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Chemical composition of silver tarnish
I had taken as a given that silver tarnish is a layer of silver sulfide on the surface of a utensil, and that cleaning it using a thiourea based agent like Tarn-X dissolves the sulfide layer.
However, while recently cleaning a variety of tarnished sterling and silver plated utensils, I noticed (1) the sulfureous odors emitted during the cleaning process varied considerably from one piece to the next; and (2) the tarnish on a few pieces could not be removed completely via the non-abrasive chemical process.
Knowing that many foods such as members if the cabbage and onion families have various sulfur compounds in them as flavor and aroma components, and that the sulfur containing amino acids cysteine and methionine are almost universally present in living organisms (including those we eat), I wondered: is silver tarnish a complex mixture of silver complexes that might reveal a history of the foods a person has eaten. Sherlock Holmes would know for sure.

The affinity of silver for sulfur is so high that silver may finally react with the S atom from thiourea. That is why I would not use thiourea for cleaning silver surfaces. Sodium bicarbonate works much better. .

Note that thiourea is a carcinogen. This is not the best cleaning approach, especially if you are smelling sulfurous compounds. You should search X-ray photoelectron spectroscopy of silver tarnish, such as https://www.researchgate.net/figure/XPS-spectra-of-sterling-silver-coupons-tarnished-and-cleaned-after-6-cycles-by-chemical_fig5_281703530 or https://www.chemistryworld.com/news/simulations-solve-mystery-of-why-silver-tarnishes/3010299.article

Upvote, Silver chemistry is always interesting!

Please see What happens during cleaning silverware?.

The composition of Silver tarnish is Silver sulfide, to quote from Wikipedia:

Silver sulfide is an inorganic compound with the formula $\ce{Ag2S}$. A dense black solid, it is the only sulfide of silver. It is useful as a photosensitizer in photography. It constitutes the tarnish that forms over time on silverware and other silver objects. Silver sulfide is insoluble in most solvents, but is degraded by strong acids.

The last sentence is important as Silver cleaning is not accomplished with harsh chemicals, but via electrochemistry which leads to the hydrogen atom radical formation, which then reduces the $\ce{Ag2S}$ liberating $\ce{Ag}$ and $\ce{H2S}$.  Here is a short summary of likely mechanics:
$\ce{Al -> Al3+ + 3 e-}$
$\ce{H2O ⇌ H+ + OH-}$
$\ce{HCO3- ⇌ H+ + CO3(2-)}$
$\ce{H+ + e- ⇌  •H }$
Effectively, $\ce{•H}$ can operate as a $\ce{(e-, H+)}$ pair:
$\ce{Ag2S + 2 •H -> 2 Ag + H2S (g)}$
Here is a confirming source, which states to quote:

The nascent hydrogen may also act as a reducing agent as it evolves from the metal surface.

And also, per the same source, the formation reaction of Ag2S can be reversed:

In that case silver suffers attack by dissolved oxygen and present sulphide species according to Reaction (5):
$\ce{4Ag (s) + 2HS- (aq) + O2(g) ⇌ 2Ag2S (s) + 2 OH- }$ (5)

So working in air, it is advised to employ a base to drive the above equilibrium to the left. Also, in alkaline conditions expect the reaction:
$\ce{H2S + 2 OH- -> 2 H2O + S(s)}$
Which is also noted in the referenced source, to quote:

However, after the cleaning of heavily tarnished samples (immersed for 30 and 60 min) some small quantities of sulfur (~0.3wt.%) are detected on the surface by EDS analysis, and the colour is changed to light yellowish.

[EDIT] With particular reference to the thiourea system, it is described in this hydrometallurgy source.  A cited reaction equation is:
$\ce{ Ag_2S + (SC(NH)(NH)2)2 + 2 H^+ + 4 SC(NH2)_2 -> 2 Ag(SC(NH2){_2}){_3}^{+} + S_o }$
Silver metal can then be recovered with reductive electrolysis.

I don't dispute the chemistry in your answer, but it addresses how tarnish--specifically, silver sulfide-- may be removed electrochemically or via sodium bicarbonate in the presence of Al metal, from a silver surface. My question referred to the use of thiourea - admittedly a controversial method since it strips off the tarnish rather than restores it to silver metal, and because of its adverse health effects. I am certain that other sulfur species besides hydrogen sulfide were released during the process, based on their unusual odors. Hence there was more than silver sulfide in the tarnish.

I have added a hydrometallurgy reference on a thiourea system.


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How does Aluminum react with nitric acid at different concentrations?
I am unable to find proper information. My book just says that conc. Nitric acid renders it passive.Could anyone please give the different reactions at different concentrations

No...I want the different reactions at different concentrations

Al don't react with HNO3 at any concentration. Asking same question about Fe would make more sense

It does react. I have read it somewhere

You want an exact answer where there is none. The passivation can be broken, but then there will be a wide variety of products. There is nothing exact about it.

Anymore problems in the language of question for anyone please let me know

Actually, the answer per this source: 'Reaction of aluminium with diluted nitric acid containing dissolved sodium chloride', is apparently no with respect to both dilute and concentrated nitric acid, except... to quote:

Metallic aluminium was found not to react with either concentrated or diluted nitric acid. Providing the diluted acid contains dissolved sodium chloride and traces of copper(II) cations, a vigorous reaction occurs. The product is basically nitrous oxide (possibly containing some elemental hydrogen and nitrogen gases), and was identified by its IR spectrum.

The qualifications for dissolution is interesting as it amounts basically to a galvanic cell anodic dissolution (think 'Bleach Battery' variation with HNO3 in place of HOCl and a copper metal electrode, the latter created from Aluminum displacing Cu(II) ions forming a Cu cathode on the Al surface).
[EDIT] Note: The above electrochemical viewpoint may apparently be a unique perspective on my part relating to the underlying reaction mechanics in conjunction with the role of NaCl in penetrating the protective Al2O3 layer (however, the author does also employ the phrase 'dissociation of the electrolytes'). Otherwise, the author does not explicitly mention 'electrochemical' anywhere in the article except in reference to the 'electrochemical series of elements'.

According to ASM ,(8 th ed. , vol 2), 5 to 20 % nitric can be used to clean aluminum for spot welding It is not intended to passivate or anodize. It was one of very many cleaning options.

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What is the heat needed for an isothermal expansion of gas. Why do physics and chemistry yield different answers?

One mole of a certain ideal gas is contained under a weightless piston of a vertical cylinder at a temperature T. The face of the piston opens into the atmosphere. What is the heat supplied in the process to expand the gas from volume $V_1$ to $V_2$ isothermally ? Friction of the piston against the cylindrical wall is negligible is small

Now thermodynamics is a common topic to both physics and chemistry and as per my understanding of thermodynamics I am getting different results for the heat that will need to be supplied.
Chemistry says $$∆U =∆Q +∆W $$ and defines work as the work done on the gas and this will give $$W = -P_{atm}(V_2-V_1) $$ and as for isothermal process$$ ∆U =0 , ∆Q =P_{atm}(V_2-V_1)$$
According to physics on the other hand $$∆Q =∆U+∆W$$ and here the work done is the work done by the gas. That is
$$W = \int_{V_1}^{V_2} PdV=\int_{V_1}^{V_2} \frac{RT}{V}dV = RT \ln V_2-RT \ln V_1$$ and as for isothermal process $$∆U =0 ,\, ∆Q =RT \ln \left(\frac{V_2}{V_1}\right)$$
I am also uncertain that maybe neither of these is the correct expression and the network ok is actually in that work from both these forces and we have to take the sum of these to yield $$∆Q =RT\ln \left(\frac{V_2}{V_1}\right)-P_{atm}(V_2-V_1)$$
Now the value of Q should be unique because in the real world when we perform an experiment, only one value of Q would be supplied.
So which one of these is correct and where am I lacking in my understanding of thermodynamics from a chemistry  standpoint ??
Now as this question links both physics and chemistry, I want to post it on both sites (I hope it's fine) and the link to it is here

It's not particularly fine - crossposting is frowned upon.

Sorry, but actually it doesn't... I'm not asking exactly why they are different , but just that due to different conventions, what I understand is we take different pressure (ie internal P v/s external P) and thus get different values of Q, whereas there should only be one value(As far as I understand)

Hmm. I would suggest you to read the definitions of first law of thermodynamics again as you are making a common misconception! BOTH equations are equivalent. The ∆W in physics is work done by the gas, whereas ∆W used in chemistry is work done on the gas. As these are always equal in magnitude and opposite in sign you always have the exact same equation!. You may use whichever definition you prefer in whichever subject you wish! Just ensure you don't mix up the sign anywhere/ calculate one quantity instead of the other.

At last, some questions they may ask just to calculate "∆W" without mentioning work done on/by the gas. Only in this(annoying) case do you have to answer as per the "subject" this question is from.

I understand they only differ in sign , but I want to know , which one is correct ?

@RandomAspirant - RE: "... which one is correct?" – Are you a chemist or a physicist?

The problem is that both the problem statement you presented and the chemistry solution you showed are fallacious.  The fallacy is that, for an ideal gas $V=nRT/P$, so that, if the pressure and temperature are both constant, the volume can't change.  If the volume does change, then either the pressure must change or the temperature must change.
In the "chemistry solution" you presented, the pressure is taken as constant (atmospheric).  So the temperature must change.  But if the temperature changes, then $\Delta U$ is not zero (as they assume).  So the solution is incorrect.  The correct solution, based on the 1st law, is $$\Delta U=nC_v\Delta T=Q-P_{atm}(V_2-V_1)$$or equivalently, $$\Delta H=\Delta U+P_{atm}(V_2-V_1)=nC_p(T_2-T_1)=Q$$Also, since $$n=\frac{P_{atm}V_1}{RT_1}$$we have $$\frac{T_2}{T_1}=1+\frac{QR}{C_pP_{atm}V_1}$$
In the "physics solution" you presented, they make the opposite assumption that, rather than the pressure remaining constant, the temperature remains constant.  For the temperature remaining constant, the solution you presented is correct.  But the final pressure is:  $$P_2=P_{atm}\frac{V_1}{V_2}$$

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Why benzenediazonium bromide and iodide are of little interest?
When we think of aryldiazonium compounds, the first compound that comes to mind is benzenediazonium chloride. Sure thing, the compound is versatile and has been used to make a vast array of organic compounds and thus serves its purpose as a useful reagent. But what about other halides like benzendiazonium fluoride, bromide and iodide? There is little to no literatureon these compounds suggesting they are of little interest. What makes them so un-noteworthy? Is it because of stability? Is it because of unfavourable reaction conditions? Or is it because BDC is so popular that other halide compound hasn't been tested or experimented with?

This paper refers to arenediazonium bromides. It is behind a paywall so I can only see the abstract  https://link.springer.com/article/10.1134/S1070428012040094

Could it be this one: https://digitalworks.union.edu/cgi/viewcontent.cgi?article=3127&context=theses   - a thesis not a paper, it also has an attempted preparation of benzene diazonium iodide which is highly unstable. There is also another google hit that mentions the instability of the iodide here  https://archive.org/stream/in.ernet.dli.2015.502084/2015.502084.A-Course_djvu.txt

There is indeed very little to be found in the literature.
A recent Russian paper (Ref.1) mentions benzenediazonium bromide as a reactant. This thesis from 1975 (Ref.2) contains a preparation of benzenediazonium bromide and an attempted preparation of benzenediazonium iodide; it notes that the iodide is unstable as does this reference here.  The discussion section of this Organic Syntheses' preparation of fluorobenzene (Ref.3) mentions an alternative preparation by formation of benzenediazonium fluoride and its decomposition referencing work from the 1900s.
My overall thought is that they are probably less stable than the well-known chlorides and undergo reaction in lower yields so they are of little interest.

References:

Yu. V. Ostapiuk, V. S. Matiichuk, N. I. Pidlypnyi, N. D. Obushak, "Convenient synthesis of $\alpha$-bromo ketones by the Meerwein reaction," Russian Journal of Organic Chemistry 2012, 48, 519–522 (https://doi.org/10.1134/S1070428012040094).
Geoffrey D. Picard, "Sulfonium Salts, Aliphatic and Aromatic Diazonium Salts, Formation and Isolation," Honors Theses 1975, 2119 (https://digitalworks.union.edu/theses/2119)(PDF).
D. T. Flood (Checked by W. W. Hartman and J. R. Byers), "Fluorobenzene," Org. Synth. 1933, 13, 46 (DOI: 10.15227/orgsyn.013.0046) and Organic Syntheses, Coll. Vol. 1943, 2, 295 (PDF).

Also, shouldn't it be cheaper to make chlorides using readily available $\ce{HCl}$ than go for expensive $\ce{HBr}$ and $\ce{HI}$? Just saying.:-)

Yes, if you're using them for synthesis, but that would not account for such extensive neglect by researchers.

Mostly it's a matter of economy.  In most diazonium salt synthesis the counterion is not involved in the organic reactions; it's just a spectator.  So we choose chloride because hydrochloric acid is a cheap and effective chemical to acidify the aromatic amine+nitrite mixture and make the diazonium ion in the first place.  Ergo an aryldiazonium chloride.
The paper (Ref.1) mentioned by Waylander in the comments and reference cited in his answer involves bromide as the counterion because, exceptionally, this ion does serve as a reactant.  In that paper the ketone reacts with the diazonium ion to give a benzylic cation, which then combines with the bromide counterion from the original diazonium salt to give the bromine-bearing product.

The case of benzenediazonium fluoride is special. In addition to the economic aspect in applications where the halide is a spectator, we cannot effectively make the fluoride salt in cases where the fluorine would serve as a reactant. Hydrogen fluoride is a strong acid as such, but in dilute aqueous solution it turns into a tightly bound ion pair, $\ce{H3O^+F^-}$, which is then too weak to effect the diazonium ion formation (see here for a discussion of this property). Instead, in the Balz-Schiemann reaction, tetrafluoroboric acid is used to form the diazonium tetrafluoroborate, whose anion then acts as a fluoride ion source to form fluorine-bearing products. Fluorobenzene and 4-fluorobenzoic acid are among the products reached via this method.
Benzenediazonium tetrafluoroborate has a safety advantage over the chloride. The latter can be explosive if isolated, apparently due to internal reaction vetween the cation and anion of the salt. The tetrafluoroborare, with a less reactive anion, is safer to isolate and store.
References:

Yu. V. Ostapiuk, V. S. Matiichuk, N. I. Pidlypnyi, N. D. Obushak, "Convenient synthesis of $\alpha$-bromo ketones by the Meerwein reaction," Russian Journal of Organic Chemistry 2012, 48, 519–522 (https://doi.org/10.1134/S1070428012040094).

Good answer by @Waylander and @OscarLanzi. They provided some literature evidence of the synthesis procedure of Benzenediazonium bromide and iodide and discussed its usage and worth and explained why they are not that important as compared to benzenediazonium chloride. My answer just revolves on the actual synthesis procedure(slightly abridged in my answer) of benzenediazonium bromide and attempted synthesis of benzenediazonium iodide sincerely taken from 1975 thesis(Ref. 2 of @Waylander's answer). Thus fully completing the discussion.
Benzenediazonium Bromide
Take 250 ml copper vessel. Add 20g ice and 20g conc. nitric acid. Maintain a temperature of -10 °C with constant stirring of the mixture. To this, add 5g of aniline. The reaction  is exothermic so keep mixture under -5 °C by pouring liquid nitrogen outside the vessel. A white to tan colored suspension is formed. To this, add 3g of finely powdered sodium nitrite in small amounts for 30 seconds till the tan colored disappears thus the reaction is complete.
3 ml of bromine is added to 100 ml water saturate with sodium bromide. Cooled at -5 °C and then added to diazonium mixture. A yellow orange crystal is formed immediately. Maintain constant stirring for 15 minutes. Then the solid is filtered and cooled to 0 °C. 5 g of wet solid is obtained(dec. at 78 °C) which was confirmed to be benzenediazonium tribromide. It was added to 100 ml acetone. 1.5 g(18% yield) of benzenediazonium bromide(dec. 108 °C) were obtained and bromobenzene as side product.
Benzenediazonium Iodide
Repeat the above procedure for making the diazonium mixture. Then add 4ml bromine to water saturated with potassium iodide. Cooled at -5 °C and then added to diazonium mixture. Reddish brown crystal were formed which decomposed immediately at -10 °C. The product has a boiling point of 188 °C and was identified as iodobenzene, not benzenediazonium iodide, in the thesis.

"Decomposed immediately at -10°C". Slide 19 here says you can go down to -23°C.  Would that buy more stability?

Point of clarification: in the iodide synthesis, the thesis reports "the product had a bp of 188°C and was iodobenzene". No iodide salt was reported.


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Retrosynthesis of spirocyclic compounds?
I have been trying to complete this retrosynthesis problem for a few days but can’t figure out how to get from Ester starting compound to a spirocyclic compound. I have tried removing the Ester but can’t figure out what best to replace it with.
I have access to any reagents and starting materials as long as the nominated substance is used.

Step 1 - Form the anion of your starting material ($\ce{NaOEt/EtOH}$ will do this nicely) and react with phenethyl bromide ($\ce{PhCH2CH2Br}$).
Step 2 - Add the product to hot polyphosphoric acid$\ce{^{[1]}}$ to cyclize it to the required product.
Reference

Cyclization of Aryl-Aliphatic Esters with Phosphorus Pentoxide in Phosphoric Acid by Richard C. Gilmore, Journal of the American Chemical Society, 1951, 73 (12), 5879-5880, DOI: 10.1021/ja01156a521


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How the attraction takes place between a hydrogen atom and the acceptor atom in a hydrogen bond
In hydrogen bonds present in water,does the hydrogen atom attracts to the lone pair of electrons present on the oxygen atom or to its partial negative charge? If former is the case are there any examples in which the acceptor atom do not have a lone pair of electrons but it attracts hydrogen because of its partial negative charge.
I got confused because I saw in YouTube video where the definition of hydrogen bond was given as: A hydrogen bond is a weak bond between a partially positive hydrogen atom and a partially negative electronegative atom

https://chemistry.stackexchange.com/questions/5503/what-exactly-is-hydrogen-bonding-and-does-it-really-need-fluorine-oxygen-or-n

Well, there has to be an orbital donating a bit of electron density, so that hydrogen bond gets some covalent character.


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How to identify if a material is perovskite or not?
This is more of a materials science question, but I believe many of you have referred NIST Inorganic Crystal Structure Database (ICSD) database.
I am collecting a database of perovskite materials and got a little bit confused on how the structure type is defined in the ICSD website.
Please find a screen shot of ICSD results on lead zirconate titanate (PZT) material.

Here, when mole fraction $\ce{Zr}=0.52$, the 'Struct. Type' is defined as $\ce{FeNbPb2O6}$. Does it mean that PZT material when $\ce{Zr}=0.52$ is not perovskite? And in general if the column 'Struct. Type' does not start with 'Perovskite,' does that mean the material is not perovskite?
Finally, sometimes I see the structure type named 'Elpasolie' or $\ce{LiNbO3}$ (disordered). Are these also perovskites? (or double perovskites?)

As MaxW indicated in the comment, perovskites are a huge family of natural and synthetic compounds. Following is a breif description of the group of compounds named perovskites:
The original Perovskite is a mineral of formula $\ce{CaTiO3}$. It was first discovered in 1839 by the Prussian minaralogist Gustav Rose in the Ural Mountains, and is named after Russian mineralogist Count Lev Aleksevich von Perovski (1792–1856). The natural crystals have a hardness of 5.5-6 and a density of $\pu{4000-4300 kgm-3}$. The crystal structure of this mineral was initially thought to be cubic, but was later  shown to be orthorhombic (Ref.1).
As with many minerals, Perovskite has given its name to a family of compounds with the same type of crystal structure, which are called perovskites that have a general formula close to or derived from the composition $\ce{ABX3}$. In these mineral type with formula $\ce{ABX3}$, $\ce{A}$ and $\ce{B}$ are usually two different sized cations, while $\ce{X}$ is an anion that bonds to both (e.g., $\ce{KNbO3}$). For instance, $\ce{PbTiO3}$) is a perovskite, crystals of which are Tetragonal (space group: $P_4mm$) with $a=\pu{0.3902 nm}$ and $c=\pu{0.4143 nm}$ of its unit cell. In contrast, the original Perovskite, $\ce{CaTiO3}$ crystals are Orthorhombic (space group: $Pbmn$) with $a=\pu{0.54035 nm}$, $b=\pu{0.54878 nm}$ and $c=\pu{0.76626 nm}$ of its unit cell (Ref.1).
To some extent, the multiplicities of phases that belong to the perovskites family can be rationalized by assuming that perovskites  are simple ionic compounds where $\ce{A}$ is usually a large cation and $\ce{B}$ is usually a medium-sized cations, and $\ce{X}$ is an anion (if the charges of the ions are written as $q_\ce{A}$, $q_\ce{B}$, $q_\ce{X}$, then: $q_\ce{A} + q_\ce{B} = 3q_\ce{X}$). At present, many hundreds of compounds are known that adapt the perovskite structure. Few examples are $\ce{NaMgF3}$ (Orthorhombic; space group: $Pbnm$), $\ce{CsPbI3}$ (Cubic; space group: $Pm \overline{3} m$),  $\ce{KCuF3}$ (Tetragonal; space group: $I4/mcm$), $\ce{LaAlO3}$ (Trigonal; space group: $R3c$), $\ce{KNbO3}$, $\ce{KNbO3}$ (Orthorhombic; space group: $Amm2$), and $\ce{(Fe,Mg)SiO3}$ (Orthorhombic; space group: $Pnma$). In fact, a perovskite structure mineral, Bridgmanite $\left(\ce{(Fe,Mg)SiO3}\right)$ is the most abundant solid phase in the Earth’s interior, making up $38\%$ of the total (Ref.1).
Figure A in following diagram illustrates an ideal cubic perovskite, $\ce{SrTiO3}$ (space group: $Pm \overline{3} m$):

The traditional view of the perovskite lattice is that it consists of smaller $\ce{B}$ cations within oxygen octahedra, and larger $\ce{A}$ cations which are nine fold coordinated by oxygen. According to this Princeton website where I abstracted the structure A (Note: I have modified to some extent):

The easiest way to visualize the structure ($\ce{ABO3}$) is in terms of the $\ce{BO6}$ octahedra (e.g., $\ce{TiO6}$), which share corners infinitely in all 3 dimensions, making for a very nice and symmetric structure. The $\ce{A}$ cations ($\ce{Sr}$) occupy every hole which is created by 8 $\ce{BO6}$ octahedra, giving the $\ce{A}$ cation a 12-fold oxygen coordination, and the $\ce{B}$-cation a 6-fold oxygen coordination. In the example shown below, ($\ce{SrTiO3}$) the $\ce{Sr}$ atoms sit in the 12 coordinate $\ce{A}$ site, while the $\ce{Ti}$ atoms occupy the 6 coordinate $\ce{B}$ site. Note that there are many $\ce{ABO3}$ compounds for which the ideal cubic structure is distorted to a lower symmetry (e.g., tetragonal, orthorhombic, etc.).

Based on these facts, your last two entries, $\ce{Pb(Zr_{0.45}Ti_{0.55})O3}$ and $\ce{Pb(Zr_{0.871}Ti_{0.129})O3}$, are of course belong to perovskite family. The first one have adapted to the structure of $\ce{PbTiO3}$ (Tetragonal; space group: $P_4mm$), which is a perovskite (Ref.1). The second one have adapted to the structure of $\ce{LiNbO3}$ (Trigonal; space group: $R3c$), which is a synthetic perovskite (Wikipedia).
There is another family of compounds with formula $\ce{A2BB'O6}$, which is called double perovskites. The crystal structures of double perovskites feature an ordered rock-salt like arrangement of corner-sharing $\ce{BO6}$ and $\ce{B'O6}$ units in the crystal structure (Figure B). This is what researchers in Max Plank Institute where I have extracted the Figure B have to say about double perovskites:

Proper choice of the metal ions $\ce{B}$ and $\ce{B'}$ and of the alkali, alkaline earth, or rare earth ions $\ce{A}$ allows to realize a large variety of physical properties in double perovskites. Prototype compounds which have found much interest are $\ce{Sr2FeMoO6}$ and $\ce{Sr2FeReO6}$ as they are half-metallic ferromagnets with Curie temperatures above room temperature and large magnetoresistance effects at ambient conditions.

The double perovskite structure is so named because the unit cell of double perovskite is twice that of usual perovskite. An example of crystal structure of a synthetic double perovskite, $\ce{Sr2FeMoO6}$, is shown in Figure C, which has a Orthorhombic (Phase Prototype: $\ce{Sr2WNiO6}$; space group: $Cmme$) unit cell (Ref.2). It has the same architecture of 12 coordinate $\ce{A}$ sites and 6 coordinate $\ce{B}$ sites, but two cations are ordered on the $\ce{B}$ site. The structure of $\ce{Sr2FeMoO6}$ shown here,the $\ce{Fe}$ and $\ce{Mo}$ atoms have ordered in a 3D chessboard type fashion.
Hence, it is safe to suggest that the compound used for first three entries, $\ce{Pb(Zr_{0.52}Ti_{0.48})O3}$, is an example of double perovskites, solely because it has a prototype structure of $\ce{Pb2FeNbO6}$ (Monoclinic; space group: $Cm$), which resembles a double perovskite ($\ce{A2BB'O6}$ type). The crystal structure of $\ce{PbFe_{0.5}Nb_{0.5}O3}$ is slightly distorted cubic ($a = b = c = \pu{4.0056 \mathring A}$; and $\alpha = \beta = \gamma = \pu{89.664 ^\circ C}$) and declaired to be a perovskite (Ref.3). On the hind site, $\ce{PbFe_{0.5}Nb_{0.5}O3}$ resembles more of $\ce{Pb(Zr_{0.52}Ti_{0.48})O3}$ than that of $\ce{Pb2FeNbO6}$.

Note: The best advice to OP is start with the book (Ref.1), which is written in basic language, understandable for any novice student learning about Perovskite materials. In addition, since OP is conveniently in England, I believe he can consult the author of the book, Prof. Richard J. D. Tilley, Ph.D., the Emeritus Professor in the School of Engineering at the University of Cardiff, Wales, United Kingdom, if OP has additional questions on Perovskite materials.

References:

Richard J. D. Tilley, In Perovskites: Structure-Property Relationships; John Wiley & Sons Ltd.: Chichester, West Sussex, United Kingdom, 2016 (ISBN: 978-1-118-93566-8).
Y. C. Hu, J. J. Ge, Q. Ji, B. Lv, X. S. Wu, G. F. Cheng, "Synthesis and crystal structure of double-perovskite compound $\ce{Sr2FeMoO6}$," Powder Diffraction 2010, 25(S1), S17-S21 (https://doi.org/10.1154/1.3478711).
Renbing Sun,  Wei Tan,  Bijun Fang, "Perovskite phase formation and electrical properties of $\ce{Pb(Fe_{1/2}Nb_{1/2})O3}$ ferroelectric ceramicse," Physica Status Solidi A: Applications and Meterial Science 2009, 206(2), 326-331 (https://doi.org/10.1002/pssa.200824432).

Excellent! This is a perfectly compiled answer that is so rich in information and I can't thank you enough for it. I believe I must do more research on perovskite materials, however, there seems to be a bit of controversy among scientists too, for defining a decision boundary between perovskites and non-perovskites. For an example, LiNbO3 may be identified as a synthetic perovskite, however, some studies differentiate between perovskite and LiNbO3 structures. Finally, if you think the question shadows this great answer, please edit the question.

@ Achintha: No need to change anything. That's why I put the infor of auther who lives in UK, so he can contribute (I assume it'd be similar in UK, alike here in US! :-) Good luck.

The HMS column is for the Hermann–Mauguin Symbol.
You can see the line for 55700 has "P4mm" which is the proper crystal symmetry to be of the Perovskite structure.
That line is also notated "Perovskite#PbTiO3#B". So of the formulations listed that is the only one that has the Perovskite crystal structure.

Sorry to ask this question again as it might be opinion based. But I found elsewhere that Pb2FeNbO6 is perovskite. Doesn't is make the first 3 materials in the image perovskite too?

Perovskite itself is the mineral $\ce{CaTiO3}$.  Exactly how the huge family of the synthetic perovskite derivatives fits together I do not know.


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On the localized screening of cerium (when compared to iron)
I was reading the paper 'A linear response approach to the calculation of the effective interaction parameters in the LDA+U method' by Cococcioni - (Paper here) .
Quoting from the first paragraph on page 9, 'The screening in metallic cerium is extremely localized..'. Even though this section deals with the calculation of the Hubbard 'U' (which might be more appropriate for a physics forum), the underlying chemistry is still intact, I understand. But I can't reconcile this from a chemistry point of view. To furnish more context, the authors compare the results of metallic cerium with metallic ion (from the previous section in the paper).
To recapitulate, the assertion here is that the screening in metallic cerium is extremely localized when compared to metallic iron. I know this is directly related to the electronic configuration, but I would be happy if someone would be able to provide an explanation for this.
PS: I added a bounty and even though it says 'answer required from reputable source', I will accept any detailed answer that solves the ambiguity, even if you don't cite or refer any sources.


text:
Qualitative Analysis- Distinguishing between two ions or two salts
I know my question is going to be quite below the level of this site but I am 1st grade in university and I can't understand anything.
I am going to give an example from the book: distinguish between $\ce{AgI}$ and $\ce{AgCl}$. Answersheet tells me to add $\ce{NH3}$; only $\ce{AgCl}$ dissolves. I don't even know why $\ce{AgI}$  precipitate and $\ce{AgCl}$ dissolves.
Or there is another question: distinguish between $\ce{Ag+}$ and $\ce{[Ag(NH3)2]+}$. And answer sheet tells to add NaCl and only $\ce{Ag+}$ gives a precipitate. Why doesn't $\ce{[Ag(NH3)2]+}$ also give a ppt.?
How do I know which cations or anions or salts are soluble in which solvent? What should I look when I am trying to distinguish ions or salts? Or how do I know if there is no reaction? I know these questions are too general but I am having a hard time understanding this class.
You can either answer the questions from the book (which I failed to understand) or help me to understand the basis of the subject by answering others. Or both. I would really appreciate any kind of answers coming from you.

Hi Ceren, Welcome to chemistry. SE! Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

AgI has extremely low $K_{sp}$ value, and $K_f$ of $\ce{[Ag(NH3)]2+}$ falls too less to compensate it. However AgCl has much higher $K_{sp}$ value compared to AgI and high $K_f$ value makes it soluble. For precipitation of AgCl, presence of free $\ce{Ag+}$ and $\ce{Cl-}$ is required. Coordination complexes, like $\ce{[Ag(NH3)2]+}$ here, don't break into free ions in aqueous solution. They are stable and exist as single entity, thus, won't give precipitation with NaCl.

Note to OP: Feel free to upvote, downvote or ignore my answer. For what is it worth, I do NOT think you should have to know such fine points! By the way, welcome to the chemistry stack exchange!

Answering this question requires a preliminary discussion of the solubility product constants of three silver halides and the formation constants of three silver complexes. First, note that $\ce{AgCl}$, $\ce{AgBr}$, and $\ce{AgI}$ are all insoluble in water, but insoluble is a relative term in the end. Their respective solubility equilibria and solubility product constants, i.e., equilibrium constants for their dissolution, are:
$$\ce{AgCl(s) <=> Ag^+ (aq) + Cl^- (aq) \quad $K_\mathrm{sp(1)} = \pu{1.8E-10}$              \tag 1}$$
$$\ce{AgBr(s) <=> Ag^+ (aq) + Br^- (aq) \quad $K_\mathrm{sp(2)} = \pu{5.4E-13}$              \tag 2}$$
$$\ce{AgI(s) <=> Ag^+ (aq) + I^- (aq) \quad $K_\mathrm{sp(3)} = \pu{8.3E-17}$              \tag 3}$$
Silver ions also form complexes with ammonia, thiosulfate ion, and cyanide ion. The formation equilibria and associated equilibrium constants are as follows:
$$\ce{Ag^+ (aq) + 2 NH3 (aq) <=> [Ag(NH_3)_2]^+ (aq) \quad $K_\mathrm{f(4)} = \pu{1.6E7}$              \tag 4}$$
$$\ce{Ag^+ (aq) + 2 S2O_3^2- (aq) <=> [Ag(S2O3)2]^3- (aq) \quad $K_\mathrm{f(5)} = \pu{2.0E13}$              \tag 5}$$
$$\ce{Ag^+ (aq) + 2 CN^- (aq) <=> [Ag(CN)2]^- (aq) \quad $K_\mathrm{f(6)} = \pu{1.0E21}$              \tag 6}$$
Now consider the following sequence of aqueous solution additions:

An aqueous $\ce{NaCl}$ solution is added to an aqueous solution of $\ce{AgNO3}$. Then $\ce{AgCl}$ precipitates, as per equilibrium (1).
Next concentrated ammonia is added in excess, i.e., well above the 2 to 1 stoichiometry of equilibrium (4). Then the following equilibrium, obtained by adding equilibria (1) and (4), occurs:

$$\ce{AgCl (s) + 2 NH3 (aq) <=> [Ag(NH3)2]^+ (aq) + Cl^- (aq) \quad $K_\mathrm{sp(1)} K_\mathrm{f(4)}  = \pu{2.9E-3}$              \tag 7}$$
Although the equilibrium constant is less than unity, adding concentrated ammonia in excess (in a hood!) results in all of the $\ce{AgCl}$ dissolving: the equilibrium is driven to the right.

Next $\ce{NaBr}$ is added. This results in precipitation of $\ce{AgBr}$ via the following equilibrium:

$$\ce{[Ag(NH_3)_2]^+ (aq) + Br^- (aq) <=> AgBr (s) + 2 NH3 (aq) \quad $1/(K_\mathrm{sp(2)} K_\mathrm{f(4)})  = \pu{1.2E5}$              \tag 8}$$
This equilibrium is simply the reverse of the addition of equilibria (2) and (4).

Next excess sodium thiosulfate ($\ce{Na2S2O3}$) is added. Then $\ce{AgBr}$ dissolves as per the following equilibrium, obtained by adding equilibria (2) and (5):

$$\ce{AgBr (s) + 2 S2O3^2- (aq) <=> [Ag(S2O3)2]^3- (aq) + Br^- (aq) \quad $K_\mathrm{sp(2)} K_\mathrm{f(5)}  = \pu{10.8}$              \tag 9}$$

Next $\ce{KI}$ is added. This results in precipitation of $\ce{AgI}$ via the following equilibrium:

$$\ce{[Ag(S2O3)2]^3- (aq) + I^- (aq) <=> AgI (s) + 2 S2O3^2- (aq) \quad $1/(K_\mathrm{sp(3)} K_\mathrm{f(5)})  = \pu{6.0E2}$              \tag{10} }$$
This equilibrium is simply the reverse of the addition of equilibria (3) and (5).

Lastly, $\ce{KCN}$ is added: in a hood, with proper safety precautions! Then $\ce{AgI}$ dissolves as per the following equilibrium, obtained by adding equilibria (3) and (6):

$$\ce{AgI (s) + 2 CN^- (aq) <=> [Ag(CN)2]^- (aq) + I^- (aq) \quad $K_\mathrm{sp(3)} K_\mathrm{f(6)}  = \pu{8.3E4}$              \tag{11} }$$
When I did this demo in lectures, I skipped the last step in order to avoid dealing with the hazardous waste issue that cyanide solutions present.
So now, at long last, the OP's questions:

Distinguish between $\ce{AgCl}$ and $\ce{AgI}$ by adding ammonia.

From equilibrium (7), $\ce{AgCl}$ can be dissolved if an excess of concentrated ammonia is added. Trust me: this should be done in a hood! But, comparing equilibria (1) and (7), $\ce{AgI}$ is more than a million times less soluble than $\ce{AgCl}$, so no amount of concentrated ammonia will significantly dissolve $\ce{AgI}$.

The OP's second question involves distinguishing between silver ion and the silver ammonia complex.

The OP's answer sheet claims that adding $\ce{NaCl}$ should result in only silver ion giving a precipitate. But equilibrium (7) shows that sufficiently high chloride ion concentrations should result in $\ce{AgCl}$ precipitation. But silver also forms $\ce{[AgCl2]^-}$, $\ce{[AgCl3]^2-}$, and $\ce{[AgCl4]^3-}$ , when the chloride concentration is high. So adding chloride will not cause AgCl to precipitate, in agreement with the OP's answer sheet. See here as well.
Last thought: I have done this demo in lecture, minus the last cyanide step, and I think the OP should not be expected to know all this stuff, especially if an exam is involved. The entire point of this kind of problem is to show that we can control equilibria to our advantage in, e.g., gold extraction via the cyanide process, and insoluble is a relative term.

Source of all solubility product constants and formation constants:
Daniel C. Harris, Appendix I In Quantitative Chemical Analysis; 7th Ed.; W. H. Freeman & Company: New York, NY, 2007 (ISBN: 0-7167-7041-5; ISBN-13: 9780716770411).

You got great answer.

@MathewMahindaratne Thanks! But the formatting appears different on a tablet than on a computer. Always something.

Hope I didn't mess up. I didn't touch the equations but the formatting on the text body.

Thank you so much, it couldn't have been any better.

@O.Ceren Thanks and happy to be of assistance!


text:
Energy and impurities balance for adsorption with a fixed bed (with description)?
Do you have any suggestion where to look for a source of information about energy and impurities (component) balance for adsorption of mixture CO2/H2O in fixed bed?
I have looked into many articles / some books (Ruthven: Principles of Adsorption and Adsorption Processes etc.) and most of them describe mass balance into a very details. I understood that, but these books do not contain information about how to express energy and impurities balance within the column.
Some articles focusing on adsorption modelling use energy, impurities, macro-micro pore balances etc. but their expression of these equation significantly differs (there are also difference in boundary and initial conditions). They also do not explain expressions used in the balances which I would very welcome while this is not my area of expertise.

What are you trying to do?

@MaxW I am trying to find an analytical and numerical solution of column dispersive mass balance and energy balance. Similar equation can be found in many articles / books, but none of them shows how to solve these equation - integration by using give boundary / initial condition to express $\partial c \big/ \partial t$ and $\partial T \big/ \partial t$. From what I have found, it is probably necessary to rewrite the balances into dimensionless form (using characteristic numbers Re, Pe, Sh, ... and other dimensionless parameters), but I did not get any further, thanks.


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Why are excess nuclei required to produce an NMR signal?
In NMR, an excess population of nuclei is required to produce a signal. If saturation is achieved (same number of nuclei in the α and β states) no net signal can be produced. A signal is produced when a radio frequency causes one state (could be either α or β) to invert. During the process of inverting, the net oscillating electric field of the nuclei is tipped to the xy plane until they undergo relaxation back to the inverted state. Even if a particular set of nuclei are saturated, they will still undergo resonance, thus producing an oscillating electric field in the xy plane. This should be detectable; however, it is not. I am guessing that this is because the oscillating electric fields cancel each other out (since they are occurring for both the α and β states). This would result in canceling out the net electric field; however, I am not sure of my guess.

"A signal is produced when a radio frequency causes one state (could be either α or β) to invert"

This isn't correct. It really isn't, even though it's often incorrectly described that way. The reality is more complicated, and it involves understanding QM as well as the fact that single spins do not necessarily exist in $\alpha$ or $\beta$ states (any superposition of them is allowed). See my answer here for more details.
Without going into too much detail, population differences* can be used to generate coherences, which are responsible for generating the detected signal. If we construct a generalised wavefunction, with two different coefficients of the two states, viz.
$$|\psi\rangle = c_\alpha|\alpha\rangle + c_\beta |\beta\rangle$$
then coherences can be (loosely) understood as an oscillation of the coefficients $c_\alpha$ and $c_\beta$. For example we could have
$$\begin{align}
c_\alpha &= \cos(\omega t + \phi) \\
c_\beta &= \sin(\omega t + \phi)
\end{align}$$
which mean that the coefficients, and hence the wavefunction $|\psi\rangle$, is oscillating with an associated frequency $\omega$ and a phase $\phi$. This generates a signal in the final spectrum at the frequency $\omega$. This sort of oscillation has nothing to do with individual spins flipping from $|\alpha\rangle$ to $|\beta\rangle$. Indeed, in NMR, individual spins have no meaning at all. Only the bulk behaviour of the $\sim10^{20}$ spins in your sample matters.
How does this relate to population differences? For a coherence to be generated, there is an additional requirement that across the entire ensemble of spins, there is a correlation between spins in the phase $\phi$.
A decent analogy would be having a collection of ($\sim10^{20}$) analog clocks on a wall. Each of these is oscillating at a rate of one revolution per 12 hours, since each clock returns to the same time every 12 hours. If you started all the clocks at the same time, then they would (largely) remain in sync with each other, and you could have a pretty good idea of what time it is right now. On the other hand, if they all started out of sync with each other, then you would have zero hope of telling what time it was.
This is the real key to understanding why there is a need for population differences. Without a population difference, you can generate oscillating spins, but each of them has a completely random phase and so there is no correlation between them. Consequently, the net signal sums to zero.
I don't want to go deeper because it will get very, very long, and also because it has been covered in a way that is better than I can ever do. See: James Keeler's book at Resources for learning Chemistry.

* As I've said before, populations should not be interpreted as the actual number of particles in a specific eigenstate, but rather the ensemble average of the coefficient corresponding to the specific eigenstate in the wavefunction. In other words, if each spin has a state $|\psi\rangle = c_\alpha|\alpha\rangle + c_\beta|\beta\rangle$, then we have $n_\alpha = \overline{c_\alpha^* c_\alpha}$ where the bar indicates an average over all spins in the ensemble.

The NMR signal is generated by magnetization oscillating in the transverse reference plane of the experiment (orthogonal to the principal field). The oscillation induces an alternating current in a pickup coil.
From a classical perspective, an oscillating transverse magnetization is generated from an initial longitudinal magnetization to which a torque is applied by coupling the magnetization to an orthogonal RF field. As the magnetization is tipped onto the transverse plane by the RF field it begins to nutate about the principal B-field at the Larmor frequency. If the applied RF frequency matches the Larmor frequency the magnetization and RF field become resonant. This resonant coupling allows the magnetization to be rotated onto the transverse plane, generating an observable in-plane magnetization. A classical treatment of the physics is possible using the Bloch equations without need to invoke QM but many important details of the NMR phenomenon are then missing.
If there is no difference in the Zeeman populations (those of $\alpha$ and $\beta$ states for spin-1/2), then there is no net longitudinal magnetization and therefore nothing to tip over or observe.


text:
SN2 reactions and inversion of configuration in compounds with more than one chiral centre
Suppose I have a compound with 2 chiral centres which are at (S,S). Now, suppose I have an iodine atom attached to the back carbon, and I wish to carry out an $\mathrm{S_N2}$ type substitution in a non-polar solvent. Obviously, the iodine atom would get replaced by the nucleophile, and there should be an inversion of configuration. But is this inversion only for the back carbon or the compound as a whole? Should the product have a configuration (R,S) or (R,R)? Someone please explain this to me briefly.

Here I have a compound with 2 chiral centers. There is only inversion at Carbon with iodine undergoing $\mathrm{S_N2}$ reaction. The second carbon does not undergo $\mathrm{S_N}$ reaction. There is no change in its configuration.
Example 1

In the example below ,

[Ref : https://en.chem-station.com/reactions-2/2016/05/neighboring-group-participation.html]
before a nucleophile attacks externally, OH of carboxylic group attacks internally via $\mathrm{S_N2}$. The 3 membred ring opens via another $\mathrm{S_N2}$ attack by hydroxide ion. The net result is retention of configuration .
Example 2
In this example , two chilral carbons are at neighboring carbons. As explained earlier , Sulphur attacks internally via $\mathrm{S_N2}$ attack. Again bromide ion attacks via another $\mathrm{S_N2}$ leading to retention of configuration ( Anchimeric assistance ).

[Ref : http://www.chem.ucla.edu/~harding/IGOC/A/anchimeric_assistance.htmll]
Example 3
This example also behaves similarly to one discissed above.

[Ref: Molecules 11(4):212-8 • April 2006   DOI: 10.3390/11040212 ]
In conclusion,

If there exists no lone pair donating group at neighboring carbon to leaving group, then $\mathrm{S_N2}$ attack takes place with inversion of configuration (Example 1).
If there exists a lone pair donating group (Example 2 and Example 3)at neighboring carbon to leaving group, then retention of configuration is seen.

Inversion will only occur in the carbon in which the substitution takes place. So if the molecule was (S,S) it will become (R,S). If you analyse the Sn2 mechanism you will understand why inversion happens and you will see that there's no reason for inversion to take place in a carbon that doesn't participate in the reaction.

As Blg90 said, the inversion only occurs on the carbon involved in the substitution. I thought it would be useful, intuitively, to mention that the rear attack on the carbon to leaving group bond instigates a motion that is very much like an umbrella blowing inside out.

You can see, in this image, the hydrogens at the 'back' (which can be swapped out for any substituents that allow $\mathrm{S_N2}$), flip through the transition state to the 'front', but are themselves unaffected. Hopefully this helps you visualise why inversion only occurs at one stereogenic centre.


text:
What would weigh more? A ice cube or water, if the ice cube is from the same amount of water
So if I frozen 50 ML of water, it turns into ice cube. Would the ice cube weigh the same as the 50 ML of water it was before?

Yes, as the amount, i.e. the number of molecules involved, does not change. Unless you look at it from a relativistic point of view, where you have to subtract the mass equivalent of the heat that is taken out of the water by cooling it down and freezing it and possibly cooling it further.


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Acidity order of xenon fluorides,oxides and oxofluorides

The acidity order of xenon oxides,fluorides and oxofluorides is:
  $\ce{XeF6 > XeO2F4 > XeO3F2 > XeO4 > XeOF4 > XeF4 > XeO2F2 > XeF3 > XeO3}$.

How to justify this order?
My thoughts were that the acidity order of xenon fluorides, oxides and oxofluorides is based on its tendency to accept an oxide(as these are the most common reactions). As fluorine is more Electronegative than oxygen, we can safely say that $\ce{XeF6}$ would be the most(As $\ce{Xe}$ would be most electron deficient and  can accept an oxide easily). $\ce{XeO2F4}$ would be the second most and the last would be $\ce{XeO3}$, based on number Fluorine+Oxygen atoms, with priority to fluorine. However evidently, "the more number of fluorine atoms" logic doesn't apply to all of them. Looking at the structure of each molecule I'm unable to deduce any logical explanation for the above mentioned order.
Also, I know for a fact that $\ce{XeF6}$ corrodes glass and $\ce{XeF4}$ doesn't. Upto which xenon compound would corrode with glass?

You have to consider the oxidation state of the xenon.  Higher oxidation state makes the central atom a stronger Lewis acid in general, xenon is just one case.
So we should consider each oxidation state separately.  Look at the ordering when we do that:
+8:  $\ce{XeO2F4}>\ce{XeO3F2}>\ce{XeO4}$
+6:  $\ce{XeF6}>\ce{XeO2F2}>\ce{XeOF4}>\ce{XeO3}$
+4, +3: Only the fluorides are given, no oxygen vs fluorine comparison possible from the list.
The +8 and +6 orderings are consistent with the expected effect of oxygen versus fluorine within a given xenon oxidation state.


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Structural isomers of hexaammineiron(III) hexacyanidocobaltate(III)

Find the number of structural isomers of $\ce{[Fe(NH_3)_6][Co(CN)_6]}$

This was a question solved by my teacher as follows:
The $\ce{CN}$ will show linkage isomerism while $\ce{NH_3}$ and $\ce{CN}$ will show coordination isomerism.
We will now consider cases of coordination isomerism and then use linkage isomerism.

$\ce{[Fe(NH_3)_6][Co(CN)_6]}\rightarrow 7$ isomers(due to linkage isomerism of $\ce{CN}$)
$\ce{[Fe(NH_3)_5(CN)][Co(CN)_5(NH_3)]}\rightarrow 2\times 6=12$
$\ce{[Fe(NH_3)_4(CN)_2][Co(CN)_4(NH_3)_2]}\rightarrow 3\times 5=15$
$\ce{[Fe(NH_3)_3(CN)_3][Co(CN)_3(NH_3)_3]}\rightarrow$ not counted.

The further $3$ cases are similar to the first $3$ and will result in the same number of isomers. Hence the answer is $2\times(7+12+15)=68$.
Why is $\ce{[Fe(NH_3)_3(CN)_3][Co(CN)_3(NH_3)_3]}$ "not counted"?
I feel that isomers due to $\ce{CN}$ showing linkage isomerism are perfectly valid. Please correct me if I'm wrong and sorry if the question is too trivial.

Firstly, there is a typo in your solution. It should be 2*(7+12+15)=68. Also I feel that the compound which you mentioned has to be counted but only once.(in the sense that it wouldn't be inside the 2. And just 44=16 isomers in the case. Giving a total of 68+16=84 isomers

We prefer to not use MathJax in the title field due to issues it gives rise to; see here for details.

@user600016 Sorry for the typo, fixed it. And for the second part of your comment, that's exactly what I want to say.

@Zenix what should be the ideal title then?

@user600016 Did you check out the meta post in my previous comment? There's a general consensus here that we try to avoid MathJax and $\LaTeX$ in titles due to URL slug. Answer by ex-moderator of site sums it nicely.

@Zenix Oh my bad. I think in MSE that is the case but not sure again

@user600016 You are right, it's not the case in MSE, rather we generally edit the titles and add MathJax. Here, \ce{^} paves problem, which is not the case in MSE ;)

@AryanSonwatikar Rule of thumb for the same chemical compound appearing both in title and in the body: whenever possible, use chemical names in the title and chemical formulas in the body. This way you not only avoid problems with the search engines incorrectly parsing MathJax, but also gain better search coverage for the future visitors, sort of kindergarten-level CEOing.

@andselisk Thank you for that! I will keep that in mind in the future!

I have found the answer:

In the following case:
  $$\ce{[Fe(NH_3)_3(CN)_3][Co(CN)_3(NH_3)_3]}$$
  Both the constituent entities are neutral and hence will exist independently, and thus will not form isomers of the original compound.

Pfft, it still could be a compound not necessarily a mixture. CN ligands are bridging, so most of reasoning in all this makes little practical sense.

@Mithoron Could you elaborate as to how $\ce{CN}$ ligands being a bridging states that the above will exist as a compound? Please keep in in mind I'm only 16 and have just started learning Coordination Chemistry- all I have learnt is nomenclature and Isomerism; no CFT and so on.

Well if the two metallic centers were connected (Fe-CN-Co) then they wouldn't exist independently. Of course formula would be different, but I'm talking about practical aspect here, maybe pointlessly as such isomer counting has little to do with realism and more with training in 3D thinking.

@Mithoron I guess that's what my teacher had in mind. This was just after we had been explained all the types of isomerizations.


text:
Vapour pressure and Boltzmann energy distribution
As theory says, there is always a certain amount of molecules in a liquid sample which have energies greater than a thresold energy and get free from intermolecular forces in liquid and produces the vapour. Theory also says that the temperature dependence of the vapor pressure is described by the following equation:
$$p = p^\circ \times \exp(-Q_\text{vap}/(RT))$$
But my question is how does this equation give such a specific value of vapour pressure while there are many molecules in the vapour phase having different ranges of energy, corresponding to different pressure ranges according to Boltzmann distribution law?

Note that the Boltzmann distribution is valid for each of 3 dimensions independently, as there are 3 degrees of freedom.

You put a gas in a cylinder and measure the force exerted on a piston. That's what pressure is about. Who cares if the molecules have different energies (which they do)? You don't measure individual molecules anyway.

There is a distribution of energy states both in the gas phase and in the liquid, while there is only one macroscopic pressure. Also, the expression you wrote is an approximation useful under a limited set of conditions.


text:
Aluminium, Carbon, Aluminium Sulphate battery. What's the reaction?
I made galvanic cells from the following items (while trying to do something else), and they give between 0.6 V and 1 V per cell.
Aluminium foil – negative terminal
Carbon (these work: charcoal sticks, graphite rods, carbon felt) – positive terminal.
Electrolyte – aluminium sulphate $\ce{Al2(SO4)3}$ in water.
With 9 cells – 3 series * 3 parallel – it lights a red LED.
Left on overnight it gets very dim, but given some recovery time, it comes back bright again.
So the current isn't high, but I wasn't expecting any.
I'm not observing any gas being evolved, or any change in the foil (yet).
The battery has been running about a week.
I know the Al sulphate reacts with water to make sulphuric acid + Al hydroxide.
Is that what's happening here?
What happens next in the reaction?
What's the final result?
Thanks for your help.
It's a long time since school chemistry lessons.

Very interesting: no gas, no change in the electrode, yet a significant electrical output. This could be the beginning of perpetual power (similar to perpetual motion!).
But no, probably the output power is barely significant, and something is happening somewhere, possibly in the solution. Aluminum sulfate doesn't usually produce a clear solution, so possibly you have missed an increased turbidity.
Possibly oxygen is reacting with the metal under the acidic conditions to produce Al(OH)3 which is removed by the acid solution (without reducing the acidity, because Al(OH)3 is not much of an alkali). The equation could be 4 Al + 3 O2 + 6 H2O --> 4 Al(OH)3. If you restrict oxygen access, the voltage and current should decrease. If you increase it, say with an aquarium pump, you might get a higher output.


text:
Honey moisture and increase in temperature?
I've been interested quite lately on how moisture would be affected as the temperature increases. My experiment methodology is that I will be manipulating temperatures of the honey between 20,40,60,80 degrees celsius and I will be using a refractometer to measure moisture in brix. Will there be a slight change in the reading of the refractometer?

If you see a change depends greatly on the quality of your refractometer. // Does the refractometer itself have the ability to vary temperature of the cell? Hot honey will cool down if applied to a cold cell.

In fact, I am using the small refractometer with a trap door on the top(as soon as you search refractometer on google you will see it). If it is good for varying temperatures then how can we avoid the uncertainty of the cooling down of honey inside the cell?

If you measure it at different than nominal temperature, you would not know, if the refraction index change is due different water content, or due different temperature.

Normally temperature affects moisture therefore there is a change in refractive index. Do you have any other idea for the reason why could there be a change in refractive index other than moisture?

It was mentioned several times. n=f(T). Moisture=f(T) only indirectly.

I suppose you would cool it back before measurement, otherwise it would be measuring of temperature dependency of honey refraction index.
There would be change, it is the way how bees evaporate the excessive water to make honey stable. Obviously, it would be function of both temperature and time. As honey is very viscous, loosing water would be very slow in bulk volume and also place dependent, approximately following the 2nd Fick's law, unless being mixed.
But $60$ or $\pu{80 ^\circ C}$ is ridiculous, as it would not be honey any more, but just thick solution of fructose and glukose + some minor components. If honey is heated to dissolve crystallized sugars, it is recommended not to heat it more than $\pu{40 ^\circ C}$.


text:
Deviations from Raoult's Law
A solution of two compounds may be an ideal solution and follow Raoult's Law, or it may show a positive or negative deviation from Raoult's Law. How to predict that whether a solution of two given compounds will show Positive or Negative Deviation from Raoult's law, or if it will be an ideal solution?

I don't think it is possible to predict whether a given mixture will behave as a non ideal solution.

For most solutions, it is nearly impossible to predict in advance whether a positive or negative deviation is observed. A deviation from Raoult's law can only be measured experimentally. However, for some solutions, you can predict whether the deviation will be positive or negative by making assumptions about the interactions between solute and solvent molecules.
If the solute-solvent forces of attraction is weaker than solute-solute and solvent-solvent interaction, then there will be a positive deviation, and if the solute-solvent forces of attraction is stronger than solute-solute and solvent-solvent interaction. For example, Acetone and Ethanol show a positive deviation due to weaker interactions due to H-bonding.

Can you expand on that example a bit


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Is the effect of hydroperoxyl radicals significant in the oxidation of organic compounds in the Fenton reaction? Why or why not?
In the Fenton reaction, a ferrous catalyst and hydrogen peroxide are used to generate hydroxyl radicals which in turn oxidise organic compounds. The reaction is commonly seen in the treatment of wastewater and operates using the following mechanism:

$$
\begin{align}
\ce{Fe^2+ + H2O2 &-> Fe^3+ + ^.OH + OH-}\tag{1}\\
\ce{Fe^3+ + H2O2 &-> Fe^2+ + ^.OOH + H+}\tag{2}\\
\ce{Fe^3+ + ^.OOH &-> Fe^2+ + H+ + O2}\tag{3}\\
\ce{RH + ^.OH &-> H2O + R^{.}}\quad\text{(chain propagation)}\tag{4}\\
\ce{R^. + Fe^3+ &-> R+ + Fe^2+}\tag{5}\\
\ce{R^. + H2O2 &-> ROH + ^.OH}\tag{6}
\end{align}
$$

In the reactions, hydroperoxyl radicals $(\ce{^.OOH})$ are produced as a byproduct of the reduction of the ferrous catalyst, among other reactions. However, literature regarding the impact of the hydroperoxyl radical on the oxidation of organic compounds is limited, with most papers ignoring the effect of the hydroperoxyl radical completely. So, is the effect of the hydroperoxyl radical significant in the oxidation of organic compounds? Why or why not?

The effect of the hydroperoxyl radical in the Fenton process can be largely ignored as the oxidising ability of the $\ce{^∙OOH}$ radical is not as effective as that of the hydroxyl radical.
According to Chang et al., the reaction rate coefficients of p-nitrophenol and $\ce{OH}$ and $\ce{HO2}$ were found to be $\pu{9.05 \times 10^6 M−1 s−1}$ and $\pu{1.50 \times 10^−6 M−1 s−1}$, respectively. Considering the orders of magnitude of these two rate coefficients, the reaction pathway involved the p-nitrophenol reaction with $\ce{^∙OOH}$ radicals can thus be neglected. This can therefore be extended to cover the oxidation of organic compounds by hydroperoxyl radicals through the Fenton process.
References

Chang, M., & Chern, J. (2010). Decolorization of peach red azo
dye, HF6 by Fenton reaction: Initial rate analysis. Journal of the
Taiwan Institute of Chemical Engineers, 41(2), 221-228.
DOI: 10.1016/j.jtice.2009.08.009

Would you please put the reference in your answer?


text:
Does motor oil get more viscous when heated & what is the meaning of the SAE motor oil code?
The motor oil in my 2004 Pontiac Vibe is 5W-30.
My (incorrect) understanding of this SAE nomenclature is that 5W means "Viscosity index of 5 in the winter" (or when really cold - like $\pu{-30 ^\circ F}$) and "Viscosity index of 30 at operating temperature" (or when the car has been running for a little while).
So how is this accomplished? I usually think of something like butter or coconut oil that solidifies at a lower temperatures and becomes fluid again as heated. To my understanding oil has a higher viscosity at lower temperatures (generally speaking).
Everyone keeps talking about "additives" that help accomplish this feat in "multi-grade" oil. What are these additives? Generally, how do they work? I'm no chemist and neither are these auto mechanics, so I figured this might be the right place to ask.
Since my understanding of the SAE motor oil code is incorrect, illumination on its exact meaning would be helpful.
A full blown chem answer and an "explain it like I'm 5" would be most appreciated :) I never studied chemistry past high school, so be easy on me!

So this is a gross oversimplification of this issue, however it is a very high level overview of the mechanism at work. From a blog that shell oil put out:

Mineral multigrades use a light weight mineral oil (the same as a
light weight monograde oil), but include an additive called a
Viscosity Index Improver. The best way to visualise this viscosity
index improver is as a long chain molecule which curl up like a ball
of string when cold, but then uncurl as the temperature increases.
Thus when an oil is cold, the presence of the viscosity index improver
has very little effect and the oil flows well as the base oil is a low
viscosity oil. As the oil heats up, the viscosity index improver
uncurls with the effect that it tends to restrict the motion of the
oil, or 'thickens' it, which to some extent counteracts the decrease
in viscosity of the base oil. This enables the oil to support more
load than would otherwise be possible.

This blog also mentions that these VII's break down when sheared (so like... in an engine) and lose their effectiveness. Time for an oil change!
Self answering as it has been a number of months; I'd still be extremely interested in the chemistry behind these VII's though I'd imagine that is incredibly complex; hydrocarbons usually are if I'm not mistaken. It's also likely proprietary, or at least the latest iterations probably are.

The numbers in the code have nothing to do with viscosity index. However, your understanding of 'W' for winter is correct. The real code for your car is actually SAE 5W-30 where SAE stands for the Society of Automotive Engineers, which has established a multi-grade numerical code system for grading motor oils according to their viscosity characteristics. I said viscosity characteristics, not viscosity index. According to Wikipedia:

The original viscosity grades were all mono-grades, e.g., a typical engine oil was a SAE 30. This is because as all oils thin when heated, so to get the right film thickness at operating temperatures, oil manufacturers needed to start with a thick oil. This meant that in cold weather it would be difficult to start the engine as the oil was too thick to crank. However, oil additive technology was introduced that allowed oils to thin more slowly (i.e. to retain a higher viscosity index); this allowed selection of a thinner oil to start with, e.g., "SAE 15W-30", a product that acts like an SAE 15 at cold temperatures (15W for winter) and like an SAE 30 at $\pu{100 ^\circ C}$ ($\pu{212 ^\circ F}$).

viscosity characteristics: Kinematic viscosity is graded by measuring the time it takes for a standard amount of oil to flow through a standard orifice at standard temperatures. The longer it takes, the higher the viscosity and thus the higher the SAE code. Larger numbers in the code are thicker.
The company called Anton Paar has analyzed these SAE coded motor oil for the parameters, Kinematic viscosity and oil density versus elevating temperatures. I have included some of graphs including SAE 5W-40, but failed to find the most common SAE 5W-30:

Finally, for your information, Viscosity Index is a dimensionless number that represents how the viscosity of a hydraulic fluid changes with temperature. The greater the viscosity index, the smaller the change in fluid viscosity for a given change in temperature, and vice versa. Synthetic oils usually have a higher viscosity index than do mineral oils. (Mobile Hydraulic Tips). The following graph illustrate how viscosity index influence the change of viscosity with temperature (Ref.1):

It is also nice to know how change of viscosity index by modifiers affect mono-grade oil to better multi-grade oil:

References:

Tiago Cousseau, “Film Thickness and Friction in Grease Lubricated Contacts: Application to Rolling Bearing Toque Loss,”  PhD Thesis; University of Porto (Universidade do Porto), Porto, Portugal, 2013.

This is a top notch answer! You really do nice work, Mathew!

@Ed V: I'm honored. I appreciate your comment.

@Mathew Mahindaratne. Your explanation is extremely well developed. But it does not  answer the question. The question was : what is the chemical substance (or the additive) whose viscosity increases when heated ?

@Maurice: the answer to that is: none, because the viscosity of motor oils does not increase with temperature.

@cbeleitesunhappywithSX Thank you! I'd be very interested in knowing what the chemistry behind multi-grade oil is that allows it to behave the way it does.

@Maurice I was thinking the same. Very well researched answer, and extremely interesting, but it doesn't address the crux of the question; how does it accomplish this? cbeleitesunhappywithSX points out that the viscosity in fact does not increase with temperature; I am hopeful that someone can help me with this quandry!

@Shrout1: I did not concentrate on chemistry part because you insist that you do not have chemistry background. Beside, these facts of how they do that are not published in order to keep them for themselves. Specifically, synthetic oils, which behave much better than natural oil are patented (e.g., Mobil 1 5W 30)..

@MathewMahindaratne I do deeply appreciate the time you've taken with this answer! You clearly have a great grasp of the concept and a willingness to help :) So thank you thank you! It also looks like I should have asked "how do Viscosity Index Improvers work" as this seems to be what I'm looking for. Honed my google in a little more and found some info :)

@Shrout1: I see your point. Keep working on it. And, it won't hurt if you post another question concentrating on that regard. So, you'd get more help from others. Good luck.:-)

The graphs from Anton Parr that you're showing of KM (kinematic viscosity) vs. T are inconsistent with the described behaviors of these oils.  E.g., compare the graphs for SAE 10W40 and 10W60.  At low T, these are supposed to have similar KM's, while at high T, the 10W60 should have a significantly higher KM.  Yet these graphs show the opposite:   At low T, 10W60's KM is about twice that of 10W40; and at high T, their KM's are about the same.

ALL motor oil gets THINNER the hotter it gets period.
Multi viscosity oils are a compromise between straight weight oils that provide the best protection and a oil that can be left in year round because of laziness of todays drivers


text:
Does sanding the oxide layer of a magnesium sample break bonds?
We all know that if we keep a piece of magnesium metal exposed in open air, an oxide layer forms on it. This layer prevents further oxidation of the bulk metal. If we then rub it with sand paper, the magnesium metal can react again because the oxide layer is removed. So my question is:
Does that mean abrasion with sandpaper can break the oxygen bonds with magnesium?

You are just mechanically removing the soft MgO on the surface of the metal.

Why is it getting removed??

Not quite soft MgO, @Ed.  Magnesia has a Mohs hardness of 6, which is as hard as some of the minerals in granite.  You're really breaking through a microscopically thin layer of the magnesia and abrading the softer magnesium metal beneath.

But @Oscar ,breaking through the microscopically thin layer of magnesia and exposing the mg metal beneath means that bonds are being broken

You are mechanically scratching the surface. When you are scratching a surface, you are essentially breaking the metallic bonds among Mg atoms.

In editing, I had to look up verbs describing the action of sandpaper: https://ell.stackexchange.com/questions/111132/verbs-for-sandpaper

I upvoted the question because it is actually very insightful. We may think of simply removing the MgO layer, and voila, there is bare magnesium. I think not.
If you took bare Mg (no oxide at all) in a vacuum, and dropped powdered MgO onto the Mg, then exposed the metal to an argon atmosphere, I think you could blow off the oxide powder because there would be no significant adhesion.
In other words, the unreactivity of the oxide-coated magnesium arises because the oxide coat is strongly bonded to the metal. This transition from metal to metal oxide occurs over a few atoms and is probably very irregular. After all, even the surface of the metal won't even be flat to within 1 or 2 atoms.
Visible scratches on the metal after sandpapering prove that some mechanical effect has removed metal. If you remove metal, you must have broken some metal-metal bonds; and in the turmoil, at least some metal - oxygen bonds must also have been broken - even tho they might soon reform with O2 in air.
A similar, but chemical, film removal occurs when stainless steel, which is likewise protected by a very thin oxide coat, is treated with hydrochloric acid. The film is removed, and the steel corrodes. Because the reaction is so thorough, removing the whole oxide coat and a few atomic layers of metal as well, we don't stop to think about the transition layer. Different thicknesses can produce different colors as well as different degrees of protection.

If you value it so much, then I suggest copy editing the question.

I would give a further example. Just an ideal piece of metal in inert atmosphere. We can certainly "sandpaper" it. This would be breaking a chemical bond. I upvoted this answer right for its incipit. In fact I've upvoted the Q as well.


text:
How to calculate mass of calcium fluoride that will dissolve in sodium fluoride solution?

The solubility product of calcium fluoride, $\ce{CaF2},$ is $\pu{1.46E-10 mol^3 dm^{-9}.}$
What mass of calcium fluoride will dissolve in $\pu{500 cm^3}$ of $\pu{0.10 mol dm^{-3}}$ sodium fluoride solution? (Molar mass of $\ce{CaF2}$ is $\pu{78.1 g mol^{-1}}.)$

I tried using the following method:
$$
\begin{align}
[\ce{Ca^{2+}}][\ce{F^-}]^2 &= K_\mathrm{sp}(\ce{CaF2})\\
x(2x)^2 &= \pu{1.46E-10 mol^3 dm^{-9}} \\
x &= \pu{3.3E-4 mol dm^{-3}}
\end{align}
$$
Solubility $s$ of $\ce{CaF2}$ can be found as such:
$$s(\ce{CaF2}) = xM(\ce{CaF2}) =  \pu{3.3E-4 mol dm^{-3}}\times\pu{78.1 g mol-1} = \pu{2.6E-2 g dm^{-3}}.$$
Finally, the mass of $\ce{CaF2}$ dissolved in $\pu{500 cm^3}$ would be
$$m (\ce{CaF2}) = s(\ce{CaF2})\times V = \pu{2.6E-2 g dm^{-3}}\times\pu{500 cm^3} =\pu{1.3E-2 g}.$$
However, my answer differs from the given answer, that is $\pu{5.7E-7 g}.$
How should I approach this question instead?

While writing the equation for $K_{SP}$ in place of [F-] concentration you have to take 2s+(0.1) instead of only 2s due to common ion effect. So equation becomes $s(2s+0.1)^2=K_{SP}$. As s is very less than 0.1 you can neglect s and equation becomes $s(0.1)^2=K_{sp}$. Then you can calculate s which is the concentration of dissolved $CaF_2$ in solution and then use the volume given to calculate it's mole and then its mass.

@Manu: Why don't you put as an answer since OR has showed his/her work (although this is very much like a homework). .

I personally like Manu's argument. By common ion effect the total fluoride ion concentration will be 0.1+2s where s is the solubility of the sparingly soluble calcium fluoride. And normally since common ion effect reduces solubility of such salts drastically, 0.1 will be much greater than s and a useful approximation is that 0.1+2s = 0.1. Then the remaining unknown term will be the s standing for calcium ion concentration and a useful answer can be got.

In the $\pu{500 cm^3}$ solution there are is already $\ce{F^-}$ dissolved coming from the $\ce{NaF}$ $\pu{0.10 mol dm^{-3}}$ that was already there. This means that the solubility of $\ce{CaF2}$ is reduced due to the common ion effect.
NB $\ce{NaF}$ can be considered completely dissociated because it has a very high $K_\mathrm{sp};$ therefore the initial concentration of $\ce{F^-}$ is $\pu{0.10 mol dm^{-3}}.$
$$
\begin{align}
[\ce{Ca^{2+}}][\ce{F^-}]^2 &= K_\mathrm{sp}(\ce{CaF2})\\
x(2x+\color{red}{\pu{0.10 mol dm^{-3}}})^2 &= \pu{1.46E-10 mol^3 dm^{-9}} \\
\end{align}
$$
The issue  here is that to find $x$ you need to solve a cubic equation…
However, $K_\mathrm{sp}$ of $\ce{CaF2}$ is very small and $x$, that is its solubility, must be much less than $\pu{0.10 mol dm^{-3}}!$ This allows you to make the approximation that $2x + \pu{0.10 mol dm^{-3}} ≈ \pu{0.10 mol dm^{-3}}.$
$$
\begin{align}
[\ce{Ca^{2+}}][\ce{F^-}]^2 &= K_\mathrm{sp}(\ce{CaF2})\\
x(\color{red}{\pu{0.10 mol dm^{-3}}})^2 &= \pu{1.46E-10 mol^3 dm^{-9}} \\
x &= \pu{1.46E-8 mol dm^{-3}}
\end{align}
$$
Solubility $s$ of $\ce{CaF2}$:
$$s(\ce{CaF2}) = xM(\ce{CaF2}) = \pu{1.46E-8 mol dm^{-3}}\times\pu{78.1 g mol-1} = \pu{1.14E-6 g dm^{-3}}.$$
Finally, the mass of $\ce{CaF2}$ dissolved in $\pu{500 cm^3}$ would be
$$m (\ce{CaF2}) = s(\ce{CaF2})\times V = \pu{1.14E-6 g dm^{-3}}\times\pu{500 cm^3} =\pu{5.7E-7 g}.$$

You did a stellar job at explaining the answer, just a couple of things (that I already corrected). First, when you copy-paste markup from the question, make sure to stick with the same notations, and, whenever possible, units: don't use two different schemes or randomly omit all formatting at all.

Second, you are not allowed to replace $\pu{500 cm^3}$ with $\pu{0.500 dm^3}.$ You don't know for sure whether zeros in $\pu{500 cm^3}$ are significant figures or not, so at best you can only guarantee $\pu{0.5 dm^3}.$ Note that normally there is a bar on top of a zero if it is significant, e.g. $\pu{5!\bar{0}\bar{0} cm^3}.$

Gc3941d has mixed the solubilities of $\ce{NaF}$ and $\ce{CaF_2}$. Let's start the calculation from the beginning.
The concentration of fluoride ion is : [$\ce{F^-}$] = $0.1$ M.
The concentration of Calcium may be calculated from the solubility product $\ce{K_{sp}}$ and [$\ce{F^-}$] according to : :$$\ce{[Ca^{2+}] = K_{sp} /[F^-]^2} = 1.46·10^{-10}/(0.1)^2 = 1.46·10^{-8} M$$ It means that in $0.5$ liter, $ 0.5·1.46·10^{-8}$ mol $\ce{Ca}$ = $7.3·10^{-9}$ mole $\ce{Ca}$ is dissolved. And of course, the same number of mole of $\ce{CaF_2}$ is dissolved .
As the molar mass of $\ce{CaF_2}$ is $\ce{78.1 g|mol}$, the mass of $\ce{CaF_2}$ dissolved in this solution is :  $$\ce{m(CaF_2) = 7.3·10^{-9} mol· 78.1g/mol = 5.7·10^{-7}g }$$


text:
Why does nitrous oxide react with bases and it is still a neutral oxide?
I have studied that N2O is a neutral oxide in my textbook(recommended by the central board of secondary education INDIA(country)). on internet and in my class notes i find its balanced reaction with NaOH. so it is neutral or is it an exception?
the reaction given is
N2O + 2NaOH -> 2NaNO +H2O

Could we see that reaction between $\ce{N2O}$ and $\ce{NaOH}?$ AFAIK there is none under ambient conditions. Also, what does "INDIA" abbreviate (countries are not written in all caps)?

You mentioned the reaction is balanced; the one you added is not, and to the best of my knowledge, doesn't occur (granted $\ce{NaOH}$ is in solution). Besides, what is "NaNO"?

i don't know the compound

Neither do I. $\vphantom{WTF}$

I agree with Andselisk, NaNO does not exist, and N2O does not react with NaOH in solution.

NaNO could be sodium hyponitrite, Na2N2O2.  Wikipedia reports that an isomer of this salt could be synthesized by a reaction between sodium oxide and nitrous oxide.

Sad to see such glaring textbook errors. Nitrous oxide solubility in water decreases with increasing concentrations of NaOH. It does not react with NaOH to "form" NaNO.
Search this paper on Google and show it to your teacher.
J. F. Nunn, Respiratory Measurements in the Presence of Nitrous Oxide, Brit. J. Anaesth., 1958, 30, 254


text:
Converting Raman Spectra to Infrared Spectra
I have data for detecting a substance using Raman spectra, but in production, I plan to be using a visible & near-infrared spectrometer. Would it be possible for me to convert from the Raman values to visible & near-infrared values? Are they the same? Thank you.

They are in general not the same. In particular, for molecules with a center of symmetry, the vibrational modes that are detected by Raman are exactly the modes not detected by IR spectroscopy and vice versa.

Not in any way that you can use in practice for getting a working predictive model (= data analysis procedure).

Depending on the molecular symmetry of your analyte(s), infrared and Raman spectra may share some bands, or not: a vibration can be infrared-active, Raman-active, or both.
Near-infrared spectra show combinations and overtones of the infrared-active bands, and that makes them far more complex and complicated to analyze.
Now, if you know your analyte(s) and the matrix and both are sufficiently simple, one may arrive at educated guesses Raman vs. infrared spectrum.
But this is in general not sufficient to develop a full predictive method for production use: typically, a successful method needs to take into account the type and size of variations that occur (in the production process) and which ones are expected and OK and which ones are not. Particularly in a production environment, relevant variations may be due to physical effects (e.g. particle size, ambient temperature, ...) rather than concentration variations of the consitituents of your mixture.
In general, you cannot expect to get anything close to a working NIR spectroscopic method without running a well-designed calibration experiment and the appropriate statistical analysis.
Also keep in mind:
In theory, there is no difference between theory and practice. In practice, there is.

Whatever your production is, please go and consult someone who knows about vibrational spectroscopy and if you plan to use near infrared spectra, also about chemometrics (statistical data analysis). If you have a third wish free: ask for someone who in addition knows about your field/domain/application.


text:
Grignard reagent will act as a base or a nucleophile
R- from grignard reagent (RMgBr) and gilmann's reagent (R2CuLi) although being a strong base reduces the carbonyl group and does not act as a base.
Why can't it abstract the alpha hydrogen forming an enolate ion?

As most of us chemists have understood and experienced, the Grignard reagent have undergone the many reactions with suitable substrates. Yet, the most familiar and probably the most used reaction is the addition to a carbonyl group to give $2^\circ$- or $3^\circ$-alcohol. The Grignard reagents' affinity for a corbonyl group in the presence of an acidic $\alpha$-$\ce{H}$ has been documented here. However, if at least $\ce{OH}$ group is present in the molecule, acid base reaction would be take over. For example, the usual Grignard reaction would not happen if your glassware is wet or not properly dried.
Albeit they were rarely published, it is well known fact that some side reactions with available acidic $\alpha$-$\ce{H}$ and some reduction reactions have been taken place during original Grignard reactions. These side reactions are frequently complicated the intended reactions, which may replace some or all of the intended addition reaction, reducing expected percent yield (Ref.1). Most importantly, the conversion of an alkyl halide to a deuterium labelled hydrocarbon has been achieved via Grignard reagent preparation (Figure A):

Keep in mind that these labelling are selective based on the alkyl halide. Recently, there was a method developed to selective labeling by using organozinc reagents (Ref.2). Although these are not conventional Grignard reagents, organozincs are close relatives (Figure B). Another example for Grignard reagents acting as a base is a method to prepare terminal alkynyl-Grignard reagents (Figure C).
Some times during Grignard preparation, you might experience no Grignard was evidently formed because you have loosen your alkyl halide reagent while substrate ketone was recovered unchanged. That is because, if your ketone is bulky like diisopropyl ketone (2,4-dimethylpent-3-one), following reaction may happen (look under an example for Grignard acting as a base):

If your Grignard reagent have an active $\alpha$-$\ce{H}$ to give, it can also acts as an reducing agent as in above case of Grignard acting as a reducing agent. This is the case of reaction between benzophenone and specific Grignard reagent (isobutyl magnesium bromde or 3-methylbutylmagnesium bromide) illustrated in Ref.1. The main products from that reaction were dibenzylcarbinol and isobutene.
References:

G. E. Dunn, John Warkentin, "An Isotopic Study of the Reducing Action of the Grignard Reagent," Canadian Journal of Chemistry 1956, 34(1), 13841–13857 (https://doi.org/10.1139/v56-007).
Aiyou Xia, Xin Xie, Xiaoping Hu, Wei Xu, Yuanhong Liu, "Dehalogenative Deuteration of Unactivated Alkyl Halides Using $\ce{D2O}$ as the Deuterium Source," J. Org. Chem. 2019, 84(21), 13841–13857 (https://doi.org/10.1021/acs.joc.9b02026).


text:
What is the state of aggregation of iodine formed during electrolysis of molten nickel(II) iodide?
Assume the electrolysis of molten $\ce{NiI2(l)}$ (inert electrodes) in a cell. The cell with a heating element is in the room; the room is at $\pu{25 °C}$ and $\pu{101.3 kPa}.$
The half-reaction for oxidation at the anode would be
$$\ce{2 I^-(l) -> I2(\text{state}) + 2 e^-},$$
whereas the global reaction will be
$$\ce{NiI2(l) -> Ni(s) + I2(\text{state})}.$$
Given that iodine is solid at room temperature, should it be marked as a solid in the equations? However, given that the $\ce{NiI2}$ is liquid, I'd think that the newly synthesized iodine would be a gas for a while?
Also, what observations could one make during the electrolysis? I.e. will the iodine accumulate on the positive electrode as a solid, or will it exit the cell in the form of a gas?

Also, is the reduction potential for
$$\ce{I2(s) + 2 e^- -> 2 I^-(\color{red}{aq})}$$
the same as for
$$\ce{I2(s) + 2 e^- -> 2 I^-(\color{red}{l}) }?$$

?!? $\ce{NiI2}$ melts at 780 °C. How are you going to have liquid $\ce{NiI2}$ in a cell at 25 °C?

The cell is not at 25 deg C. The room in which the cell is is at 25 C.

Well iodine boils at 184 °C. What do you think will happen to it?

My guess is that the iodine will be a gas.

I took a liberty to correct formatting, rewrite the title and add a brief detail that your cell has a heating element to avoid confusion with the RT and m.p. of the salt (you got me here, too). Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

The half cell reduction reactions should be: $\ce{I2(s) + 2e- -> 2 I-}$

Basically, during the electrolysis process of molten nickel(II) iodide ($\ce{NiI2}$), $\ce{I– (l)}$ ions would be oxidized to $\ce{I2 (g)}$ at the anode, and $\ce{Ni^2+ (l)}$ ions would be reduced to $\ce{Ni (s)}$ at the cathode. My state assignments were based on following facts assuming the temperature of electrolytic cell is kept below $\pu{900 ^\circ C}$:

Melting point of $\ce{NiI2}$ is about $\pu{797 ^\circ C}$. Therefore, electrolytic cell must be kept above $\pu{800 ^\circ C}$. Thus following equlibrium will be maintained during electrolysis: $$\ce{NiI2 (l) <=> Ni^2+ (l) + 2I- (l)}$$
Boiling point of $\ce{I2}$ is about $\pu{134 ^\circ C}$. Hence, $\ce{I2}$ would be released as a gas at anode.
Melting point of $\ce{Ni^\circ}$ metal is about $\pu{1455 ^\circ C}$. Hence, $\ce{Ni}$ would be deposited as a solid at cathode. If you want to keep nickel metal in liquid form, your electrolytic cell must be kept above $\pu{1500 ^\circ C}$.

I think I have given enough information you to figure out the rest.


text:
Why does silver stick to the glass in Tollens' test?
Why does it not simply form a solid powder at the bottom of the flask?
Does silver also stick to the glass in other silver precipitating reactions?
EDIT:
My question is definitely not the same as the one linked (which btw does not have an answer and is inactive for three and a half years).
Normal precipitation of silver from silver ions results in a black powder and no mirror.
As pointed in the comments, kinetics seems to play a major role here, with slow precipitation being key to achieve a mirror.
So why is it that a slow reduction of silver ions make them deposit on the surface of the glass?

Well, only logical explanation is that the reaction happens on surface, not in the bulk of solution, I do wonder why it happens like that, though.

This has some information: https://www.compoundchem.com/2017/09/06/silver-mirror/. Stannous chloride solution is also commonly mentioned as a preparation step: the glass surface is rinsed with this reducing agent.

@EdV The important hint in there might be that the diammine complex is more stable than the aqua complex, and leaving out the ammonia does lead to a black suspension.

It is qualitative but whatever reason nucleation centers being at the glass surface isn't surprising. True is that the film forming is quite peculiar of the test.

@Karl I agree: it was the lower right corner that had a little tidbit of maybe helpful insight. There is certainly no problem making silver nanoparticles or growing a silver tree: I have done these as demos. And adding zinc metal to a silver nitrate or nitrite solution just makes silver powder. So there must be something about kinetically hindering the process to ensure slow and orderly deposition. I hope someone provides a correct answer for this one!

@EdV Me too! But I guess someone has to write a new question that already clears the more obvious points. OP: Please update, then we can vote to reopen!

@Karl Sorry, I just got distracted by a message to vote for moderators in the EE stack exchange. I see nothing that I can add to elucidate how the silver mirror forms. Maybe this could be studied using things like SERS, but I have nothing substantive to go on. I know people still make relatively large silver mirrors for fun, but presumably they are following prescriptions or old recipes known to work. In any event, an interesting question that does not directly involve honey or homework. ;)

@EdV Silvered (Dewar) lab glassware is still made this way, and amateurs grinding large telescope mirrors certainly do the same. I once tried to find scientific work on how to control the thickness of such silver films, wanting a dewar with specific a.c. electrical properties. Did´t find anything really useful, however. Very mysterious. ;-)

@Karl   Just found a scan of a yellowed 1931 article from the old N.B.S. Journal of Research: https://nvlpubs.nist.gov/nistpubs/jres/7/jresv7n5p935_a2b.pdf. This must have been studied before 1931, but the available techniques back then would have been problematic. Even pH meters date from around 1931.

@Karl    Another paper from 1911: http://articles.adsabs.harvard.edu//full/1911PA.....19..327C/0000329.000.html. And it references even older work more than half a century before 1911. But no luck finding out why this process, which appears almost alchemical, works. There is a sentence in German on the first page that my college German is too long gone to translate.

@EdV Quincke says that all those methods for silvering occasionally fail, or give substandard mirrors, and one is unable to find out a proper reason for that. Alchemy. ;-)


text:
Why does ethanol have a stronger cooling effect on the body than water?
Why does ethanol have a stronger cooling effect than water (i.e. it feels cooler when dropped onto one’s hands) even though water has a higher heat of vaporization compared to ethanol?
If we drop a unit amount of both ethanol and water on two body surfaces (both at same temperature), which substance will lower the temperature of the surface more after complete evaporation?

You beg the question: What do you mean by "more cooling effect"? More cal/g? Quicker action at room temperature or at body temperature? Without details, it's meaningless. Edit the question with more specific info.

I have edited my question.

Related: Why does hand sanitizer leave your skin feeling cool?


text:
Reaction of hydrogen peroxide with sodium carbonate
In my book the following acid base reaction is written:
$$\ce{H2O2 + Na2CO3 -> Na2O2 + H2O + CO2}$$
But I found that the $\mathrm{p}K_\mathrm{a}$ value of $\ce{H2CO3}$ is less than the one for $\ce{H2O2},$ which means $\ce{H2CO3}$ is more acidic than $\ce{H2O2}.$
As $\ce{H2CO3}$ is getting formed as the product (and then decomposing to give $\ce{CO2}$), shouldn't the reverse reaction be more favourable due to greater acidity of $\ce{H2CO3}$ and the above reaction not happen significantly?
Is $\ce{H2CO3}$ maybe getting decomposed as soon as it forms allowing this reaction to occur?

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Unfortunately, the reaction is wrong and incorrect. Which textbook is teaching this non-sense?

Just some local book. So if this reaction doesn't occur, is the reverse reaction true? Do we get sodium carbonate and hydrogen peroxide by mixing carbonic acid with sodium peroxide?

Note that carbonic acid is not an ordinary reagent, it does not exist as such. You cannot buy it. So the terminology that mixing carbonic acid with something is incorrect. Yes, the opposite may be likely if you understand that carbonic acid is nothing but carbon dioxide dissolved in water (like Sprite or Coke)

The impossibility of a reaction producing simultaneously $\ce{Na_2O_2, H_2O_2}$ and $\ce{CO_2}$ can be proved by the following arguments, drawn from the Merck Index, under the reference "$8728$. Sodium Peroxide".
1) $\ce{Na_2O_2}$ is produced by burning metallic sodium in a current of air, from which carbon dioxide has been removed.
2) $\ce{Na_2O_2}$ reacts so easily with $\ce{CO_2}$ that it is used for rendering air charged with $\ce{CO_2}$ respirable as in submarines and diving bells.
3) $\ce{Na_2O_2}$ absorbs water from the air forming sodium hydroxyde and hydrogen peroxide, the latter quickly decomposed into Oxygen and water.
Ref.: Maryadele O'Neil, Ann Smith, et al., The Merck Index, An Encyclopedia of Chemicals, 13th Ed. 2001, Merck & Co Inc., Whitehouse Station, NJ, USA.


text:
Decomposition of oxalic acid
Normally, carboxylic acids on heating form an anhydride. However, Oxalic acid on heating undergoes decomposition to yield $\ce{(CO + H2O)}$.
Why does not Oxalic acid form an anhydride?

Have a look at the hypothetical molecular structure of oxalic anhydride. How stable do you think it is?

"Normally, carboxylic acids on heating form an anhydride." Who said that? Never trust textbooks that give chemical "facts" without presenting the evidence.

Btw. https://en.wikipedia.org/wiki/Dioxane_tetraketone

https://pubs.rsc.org/en/content/articlehtml/1998/p1/a803430c

Furthermore, oxalic acid produces $\ce{H_2O + CO + CO_2}$ on heating

You cannot say oxalic acid does not form self condensing anhydride for certain because there is a publication in a higher rank journal calming they have synthesized ethanedioic anhydride dimeric compound $\bf{1}$ (Ref.1). Until someone proves they were mistaken, that fact stays. The even more unusual anhydride, ethanedioic (oxalic) anhydride momomer was not isolated, but believed to be the intermediate during the decomposition of $\bf{1}$ to $\ce{CO + CO2}$. Yet, it still has a CAS number: CAS 308818-63-5, probably gotten from this sited publication (Ref.1), abstract of which states that:

Ethanedioic (oxalic) acid anhydride $\bf{1}$ was prepared for the first time and its dimeric cyclic nature established by spectroscopic and chemical evidence. The strong instability of 2-chloro-2-oxoethanoic acid $\bf{3}$ was incidentally observed. The chemical behaviour of $\bf{1}$ with $\ce{MeOH}$ and $\ce{CH2N2}$ was investigated.

The compound $\bf{3}$ mentioned in the abstract is oxalic acid monochloride ($\ce{HO2C-COCl}$). According to the authors, unsuccessful attempts have been made as early as 1908 (Ref.2) when Staudinger has evidently dealt with likely intermediate $\bf{2}$  in the decomposition process.
During the present study, $\bf{1}$ was achieved by adding  ethanedioyl dichloride ($\ce{ClOC-COCl}$) to silver salt of oxalic acid ($\ce{AgO2C-CO2Ag}$) at $\pu{-15 ^\circ C}$. The evidence for formation of $\bf{1}$ was the sharp single peak at $\pu{144.94 ppm}$ in $\ce{^{13}C}$-NMR spectrum run at $\pu{-15 ^\circ C}$ (Ref.1). The calculated value of $\ce{^{13}C}$-NMR signal for $\bf{1}$ is $\pu{154 ppm}$. Also, the authors noted $\bf{1}$ is indifinitely stable at $\pu{-30 ^\circ C}$ (Ref.1).
References:

Paolo Strazzolini, Alberto Gambi, Angelo G. Giumanini, Hrvoj Vancik, "The reaction between ethanedioyl (oxalyl) dihalides and $\ce{Ag2C2O4}$: a route to Staudinger’s elusive ethanedioic (oxalic) acid anhydride," J. Chem. Soc., Perkin Trans. 1 1998, 2553-2558 (DOI: 10.1039/A803430C).
H. Staudinger, "Oxalylchlorid," Chem. Ber. 1908, 41, 3558-3567.

The story has progressed with Professor Rzepa's recent blog post https://www.ch.imperial.ac.uk/rzepa/blog/?p=25270 suggesting that the compound is more stable than experiment suggests.  It's instability may be due to acid contamination.


text:
Why is acetate a monodentate ligand?
Why is the acetate ion considered a monodentate ligand as opposed to a bidentate ligand considering both oxygen atoms can donate their respective lone pair of electrons?

There is a delocalization of the negative charge over the two Oxygen atoms of the carboxylate group $\ce{-COO^-}$. These Oxygen atoms are not independent from one another. They behave together as a mono dentate ligand.

Acetate ion is considered a monodentate ligand because it is a monodentate ligand (except when it isn't). Then come the explanations.

This depends a lot on the compound in question.
In general, a bidentate acetate would lead to a four-membered ring with the central metal; since the $\ce{O-C-O}$ angle is relatively fixed at approximately $120^\circ$ and since we expect the resulting compound to be at least somewhat symmetric, it would typically lead to unusually small bond angles or a $\ce{C\bond{...}M}$ distance that is too short. If, for example, we imagine $\angle(\ce{O-M-O})=90^\circ$, that means each $\ce{M-O-C}$ angle would be $75^\circ$ which is not what we usually observe in such compounds without a significant force holding it together. Coordinate bonds are usually not very strong, do not liberate enough energy and it would thus be more favourable for one of the two $\ce{O-M}$ bonds to be broken leading to acetate as a monodentate ligand.
There are, however, exceptions. My favourite is probably $\ce{[Cr2(OAc)4(H2O)2]}$ or chromium(II) acetate monohydrate. In this compound, four acetate ions arrange around two chromium atoms in a bidentate fashion; however, each individual acetate bonds to two different chromium ions via its two oxygens. In addition, a water molecule sits on each end of the $\ce{Cr-Cr}$ axis and it turns out the bond between the two chromium ions is, in fact, a quadruple bond. The quadruple bond permits the chromium atoms to come close enough to each other so that the acetates can arrange in a preferable configuration.

(Image taken from Wikimedia, where a full list of authors is available)

My favorite is uranium. With it, acetate works as a bidentate ligand to one atom.

@IvanNeretin Ah, but then it can’t compete with a quadruple bond, can it? ;)

Yeah, that's a rare gem indeed.

Searching for acetate metal coordination turned up a few compounds with similar coordination to your chromium(II) acetate: Palladium(II) acetate, Rhodium(II) acetate, Copper(II) acetate, Iron(III) acetate ... it does not seem exceptional at all, except I assumed these are solids and that generally speaking the acetate bonds to two metal atoms. So it seemed contrary to the spirit of "bidentate" in the original question. Ivan's uranyl acetate seems closer to the real thing.

Not the same thing, but fun to see, are basic acetates of zinc and beryllium.  Six acetate ligands bond to all the edges of a metal tetrahedron with a four-coordinate oxygen in the center.

@BuckThorn You’re raising a good point, namely whether denticity refers to bonds made to a single atom or bonds made to any dative bond acceptor. I understood it to be the latter, but the definitions I can find in two seconds of Googling are not clear enough to exclude either meaning. That said, I prefer chromium(II) over the other metals because it features a 3d metal in a quadruple bond which never ceases to fascinate me.