text:
Hybridisation of ClO2

Find the hybridization as well identify the pπ-pπ as well as pπ-dπ bonds in $\ce{ClO2}$.

$\ce{ClO2}$ has 2 $\sigma$ bonds, 1 lone pair, 2π bonds and 1 odd electron.
Hybridisation is equal to number of $\sigma$ bonds + lone pairs. Since we consider odd electron a lone pair like in $\ce{NO2}$ therefore hybridisation is coming to be $\ce{sp^3}$. There will be no pπ-pπ bonding as all p orbitals are hybridised and there will be 3pπ-dπ bonds
But this is given wrong according to my textbook. What is my mistake?

Same question, but no answer: Structure and bonding in ClO2

I actually take issue with the question (which was asked of you not by you), as I think this oversimplifies a molecule, that still isn't well understood in the first place.
That is why I find this question incredibly difficult to answer and why I find that Nuclear Chemist's answer dangerous, because it propagates myths about bonding that have been disproved already.
But enough of the commentary, let's clear up some issues.

Find the hybridization as well identify the pπ-pπ as well as pπ-dπ bonds in $\ce{ClO2}$.

$\ce{ClO2}$ has two $\sigma$ bonds, one lone pair, two π bonds and one odd electron.

There comes a time, where you should realise that hybridisation is not always a useful concept to understand bonding. This time is now.
Your analysis of the bonding is not wrong, but maybe a bit too simple.
The reason for this is that the molecule has a very high symmetry, but it also has an unpaired electron. The problem here is that orbital symmetry cannot match with spacial symmetry, which means that hybridisation doesn't actually work here.
A bit more precise: The electron density of α-spin electrons is different from that of β-spin electrons. The hybridisation of the orbitals to produce either electron density need to also be different. In the very least you would have to consider two different hybridisation schemes at once.
All possible states (that is independent of any hybridisation model) then will be in superposition to form the actual ground state. This ground state will have a complicated bonding situation.
Of course you can use VSEPR theory to predict the shape of the molecule, that is a very good starting point. You will find out that the molecule is bent, and that matches the crystal structure. A bent molecule of the form $\ce{B-A-B}$ always has $C_\mathrm{2v}$ symmetry, it therefore always has π-orbitals. Whether those orbitals are essentially bonding or not, is an entirely different question.
It also does not tell you which atomic orbitals will be involved.
One obvious fault in the question is that there is no involvement of d-orbitals whatsoever in the molecules. The d-orbitals of chlorine are simply not accessible energetically; at least not in the way that it would form stable bonds. Anyone who is jibber-jabbering about 'octet-expansion' has stopped learning about chemistry about 20 years ago. (There are plentiful discussions about this on chemistry.se alone.)
As long as that last part of the question is not meant as a trap (which would be even more awful), it alone shows that the exercise is not well-thought about.
The π-orbitals will therefore be entirely composed out of p-orbitals of the involved elements.
If you look close enough you will realise, that writing a valid Lewis structure of the molecule is not that hard, and you don't even need π-bonds in the first place.
These are solely dictated by symmetry.
In principle that molecule is of a similar system like allyl-systems (with two extra electrons), or ozone (with one more electron). You will need a better than rudimentary grasp at molecular orbital theory to understand it.
Let's assume that things would be easier, and try tor roll with that for as long as possible. I have written a couple of times that terminal atoms can be (as a rule of thumb) regarded as having sp-orbitals.
Therefore the two oxygen will use one sp-orbital to form a σ-bond towards chlorine, the other one is an in-plane-σ-lone-pair. That leaves for every oxygen one in-plane-p-lone-pair, which we will (in the spirit of the approximation) disregard as having any effect on bonding. There is one remaining out-of-plane-p-orbital per oxygen. These will, due to symmetry, participate in π-bonding.
From VSEPR-theory we know the molecule is bent. The approximate hybridisation of the orbitals of chlorine that match that would be sp². Let's just assume that this is the case. Two of these will be the counterpart for the σ-bonds. The remaining will be either a lone-pair, or singly occupied. Already in this very simplified view, the problems start here. Let's further assume, that because it has higher s-character, the orbital will be lower in energy and therefore be doubly occupied. That leaves the odd electron in an out-of-plane-p-orbital, which will participate in π-bonding.
In this simplified and assumption riddled thought-experiment, the π-bonding will form three molecular orbitals pictured below.

Essentially the stabilisation from the π-bonds itself should be small, and could be described as π-backbonding, or is similar to negative hyperconjugation; maybe negative π-backbonding (!?).
This all was deduced from a very reduced and simplified set of theory; the reality is probably by far more complicated. Some of the greatest minds of (theoretical) chemistry have been puzzled about $\bf\ce{ClO2}$, and we have not even started to discuss the three-electron bond they assumed in between. Let's just say that this is not an undergrad study problem.
I know this doesn't really answer your question, and this is not very satisfactory either, but this is science: It often gives you more questions than answers.

What I would suggest to you is to start with working out how many electrons (valence electrons) are present in the molecule.
Each oxygen atom ($\ce{O}$) brings six electrons and the chlorine ($\ce{Cl}$) has seven electrons. This gives us a total of 19 electrons.
We should aim for having 8 electrons around each atom, keep in mind that the more heavy p block elements can expand above an octet.
If we look at the crystallography of chlorine dioxide ($\ce{ClO2}$) (A.Rehr, M.Jansen, Investigations on Solid Chlorine Dioxide ($\ce{ClO2}$)): Temperature-Dependent Crystal Structure, IR Spectrum and Magnetic Susceptibility, Inorganic Chemistry (1992) 31, (*) p4740-p4742) we find that the chlorine dioxide ($\ce{ClO2}$) molecule is clearly bent. The O-Cl-O angle is 116.2 degrees. Now this is an interesting finding.
Normally if we have a perfect $sp^3$ molecule such as methane ($\ce{CH4}$) we get an angle of 109.5 degrees between the bonds. But in the case of ammonia ($\ce{NH3}$) or water ($\ce{H2O}$) the increased ability of the lone pairs to repel electron pairs decreases the angle.
Thus it looks like a distorted $sp^2$ to me, I think that the bonding is likely to be two lone pairs on each $\ce{O}$, one lone pair on the $\ce{Cl}$.
Two $\sigma$ bonds each holding two electrons connecting oxygen ($\ce{O}$) atoms to the chlorine ($\ce{Cl}$).
Two $\pi$ bonds each holding two electrons connecting oxygen ($\ce{O}$) atoms to the chlorine ($\ce{Cl}$).
A $\pi^{*}$ (antibonding) orbital will have one electron in it, think it would be interesting to do ESR on $\ce{^{17}O}$ enriched chlorine dioxide ($\ce{ClO2}$) and compare the results with ESR on normal chlorine dioxide ($\ce{ClO2}$). This could allow us to get a better idea of where the unpaired electron is.

-1 for expanding octet, -1 for increased ability of the lone pairs to repel. This answer is propagating those myths in chemistry which gives thousands of students the wrong idea of chemical bonding. The most important part is that hybridisation (if applicable at all) follows the molecular structure. From Coulson's theorem we can actually calculate what hybrid orbitals are necessary.

Martin, it appears that you do not like VSEPR. The problem is that VSEPR does work for a lot of compounds. In this case the use of MO theroy to deal with the pi bonds and VSEPR for the sigma bonds does provide an answer which can be reconciled with the crystallographic evidence.

No you are wrong, I do like the VSEPR model very much. I also know that it works for many compounds, and I know that it offers valuable insights into geometries. However, the claim of expanded octets and electron clouds being more repelling is fabricated garbage to make weak theory fit beyond its means. In this case, if you are already using MO theory, there is no reason to use VSEPR in the first place; it also, while at first glance looks correct, produces the 'right' answer/ prediction, for the wrong reasons.

@Martin-マーチン I agree. I would go so far as to shame whomever made the task. ClO2 needs MO, and 3-electron-2-center bonds to make even close to sense, and the woefully inadequate vsepr theory or hybridisation will not be enough. And don't get me wrong, VSEPR is a great model, but as with Newtons Law - there comes a point where it just doesn't explain what is going on. For some reason, many chemists in teaching positions keep VSEPR as a holy cow for far too long up the grades.

Well I still love VSEPR, but it has its limits. If we take the crystallogrpahic data then it is clear that the chlorine has three sigma bonds / lone pairs (or single version of lone pairs). I reason that the angle of 116 degrees makes it look like we have a distorted trigonal arrangement of sigma orbitials around the chlorine. I am sure that for the pi system we need to use MO theroy and the the best I have come up with is that a pi star orbitial has a single electron in it. I was always taught to use VSEPR for the sigma bonds and then MO for the pi system.


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Malic acid 3-step synthesis
I was quite puzzled by the answer for this question in my Carboxylic Acids and Derivatives tutorial question set. It is question 6 (a), on the synthesis of malic acid, as shown in the image below.

The idea for the synthesis is to convert the $\ce {Br}$ to $\ce {CN}$ and the carbonyl group to a cyanohydrin, then convert the $\ce {CN}$ to $\ce {COOH}$. I have no disputes about the last step, which is to heat with aqueous hydrochloric acid, or some other mineral acid. However, what I proposed was to perform the two initial conversions in one step, which is to heat the compound with $\ce {KCN (aq)}$ and ethanol. Would that not allow for both the nucleophilic addition, as well as the substitution, to take place, all in one step?
The proposed answer for steps 1 and 2 was to first heat with reflux, with ethanolic $\ce {KCN}$, then react with $\ce {HCN (aq)}$, with trace $\ce {KCN}$, or $\ce {NaOH}$ as catalyst, under temperature of 10-20 degrees C. Is it really necessary to perform the initial two conversions in two separate steps?

Cyanohydrin formation is a net addition of HCN. KCN + EtOH doesn’t generate any HCN, crucially it doesn’t provide any protons so the anionic RR’(CN)(O-) will just collapse back to the aldehyde.

@orthocresol But isn't the attacking nucleophile cyanide? The protonation can occur as long as there is a proton source available. In this case, water can provide that so we don't really need to add HCN specifically right?

There are two reasons why this is a bad question, which is obviously not your fault. (1) Chloroacetaldehyde exists as the hydrate, and likely, bromoacetaldehyde does the same. This is an important fact that the problem solver should know about. (2) Malic acid is a commodity chemical made from oxidation of maleic anhydride, which is produced from oxidation of n-butane. The suggestion that we should take this route is, therefore, a terrible idea but nevertheless useful as an exercise.

I have clarified with my chemistry tutor and it appears that performing the substitution and the addition as two separate steps with two separate sets of conditions is absolutely necessary. The most important reason would be the temperature, which is also the most stark difference between the two different set of conditions.
Due to entropy considerations, addition should ideally be carried out at low temperatures. Thus, if we would like to have the addition of $\ce {HCN}$ to the substrate, it should be done at much lower temperatures as the reaction has a negative entropy change. However, the substition of $\ce {Br}$ by $\ce {CN}$ does not need such low temperlatures and should be done at higher temperatures to provide enough energy of activitation to break the carbon-bromide bond.

Did you count the number of carbon atoms in the product? Adding KCN only gives you 3 carbon atoms.

@ringo I think the point is that a cyanide equivalent would add into the aldehyde and another would substitute the bromide, increasing the length of the chain by 2.

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How to name this hydrocarbon
I saw a hydrocarbon bond line diagram. That have a methyl on top and a complex group down. I don't know from where to start numbering.Ihave posted the image.Can any body help me.Thanks in advance.

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

But what about methyl on the top.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the ring is selected as the parent structure to construct the preferred IUPAC name.

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

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

Thus, the compound given in the question is named as cyclohexane. The substituents are expressed as prefixes, i.e. x-(butan-2-yl)-y-methylcyclohexane.
The most important simplified criteria for the numbering of the locants x and y are:

lower locants for the group that is expressed as suffix
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current IUPAC Recommendations reads as follows:

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

The compound that is given has no principal characteristic group that would be expressed as a suffix and no multiple bonds. Thus, Rules (c) and (e) are not relevant in this case.
Rule (f) does not lead to a decision since the locant set (all considered together in a series of increasing numerical order) is always ‘1,4’.
Finally, Rule (g) stipulates that the lowest locant is assigned to the substituent that is cited first as a prefix in the name. Therefore, the correct name for the compound that is given in the question is 1-(butan-2-yl)-4-methylcyclohexane.

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Electrophilic aromatic substitution of napthalene
It is known that the $\alpha$-position (or 1-position) for napthalene is the preferable position for substitution to occur at. However, I am getting two different explanations as to why this is the case, one of which I am not able to make sense of.
The first explanation which I have come across and which is also the one I am able to make sense of is given by Hepworth, Waring and Waring (2002). They explain it by using the concept of Clar's rule. With an attack at the $\alpha$-position, there would be two resonance structures in the intermediate with preservation of the aromatic sextet. With an attack at the $\beta$-position, there would only be one resonance structure in the intermediate with preservation of the aromatic sextet. As such, the $\alpha$ attack would be more preferable since the resultant intermediate would be more stable.
The second explanation which I have come across just today comes from Carey and Sundberg (2007). An image of the section of the text discussing the issue is shown below.

I am not able to understand why they have mentioned that the intermediate resulting from $\beta$ attack is not under allylic stabilisation because it seems to me that it is under such stabilisation, after drawing out the resonance structures. I would like to clarify if this explanation is accurate.
References
Carey, F. A., & Sundberg, R. J. (2007). Advanced Organic Chemistry Part A. Structure and Mechanisms (5th ed.). Springer.
Hepworth, J. D., Waring, D. R., & Waring, M. J. (2002). Aromatic Chemistry. United Kingdom: The Royal Society of Chemistry.

Related: Thermodynamic vs Kinetic Sulphonation of Naphthalene

Essentially, both descriptions are identical. However, they use different phrasings.
What Carey and Sundberg mean when they say no allylic stabilisation is essentially that the cation itself is not in an allylic position; that there is no single (non-phenylic) double bond in α-position. To reach the allylic double bond in the β-attack case, you would need to resonate across at least part of the phenyl ring (meaning you’re already making use of benzylic stabilisation) while in the case of α-attack you have access to allylic resonance without interrupting the phenyl resonance.

Your answer has really cleared things up for me. So there is no allylic stabilisation from a non-phenylic double bond in the case of the intermediate resulting from beta attack. That makes sense.

The best explanation is a hybrid of the two: the more stable intermediate resonance form has allylic stabilization and preserves the Clar sextet. You can only draw resonance forms with one or the other in the beta substitution intermediate.


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How was it possible to accurately measure the mass of gases in the 19th century?
I have seen periodic tables of Mendeleev (1834-1907) and Newland (1837-1898) in which masses of gases such as oxygen and nitrogen were shown. I don't think there were instruments to measure mass of gases at that time. So how were they able to predict the mass of gases at that time accurately?

Hm? Of course the molar mass of gases was known at the beginning of the 19th century. Heard of Avogadro? One liter of air weighs over one gram, that's not exactly difficult to measure.

but how were they able to know the molar mass of gases at that time?                                     And your statement "One liter of air weighs over one gram" is only telling mass of one liter of air and I can't relate that to mass of gases.

Related question on Physics Stack Exchange: Why do gases have weight?

There was no moles and no molar masses in the beginning of the 19th century. But gases could be weighed. If you have a vial or an ampule with a stopcock, you can make a vacuum in it (or nearly) with a water pump, and weigh it empty. Now you fill it with a given gas, and weigh it full of this gas. Then repeat the same operation with another gas. By difference, you can determine the masses of your different gases. Comparing these measurements, you will find that the gaz oxygen contained in your vial is 16 times heavier than the hydrogen, whatever the exact volume of the ampule or vial. With the same experiment, you will find that the gas we call carbon dioxide is 22 times heavier the hydrogen. All gases could be weighed and compared to the lightest of all : hydrogen. Even though the notions of atom and molecule were not accepted everywhere, all chemists could state that in the same volume, oxygen is 16 times heavier than hydrogen.


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Which chemicals can counter the corrosion of tamperproof hardware encryption chips?
I'm moving from software implementations to hardware. I need to reverse engineer silicon chips, though the tech that I want to study may have a corrosive chemicals that's supposed to destroy the chip upon touching.
What chemical gets sprayed onto the hardware and which chemical reactions occur?
If there's one substance that causes destruction then there must also be one that prevents it. Does anyone know what this chemical is?
This question in currently going on in the back of my head. Chemistry isn't really my thing; but I know enough to understand the chemical reactions to make various drugs.
I'm not funded by government. This is independent research. The field of cryptography is somewhat of a front line of engineering. If there are no answers to the main question, chemistry resources in the context of silicon and the relating chemicals used in production as well as the securing in the post production will be accepted as the answer.

The chemical that first comes to mind when talking about destroying silicon is hydrogen fluoride which is used as an etchant in chip manufacture.

this brought me here https://www.cl.cam.ac.uk/~mgk25/sc99-tamper.pdf much appreciated!

Hydrogen fluoride protects the chip by converting into hydrofluoric acid upon contact with moisture or human tissue.
I was able to conclude this through this wiki article:
Hydrogen Fluoride

Upon contact with moisture, including tissue, hydrogen fluoride immediately converts to hydrofluoric acid, which is highly corrosive and toxic, and requires immediate medical attention upon exposure.[31] Breathing in hydrogen fluoride at high levels or in combination with skin contact can cause death from an irregular heartbeat or from fluid buildup in the lungs

The substance that can neutralize the Ph levels of the hydrofluoric acid is calcium gluconate.
Calgonate 2.5% calcium gluconate gel neutralizes HF
This can be used to help clean up accidental contact with hydrofluoric acid and could possibly be used to prevent any chemical reactions from occurring upon tampering with the chip. The best bet in this situation is to conduct experiments in a dry lab environment and wear gloves. You want to have a fume fan as well as the caluium gluconate in the event of failure.

One idea is working in a closed environment that has a persistent calcium gluconate fog/gas at all times.


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What is the relationship between emission and fluorescence?
A molecule exposed to a photon with some energy is put into an excited state, and emits a photon of some energy when it returns to ground state.
The photon that provided the energy for excitation corresponds to a particular wavelength, at which we see a peak in an absorption spectrum.
The photon that was released when the molecule returned to its ground state also corresponds to a particular wavelength, at which we see a peak in an emission spectrum.
Fluorescence is when a molecule emits a photon of lower energy when it returns to ground state than what it absorbed initially to be put in an excited state. That photon corresponds to a particular wavelength, at which we see a peak in the fluorescence spectrum.
But isn’t the emission of a lower energy photon the case for all molecules, since all molecules lose some energy in form of vibration and rotation before they reach ground state? How is fluorescence distinguished from emission?

Emission is the generic term for fluorescence (from excited singlet state) and/or phosphorescence (from triplet) to singlet ground state. Look up Jablonski diagrams for molecules in gas phase and in solution to see the effect of  solvent and vibrational/rotational effects

You can google all of this for more detail...
Basically an "emission spectrum" is the most general term for detecting photons as a function of energy or wavelength. All the term implies is that your measuring photons.
A "fluorescence spectrum" is a little more specific. It means that a photon was used to excite the atom/molecule and that the atom/molecule decayed very rapidly with the emission of a photon of longer wavelength (lower energy).


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Does alumina become transparent at high temperatures?
Consider a normal opaque alumina crucible. I've heard that they become transparent at higher temperatures and therefore, are not appropriate for high precision DSC (caloric) measurements. The reason is, that there might be a lack of energy due to radiation losses (in addition to the bad thermal conductivity).
I was not able to find at least one quotable source for this claim. So my two questions are:

Is this actually right?
Does someone know of a quotable source for this (scientific papers, textbooks, etc...)?

Thanks!
Edit: Based on my research, the absorption rises with temperature, so the transparency should degrease, what is a somewhat contradictory to A.K. answer, isn't it?

Aluminium oxide does form large transparent crystals e.g. in the Ruby laser, in which the crystal is doped with a few chromium ions, so it is possible that at high temperature the micro crystals are combining to become more transparent and less light scattering.  At high temperature the radiation losses would be that due to a 'black body, as with any other material. Have you tried to heat a crucible up with an acetylene torch to see if transparency is changed?

What temperatures are we talking about?!? What does the manufacturer say about accuracy at high temperatures? Why don't you test this, by measuring a standard substance?

between 25 °C and 1500 °C. The manufacturer says that it should be avoided but not why. I cannot test this at the moment and I want to understand the reasons.

Well, alumina is sapphire so stand to reason that it is transparent at a low temperature too.  I think what is meant by what you heard is that the higher intensity of infrared light is transmitted by the alumina causing thermal loss.
Alumina is transparent to UV, visible and infrared light. To our visible light vision, the scattering and reflection of polycrystalline alumina make it appear as opaque white. In the IR spectrum though it may appear transparent. This is because scattering is proportional to the inverse of the wavelength to the fourth power ($\alpha _{\text{scat}} \propto \lambda ^{-4}$).  Thus a wavelength just twice as long a red light (the longest wavelength of visible light; $\lambda = \pu{750nm}$ ), in this case, $\pu{1.5 \mu m}$, which is near infrared, would scatter one-sixteenth ($6.25 \% $) as much as the red light would. A wavelength twice as long as that wavelength ($\pu{3\mu m}$) would be in the thermal infrared spectrum and be scattered one-two-hundred-fifty-sixth as much as the red light ($0.39\%$) This reduced scattering allows more transmission. At high temperature when infrared radiation is more intense, the energy loss due to transmission becomes more significant, but the alumina was always transparent.

Thanks, this answers question #1. But I need a source for this, wikipedia and websites are not really quotable. Do you (or anybody else) have a valid source?

Rayleigh light scattering is considered common knowledge and should not require a source.  Here is an article for the transmission of  sapphire: https://www.osapublishing.org/ao/abstract.cfm?uri=ao-27-2-239

Ok, I've skimmed through the paper and some other sources. They all write, that with rising temperature the absorption coefficient also rises. So shouldn't the losses be lower? Despite the fact, that your explanation is understandable, the larger absorption coefficients are confusing me again. This might be due to some misunderstandings between absorption, transmission, emission and scattering...

The absorption can rise, but the intensity of incident light from the sample increases at a faster rate.  Thus the relative amount of ligt (the ratio) is less but overall the absolute amount of radiation that transmits through the alumina is greater

Ok, but why is there almost no literature which describes this? Maybe this is only an indirect conclusion form basic heat transfer and radiation theory?


text:
Are there any process and catalyst for convert CO2 and H2O into CH4 and O2
The equation is simple
$$\ce{CO2 + 2 H2O -> CH4 + 2 O2}$$
Yet, is it possible? Are there processes like heat, pressure, electricity, laser and/or any catalyst that could do this process directly?
And if it not possible then why it isn't?

Are you sure you are not looking for this?

See more: https://www.popularmechanics.com/science/green-tech/news/a27412/catalyst-turn-co2-into-methane/
https://phys.org/news/2017-07-carbon-dioxide-methane-iron-sunlight.html

@NilayGhosh as closely as it seem, no, I was specifically want CarbonDioxide and Water to generate hydrocarbon and oxygen

It can be done but not in a way that makes much economic sense
The trouble with this reaction, however attractive it looks as a neat way to use up carbon dioxide, is that is requires a lot of energy. It is basically the reverse of the normal reaction of burning methane, which releases a lot of heat (this is what your cooker is doing if supplied by natural gas). So going in reverse requires an input of at least the same amount of heat.
The Sabatier reaction (which is one way to force the required process to happen) has been known since 1910. It requires both a catalyst (nickel or ruthenium based) and both high temperatures and pressures. This means there is little incentive to do it in most circumstances. It has been used in exotic locations (the International Space Station, for example, but there it is used to use up excess carbon dioxide and regenerate oxygen not to produce methane which is discarded).
So it is possible but not sensible to do the reaction in most circumstances.
If you really want to use up carbon dioxide and turn it into something more useful, then plant a tree. Plants have been turning carbon dioxide into useful products for more than a billion years using the sun's energy to drive the reaction.

Using a lot of energy is one thing I hope for the process. It means it could be used to store energy that highly fluctuate such as solar farm, if it could be economical sufficient

@Thaina There are lots of well known ways to store energy from fluctuating sources, some of them based on chemical reactions some on simple physical systems. Most are already far more efficient that a complex reaction like the Sabatier process could ever be.


text:
What is this golden looking substance?
I am currently messing around with gallium compromising aluminum. I am using a titalium lock which is just a brand of aluminum padlock and I put a gallium-indium euctic alloy on the surface of it. It’s been around 10 minutes and on the surface of the liquid is a gold colored substance. What is this? Any idea why it’s happening??

By the way this happens every time i expose aluminum to the gallium-indium alloy but doesn’t happen when it’s just gallium

It also does not react with water when I put it in water.

If anyone else has done experiments with gallium and aluminum please leave feedback on what you think this is

Do you see an attack of the aluminium surface? Could it be something similar to the aluminium mercury amalgam reaction, where the passivation of the aluminium surface is broken, and the oxide is forming at the surface of the liquid metal? The color then just being the result of impurities or small size of the aluminium oxide particles.

I think you are correct that this is small particles of aluminum oxide. Aluminum oxide appears slightly yellow in my opinion.

\begin{array}{c}
\ce{\phantom{H_3}CH_3}\% phantom to get the bond aligned with the C
| \
\ce{CH3-C-CH3}\
| \
\ce{Cl}
\end{array}
\

Low oxidation state indium oxide
It cannot be pure aluminum oxide.
Indium has stronger tendency to form mixed valence compounds than aluminum alone.
I have also seen similarly golden compounds on other surfaces of tin alloys that were exposed to air while still liquid. There are many golden substances so You need to consider other components from the aluminum alloy of the lock, otherwise You may not be able to tell the difference just by looking.


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Action of hot alkaline KMnO4 on tert butyl benzene

Side chain oxidation of aryl alkanes takes place if there is at least one alpha hydrogen present by all accounts I presumed that there would be no reaction. However the answer given was trimethyl acetic acid. Is the answer key wrong?

https://chemistry.stackexchange.com/a/86/51553

Permanganate oxidises double bonds, even those in aromatic systems if you use forcing conditions.

My teacher has taught me same rxn. but acc. to him it happens only in acidic conditions not in basic medium

A similar reaction with chromium oxide in alkaline solution has been documented.

The acidic chromium oxide serves the same role as alkaline KMnO4 i.e a strong oxidizing agent. As pointed out by @Waylander in the comments permanganate oxidizes double bonds even in aromatic compounds.
Reference:
$\large{•}$ Chromium Oxidations in Organic Chemistry; By G. Cainelli, G. Cardillo

For reaction  with $\ce{KMnO4}$, the given compound must have a carbon atom attached directly to the ring and that carbon atom must have at least one hydrogen atom.
But in case of tert-butylbenzene this condition isn't  followed and hence is a special case.
The final product obtained on reaction of tert-butylbenzene with $\ce{KMnO4}$ is Pivalic acid $\ce{(CH3)3CCOOH}$
(Common name: Dimethylpropanoic acid).

Under normal circumstances, kmno4 will not oxidise tertiary butyl benzene. However, under drastic conditions, oxidative cleavage of bonds occur and a mixture of acetic acid and trimethyl acetic acid is formed.
Hence the answer is incorrect. They should have mentioned the word 'drastic conditions'.

What are "drastic conditions?"  If something lives in the hydrocarbon lakes of Titan then liquid water would seem drastic to it.


text:
Is there a thermodynamic explanation for how a substance behaves based on its pKa and the solution’s pH?
If the pH of the solution is lower than the pKa of the compound, then the compound will be protonated, because pH is a measure of the concentration of protons, and low pH means a high proton concentration, so the compound is “driven” to act as a base and take up protons because there’s so much of them in the solution.
By the same logic, if pH is higher than pKa then the concentration of protons in solution is not very high, so the compound is “driven” to act as an acid and donate protons to the solution.
What is the “driving” force (I know it’s not technically a force) that causes this behaviour? Is there a thermodynamic reason involving changes in enthalpy and entropy and an increase in stability (decreasing Gibbs free energy) in the system?
My own thought is that perhaps this spontaneous exchange of protons between the compound and solution based on pKa and pH values takes the system closer to an equilibrium... but I don’t know where to go from there.

It is necessary to consider both the equilibrium of water and of your acid and base as the system moved towards equilibrium which will gave the lowest free energy.  The equations to describe the equilibria  are given in the answer to this question.https://chemistry.stackexchange.com/questions/60068/how-to-set-up-equation-for-buffer-reaction/95566#95566


text:
How is polarization of light measured?
Chiral molecules are classified as dextrorotatory (+) or levorotatory (-) depending on whether they rotate plane-polarized light (ppl) to the right or left.
My question is, how is such a measurement made? What instrumentation is used and how would you determine that ppl has been rotated right or left?

https://chemistry.stackexchange.com/questions/35008/how-do-we-identify-left-handed-and-right-handed-crystals

The instrument used is called a polarimeter. You basically shine monochromatic linear polarized light through a solution of known concentration and you measure the rotation of the light on the other end of a tube with defined length. You then normalize the rotation according to concentration and length of your tube to get the specific rotation which is then independent from concentration and length and only depends on temperature, solvent and wavelength of the light.
You can determine if it's rotated left or right by changing the concentration since the observed rotation is directly proportional to concentration. You also need to do this to determine if your rotation is more than a full 360°, since a measurement of 10° could also be 370° or 730° or 1090° or also −350°, −710°, and so on.
Let's look at this using a simple example and let us assume our tube has a length of 10 cm (which is the length used for the specific rotation, so we don't need to worry about this). Specific rotation is also normalized to a concentration of 1 g/100 mL.
To determine our specific rotation we could do 3 measurements at 1 g/100 mL, 2 g/100 mL and 0.5 g/100 mL.

1 g/100 mL : +160°
2 g/100 mL : −40°
0.5 g/100 mL : −100°

Now what does this tell us? If the specific rotation would be +160° then double the concentration should be 320° and half should be +80°, so that's clearly not the case. If it would be −200° instead of +160°, then double the concentration would be −400° which would show up as −40° and half the concentration would be −100°. So our specific rotation in this case is −200°.

Can you give an example of a rotation of more than $360^\circ$ at 1% concentration?

@RaoulKessels according to 10.1002/cber.190704002126 D-Hexol bromocamphorsulphonate has a specific rotation of 2640.

Your reference is wrong. I have checked the original article (published in 1907) and no optical resolution is done. I have researched a bit more and it was not until 1914 that Werner resolved the hexol. Then, it is true that it has an amazing rotation with a maximum at 560 nm. It might be good to state that rotations above $360\circ$ are uncommon, especially in organic chemistry where the optical activity is most abundant.

@RaoulKessels ok sorry, didn't knew that. I found it referencedquite a few times so I assumed it's in there. And yes it might be uncommon but it's possible (especially if you go to higher concentrations) so one should look out for that.

Helicenes would be an example of organic molecules with very high optical activity


text:
Properties of the products of styrene chlorification
Background: When chlorinating styrene with an electrophile two products can be formed by substitution of the benzene ring. That is because the ethylene group is ortho and para directing, from the +M and the ± effect.
Questions: Which is the more stable product? Is the difference in stability related to the steric repulsion between the neighbouring groups in the ortho (1,2) product?
Also does the combination of the aromatic system and the vinyl group have any effect or interaction?


text:
Why does the outside of a plastic cooking oil bottle get sticky over time?
Cooking oil (I've observed it with canola oil, if that matters) is not ordinarily sticky. When I buy a plastic bottle of it, the outside is not sticky.
But when use some of it, then return after a few weeks, I find that the outside of the bottle is sticky. Why is this?
It does not appear to happen with olive oil in glass bottles.

Do you keep the bottle in the open near where you cook?

Some drying oils, such as linseed oil, react with oxygen in air to polymerize, eventually forming a hard film. When the oil is in the midst of change, i.e. short-chain polymers, it's a sticky mess (don't touch wet paint).
Most cooking oils are unsaturated. Having double bonds makes the oil easier to digest and less likely to deposit in the circulatory plumbing. It also makes the oil more chemically reactive. When in a closed bottle, it reacts fairly slowly with residual air, eventually becoming rancid. Spread over the surface of a bottle, the oil is exposed to more air and oxidizes and polymerizes more rapidly.
An experiment to test this would be to coat two pieces of the plastic used for bottle, usually PETE, one with cooking oil (unsaturated) and one with mineral oil (mostly saturated) and check if the cooking oil becomes sticky, while the mineral oil should not.
One other factor, though, is that oils will cause dust in the air to adhere, so that might also cause some stickiness. Place the test swatches in a container covered with paper towels to keep off dust.

Canola oil contains about twice as much unsaturation as olive oil ("composition of canola oil": Wikipedia). That could be part of the reason.
But I think another significant reason is the bottle top. The plastic bottle tops on canola oil are typically larger than the tops on olive oil bottles, so might drip more after being closed.


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About nitrogen acidity in function groups
I have a compound and it has several nitrogen atoms. I wish to know how do we differentiate between the basic and acidic nitrogens?
This is an example compound:

Acidic nitrogens will have stabilized conjugate bases, and basic nitrogens will have stabilized conjugate acids. Trying drawing them out for each N.

The presence of so many electronegative atoms around one of the nitrogens clearly tells us something about the relative acidity of that N-H group.

First of all, a terminology correction. While a nitrogen atom may be basic, a nitrogen atom itself won't be acidic, rather, the hydrogen atom attached to that nitrogen atom would be acidic, if at all.
Now, whenever given an arbitrary compound, and this also holds true for atoms like carbon, oxygen, etc., here's what you need to do to determine the acidic and the basic atoms.
What does acidity mean? Acidity means the ability to lose a proton from the compound. The higher the ability to lose a proton, the higher the acidity. If a compound, on losing a proton, forms a more stable conjugate base, it is considered more able to lose a proton. Hence, it is also more acidic (than other similar protons in that compound).
What does basicity mean? Basicity means the ability to donate a lone pair of electrons to a proton. The higher the ability to donate a lone pair to a proton, the higher the basicity. If a compound, on donating a lone pair, forms a more stable conjugate acid, it is considered more able to donate a lone pair. Hence, it is also more basic (than other similar lone pairs in that compound).
They both sound similar, don't they? That's because they're on the same underlying principle: a chemical species would want to accept a proton or donate a lone pair if the resulting species is more stable.
Now, you can simply check for this in any given compound by deprotonating every H-atom and protonating every lone pair, while applying the simple rules of aromaticity, resonance, hyperconjugation, and inductive effects individually to each.


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Is the electrophile in sulfonation of benzene SO3 or SO3H+?
What's the electrophile in sulfonation of benzene? Is it $\ce{SO3}$ or $\ce{SO3H+}$?
According to what my professor taught me, both $\ce{SO3}$ and $\ce{SO3H+}$ may be the electrophile involved - and mechanisms with both species can be successfully written.
I'm looking for some source(s) that confirms the existence of both electrophiles in sulfonation reaction of benzene.

SO3H+ seriously? That's equal to Lewis superacid able to bind almost anything, and even best non-coordinating anions would probably fail to keep separate from it.

From Clayden, under the heading "Sulfonation of Benzene" (electrophilic aromatic substitution):

The cationic intermediate can
  also be formed by the protonation
  of sulfur trioxide, $\ce{SO3}$, and
  another way to do sulfonations is
  to use concentrated sulfuric acid
  with $\ce{SO3}$ added. These solutions
  have the industrial name oleum.
  It is possible that the sulfonating
  agent in all these reactions is not
  protonated $\ce{SO3}$ but $\ce{SO3}$ itself.

From March's $7^{th}$ edition, under Sulfur electrophiles:

Sulfonation is often accomplished with concentrated sulfuric acid, but it can also be
  done with fuming sulfuric acid, $\ce{SO3}, \ce{ClSO2OH}, \ce{ClSO2NMe2/In(OTf)3}$ or other reagents.
Sulfur trioxide reacts much more rapidly
  than sulfuric acid with benzene it is nearly instantaneous
A great deal of work has been done on the mechanism, chiefly by Cerfontain
  and co-workers. Mechanistic study is made difficult by the complicated nature of
  the solutions. Indications are that the electrophile varies with the reagent, though
  $\ce{SO3}$ is involved in all cases, either free or combined with a carrier. In aqueous
  $\ce{H2SO4}$ solutions, the electrophile is thought to be $\ce{H3SO4+}$
  (or a combination of
  $\ce{H2SO4}$ and $\ce{H3O+}$) at concentrations below $\sim 80–85%$ $\ce{H2SO4}$, and $\ce{H2S2O7}$ (or a
  combination of $\ce{H2SO4}$ and $\ce{SO3}$) at concentrations higher than this. (the changeover
  point varies with the substrate)
In fuming sulfuric acid ($\ce{H2SO4}$ containing
  excess $\ce{SO3}$), the electrophile is thought to be $\ce{H3S2O7+}$
  (protonated $\ce{H2S2O7}$) up to $\sim 104% \ce{H2SO4}$ and $\ce{H2S4O13} (\ce{H2SO4+ 3SO3})$ beyond this concentration.
Finally, when pure $\ce{SO3}$ is the reagent in aprotic solvents, $\ce{SO3}$ itself is the actual
  electrophile. Free $\ce{SO3}$ is the most reactive of all these species, so that attack
  here is generally fast and a subsequent step is usually rate determining, at least
  in some solvents.


text:
How does alcohol infusion (maceration) work chemically?
To make liqueurs (e.g. lemon liqueur) you take some fruit and/or plant and let them macerate in pure alcohol to make the alcohol have the fragrance and color of the original ingredient.
How does it work from a chemical standpoint?
Is it simply a process in which the aromatic substances are soluble in alcohol and so the solution inherits their characteristics, or are there (as I expect) more complicated processes at work?

There are more complicated processes at work (as you expect), but you are correct in that the aromatic (and other) substances are soluble in the alcohol and thus are EXTRACTED by the alcohol. This is simply an extraction process.


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Group 15... Lewis Acid, Lewis Base or Not At All?
I am a long-time reader of this forum, but this is my first time post. I am a chemistry undergrad with a question about Lewis acidity or basicity of Group 15 compounds.
I know Lewis acids act as an electron pair acceptor and a Lewis base as an electron pair donor. Then sometimes a dative bond is formed between the two, forming a Lewis adduct (as per classic $\ce{NH3}$ and $\ce{BF3}$ reaction). But this is where my knowledge becomes a little sketchy, and hence my question:
When considering group 15, is it possible for the elements of this group to form compounds which are either Lewis acids or bases?
Firstly, the requirements for Lewis acids of group 15, is it empty d-orbitals or an incomplete octet which give a compound the ability to accept two electrons?
Therefore, would this make $\ce{PF5}$ an acid (d-orbitals)? Would $\ce{AsF5}$ be classed as an acid? (If so, is this due to $\ce{As}$'s ability to violate the octet rule coupled with the fact $\ce{As}$ is very positive in this compound).
Then are the requirements for something to be a Lewis base purely the presence of two non bonding electrons? Would this then make compounds such as $\ce{AsF3}$ and $\ce{PF3}$ bases?
I am a little lost, would like to know if the above is along the right lines. Do things like electronegativity and atom size play a part?
I think this is a base?

I think this is an acid?

thanks to those who edited my post earlier, I am new to posting on this site. I also realised my tags were the wrong way around on my picture, hence the repost with correct tags on picture

For anyone else with a similar question above... I have been researching this for the last day or two, I have found Group 15 should be spilt into two groups - either nitrogen or heavier G15 elements when considering basicity or acidity.
Nitrogen - Trihalides
NF3 < NCl3 < NBr3 < NI3 (with NI3 as the most basic due to increasing relative electronegativity of the nitrogen). (These compounds are generally only weak bases.)
In my research I have found the greater the electronegativity of the lone pair atom in comparison to its ligands (i.e. the halides in this instance), the more free the lone pair is to participate in a dative bond to form a Lewis adduct with a Lewis acid. The more free the lone pair is the stronger the base is, hence NI3 is the strongest base out of the trihalides.
This is why NH3 is such a great base as the lone pair on N is not inductively pulled towards the ligands because H is relatively less electronegative. Whereas in NF3 the lone pair is so drawn away from nitrogen this compound is sometimes not even classed as a base, hence it is the weakest of the trihalides.
The reason nitrogen is on its own is because it has max coordination number 4.
P, As and Sb - Trihalides or Pentahalides
I have found halides, either tri or penta, are generally classed as Lewis acids.
PX3, AsX3, SbX3 (X - F, Cl, Br or I)
First the trihalides, these act as mild Lewis acids towards trialkylamines and halides. This is even though the P, As or Sb will have a lone pair. As before,  this is partly due to electronegativity reducing the availability of the lone pair.
My textbook points out PF3 is probably a little bit unusual in this group because it has empty d-orbitals (making it a weak sigma donor yet a strong pi acceptor.)
The pentahalides of heavier group 15 elements are pretty strong acids
PX5, AsX5, SbX5 (X - F, Cl, Br or I)

The trihaldies of nitrogen are explained nicely here -
https://courses.lumenlearning.com/introchem/chapter/trihalides-boron-halogen-compounds/
Also a clever link about NCl3 -
http://www.chm.bris.ac.uk/motm/ncl3/ncl3h.htm
The heavier G15 information and pictyre came from a textbook Inorganic Chemistry (Weller, Overton, Rouke, Armstrong)

Lewis Acid-Base Theory
Simply put, if, in a reaction, a compound is donating electrons, then it is basic while if a compound is accepting electrons, then it is acidic. Do note that this requires some kind of electron transfer. For example, there is no acid or base in the reaction $\ce{N +N->N2}$ because neither nitrogen is more or less acidic than the other.
Group 15 halides
Now, the Group 15 compounds, interesting bunch of troublemakers, with lone pairs, empty orbitals, a wide range of oxidation states, and with the infamous element $\ce{Sb}$ known to form the most acidic compound $\ce{HSbF6}$ or fluoroantimonic acid, which I talked about a little bit here. I will focus on group 15 trihalides and discuss the relative Lewis acidity or basicity.
Frontier Molecular Orbital Acid-Base Theory
In a reaction, if a highly energetic highest occupied molecular orbital (HOMO) of a compound readily donates electrons to a lowly energetic lowest unoccupied orbital (LUMO) of another compound to form a bond, then the HOMO-compound is termed basic and the LUMO-compound acidic. Thus, highly energetic HOMO hints towards basicity while a lowly energetic LUMO hints towards acidity. Since a compound has both HOMO and LUMO, it may act as a base and an acid in different reactions.
I couldn't find a study that corresponds to this directly, so I sought out to carry it myself. If someone could corroborate this study, I would greatly appreciate it. All calculations were performed at RHF level of theory using def2-SVP basis set with geometry optimization.
Group 15 Trihalides

Fixed Group 15 Element
Since Homo energies are increasing and LUMO energies are decreasing, group 15 trihalides show a regular trend of increasing basicity and acidity (both) in the order:
$$
\ce{EF3<ECl3<EBr3<EI3}
$$
I doubt if $\ce{NX3}$ will act as a Lewis acid though, because I haven't seen $\ce{N}$ form compounds with an expanded octet. However, the LUMO energies fall rapidly from $\ce{NF3}$ to $\ce{NI3}$, and the latter might just be able to act as Lewis acids as well.
Fixed Halide
More clearly visible in the graphs, the compounds show an irregular trend of slight increase in basicity (refer data below), since HOMO energies are increasing slightly, and a significant increase in acidity, since LUMO energies decrease rapidly except for chlorides, bromides, and iodides of $\ce{N}$ and $\ce{P}$:
$$
\ce{NX3<PX3<AsX3<SbX3<BiX3}
$$
Data
All energies are reported in $\pu{eV}$.

Group 15 Trihalides
HOMO
LUMO

$\ce{NF3}$
-14.1362
5.3066

$\ce{NCl3}$
-11.2383
1.3535

$\ce{NBr3}$
-10.4263
0.7615

$\ce{NI3}$
-9.4958
0.1229

$\ce{PF3}$
-13.0271
3.8809

$\ce{PCl3}$
-11.2461
2.1658

$\ce{PBr3}$
-10.4584
2.1765

$\ce{PI3}$
-9.7250
0.7336

$\ce{AsF3}$
-14.0260
1.5621

$\ce{AsCl3}$
-11.5523
1.2755

$\ce{AsBr3}$
-10.9275
0.8633

$\ce{AsI3}$
-9.7864
0.3888

$\ce{SbF3}$
-13.3738
0.7485

$\ce{SbCl3}$
-11.2624
0.6446

$\ce{SbBr3}$
-10.6888
0.3214

$\ce{SbI3}$
-9.8524
0.0453

$\ce{BiF3}$
-13.7663
0.4729

$\ce{BiCl3}$
-11.4354
0.3244

$\ce{BiBr3}$
-10.8085
0.0180

$\ce{BiI3}$
-9.6861
-0.2291

P.S. These calculations take a lot of time, and though I am extremely curious, I am not sure if it is very useful or if I am on the right track. Some feedback or criticism would help. I will do a similar analysis with pentahalides if you find this useful.

This q. is 5 years old, with OP "Last seen more than a month ago", so I'm not sure she'll respond. Pentahalides are obviously Lewis acids, and OP got that^

Nitrogen triiodide is indeed Lewis acid so you got something there.


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Phenol and Deuterated Sulphuric Acid
What happens when phenol reacts with $\ce{D2SO4}$ in presence of $\ce{D2O}$?
How many hydrogens in total are replaced by deuterium?
Is the hydrogen of the hydroxyl group and three hydrogens ortho and para to the -OH group replaced by deuterium, and the remaining two unaffected?
OR are all hydrogens replaced by deuterium?
I'm not sure what really happens, but as far as I know - even though ortho and para positions are greatly activated, some small percentage of meta attack may also take place. In fact, if it does take place, the final compound will be totally deuterated, as bonds with deuterium are stronger than those with hydrogen, and backward reaction will not be thermodynamically favoured.
It'd be great if someone could explain what's going on, and if possible please quote sources supporting your answer. Thanks a lot.

Check what Phenol-d6 costs. If you can buy it and it is not prohibitively expensive, that would answer your question I guess.

The deuteration of phenol and aniline hydrochloride has been investigated by Werstiuk and Kadai by the High Temperature/Dilute Acid (HTDA) method. For phenol, ortho/para substitution is faster than meta (2.7 D/3.0 H vs. 1.3 D/2.0 H). OD/OH exchange was conducted at the end of the reaction. The presence of conc. HCl undoubtedly lowers the efficiency of incorporation of deuterium.
Incorporation of deuterium without the presence of conc. HCl allows for efficient deuteration of aniline hydrochloride. [RND3+ to RNH3+ exchange was conducted.] The authors note that conversion of aniline hydrochloride-d5 to phenol-d5 is efficient via diazotization.

Werstiuk, N. H.; Kadai, T., Can. J. Chem., 1974, 52, 2169.

Since phenol exhibits acidity ( phenol is a weak acid ), the first hydrogen atom to be replaced by deuterium will be the one in the hydroxyl group. Then, ortho and para positions relative to the hydroxyl group will be deuterated. You would obtain O,2,4,6-tetradeuterophenol. Here is an article which gives a slightly different example:
Banijamali, A. R., Charalambous, A. , van der Schyf, C. J. and Makriyannis, A., Specific deuteration of phenols and aromatic ethers using boron trifluoride and deuterium oxide. J. Label. Compd. Radiopharm. 1987, 24 1479-1482. doi: 10.1002/jlcr.2580241210.

We're more concerned with the final product.

@schrodinger_16 Will probably depend heavily on your acid concentration and temperature. If you can't find literature about this specific reaction, you'll have to try it out. Model reaction to run in an NMR tube. ;-)


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I mixed Miracle Gro fertilizer and Soil Mender Liquid Fish in my hose-end sprayer to save time. The mixture became very cold. Why?
Miracle Gro = Ammonium Sulfate, Potassium Phosphate, Potassium Chloride, Urea, Urea Phosphate, Boric Acid, Copper Sulfate, Iron EDTA, Manganese EDTA, Sodium Molybdate, Zinc Sulfate.
Soil Mender Liquid Fish = hydrolyzed fish.
Why would mixing these together in the jug of my hose-end sprayer cause the mixture to become very cold? Nothing else added.

Endothermic dissolution.


text:
Thermoplastic & Thermoset adhesives:
I have a question that is bothering me since few weeks. I'm not a chemist, nor I have any experience with polymers theory, but:
Epoxy - is defined on Wikipedia as a thermoset polymer as a process of polimerization occurs that can not be reversed.
PVA - Is defined (again on Wikipedia) as a thermoplastic polymer.
So why the cured epoxy has a glass transition temperature, and therefore when heated above that temperature it softens and its shape can be modified. But when PVA glue hardens (by loss of humidity) it can not be softened by heat?

First to clear up some terms. Glass transition just means that the intermolecular bonds are weak enough for the material to continue crystallizing—it is not the melting temperature.  This can be accompanied by softening of the material but that does not mean the polymer will flow.  Thermoplastics can flow at elevated temperature but thermosets (which includes cross-linked thermoplastics) do not by definition.
Epoxy has a glass transition due to a loss of crystallinity but is not usually shapeable unless formulated to or poorly prepared.  PVA is a thermoplastic that melts around $\pu{200^\circ C}$  which is also the temperature that it starts to pyrolyze and char. With careful temperature control you can melt and form PVA and there are 3D filaments dissolvable in water based on PVA.  Without careful control, you decompose the plastic before melting it.

By shapeable (heated epoxy), I ment, that can be bent with a plastic deformation, I assume this is not poorly prepared, as I noticed it on any epoxy I use. Don't you agree?

Yes thermoplastics can be plastically deformed.  One example of a room temperature deformable thermoplastic is PVA slime which is PVA cross-linked with borate groups. The length of the polymer units between cross-linking groups also has a large effect on the amount of strain a polymer may experience without failure, with shorter chains being more rigid.


text:
Tunneling in chemical reactions
We know that quantum tunneling is the reason behind several natural phenomenon like alpha decay and thermonuclear fusion inside the stars. How can it influence chemical reactions by tunnelling a species through the activation energy? If so how does it influence the kinetics and fraction of molecules taking part in the reaction.

Bit of a broad question. Relevant recent article: https://pubs.acs.org/doi/abs/10.1021/jacs.7b06035

It is not a broad question. I just wanted to know the influence of tunneling on a chemical reaction and it's parameters.

https://chemistry.stackexchange.com/questions/53763/tunneling-corrections-in-reaction-rates https://chemistry.stackexchange.com/questions/20377/significance-of-proton-tunneling

@orthocresol thanks for the article but I'm unable to access the complete article

Random primer: https://www.princeton.edu/chemistry/macmillan/group-meetings/DEC_tunneling.pdf

The probability of tunnelling at an energy $E$ is given by $p(E)\approx e^{-bA\sqrt{m}}$ where $A$ is proportional to the area of the potential energy barrier above  energy $E$, i.e. the top part of the potential barrier, $m$ the mass and $b$ some constants, $\pi, \hbar$. etc. Thus for a given mass and energy if the barrier is narrow, so $A$ is small, tunnelling is more likely than if the barrier is wide. At a given energy for the same barrier if the mass is large tunnelling is small. Thus we tend to see tunnelling only with H and D  and not Cl atoms for example. Tunnelling is also important in electron transfer reactions.
As there is not just a single energy in a reaction but a distribution of energies, according to the Boltzmann distribution, it is necessary to modify the expression above to average over the energy but the basic result is the same, which is that the reaction rate constant is reduced by the factor $p(E)$.
[If the potential energy barrier is $V(x)$ then $\displaystyle p(E)=\exp\left(-\frac{2\sqrt{m}}{\hbar}\int_{x_1}^{x_2}\sqrt{V(x)-E}\right)$ where $x_{1,2}$ are the points either side of  the barrier with energy $E$.]

Quantum tunneling in chemical reactions will only play a role if the collision energy and the spread in the collision energy of the reaction are very low. Just like in nucleosynthesis, the tunneling probability of a particle through a barrier only gets an appreciable value if the collision energy is very close to a quasi-bound state of the reaction complex (as in the triple alpha process). At this energy, this results in a large increase of the cross section (a scattering resonance), and thus the reaction rate (which is the product of the cross section and the relative velocity of the reaction partners). At high temperature, however, you can no longer speak of a collision of a single energy because different collision energies are mixed (to form a wave packet, if you like). This mixing of collision energies results in a mixing of the cross sections and washes out the sharp resonance of the single reaction channel. Resonances do play a role in astrochemistry where reactions happen at very low temperature and have been observed in the laboratory under controlled conditions, but are in general not important under normal conditions.


text:
Regarding interaction with metal sword and white gas
I have a large sword that is comprised of over 1000 layers of Titanium and High Carbon Steel. I wrapped it in Kevlar Wick and I set it on fire using Coleman White Gas (Camp Fuel) for performances. What I want to do is keep my sword in a tube or PVC pipe that has a reservoir of the white gas liquid fuel I use.
My question: will I harm my sword if I let it sit in the liquid gas for a long time?

What is in white gas?

This seems a very broad question, I do wounder if it is broadsword

It wont harm it more the putting on fire, which unless you're some kind of Thoros of Myr, isn't particularly good idea.

I certainly appreciate the joke, sorry for the vagueness of my question. This is a link to the exact fuel I have that I want to keep my sword with wick submerged in.
https://www.walmart.com/ip/Crown-Camp-Fuel-Gallon/51741744

also when it is on fire, the metal doesn't get hot at all. the wick is heat resistant and you swing the sword around, preventing it from heating even the wick. After the flame extinguishes you can immediately put your hand on the wick and feel only a gentle warmth

what I really was looking for specifically was if the liquid gas would begin to break down, structurally weaken the integrity of, or corrode my steel/titanium metal sword. if there is any more information I could provide I would be happy to try my hardest

I have looked at the fuel, it looks like some version of gasoline (US) or petrol (UK) which is for some special camp stoves. I wounder how the heating of the sword in the fire display will be any worse than the heating of the sword in a furnace during the forging will be. I am glad the OP liked the joke.

So from what I know about my fuel, is that it is different from lamp oil in the sense that my fuel burns brighter, bigger, and for less time than lamp oil would. When you use this fuel and wick the fire obviously is hot, but like I stated earlier when you swing the sword and keep the fire in motion it never actually get the sword or even the wick hot. it has a gentle warmth to it and when the flame is extinguished you can put your hands on it. The sword itself is wrapped in several layers of wick, and I am to reaching out to see if anyone knew if the gas would corrode the metal over time

https://www.facebook.com/photo.php?fbid=1031670893591597&set=a.105886432836719.12383.100002460889705&type=3&theater

If it is similar to gasoline i.e. just a mixture of alkanes, then I cannot see how it can break down to form something that will damage the metal surface. If anything it will be protective because you're going to want to keep water away from the surface

on the back of the can of fuel it does say it has properties added to prevent corrosion, so I would certainly think it would be safe, but I planned to have the blade and wick submerged in an airtight sheathe that would contain fuel, and it would be covered in liquid gas indefinitely until I unsheathe it and light the sword

I don't think the fuel will harm your sword. The metal won't react with the fuel.
Does your sword have any parts that aren't metal though? If there's any plastic, leather, fabric etc. then those could be dissolved or harmed by the fuel. I also don't know what will happen to your Kevlar wick if it's left in camp fuel for a long time. It might contain softeners or other chemicals that might leach out.
Most importantly, I am not comfortable with your tube or PVC pipe that has a reservoir of white gas in it. Camp fuel is flammable, and if you get a leak and it evaporates and mixes with air, it's quite possible to fill a car with an explosive mix. (Link to 1st search result on Youtube.) I know this sounds like a prosaic concern, but fire is dangerous, and I'm pretty sure more people get injured or die from stupid accidents with flammable substances than swords.
Please keep your fuel in a suitable, dedicated container. Read and follow the warnings on the container that your fuel comes in. Respect fire.

The tube/PVC pipe was just an example. I will be doing more research before I ever put the idea to practice and make a perfectly sealed (blade in and when the sword is out and I am performing) container and test it with other liquids before the white gas, then put the gas in the container and let it sit for a while, move it into different positions here and there, and make sure my container doesn't leak or degrade. I am no fool, I will take every reasonable precaution and test as much for as long as I can before I dedicate this idea into practice. Thank you so very much for your assistance


text:
Does ICP-OES detect molecules in suspension?
I'm curious whether ICP-OES has the ability to detect molecules in suspension?

https://en.wikipedia.org/wiki/Inductively_coupled_plasma_atomic_emission_spectroscopy

@Mithoron already read. provides no direct information regarding this question.

It will depend on the size of the particles which are in suspension. If the particles are very large compared with the capillary feeding the solution into the spray chamber then the capillary will block up and you will detect nothing and end up very unhappy with a broken machine.
If the particles are much smaller than the capillary then they will be able to enter the spray chamber and then be present in the small droplets which enter the plasma torch. As a result it will be possible to observe the particles by means of the ICPOES machine.
With a normal solution a droplet enters the torch.
The droplet dries out to make a solid particle
The solid particle melts
The molten droplet then boils in the heat of the plasma
The molecules formed are broken down to atoms in the heat of the plasma
The atoms are then ionized.
The inclusion of solid particles in the liquid fed into the ICPOES will give as the droplets boil to dryness additional solids particles. These solid particles may modify the behaviour of the other elements in the torch. I suspect that it may be similar in some ways to the processes which occur in the fireball as it cools after a low altitude nuclear detonation where things can adsorb onto existing particles. If you look in the literature at studies of fallout particles, the particles from the nevada tests have radioactivity deposited on minerals which are similar to those found at the test site.
I have found that if you increase the salt content of what you are feeding into either a ICPOES or ICPMS that the intensity recorded for a given concentration of an element (not found in the added salt) will decrease. This is a possible sign that some of the element is being lost onto the particles formed by the dehydration of the salt solution.
It is best if the ICPOES machine is only ever fed with samples which have been filtered through a very fine filter. I tend to use a syringe filter with holes which are about 0.5 micrometers. This helps to keep the fine particles away from the machine.

I concur with Nuclear Chemist.  You absolutely can see elements present in solids with ICP-OES: I've done it myself, in particular with suspended niobium and tantalum oxides.
If I aspirate a dilute suspension of tantalum oxide particles, I see a clear, albeit somewhat weak, peak at the $263.558\,\mathrm{nm}$ line for Ta:

    (Click to enlarge)
If I centrifuge the suspension and analyze the supernatant, though, the signal is appreciably reduced:

    (Click to enlarge)
I was only looking to test qualitatively for tantalum in these samples, so the above results were sufficiently conclusive.  Had I wanted to quantitate, it would have been quite challenging...ensuring a consistent method for (a) delivering the particulates into the plasma and (b) referencing against some sort of calibration curve would likely have been extremely challenging.


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How to calculate pH of the Na2CO3 solution given ambiguous Ka values
This is the exact question I faced on an exam:

Calculate the $\mathrm{pH}$ of $\pu{0.05 M}~\ce{Na2CO3}. \\(\ce{H2CO3}: K_\mathrm{a_1}= 4\times 10^{-7},~ K_\mathrm{a_2}= 4.7\times 10^{-11})$

Solution
$$\ce{Na2CO3 ->2Na+ + CO3^2-}$$
I suppose nothing that can contribute to the $\mathrm{pH}$ of a solution happens to $\ce{Na+}$ ions and we proceed with $\ce{CO3^2-}$ which has a concentration of $\pu{0.05M}$
$$\ce{CO3^2- + H2O <=> HCO3- + OH-}$$
To calculate $\mathrm{pH}$, I need to first figure out concentration of $\ce{OH-}$ ions, and to do so I have to know $K_\mathrm{b}$ dissociation constant for $\ce{CO3^2-}$
On equilibrium concentration of species are as follows: $$[\ce{CO3^2-}] = \pu{0.05 M}-x,[\ce{HCO3-}] = [\ce{OH-}] = x$$
So, we have:
$$\dfrac{x^2}{0.05-x}=K_\mathrm b$$
All I need to know is $K_\mathrm{b}$ value for $\ce{CO3^2-}$ ion. I tried to derive $K_\mathrm{b}$ from $K_\mathrm{a}$ values using $K_\mathrm{a} \times K_\mathrm{b} = 1\times 10^{-14}$ but apparently obtained the incorrect answer. What method should I use to do it the right way?

You should use that method, but you messed up somewhere. Without more work, we can't tell you what you did wrong.

@Zhe Thank you,i rechecked my work,found mistake and got the correct result.

The  carbonate ion is the Conjugate base of the weak acid
$\ce{HCO_3^-}\ (K={4.7\times10^{-11}})$, so this solution will alkaline. Given the concentration of this solution ,the pH should be sufficiently high to preclude the formation of any significant amount of $\ce{H_2CO_3}$ , so the solution of this problem as a solution of a monoprotic weak base:
$\ce{CO_3^{-2} + H_2O <=> HCO_3^-  + OH^-}$
$$\ce{K_b}=\frac{[OH^-][HCO_3^-]}{[CO_3^{-2}]} =\frac{K_w}{Ka}=\frac{10^{-14}}{4.7\times 10^{-11}}=\ce{10^{-3.7}}$$
Neglecting the $\ce{OH^-}$ produced by the autoprotolysis of water, it is valid to make the usual assumption that $\ce{[OH^-]}={[HCO_3^-]}$,and thus
  $$\dfrac{[OH^-]^2}{0.05-{[OH^-]}}=K_b= \ce{10^{-3.7}}$$
The equilibrium expression must be solved as a quadratic and yields the root $\ce{[OH^-]}$=0.00306 Which corresponds to pOH = 2.5 or pH = 11.5


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What is the proof of complete ionic dissociation in water?
Is there any way to prove that ionic compounds, for instance $\ce{NaCl}$, completely dissociate into singular $\ce{Na+}$ and $\ce{Cl-}$ ions within water by hard experimental evidence?
Besides adhering to the perfect solvation model in theory, I am interested in definitive proof of monoatomic solvation in practice.
How do we know that the ionic crystal lattice isn't only partially dissolved? When the superficial reaction between water and the ionic compound takes place, could it stop at say 10-100 atom microcluster lattices ($\ce{Na/Cl/Na/Cl}$ etc), giving the appearance by standard means that it is completely dissolved?
I am only interested in experimental means of verification, not concepts taken from books.

Osmotic pressure?

Possibly, but I am not sure if this thoroughly ensures that there aren’t still ultra small lattice clusters emulating ions to a sufficient degree. For example, what if the cluster had only 4 atoms, Na-Cl-Na-Cl? This particle is ridiculously small. How do we know that it doesn’t exist? Or rather, larger ones?

You wanna check out if Earth isn't flat too? Salty water conducts electricity way better and it wouldn't if salt wasn't dissociated.

Is that your experimental evidence? I am discussing whether or not it is completely dissociated vs. partially. Wouldn’t it follow that a partially dissociated salt would still conduct electricity in an electrolysis setup? The ions that are free could conduct electricity while not disproving the existence of micro lattices, which could still exhibit similar effects.

It strikes me as a statistical probability that there will be one Na+ ion next to a Cl- ion without a water molecule in between for some number of femtoseconds. But real "particles" of NaCl floating around... No. Particles of NaCl could be detected by x-ray diffraction, but it is a S/N problem. One cluster of four NaCl "molecules" floating around in 1 ml of water would be undetectable.

Precisely, therein lies the problem!

@MichaelS. - Yes, using statistical arguments you could calculate the probability of a 4 "molecule" cluster existing. But chemical properties are about "average" behavior not some probability like winning the lottery 5 times in a row. The chemical behavior of ionic solutions isn't determined by the concentration of the ions, but rather their activity. Within experimental error the plot goes through the origin. You have to realize with $6.023\times 10^{23}$ molecules per mole the state of  3 or 4 molecules just isn't important chemically.

If it is undetectable, how do we know it is insignificant? Also, we haven’t studied the nature of partially solvated micro lattices and their unique behaviors, which may chemically emulate the supposedly free ions for all we know. For instance, electrostatics and energy flow. For example, colloids can help with conductivity and gold nanoparticles can even undergo electrodeposition onto an electrode. My point is that we really don’t know, until someone comments and tells me they do. I just hope the logic is valid and not a repetition of a mantra. Please keep brainstorming this!

Given that Kohlrausch laws hold up remarkably well for ions coming from very different ionic solids, yup, you can be fairly sure that there isn't any nefarious "microlattice" masquerading as completely solved ions.
There are some clustering and counterion behaviours in completely dissolved ionic solids, but these are no different from the hydrodinamics of any other ion in solution and therefore separate from the dissolution process.

Then you should ask about finding traces of undissociated salt. Your question wasn't exactly clear.

One method is by making electrical conductivity measurements on aqueous solutions of salts. If we take four strong electrolytes:

Sodium chloride
Potassium chloride
Sodium perchlorate
Potassium perchlorate

We can measure the electrical conductivities at very low concentrations which allows us to obtain the limiting specific conductivties of the salts (extrapolation to zero concentration).
We will then find that the sum of the limiting conductivity of sodium chloride and potassium perchlorate will be the same as that of potassium chloride and sodium perchlorate.
This is because the limiting conductivity of a strong electrolyte salt is the sum of the contributions of the cations and anions. This allows us in an indirect way to show that some salts can totally dissociate when very dilute.

I think this is a good answer, but I was wondering whether limiting conductivities imply true ionic dissociation in practice even if the measurements are the same for different species. I understand logically the sum of the parts equals the whole, I am still trying to confirm the true nature of the parts. So, far this is a good answer. I am nearly satisfied. (Upvoted)

Another method is to look at boiling point elevation (using this to study dissociation is called ebulliscopy).  In theory (think ideal liquid), boiling points increase linearly with the molality of dissolved non-volatile components.  Thus change in boiling point of water is equal to the ebulliscopic factor times the solution molality.  This allows the estimation of a value 'i' that denotes the moles of dissociated ions vs the moles of the dissolved substance.
For instance, i=1 for sugar (no dissociation), i=1.9 for NaCl, showing that it is highly dissociated, i=2.3 for calcium chloride, etc..
This would indicate the likely conclusion that NaCl does not completely dissociate with the resulting ions randomly dispersed throughout an completely disordered solution (there is probably some local structure around the ions), but it's close.
(NOTE:  Alternatively, you can look at freezing point depression, which for low concentrations of solute is a linear function of molality, also.)

Taking aside modern methods on nanochemistry, many experiences can be used. The mixture will have very different properties depending on which is the final particle size, from entropy changes to conductivity changes.
You can argue in terms of conductivity just by the verification of the law of independent ionic molar conductivities. Salts must dissolve completely for the law verification.
The halfwave potential (polarography) is quite independent of the counterion.
In essence, almost anything that turns to be independent of the counter ion is a good proof (from electric methods above, to ionic chromatography, to selective electrodes etc.)
One of the simpler evidences can be the diffusion speed, which is easily modeled and measured and turns to be compatible which small particle size. Of course you can say that the models are not exact and small differences are hard to predict. Let's consider we just bounded the possible sizes to very few ions clusters. In such case you can measure (for example) frequency dependent dielectric permittivity and confirm that there are not relaxations at frequencies compatibles with molecular rotations.
By the way, thousands of good predictions are made monthly requiring such consideration (dissolution to individual ions), and do not forget nanochemistry.


text:
Find the boiling point of a compound given a pressure
I need to find the boiling point (in degrees centigrade) of ethanol on a day when the atmospheric pressure is $780\text{ torr}$.
I know that the molar mass of ethanol is $46.07\text{ g/mol}$ but I have no idea how to continue. Can you tell me what equations should be used or how is it solved, please?
Thanks!

Antoine Equation is fitted to experimental data if you just need a number.

@MaxW I appreciate that but I don't know how to find the unique solution. It seems we need some more data... right? Because I have the answer, and it's unique. With your link I don't understand where the solution is.

Huh? $$P = 10^{A-\frac{B}{C-T}}$$ is the solution. A,B,and C are constants. You can solve for T knowing P, or P knowing T.

Care, @MaxW, the equation is $$P = 10^{A-\frac{B}{C{\bf{\color{red}+}}T}},$$ with a plus :). Cheers

You could use the Clausius Clapeyron equation:
$$ \frac{\mathrm{d}p}{\mathrm{d}T} = \frac{\Delta H_\mathrm{v}}{\Delta V \cdot T}$$
Since the change in pressure is rather small, it is fair to assume the molar enthalpy of evaporation to be constant and use the value given for standard conditions, which should be easy to obtain.
The other simplification you can make is to assume that the volume of the liquid is very small compared to the volume of the gas:
$$ V_\mathrm{l} << V_\mathrm{g}$$
so that
$$\Delta V = V_\mathrm{g}-V_\mathrm{l} \sim V_\mathrm{g}$$
Now by using the ideal gas law and expressing volume as molar quantity,
$$v=\frac{p}{RT}$$
you get the ODE
$$ \frac{\mathrm{d}p}{\mathrm{d}T} = \frac{\Delta H_\mathrm{v}\cdot p}{T^2}$$
from whose solution you get a nice equation to solve your problem.

A good estimate of the boiling point of a material at 760 mm Hg can be made from other data, using a nomograph at the back of the Aldrich catalog (under Tables and Graphs) and online at https://www.sigmaaldrich.com/chemistry/solvents/learning-center/nomograph.html.


text:
Sephadex and purification of products
Sephadex is useful for the separation and purification of many natural products.
 Is it also useful for the separation and purification of artificial chemicals (in organic synthesis ) or not and why ?

I have not used Sephadex but here's my educated guess - the molecules produced in organic synthesis are mostly too small to separate from each other by this method, they will all elute together. The other issue is unfamiliarity. It is not routinely used in synthesis labs and no-one wants to risk losing their precious product using a technique they are not familiar with.

First, a conceptual but important point: there is nothing that sets artificial (a better word is synthetic) chemicals from natural products, chemically speaking. What we call "natural products" tend to be larger, complex molecules largely because we reserve the term for compounds for which extracting from natural sources is easier than synthesis, and these tend to be very large organic or bioorganic macromolecules, but there's no fundamental difference in behaviour between natural and synthetic molecules of those compounds, and in principle we could synthesise any natural product and reproduce exactly the same chemical properties that the same compounds have when extracted from natural sources. I state this because your use of the categories of natural and artificial can suggest some essentialist distinction while there is none.
Now, as to whether Sephadex can be used to separate and purify synthetic compounds, the short answer is yes, if these compounds are suitable. As @Waylander mentioned, and as with any separation medium, the main question is whether the compounds you want to separate are in the fractionation range of the medium. Sephadex can be manufactured with different bead sizes, which result in different separation ranges (these are marketed as G-10, G-25, ..., G-200). Different Sephadex gels can are used to separate proteins in the range from a few kDa to hundreds of kDa - and in principle they will be useful to separate synthetic organic compounds in the same molecular weight range through size exclusion. Commercial products are generally considered capable of discriminating molecules above ca. 700 Da.
Another factor to consider is functionalisation: Sephadex is highly functionalisable, which can produce specific affinities and retain desired compounds. This allows for a large range of possibilities for purification or separation, depending the composition of your mixture, what you want to retain and what you want to remove.
In practical terms, it is probably wise to begin by looking at separation procedures involving compounds of similar molecular weight and functional groups to those of your compounds of interest that are already described in the literature and work from there.


text:
Calculate the third and fourth nearest neighbours in bcc
My textbook has given data for third and fourth nearest neighbours to be 12 and 8 with distances $\sqrt{2}a$ and $\frac{\sqrt{11}a}{2}$.
I have been able to calculate for  the first and second nearest neighbour but it has become difficult to visualise for the other two to calculate. Can you help me with hints on how to proceed preferably with a diagram.

With $\sqrt3\over2$ being that close to $1$, BCC packing is better not looked at in terms of coordination spheres. But if you insist...
Say you are sitting in the center of a cell. Then:

Your first neighbours are at the corners of the same cell.
Second neighbours are at the centers of the nearest adjacent cells.
Third neighbours: centers of the next adjacent cells (those which share two corners with your cell).
Fourth neighbours: far corners of the nearest adjacent cells.

On a side note, there are more than just 8 of the latter.

"BCC packing is better not looked at in terms of coordination spheres" then in what terms should we look at it instead?

@GaurangTandon Two interpenetrating cubic lattices would do.

...as the one mentioned in Oscar's answer? ("two simple cubic lattice, one with points at coordinates $(ma,na,pa)$ where $m,n,p$ are integers, the other with points at $((m+(1/2))a,(n+(1/2))a,(p+(1/2))a)$")

Yeah, that one.

Although it is quite common to look at the radial distribution function from x-ray diffraction and find the peaks corresponding to the nth nearest neighbors. From that perspective, 'coordination spheres' make a lot of sense.

@Jon You know full well that the peaks have widths. How far shall we go before we stop being able to tell one peak from the next?

Getting out to 3rd and 4th nearest neighbors is not a problem, particularly in something like bcc (triclinic, now that is a different story). I don't think anyone is going to ask you about 20th nearest neighbors, since they aren't exactly near!

@Jon Not a problem? In a nice crystalline bcc, sure; but then you'd probably do a single crystal job (or Rietveld, for that matter, since there aren't awfully many parameters to refine) and never bother about RDFs in the first place. In a distorted nanoparticle stuff... not so sure anymore.

Sure, but in polycrystalline material that isn't distorted nano you can get out pretty far. Fine grained iron alloys for example...

Yeah, in that case I guess you can. I just wanted to point out that there are cases when you can't.

What does your first line mean by the way?

Just what it says. You can easily tell 1.5 from 1 and build a theory on that, but with 1.15 against 1, you won't be so sure.

@IvanNeretin Thanks for your answer. Could you please tell me how to spot the 4th nearest neighbour using the atom on the corner (instead of the centre one)?

@Abcd Use my construction, write down the vectors from the centre to the 4th neighbours, then apply the same vectors to the corner atom.

You can think of the body centered cubic lattice as two simple cubic lattice, one with points at coordinates $(ma,na,pa)$ where $m,n,p$ are integers, the other with points at $((m+(1/2))a,(n+(1/2))a,(p+(1/2))a)$.  If you work out increasing distances for both omponent cubic lattices you get, in units of $a$:
$\sqrt{0^2+0^2+1^2}=1$
$\sqrt{0^2+1^2+1^2}=\sqrt{2}$
$\sqrt{1^2+1^2+1^2}=\sqrt{3}$
$\sqrt{0^2+0^2+2^2}=2$ -- generally square roots of whole numbers
and also
$\sqrt{(1/2)^2+(1/2)^2+(1/2)^2}=\sqrt{3}/2$
$\sqrt{(1/2)^2+(1/2)^2+(3/2)^2}=\sqrt{11}/2$
$\sqrt{(1/2)^2+(3/2)^2+(3/2)^2}=\sqrt{19}/2$
$\sqrt{(3/2)^2+(3/2)^2+(3/2)^2}=3\sqrt{3}/2$ -- generally half the square roots of numbers that are 3 more than multiples of 8.  Each component square under the radical is one fourth of an odd square, and each odd square is one more than a multiple of 8.
Incidentally, there are not eight fourth-nearest neighbors.  There are 24.  The $3/2$ coordinate may occur along any one of three axes and each nonzero coordinate can independently have a negative sign.


text:
Why does 2-pyridone with chlorine gives 3,5-dichloro-2-pyridone?
Why does 2-pyridone react with chlorine to give 3,5-dichloro-2-pyridone? I thought that 2-pyridone is resistant and will yield only 5-chloro-2-pyridone or even a combination with 3-chloro-2-pyridone

resistant does not mean it does not react


text:
How did early chemists measure mass of atoms?
Here a book says that Berzelius measured atomic masses using a simple lab and his measurements was so close to modern results. Can someone explain how a person living in 19th century was able to measure atomic masses? Like he took this and those, observed this and that and did something else. Can I repeat his experiments at home too and get his results?
Do I actually understand atomic mass right? For me it is a mass in gramms of one atom of an element.
Also, I can't understand how Gay-Lussac measured the volumes of gases reacted with each other. What does the book mean by saying 2 volumes of Hydrogen? How much are 2 volumes? Are they measured in m^3 or what?
Please try to avoid the use of moles in answers as I don't understand why we need moles actually. So, I think I need to understand questions above first before moving to moles.

Related:  How was it possible to accurately measure the mass of gases in the 19th century? Actually, I think it is an exact duplicate. Could Turkhan please confirm? Welcome to Chem.SE btw!

I guess two volumes of hydrogen are twice the volume of another gaseous reactant.

Many people seem to believe that people were stupid in the old days. One of my favorite counterexamples is from 1932. It was the early days of nuclear physics (the neutron had just been discovered), and Cockroft and Walton used their newly-invented accelerator to drive the $^{7}$Li(p,$\alpha$)$\alpha$ reaction - the first human controlled nuclear reaction. Based solely on the observed range of the $\alpha$ particles they concluded a reaction Q valued of 17.2MeV. The current accepted value is 17.3468MeV. People were smart 100 years ago.

@JonCuster, why I ask this is not to understand how stupid/clever they were. Instead, I want to understand why I am not able to conduct such experiments and make such measurements

Perhaps a quote from the book would be helpful here. If you're not a chemist, your paraphrasing of the book may be incorrect. A better question is why are you bothering to repeat this? First, these values are well known, and second, you should not attempt to conduct experiments at home without knowledge of chemistry. No, moles are much more important than measuring atomic mass.

@GaurangTandon If it is a duplicate it is more like https://chemistry.stackexchange.com/q/81873/81 which explains how the weights of gases were measured in an answer.

Careful measurements were the foundation of chemical theory
Let's take the questions out of order to simplify the answers.
It doesn't matter what yours you use to measure volume as long as you stick to the same units in all your calculations. Cubic widgets is fine as long as you do all your calculations using them.
The key to understanding how chemists came to know about atomic mass is careful measurements of the widths and volumes involved in reactions. Chemists observed that certain reactions always involved the same mass ratios. More interestingly, when gases are reacted they also involve the same volume ratios. When hydrogen and oxygen react to for  water you start with two volumes (2 cubic widgets if you want) of hydrogen and one of oxygen react together to give water. How do we know it is just that ratio? Do the experiment with 3 volumes of hydrogen and one of oxygen and observe that there is still one volume of hydrogen left after the reaction is done.
Those experiments showed that the mass ratios were constant but different to the volume ratios if the reactants were gases. Oh, and we know the weight of the gases because we can measure the weight of a given volume of gas using a vessel of known volume containing a gas at a specific pressure using a carefully constructed vessel. A litre of oxygen weighs about 1.5g which is easy to measure using a good balance and good experimental conditions even in the 19th century.
Chemists hypothesised (and later proved) that the volume of gases were proportional to the number of molecules they contained so the ratio of gas volumes told them something about the number of units of substance (or molecules) involved in reactions. Comparing many reactions led to an understanding of more complicated reactions than hydrogen burning in oxygen. Comparing the known results from gas volume ratios to the known weights of substances involved in liquid and solid reactions extended this idea (with an early glitch caused because chemists didn't realise that many gases consist of diatomic molecules).
The actual number of molecules or atoms in a give volume took a lot longer to determine than the basic principles involved in chemistry (that reactions always involved specific ratios of inputs and outputs). Initial methods involved atomic theory applied to gasses or measuring the charge on a single electron. Later methods involved measuring the mass of a very pure crystal and comparing it with density derived from x-ray measurements to give more accurate values. (see this).
The basic experiments about atomic ratios can be repeated with good experimental technique and apparatus. But you need a very good way to measure weights of gases and volume. Repeatable, but not easy. You won't be easily able to measure the number of atoms in anything with household apparatus, though. That requires fairly sophisticated equipment.

How they were able to put the required gas into cubes? Just imagine I am at home. How can I put all neded gases into some cube knowing that the cube contains exactly the gas I need assuming I am in 18-19th centuries?

@TurkhanBadalov Vessels have fixed volumes even if they are not cubes. And that volume is easy to measure.

okay. But how do I collect the required gas? How did they obtain all required pure gases, especially hydrogen?

@TurkhanBadalov teaching chemistry from scratch (which is what you are effectively asking for) is a little beyond the scope of a single answer here.

you mean it is a large topic how to fill a vessel with a gas assuming we are in 18-19th century? I will be thankful if you share some link related to this

@TurkhanBadalov How to fill a vessel is basic stuff you can look up on the internet. Ways of making a range of gases suitable for experimentation is a big topic, far more than a single answer's worth on this site.


text:
What is the best way to strip ammonia from water at a mining site?
Working at a mine and have excess ammonia in the water. Was going to run the water through an ammonia wood chip bio filter, was wondering if this is the best technique or should I approach it from a different way?

You need an environmental chemist and a lot more information. How much water? What else is in the water? Concentration of ammonia? Where is water going to be discharged? What are appropriate regulations about discharge water?

In industry , the water would be boiled in a vessel and the vapors treated for specific components. Mine waters are notorious for containing a variety of stuff.

Interesting question, worrisome context... Sounds a bit like, "hi all, so I'm running this nuclear reactor and I'm wondering what to do about this 'core overheating'  error message that keeps popping up?"


text:
Mechanism of dimedone preparation
From the Wikipedia article on dimedone,

Dimedone is prepared from mesityl oxide and diethyl malonate.

I can't really figure out the mechanism for the reaction. Could someone please help me out?

Diethyl malonate is a very good nucleophile and mesityl oxide is a good Michael acceptor - does this give you a start?

Good question. Don't understand why they closed it.

@Abcd because it clearly meets the criteria for "homework question."

You will need to start in basic conditions. Dimethyl malonate is a good nucleophile on account of the two ester groups on either side of the central carbon atom. It has a $\mathrm{p}K_\mathrm{a}$ of 12.9, which is extremely acidic for a $\ce{C-H}$ bond. The first step is a Michael addition.

The resultant enolate will equilibrate, enabling it to form a much more stable 6 membered ring. We then have a Claisen condensation.

Lastly, decarboxylation will yield dimedone.

As I was writing out the mechanism, I had essentially the same steps. However, in your second step (equilibration of the enolates), the malonate is so acidic, I wasn't sure if the enolate you have on the right would ever really be able to form. However, assuming some excess base, it would probably happen eventually, and that attack to make the six-membered ring would be fast after deprotonation, so you wouldn't need it to happen much. Okay, I feel better about it now.

The Wikipedia page said diethyl malonate, instead of dimethyl malonate (the one you've used). I'm sure there's no huge difference and the product is still obtained - but which path do you think gives better yield and why?

Methanol is more acidic than ethanol making methoxide a better leaving group in a polar solvent, so probably dimethyl malonate, but I’m sure the difference is marginal.