketone (with a lone pair on the alpha position carbon) or the Gilman reagent, it attacks the carbon in a vinylic ketone in a 1,4 addition.

However, if we used a Grignard reagent, than it attacks the carbonyl carbon in a 1,2 addition.
How can we use HSAB theory to justify this? I've figured out that the carbonyl carbon's LUMO is the pi-star orbital while the vinylic carbon center's LUMO is also the pi-star orbital except with the carbon bonded to another carbon, thus it should be higher in energy than a pi star orbital bonded with an oxygen.
I've also figured that the carbon nucleophile center on a Michael donor (such as that of the diketone) should be "softer" than the nucleophile center on a Grignard reagent's carbanion because the charge is less concentrated due to resonance and thus the HOMO of the Michael donor is not as high as the HOMO of a Grignard reagent.
However, I do not know where to go from here. Can someone help me what I'm doing wrong?

Your train of thought seems reasonable and I think it does lead to the expected conclusions. The soft nucleophiles (e.g. diketone enolate) adds to the soft Michael acceptor in 1,4 manner. The hard nucleophiles (e.g. RMgX) adds to the carbonyl group in direct fashion.


text:
Metallic character of bonds?
Why in discussions of percent character of bonds, are only ionic and covalent bondings discussed? Do bonds not have a partial metallic character, and are either metallic and ionic-covalent?

related https://chemistry.stackexchange.com/questions/6429/which-has-more-metallic-character-aluminium-or-magnesium Metallicity can be viewed as very strong delocalisation, so it's relevant only for very big molecules/networks/phases.

One could think of the Fermi surface and conduction bands as an expression of the degree of metallic bonding, where metals such as aluminum or silver have overlapping empty and filled bands, allowing for electrical conduction, while semiconductors have a small gap between filled and conduction bands, offering more resistance.
As @Mithoron states, metallic bonding is a bulk property. Metals behave differently as nanoparticles -- for example, bulk silver is reflective, but nanometer particles of it are black, as in photographs.


text:
Can gas be made to block radiation better?
Can any gas block radiation? I ask this because I would like to know if the properties of any gas element would chemically react differently with radiation from adding an electrical current.

This picture shows an Earth magnet's field deflecting plasma naturally in a vacuum.
Does any gas extend the electromagnetic shielding properties of a magnetic field?
I am interested if the lack or composition of an atmosphere's chemical properties would inhibit or complement the deflection and absorbing of cosmic radiation when coupled with a magnetic field.
Can contained pure $\ce{O2 O1 O3}$ be cycled back with electricity to replenish $\ce{O3}$ broken down by radiation while giving the gas an electromagnetic field?

Ozone doesn't absorb "radiation" in general: it absorbs specific parts of the UV spectrum (which is why a layer of it in the upper atmosphere provides useful protection for us from UV). If we need additional protection from UV we can use plenty of other chemicals (like sunscreen or UV filters on glass).

Your question seems to be about the ozone layer, but shows some misunderstanding.
First, ozone, $\ce{O3}$, absorbs some "radiation", specifically electromagnetic radiation, e.g. visible light or ultraviolet light (UV), as do many other gases. This absorption is not the same for all wavelength ("colors"), but in the UV region peaks about 250 nm, which is a good thing for us, because plants and animals have not evolved good protection from that light, and the small amount of $\ce{O3}$ in the upper atmosphere is enough to block most of that UV. Sulfur dioxide, $\ce{SO2}$, released in volcanic eruptions, is also able to absorb 250 nm UV. Molecular oxygen, $\ce{O2}$, absorbs "vacuum UV", 10-100 nm, which otherwise would be a problem.
Second, Earth is continually bombarded by another type of "radiation", particles such as cosmic rays and solar energetic particles. Rather than electromagnetic energy, these are made of particulate matter such as protons or nitrogen nuclei. All gases, in fact all matter, help block these particles. Magnetic fields such as that of the Earth help divert them around the planet, too, but there is some question whether a magnetic field is essential to blocking particulate radiation. Here at the bottom of the atmosphere there is a column of air weighing about a kilogram over each square centimeter, which is sufficient to block primary cosmic rays, though a showers of weaker secondary particles do reach the surface. There is no advantage to a conductive Faraday cage to protect the Earth.
Third, all gases conduct electricity when ionized. The majority of air is nitrogen, oxygen and argon, which produce various colors in an aurora. Though magnificent to watch, aurorae are simply evidence of energetic particles trapped in a magnetic field, ionizing the upper atmosphere. This layer of the atmosphere is called the ionosphere, and is responsible for long-distance radio communication. It's height varies from day to night and with solar storms.

A Faraday cage blocks low-frequency electromagnetic radiation, e.g. power-line 50-60 Hz, LF broadcast or UHF, but does nothing for visible light or UV -- the mesh would need to be on the order of molecular size. Might as well use a solid sheet of metal.

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


text:
Can diamond be grown on copper wires using chemical vapor deposition techniques?
I am asking this as a follow-up to another question I asked earlier. It is regarding how to make higher voltage transformers. Diamond is a good dielectric material and would be good for using as a dielectric barrier for wires wound into transformers. However, diamond is much too brittle to coat the wire and then wind, so my question is this:
Can diamond be grown on copper wires wound into a transformer shape using chemical vapor deposition techniques?

If it could be grown, would it adhere through a thermal expansion-contraction cycle??


text:
Is it possible to have negative bond orders?
In the MO theory, is it possible to have a molecule with bond order less than 0? The least bond order I have come across as of now is 0 (for di-atomic noble gases), indicating that the molecule cannot exist and the atoms get pulled apart.
Is it possible to create a scenario (at least theoretically) where we deal with a negative bond order by manipulating the system's conditions or any of that sort?

No, it's not possible, theoretically either.
If you see the Molecular orbital theory, maximum number of anti bonding electrons = number of bonding electrons.
And bond order is (Number of bonding electrons - number of anti bonding electron)/2
So, the minimum value that you can get is 0 and not negative, as anti bonding electrons can't exceed the number of bonding electrons.

There is more to it.

What more to it?

A molecular orbital is not necessarily associated with particular bond. It may spread over many atoms. In this case the number of bonding electrons of one bond ceases to be well-defined.

@IvanNeretin could you elaborate?

You'll need a different definition of bond order, like that of Huckel method or something, that's what I mean. With that definition, I see no fundamental reason to believe that negative bond orders are forbidden.

As a ground state, I don't think so. For the example of diatomic molecules, every combination of atomic orbitals forms equal numbers of bonding and anti-bonding orbitals, and the bonding orbitals are by definition lower in energy. To get to a zero bond order configuration, you have to fill all the anti-bonding and bonding orbitals, so the next orbital up would have to be dominated by higher energy atomic orbitals, and so would be primarily a bonding orbital, and then the bond order is positive again. Adding more atoms might change things, but I'd be surprised. Of course, excited states could have a negative bond order.

I don't think that bond order can be zero. Think of bond order as the number of bonds an atom is forming in the molecule. So, least, it can form no bond . There would be no physical significance of an atom forming negative number of bonds. That would be absurd.

Di atomic noble gases have a bond order of 0

Yes I completely agree. But zero here signifies that they do not form bond with each other. If they did it would be unstable. This shows that diatomic noble gases donot exist that is why their bond order is zero.

MO theory is full of seemingly absurd things.


text:
How would you separate sodium chloride and copper sulfate?
I've tried dissolving both compounds in water and evaporating the mixture, but this leaves a combination of brown and green crystals. Is there was any way to cleanly separate both compounds? Any advice would be much appreciated. :)

They are not there anymore. You can't separate them without chemically transforming at least some of the ions involved. Typically, when you separate a mixture, you need just one of the components. Which one is it?

Suppose you have a solution of sodium chloride $\ce{(NaCl)}$ and copper sulfate ($\ce{CuSO4}$), in a beaker/conical flask. First, you treat this mixture with excess potassium ferrocyanide $(\ce{K_4[Fe(CN)_6]})$. Due to this addition, a choclate brown precipitate of $\ce{Cu_2[Fe(CN)_6]}$ (cupric ferrocyanide) is formed while the $\ce{NaCl}$ remain unreacted.
$$\ce{CuSO_4 + K_4[Fe(CN)_6] -> Cu_2[Fe(CN)_6](ppt) + K_2SO_4}$$
Then you filter the solution, and you will have the precipitate separated from the solution, which now only contains $\ce{NaCl}$. Now take the cupric ferrocyanide and treat with excess $\ce{NH_4OH}$, and it will form a precipitate of cupric hydroxide $(\ce{Cu(OH)_2})$.
Again, filter the precipitate and treat the compound with a concentrated solution of sulfuric acid ($\ce{H_2SO_4}$). With this, you will get back original copper sulfate, and now, it is seperated from $\ce{NaCl}$.

Thank you for your answer. I suspected that separating this mixture would require a little more than physical separation techniques. Very informative.


text:
Osmium tetroxide production
I know osmium powder will produce osmium tetroxide when exposed to air,but what about solid piece of osmium,such as pellet or ring,finely polished to minimize surface area,will this produce significant tetroxide at 24 degrees celsius air?
How much tetroxide will 1 cm2 surface area of polished osmium produce in 24 hours when exposed to air at 24 celsius?

Osmium tetroxide will be formed at an osmium surface exposed to air. The question is whether enough is formed to be harmful. Perhaps not, but perhaps nobody has ever tried so if you want to be the first...

As one MSDS indicates: "Potential Acute Health Effects:
Extremely hazardous in case of ingestion. Very hazardous in case of inhalation. Hazardous in case of skin contact (irritant), of
eye contact (irritant). Slightly hazardous in case of skin contact (permeator). Prolonged exposure may result in skin burns and
ulcerations. Over-exposure by inhalation may cause respiratory irritation. Severe over-exposure can result in death." As @matt_black says, there will be an oxide on the surface of the metal, and it will be right on your skin.

I would like to add one sub-question,how much of the osmium oxidation at 24 C air is actualy tetroxide? There is also osmium dioxide,is the oxidatation at living room temperatures tetroxide only,dioxide only or some mixture of the two? Is the type of oxide depending on temperature? Becose if osmium is producing mainly tetroxide,that means solid piece of osmium such as pellet or ring will spontaneously disappear over long period of time becose osmium tetroxide is volatile,the oxide layer will continuously evaporate,uncover naked osmium,form & repeat till nothing is left,at 24 C seems unlikely.

A block of Osmium is quite immune to atmospheric attack, due to the low surface area of the osmium, but when heated it can form Osmium tetroxide. I wouldn't like to risk wearing an osmium ring just incase Osmium tetroxide is formed. The oxide has a pungent smell so if you do wear a ring of Osmium, at least you'll have some form of warning to take it off. So really, I wouldn't recommend it, but do it at your own risk.

I wouldn't rely on the smell. Sister wiki says, that inhalation at concentrations well below those at which a smell can be perceived can lead to pulmonary edema and subsequent death.


text:
Complete this series of reactions

Cyclohexene is treated with $\ce{RCOOOH}$ and 'A' is formed.  A is treated with $\ce{CH3MgBr}$ and 'B' is formed. On hydrolysis of B we get 'C'.  Identify A,B,C.

I identify A as cyclohex-1,2-diol. Could someone give me assistance after that please.

I suggest you read "the reaction of alkenes with peracids" first, then Grignard reaction (assuming you are in OCHEM II class).

First of all, A is not Cyclohexane-1,2-diol. Because, when you treat a peroxy acid with any alkene, first an epoxide is formed, and it is stable under slightly acidic or basic medium, as it is an ether. But if you hydrolyse the compound under highly acidic conditions then cleavage of this ether happens and the ring opens to form trans-Cyclohexane-1,2-diol. But, here no hydrolysis is being performed under any acidic conditions. So, the treatment with peroxy acid will yield only Cyclohexene-oxide. So, A) is Cyclohexene-Oxide
Now when you treat this with $\ce{CH_3MgBr}$, there is a nucleophilic attack( $S_N2$ ) by $\ce{CH_3^-}$ at one of the two carbons to which oxygen is attached, and the ring opens to form 2-methyl cyclohexanoxide Magnesium Bromide.As, in this case the substitution mostly occurs through $S_N2$, it is most probable to get a trans product.  So, In B), major is the following compound
Now, when you hydrolyse it, you will get trans-2-methylcyclohexanol as a major product in C)

A cautionary note: With Grignard reagents and other coordinating metals, beware of rearrangement of the epoxide to a carbonyl group. In this case cyclohexanone may be formed and add the Grignard reagent to form 1-methylcyclohexan-1-ol.

How can Cyclohexanone be formed ? I didn't get what you said!!

Mg++ coordinates with oxygen and opens C-O bond leaving cation on carbon. Hydride migrates 1,2 to give cyclohexanone (as enol) coordinated to Mg. Mg++ is lost. Mg++ can be MeMgBr as well. 1,1-Carbon substituted epoxides are notorious in this regard giving aldehydes as intermediates. I do think that your answer is correct as the major product. PS: The mechanism is similar to this. Substitute -H for -CO2H. https://chemistry.stackexchange.com/questions/86726/epoxide-ring-opening-via-decarboxylation/86729#86729

Oxygen can still coordinate to $Mg^{2+}$ being in that epoxide (ether) form, why does the ring need to open up for that??

I don't know the intentions of an epoxide. Just competitive rates, direct addition vs. rearrangement.


text:
Why does fluorine form stable oxides while chlorine form unstable oxides?
The fluorine however it says my text book doesn't combine well with oxygen, nitrogen and itself but when about its oxides they are quite stable as $\ce{O3F2}$  has a melting point of $\pu{363 °C}$ means its quite stable.It also says that oxides of chlorine are unstable.
Now I have  two questions:

Why does  Fluorine form stable oxide even though it's as electronegative as oxygen?
Why does  chlorine form unstable oxides?

Where did you even hear about something as unholy as O3F2 ?! Even if it does exist, it would be like most reactive thing on Earth! And by no means having such m.p.

Technically, fluorine does not form oxides.  It forms oxygen fluorides.  Fluorine tops oxygen in IUPAC's  electronegativity pecking order.

Are you sure of that melting point? This book states that O3F2 melts at 84K, or -189 C. It also reports "mild explosions" in reactions of O3F2 at 90K, which is not the hallmark of a stable compound.

Late edition:
To our surprise (most of us, evidently), the compound $\ce{O3F2}$ exist, at least in lower temperatures. The synthesis of $\ce{O3F2}$ was claimed
by two Japanese scientists during World War II [Ref. 1], but their claims have not been accepted due to lack of quantitative analysis of the compound to prove the molecular formula. After about 20 years later, its existence was confirmed by a publication in the Journal of American Chemical Society in 1959 titled Ozone Fluoride or Trioxygen Difluoride, $\ce{O3F2}$ [Ref. 2], which gave detailed analysis of the compound. The abstract of the paper states that:

The existence of ozone fluoride, $\ce{O3F2}$, has been put on a solid foundation by isolating the pure compound and analyzing it. $\ce{O3F2}$ is a deep blood-red liquid, solidifying at $\pu{83 ^\circ K}$. and decomposing at about $\mathrm116^\circ$ or higher in a clean cut reaction to $\ce{O2}$ and $\ce{O2F2}$. It is an endothermic compound and is one of the most potent oxidizers known. It is more reactive than either $\ce{F2}$, $\ce{OF2}$ or mixtures of $\ce{O2}$ and $\ce{F2}$.

The description of physical properties states that:

It is a blood-red viscous liquid which can be refluxed and distilled with only slight decomposition in the range of $96$ to $\pu{114^\circ K}$ and at a pressure of $0.1$ to $\pu{1.5 mm}$. It remains liquid at $\pu{90^\circ K}$ and can thus be easily distinguished from $\ce{O2F2}$.

Note: $\ce{O2F2}$ is an orange solid discovered in 1933, which melts at $\pu{109.7^\circ K}$ to a red liquid. Thus, at $\pu{90^\circ K}$, it should be still a solid.
The paper also stated that, alike $\ce{O2F2}$, with heat evolution, $\ce{O3F2}$ also decomposes quantitatively to $\ce{O2F2}$ and $\ce{O2}$ at about $\pu{115^\circ K}$ ($\ce{2 O3F2 -> O2 + 2 O2F2}$).
I believe this reaction might be a one reason to ban Freon and other fluorinated compounds from common use to save the ozone layer. But its claimed melting point $\pu{363 °C}$ by Pakistan Zindabad is not quite right. According to the discoverer, the compound is not existing after $\pu{115^\circ K}$. At that temperature, it converts to $\ce{O2F2}$ (and $\ce{O2}$). At about $\pu{200^\circ K}$, $\ce{O2F2}$ will further dissociate quantitatively, to $\ce{O2 + F2}$, again under heat evolution.
References:

Gmelin Handbook of Inorganic Chemistry: Fluorine - Compounds with Oxygen and Nitrogen; Suppliment Vol 4, 8th Edn., Susanne Jager, et al., Eds., Springer-Verlag: Berlin, Germany, 1986, p. 103-104 (Chapter 3. Compounds of Fluorine: Fluorine and Oxygen: 3.13. Trioxygen Difluoride, $\ce{O3F2}$).
Ozone Fluoride or Trioxygen Difluoride, $\ce{O3F2}^1$: A. D. Kirshenbaum, A. V. Grosse, J. Am. Chem. Soc., 1959, 81(6), 1277–1279 (https://pubs.acs.org/doi/abs/10.1021/ja01515a003).


text:
Textbook suggests secondary alkyl halides do not undergo SN1 reaction
The 8th edition of Organic Chemistry by Paula Bruice suggests that secondary alkyl halides do not undergo SN2 reactions. However, many other sources state the opposite. So, what is the correct answer?

Tertiary alkyl halides undergo SN1 solvolysis reactions with poor nucleophiles such as water and alcohols, but methyl halides and primary alkyl halides do not. A recent investigation of Huges and Ingold's data showed that secondary alkyl halides also do not undergo SN1 solvolysis reactions.* Thus, methyl halides, primary alkyl halides, and secondary alkyl halides undergo only SN2 reactions.

The textbook refers to Murphy, T.J. J. Chem. Ed. 2009, 86, 519-524.
I'm curious as to whether this assertion has been upheld or disproved by other members of academia.

Possible duplicate of SN2 vs SN1 in Secondary Alkyl Halides

@Mithoron That seems to be a way more generalized question. I think analysing this specific quote from Paula Bruice may be of help to future readers of this book. Also, that "Hughes and Ingold's" data seems interesting, and isn't covered in that other post. So, I think Leave Open is a better option here.

There seems to be the general understanding that it has to be one or the other, which is truish only for the most extraordinary cases. The truth lies somewhere in between and is heavily dependent on all parameters of the reaction, to name a few: solvent, temperature, pressure, nucleophile. I don't think this question should be closed, but  I would like to lower your expectations as to what kind of answer you'll get.

I'm rather curious about this too; if this assertion is true, then how did so many sources and educators manage to be wrong for so long? And if it is false, did Murphy mess up an experiment, or did he draw incorrect conclusions?

@Righter These mechanisms are models, ideal cases. They don't happen in reality. Everything depends on the conditions, too. Even for the same reaction, if the parameters are different, the kinetic will be different. In beginners education it is reasonable to utilize these approximations or model reactions. It's easier to understand. The problem is that some models are used beyond their means and propagated as the only truth. Change of these opinion is generally slow.


text:
How to test for methanol?
I wanted to make some ethanol from baker's yeast and I used sucrose. I've made it and put it in a glass jar. I left it on a warm place (near the radiator) until the yeast died and after they died I distilled the yellowish liquid using simple distillation. When I smelled the product in the receiving flask I was confused. It resembled ethanol a little. The smell was like sweet and medium-intense ethanol. I think that it might be methanol. If there are any reagents or tests other than smelling that I can use at home, please do tell.

Do you want to know whether there are impurities at all, what the impurities are, or the concentrations of the impurities? If you were careful about measuring your inputs, you may be able calculate expected refractive index and density and compare those to the actual for under $30 of equipment.

I just wanted a simple test for presence of methanol (don't need to measure the quantity).

Do you have iodine solution and sodium hydroxide? http://www.rsc.org/learn-chemistry/resource/res00000548/a-test-to-distinguish-between-ethanol-and-methanol?cmpid=CMP00003293

Yes, the NaOH is at hand and I can buy the iodine at the pharmacy. Thanks!

Sorry @ JavaScriptCoder, but I've to agree with aventurin. I'd say idoform test should have been done in a certified laboratory because of safety concern. However, I think fractional distillation might be helpful for this identification (or simply verification). OP can do this at home since he/she has done simple distillation at home. Ethanol/water binary system give an azeotrope, which boils at $\pu{78.1 ^\circ C}$ and gives $95\%$ ethanol. Methanol, on the other hand, make no azeotrope with water, and boils at $\pu{64.7 ^\circ C}$. The temperature range of $\pu{13.4 ^\circ C}$ need only about ≤20 theoretical plates to separate. If you get first distillate below, say, $\pu{67 ^\circ C}$, then you got methanol or something other than ethanol. So, don't drink it.
For more information, read the Wikipedia page having Azeotrope tables.

There's probably no reliable chemical test for the presence of methanol in alcohol mixtures that could be carried out at home without decent laboratory equipment.
Qualitative tests such as esterification with boric acid and subsequent flame test or oxidation to formaldehyde and its detection are unreliable, especially in the presence of ethanol. I advise against using such tests to decide about the presence of methanol in distilled liquors.

I have decided to format my comment into an answer.

To distinguish the ethanol from methanol, have three test tubes. In one of them, put $\pu{2 mL}$ known methanol, in one of them $\pu{2 mL}$ known ethanol, and one of them $\pu{2 mL}$ your mystery solution. Then, apply about $\pu{5 mL}$ of $\pu{0.5 M}$ of $\ce{I2}$, and about $\pu{10 mL}$ $\pu{1 M}$ $\ce{NaOH}$ to each test tube in that order. If your chemical is ethanol, the reaction will produce $\ce{CHI3}$, which is harmful (make sure to dispose of it properly). However your alcohol won't kill you unless you drink to excess. Otherwise, your chemical is methanol, which is poisonous (make sure to dispose of that properly also). Definitely don't drink that.

Source is from the Royal Society of Chemistry. It also linked this source.

This does not answer the question since the test can not show the presence of methanol in the distilled liquid.

@aventurin he simply asked how to distinguish methanol or ethanol. I provided a distinguishing test.

The question is "how to test for methanol?"


text:
Good Alternative for Plasticized PVC
In my current job we use gloveboxes, but unlike most we use disposable PVC bags on the entry ports instead of fixed gloves. The bags have an O ring which is stretched onto a spigot proud of the porthole and are then clamped into place. They get disposed of and replaced after every use.

^ Like this but our glovebox is permanent ^
The problem with the PVC bags is that they are releasing hydrogen chloride. I'm in a team investigating a replacement material; however none of us are chemists or material scientists!
The material needs to be able to fit over 6” and/or 10" spigots, allow hydrogen to pass, and contain no halides.
So far we have considered using HDPE + plasticizer, and are trying to find a way to make it, although we are not sure it will work until we try it. We have also looked into using polyester, but we couldn’t find a clear and stretchy version.

What good is speculating about a useful material if you can't find a supplier for useable gloves or bags? Try out what you can get your hands on! And sorry, stupid question: What or whom is this supposed to save from what kind of danger?

Hi @Karl We are doing that as well, but the materials local suppliers can give us might not be the best, or right for the job.
Unfortunately I can't tell you what the hazard is, but I can tell you it's big enough that if the best material isn't available in bag form, we'll have it made.

If your PVC gloves are releasing HCl then your problem is not PVC. PVC does not, under normal circumstances, release HCl (unless you expose it to excessive heat of very nasty chemicals).

Yes the PVC would not normally produce HCl but it does due to the way we're using it, it's a pretty weird situation. We can't change their process so we have to change the material.

An issue is that you expect it to block HCl but to pass H2. It would make more sense to change the process to eliminate the HCl fumes, e.g. absorb it.

I don't want it to block HCl, I want it to stop producing it. The hydrogen passes through the PVC and combines with the chlorine in the PVC to form HCl.

I think what you should try is thermoplastics elastomers. For general descriptive overview, read: https://en.wikipedia.org/wiki/Thermoplastic_elastomer
DUPONT has a long line of Chemical Resistant Elastomers, but may be expensive for your use, yet I mention it here because you said your team may consider custem made units to fix it in your glove boxes, so it could be quite expensive. Following is a cheaper option, but you can find more by just googling under "thermoplastics elastomers":
Kent Elastomer Products, Inc., PO Box 668, Kent, OH 44240 (1-800-331.-4762); email: info@kentelastomer.com (https://www.kentelastomer.com/cost-of-cheap/)

Thanks, I have emailed DuPont and currently waiting for an response.

I think what you want is a fluoropolymer, like Teflon PTFE or Viton. These are used to coat magnetic stir bars, they're that unreactive. That is, as long as you don't raise their temperature over 300c - otherwise they'll decompose into HF (not so good).
This company seems to sell COTS fluoropolymer bags (no affiliation, just found them on Google):
http://www.welchfluorocarbon.com/custom-manufacturing/standard-products/teflon-bags/
You can also buy plain Teflon sheets in a variety of thicknesses from standard suppliers like McMaster-Carr (8569K74).
Viton fluorocarbon is a little stretchier, and might be easier to work through - it's also very chemical resistant. Again, sheets are easy to buy.
If you absolutely need a "glove" shape, you can try vacuum-forming or blow-molding - in both cases, just heat an area of the material you want to re-shape (with, say, a heat gun), then either push a glove shaped mandrel through or try injecting a stream of air. This likely won't work with many fluorocarbons, though.

Doesn't Teflon contain Fluorine? Wouldn't that mean it includes halides?

@SnrSrub Whoops, sorry, I totally missed that requirement.


text:
What is the stability of Pregnenolone in various mediums?
I have some pregnenolone powder that I want to dissolve it in to a liquid medium for later human ingestion, some potential medium choices would be: alcohol, fruit juice, water, olive oil. I am trying to find out what mediums it will be most soluble and stable in. Please help, I am not a chemist and have not been able to find the answer to this anywhere, the best I have found is that it should mix in oil well - but I do not know how that would effect the shelf-life. I just have this powder that I want to distribute in to a medium for easy supplementation. If you can help, then thank you!

Pregnenolone has a water solubility of less than 1mg/ml so forget fruit juices and water if you want it dissolved. According to this article it is soluble in ethyl alcohol at 19mg/ml http://www.jpharmsci.org/article/S0095-9553(15)32106-5/pdf
I'm not finding any data that suggests it is chemically unstable in neutral solution.

Thank you Waylander! That table is exactly what I needed.


text:
Reaction of conjugated alkene with KNH2
I could not think of this reaction fitting into any of the reactions I know except the aryne mechanism. And hence I came up with the following mechanism after seeing the answer (which I've written in the picture next to question). Please tell if my proposed mechanism is correct.
Also, I am highly unsure of the fact that the last species formed will be able to extract a hydrogen atom from somewhere in the solution.
Question:

This is my proposed mechanism:

Source: Joint Entrance Exam (JEE) 1997 India

If the product is indeed correct (diphenylacetylene), I would suggest alpha-elimination to form a carbene which rearranges with phenyl shift to give the alkyne product

I have never seen any alkene form a carbene before. Could you please give an example of a reaction where you have seen such a thing. Or tell, why would it happen.. Please..

@orthocresol I believe you are correct. Deprotonation of the bromo-alkene to give the C(K)Br species will certainly give rise to a carbene (or metallocarbene) see https://en.wikipedia.org/wiki/Carbene   and the phenyl shift seems the obvious route to the product

Here is a flowchart of what the commenters have stated. An α-elimination occurs stepwise 1 --> 2 --> 3. There is a phenyl migration (bridged?) to afford tolane 4. Look here for this chemistry.

Try this: J. Am. Chem. Soc., (1936), 58, 2310. https://pubs.acs.org/doi/abs/10.1021/ja01302a066?journalCode=jacsat&quickLinkVolume=58&quickLinkPage=2310&selectedTab=citation&volume=58

Can this mechanism be concerted? That is, the abstraction of proton, migration of phenyl and elimination of Br all in a single step?

Is the reaction concerted? Perhaps. I haven't researched it. I suspect that the deprotonation may be reversible. There are alkylidene carbenes that undergo intramolecular C-H insertion. http://www1.udel.edu/chem/valhalla/alkylidene.html

I have a doubt in the migration step. I would have thought that the bottom phenyl migrates, not the top one - similar to how in Beckmann rearrangement, the group trans to the leaving $\ce{-OH2^+}$ migrates - isn't this reaction similar with $\ce{-OH2^+}$ replaced by $\ce{-Br}$, and a carbon in place of the nitrogen?


text:
What is the stability order for these octahydrobiphenylene structures?
According to me most stable is Z because it has maximum alpha hydrogen,
Between X and Y both have same alpha hydrogen, but X has conjugation. So it is more stable than Y.

So Order is Z>X>Y.
Is my explanation correct?

You are ignoring strain in a cyclobutene ring which has to count for something in stability.

Yes Sir Your Right, Can Please help me with what other factors I should Look For

This is all about ring strain, you do not need to look for ther factors

...and the antiaromaticity of cyclobutadiene X. Stereochemistry would help. Y>Z>X


text:
Second Ionisation Energy in d Block elements

The second ionisation energy is expected to be highest for element
  with atomic number:

22
24
25
30

I had expected the answer to be option 2 because the second electron has to be removed from the stable half filled $\ce{3d}$ configuration. But the answer is option 4.