text:
Is there a difference between dipole and dipole moment?
I’m confused as to whether both terms “dipole” and “dipole moment” are the same or different, does the moment have something to do with the molecular geometry? I know the vectors of the charges cancel out however carbon dioxide has no dipole moment because of 180 degrees, this makes it non polar, but it still has partial charges right? So isn’t it still polar?

There is no difference. CO2 is pretty polar, despite having zero dipole moment.

Well, being polar is not the same as being a dipole. A dipole is a phenomena of charge displacement and dipole moment is its quantification. To be polar, a molecule must be a dipole, either as the whole as HCl, or its parts, like 2 polar C=O bonds in CO2 are dipoles, that cancel each other. For polarity in sense of relative permitivity, both dipole moments and polarizability count.

As @Poutnik writes the electric dipole is the separation of equal and opposite charges $q$ in a molecule and the dipole moment is this difference multiplied by their separation or $ q \times d$. As a vector, conventionally it points from negative to positive charge. The dipole experiences no net force in a uniform electric field but does experience a torque which rotates it to align with the field.

Note that (permanent and oriented) dipole-dipole electrostatic interaction is proportional to 1/r^4. If one of dipoles is induced due molecule polarizability, the distance dependence is even stronger.

The (electric) dipole moment is a mathematically clearly defined quantity, the product of charge difference and distance
A dipole is anything (e.g. a molecule, or a part of a molecule), that has a non-zero dipole moment.
And a polar molecule is one that has at least local dipoles. CO2 is polar if you get close (i.e. in a condensed phase). From a slightly larger distance, it's unpolar, because the centers of the positive and negative charge distributions in it are identical, the two local C=O dipoles cancel each other out.

The word polarity is also discouraged by IUPAC because polarity has no specific meaning. It is an umbrella term for several types of interactions. The proper term is dipolar or dipolarity (in a strict sense).

@M.Farooq Yea, I heard about that. I don't think that there can be a context in which "polar" or "polarity" run any immediate danger of being misunderstood. Think IUPAC is slowly running out of things that need a proper new definition. ;-)

Actually, I am in favor of IUPAC's suggestion. Come to the field of analytical separations, the usage of solvent polarity will only disappoint you. I am talking about advanced level/ research papers. Yes organic chemists are free to define whatever solvent polarity is, but to a physical chemist, it has no meaning whatsoever.

@M.Farooq You mean it's not a quantitative measure? Right. But there are ways to quantify solvent polarity. Changing the name to dipolarity is just a new label.

Sort of, but polarity is an umbrella term for so many interactions. One can certainly use solvatochromism to assess "polarity." For example see "dipolar aprotic solvent: A solvent with a comparatively high relative permittivity (or dielectric constant), greater than ca. 15, and a sizable permanent dipole moment, that cannot donate suitably labile hydrogen atoms to form strong hydrogen bonds, e.g. dimethyl sulfoxide. The term (and its alternative 'polar aprotic solvent') is a misnomer and is therefore discouraged."

I’m confused as to whether both terms “dipole” and “dipole moment” are the same or different, does the moment have something to do with the molecular geometry?

Yes, in colloquial usage you will hear dipole, dipole moment used interchangeably. I think you will feel better if you use the full name: electric dipole moment. This is what the mother of all chemistry related terminologies, IUPAC, tells us to do so.
Dipole moment is a general term because in advanced physics classes you will hear magnetic dipole moment. There you would have to differentiate between electric vs. magnetic dipole moments.
In order to understand O=C=O, you can say that this molecules has two bond dipole moments, since these vectors are 180 degrees apart, there is no net electric dipole moment.
Regarding molecular geometry and dipole: Yes the molecular geometry will let you decide whether there is a net electric dipole moment or not. Nobody in the real world determines the molecular structure on the basis of bond electric dipole moments or the presence of so-called lone pairs. This is good for first year textbooks only. Both molecular geometry and electric dipole moments are experimentally determinable parameters. Electric dipole moments will help to eliminate certain possible molecular geometries. For example, one can be confident that O=C=O is not bent. If it were, it would have a net electric dipole moment.

Dipole moment refers specifically to the second moment in a multipole expansion. An electric dipole moment appears in the expression for the second moment in the multipole expansion of the electric field (or potential) produced by a distribution of electrical charges. "Dipole" is generally a casual abbreviated way to refer to the same thing$\dagger$.
In the case of the distribution of electrical charge in a molecule, the field is divided into a sum of contributions from various multipoles. At large distances from the molecule the sum over the contributions of a small number of terms often suffices to describe the molecule's electric field. The multipole expansion may be applied to the field generated by charges in a local region involving only a pair of bonded atoms, in which case the dipole moment can fairly be referred to as a bond dipole moment. It will differ from the dipole moment for the entire molecule.

$\dagger$ For instance, in the wikipedia entry for electric dipole moment it is stated that

Theoretically, an electric dipole is defined by the first-order term of the multipole expansion;

In this example, "dipole" is used where "dipole moment" is meant. An electric dipole in strict usage consists of two charges separated by a finite distance. The dipole moment is a property of any distribution, including a real dipole, obtained when attempting to approximate the field generated at large distances from the charge distribution by performing a multipole expansion. The higher order multipoles of a true dipole are nonzero, that is, such a dipole will have a dipole moment and other non-zero moments. This is because a dipole moment is a theoretical abstraction used to help represent the field generated by a real distribution.
The wikipedia is a little more careful in making this distinction in the description that follows the above statement:

it [the dipole] consists of two equal and opposite charges that are infinitesimally close together. This is unrealistic, as real dipoles have separated charge. However, because the charge separation is very small compared to everyday lengths, the error introduced by treating real dipoles like they are theoretically perfect is usually negligible. The dipole's direction usually points from the negative charge towards the positive charge.

I challenge you to make a sensible statement in proper English, casual or not, in which you can leave out the "moment" and it's still correct.

@Karl challenge accepted. The wikipedia was sufficient to provide an example.

Care to be specific? ;-) I looked in there, and didn't find one. No surprise, you can't replace something that describes an abstract property with a name for a real object and hope to still make sense.

If you mean the sentence you cited above: "Theoretically, the electric dipole moment is defined as the first-order term of the multipole expansion; " is quite a different statement.

@Karl Yes, I meant exactly that example. My point is that people will often use words without literal intent. Someone might write "Theoretically, an electric dipole is" and mean "Theoretically, an electric dipole moment is". It is incorrect usage, therefore my explanation that this is "casual" usage, not strict.

Sometimes you'll read the "dipoles of the two molecules interact leading to alignment" and so forth, where the authors are referring to interaction of a particular term of the multipole expansion of the electric field.

Well, I find it terrible to write down such sloppy language, esp. on wikipedia. For lab-talk, nevermind. That last example is weirdly put. I would write "dipoles interact, leading to alignment", or, more formally, "dipole moments of the two molecules interact, ..".


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What is the major product of the reaction of a geminal dibromide with silver nitrate?

$\ce{Ag+/H2O}$ is used for peroxide formation, then followed by hydrolysis to yield an alcohol. But in the question it is a geminal dihalide and I don't know if the mechanism follows through a carbocation intermediate.
According to the solution it should go through carbocation rearrangement, because the given answer is (b) which is obtained from ring expansion.

possible duplicate of https://chemistry.stackexchange.com/questions/41500/reaction-of-furan-with-chloroform-in-base .Check the answer in the link.That might help you

cant relate reimer tiemann reaction with this

check ring exoansion of gem dihalo molecule in the same post.

@ChakravarthyKalyan - I'm not an organic chemist, but that carbene/carbanion mechanism looks very different to the carbocation mechanism here.

@ Karsten Theis ,yes mechanism is via carbocation assisted by Silver ion.One above is base catalysed via carbanion.I have answered this to give relevant mechanism. You are pretty good in understanding reaction  mechanism.Pl.check my answer below.

This is a specific example of the Silver Nitrate Test, one of the experiments used in organic chemistry teaching laboratories. Usually, the Silver Nitrate Test allows for the identification of alkyl halides by observing them in an alcoholic silver nitrate environment. For change, here used aqueous environment instead. In general, You will test the reactivity of several alkyl halides in a $\mathrm{S_N1}$ reaction. The silver ion (a halophilic reactant) complexes with the halide and precipitates out of solution first (note that silver nitrate in aqueous or alcoholic solution promotes ionization of the alkyl halide), and resulting carbocation react with ethanol to form ethyl ethers (in your case, it forms alcohol with water). The rate of the silver halide salt precipitation is characteristic of different types of alkyl halides ($3^\circ \gt 2^\circ \gt 1^\circ$).
When silver nitrate is used with $1^\circ$- or $2^\circ$-alkyl halides, a rearrangement may occur before the product formation stage. For example:
$$\ce{(CH3)3CCH2-Br + H2O + AgNO3 -> (CH3)2C(OH)CH2CH3 + AgBr + HNO3}$$
Rearrangements will only occur when the resulting carbocation is more stable than the initial carbocation. For example, see a relevant reaction below (Ref.1):

In your case, $\ce{Ag+}$ ion act as a Lewis acid and abstract one of bromide to leave bromocyclopropyl carbocation which is also comparatively stable cation (see the ring expansion in the diagram). The driving force here to rearrange the structure is releasing the strain energy, which gives you a relatively stable allyl cylohexenyl carbocation. It would then react with a water molecule to give the compound 2 as the final product. The same result was obtain from this reaction and has been reported (Ref.2), abstract of which states that:

Several ring expansion products carrying vinylic bromo functionality were synthesized by opening of the geminal dibromobicyclo[n.1.0]alkanes ring. Dibromocarbene was formed from bromoform and potassium tert-butoxide in hexane. Its reaction with various cyclic alkenes was the resultant dibromobicyclo[n.1.0]alkanes. Then, opening was performed using $\ce{AgNO3}$ in various solvent systems, such as acetic acid/DMSO, acetic acid/DMF, $\ce{CH3OH}$/acetone, and $\ce{H2O}$/DMF.

References:

Russell J. Hewitta, Joanne E. Harvey, “Synthesis of C-furanosides from a D-glucal-derived cyclopropane through a ring-expansion/ring-contraction sequence,” Chem. Comm. 2011, 47(1), 421–423 (DOI: 10.1039/C0CC02244F).
Mesut Boz, Hafize Çalişkan, Ömer Zaím, “Silver Ion-Assisted Ring Expansions in Different Solvent Systems,” Turk. J. Chem. 2009, 33(1), 73–78 (DOI: 10.3906/kim-0807-3).

@ Mathew Mahidaratne ,quoting from your post "The driving force here to rearrange the structure is releasing the strain energy" , is logical explanation for ring expansion from 5-6.In example, in Ref.1,ring contraction from 7-5 seems opposite.Any mechanism for this change would help.

The ring expansion I mentioned is cyclopropyl ring. Ring opening gives more stable cyclohexene instead. In *Turk. Chem. examples, appropriate ring opening to give even larger ring, because releasing strain energy is important.

The question is  summarised in the paper, Silver Ion-Assisted Ring Expansions in Different Solvent Systems.$\ce{^1}$
The same reaction is reported$\ce{^2}$,which states that:

Opening 6,6-dibromobicyclo[3.1.0]hexane (7) in a methanol/acetone solvent system resulted in 1-bromo6-methoxy-1-cyclohexene (9) 16 and 1,6-dibromo-1-cyclohexene (10).

Another solvent system (H2 O/DMF) was also used for ring-opening reactions of the dibromocyclopropane ring that was synthesized from branched alkenes and only gave the alcohol derivatives, 2-bromo-1-methylcyclohept-2-en-1-ol (13) and 2-bromo-3-methylcyclohept-2-en-1-ol (14).

The possible mechanism for this ring expansion is shown below.

Referances
1 . Mesut Boz, Hafize C¸Alis¸Kan, Omer Zaim ," Silver Ion-Assisted Ring Expansions in Different Solvent Systems,” Turk. J. Chem. 2009, 33(1), 73–78.
2 . Mesut Boz, Hafize C¸Alis¸Kan, Omer Zaim ," Silver Ion-Assisted Ring Expansions in Different Solvent Systems,” Turk. J. Chem. 2009, 33(1), 73–78.

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Is it possible to burn things in a halogen atmosphere?
Fires can burn in oxygen. But, can they burn in halogen atmospheres (fluorine and chlorine only)? I am wondering this because combustion reactions involve oxygen and carbon and other elements. But, can they burn in fluorine and/or chlorine? If they do, what color will the flame of paper and wood be in fluorine and chlorine atmospheres respectively?

Certainly can in fluorine, it has been used as a rocket fuel despite the obvious difficulties in handling

Most things burn with selfignition in ClF3, including glass, sand, asbestos and human body

See this answer: https://chemistry.stackexchange.com/questions/27465/can-paper-burn-without-oxygen-or-air/104156#104156

Also see: https://chemistry.stackexchange.com/questions/64541/is-oxygen-necessary-for-burning/64546

Titanium metal will ignite and "burn" in bromine, so I expect it also will in F and Cl.

Here's a video of a BRICK burning in fluorine gas

I saw titanium ignite and burn in liquid bromine at room temperature; I didn't see much as we were running to the lab door.

They certainly can. Put some water in the bottom of a beaker, fill the beaker with chlorine gas, and drop in bits of calcium carbide. The carbide will produce bubbles of acetylene, which will spontaneously burst into flame when they hit the chlorine.
Flame colors depend on many things, even in oxygen. In general, flames in a chlorine atmosphere will probably look slightly more greenish than flames in air or oxygen, simply because the gas itself is green. Fluorine isn't as strongly colored, and flames from combustion in fluorine will be VERY VERY BRIGHT.

Fluorine gas definitely supports flaming combustion.  And does so more than oxygen.  Very dramatic, actually.  Chlorine, less so than oxygen or air, but still possible in right setting.
See below for examples of fluorine flames.  Examples are not in $100~\%\ \ce{F2}$ environment, but should be obvious that $\ce{F2}$ makes more flames than air.  In many cases, no spark is required for flames to start.
https://www.youtube.com/watch?v=vtWp45Eewtw
Here are some demonstrations of flames in straight $\ce{Cl2}$ (no oxygen):
https://www.youtube.com/watch?v=MtygiCwnEzw
https://www.youtube.com/watch?v=aK85PZX2xNE

Note that forming  flame is not limited to the particular environment like gas containing enough of oxygen, even if we are used to that.
I remember an experiment in the high school chemistry class, where the lecturer presented oxygen being burnt in flame in atmosphere of hydrogen.
For a flame, several conditions must be met:

There is ongoing fast exothermic enough chemical reaction ( or there is provided intensive external energy like MW powered argon plasma in ICP )
Reaction must provide gas, or occur in  already existing gas.
Gas must be heated by reaction enthalpy enough to form ions, radicals, excited molecules or atoms, actively emitting light by releasing excitation or recombination energy in form of visible light.
The flame occurrence depends on scenario. Oxidizing and oxidized stuff may come into contact at contact area (flammable gas output lighted to prevent gas accumulation), in whole volume (welding by acetylen - oxygen flame ), or a chemical stuff burns itself, being a mixture(black gunpowder), or containing oxidative groups ( explosives and smokeless gun powder staff )
Flame color depends on emission lines of above mentioned excited particles and there is no general answer.

Just to keep things confusing, it's possible to have flame without gas as a reactant or product; "solid flame" is a thing. Tantalum and carbon, mixed thoroughly enough, sustain a propagating exothermic reaction that does not melt or vaporize either reactant.

@jeffB Well, it is still possible to create a flame within non involved gas.

Most chemists understand that Pyrex glass is an extremely stable formulation which is part of the reason it is used in chemistry. But if you introduce a hot piece of Pyrex borosilicate glass into an atmosphere of fluorine, the Pyrex will actually burn. Try to imagine a more unlikely scenario than burning glass. An elemental fluorine atom's need for an electron is overwhelming.

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Are glycosides more soluble in dichloromethane than in water?
I'm designing an experiment to extract the amygdalin from apricot leaves and kernels. I want to steam the kernels and leaves, grind them, add water and then filter it using a filter paper. Then transfer the solution to a separatory funnel, add dichloromethane (DCM), shake and collect the bottom DCM layer that I'm hoping would contain the amygdalin.
This is the process I used for extracting caffeine from coffee beans and it worked because caffeine is much more soluble in DCM than in water. However, I'm not sure about the solubility of glycosides and whether this process would even work or not which is why I'm making this post.
How soluble are glycosides in DCM, is it enough to make this work? If not, what other solvents do you recommend?
Also, how can I increase the efficiency of this process in general? I've been thinking of using vacuum filtration, but I'm not sure. Tell me what you think.

Possibly related: https://chemistry.stackexchange.com/questions/24528/how-do-i-extract-cyanide-from-apple-seeds, with the cited paper comparing three extraction methods: https://academic.oup.com/chromsci/article/43/7/383/384427

Anybody with glycoside isolation from bio-matrices would tell you that your plan is not going to work. Amygdalin is classified as a cyanogenic glycoside because it contains nitrile moiety. It is contained in the seeds of bitter almond, apricot, apple, black cherry, peach, plum, etc. Amygdalin was first isolated in 1830 from bitter almond seeds (Prunus dulcis) by French chemists, Pierre-Jean Robiquet and Antoine Boutron-Charlard. Traditionally, the extraction method of the chemical ingredients in the medicinal materials is decoction in boiling water, which was the method French chemists has used to isolate amygdalin. Presently, it has been isolated using several other methods and its structure has been characterized by several groups around the world.
Amygdalin can be dissolved in water and ethanol easily. Solubility of amygdalin in water and ethanol is $\pu{83 g/L}$ and $\pu{1 g/L}$, respectively (Ref.1). Ref.1 described the isolation of amygdalin from plum seeds for the first time, has used ethanol to extract soluble compounds, and finally, used diethyl ether to isolate amygdalin from the complex ethanol extraction mixture. Before extract, the plant material was dried to the moisture content of 6% in a dark place and then ground in a blender to obtain a fine powder with average particle size of $\pu{0.3 mm}$. The ethanol extraction and diethyl ether isolation procedure used in Ref.1 have been described as follows:

The powdered solid plant material ($\pu{2.0 g}$) was transferred into the Plant material (2 g) was transferred into the $100$-$\pu{mL}$ round bottom flask and treated with exactly defined volume of ethanol. The extractions at higher temperatures were carried out under reflux, while the extractions at lower temperatures were carried out in closed vessels. The system temperature was maintained using a water bath. After extraction, the extract was separated from the solid plant material by filtering process. Plant extracts were evaporated in rotary evaporator under reduced pressure to remove the solvent, and then stored in desiccator until it reached the constant mass. After addition of diethyl ether ($\pu{10 mL}$) to the complex extraction mixture, amygdalin was precipitated, while fatty compounds were dissolved. The dissolved fatty compounds were removed by decanting process. The solid phase was dried at $\pu{30 °C}$ in order to remove the residual solvent. In this way, the sample was prepared for determination of its purity using HPLC analysis and structural characterization by FT-IR, UV, and MS methods.

The effect of extraction time ($\pu{10–120 min}$), ethanol concentration ($20–100\%,\: v/v$), solid-to-liquid ratio ($1: 5\: – 1: 25,\: m/v$) and extraction temperature ($\pu{22–78°C}$) on the content of amygdalin in the dried extract has been studied. As a result, the optimal conditions were achieved: The extraction time of $\pu{120 min}$; $100\% \:(v/v)$ ethanol; solid-to-liquid ratio of $1:25\: (m/v)$; and extraction temperature of $\pu{34.4°C}$. The optimal yield of $\pu{25.30 g}$ of amygdalin content in the extracts per $\pu{100 g}$ of dried extract was reported.
Reference:

Ivan M. Savic, Vesna D. Nikolic, Ivana M. Savic-Gajic, Ljubisa B. Nikolic, Svetlana R. Ibric, Dragoljub G. Gajic, “Optimization of technological procedure for amygdalin isolation from plum seeds (Pruni domesticae semen),” Frontiers in Plant Science 2015, 6, Article 276 (11 pages) (DOI: 10.3389/fpls.2015.00276).

I understand that your answer explains what has been done before, but you don't mention what is wrong with DCM as a precipitant.

If amygdalin is 80-times more soluble in water than in ethanol, do you think it would be soluble in dichloromethane? The OP envisons extrcting with DCM (like caffeine), not precipitating. But you are right, the solubility of amygdalin in DCM is not addressed in this answer. @BuckThorn

Ok, then I misunderstood how the DCM was to be used. I thought it might be used to precipitate amigdalin.

@Buck Thorn:  Karsten Theis was right on spot answering your question. DCM may be good enough to precipitate amigdalin, but not for solvent extraction. I'd even use ethyl acetate instead of DCM for environmental concern. When you look at solubility difference of amigdalin in water and ethanol, it is obvious that it won't dissolve in DCM preferably, because hydroxy group interactions are seemingly playing important role here.


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Reactivity of Grignard reagent
I had the following question at an exam:

What is the fastest reacting compound towards nucleophilic addition with Grignard reagent?
(a) Acyl chloride
(b) Aldehyde
(c) Ester

It is obvious that ester would be the slowest, but among acyl chloride and aldehyde, acyl chloride would have a highly electrophilic, but a bit sterically hindered carbon atom and the resonance will also decrease the electrophilicity, whereas the aldehyde will have a less sterically hindered and moderately electrophilic carbon atom.
Which one will prevail and how to judge?
What would happen if instead acyl chloride there were acyl bromide?

IMO,the approach at the Burgi-Dunitz angle will be having quite less steric hindrance both in the case of (a) and (b) (although there will be relatuve differences). But the increased electrophilicity of the acyl chloride(in MO terms it essentially means that the LUMO has gone down in the energy profile) means that the reaction rate should be faster for the acyl chloride as compared to the aldehyde(assuming all the three options to be approximately homologous with similar substituents)

@OscarLanzi Gringard being a nucleophile attacks the Carbon of the Carbonyl , which on hydrolysis yield a 2° Alcohol (3° in case of Ketones)

Oh , Oops , Sorry , Nucleophilic Addition, My bad , so sorry..

Tech note: please avoid unnecessary edits that lower the quality of your post, namely in written English there is no need for a white space before punctuation marks (with rare exceptions). For your case with ? see Is it ever correct to have a space before a question or exclamation mark?.

@andselisk Non Native English Speaker , I will takes a note of that

@user232243 Me neither, also there is no white space before the comma ,:)

The answer will be (a) Acyl chloride.
The acyl carbon in an acyl chloride is connected to the chlorine atom. While you have acknowledged this fact, you have missed the point that the +M (resonance) effect of Chlorine is overcome by the -I (inductive) effect. Thus, the electrophilicity of the acyl carbon is actually greater than the electrophilicity of the carbonyl carbon in an aldehyde. This jump in electrophilicity is so significant that the difference in steric hindrance is neglected.
In fact, this the major reason why acyl chlorides are considered to be one of the best electrophiles in all of organic chemistry.
If Acyl bromides were used, then they would have an even greater electrophilicity and it would be the answer to this question.
Hope this answers your question!

order of relative re-activities  of aldehydes and ketones and carboxylic acids is studied by C.E. Entemann Jr and John R. Johnson of cornell univeristy in 1933.
(J. Am. Chem. Soc. 1933, 55, 7, 2900–2903)

They have taken equal quantities of the reactants in dry ether in the same vessel, which are to be compared. PhMgBr was slowly added over a period of two and half hours. Based on the amount of unreacted (recovered) of reactants, they have established the order.

That is the reverse order in running Grignards usually the reagent is generated in excess  and the electrophile added slowly to prevent an explosion.


text:
Redox titration or iron(II)

The concentration of $\ce{Fe^2+(aq)}$ can be determined by a redox titration using

A. $\ce{KBr}$
B. $\ce{SnCl2}$
C. $\ce{KMnO4}$ (basic)
D. $\ce{KBrO3}$ (acidic)

Can anyone please help me on this question? I know the answer is between C and D, since those two are the only redox reactions. However, I don't know how to determine between C and D, as they will both act as an oxidizing agent and Fe will be reduced in both cases (the correct answer is D).

This question requires some background chemistry knowledge. First of all, think whether the presence of base will interfere or not. Will iron(II) precipitate out as a hydroxide? We wouldn't like to do titration in a heterogeneous mixture. The next to keep in mind is electrode potentials of permanganate and bromate half cells.

May I ask for clarifications to electrode potentials? do you mean I should calculate for the overall E value of the reaction?

Can you please elaborate on that part? I know that the E value of Mno4- is +1.51, and the one for BrO3- is +1.48. What does that suggest??  that Mno4- will more likely to be reduced by Fe+ than Bro3-?? please explain

Did you understand what is the reductive electrode potentials are? By the way, $\ce{Fe}$ won't reduce in any case (as you suggested).

Yes i do. I guess I dont understand the role that acid and base play in redox reactions. When doing a titration, how do I know when to use an acidic medium and when to use a basic one? I dont understand how looking at half cells potentials can allow me to determine this

It all depend on your reaction. First tell me what is $\ce{Fe^2+}$ going to do during the titration. Is it going to oxidize? or reduce?

oxidize. I made a mistake when I typed the question out

Can you please elaborate on that part? I know that the E value of
  Mno4- is +1.51, and the one for BrO3- is +1.48. What does that
  suggest?? that Mno4- will more likely to be reduced by Fe+ than
  Bro3-?? please explain

The first and foremost hint, even before you calculate the electrode potentials is the solubility rule taught in general chemistry. Hydroxides of most metals are insoluble. Write the equations : Fe(II) + hydroxide ions = iron (II) hydroxide. This is why alkaline permanganate is not a good choice. However, this is not the final answer because you would like to further confirm whether C or D is the correct choice.
Next determine the E_cell= E(cathode)-E(anode) > 0 V or not.
Now you need to oxidize Fe(II) to Fe(III), so the iron half cell will be the anode. As a mnemonic, anode and oxidation both start with vowels; cathode and reduction don't. You would like to reduce the permanganate in an alkaline medium, this half cell is the cathode.
Repeat the same for bromate half cell in acidic medium and Fe(II). Both (C) and (D) would turn out to be positive E_cells. However, the (C) is not an answer because you don't want iron to precipitate out as iron(II) hydroxide.

In real life analysis, Fe(II) can be routinely titrated with potassium permanganate but in highly acidic medium. For practice, check the E_cell after writing a balanced reaction.


text:
Optical activity of (1r,3R,5S)-3,5-dimethyl-4-methylidenecyclohexan-1-ol
Could anyone explain why compound L is not optically active?

Draw the mirror image of the compound. Can you prove the mirrored version is different from the original?

For the same reason why meso-tartaric acid does not manifest chirality. Its mirror image can be identified with the original, even if it has 2 chiral centers.

Because it has a quite obvious plane of symmetry, period. That's enough of a justification, no matter what the chiral centers are trying to say.

The given molecule,(1r,3R,5S)-3,5-dimethyl-4-methylenecyclohexanol , is optically inactive .This molecule has a plane of symmetry (as shown below) and, therefore, is achiral.

Imagine yourself cutting the molecule in half along the axis determined by the OH group and the double bond, perpendicular to the plane of the cyclohexane ring. Observe that the 2 halves match each other perfectly, in other words you found a plane of symmetry. For any molecule to be chiral, and thus be optically active, it must not have any plane of symmetry.


text:
Why don't ionic compounds show stereoisomerism?
When I was reading about crystalline nature of ionic compounds, I came across the statement that ionic compound doesn't show stereoisomerism. What does that mean and can anybody explain the reason with an example?

If any, it will not even called stereoisomerism. Rather one would say that the crystals have opposite light polarizing properties. Isomerism refesr to molecular compounds, because is where it has sense to be discussed.

They can, and they do.  The secret is to have a chiral cation and separately a chiral anion.  Letting $D^+$ be the dextrorotatory form of the cation, $L^+$ be the levorotatory form, and anologously for the anion, we then have four isomeric salts:
$D^+D^-$
$D^+L^-$
$L^+D^-$
$L^+L^-$
The first and second are diastereomers because only the anion is mirror-reflected; the cation is not mirror-reflected.  Similarly for three of the other five possible pairs; only two of the six possible pairs are enantiomeric.
Being diastereomers instead of enantiomers, the first and second salts, for instance, may have different physical properties such as solubility in a given solvent.  This may be used for separating different enantiomers of the anion (or, with a different pair such as first and third, separating different enantiomers of the cation).  And it's actually done.  See here for a brief summary and here for an example of diastereomeric salt formation.

Interesting! Can you suggest me a source for further reading on stereoisomerism in ionic compounds?

This goes into more detail and provides a textbook reference.


text:
What happens to copper(II) hydroxide when it's heated? what's inside its decomposition?
So just wondering, copper(II) hydroxide undergoes thermal decomposition when it's heated. But isn't it a solid by itself too? Why does it turn into another solid and water all of a sudden?
Can someone please explain to me the reason behind it, thank you!!

Well, chemistry is like that: compounds just up and turn into other compounds all of a sudden.

Well,this is a kind of dehydration reaction,and you will come across similar such reactions plenty of times as you read about the effect of heating on various compounds.
I am going to approach this in a sligthly different manner: how the copper(II) hydroxide was generated from the copper(II) oxide.
According to Wikipedia

Copper(II) oxide belongs to the monoclinic crystal system. The copper atom is coordinated by 4 oxygen atoms in an approximately square planar configuration

So we can think of it this way,that when water was introduced in the copper(II) oxide powder,the water molecules entered into the voids(in a stoichiometric fashion)of the oxide lattice.Mind you,this is just a way of visualization,but it can help a lot. So now,when you are heating the hydroxide in a controlled manner,then we can imagine that the water molecules residing in the voids escape out,leaving the oxide in the same oxidation state. (Also,now you should keep in mind that most hydroxides lose one or more molecules of water to form their respective oxides)
Note: if heating is done excessively,then the oxide itself can decompose to a variety of products


text:
Book Recommendations for Inorganic Chemistry Mechanisms
I am looking for a book which has details or basics to reaction mechanisms for inorganic chemistry for highschool level(preferably) or maybe undergrad level.
Basically I am looking for a book which can help teach me inorganic chemistry in such a way which decreases the amount of rote learning and memorising , but gives concepts/explanation for things. Like in organic chemistry we have mechanisms , and different things, something like that.
Is there any such book out there ?? Please help !!
I was reading this link which has basic resources, but none meet such criteria , please help
Edit: Regarding the question being duplicate: The posts there talk about in general getting started with chemistry. What i am looking for is something more specific which is not covered there in any detail. The book on arrow pushing mostly goes over the reactions of organic chemistry which involve inorganic reagants.
I am looking for something that will teach me inorganic ,not just memorise it.
Eg ,Reasons for the different trends shown by the elements, Reaction Mechanisms , etc ..
PS:For any of my Indian Friends, I am looking for a book helpful for JEE Advanced

This would surely meet your criteria

@YUSUFHASAN  Actually it covers a lot of Organic reactions with the inorganic reagants , I want a bit more purely inorganic type.....
For any of my Indian Friends, I am looking for a book helpful for JEE Advanced

Some very famous reactions are also covered here from the point of view of JEE Advanced and in general,inorganic chemistry in India if you will look carefully. You will find mechanisms(arrow pushing) of the reaction of P4 and NaOH to give PH3,thermal decomposition of NH4NO3 and NH4NO2,and reaction of S2Cl2

These are the few reactions which I remember from this book... You will just have to look carefully

@YUSUFHASAN Okay!! , I had Just skimmed through the book ...

@YUSUFHASAN But I think you would agree , that these are very  scanty

No in this book,you will find a list of indicators for drawing out inorganic mechanisms yourself,and then a few famous reactions are made by the authors... The idea is to learn the art of drawing inorganic mechanisms,so that you can draw out mechanisms for other reactions you come across in your chemistry course


text:
Equilibrium constant for a solid-solid equilibrium
Suppose there are two equations:
$$
\begin{align}
\ce{A(s) &<=> B(g) + C(s)} &\quad &K_1 \tag{R1}\\
\ce{B(g) &<=> D(s) + E(s)} &\quad &K_2 \tag{R2}
\end{align}$$
These reactions have equilibrium constants $K_1$ and $K_2$, respectively. Both of these reactions are occurring simultaneously inside a container.
When I add these equations, I get
$$\ce{A(s) <=> C(s) + D(s) + E(s)} \tag{R3}$$
Since I added the reactions, I must multiply the equilibrium constants to get the equilibrium constant for above reaction which essentially is the net of the two reactions.
I have two questions:

How do I even write equilibrium constant for solid solid equilibrium?
Since $\ce{B}$ doesn't appear in equilibrium constant, is it safe to say that the concentration of $\ce{B}$ doesn't affect the equilibrium concentrations of any other species?

Where do you get "using the adding of reactions argument K=K1K2 , which is a different value."? Write out the multiplication and you should find that it's the same value of 1, as the B terms cancel.

yes indeed , I realised that after I posted the question , but still , the other two questions were my main concern

I excluded that part

How do I even write equilibrium constant for solid solid equilibrium?

First, let's write the expression for the reaction quotient:
$$Q = 1$$
Q is a constant, always one, as long as all reactants and products are present. If you run out of one, it is zero or infinity.
Now, there are two cases. If the reaction is at equilibrium, K is also 1. If not, you can't measure the equilibrium concentrations (the reaction will go to completion, that is not an equilibrium). If you want to obtain the equilibrium constant, you would have to determine the Gibbs energy at standard state, and calculate K from that.
This is certainly quite confusing when we contrast it with other reactions (like R1 and R2), which will reach equilibrium. It gets less confusing if we rewrite R3 as:
$$\ce{A(s) -> C(s) +D(s) +E(s)}$$
An example you might be familiar with is ice turning into water. The reaction quotient is always one. Depending on the temperature, the equilibrium constant is smaller than one (all water freezes), is larger than one (all ice melts) or is exactly one (water and ice coexist in an equilibrium). You might think it is unlikely that we adjust the temperature exactly right to hit the melting point, but in fact it self-adjusts because melting ice is endothermic.

Since B doesn't appear in equilibrium constant, is it safe to say that the concentration of B doesn't affect the equilibrium concentrations of any other species?

It is best to un-ask the question, for two reasons.

There is no equilibrium
The concentration of the other species are constant, so they are not affected by the reaction at all (unless the species is used up)

For reactions where at least one species is a solute or a gas, i.e. a species whose concentration changes continuously and where there is some entropy of mixing, the Gibbs energy will have a minimum somewhere along the way between all reactants and all products. In this special case, everything is a bit different.

So if in a reaction there is no entropy of mixing then we will have either the reactants or the products depending on their relative stability?


text:
Why the bond angles of crystalline compounds uniform while in the amorphous state the same compound has non uniform bond angles?
For example, let us take amorphous and crystalline quartz. While in crystalline quartz all bond angles are uniform, in amorphous form bond angles vary even though quartz has the same valency. Why is it so?

One word: symmetry. Note that using all CAPS letters is inappropriate and is considered offensive in virtually any context; also use the punctuation marks and don't drop a wall of text without any of them!

I think you really want to ask why quartz can have these odd bond angles, right? If quartz glass had a periodic structure, it'd be crystalline. Contradiction in terms.

I object. Amorphous quartz mostly has normal bond angles, just as well as crystalline.

Source of figure e.g. http://www.studiestoday.com/sites/default/files/rdsharma-solutions/NCERT%20Exemplar%20Solutions%20Class%2012%20Chemistry%20Chapter%201%20Solid%20State.pdf

As a primer:
A requirement to qualify as crystalline, you need predictive patterns both in short, as well as in long distance range.  In other words, there is a smallest structural pattern (unit cell) describing all atom's relative position well enough, which then is only replicated by translation.  Which is the case in crystalline state (e.g., quartz, or sodium chloride), but not in the amorphous state (like glasses in general, not only quartz glass or glass in the windows).
However, it seems good to add some grains of salt:

The property "crystalline" is about order.  It is not restricted to "solid state".  A solid sample may be crystalline, but needn't be -- glasses are an example.  On the other hand, a crystalline material needn't to be solid -- think about examples of liquid crystals seen in LCDs.
Mentioned "translation" may occur in all three directions in space.  But putting an other unit cell just next to an other unit cell to build a crystal need not to occur this way, there equally are 2D (think about tiles in your bathroom) and 1D crystals, too.
There are cases outside this strict discern "crystalline or armorphous".  There are polymers neither fitting into one, or the other category with quite some impact on the physical porperties (cf. for example  this reference). An other exception to the simplification above are quasicrystals, which are ordered yet not periodic; and still crystalline.

Quartz can be amorphous or crystalline, and both in close proximity. So do polymers, nothing really special there imo.

The variation in bond angles is much less than in the OP's diagram. Here is a bilayer of quartz observed under a scanning microscope:

Source: https://phys.org/news/2014-06-thin-silicate-atomic-glass.html
It shows a transition from crystalline (left) to amorphous (right). In the transition, the number of rings with six silicon atoms decreases in favor of some with seven or five, mostly.
The amorphous form has higher potential energy (some bond angle strain) and I guess higher entropy. Glasses are metastable, meaning that the crystalline form is thermodynamically favored, but the kinetics of reaching it are very slow.


text:
DNA-Sequencing (detection of the change of pH-value) after emulsion PCR
I have troubles with understanding the exact order of the steps for DNA-sequencing after the Emulsion PCR (EmPCR). The EmPCR fills each bead with the same DNA strand (copies it $n$ times, till the bead is entirely filled up). But how is then the sequencing done with these beads?
The sequencing is described as follows (in easy words): Each nucleotide is provided to the DNA strand. If a nucleotide is accepted, the $\ce{H+}$ concentration rises and can be detected. Hence, the complementary nucleotide was present at the DNA strand.
But to be able to attach a nucleotide to the DNA, it has to be broken up. But the EmPCR already filled up every copied DNA strand. Where is it then broken up again?

May I suggest migration to Biology.SE? Please use the biotechnology and/or biochemistry tag there if you do migrate.

Are you using Roche-454 pyrosequencing machine?

In molecular biology, it is a very important that DNA should be replicated in order to facilitate sequencing. Basically, this is a very important concept that DNA should be replicated, simply in order to amplify signals. No matter the method of sequencing, without a proper amount of amplification, it's near impossible to detect each base call (signal). All Next-Generation Sequencing (NGS) techniques replicate DNA before sequencing is done.
Emulsion PCR is a PCR variation that some NGS technologies (e.g., NGS flatforms: Roche/454’s GS FLX, Titanium; Life/APG’s SOLiD 3; and Polonator G.007) use to replicate DNA sequences. It is conducted on a bead surface within tiny water bubbles floating on an oil solution.
Procedure (Source: https://binf.snipcademy.com/lessons/ngs-techniques/emulsion-pcr):

Fragmentation of DNA: The library is first fragmented either by sonication (high sound energy) or nebulization (forces DNA through a small hole) to fragments ranging from 300 to 800 base pairs.
Adapters are ligated: Adapters are then ligated onto the DNA fragments. These allow the strands to bind to the emulsion beads.
Denature to single strands: The double stranded DNA's with adapters are then denatured by heating the DNA up to $\pu{95 °C}$. Denaturing DNA simply means to go from double stranded DNA (dsDNA) to two single strands (ssDNA) - the hydrogen bonds keeping the two together are broken.
Formation of clonal bead populations: Each bead coated with streptavidin (e.g., Streptavidin-coupled Dynabeads), which is resistant to organic solvents, denaturants, detergents, proteolytic enzymes and extremes of temperature and $\mathrm{pH}$. Over a billion beads are used with a primer that matches the adapters attached earlier. The ssDNA is then attached to these beads. Each bead is emulsified in a water-in-oil droplet with PCR reagents (DNA polymerase, primers, buffers, dNTPs):

Emulsion PCR (emPCR) amplifies DNA strands on beads: Within these droplets, PCR is conducted. This involves the steps Denaturation, Annealing, Elongation. Firstly, the strand is elongated with DNA polymerase and dNTPs. Then the double-strand is denatured, allowing for the strand to ligate to another site on the surface of the bead. Eventually, 1 million copies of the target is amplified on the surface of each bead. The water-in-oil droplet is approximately 1-um. Follow the figure to see how each bead is able to replicate DNA on its surface.

Emulsion Breaking: After the DNA strands are amplified, the emulsion from the preceding step is broken using isopropanol and detergent buffer. The solution is then vortexed, centrifuged, and magnetically separated. The resulting solution is a suspension of empty, clonal and non-clonal beads, which will be filtered in the next step.
Bead enrichment: After PCR is conducted, you are left with a mixture of some beads that have amplified DNA attached on its surface, and some that do not. We may take out the enriched beads by attaching streptavidin coated magnetic enrichment bead. With a magnet, we can then pull out the beads with amplified DNA. There are other methods of bead enrichment that include using larger beads that are able to bind to beads with amplified DNA. After centrifugation, the beads with amplified DNA and without can then be separated.

Bead Capping: Attach a capping oligonucleotide to the 3' end of both unextended forward EmPCR primers and the RDV segment of template DNA. This helps in cover-slip arraying, which is used to polony sequencing, and prevents fluorescent probes from ligating to the ends.
Result: The beads with amplified sequences are then placed on a slide and are sequenced. Due to their high density of the same DNA molecule, the signal is amplified, allowing computers to read the sequencing data.

For example, the method used by the Roche/454 sequencer to amplify single-stranded DNA copies from a fragment library on agarose beads. A mixture of DNA fragments with agarose beads containing complementary oligonucleotides to the adapters at the fragment ends are mixed in an approximately 1:1 ratio. The mixture is encapsulated by vigorous vortexing into aqueous micelles that contain PCR reactants surrounded by oil, and pipetted into a 96-well microtiter plate for PCR amplification. The resulting beads are decorated with approximately 1 million copies of the original single-stranded fragment, which provides sufficient signal strength during the pyrosequencing reaction that follows to detect and record nucleotide incorporation events. sstDNA, single-stranded template DNA (Ref.1):

Keep in mind that in pyrosequencing method, each incorporation of a nucleotide by DNA polymerase results in the release of pyrophosphate, which initiates a series of downstream reactions that ultimately produce light by the firefly enzyme luciferase. The amount of light produced is proportional to the number of nucleotides incorporated (up to the point of detector saturation). Above diagram shows an example of four-color mapping sequence. For additional info, should read all of Ref.1-3.

References:

Elaine R. Mardis, “Next-Generation DNA Sequencing Methods,” Annu. Rev. Genomics Hum. Genet. 2008, 9, 387–402 (doi: 10.1146/annurev.genom.9.081307.164359).
R. Kanagal-Shamanna, “Emulsion PCR: Techniques and Applications,” Methods in Molecular Biology, Volume 1392: Clinical Applications of PCR;  R. Luthra, R. Singh, K. Patel, Eds.; Humana Press: New York, NY, 2016, pp. 33–42 (https://doi.org/10.1007/978-1-4939-3360-0_4).
Michael L. Metzker, “Sequencing technologies —the next generation,” Nature Reviews Genetics 2010, 11, 31–46 (doi: 10.1146/annurev.genom.9.081307.164359).

Does the "Emulsion Breaking" mean, that the DNA strands on the Beads are again broken that only the single strand DNA remains on the Beads?

No, it's double strand DNA, which you are targeting. Based on method of breakup, only  streptavidin-coupled beads or those with PI-adapter (linked to streptavidin by biotin link. Your targeted DNA would remain in the solution and ready to sequence after specific separation (or go to DNA library).

@vogs: Also, you must read those articles and get some understanding. And, it would have been very helpful if you had provided your protocol.

I'm trying. But somehow every explanation of the sequencing (let's take the Roche/454) starts with ssDNA on it's strands but the emPCR produces dsDNA on the Beads.

That's correct.

So where is the step between the dsDNA of the emPCR and the ssPCR used in the sequencing? And what happens in this step? Simply a denaturation again?

It is a broad subject. It can be pyrosequencing (see last paragraph of my posting) or  4-color sequencing or etc. All happen during beads in microreactor (see the last diagram) and when ssDNA attached to beads get complementary nucleotides in order to making ddDNA. It's a complex process. Please read the review articles.

I am inerested in ion torrent. I know that it is a wide field but you write yourself, that all mentioned methods rely on ssDNA on the Beads. But you always miss explaining how it gets ssDNA again after dsDNA were produced by emPCR

Before the sequencing starts, the DNA on the bead is denatured (the strand that is covalently boudn remains, the other is washed away). Polymerase and primer are added. Then, the four nucleotides are offered one by one, leading to extension or not. The pH change not only shows whether a nucleotide is attached, but also how many (in case of runs of a given nucleotide in the template strand. This picture illustrates the process:

Source: https://users.ugent.be/~avierstr/nextgen/nextgen.html

Thanks for the answer. This means, that the denaturation is done in the emPCR step (several times) as well as before the sequencing starts? I am always missing this part in the explanations. E.g. when looking  here on slide 16 the Bed is filled with full DNA strands but in slide 25 a single strand DNA is used for sequencing. But no one explains the step in between.

@vogs I guess so. In the classic PCR protocol, denaturation is critical because it supports the other two steps - primer annealing and primer elongation. Without denaturation (or melting), you would not get exponential amplification. If you like the answer, you can upvote it or accept it.


text:
Why does hydrogen burn spontaneously in fuel cell at room temperature?
It's well known that if you mix $\ce{H2}$ and $\ce{O2}$ in ambient conditions, the $\ce{H2}$ will be stable indefinitely, and will only react with $\ce{O2}$ to form water when ignited by a flame. The reason being that for the reaction
$$\ce{2H2 + O2 -> 2H2O} \label{rxn:1}\tag{1}$$
to occur, first the bonds must be broken in $\ce{H2}$ and in $\ce{O2},$ which is an insurmountable activation energy at room temperature.
So I am confused why a hydrogen fuel cell works at room temperature without ignition, as it is powered by the same chemical reaction \eqref{rxn:1}.
In particular, why can the strong double bond in $\ce{O2}$ suddenly be broken in this context without a flame?
My best guess is that somehow the electron sea in the copper wire acts as a catalyst. If this is true, what are the details of this mechanism that allows the $\ce{O2}$ bond to break at room temperature?

My guess would be that the reaction you've stated isn't the actual reactions occurring but the overall reaction of the two half equations that take place in an alkaline hydrogen-oxygen fuel cell. $$\ce{H2 + 2OH- ->2H2O + 2e- }$$ $$\ce{O2 + 2H2O + 4e- ->4OH-}$$

You guessed it right that hydrogen and oxygen mixtures will remain stable. Someone in Harvard waited for >30 years and found very little water if any. As you said, there is a energy barrier. What type of materials can lower this barrier? Catalysts.
The electrode materials require some catalysts such as platinum or nickel. Of course, making cheaper and sustainable catalysts is an extensive area of research. If a person discovers a better catalyst which is far more common than Pt, then that person certainly deserves a big award. Copper does not act as a catalyst here.

Thanks! I was missing the connection to Platinum. After reading more about this, I feel like I am still missing the magic of how metals are able to break the O2 bond on their surface. Posting in a new question here: https://chemistry.stackexchange.com/questions/116602/how-do-metals-split-the-oxygen-double-bond-during-corrosion

Search the key word adsorption of hydrogen on platinum. People must be studying this phenomenon for ages.


text:
Cause of uniform glow in cathode ray tubes
I got the following extract from a chemistry book (emphasis mine):

It is observed that current does not flow through the gas at ordinary pressure even at high voltage of 5000 volts. When the pressure inside the tube is reduced and a high voltage of 5000–10000 volts is applied, then an electric discharge takes place through the gas producing a uniform glow inside the tube. When the pressure is reduced further to about 0.01 torr, the original glow disappears. Some rays are produced which create fluorescence on the glass wall opposite to the cathode.

This excerpt is about cathode rays discovery in a chapter on atomic structure. Here, the author is talking about a uniform glow appearing when the pressure was reduced. Then he says that the rays (which he then refers to as cathode rays) appear after the disappearance of the glow. I have never noticed this glow in any experiment. Does anybody know what this is?

To close voters, I'm not sure what is unclear about the question. The OP points to a specific phenomena reference in a book (which should ideally have a citation @MShehzad) and it had a relatively straightforward answer.

I believe the quote is from the Chemistry 11 (p 119), available via the Internet Archive: https://archive.org/details/ChemistryPart1/page/n119 (cc @Tyberius)

is this question from PTB_Fsc_chemistry ?

@user106509 Yes.

The uniform glow is due to ionization and recombination of the residual gas; it's called a glow discharge. At higher pressure, a spark or arc discharge occurs at much higher current density. Fluorescent lamps and neon lamps operate in the glow-discharge region, and high-pressure xenon lamps use an arc discharge.
Elements of the residual gas can be identified by the color of the glow discharge. Nitrogen and argon in air, for example, glow purplish blue.
As pressure decreases further, density is too low (and path too short) to produce a visible glow discharge, but electrons, AKA "cathode rays", hurled off the electrodes impact the walls of the container, which may glow green in borosilicate glass, and in very low-pressure vacuum tubes, such as the 1G3GT high-voltage rectifier, the electrons impacting the anode produce X-rays (high energy photons), which may also create fluorescence in the glass shell.
The difference is easy to see in this video, which demonstrates the change in appearance from an arc discharge to glow discharge, and finally to the fluorescence of the glass envelope due to X-rays or electron bombardment. The arc starts at ~20 seconds, the glow discharge at ~26 seconds, striations form ~36 seconds, and by ~46 seconds, the glass envelope glows green from electron bombardment (though it might rather be the green of residual oxygen). This was one of my favorite demonstrations to show students the basis for spectroscopy!


text:
How do metals split the oxygen double bond during corrosion?
For a metal like $\ce{Mg}$ to corrode into $\ce{MgO}$, the double bond in $\ce{O2}$ must break. Since the dissociation energy of $\ce{O2}$ is 500 kJ/mol, using estimates from the Boltzmann distribution, it looks like there would not be any $\ce{O2}$ molecules in the entire atmosphere  with enough energy to break this bond at room temperature:
$$\text{Fraction of $\ce{O2}$ with at least $\pu{500 kJ}$ energy} \approx \mathrm e^\left(\dfrac{-500\,000\ \mathrm{J/mol}}{298\ \mathrm K(8.3145\ \mathrm{J/(mol\ K)})}\right) \lt 10^{-261}.$$
But my understanding is that $\ce{MgO}$ does form at room temperature. In fact, it seems that even stable $\ce{Pt}$ can break the $\ce{O2}$ bond at room temperature (for example in the context of catalyzing the ignition of hydrogen).
It seems that something like this must be happening: an $\ce{O2}$ molecule collides with $\ce{Mg}$ metal and gets stuck onto the surface, temporarily forming $\ce{MgO2}$. And once this bond is formed, the remaining bond between the two oxygen atoms suddenly becomes easy to break, and one of the oxygen atoms falls off and bonds with a different $\ce{Mg}$ atom. Is this the case?
I'd like to understand this mechanism better. How weak does the $\ce{O2}$ bond become after latching onto magnesium, and why does it become weaker? Why are metals able to split the $\ce{O2}$ bond this way, compared to other materials like hydrocarbons?

Pt is not breaking the oxygen bonds. Pt only interacts by adsorbing hydrogen. I am sure the mechanism has been very very well studies. Use Google Scholar this time. Pt is just helping by lowering the kinetic barrier of the reaction.

Humidity is the catalyst. You need water, then it happens electrochemically, see DOI: 10.1039/QR9672100029 'The Mechanism of Rusting'. Or more specifically: https://doi.org/10.1002/(SICI)1527-2648(199909)1:1%3C11::AID-ADEM11%3E3.0.CO;2-N 'Corrosion Mechanisms of Magnesium Alloys'

The corrosion of metals like iron is essentially an electrochemical process.
In corrosion, a metal is oxidised by loss of electrons to oxygen and formation of oxides. Corrosion of $\ce{Fe}$ (commonly known as rusting) occurs in presence of water and air. The chemistry of corrosion is quite complex, but it may be considered essentially as an electrochemical phenomenon. At a particular spot of an object made of iron, oxidation takes place and that spot behaves as anode and we can write the reaction.
\begin{align}
\tag{Anode}
  \ce{2Fe (s) &-> 2 Fe^{2+} + 4 e-} &
  E^\circ_{\ce{Fe^{2+}/Fe}} &= \pu{– 0.44 V}
\end{align}
Electrons released at the anodic spot move through the metal and go to another spot on the metal and reduce oxygen in presence of $\ce{H+}$ (which is believed to be available from $\ce{H2CO3}$ formed due to dissolution of carbon dioxide from air into water. Hydrogen ions in water may also be available due to dissolution of other acidic oxides from the atmosphere). This spot behaves as cathode with the reaction
\begin{align}
\tag{Cathode}
  \ce{O2(g) + 4 H+(aq) + 4 e- &-> 2 H2O (l)} &
  E^\circ_{\ce{H^+/O2/H2O}} &= \pu{1.23 V}
\end{align}
The overall reaction being:
\begin{align}
  \ce{2Fe(s) + O2(g) + 4H+(aq) &-> 2Fe^{+2}(aq) + 2 H2O (l)} &
  E^\circ_{\text{cell}} &= \pu{1.67 V}
\end{align}
The ferrous ions are further oxidised by atmospheric oxygen to ferric ions which come out as rust in the form of hydrated ferric oxide $\ce{(Fe2O3. x H2O)}$ and with further production of hydrogen ions.

I think more important is the reaction $\ce{O2 + 2 H2O + 4 e- -> 4 OH-}$ and consequently the formation of iron hydroxides, which may age to iron oxides.

MgO2 is easy to visualize, but is probably not an intermediate in the oxidation of magnesium. In fact, magnesium is not well described for this situation as Mg, but rather as (Mg)n, because we are looking at bulk material, with different surfaces and facets and irregularities on the surface.
If the corrosion continues in the presence of water, then an electrochemical explanation is appropriate.
But magnesium does not continue to corrode in dry O2. A surface oxide forms which is quite inert; further oxidation is stifled. Aluminum is similar, but continues to oxidize slower and slower, building up a thick oxide coat. Stainless steel oxidizes to a very passive state, with a very thin - very thin! - oxide layer. Zinc is similar to aluminum.
Atoms which are adsorbed on the surface of metals may be bonded securely, as in the case of stainless steel, or less so, and in the case of metals which continue to oxidize, the adsorbed atoms are evidently moving around.
You could make the adsorbed atoms move around faster by heating the system. Magnesium will ignite in hot O2. Many metals will. If you get a magnesium fire going, it will even burn in nitrogen (https://video.search.yahoo.com/yhs/search?fr=yhs-Lkry-SF01&hsimp=yhs-SF01&hspart=Lkry&p=magnesium+fire+nitrogen#id=2&vid=eaf040c65f5c489889407fb36f263ece&action=click).
Combustion is a free radical process; perhaps oxidation of metals could be viewed in the same light. Things that burn develop into pieces of a larger molecule, but with a free electron. Oxygen is electronegative; it likes to grab electrons, even if only one at a time. Then the oxygen becomes part of a chain --O-O-, and the O2 bond is weakened. On a magnesium surface, this could lead to a splitting of the O2 bond and formation of MgO, except that this "MgO" molecule is still part of the bulk metal, until you scrape it off.
So corrosion of metals in dry air is not electrochemical, but could be some type of chain reaction which has a fairly high, but not super-high activation energy. Older cars had lots of iron in the engine compartment which corroded by the high temperatures there; parts of grills used for cooking may similarly corrode at high temperatures.


text:
How do fuel cells work?
I have been reading an introductory book on fuel cells, and it's indicated that SOFC need to be brought to $\pu{600\! -\! 1000 ^\circ C}$ to be more efficient.
So, my understanding is that if I want to have a working fuel cell, I would need to come up with a system that could "heat up" my fuel cell to $\pu{750 ^\circ C}$?
Most of the articles I have been reading explain how fuels cells work, but what would be the actual set-up? I would have to put the fuel cell in some sort of oven?
Thanks for your input!

There is no physical or chemical law that says all fuel cells work better at 600-100°C. For SOFC's that is generally true, but the actual optimal temperature is different for every model.

The actual setup is a well insulated fuel cell, a heat exchanger for fuel and exhaust gas/water, and an additional electrical heater for startup. Or so i would guess.


text:
Mechanism of Acid Hydrolysis
I wanted to ask a question about acid catalysed hydrolysis.
I learnt in an introductory course to Physical Organic Chemistry that there are different ways that an ester can be hydrolysed acidically:

$\ce{A_{AC}1}$
$\ce{A_{AC}2}$
$\ce{A_{AL}1}$
$\ce{A_{AL}2}$

There are different ways that esters can be hydrolysed:
– catalysis by acids (A) or bases (B)
– cleavage of acyl-oxy (AC) or alkyl-oxy (AL) bonds
– the molecularity of the key step (1 or 2).
Using Acids (A) there are 4 different ways to hydrolyse an ester using acid catalysis:

$\ce{A_{AC}1}$ Cleavage of Acyl-Oxy Bond Unimolecular
$\ce{A_{AC}2}$ Cleavage of Acyl-Oxy Bond Dimolecular
$\ce{A_{AL}1}$ Cleavage of Alkyl-Oxy Bond Unimolecular
$\ce{A_{AL}2}$ Cleavage of Alkyl-Oxy Bond Dimolecular

An explanation can be found here.
But consider this question, with the marking scheme given below:

Our professor did not mention anything regarding how to choose which type of hydrolysis takes place. From this question, he states and goes further onto mention that either $\ce{A_{AC}1}$ or $\ce{A_{AC}2}$ can take place, but no other.
I'm looking at these structures and thinking
Why can't $\ce{A_{AL}1}$ or $\ce{A_{AL}2}$ take place for these two? Why must it be the two given in them marking scheme?

Perhaps you could explain in a few words what those mechanisms mean, or provide a reference. I think that's not too widely known.

@Karl Edit made. Reference added.

+1 , but I wouldn't add this as a P.S. People have a short attention span, they read five lines, find they don't get it, and move on. ;-)

@Karl edited it further so now it appears in the introduction rather than the end!

You should have read your explanation material more carefully! It clearly states that $A_{AL}2$ has not even been observed in acid hydrolysis due to water being a poor nucleophile in an $S_N2$ process. As for $A_{AL}1$, that only happens when the esterified radical can leave as a stable carbocation which, in this case methyl is a very very unstable carbocation.

Well if I am such a fool then I will learn today! Thanks very much this isn't actually my material but I read it so fast I couldn't actually make the point clear to myself! My apologies!


text:
4-Bromophenol vs 4-aminophenol pKa in water comparison confusion
Recently the other week, I was asked the following question:

Would you expect 4-aminophenol to have a higher or lower pKa than 4-
  bromophenol when measured in water?

And I thought about this on the way home, and decided to try and justify this using resonance.

My feeling was that Nitrogen couldn't handle 4 bonds and an extra lone pair on its atom, so the negative charge won't leave the benzene ring, whereas the Bromine could handle the double bond (I might have been mistaken with the negative charge on the Bromine - if so, do let me know please.)
Using this, I thought that this is the reason why 4-aminophenol would have a higher pKa than 4-bromophenol as its resonance form is not as extended as Bromine's, which is at a lower energy level to be able to handle the extra negative charge, plus all the carbons in the ring have 4 bonds as normal.
Is this a correct answer and justification?

Forget about octet expansion, it's a myth. The answer is simpler than that, anyway. NH2 donates electrons to the ring, whereas Br withdraws electrons. The phenoxide conjugate base is stabilised by donating electrons to the ring, so it's much easier if the other substituent helps it out by taking electrons away (Br), and it's much harder if the other substituent fights back instead (NH2).

Answer to this question is, as orthocresol pointed out, simpler than what you think. As you know, $\ce{-Br}$ is deactivate the ring (even though it is o,p-directing) while $\ce{-NH2}$ is great ring activator. What that means is regardless of the mesomeric effect, compared to $\ce{-H}$, $\ce{-Br}$ is overall electron withdrawing and $\ce{-NH2}$ is electron donating (remember, $\ce{-NH2}$ is a Lewis base).
Said that, I'd say basically, $\mathrm{pH}$ of a phenol ($\mathrm{p}K_\mathrm{a} = 9.95$) is depended on how stable phenolate ion is after it donated $\ce{H+}$. Electron withdrawing groups stabilize phenolate ion (e.g., $\ce{-NO2}$ group, p-nitrophenol: $\mathrm{p}K_\mathrm{a} = 7.1$) while electron donating groups destabilize it (e.g., $\ce{-OCH3}$ group, p-methoxyphenol: $\mathrm{p}K_\mathrm{a} = 10.2$). Thus, one can expect that p-bromophenol should have a lower $\mathrm{p}K_\mathrm{a}$ than that of phenol (meaning more acidic than phenol) while  p-aminophenol should have a higher $\mathrm{p}K_\mathrm{a}$ than that of phenol (meaning more basic than phenol). This is true in reality. Actual $\mathrm{p}K_\mathrm{a}$ values of p-bromophenol  and  p-aminophenol are $9.34$ and $10.3$, respectively ($9.34 \lt 9.95 \lt 10.3$).
Sources for $\mathrm{p}K_\mathrm{a}$ values: Evans pKa Table and Wikipedia.

In this situation, considering the distance between the $\ce{-OH}$ group and the substituents we can ignore inductive effects and focus on the mesomeric ones. $\ce{-Br}$ has a $+M_s$ effect, same as the $\ce{-NH2}$ group. To make an $\ce{-OH}$ group more acidic we need electron withdrawing groups as the slight positive charge built on the oxygen atom will facilitate the proton leaving. In order to find the more acidic molecule we look for the one with the most electron withdrawing substituent(or the least electron donating in this case). Consider the strength of the substituents' conjugation with the ring. Bromine is a huge atom, this difference in atomic radius compared to carbon gives their orbitals a hard time overlapping(I'm simplifying here for convenience's sake) and consequently a weaker $+M_s$ effect than $\ce{-NH2}$. Considering that, we can argue that 4-bromophenol has a lower pKa(is more acidic) than 4-aminophenol. And indeed, a quick google search validates my answer, the pKa of 4-bromophenol is 9.17 and the pKa of 4-aminophenol is 10.30 for the $\ce{-OH}$ group.
Sources:
https://pubchem.ncbi.nlm.nih.gov/compound/4-bromophenol#section=Dissociation-Constants
https://en.wikipedia.org/wiki/4-Aminophenol

[sorry for posting this as an answer, but it isn't possible to include an image when posted as a comment]
Do we also need to consider hydrogen bonding effects in the conjugate base, since the measurement is in water? Presumably forming the phenolate would increase the basicity of the amine through the inductive effect, and this could be stabilized by hydrogen bonding to solvent. I have no idea the strength of this effect, but thought it worth considering.

"it isn't possible to include an image when posted as a comment": this isn't true. One can upload an image via the uploader within an answer field, copy obtained URL and attach inline link to the image using [<image name>](<URL>) syntax (discard the answer afterwards).

Or you can ask a new question by clicking Ask Question. Include a link to this question if it helps to provide context.

My bad, didn’t realise


text:
The effect of nitric acid on the rate of corrosion of copper
I conducted an experiment concerning the rate of corrosion of copper when submerged in three solutions. In the first solution, 5ml concentrated nitric acid and 10 ml distilled water was added. In the second solution, 5ml concentrated nitric acid, 7.5 ml distilled water and 2.5 ml 5 molar NaCl solution was added. In the last solution, 5 ml nitric acid, 5 ml distilled water and 5 ml NaCl solution was added. I expected the rate of corrosion to be the fastest when submerged into a solution where more NaCl solution is evident, but this was not the case. My results were quite the opposite. The rate of corrosion was the fastest in the '5ml concentrated nitric acid and 10 ml distilled water' solution, which confused me. I repeated the experiment multiple times, but the results came out the same.(I used colorimetry to determine the rate of corrosion)
I proceeded to take the temperature of the three solutions and that's when I discovered that temperature played a huge role in the rate of corrosion. The solution with an addition of only distilled water had the highest initial temperature, while the solution with the addition of only the NaCl solution had the lowest initial temperature.
However, I don't quite understand why my results are the way they are. Isn't salt supposed to accelerate the rate of corrosion?
(the color of the distilled water solution after corrosion was blue, while the color of the NaCl solution was green).

What was your reasoning into thinking salt would speed up the corrosion? HNO3 is the one attacking the copper. When you add salt, what I believe is happening is that part of the nitric acid is reacting with NaCl, thus not all of it attacks the copper

How did you discover that temperature played a huge role in the rate of oxidation? Did you measure the rate of oxidation while keeping the solution composition constant and varying the temperature?Did you then perform the original experiment again, being careful to keep all solutions at the same temperature?

Another contributing factor may be that increasing the ionic strength decreases activity of ions.

Please correct the title because no distilled water is at play.


text:
Why does steam distillation work?
I'm aware that there is a similar question here but it didn't really answer my specific questions.
I understand that steam distillation is used to separate an insoluble liquid from an aqueous solution, due to passing steam through the reaction mixture so they both evaporate etc...
I don't however understand what is so significant about this. If we started with a mixture of, let's say, 50% lavender oil and 50% water, wouldn't we also end up with a mixture of lavender oil and water, if both are evaporating below their boiling point? How is that useful - aren't we just ending up with what we started with? Or is it that the distillate will contain a greater proportion of oil? eg 60% to 40% - but then I don't see how that works, as shouldn't water should be more easily evaporated than the oil?
Furthermore, surely the two immiscible liquids form two seperate layers that can be put into a separating funnel and poured into separate containers?

Separating water from an organic compound is not the purpose of steam distillation. See the Wikipedia entry for "steam distillation" for a good background.

Possible duplicate of Steam distillation- how it actually works?

I think that the main purpose (/ historical advantage) of steam distillation was to avoid the exposure of sensitive compounds to sustained heat by either lowering the required distillation temperature, or the distillation duration (at equal temperature), or both, by using steam as a carrier.
The solution that contains the compound to be distillated doesn't necessarily initially contain water itself, though even if it does, there is a significant advantage in that the duration of exposure to high temperatures is as short as the transport time until it is condensed (since the water is boiled separately, and not together with the compound). In traditional distillation the compound has to be exposed to heat for the time necessary to reach the solvant's  evaporation point AND the transport time, which is significantly longer than the transport time alone.


text:
"Hairy" corrosion on anodized alu-part in water
I got a water tank with a mechanical assembly in.
The assembly consist of parts of anodized aluminium mounted on stainless steel plates (AISI 304). This is submerged down in fresh water. The tank is open to air, and there is very little movement/current in the water.
After a few weeks there has developed some "fuzzy" dots on the surface of the anodized aluminium parts. See image.
Question is what this is?
My first thought was that this is organic material from the water that fastened easier to the rougher alu surface than the steel surface?
However, now I think it is more a chemical reaction due to galvanic currents due to material difference in aluminium and stainless steel. This should dissolve the aluminium and transfer the electrones to the stainless steel plate. But would this make it look like the fuzzy growth on the alu-part?
Or could it be something totally different I haven't though off?
One thing I wonder about in this connection is that the fuzz is not more concentrated close to where it is connected to the AISI 304 plate?
! Note: The alupart is anodized, which should make it electrical isolated on all surfaces. However, the Alu-part and the steel plate is connected with a few screws, and through this has electrical connection.

Caption: Photo of the "fuzzy" anodized alu part ("Eloksert" means anodized)

Why did you reject the "organic material" hypothesis? That fuzz sure looks like bacterial growth (no water movement). Can you add a biocide? O3?

I thought it be ut of the question since it only adhered to the aluminium and not any other parts in the pool. (Stainless steel and POM plastic and a rubber band is also in the pool).

Although there is definitely a biofilm all over. I guess I could add a biocide, but not sure if ozone would be the best. Pool is for testing equipment, and the water should be as "dead" as possible.


text:
Silver (111) surface structure, and is bulk structure body-center or face-center cubic?
I'd like to understand the atomic arrangement is on a silver (111) surface.
Step 1 would be to find out the bulk crystal structure.
Wikipedia lists silver as face-center cubic (fcc) here and here and the page Silver: crystal structures also shows fcc and links to Lin-gun Liu and W. A. Bassett, J. Appl. Phys., 1973, 44, 1475:

But the page Ag(111) and Ag/Si(111): similarities and differences begins

The crystal structure of silver is body-centred cubic (Figure 4.4a), and so the Ag(111) plane exhibits a close-packed, hexagonal structure (Figure 4.4b). Ag is a stable transition metal, and one of the few metals found abundantly in nature as a pure native element. This stability is due to its relative inertness, and the Ag(111) close-packed surface has the lowest energy of its high-symmetry surfaces.

While this page tells me that the silver (111) surface will be a close-packed hexagonal arrangement, I am uncertain I can trust this because it begins by stating that silver is bcc and Wikipedia tells me silver is fcc.

Figure 4.4: (a) Schematic model of the silver body-centred cubic structure, showing the (111) cleavage plane. (b) Model of the close-packed Ag(111) surface, with the principal directions indicated.

Question(s):

Is bulk silver bcc or fcc?
What is the structure of the silver (111) surface?

companion question: Lattice matching of graphene on silver (111) crystal surface, what is the meaning of the √3 x √3 lattice?

Silver is most definitely fcc (face centered cubic). Now, what the surface looks like as a clean surface is down to the surface reconstruction (which your companion question is getting at).

It's been two years, it seems I can finally answer my own question.

Is bulk silver bcc or fcc?

Per Wikipedia's Silver and @JonCuster's comment its structure is face-centered cubic.
 Source

What is the structure of the silver (111) surface?

This has two parts;

What does a mathematical fcc lattice look like if you slice it along a (111) plane?
What happens to atoms on a real Ag(111) surface; do they stay put or do they move around into an energy minimizing surface reconstruction the way Au(111) and Si(111) famously do?

What does a mathematical fcc lattice look like if you slice it along a (111) plane?

If you stand an fcc structure up on one corner and look from the side, then you see the atoms fall on three planes, and the arrangement in each plane is hexagonal with each atom touched by six coplanar atoms. Each plane is offset so the atoms of one fit into the voids in the layers above and below.
The fcc(111) planes are arranged in an ABCABC fashion. From one corner (in the "A" plane) the three adjacent corners and sides make the next "B" plane. The opposite corner is also "A" and the three adjacent corners and sides can be called "C".

above left: same image as abo e but with "A", "B", and "C" circled in red, green and blue. right fcc structure tipped on one corner showing ABCA planes, from https://www.chemtube3d.com/ccp-cubic-close-packing/ click for larger

above: Screen shots from slide 8 of 5/22/2013 L. Viciu| ACII| close packed structures click for larger

What happens to atoms on a real Ag(111) surface; do they stay put or do they move around into an energy minimizing surface reconstruction?

While Au(111) reconstructs into a very elegant $23 \times \sqrt{3}$ super cell that can have three possible rotations and these join up into a herringbone pattern, the Ag(111) surface remains flat, almost exactly like it was a (111) plane in the bulk. There is a very slight readjustment of the atomic locations for energy minimization, the inter-plane spacing slightly expand, but there is no reconstruction.
This flat hexagonal metal surface can then be used as a "workbench" or template on which to grow a wide variety of materials.

Atomic and Molecular Adsorption on Ag(111)
Exploring Ag(111) Substrate for Epitaxially Growing Monolayer Stanene: A First-Principles Study

See also this answer to Why does FCC crystalline forms form hexagonal crystalline structures during CVD?


text:
How do I calculate the proportion of respective conjugates of one base mixed with several acids (and/or conversely) at different temperatures?
How do I calculate the proportion of respective conjugates of one base mixed with several acids (and/or conversely) at different temperatures, and is there a free software tool that would allow me to interact with the computations therefor (e.g. search the temperature or portion that maximises one conjugate over another)?

We have a policy which states that you should show your thoughts, effort and attempts to understand underlaying principles. It'll make us certain that we aren't doing your homework for you.
Basically any question with the wording your question has is considered homework; it needn't be literally one. Self-study questions, puzzles, worked examples etc. also count as homework.
Otherwise, such a question may get closed. Please [edit] in your full reasoning or thoughts on this.
See Homework

@Poutnik Hello! Thanks for explaining this. This is not a homework question. I genuinely have no idea where to start to answer the functional need I have to be able to calculate this and dynamically interact with the computation. I run a google and Quora search to no avail, which leads me to asking the question here.

I'm not looking for a specific result but for (1) the laws / computations necessary to produce the result, and for (2) which free software would allow me to both interact with such computations in an advanced way and to have a library of molecular characteristics as input.

If you want a general solution for complex mixtures, where simplified formulas do not apply, you may need to involve numerical mathematics.
As solving the equation set may lead to polynomial equation of higher order you may like.
Enumerate equations of the known.
First, charge balance.
Then, mass balances.
Finally, dissociation equilibriums.
Involve a function of [H+] with a minimum at the value being searched for.
About temperature, there is problem the  constants are rarely provided for other but the standard temperature $\pu{25 ^{\circ}C}$

Is there a software that could dynamically factor in constants vs temperature?

Rather, is there software with available thermodynamic data of hydrated species like Gibbs energy,enthalpy and entropy ?

Though I cannot say for sure, yes I suppose this is what I am ignorantly looking for (free if possible). Do you know of such a software?


text:
Titration of aspirin
I have performed an experiment to test how temperature affects how fast aspirin dissolves by titrating solutions at various time intervals. According to this website, aspirin reacts with water as such:
$$\ce{C9H8O4(aq) + H2O(aq) -> C9H7O4-(aq) + H3O+(l)}$$
However, I've also read that aspirin undergoes hydrolysis with water, according to this site, where Aspirin (acetylsalicylic acid) hydrolyses to produce 2-hydroxybenzoic
acid and ethanoic acid.
So, I'm confused — when I'm dissolving the aspirin tablets in water, am I just making acetylsalicylic ions and hydronium ions which I can then titrate to find the concentration, or am I making two other acids as per the second site suggests? I have all my titre values, I just need to calculate the concentration of the aspirin the solutions, and I'm now confused about what chemical I'm actually finding the concentration of, e.g.
$$\ce{NaOH + HCl -> NaCl + H2O}$$
Here, sodium hydroxide reacts with hydrochloric acid. In my experiment, does the $\ce{NaOH}$ react with acetylsalicylic acid, or with 2-hydroxybenzoic acid and ethanoic acid?

I am afraid, direct acid-base titration is not the right way to analyze aspirin exactly because of the you stated - hydrolysis. The rule number no of any titration is that there should be no side reaction and it should go to completion almost instantly. The direct titration of aspirin is problematic because hydrolyzes pretty fast to salicylic acid- an unwanted side reaction which may or may not go to completion.
In order to circumvent this, analytical chemists do a back-titration. You add an excess of base, heat the mixture so that neutralization as well as hydrolysis are complete. The remaining base is titrated. The rest of the calculations are normally done.

Thanks for your response. This is a time-critical project that I can't repeat, but I can mention flaws and improvements. Would the results of the experiment be completely invalidated? I calculated the concentrations using a 1:1 ratio and the trend is still obvious - the heated aspirin mixture was ultimately much more acidic than the cooled mixture after stirring for the same amount of time. Do you think most of the acid I titrated would be aqueous aspirin, or would it be salicylic acid?Your help is much appreciated.

I am afraid your results will be overestimated because more base is consumed (neutralization +hydrolysis). If you heated, it would be mostly salicylic acid.


text:
Understanding lattice matching of graphene on silver (111) crystal surface and meaning of the √3 x √3 lattice?
Figure 1 in Ivor Lončarić and Vito Despoja, Phys. Rev. B 90, 075414 Benchmarking van der Waals functionals with noncontact RPA calculations on graphene-Ag(111) (available in Researchgate) shows a schematic of the graphene structure on what I believe is a silver (111) surface.
The caption reads:

FIG. 1. (Color online) Geometry of the supercell used for our vdW DFT calculations. Upper part: 2 × 2 graphene lattice matched to √3 × √3 Ag(111) lattice. Lower part: Geometry of the supercell in perpendicular direction.

What exactly is a "√3 × √3 Ag(111) lattice" and how exactly does the square root of three come in to this?

companion question: Silver (111) surface structure, and is bulk structure body-center or face-center cubic?

The $\sqrt{3}$ is the distance shown as black lines between every other C atom (assuming bond length is one unit). Twice this is the equivalent length between the Ag atoms  shown as silver/grey. So I'm assuming this is what they mean by lattice matching?

@porphyrin I'll take a look based on that, thanks!

See http://ww2.sljus.lu.se/staff/anders/Compressed/Notation_Edvin.pdf for a general overview (googled 'silver 111 surface reconstruction').

@JonCuster I've been checking that out, it seems everything I need to know is in there; bottom left of slide #22 suggests Ag(111) has a vertical reconstruction but the lateral positions are same as truncated bulk, slide #16 and Masel's Principles of Adsorption and Reaction on Solid Surfaces helps me understand the notation for graphene's orientation, and slide #21's illustration of fcc vs hcp notation preempts my next graphene on Ag(111) question! Nice find, thank you!

Good news! Classic surface science seems to have faded away if late. Recall that most of the original surface reconstruction information was from techniques like LEED that gave relative symmetry and rotation, but we had to wait for AFM to see actual atom positions.

@porphyrin I'm sorry but I still don't understand. Is it possible for you to post an answer and include at least a crude drawing? I don't know what "distance shown as black lines between every other C atom" means, nor what "equivalent length between the Ag atoms" means on a fcc(111) surface. Can you just take my image and overlay a few arrows? Thanks!

Start at the left bottom most blue atom (crossed by 2 black lines), go north by 1 atom and north east  by 1 atom. The distance as the crow flies is $\sqrt{3}$ times bond length, i.e. length of black line connecting atoms. Now look at underlying Ag atoms; twice this length is equivalent length in this lattice so I guessed that this is connection between lattices and why one fits over the other. This could be rubbish, however, as I'm not an expert in surface science.

"Root 3" or "$\sqrt{3}$" or "$\sqrt{3} \times \sqrt{3}$" are slang for the Wood's notation (1, 2) $ \left( \sqrt{3} \times \sqrt{3}\right)R30$, a commonly occurring configuration of one hexagonal 2D lattice with respect another.
In this example the substrate is the (111) face of silver. Since it is face-centered cubic with lattice constant of about 4.08 Angstroms the (111) surface is hexagonal with a lattice constant 4.08 / $\sqrt{2}$ = 2.88 Angstroms.
The figure caption says "2 × 2 graphene lattice matched to √3 × √3 Ag(111) lattice."
The annotation in the OP's drawing copied below shows solid arrows for the Ag(111) lattice vectors the dashed arrows are rotated by 30 degrees and $\sqrt{3}$ longer, and they match the periodicity of a $2 \times 2$ graphene cell.
The lattice constant of free graphene is about 2.46 Angstroms and double that is 4.92. Compared to $\sqrt{3}$ times 2.88 = 4.99 Angstroms it's a fairly good lattice match.


text:
What does "lit." mean in boiling point or melting point specification?
For example, this Xanthene 99% specification from Sigma Aldrich states:
\begin{array} \\\hline
\text{bp}  & 310-312 ~^\circ\mathrm{C~ (lit.)} \\\hline
\text{mp}  & 101-102 ~^\circ\mathrm{C~ (lit.)} \\\hline
\end{array}
My guess is that it's from literature.

Yes, you are correct: “lit.” refers to the literature value.
Usually one would also expect the corresponding reference, but it looks like Merck doesn't attach any to the provided data directly in place.
From The ACS Style Guide, section Reporting Analytical Data [1, p. 274] (emphasis mine):

Melting and Boiling Points
mp 175.5 °C (lit.25 mp 175–176 °C)
mp 225 °C dec
bp 127 °C
Abbreviations: mp, melting point; bp, boiling point; lit., literature value; and dec, decomposition.
A full space is used between the number and the unit °C; the degree symbol is closed up to the C.
A superscript number after “lit.” denotes the number of the reference.

The CSE Style Guide also recognizes “lit.” as abbreviation for literature in Appendix 29.2 Common English words used in bibliographic description and their abbreviations [2, p. 645], further referring to ISO 832:1994.
References

The ACS Style Guide: Effective Communication of Scientific Information, 3rd ed.; Coghill, A. M., Garson, L. R., Eds.; American Chemical Society; Oxford University Press: Washington, DC; Oxford; New York, 2006. ISBN 978-0-8412-3999-9.
Scientific Style and Format: The CSE Manual for Authors, Editors, and Publishers, 8th ed.; Council of Science Editors, Ed.; Council of Science Editors; London: in cooperation with The University of Chicago Press: Chicago, 2014. ISBN 978-0-226-11649-5.


text:
Any method for testing Nitrate and Nitrite in biological fluids like aqueous humour?
If anyone knows the methods of detection of nitrate and nitrite in biological fluids such as aqueous humor, saliva of animals, please let me know. Any relevant literature for the same is available, kindly share.

Welcome to SE.Before we answer , the extant of your efforts to research the question is lacking.Edit your post to include these .

Did you search the term ion-chromatography?

There is 1982 publication describe nitrate, nitrite, and $\ce{[^{15}N]}$-nitrate in biological fluids. It's abstract describe the procedure briefly:

Abstract: A new automated system for the analysis of nitrate via reduction with a high-pressure cadmium column is described. Samples of urine, saliva, deproteinized plasma, gastric juice, and milk can be analyzed for nitrate, nitrite, or both with a lower limit of detection of $\pu{1.0 nmol} \; \ce{NO3−}$ or $\ce{NO2−/ml}$. The system allows quantitative reduction of nitrate and automatically eliminates interference from other compounds normally present in urine and other biological fluids. Analysis rate is 30 samples per hour, with preparation for most samples limited to simple dilution with distilled water. The application of gas chromatography/mass spectrometry for the analysis of $\ce{^{15}NO3−}$ in urine after derivatization to $\ce{^{15}NO2}$-benzene is also described.

Thousands of samples of human and rat urine, human saliva, and plasma have been analyzed for nitrate and nitrite using this system. Typical ranges of nitrate and nitrite levels in biological samples are shown in the Table.
$$
\begin{array}{ccc}
\hline
\text{Biological fluid} & \text{$\ce{NO3-}$ Concentration, $\pu{\mu M}$} & \text{$\ce{NO2-}$ Concentration, $\pu{\mu M}$}  \\ \hline
\text{Urine} & 250-2000 & \text{ND}^a \\
\text{Saliva} & 200-600 & 30-210 \\
\text{Plasma} & 15-60 & \text{ND}^a \\
\text{Gastric juice} & 50-85 & 0.4-60^b \\
\text{Milk} & 20-30 & \text{ND}^a \\\hline
\end{array}
$$
$^a$Not determined. $\; \; ^b$Nitrite concentrations at the high end of this range are from people with gastric pathologies.
In general, for methods of the detection and determination of nitrite and nitrate, read Ref.2  (a review) and 3. However, Ref.1 claimed that the method in Ref.2 did not work for biological fluids.

References:

Laura C. Green, David A. Wagner, Joseph Glogowski, Paul L. Skipper, John S. Wishnok, Steven R. Tannenbaum, “Analysis of nitrate, nitrite, and $\ce{[^{15}N]}$nitrate in biological fluids,” Analytical Biochemistry 1982, 126(1), 131–138 (https://doi.org/10.1016/0003-2697(82)90118-X).
Qiu-Hua Wang, Li-Ju Yu, Yang Liu, Lan Lin, Ri-gang Lu, Jian-ping Zhu, Lan He, Zhong-Lin Lu, “Methods for the detection and determination of nitrite and nitrate: A review,” Talanta 2017, 165, 709–720 (https://doi.org/10.1016/j.talanta.2016.12.044).
M. P. Stainton, “Simple, efficient reduction column for use in the automated determination of nitrate in water,” Analytical Chemistry 1974, 46(11), 1616–1616 (https://doi.org/10.1021/ac60347a051).


text:
What is the term for passing an eluate back through an adsorption column?
In my experiment I am passing a solution through an adsorption column, collecting the eluate, then passing the eluate back through the same column.
Is there a concise term which describes this process?

reapply the eluate to the column?

Why are you doing this?

@Karl The solution effectively is influent, and the adsorption column removes a pollutant from it. The eluate (i.e. effluent), when passed back through the same column, further decreases the pollutant's concentration. Therefore I'm trying to find a term to describe the process.

This process is similar to re-cycling chromatography or recycling HPLC. In a more strict sense, it does not do any good, at least theoretically, if you keep on recycling all the eluate and transfer it back to the column.
What makes more sense is that you collect fractions and choose certain fractions for recycling.

Indeed.  It is like to perform multiple chromatography columns. Differently phrased and at least changing the second column is not too unusual.


text:
What is the mechanism of ring contraction of 6-bromo-7-methoxy-2,3,4,7-tetrahydrooxepine from seven to five?
I have recently come across a paper from here.I have tried to workout the mechanism for the following reaction presented in the paper.

I was able to workout mechanism till acetal formation.It is a ring expansion from six to seven. However,the next step is a ring reduction from seven to five upon heating.
Can someone please explain what is actually happening during ring reduction, and why it happens?

It's in scheme 3 of the paper you cited! The 7-membered acetal undergoes ring-opening to the α,β-unsaturated oxocarbenium ion, then conjugate addition to give the 5-membered tetrahydrofuran. A bit of arrow pushing and addition of solvent gives the other THF product
Not sure the legality of posting screenshots of papers, so I redrew the scheme:

@ PCK I hit a pay wall and could not open this paper.

@ PCK it is obvious, after protonation ,C-O bond breaks,followed by 1,4 addition, just great.


text:
DCC coupling of N-hydroxysuccinimide and 2,2-dimethoxyacetic acid
I am forming N-hydroxysuccinimide ester (1) by DCC coupling with N-hydroxysuccinimide and 2,2-dimethoxyacetic acid using distilled THF as solvent.

What I see on $\ce{^1H}$-NMR is three major singlets:
$\delta~\pu{4.840 ppm}~(1\ce{H}),$ $\delta~\pu{3.432 ppm}~(7.26\ce{H}),$ $\delta ~\pu{2.731 ppm}~(6.36\ce{H})$ and some impurities.
While the chemical shifts are well-aligned with theoretical predictions (in my opinion), integration ratios do not quite make sense. I wish I could purify it to have a clearer NMR spectrum and see whether I actually have the desired product, but I don't think my ester will stay stable in silica in column chromatography. Any suggestions?

Maybe take or get an NMR of the reagents and theoretical side products and see if there can be any overlap with the signals you see. I'm not sure I understand what your doubt is, though. NMR integration is not always very precise, especially for acidic protons.

What is your next reaction with this product?

@user6376297 I have already taken the NMR of 2,2-Dimethoxy Acetic acid:
$\delta : \pu{4.867 ppm}$ (1H) and $\delta : \pu{3.4615 ppm}$ (6.6H)
(c.f. $\delta : \pu{4.840 ppm}$ (1H) and $\delta : \pu{3.432 ppm}$ (7.26H) in my product NMR)
Not sure whether I can argue that peaks have 'shifted' in the product NMR

If peaks for "some impurities" are appear around $\delta \: \pu{1.8 ppm}$, I assume you have residual THF in your crude product. The other multiplet of THF is probably burried under $\delta \: \pu{3.43 ppm}$ resonance. That would confirmed by your impurity integration, which should be equal to 1H unit equivalence. If I'm right, your 6.36H unit equivalence is actually overlap one of $\ce{CH3}$-resonance and one of $\ce{CH2}$-resonance.
On the other hand, the two resonances at $\delta \: \pu{3.43 ppm}$ and $\delta \: \pu{2.73 ppm}$ would very well be representing two $\ce{CH3}$-groups and (two $\ce{CH2}$-groups + part of THF impurity), respectively, as well.

Thanks. but I have thoroughly vacuum dried my product and there is no peak at δ1.8 ppm.