text:
Horizontal symmetry plane of cis-dichloroosmiumtetracarbonyl
Consider the molecule $\ce{Os(CO)4Cl2}$ with a drawing given below via Chemtube3d:

I thought in this molecule, if I drew a horizontal plane, then a $σ_\mathrm{h}$ would be present i.e. A ‘horizontal’ mirror plane is perpendicular to the principle axes seeing as the mirror would, in my view, cut the atoms in half so the top would be the reflection of the bottom.
However, my professor said in a lecture that this molecule shown here does not have $σ_\mathrm{h}$ and I'm slightly puzzled as to why.
Why does this molecule not have $σ_\mathrm{h}$ plane?

Your compound has only one proper axis of rotation and it is $C_2$ as shown in the diagram. Therefore, it is the principle axis of rotation. By definition, $\sigma_\mathrm{h}$ is a plane of symmetry perpendicular to the principle axis of rotation. However, the plain you were talking about is paralleled to the principle axis of rotation. Thus, it is not a $\sigma_\mathrm{h}$ plane. Nonetheless, it is a $\sigma_\mathrm{v}$ plane, by definition.


text:
Water colour changes to yellow after electrolysis
A salesman recently conducted electrolysis experiment on 4 samples of water-- river water,tap water, his own water and water from Aquaguard RO filter. The colour change after electrolysis on different samples of water is given in the pic given below? Is tap water ,river water and Aquaguard RO water can be consumed for drinking purposes?

This is pure scam! All he is doing is using an electrode which is corroding during electrolysis. The flocculated material you see in tap water is just a metallic hydroxide. If tap water had this much of metals, nobody will survive. Such a fraudulent salesman should be shooed away!

What can be the possible explanation why aquagyard RO water changes yellow in colour?

corroding electrodes. Most likely he used iron anodes with some traces of salt. He is using different metallic anodes, hence different colors.

He used the same electrodes in the whole process.

Was he adding anything? Most likely it is an iron electrode.

It suffices that the same inert el. is placed deeper to bring a metallic part in contact to the water... I agree with this, likely iron. Note the different colour in tap water, meaning tap water is chlorinated. Just I disagree with the fact that water from which things can be precipitate or flocculated must necessarily kill.

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text:
Will 1-[2-(chloromethyl)phenyl]propan-2-one, follow a Williamson ether synthesis route or, a Diels-Alder reaction?
I was asked to find the product of potassium ethoxide reaction with 1-[2-(chloromethyl)phenyl]propan-2-one (1).
I came up with two different paths, each leading to the same product.
1-[2-(chloromethyl)phenyl]propan-2-one (1) on reaction with potassium ethoxide gives a carbanion 2.
This carbanion 2, in conjugation with $\ce{C=O}$ gives enolate 3. This enolate can follow path 1, an $\mathrm{S_N}{2}$ reaction (Williamson ether synthesis) to give 4.
The same carbanion 2 in conjugation with benzene ring gives 6 via allylic rearrangement(Path 2).6 could undergo a possible Diels-Alder reaction giving again 4.
My question is, will the reaction follow Path 1, a Williamson ether synthesis route or will it follow Path 2, a Diels-Alder reaction.

Path 1 looks the more probable as path 2 breaks the aromaticity of  the system

@ Waylander , got it.However , is path 2 at all possible , a diels alder route.That would be a different route

Unrelated to the actual issue of which path is more likely, but I don’t think path 2 can be called a Diels-Alder reaction. It resembles more an 6-pi Electrocyclic Ring Closing reaction.

Wouldn't you get a different product (2-oxa~ instead of 1-oxa~) via Path 2? The O and a C seem to have changed places in the last step, between 6 and 4

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text:
What causes the blackening of light bulbs
in 1878, Thomas Edison (or rather William Joseph Hammer) added an anode in a light bulb in order to catch the flux of electrons coming out from the carbon filament and reduce the phenomenon of blackening of the glass bulb.
However, am I right in saying that this was the wrong course of action? I understand the blackening is due to the accumulation of carbon oxides (such as CO and CO2) produced by the reaction between carbon (and, more recently, tungsten) vapour and the remainder of air present into the glass bulb. Thus, capturing electrons did not solve the problem of blackening (and that is why Edison did not find any use for the extra electrode).

I am quite sure that the problem is the evaporation of tungsten (even tungsten, at those temperatures, has a non-negligible vapour pressure). The evaporation of tungsten is the (unwanted) process which lead to the development of halogen lamps.

Esprit de l'escalier: sublimation

And even before the blackening should have been carbon in some form rather than carbon (di)oxide, which are both colorless gases.

A 14 page article (https://www.jstage.jst.go.jp/article/jieij1917/6/4/6_4_398/_article) describes the blackening of incandescent light bulbs as being totally due to evaporation of the filament. Platinum had been used as far back as 1840, and Edison's first patent was for a carbon filament (1879), but by this time (1922), tungsten was the metal used in almost all filaments.
Blackening of carbon-filament bulbs has not been a significant problem, due to the short commercial lifetime of carbon filaments. Any carbon deposition would likely be from sputtering of the filament, not evaporation. Blackening is a problem with tungsten filaments, and not so much from evaporation of metal, but from evaporation of tungsten oxides formed from adventitious water or oxygen, followed by deposition on the cool bulb surface. Inert gas is used instead of vaccuum to impede the mobility of tungsten atoms. But then, tiny amounts of oxygen or water vapor in the bulb can cause severe blackening; getters such as zirconium are used to combine with oxygen. Large bulbs used for theater lighting often contained some tungsten powder; when the bulb darkened, the bulb was shaken to abrade off the tungsten and increase the brightness. https://en.wikipedia.org/wiki/Incandescent_light_bulb#Bulb_blackening
Halogen bulbs use iodine to react with the deposited tungsten, which vaporizes as the iodide and is redeposited on the hot filament, where it decomposes. Halogen bulbs are smaller than the usual incandescent bulb and they run hotter filaments to keep the glass hotter, so that iodine can react there with the deposited tungsten and carry it back to the filament.
Your description of Hammer adding an anode to catch electrons emitted by the hot filament is correct, but I suspect that the original thinking went something like this: "We know that electrons are emitted by the hot filament. As a result, the filament must become positive; thus some metal atom at the surface might be repelled, carrying a positive charge (which would be neutralized as it passes thru the electron cloud). This metal atom might deposit on the cool bulb, producing a darkening. Perhaps we can prevent this repulsion/evaporation by making the filament negative so as to retain the positive metal ions." Apparently, this did not work, but only increased the electron emission, which could then be described as a bug or a feature.  OK, let's call it a feature and name it after Edison.


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How would one carboxylate CBG into its acid form, CBGA?
I have a project that would require cannabigerolic acid (CBGA) as a synthase substrate, but the only available chemical I can find is cannabigerol (CBG). CBG is the product of CBGA through decarboxylation at about 105 °C. Is this reaction reversible? How can I carboxylate the CBG I have available into CBGA?

Looks like you need to exploit the reactivity of the bis-hydroxylated aromatic ring. Reimer-Tiemann variation with CCl4 perhaps? But not sure, the molecule looks a bit delicate.

The substrate is a resorcinol derivative: a strongly activated aromatic ring, which upon reaction with a base delocalizes its negative charge on the ring, making it reactive towards acids (Eg. CO2).
Here (entries 130-132) good conversion is claimed (85%) in a high pressure vessel with magnesium methoxide and dry ice, or a lower conversion (40%) is reported at ambient pressure with magnesium methyl carbonate


text:
Addition of aqueous KOH to 3-bromobut-1-ene
On addition of aq $\ce{KOH}$ to 3-bromobut-1-ene, substitution reaction should take place via $S_N1$ mechanism but in this an allylic carbocation forms which could rearrange.
In that case which should be the major product?
But-3-en-2-ol or but-3-en-1-ol, because in the first case the carbocation would be stable because of being secondary but the double bond would have less number of hyperconjugated structures and will be less stable whereas in the latter case, the carbocation would be less stable because it would be primary but the double bond would be much more stable because of 5 $\alpha$ hydrogen available

In my opinion But-3-en-2-ol will be be the major product if the reaction is favored by Sn1 (due to stable carbocation )
Nevertheless, the product in which the OH group is on the primary atom is minor.


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Compound of fluorine and tellurium
As an exercise I have to predict the formula of the compounds that fluorine and tellurium can build using group properties.
My idea was to use the octet rule and therefore the formula would be $\ce{TeF2}$ so both atoms have eight valence electrons.
However, the correct solution given by the exercise is $\ce{TeF4}$ and $\ce{TeF6},$ and I have literally no clue how to get these solutions. Can you give me some advice on how to retrieve these solutions?

That is actually a very sensible answer, unfortunately Mother Nature has conspired against you here. As to what is the reason for the answer it is difficult to say without knowing how much you know already - what compounds of fluorine and sulphur do you know?

@IanBush I'm new to chemistry so my knowledge is very limited...
I know $GeF_2$ and $BiFe_3$ as other fluorine compounds.
For sulfur I know $SO$, $SO_2$, $S_2O_2$ off the top of my head.

@IanBush I was going to write an answer about how it was due to the presence of d and f orbitals on elements of higher groups and how those could be involved in bonds even being higher in energy. Than I found out about diazomethane ($\ce{CH2N2}$) and it all went to space lol. Wikipedia article about Hypervalent Molecules might be an interesting read for the OP though.

@JohnyDow I believe Ian was hinting at that Tellurium could/should form the same structures with Fluorine as Sulfur (or Selenium).

From my research, the answer seems much more complicated than it could be conveyed in a simple answer. At first I thought it was due to the d and f orbitals being involved in bonding. But apparently this explanation (dated about 1950) was incorrect, since experiments showed the contribution of d and f orbitals is very low on those hypervalent molecules. It seems the more up to date explanation is the (3c-4e) bond model, which I'm still reading about.

I guess the most simple answer that can be given to the OP without further complicating things is that elements in lower rows of the periodic table present this feature of not complying to the octet rule by having more bonds than it would be predicted. This almost never happens with elements with valence shells of quantum number 2 (the previously cited diazomethane being the only exception I found to this).


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Coupling of silyl imine and aldehyde

LDA is a strong base and probably deprotonates somewhere which I'm not sure about. It I would to make a guess it would probably be beta to the imine. This is following by nucleophilic attack of the carboanion onto the use of the second reagent Pr-CHO to generate an alcohol. But I'm not sure the purpose of the 3rd reagent (oxalic acid).

This is a Peterson olefination, specifically this one is from the synthesis of Roseophilin published in J. Am. Chem. Soc. (2001) 123 8509. The LDA deprotonates next to the silicon as both it and the adjacent imine stabilise the anion.
The Wikipedia page gives a good summary of the mechanism and why an acidic work up is used in this case to give the required double bond geometry (Wikipedia).

(Image from www.chem.wisc.edu)

@Waylander has nailed it.
But, an interesting part of mechanism is that it proceeds via chair form of six membered transition state as shown here. Among two possible transition states, only one is favored. The other transition state, disfavored has isopropyl group (from LDA) and trimethylsilyl group in axial positions leading to 1,3 diaxial repulsions (shown in the figure below).

source
In this mechanism, the starting material is an ester. You could use starting material in the post instead.


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In an emission spectrum, the limit of convergence at higher frequency corresponds to the first ionization energy
This is one of the syllabus points in the IB syllabus 12.1. We are advised to use the formula for the energy to remove one electron:
$$E = hν$$
which apparently is the same as
$$E_2 - E_1$$
where $E_2$ is the energy of the convergence limit and $E_1$ is the energy of the energy level where you remove the electron from.
However, isn't $ν$ the frequency of the emitted radiation when the electron falls from $E_2$ to $E_1$ (so with energy same as the energy difference between $E_2$ and $E_1),$ and not the frequency with the energy $E_2,$ i.e. the convergence limit?
I think I've misunderstood something and I'd appreciate if someone could explain this to me. Please use simple language because I'm only in high school.

$\nu$ is but a letter of the Greek alphabet. Whatever you denote by it is up to you.

$E=h\nu$ defines the energy of any photon of frequency $\nu$. It would be better to write $\Delta E=h\nu$ as the energy of a photon emitted from one energy level to another ( preferably with subscripts such as $\Delta E_{1,2}$). The convergence limit does not have to come into it, it just depends how levels 1 and 2 are defined.

Thank you for your answer! Still, in this question, for instance: "The convergence limit for the sodium atom has a wavelength of
242nm. Calculate the first ionization energy of sodium from this
data", shouldn't we be given the wavelength/frequency of the radiation emitted when the valence electron (the eleventh electron) falls from n = infinity back to n = 3? What I mean is that the frequency of the convergence limit and the radiation emitted from n = infinity to n = 3 is not the same, and shouldn't we use the latter? Thank you in advance!

This is an excellent question! Just to elaborate upon what @porphyrn and @M. Farooq have said, imagine that the universe is entirely empty except for one sodium atom and one photon (particle of light) that has a wavelength of $\pu{242 nm}$. If the photon bumps into the sodium atom, it might just bounce off: this is called scattering. But another big possibility is that the photon will be absorbed (i.e., ‘taken in’, as it were) by sodium’s 3s valence electron. The energy of a $\pu{242 nm}$ photon is $E_{\pu{242 nm}} = h\nu = \frac{hc}\lambda$, where $h = \pu{6.626 x 10^{-34} Js}$ and $c = \pu{2.9979E8 m/s} = \pu{2.9979E17 nm/s}$. This energy is $\pu{8.21E-19 J}$.
Then the sodium’s 3s electron would have just enough energy to escape, i.e., ‘break free’, from the sodium atom. The result would be a sodium ion, $\ce{Na+}$, and an electron, $\ce {e-}$, that were free from one another: the sodium atom has been ionized, i.e., has become an ion, so the energy of the $\pu{242 nm}$ photon is the ionization energy of the 3s electron in sodium. Using the $h$ and $c$ values, the required ionization energy, for one sodium atom, is $\pu{8.21 x 10^{-19} J}$. For a mole of sodium atoms, multiply by Avogadro’s number, resulting in $\pu{494 kJ/mol}$.
So where does the infinity enter the picture? Well, a positively charge sodium ion and the electron that escaped from it attract: opposite charges attract. The Coulombic force of attraction is an inverse square law, similar to gravity, so total absolute freedom would mean that the electron was infinitely far away from the sodium ion and had zero kinetic energy, i.e., zero energy of motion. But, as a practical matter, the separation can be relatively small, not infinity! As a practical matter, it is known experimentally that an isolated sodium atom, doing its thing in a vacuum, can be ionized by shining $\pu{242 nm}$ light on it.
In your example, suppose a sodium ion, in vacuum, happened to recombine with a free electron. Assume the sodium ion and electron are moving very slowly relatively, so there is negligible excess kinetic energy: we don't want the electron to 'harpoon' the poor sodium ion! The electron could drop down into any empty orbital of the sodium ion. Potentially, that is an infinite number of possibilities, hence M. Farooq's answer (and sweet Grotian diagram). The result would be release of a photon with wavelength longer (numerically 'larger') than $\pu{242 nm}$. This excited (it has more energy than a 'ground state') sodium atom must eventually drop to the 'ground state' (the only state where it is indefinitely stable) by emitting one or more additional photons or perhaps getting 'mugged' of its excess energy, via collision with something else. This is collisional deactivation.
But what if the electron dropped all the way down, in one fell swoop, to the 3s orbital? Then the resulting sodium atom would be in its 'ground state', i.e., lowest energy (and only indefinitely stable) state, and the emitted photon would be at $\pu{242 nm}$.

Very nice answer. Dr. Ed, May be your are the right person to enlighten us. For almost ten years, I have been searching a simple example, as to how to construct Grotian diagram from an experimental spectrum of Na I or any other complex atom? I've got his original German book, but like a typical German Handbuch, it is a compilation of spectra. The question is here: https://hsm.stackexchange.com/questions/8346/how-did-early-physicists-experimentally-assign-electronic-transitions-in-atoms. I have asked at so many places, everyone has a hand waving answer.

Please consider giving the green checkmark to the most helpful of the posted answers. It encourages people to put some thought and time into crafting answers that are factually correct, relevant, understandable and likely to be of benefit to those, in future, who encounter the question and accepted answer. It is a small reward for those who volunteer their considerable time, effort and experience to aid others and they might well look favorably at future questions from the same person. Thanks for considering this!

good point, at times I feel we tend to overfeed the OP who are usually in their first year and they are often looking for an easy answer. Although we can appreciate each other's answers and the depth of the answer, the beginner is at times clueless. There is a nice biblical proverb but I will not write it here.

I am thinking of posting this exact comment whenever I see that it would be beneficial. I very frequently upvote your answers, but it is distressingly common for OPs to just vanish away without awarding that green checkmark. It needs to be more of a two way street, helping each other. Anyway, this is what I plan to do from now on. The OP is under no pressure to do anything. But maybe they will think about how nice it is to have experts willing to help them.

I don't mind the green tick. We don't need an amateur's endorsement to support the veracity of the answers. However I am happy to receive feedback (positive or negative). The worst part of the story is when OPs don't even respond, vanish or when people cowardly down vote a well written answer without a reason.

shouldn't we be given the wavelength/frequency of the radiation emitted when the valence electron (the eleventh electron) falls from n = infinity back to n = 3? What I mean is that the frequency of the convergence limit and the radiation emitted from n = infinity to n = 3 is not the same, and shouldn't we use the latter? What I mean is that the frequency of the convergence limit and the radiation emitted from n = infinity to n = 3 is not the same, and shouldn't we use the latter? Thank you in advance!

Interesting point, but transitions from n= infinity to some value of n give rise to a continuum. This type of emission might happen at extremely high temperatures. A free electron in a plasma or a gas phase can have a continuum of kinetic energies (i.e. it is not quantized) unlike a bound electron to a nucleus. When free electrons recombine an ion, this recombination will lead to an emission of a range of wavelengths. You cannot tell a free electron to "fall" to a particular energy level. On paper it may sound easy, but experimentally is very difficult to determine a transition from infinity. Although Grotrian diagrams are not taught in chemistry, they are quite useful in visualization of emission spectrum. Have a look at the Grotrian diagram of Na. You can visually see convergence limits and corresponding wavelengths. This image is from his original book in German.


text:
Which reaction is useful to prepare odd number of hydrocarbons?
Which reaction is useful to prepare odd number of hydrocarbons:

Wurtz reaction
Kolbe electrolysis
Corey–House synthesis
Oxymercuration

Wurtz reaction produces even products except if we take mixtures. Kolbe ones give even ones too. Oxymercuration merely involves addition, so I think it's Corey–House.
But how does one justify that exactly? Is it because of $\ce{Cu(CH3)2Li}?$

You are quite correct. Wurtz works best (and even that isn't great) when dimerising, same for Kolbe (which works better). Oyxymercuration functionalises a double bond but does not introduce alkyl groups. Corey-House works particularly well with Me for extending the chain by one unit, but also works for propyl, pentyl etc as well as benzyl.

The Corey–House reaction constitutes a general and high yielding method for the joining of two alkyl groups or an alkyl group and an aryl group.

Step one involves preparation of the Gilman reagent from an alkyl halide.

$$\ce{R–X + 2Li -> RLi + Li+X–}$$

Second step, a lithium dialkylcuprate (Gilman reagent) is prepared from the alkyllithium by treatment with cuprous iodide (CuI).

$$\ce{2RLi + CuI -> Li+[R–Cu–R]– + LiI}$$

The third step ,involvs reaction between lithium dialkylcuprate (Gilman reagent) and a second alkyl halide ,resulting in the formation of a C–C bond between the two organic fragments:

$$\ce{Li+[R–Cu–R]– + R'–X -> R–R' + "RCu" + LiX}$$
Example 1
In the following example, gem dihalides react with $\ce{Li+[CH_3–Cu–CH_3]– }$ to give the final product. You could read about it in wikipedia

Example 2

source
Reference
1.https://en.wikipedia.org/wiki/Corey%E2%80%93House_synthesis
2.https://www.synarchive.com/named-reactions/corey-posner-whitesides-house-reaction


text:
Polar Protic and Aprotic Solvents
I know $\mathrm{S_N2}$ is favored polar aprotic solvents, which means that they may have hydrogens on them somewhere, but they lack $\ce{O-H}$ or $\ce{N-H}$ bonds, and therefore cannot hydrogen bond with themselves.
But I found that $\mathrm{S_N2/E2}$ is favored by $\ce{OH-, RO-}$ (a strong base); if I look at $\ce{OH-}$, doesn't it contain $\ce{O-H}$ and form hydrogen bonds?
Can someone please explain what aprotic solvents then?
Are aprotic solvents a strong base?
What's the difference between  aprotic solvents and a strong base?

OH- in what solvent? Powdered KOH in THF reacts in a very different way from KOH in MeOH.

This was not mentioned .


text:
How do we know if a compound is amphoteric or not?
My teacher told me if it reacts with an acid and a base, then the compound is amphoteric.
However, I am not satisfied. Is there any other way of telling? For example we are give a compound say BeO, how can we say if it's amphoteric or not?

Examine the structure? Do you mean, experimentally?

https://socratic.org/questions/what-is-a-substance-that-can-behave-as-an-acid-or-a-base $$$$ If the substance can act like both an acid and a base, i.e., both give and accept protons, then it's amphoteric.

Sorry to tell you but your teacher's opinion is not right, just because the "react" is not enough. You may say that if an oxide reacts with an acid and forms a salt or if an oxide reacts with a base forms a salt then you can say that it is an amphoteric oxide. Take an example of aluminum oxide, when dissolved in HCl it will form aluminium chloride, but we can also dissolve it in NaOH, where it forms sodium aluminate.

Sorry to tell you but your teacher's opinion is not right, just because the word "react" is not enough. Usually you label oxides as "amphoteric". You may say that if an oxide reacts with an acid and forms a salt or if an oxide reacts with a base forms a salt then you can say that it is an amphoteric oxide. Take the example of aluminum oxide, when dissolved in HCl it will form aluminium chloride, but we can also dissolve it in NaOH, where it forms sodium aluminate.
However, we can broaden this definition further (good for advanced classes). One formal definition is "A chemical species that behaves both as an acid and as a base is called amphoteric. This property depends upon the medium in which the species is investigated: H2SO4 is an acid when studied in water, but becomes amphoteric in superacids."
See: https://goldbook.iupac.org/html/A/A00306.html

Two rules that partially solve the problem:
1)  Predominantly ionic oxides, such as those of magnesium and heavier alkaline earth metals, will be only basic in water, even if only weakly so due to limited solubility.  You need significant covalent bonding of the metal to oxygen to get the metal to take on additional hydroxide ions in strong bases.
2)  Notice I say "metal" above.  Nonmetals generally form oxides that are only acidic ($\ce{CO2}, \ce{SO2}$) or inert ($\ce{CO}, \ce{NO}$).  You need a metal for the oxide to react with water or acid as a base would.


text:
Bond dissociation energy among halogens
Among the halogens, why is it that the fluorine has the lowest bond dissociation enthalpy, considering the fact that fluorine is the smallest and the internuclear distance between the fluorine molecule is least and $2p-2p$ overlap should also very very strong ,the bond dissociation enthalpy should have been very high but it is lowest among the halogens. Why is it so??

https://chemistry.stackexchange.com/questions/43200/why-is-fluorine-more-reactive-than-iodine-despite-the-weaker-i-i-bond


text:
Order of acidic strength of oxides

Select the correct order of acidic strength of oxides:
(A) $\ce{SO2} > \ce{SO3} > \ce{CO2}$
(B) $\ce{Cs2O} > \ce{K2O} > \ce{Na2O}$
(C) $\ce{CO2} > \ce{N2O5} >\ce{B2O3}$
(D) $\ce{SO3} > \ce{CO2} > \ce{B2O3}$

The answer given is D, but I think it should be B because I  have been taught that acidic strength increases as we go down the group. But since these are oxides, I do not know if this trend is generally true. Can you please help me what exactly is a trend to be followed in such cases?

You didn't notice that in "B" are "basic" oxides?

@Mithoron Facepalm, How to decide for oxides. PS , Since the community asks to show your working or else it closes the question, I wrotee what came to my mind. I noticed it but nonetheless wrote in the hope that the question won't be closed , and I would come to know how to decide strength of oxides

Well those substances are definitely not acids , they are bases.Draw the molecule of its substance and the charge of every region and you will understand which is which.

The first group in the periodic table has metals. Metallic oxides are predominantly basic. As we go across the period the basic property decreases,  trend changing to first amphoteric then acidic elements.
The oxides mentioned above in the question do now fall into a particular group or period thus requiring brainstorming.
My suggestion is to convert the elements into their respective acids and hence comparing their acidity. Coming directly to the D option, $\ce{SO3}$ forming $\ce{H2SO4},$ $\ce{CO2}$ forming carbonic acid, and $\ce{B2O3}$ forming boric acid.
Now I guess it's obvious that which oxide is more acidic.

I think your answer could be significantly improved by reviewing all four choices (not focusing on the one that is marked as answer) and writing the corresponding reactions between oxides and water and supplying the $\mathrm{p}K_\mathrm{a}$ values for the products. Also, please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.


text:
How to remove stain from pavement after having dropped sulfuric acid on it?
I had this car battery that flipped in my car trunk.
When I got it out of there it spilled some acid on the pavement. It's a sulfuric acid battery. I added some baking soda on it and poured some more water+baking soda to make sure the acid was neutralized.
However there is still a brown-rusty stain on the pavement. I tried water and a big broom brush but it still there.
Any idea what I can use to clean up the pavement to its original color?

Sorry about your mishap. I don't think the organosulfur tag applies.

"a big broom brush" - A short bristle/wire hand brush would be far more abrasive.

You did the right thing by neutralizing with a base.
The best solution is to let time heal the stain. The surface is obviously damaged, but there is no way to reverse that damage. The stain will only go away if someone keeps on scrubbing the area until a new surface is exposed. The good news is that dirt, soil, and water will slowly remove and even scrub away the (temporary) surface stain. Maybe keep a flower pot covering the area for the time being.

The brown stain may be iron oxides that were dissolved from the stone and precipitated by the sodium bicarbonate, or possibly a sulfonation product of the asphalt which is not soluble in water, but coats the asphalt. Three techniques could reduce the staining:
1) Put a few spoonfulls of vinegar on the spot and let it stand several minutes before washing it off with water. If it works a little, you can do it again.
2) Put some 3% H2O2 on the spot and let it stand. If it foams quickly, insoluble ferric compounds will be reduced to soluble ferrous compounds. H2O2 and acetic acid (vinegar) are compatible in this application.
3) Make a fairly concentrated solution of a laundry detergent, put it on the stain, and abrade with a stiff broom or brush. If you are able to brush thru the stain layer, the detergent may help suspend the brown stuff. Warm water should be more effective.

I wouldn't put any more acid on the concrete. Even though vinegar is a weak acid, it will pit the concrete.

James, I am afraid that point no 2 is not correct "% H2O2 on the spot and let it stand. If it foams quickly, insoluble ferric compounds will be reduced to soluble ferrous compounds". H2O2 will not reduce ferric to ferrous, nor it will help in dissolution.

Sure it does. It does it for blood and for ferric oxides. You may have to acidify the H2O2 to keep the ferrous ions in solution, but H2O2 does reduce Fe+++.

If the sulfuric acid hasn't pitted the black concrete already, vinegar won't do any more. It seems that the driveway is asphalt concrete, not portland cement concrete, like the much lighter colored garage floor visible in the upper right. If it were my driveway, I would use vinegar + H2O2. You couldn't hurt it any more, and could help. Of course, time is the easiest cure, or just forget about it, but that requires patience or wisdom.


text:
Which of the following statements is true if HI is formed in the following reaction?

Which of the following statements is true if HI is formed in the following reaction?
$$\ce{H2(g) + I2(g) -> 2HI(g)}$$
a. There are no collisions between the reactants in the reaction.
b. All of the collisions between reactants have enough energy to overcome the activation energy.
c. Some of the collisions between reactants have enough energy to overcome the reaction mechanism.
d. Some of the collisions of the reactants have enough energy to overcome the activation energy.

I think it is d because I think it has enough energy to overcome the AE due to the equation, but I am not sure.

Why do you think D is correct? Please include that in the question as well. Welcome to our site!

I think it has enough energy to overcome the AE due to the equation.

@Athena You really can't judge the AE by looking at the equation; but option d is still a safe bet,as per the modern collision theory.

Yes i know the equation is not the safest but im wondering is that how you would answer it? If not can you teach me why not. Thank you

Both B and  D can be true, C does not make sense; 'energy to overcome reaction mechanism' is nonsense. So the question is ambiguous, B, D could each lead to product


text:
Quantify the effectiveness of baseline correction in Raman spectra
I've produced several thousand Raman spectra using hyperspectral imaging techniques. These suffer from significant fluorescent interference which produces a curved, sloped baseline.
Using Matlab, I have attempted several baseline subtraction techniques but I can't say quantitatively which is performing best - beyond a visual inspection I'm not sure how to compare them.
Is there any quantitative way of assessing the performance of a baseline subtraction? Or is visual inspection really the best we can do?
Thanks for taking the time to consider this.
Edit:
I've added an example spectrum along with one with a corrected baseline.
Thank you for your suggestions as regards methods for baseline correction, however, I have already used several different methods successfully. My intention was to discover if there was a way of measuring the performance of various techniques in order to asses which is the most effective.
Thanks again for your consideration.
Edit 2:
I just wish to clarify that I am not looking for methods for baseline correction. I have used several which have given me varying results - these include polynomial fitting, piece wise fitting, ModPoly, subtraction of iteratively smoothed savitzky-golay baselines, and WPLS.
What I'm looking for is a way to quantify their performance, is there any metric that can be used to do this?

Being the fluorescence due to the measurement, I do not think one can (easily) compute the fluorescence spectrum to be subtracted. (Also, would be useful if you tell us the class of molecules/materials).

If you are using hyperspectral techniques, why aren't you letting the algorithm sort out what the principal components are?

Generally, Raman spectral lines are very narrow in frequency compared to those bands from fluorescence. Why not try to separate by Fourier transforming then removing lower transform frequencies and transforming back.  Alternatively try a moving average filter to remove Raman lines then subtract the result from the original spectrum. This should leave just the Raman lines. At the  least these methods should give some improvement.

@porphyrin, FT sounds very nice on paper, but based on my experience, mathematical manipulations in the Fourier domain lead to an increase in noise, and they key question would be how to determine the cut-off frequencies? Also a very large moving average window will also distort the baseline. I think the easiest approach is to fit a higher degree polynomial (n=6) by manually selecting the baseline in OriginPro, just for the sake of trial and then subtract the model of the baseline from the raw spectrum. The poster should show the spectrum first before we speculate too much.

@M.Farooq I have found that the Fourier method works ok, as to the cut off frequency of course this will depend on the data but once determined could be automatic. As to the moving average, the idea is to get this done and then subtract from the spectrum. Polynomial fitting leads  much oscillation about the data and has to be avoided. The issue in not about the baseline, as I understand it but removing a contaminating fluorescence spectrum. If the OP would post a spectrum this would help as we are all stabbing in the dark otherwise. I assume that Savitsy-Golay has not worked

I agree, I often mention the poem in talks "The blind men and the elephant" when we don't have a clue how the original data looks like. https://en.wikisource.org/wiki/The_poems_of_John_Godfrey_Saxe/The_Blind_Men_and_the_Elephant

@Alchimista The spectra were taken from a tissue sample.

@JonCuster Which algorithm are you referring too? I will likely run the image through PCA as well as PLS-DA, LDA and SVM at a later stage, however, I would prefer to have an optimally corrected baseline first to minimise any effect it might have on classification.

@JonCuster: to a certain extent I think this is sound advise, but letting the algorithms sort out things by itself inevitably "costs information", and if the available sample size is already (too) small - which I'd say is the normal state for such studies - the prediction may overall profit from the external information (i.e., taking out surely irrelevant data) that the baseline correction can contribute.

Could you also mention the correction method? Your auto-fluorescence background is really huge and your signal to noise ratio is low. One of the most cited methods in automated background correction is indeed based on polynomial fitting, albeit little but more sophisticated than just least square fitting. This article is worth reading.
Automated Method for Subtraction of Fluorescence from Biological Raman Spectra
https://journals.sagepub.com/doi/abs/10.1366/000370203322554518

Another simple option is that you can choose a spectral range where to apply a polynomial baseline. Say from 0 to 1200 you apply a polynomial of degree 5, but 1200 onward you choose polynomial of degree 6. However, you have thousands of spectra and it is not clear whether your background shape is similar in each case or not.
Addendum:  I am afraid there is no solid way to compare the baseline correction approaches. The reason is that you do not know the "true" model of the autofluorescence. This is also stated by Ed in comments. He is a leading spectroscopist. In such cases, you should simulate a baseline like these authors and do a comparison. Make a synthetic baseline by polynomials with say n=5, make it more complicated by mixing certain regions with polynomial  n=6, 3, 2, and then try your approaches. Don't add any peaks to the baseline, and just check which technique is showing least residuals. Is there a trend in the residuals? The trend is also important. The residuals should be scattered. What technique did you use in your case?
Alternatively, you can try a known fluorescent compound, Rhodamine G, and collect a fluorescence spectrum. In this case you will know the true shape of the spectrum. Use that as a true model and try your baseline corrections.
One general advice given to us by a professor, who was an author of Raman spectroscopy monograph, was that a better and better fit does not indicate reality. The model may be completely wrong but may have a correlation coefficient close to unity. Thus fitting a polynomial or any experimental baseline correction is just a utilitarian approach, which is changing the aesthetics and cosmetic appearance of a Raman spectrum. There is nothing fundamental going on.

Yes, that is an excellent technique for dealing with fluorescence in biological samples. Are you also aware of the related paper by Zhao et al.? [link]https://doi.org/10.1366/000370207782597003[/link]  However, my question was not how to go about correcting the baseline but how to measure the performance of various techniques used for doing so.

Apologies, I appear to have used the wrong kind of markup. Here is a working link to the paper.

@Tuirseach, I added more suggestions above, see if they are of any help.

Bit late to the party but here we go:
I fully agree with @MFarooq that we don't have the reference information of how the corrected spectrum should look like to directly compare baseline correction approaches (that is, outside very special and often rather artificial scenarios).
We can say, though that the proof is in the pudding if we don't do automated baseline correction for the sake of baseline correction but rather as preprocessing step to remove a known distortion of the spectra (the background) before a predictive model. In that case, we can compare the predictive ability of the models we get based on the various baseline correction approaches.
This basically means that we treat the choice of baseline correction as a hyperparameter of the overall model. Unfortunately, the sample sizes (numbers) I've had for these biospectroscic studies were not sufficient to allow a meaningful comparison of the predictive quality which is the basis for such a hyperparameter optimization. (Many people happily do rather massive numbers of model comparisons though, being blissfully unaware that their sample size allows barely to distinguish "practically perfect, you're going to get the best classifier award and a spin-off founding stipend" from "I barely dare to show this in the group meeting" models).
Based on that, I'd recommend to

use your spectroscopic expert knowledge in judging the baseline correction and
be very conservative in terms of the complexity of the baseline.

Of course, if you are in the happy situation to sufficient sample numbers to allow such comparisons: go ahead.
(I think I see a bit of light at the end of the tunnel here, but am not yet fully there.)

That being said,

I think it is important to realize that for many data analysis models, baseline correction isn't all that important because they can incorporate their own baseline correction: if you consider bilinear models (i.e. $\mathbf Y = \mathbf B \mathbf X), they can contain their own polynomial baseline correction.  As a very easy example, consider a coefficient pattern -1/2, +1, -1/2. This pattern measures intensity at the middle position over a linear baseline through the first and last point. This corresponds to @MFarooq's observation that baseline correction is often rather cosmetic.
In addition, in order to be good (helpful), the baseline correction must be sufficiently certain so that we have a net information gain. IMHO this is far from guaranteed: I've done very simple simulations of a "Raman band" on top of a perfectly flat background with the corresponding Poisson noise. Fitting and subtracting a linear baseline outside that band in such a situation that was a) ridiculously easy by eye in terms of saying what is signal and what is baseline, and b) using far more data points to fit the baseline than we  typically do, I still observed that subtracting this fitted baseline decreased the SNR of the integral intensity of my "Raman band" easily by a factor 2 compared to subtracting the true baseline!
My very personal opinion about the automated polynomial baselines for biological* tissues is that consider much of what I've seen in that respect overfitted: the resulting baselines are too wiggly compared to what structure I'd (not) expect in the background below the Raman spectrum, and far too wiggly compared to what I think we can afford considering the level of shot noise that 200 counts signal on 4000 counts background mean.
Still, I do use polynomial baselines if I don't have anything better (EMSC, see below), and often find that local polynomials allow baselines with overall fewer (equivalent) degrees of freedom: they can "concentrate" curvature in one region which is sometimes useful.
* I'm talking about biological tissues because that's where my Raman experience is. However, I do not expect other tissues (thinking cotton) to be totally different with respect to the general aspects I consider here.
I did compare several preprocessing approaches for classification of mid-IR spectra of biological tissues by spectral region selection followed by LDA in my Diplom thesis: baseline correction didn't matter.
Keep in mind that not all background that disturbs Raman measurements is fluorescence. While that doesn't matter for the algorithmic correction, it may offer optical ways of getting rid of some part of the background optically. That would have an additional benefit on the signal to noise ratio: the background is light, so it does lead to shot noise, and that shot noise does increase (absolutely) with the total light.
So even if you could perfectly subtract the correct background spectrum, you'd still be left with the additional noise caused by that additional light. Thus a mathematical correction of the measured background can never be as good as (optically) avoiding that background in the first place.
Two possibilities than in my experience can make a whole lot of a difference here are measuring in immersion (also gives more collection efficiency with higher NA) and higher confocality: both reduce stray light, higher confocality can also help against fluorescence. The former by having smaller changes in refractive index and the latter by excluding light coming from the wrong directions.
Iff the background is fluorescence, it may be possible to bleach it by the excitation laser. If you collect spectra while doing this, you may be able to get a decent estimate of the fluorescent background (at least of the component that does bleach).
This finally leads me to one baseline correction approach which I think stands out of the many entirely data-driven baseline fitting algorithms (sorry, I cannot keep this opinion back even though I perfectly understand that you are not asking for recommendation which baseline correction to use): EMSC describes the spectrum as being composed of polynomial or other "mathematical" terms plus reference spectra. So you can very elegantly combine polynomial baseline terms with substrate or fluorescence spectra (you can even include spectra of what should not be subtracted as baseline). The huge advantage of this approach is that it allows to incorporate far more external  nowledge/information than "the baseline should be a polynomial".


text:
Synthesizing amino acids - why sodium hydroxide goes after certain bonds
everyone. In the following reaction, what is it that causes NaOH to attack either bond A or B over all the other bonds available? Subsequently, why does NaOH not continue with bond B once bond A has been reacted with? Thanks!

I believe this is a matter of manipulating the stoichiometry and concentrations of the reactants such that only one of them reacts. Note that bonds A and B are equivalent actually. To make only one of them break, I suppose the concentration of the base could be limited.

Imide carbonyl groups are more reactive toward nucleophiles than amides. That is why hydrazine is used in the removal of the phthaloylimide moiety in  the Gabriel amine synthesis. The second bond, the amide bond, is cleaved intramolecularly by the hydrazide formed in the intiial hydrazine nucleophilic attack on the imide.

The above reaction is a modification of Gabriel phtahlimide synthesis from here:

This procedure, known as the Gabriel synthesis, can be used to advantage in aminating bromomalonic esters.Since the phthalimide substituted malonic ester has an acidic hydrogen (colored orange), activated by the two ester groups, this intermediate may be converted to an ambident anion and alkylated. Finally, base catalyzed hydrolysis of the phthalimide moiety and the esters.

The reaction you show is base catalyzed hydrolysis of the phthalimide shown in the link.
The mechanism of this reaction is shown below:

Both amide bonds(2 to 3 and 5 to 6) cleaves.
References

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(Vollhardt_and_Schore)/26%3A_Amino_Acids%2C_Peptides%2C_Proteins%2C_and_Nucleic_Acids%3A_Nitrogen-Containing_Polymers_in_Nature/26.02%3A_Synthesis_of_Amino__Acids%3A_A_Combination_of_Amine__and_Carboxylic_Acid__Chemistry
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Amines/Synthesis_of_Amines/Gabriel_Synthesis

Yes, but why that particular bond (either A or B)? Why does the reaction happen at one of those two bonds than, say, at one of the carboxylate esters, for example? What makes the resulting intermediate more stable at A or B than anywhere else?

The esters are hydrolyzed also, in your reaction scheme. The scheme shows both esters being cleaved, along with one of the amide bonds. Your question is really about why is one amide bond cleaved in the imide, rather than both


text:
Plutonium element configuration
For Plutonium 94 when I do it's element configuration I get
1s2-2s2-2p6-3s2-3p6-4s2--3d10-4p6-5s2-4d10-5p6-6s2-4f14-5d10-6p6-7s2-5f6

I found on this site that I'm correct however when I look it up on wikipedia it says
[Rn] 5f6-7s2
I know this is the short way to write it but I thought it would
[Rn] 7s2-5f6

What am I not understanding?

https://en.wikipedia.org/wiki/Electron_configuration may help some. Yes, a different page on Wikipedia...


text:
Polyethylene glycol
I used a product on my family containing 90 percent Polyethylene glycol. Now I’m reading how Polyethylene glycol can be contaminated with ethylene oxide and also the lower the weight the higher the chance. The Polyethylene glycol in the product I used has no number following and it was a soft cream used for skin care.
I was wondering if someone could explain to me how Polyethylene glycol is safer than ethylene oxide and how exactly is Polyethylene made? I’m very worried that I put a toxic cream on my children one of them who had eczema at the time.
Thank you

Do not trust random websites. I suspect that widespread misinformation is made for the interest of "bio products" companies. Polyethylene glycol can be synthetized by reaction with ethylene oxide. But ethylene oxide, being far from healthy, is carefully removed from polymers suitable for human use (and it's not hard to do, being it a gas). I would say not to worry

@The_Vinz: I like your comment. Also, any remaining ethylene oxide may be converted to ethylene glycol and would be okay to external applications. Lynn has nothing to worry about.

Agree with @The_Vinz, also, polyethylene oxide (or polyethylene glycol) has nothing to do with polyethylene and is only formally related to ethylene glycol as well. The name polyethylene glycol comes from the formal polycondensation of ethylene glycol (which does not work in practice), while the name polyethylene oxide comes from the anionic polymerisation of ethylene oxide (this is the synthesis route used in practice).

While polyethylene glycol [PEG] is safe for human consumption, and the worst part of a colonoscopy,  ethylene glycol is not. It is metabolized to oxalic acid which is toxic.

@user55119 Oxalic acid is not toxic in such trace amounts. It is present in a lot of vegetables which are safely edible. Rhubarb, chocolate, spinach, ...

@Karl: Agreed! I'm well aware. I like all three of them. I intended to point out that the lowest homolog of PEG does not have the same properties as the polymer.

@user55119 Yes, that's basically a fundamental principle for polymers.

Thanks you for your responses. After using this cream and then reading many scary articles on polyethylene glycol I have been very worried . I’m ok if it’s one of many ingredients in my products but because it was 90 percent I will not be using it again.  It still scares me that it is made from ethylene oxide though

Lynn, i'd like to know where you are reading about the ethylene oxide.

I read a article online,  I believe it was  “ the dirty dozen” and it said PEGs may be contaminated with ethylene oxide during its production and more. There are other articles that said the same thing.

Poly(ethylene glycol), commonly known as PEG is non-ionic, hydrophilic, non-toxic, and the most used polymer in wide variety of applications. For example, PEG is the gold standard for stealth polymers in the emerging field of polymer‐based drug delivery (Ref.1). You may find mostly unbiased information about PEG here, which is generally considered biologically inert and safe according to the clinical studies. However, drawback is these studies of clinical safety of PEG are generally based on adults, not children. Nonetheless, at least one study using children have proven it is safe for children as well (Ref.2).
So far, I addressed the suitability of PEG for human consumption. Note that these PEG molecules suitable for human intake have specific lengths such as PEG 3350, which has an average molecular weight of $\pu{3350 g/mol}$ (The structure of PEG is commonly expressed as $\ce{H−(O−CH2−CH2)_n−OH}$). Yet, my argument is if PEG is clear to human intakes, then it should be safer for external applications such as skin creams. To my understanding, among available PEG molecules ($n=1, 2, 3, ..., n$), only toxic one to human is its monomer,  $\ce{H−O−CH2−CH2−OH}$ (when $n=1$), commonly known as ethylene glycol or antifreeze. The apperent toxicity of that molecule is mainly due to its enzymatic oxygenation inside the body to oxalic acid, which produce insoluable calcium oxalate crystals.
As you concern, PEG is the basis of many skin creams. An example of PEG derivative used in skin creams is cetomacrogol 1000, which is the trade name for polyethylene glycol hexadecyl ether ($\ce{-O-C16H33}$) with the general formula $\ce{HO(CH2CH2O)_nC16H33}$ (where $n= 2-20$). When $n=10$, it is named Brij56, and when $n=20$, it is Brij58, etc. Another polyethylene glycol ether commonly used in cosmatic industry is Isoceteth-20, which has the same general formula $\ce{HO(CH2CH2O)_nC16H33}$ where $n$ has an average value of 20. The difference between Isoceteth-20 and cetomacrogol 1000 is use of iso-cetyl alcohol instead of cetyl alcohol. Few examples for cosmetics containing Isoceteth-20 are: Nivea, Aveeno, and  Nu Skin products, among many others. So, I'd be comfortable with using these products, just based on their reputation.
Now, how PEG is synthesized?

The production of polyethylene glycol was first reported in 1859. Both A. V. Lourenço and Charles Adolphe Wurtz independently isolated products that were polyethylene glycols. Polyethylene glycol is produced by the interaction of ethylene oxide with water, ethylene glycol, or ethylene glycol oligomers. The reaction is catalyzed by acidic or basic catalysts. Ethylene glycol and its oligomers are preferable as a starting material instead of water, because they allow the creation of polymers with a low polydispersity (narrow molecular weight distribution). Polymer chain length depends on the ratio of reactants (Wikipedia).

For example, branched PEG haas been recently prepared by Copolymerization of ethylene Oxide and glycidol (Ref.3):

$\mathrm{^{13}C \; NMR}$ of the product mixture shows no presence of unreacted ethylene oxide and glycidol:

References:

Katrin Knop, Richard Hoogenboom, Dagmar Fischer, Ulrich S. Schubert, “Poly(ethylene glycol) in Drug Delivery: Pros and Cons as Well as Potential Alternatives,” Angew. Chem. Int. Ed. Engl. 2010, 49(36), 6288–6308 (https://doi.org/10.1002/anie.200902672).
Dinesh S. Pashankar, Vera Loening-Baucke, Warren P. Bishop, “Safety of Polyethylene Glycol 3350 for the Treatment of Chronic Constipation in Children,” Arch Pediatr. Adolesc. Med. 2003, 157(7), 661–664 (https://doi:10.1001/archpedi.157.7.661).
Daniel Wilms, Martina Schömer, Frederik Wurm, M. Iris Hermanns, C. James Kirkpatrick, Holger Frey, “Hyperbranched PEG by Random Copolymerization of Ethylene Oxide and Glycidol,” Macromol. Rapid Commun. 2010, 31(20), 1811–1815 (https://doi.org/10.1002/marc.201000329).


text:
Burning magnesium ribbon
I know that this might sound wrong but, in the reaction
$$\ce{2 Mg(s) + O2(g) → 2 MgO(s) + energy}$$
isn't oxygen supposed to be oxidised as it has excess electrons $(\ce{O^2-})$ and can donate a few to magnesium as it has less electrons $(\ce{Mg^2+})?$
I have searched for it in almost all books and it is written that magnesium gets oxidised. Please clear my misconception. It would be helpful if I can get a reaction mechanism for this, because it includes transfers of electrons.

In layman's terms both magnesium and oxygen within MgO are "happy" since both $\ce{O^2-}$ and $\ce{Mg^2+}$ have configuration of neon, a noble gas. For a more in-depth explanation probably have a look at the elements' electronegativities or dive into physical chemistry (review free Gibbs energy for the burning process).

But MgO is formed after the reaction. I am talking about their reaction to achieve the stable state of MgO.

But, oxidation means loss of electrons. So, if magnesium is oxidized it has to lose electrons, but it clearly has less electrons, then why will it lose them?

Well, then why did magnesium lose 2 electrons and oxygen gain 2 electrons? I mean why did Oxygen take 2 electrons from magnesium?

The reactants are neutral Mg and neutral O2. The magnesium doesn't turn into Mg++ until after reaction, and it gets that way by being oxidized. The O2 doesn't get extra electrons and go to 2 O-- until after reaction. The MgO is the product and doesn't do any oxidizing or reducing, so we imagine it to be happy as a stable lattice of Mg++ and O-- ions.

How can we tell that the reactants are neutral? I mean is there any sort of test or something like that, you know?

I'm unsure the level which this question is being asked, but hopefully this answers your question.
As you pointed out "why did Oxygen take 2 electrons from magnesium" well if you would allow me to be rather 'hand-wavy' about the whole thing the reason is due to a activation energy vs ionization energy and bond-dissociation energy.
If you stick some fresh magnesium in an oxygen rich environment it won't all spontaneously react. This is because there is an amount of energy required for the reaction to happen (I will refer to this as activation energy). Now to make the reaction between magnesium and oxygen gas happen two things need to happen, the magnesium needs to lose two of its electrons (ionization energy) and the oxygen needs to break into two individual oxygen atoms(bond-dissociation energy). The misconception here is that oxygen gas will split into two -2 ions. If you count the number of electrons in one molecule of oxygen gas you get 12 electrons. Two -2 oxygen ions come out to total 16 electrons.
In reality, O2 would split into two neutral oxygen atoms as the oxygen gas is a neutral species not an ion. This is obviously incredibly unstable which is why it will look for the nearest source of electrons to pull from and magnesium's electrons are relatively easy to grab vs the oxygen. Reply if you want any more detail on anything I've said so far (as a lot of it is rather hand-wavy on my end), but if this is for a intro chemistry class I don't want to overload you with more information than you need.

No, no, don't think so. Actually, I got this question wrong on a test and wanted to find out the detailed reason about it. So, if you think that there is a detailed reason, then feel free to answer.

What part of my explanation do you want more detail about? If it is on how the Oxygen is a neutral it's simply because there is no charge on either species. You can determine this by finding the oxidation state of each element on the reactants side vs the product side.
Here are some general rules for you to follow to determine this:
https://socratic.org/questions/how-do-you-calculate-the-oxidation-number-of-an-element-in-a-compound


text:
How to calculate S² value of a broken-symmetry wave function?
$S$ represents spin, signifies the number of unpaired electrons in the system. For example, if the number of unpaired electrons is $1$, then $S=1/2$. $S^2$ is calculated as $S(S+1)$.
From what I have read, the $S^2$ value of a broken-symmetry singlet (contaminated by a triplet) is $1.0$, which is calculated to be the average of the singlet and triplet $S^2$. Similarly, the $S^2$ of broken-symmetry doublet (contaminated by the quartet state) turns out to be $1.75$ (average of a doublet and a quartet). I would like to know how these average values are being calculated. I understand that these are weighted averages (as $1.75 \neq 0.5\cdot(0.75+3.75)$, where $0.75$ and $3.75$ are $S^2$ values of doublet and quartet, respectively), but I don't know how that weighting is being done. I would be grateful if someone could provide a detailed explanation (mathematical derivation)  of this.

Cross-posted to CCL: http://www.ccl.net/cgi-bin/ccl/message-new?2019+06+18+001

The spin operators for the $p^\textrm{th}$ MO are defined as follows:
\begin{align}
S_{x,p} &= \frac{\hbar}{2} \sum_{\tau,\tau'} c_{p,\tau}^\dagger \hat{\sigma}^x_{\tau,\tau'} c_{p,\tau'}, \\
S_{y,p} &= \frac{\hbar}{2} \sum_{\tau,\tau'} c_{p,\tau}^\dagger \hat{\sigma}^y_{\tau,\tau'} c_{p,\tau'}, \\
S_{z,p} &= \frac{\hbar}{2} \sum_{\tau,\tau'} c_{p,\tau}^\dagger \hat{\sigma}^z_{\tau,\tau'} c_{p,\tau'}, \\
\end{align}
where $\sigma_{\tau,\tau'}^{x}$ denotes the matrix elements of the Pauli matrix $\sigma^x$ and so on. Any spin-observable can then be constructed from these basic blocks. For instance, $S^2$ operator takes the form
\begin{equation}
S^2 = \sum_{p \in MO} \frac{3}{4}\left( c_{p,\uparrow}^\dagger c_{p,\uparrow} +
c^\dagger_{p,\downarrow} c_{p,\downarrow} - 2c^\dagger_{p,\uparrow} c^\dagger_{p,
\downarrow} c_{p,\downarrow} c_{p,\uparrow} \right )
\end{equation}
All one need is to evaluate the expectation value of these operators over the wavefunction of interest.

I don't think this formalism is relevant to the question.


text:
Which salts can be used to precipitate chloride ions?
We don’t have AgNO3 in the lab. I have no idea what else can be useful.
I am washing a product and I want to make sure the chlorides are washed away.

AgNO3 is not really used to wash products. You remove the chloride ions by precipitation but you replace them with nitrate ions, and the sodium ions are still there too, so the product would still be impure.

The idea is to see if the filtrate precipitates.

Well, Lead(II) nitrate is an option, it precipitates in the cold, and dissolves upon heating. But even in the cold it is far more soluble than AgCl, so this is not a very sensitive method. You can concentrate the solution beforehand. But why have you no AgNO3, and why don't you just buy some? It's not really expensive.

Hg(I) cations would also precipitate as Hg2Cl2, their low solubility is the whole idea behind Group I of cation analysis. But then again, it wouldn't be really washing a product since you'd be only removing the chloride ions and replacing them with whatever anion you had on your added salt. Not sure what is the OP goal. If it's removing only the chloride because of some reaction interference then I think it's fine, but as product purification this doesn't sound much useful..

@IanC OP wants to check if he washed his product chloride-free, by testing the water with silver nitrate. At least that's what he says, and it's a very common procedure. ;-)

Then the OP wants to detect chlorides in truly minor amounts, and neither Hg(I) nor Pb are good for that.

@Karl Oh thanks, I misunderstood the question! It's much clearer now after the edits! :)

Another option could be a ion electrode, selective for chlorides, but it  would make sense for regular checks, as there would be  high one time investment.

@Poutnik How do these work? That'd be a nice alternative answer.

The only well known method to check for (the absence of) chloride ions is precipitation as AgCl (solubility 2 mg/l at 20°C).
Mercury(I) chloride is similarly badly soluble, but toxic (and unstable, disproportionation) and therefore bad practice.
Lead(II) chloride is already far more soluble (4 g/l) than AgCl, and practically all other simple chlorides are just very well soluble in water.
Silver nitrate is cheap, the internet sells 10 g high purity $\ce{AgNO3}$ for 25€. Just get some.

Another option could be a ion selective  electrode (ISE) selective for chlorides, but it  would make sense rather for regular checks, as there would be  high one time investment.
The principle is the same as for the potential reference electrode ...
 $$\ce{Ag | AgCl | KCl}$$
... but this time with mono or polycrystallic $\ce{AgCl}$ exposed to the solution. It is usually integrated with a true reference electrode, forming the complete cell with the voltage dependent of $\ce{Cl-}$ activity.


text:
Why is there visible liquid (possibly water) on matches when they're burning?
I was just burning a normal standard match and I observed visible liquid running down the wooden shaft as the fire burns and consumes it, what could explain such phenomenon ? could it be that the wood "sweats" when it's burning ?

In most cases this is just [molten] paraffin which the match stick is impregnated with. Paraffin is used to stabilize and prolong burning.

Wood does sweat in certain circumstances, but it has to be wet in the first place, which the match wood is definitely not. No, this is just molten wax coating, like andselisk said.

oh thank you. I thought somehow the little water inside the wooden components is coming out to the surface and vaporizing as it's burning so for a split second it's visible.

If the liquid on the match sizzles like a drop of water on a stove, I'm prepared to bet it is indeed water. ;-)

Water surely leaves burning wood and recondense.


text:
Comparing the emission spectrum of ethanol and mixture of alcohols
Sorry if this question has a simple answer, but I've been unable to find it elsewhere.

Using a diffraction grating, burning ethanol (99.9%) will produce a rainbow of colours. However, when burning a mixture of alcohols (methanol 76.6%, ethanol 21.4%, propanol 0.3%) the rainbow pattern is missing. Could someone explain?

If methanol only emits a small range of wavelengths, why wouldn't there be a dimmer  ethanol rainbow?

Please clarify "However, when burning a mixture of alcohols (methanol 76.6%, ethanol 21.4%, propanol 0.3%) the rainbow pattern is missing". What do you see instead, and how is ethanol being burnt? Is there a wick?

Is this the X-ray emission spectrum of ethanol you are asking about?

This is hard to believe.

Sorry, to clarify. I'm working in an elementary school. We are simply holding up a diffraction grating between the flame and the naked eye. With ethanol, the students would see 8 elongated rainbow patterns. But with the mixture (bought as fuel for spirit lamps), there would simply be 8 flames (blue/orange) without the elongation or rainbow pattern. There is a possibility that the rainbow is present but simply too faint to be detected by the naked eye. When the students are looking at the flames, they are in spirit lamps. However, when I was testing, I poured some in a petri dish and ignited it

Do exactly what you will do in class. There might be a difference in flame temperature but I would expect to see more or less the same. Also avoid the term rainbow if not specifying other characteristics and taking it as synonymous of dispersed light. Otherwise they might be confused between refraction reflection diffraction. ...


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What chemical reaction could be used to make copper pennies a vibrant red?
I’m an artist working on a piece using pennies to illustrate my drawing and I need my pennies to be red in a certain portion. I’m trying not to use paint and was wondering if there was a way to make them a brownish-red color.

Note: copper itself is brownish-red. Welcome to our site!

“The Colouring, Bronzing and Patination of Metals: A Manual for Fine Metalworkers, Sculptors and Designers” by Richard Hughes and Michael Rowe (Thames & Hudson, 1991) is the definitive reference. Pages and pages of photographed samples listing experimental conditions for each - chemical, recipe, temperature, duration. Even things like whether you dip in the solution, paint it on, or saturate sawdust in the solution and leave your item in there for a week. Still in print after 28 years, deservedly. Horribly expensive but you will be able to get it second hand.

@Martin Kochabski I think your comment is informative enough to be an answer! :)

@MartinKochanski Yes, do put it into an answer.


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Solvents and Enzymes
I was reading recently about environmental remediation being done, using an esterase to break down / neutralize cocaine that had contaminated waterways in Europe. Are there similar enzymes that can be used to safely break down pharmaceutical drugs, like surplus SSRIs and neuroleptics, that medical staff might otherwise just flush down the toilet thus causing more pollution? A way to neutralize them into safe inert ingredients first would seem beneficial to educate people about.

One of the great properties of enzymes is their specificity. Even broadly specific enzymes such as lipases and proteases are picky compared to, say, a general base catalyst. As a result, if you are interested in degrading a single specific compound (cocaine in your example), it is very likely that you can find or engineer an enzyme to do the job. But if you want to degrade a wide variety of compounds with different structures, you will need a diverse set of enzymes, each specific for one or a few of the compounds.


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Energy absorbed by a system in isothermic process in which the phase changes from liquid to gas
A system contains $x$ mole of material, with surface area $S$ is pressed by a mass $M_1$ (no fraction and no any other outer forces involves). Let $\Delta H_\mathrm{l\rightarrow g}$ be the transition enthalpy from liquid to gas. The system is position in $(T,P_1)$ such that the material phase is liquid for those temperature and pressure values ($T$ and $P_1$). Now instantaneously the mass $M_1$ is decreased to to an unknown value which we denote $M_2$ so the pressure decreases. In case that the temperature stays constant until the inner pressure is equal to the outer pressure $(T,P_2:=\frac{M_2g}{S})$ system gets to back to equilibrium, and the material phase in $(T,P_2)$ is gas, I wish to calculate how much energy was absorbed by the system till it gets to pressure equilibrium $(T,P_2)$.
In the question is says that the volume of the liquid can be neglected when compared to gas' volume. And that the volume of the water is independent of the pressure.
What I did so far:
The chemical equilibrium pressure (i.e the pressure in the phase transition point) can be calculated because the exercise gives the triple point $(T_\triangle , P_\triangle)$ so using the transition enthalpy constant and the constant $T$ of the process, one can calculate $(P_\mathrm{l\rightarrow g, eq},T)$ (using Clausius–Clapeyron relation). And by $PV=nRT$ one may calculate $V_\mathrm{eq} = \frac{nRT}{P}$ the volume in the chemical equilibrium point. By the unstruction $V_\mathrm{eq}$ is the total volume of the material (we neglect the liquid volume)
We know the starting inner pressure which is $\frac{M_1g}{S} \text{[Pa]} $ and by the given instruction the pressure doesn't change the liquid volume we learn that until the phase transition point, the volume of the system stays constant $V_\mathrm{eq}$ which we calculated above.
When the system gets to pressure equilibrium point $(T,P_2=\frac{M_2g}{S})$ we don't know the volume or $P_2$ because $P_2$ is unknown how every we not that its larger then $V_\mathrm{eq}$.
I'm pretty much stuck here, I don't see how to translate this information to evaluate the amount of energy absorbed. It seems to me that maybe one more piece of information is needed, but the exercise doesn't give one.

Would it be correct to say that you started with all saturated liquid?

I thinks so, if saturated means the whole material is in liquid phase @ChetMiller

I assumed this.  See my answer.

As indicated by @Buck Thorn,the work done by the substance on its surroundings is $P_2(V_2-V_1)$.  So, from the first law, $$U_2-U_1=Q-P_2(V_2-V_1)$$Solving for Q then gives:
$$Q=(U_2+P_2V_2)-(U_1+P_1V_1)-(P_2-P_1)V_1=H_2-H_1-(P_2-P_1)V_1$$Here, $$H_2=H_1+(H^{sv}_1-H_1)+(H_2-H^{sv}_1)$$where $H_1$ is the enthalpy of saturated liquid in State 1, and $H_2$ is the enthalpy of the superheated vapor that exists in State 2 (i.e., at a pressure lower than the equilibrium vapor pressure at the constant temperature T).  For an ideal gas, the change in enthalpy at temperature T from the saturated vapor in State 1 to the superheated vapor in State 2 is zero.  So the previous equation reduces to $$H_2=H_1+(H^{sv}_1-H_1)=H_1+x\Delta h^{vap}$$where $\Delta h^{vap}$ is the heat of vaporization at temperature T.  Substitution of this into the equation for the heat added gives:
$$Q=x\Delta h^{vap}+(P_1-P_2)V_1$$where $V_1$ is the volume of the saturated liquid at T. The second term on the rhs will typically be negligible.

Hmm, I assumed that there was no headspace above the liquid so no vapor in equilibrium with the liquid at 1. This does not change that my answer had a flaw but I still wonder if it would make a difference in the case of your answer? Somehow I think I'm visualizing the problem the wrong way.

@Buck Thom Once the pressure is dropped, the volume starts increasing, and you start forming vapor.  This continues until there is no liquid left (assuming that enough heat is being added to maintain the temperature).

Thanks for your answer, I have some questions:  1. what does $H_{1}^{sv}$ means in the equation $H_2 = H_1 +(H_1^{sv} - H_1) + (H_2 - H_1^{sv})$  (I think maybe $H_1^{sv}$ is super vapor and $H_1 = U_1+P_1V_1$ as mentioned before) 2. What does the $x$ mean in $x\Delta h^{vap}$.  3. At the end one needs to know $P_2$ to calculate the heat, but I don't see how to calculate the values of $P_2$ without knowing the new mass $M_2$. @ChetMiller

$H_1^{sv}$ is the enthalpy of saturated vapor at temperature T.  x is the number of moles of substance, as was indicated in the problem statement.  With regard to item 3, you are correct.  I guess they are assuming that, if the 2nd term on the rhs is negligible, it doesn't matter.

I wonder to what extent the result is path-dependent? (I am still not sure what might be wrong with my answer, but it certainly isn't as elegant a derivation as yours, and a whole factor of RT off...)

@Buck Thorn You were given the (irreversible) path, and you handled it perfectly, flawlessly calculating the work done by the system on its surroundings. Where you went wrong was in the way you applied the first law of thermodynamics. All you needed to do was to set the heat added Q equal to the change in internal energy (a function of state, not path) plus the irreversible work done by the system on its surroundings (which, again, you determined correctly).

@BuckThorn If the process were carried out reversibly, by first adding heat to vaporize all the liquid at the (constant) initial pressure $P_1$ and then easing up on the pressure gradually to $P_2$ while adding further heat to hold the temperature constant, the second term on the right hand side would be $xRT\ln{(P_1/P_2)}$

Yes, you got the right answer afaik. Thank you for your comments, I was looking for a shortcut and got lost along the way. I'm not sure why the OP accepted my answer. I'm assuming that you could also add a term for the heat associated with taking the liquid from the initial to the saturation pressure (while remaining liquid).

I made the assumption that since it's an ideal gas and an isothermal process, for the expansion of the gas $q=-w=P_2 \Delta V$ as I wrote in my answer. I see however how you used computation of $\Delta H$ to arrive at an expression for q, which is the right way to do things (assuming you know w).

@BuckThorn If you think my answer is right and yours isn't, you should delete yours.  In getting the work, there was no need to assume ideal gas an isothermal process.  If the external pressure was P2 throughout the expansion, the work would always be P2(V2-V1), irrespective of whether the gas is ideal and/or isothermal.

@BuckThorn Anyway, with a phase change, the internal energy change is not zero even if the temperature is constant.

Let us continue this discussion in chat.


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Rearrangement considerations in feasibility of SN1 in given reactants
I have been asked to compare the feasibility of Unimolecular Nucleophilic Substitution ($\mathrm{S_N1}$) in the following question.

I have received instruction that the reaction will be feasible only in the first compound (which results in a 3° stable carbocation) since 1-chloro-1-methylcyclopropane is angle strained and the resulting carbocation in this case will be unstable.
However, if we consider the possibility of hydride rearrangement in the second molecule, after formation of the carbocation, the following rearrangement should stabilise the carbocation formed, through the delocalisation of sigma electron pair as is seen in the case of cyclopropyl methyl carbocations.

Where exactly am I going wrong?

Related: https://chemistry.stackexchange.com/q/19090

The proposed rearrangement would produce a primary carbocation which, relatively lacking in hyperconjugation, is itself unstable.  The proposed rearrangement therefore is not really plausible.
The "instruction" that the highly strained 1-methyl-cyclopropyl cation cannot be formed may be wrong anyway.  This paper reports a substituted cyclopropyl cation as an intermediate in a reaction between a more complex chlorocyclopropane and an alcohol.


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Priority to longest carbon chain or double bond?
Priority to longest carbon chain or double bond?

Above structure and its nomenclature is given in my notes. However, I suspect it is wrong because the Carbon chain of 8 instead of 7 is available. So my question is do we give preference to double bond over the number of carbon in a chain during nomenclature?
I found this question here but it only makes me more confuse as its answer states that we give preference to the carbon chain. So is my notes are outdated(which is possible) or is there any other rule which I am unaware of?

The answer by orthocresol should apply to your current one. I am on app and can't see the sketch to see if you are right and the book wrong. Perhaps the authors refer to prior 1979...


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Iron(III) nitrate and hydrogen peroxide reaction
When iron(III) nitrate is added to hydrogen peroxide as a catalyst, the temperature initially decreases and begins to increase after 30 seconds. Why is this the case? (is there any intermediate reactions of $\ce{H2O2}$ and $\ce{Fe(NO3)3}?)$

Nitrates tend to lower the temperature on dissolving, having a positive enthalpy change of solution. For example, $\ce{NH4NO3}$ has an enthalpy of solution in water of ~26 kJ/mol at ~300 K, and it is used in self-cooling cold-packs for that reason.
Sorry, I could not locate the enthalpy change of solution for $\ce{Fe(NO3)3}$, so this is not a definitive answer. You could measure that value, though, with some  $\ce{Fe(NO3)3}$, some $\ce{H2O}$ and a calorimeter.

To quote from Science Direct on the Fenton process:

The Fenton reaction was first reported by H. J. Fenton (1894) and it is described as the enhanced oxidative potential of H2O2, when iron (Fe) is used as a catalyst under acidic conditions. The reactions involved in Fenton processes are (Haber and Willstätter, 1931):

$\ce{Fe++ + H2O2 -> Fe+++ + .OH + OH-}$ (3.1)

$\ce{H2O2 + .OH -> .HO2 + H2O}$ (3.2)

$\ce{Fe++ + .OH -> Fe+++ + OH-}$ (3.3)

$\ce{Fe+++ + .HO2 -> Fe++ + O2 + H+}$ (3.4)

Also further important comment:

Since the reaction of H2O2 with Fe3+ is significantly slower as compared to Fe2+, the catalyst regeneration phase is considered the most critical stage of the process as it reduces the radical production rate and could end up depleting the iron available for the reaction (Eqs. 4.3–4.4) (Malato et al., 2009).

So, there are, indeed, intermediate reactions with ferric cycling to ferrous.


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Evaluate indicator choice (phenolphthalein)

A student wishes to find the concentration of a $\pu{50 mL}$ solution of acetic acid $(K_\mathrm{a} = \pu{1.8e-5}).$ He performs a titration using a solution of $\pu{0.2 M}$ sodium hydroxide. The student selects phenolphthalein as his indicator choice which has a $K_\mathrm{a} = \pu{3.98e-10}.$ Justify the student’s choice of indicator.

My attempt: I found the $\mathrm{p}K_\mathrm{a}$ of the indicator to be $9.4$ and the fact that the titration was weak acid + strong base meant the $\mathrm{pH}$ at equivalence would be around that too, so the indicator was justified. In saying this, my justification isn’t very mathematical.

Easy marks for phenolphthalein usability are:

$99.9\%$ of acetic acid is titrated at about
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log{1000} = 7.75$$

$100\%$ of acetic acid is titrated at
$$\mathrm{pH} = 7 + \frac12 \cdot (\mathrm{p}K_\mathrm{a} + \log c) = 9.385 + \frac12 \log c$$

Phenolphthalein color transition is $8.2-10.0.$


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Temperature and pressure for three phase equilibrium of aqueous sodium chloride solutions
I am currently investigating the effect of adding Salt(NaCl) to water on its triple point. Let us assume the mass ratio of water to NaCl is $1:1$. I asked a similar question previously. But, I still haven't been able to find the temperature(s) of the triple point(s).
I understand that according to the Gibbs equation $f = C-P+2$ Therefore, in my case of Salt water, $f = 1$. Thus we have a range of temperatures/presures where the triple point exists.
But how can we find this range of temperature or pressures? IS there an experimental methods?

This problem must have been solved in the beginning of the 20th century.

But how can we find this range of temperature or pressures? IS there an experimental methods?

The range of temperatures is fairly easy. Mix saturated aqueous NaCl (liquid solution, concentration about 5 mol/L) with ice (pure water, solid state), and measure the temperature. It will be about equal to the temperature of a household freezer (no coincident, freezers were using salt/water mixtures in the old days), i.e. -20 $^\circ$C. If you keep measuring while the ice melts and the NaCl concentration drops, you will get the temperature ranges, which will have the normal freezing point of water at the high end.
Partial pressure of water measured at atmospheric pressure
Using the same setup, you could measure the humidity above the ice/solution slush. In the experiment, water exists in the solid, liquid, and gas phase, just like pure water at the triple point. The gas phase contains mostly nitrogen and oxygen, so you would have to measure the partial pressure of water rather than the total pressure (which is about 1 atm and depends on the weather). The partial pressure will be a bit lower than that of pure water at the same temperature (the activity of water is less than one in a solution, so the partial pressure will be lower relative to pure water, Roult's law).
Pressure of water measured with the aid of a vacuum pump
In a slightly more involved setup, you would place the slush into a vessel that you evacuate with a vacuum pump. This removes all gas components (nitrogen, oxygen) from the gas phase, and replaces them with evaporating water until the water boils and there is pure water in the gas phase. At that point, you can close off the system and measure the pressure. The result should be the same as the first experiment.


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Do temporary dipoles form in silicon(IV) dioxide?
Textbooks only ever mention covalent bonds between silicon and oxygen atoms in silica. However, as electron movement is random, one end of a silicon atom in a given instant could be more positive than the other (setting up a temporary dipole) and this temporary dipole could induce dipoles in neighboring oxygen/silicon atoms. Could somebody confirm if this is indeed the case?

Temporary dipoles form everywhere.

Temporary dipoles are induces even by other temporary dipoles, not talking about permanent dipoles.


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is every molecule either a base or acid
I read today that water is a base because it can receive a proton when mixed with some acids. That got me thinking: is every single molecule capable of either donating or receiving a proton -- when mixed with just the right complementary molecule? Thus, is every molecule either a base or an acid to some degree?

I'll tell you more: water is an acid too. There is no "either".

Receiving/donating a proton is just one definition of an acid or a base, made by Bronsted. The Lewis definition is different, though it won't answer your question.

If it doesn't have a hydrogen, it can't donate a proton. If it doesn't have valence electrons, it can't accept a proton. Often, it is more useful to think about molecules acting as acid and/or base than being an acid and/or base.

In theory, any molecule that contains a H atom ("proton") can act as a Bronsted acid - it can donate that proton. How likely this is to happen is described by that proton's pKa value, the lower the value, the more likely this is to happen (the more stable the conjugate base that forms will be). You can look up pKa values for different types of proton on an inorganic or organic molecule by searching for pKa tables like this:
Different proton environments on the same molecule will have different pKa values:

The pKa describes the degree to which that proton will dissociate in a solvent (normally water, but other solvents are sometimes used) and high values mean that this essentially doesn't happen, or only to a tiny extent (in the figure above, these are labelled as "not acidic"). However, a very strong base could abstract these protons, so the molecule could still act as an acid.
To act as a base, a molecule normally needs to possess a lone pair. The Bronsted definition of a base only requires a molecule to accept a proton, but a lone pair is generally required in order for this to happen.

The example in the figure above shows ammonia acting as a Bronsted base, using its lone pair to bond to a proton. Any lone pair could (potentially) do this, but some functional groups (such as amines) have more propensity to do this. The pKb value for the functional group will quantify this propensity.
Molecules often contain both protons to donate and lone pairs, so can act as Bronsted acids and/or bases.
Lewis acids and bases are defined slightly differently. The definition here involves the electron pair, rather than the proton. So you can have Lewis acids (electron pair acceptors) that have no protons at all. These tend to be electron deficient species such as the $BF_3$ in this figure below:

Other examples of Lewis acids include small, highly charged cations that can accept coordinate bonds.
Some molecules can accept protons by using electrons from unsaturated moeities rather than lone pairs. An alkene is a classic example, which can use the pi electrons in the double bond to accept a proton, forming a carbocation.

So an alkene could be described as a base, but we don't normally categorize it in that way.
It is also worth looking at the distinction between acid/base and electrophile/nucleophile. There are some good posts on this site like this one


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Why are Born-Haber enthalpy calculations inaccurate for transition metal compounds?
A senior high/secondary school level exam question is about discrepancies between calculated and empirical enthalpies of formation. It asks why a discrepancy “usually” exists for d-block compounds such as copper (II) chloride.
For something like AgCl or AlCl3 the answer is covalent character. But that’s not true for CuCl2, is it?


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How do I go about either finding or calculating Aluminum-Oxalate salt solubility values in aqueous medium?
Water would contain other salts which may complicate things a bit. But finding out the Aluminum-Oxalate salts solubility values in pure water would be a good indicator already. It's for temperatures from 10-50C. Wikipedia doesn't have a page for Al-Oxalate but it has a page here: https://pubchem.ncbi.nlm.nih.gov/compound/Aluminum_oxalate


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How to improve efficiency of the diamine synthesis from fluorescein?
In this work, fluorescein was incorporated with an equal molar of 1-chloro-4-nitrobenzene and double potassium carbonate (molar ratio), the solvent of this reaction is dimethylacetamide (50 ml for 5 g of fluorescein).
The solution was stirred in at 110 °C for 24 h, after that cooled to the room temperature and precipitated in cold water, but a solid, similar to potassium carbonate, remains at the bottom of the reaction vessel, and the efficiency is very low (25%). I don't know the cause of this problem.
This is the structure that I expected for first stage:

In the next step the solid that was obtained from the previous stage, incorporated with 10% Pd/C and methanol. 30 eq hydrazine monohydrate was added dropwise and the solution refluxed at 50 °C for 12 h. After that the solution was filtered to separated catalyst and the crystals of diamine was obtained, but the efficiency of this stage is very low too.

What did the tlc of the first reaction mixture look like? Did you lose the chloronitrobenzene?.

@Waylander, Unfortunately, I did not use TLC to control the end of the reaction.

Then you'll know better next time you run this. Can you post the structure of your required product

@Waylander, thank you so much. I added the structure of the first stage's product.

How effective was the stirring? Did you get evolution of CO2 during the early stages of the reaction?

@waylander, the reaction was carry out in 600 (rpm), Because we work on a small scale and the solvent level is low, when the strring becomes more intense, the materials stick to the wall of the container and burn it.
No, How should I do this?

What I think is happening is that you are not getting enough mixing between your reagents, the solid remaning is unreacted potassium carbonate. I think you can reduce the amount of potassium carbonate you are using - you only need 1 eq as it is dibasic, and use finely powdered if you have access to it. Run the reaction more dilute with a more powerful stirrer (have you an overhead stirrer?) so that material does not stick to the side and follow it by TLC. If this does not improve the outcome I would consider using a strong organic base such as DIPEA or DBU in place of the potassium carbonate so that everything is in solution.

Could it be necessary to use a stronger solvent like DMSO?

It might help. DMSO or DMF but I think the problem is not enough solvent and ineffective mixing

If the amount of hydrazine monohydrate used to synthesize diamine is too high, what problems might it occur?

Given that the procedure uses 30eq of hydrazine, I don't think using more will be any problem. It is decomposed by the Pd/C catalyst. If the reduction is going poorly, I would look firstly at the quality of the catalyst and secondly investigate alternative reduction conditions e.g. Ammonium formate, Pd/C in reluxing EtOH or iPrOH


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Does the radius of the atom change when an electron is excited?

The radius of the atom increases when the electron in a hydrogen atom undergoes a transition from the 1s-orbital to a 2p-orbital.

How is this possible? Why does exciting an electron to a higher energy state increase the radius of the atom?

Related: Angular and radial nodes Why is the 2s orbital lower in energy than the 2p orbital when the electrons in 2s are usually farther from the nucleus? Atomic radius definition disproved - Bohr's model to Schroedinger's wave equation What is the difference between the mean radial distance and the most probable radial distance of an orbital?

This can be anything from a very short to a very long story, depending on how much you already know. What is a hydrogen atom, to begin with? What is it made of?

Atomic radii are not very well defined. tl;dr: the 'average radius' / most probable distance is larger for a 2p orbital than for the 1s.

If an atom is excited by a large amount of energy to close to its ionisation energy, its size becomes vast compared to that at its lowest energy. In the Bohr model the radius increases as $n^2$ when $n$ is the principle quantum number.  Search for Rydberg atoms.

See also Rydberg atoms. E.g for an atom of potassium, and an electron excited to n=600, the electron occurs up to 0.1 mm far from the kernel.

As for the just above, you can get a sense of it already with rudimentary models.


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Entropy change in non-ideal solutions
We know that some solutions show positive deviations and some show negative deviation from Raoult's law.The former is endothermic and the latter is exothermic. And again, the volume change show positive and negative terms.  In which case entropy change is more? And why is that?

The detailed calculation for $\Delta G$ is here,  Derive expression for internal energy of mixing and entropy of mixing using statistical thermodynamics  which gives $$\Delta G = RT[n_1\ln(x_1)+n_2\ln(x_2)] +(n_1+n_2)x_1x_2w$$
and as $\Delta S= -d\Delta G/dT$ then
$$\Delta S = -R[n_1\ln(x_1)+n_2\ln(x_2)] - (n_1+n_2)x_1x_2\frac{dw}{dT}$$
where $w \approx (2E_{12}- E_{11} - E_{22})$ and $E_{12}$ is the interaction energy between molecules of type $1$ and $2$ when they are adjacent and similarly for $E_{11}$ and $E_{22}$ all of which are negative. From a consideration of the energies involved $w$ is expected to be positive.
The first term is the  Raoult's law expression for an ideal solution and the second term the deviation, which is sometimes also called the excess  entropy of mixing. For benzene cyclohexane when $n_1=n_2=x_1=x_2=1/2$  the excess entropy  $\displaystyle -\frac{1}{4}\frac{dw}{dT}\approx 1.7$ J/mol/K at $298$ K and $1.2$ at $343$ K.


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Products of the reaction between chromium(III) chloride, zinc and sulfuric acid
The aim of my experiment was to make a Cr(II) aquacomplex in situ, and for that I have the reaction:
$$\ce{CrCl3·6H2O + Zn}$$
and I add concentrated sulfuric acid to it. I know that chromium(III) reduces to chromium(II), but I'm not sure about the reaction. Would it be
$$\ce{CrCl3·6H2O + Zn -> Cr(H2O)6Cl2 + ZnCl + H2}?$$

6H2O is irrelevant, since we are dealing with water solutions anyway. As for the rest, the reduction can be performed by Zn, or it can be done by the nascent hydrogen. I strongly suspect the latter (otherwise you won't need acid around).

@IvanNeretin The instructions were to use fine Zn powder so we didn't use nascent hydrogen

But you added acid, didn't you? That was my point.

The equation you have given is not balanced.

@NilayGhosh I wanted to make sure the products were correct before balancing. The question was if the products were correct.

ZnCl is not a product. It does not exist at normal conditions.

@NilayGhosh $\ce{ZnCl_2}$ would be one of the products

I know that chromium(III) reduces to chromium(II), but I'm not sure about the reaction. Would it be $$\ce{CrCl3·6H2O + Zn -> Cr(H2O)6Cl2 + ZnCl + H2}?$$

Your concept is right and I am glad nobody taught you the concept of nascent hydrogen. It is an obsolete idea of the 1920-40s. Please don't use it anywhere. Wikipedia has a nice summary https://en.wikipedia.org/wiki/Nascent_hydrogen and of course there are several articles which negate this very concept.
You can just simply write the ionic equation involving Cr(III), Zn, resulting in Cr(II) and Zn(II). Hydrogen is just a side-product, which is not participating in the reaction. Write the conditions over the arrow.
Alternatively, write the aquated forms as a complex. Your current equation has incorrect formula of zinc salt. Check what it should be!

so the full formula would be: $\ce{3CrCl3 + 3Zn + 12H_2O-> 2[Cr(H2O)6]^2+ + 3ZnCl_2}$. I kept the water molecules to be able to see the metal-aqua complex that builds in the end

No need to add water, the starting product is most likely [Cr(H2O)6]^3+ . Chloride is just a spectator ion. Show the acid over the arrow. I would personally write, Zn and acid over the arrow, [Cr(H2O)6]^3+  ->  [Cr(H2O)6]^2+, because nobody knows the exact mechanism.


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Over time what happens to soluble liquids with different densities
I have a case with two liquids: Water and another chemical which is miscible in water.
The chemical has a density of 800 g/l so it's 20% less dense than water.
Initially the chemical was mixed with the water but now it has been sitting stagnant for 20 years.
Since such a long time has passed what could I expect to see in terms of the mixing of the liquids? Will the different densities have caused the two liquids to separate? Will there still be a trace of the chemical dissolved in the water?

Vodka is pretty much a binary mix of ethanol $(ρ = \pu{789 g L-1})$ and water. I have a 1970s bottle of Stoli from my grandfather and it's still a homogeneous ready-to-drink solution. So no, unless there were some chemical reactions involved resulting in building up of a residue of third component, there is likely no reason for a mix of miscible liquids to split.

If the liquids are miscible, I would expect no separation at all.

If they were gasses then gravity will have an effect, but a column of gas kilometres high will be needed. Density will vary as $\rho=\rho_0 \exp(-Mg\Delta H/(RT))$ with $g$ acceleration due to gravity, $M$ molar mass (kg/mole) and $\Delta H$ elevation ( metres). You could increase the force used to separate  by using a centrifuge.