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Creating Oxalic Acid Gel for rust removal
While this question relates to an automotive application I consider that it is liable to be best asked on this site.
Summary:

I wish to create a low cost gel suitable for mixing with Oxalic acid to produce a material that was adequately thixotropic to remain on the underside of horizontal metal surfaces in a thin layer for maybe an hour.
Longer would be good.

It would be a significant bonus if it also was suitable for use with phosphoric acid.

I'm investigating various means of automotive chemical rust-conversion and rust-removal.
There are many weird and wonderful DIY solutions (potatoes actually do work ! :-) ) but the two that seem most worthwhile are

Phosphoric acid as a rust converter.
This is a very common and well respected means of converting rust to insoluble and stable iron-phosphate.
Concentrations in the order of 25-50% seem typical.

Oxalic acid as a rust remover.
This seems to be slightly less well known but is a technically sound method of converting rust to water soluble iron-oxalate which can then be removed by hosing or waterblasting.
Supplied as a water soluble powder mixed with maybe 3 to 10 parts by volume with water (may have much more water).

I have tried a commercial rust removal gel whose formula is intractably secret (I tried) but which works superbly to convert serious rust to a soluble state which could easily be removed with a waterblaster.  The product is excessively costly for the area that I wish to deal with.
Oxalic acid (solid powder dissolved in water to suit) also works well BUT, as it is about as viscous as water, has the nasty habit of dripping off rather than remaining on the surface for the desired periods of tens of minutes or more. Especially so on vehicle undersides :-).
I would like to produce an Oxalic acid gel carrier to solve the above requirement. Ideas encountered for possibly suitable gels include agar, gelatin, cornstarch & vinegar, seed based gels - with boiled rice "porridge" a likely candidate, and more.
I'd be interested in knowing if anyome else has experience in either DIY gels or other products that might allow me to formulate a suitably thixotropic rust remover.

Polyethylene glycol could be a candidate. There are commenrcially available preparations of KOH in polyethylene glycol gels for oven cleaning and I don't see why oxalic acid would not work.

Acids and alcohols can react to form esters? Do you think this suspension would be stable in the long term?

Depends on storage conditions and how you define "long term". This SE answer suggests that in the absence of catalyst the mixture is stable.  https://chemistry.stackexchange.com/questions/68662/will-oxalic-acid-react-with-ethanol

@AChem Ester formationg is not a very exenthalpic process, to say the least. Unless you remove the water, it just doesn't proceed.

@Waylander I can buy polyethylene glycol at acceptable prices. Can viscosity be modified in some manner?

@RussellMcMahon  I'm out of my area of expertise here, but I guess that you could modify the viscosity by adding shorter chain alcohols or even water

I made an oxalic acid gel with xanthan gum. I’m trying to make an alternative to barkeepers friend, like a soft cleanser, but with fewer ingredients. It sounds like what I make would work for you- blend a few teaspoons of xanthan gum in a blender with water and in a separate container, mix oxalic and hot water til dissolved. Add to blender and mix with xanthan gum mixture. The result will be a gel like consistency.


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Why are sulfites often ignored in soil studies?
From a document on soil chemistry that I am reading:

The acid-base potential of a soil takes soil pH, total sulfur (sulfides and sulfates), and neutralization potential into account to determine the potential ability of a soil to generate acid.

Also, resources on well water quality usually mention only sulfides and sulfates, see, e.g., Indiana Department of Health — Sulfates & Sulfides in Well Water.
Why do these studies include sulfides $(\ce{S^2-})$ and sulfates $(\ce{SO4^2-}),$ but omit sulfites $(\ce{SO3^2-})?
$ Don't sulfites occur in soil? Or is this just a question of terminology?

I think it's generally assumed that sulfite has a short lifetime in soil, decomposing to SO2 gas or getting oxidized to sulfate. See for example https://www.jstor.org/stable/42935238 (preview can be viewed for free).

This article(ref.) gives you your answer:

While techniques are available for determining the concentration of most S-oxides in soil, determination of soil sulphite has been largely neglected. This is probably because the ion is considered too labile to make a significant contribution to the S-content of most soils. An exception may be soils exposed to heavy atmospheric pollution where sulphite enters in rainfall or directly by absorption of $\ce{SO2}$

Ref.: WAINWRIGHT, M., and J. JOHNSON. “DETERMINATION OF SULPHITE IN MINERAL SOILS.” Plant and Soil 54, no. 2 (1980): 299–305. http://www.jstor.org/stable/42935238.

I wrote the answer then I noticed that the article was already mentioned in comments. So, I made the answer communtiy-wiki.

I think Andrew is correct in his statement:

I think it's generally assumed that sulfite has a short lifetime in soil, decomposing to $\ce{SO2}$ gas or getting oxidized to sulfate.

Sulfur is found in natural soils as sulfide minerals like pyrites, marcasite, and greigite, and in sulfate forms like gypsum, anhydrite, barite, and jarosite. The S available for plants in agricultural ecosystems is in a dynamic storage, which can be schematically represented as the flow of sulfur in soil (where APS is adenosine 5′-phosphosulfate; Ref.1):

As correctly stateed in Ref.1:

In agricultural practice, sulfur has broad use in the form of sulfate fertilizers and, to a lesser extent, as sulfite biostimulants. In agricultural practice, sulfur has broad use in the form of sulfate fertilizers and, to a lesser extent, as sulfite biostimulants. When used in the form of bulk elemental sulfur, or micro- or nano-sulfur, applied both to the soil and to the canopy, the element undergoes a series of changes in its oxidation state, produced by various intermediaries that apparently act as biostimulants and promoters of stress tolerance. The final result is sulfate $\ce{S^6+}$, which is the source of sulfur that all soil organisms assimilate and that plants absorb by their root cells. The changes in the oxidation states of sulfur $\ce{S^0}$ to $\ce{S^6+}$ depend on the action of specific groups of edaphic bacteria. In plant cells, $\ce{S^6+}$ sulfate is reduced to $\ce{S^2-}$ and incorporated into biological molecules. $\ce{S^2-}$ is also absorbed by stomata from $\ce{H2S}$, COS, and other atmospheric sources. $\ce{S^2-}$ is the precursor of inorganic polysulfides, organic polysulfanes, and $\ce{H2S}$, the action of which has been described in cell signaling and biostimulation in plants. $\ce{S^2-}$ is also the basis of essential biological molecules in signaling, metabolism, and stress tolerance, such as reactive sulfur species (RSS), SAM, glutathione, and phytochelatins.

It is common that when sulfur containing minerals such as pyrites (a sulfide) exposed to atmosperic oxygen, they undergoes oxidation process and exists in oxidation states ranging from +6 to −2 (Table 1), with the most oxidized state in the form of sulfate $(\ce{SO4^2-})$, which is the chemical form that plants absorb from the soil to feed themselves with $\ce{S}$. That oxidation process may undergoes  through sulfite formation, but it may only act as an intermediate in the presence of abundantly excess oxygen.
On the other hand, when sulfite is introduced to soil as $\ce{CaSO3}$, laboratory experiments has confirmed that two transformations of $\ce{CaSO3}$ occurred in soil systems in two ways: (1) decomposition to produce $\ce{SO2}$ gas, and (2) oxidation to calcium sulfate. Conversion to $\ce{SO2}$ occurred in solution and soil at low $\mathrm{pH}$, and acid soils treated with $\ce{CaSO3}$ were initially toxic to seedling root growth. The degree of toxicity was time-dependent, with reduction in toxicity occurring as $\ce{CaSO3}$ oxidized to calcium sulfate. Soil reaction also influenced toxicity, and at soil $\mathrm{pH}$ levels above 6, little seedling toxicity was evident (Ref.2). That may conclude this answer.
References:

Laura Olivia Fuentes-Lara, Julia Medrano-Macías, Fabián Pérez-Labrada, Erika Nohemí Rivas-Martínez, Ema Laura García-Enciso, Susana González-Morales, Antonio Juárez-Maldonado, Froylán Rincón-Sánchez, and Adalberto Benavides-Mendoza, "From Elemental Sulfur to Hydrogen Sulfide in Agricultural Soils and Plants," Molecules 2019, 24(12), 2282 (17 pages) (DOI: 10.3390/molecules24122282).
K. Dale Ritchey, Thomas B. Kinraide and Russell R. Wendell, “Interactions of calcium sulfite with soils and plants,” Plant and Soil 1995, 173(2), 329-335 (DOI: https://www.jstor.org/stable/42947538).

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How many m values are there for the 4f sublevel. In high school so no fancy words please!
How many m values are there for the 4f sublevel? In high school so no fancy words please!

With no fancy words:
The orbital angular momentum quantum number $ l $ is given by the letter you see there: f. Remember that we link this quantum number with integers in the following fashion:

$ l = 0 $: sublevel s.
$ l = 1 $: sublevel p.
$ l = 2 $: sublevel d.
$ l = 3 $: sublevel f. So we got it, $ l = 3 $.

How many orbitals we have? Well, we have $ 2l + 1 = 2\cdot 3+1 = 7 $. These seven orbitals are numbered from the value $ -l $ to $ l $, so you can list them in the following way:
$$ m = \{-3, -2, -1, 0, 1, 2 ,3 \} $$
Remember these $ m $ values are the magnetic quantum numbers. Sometimes you may find them with the symbol $ m_l $.

Huh, "orbital angular momentum quantum number" could be some "fancy words". You sure you wanted answer this thing in particular ;)

Everything is written in fancy words, except the names for $ l $ and $ m_l $... I didn't put those names! But that's how the OP will see them in the exam, for sure!

I was being snarky about OP's question, not the answer. This looks like one that a mod could have already closed if he saw it. But, well, you can answer what you want... I just recommend being a bit more picky in the future ;)

I will, I am new to the site, but I just like to help. Thanks

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Could there be any reaction between piperine and Pink curing salt #2
EDIT: I originally (incorrectly) asked this question re: capsaicin; but found that the actual compound of concern in black pepper is "piperine".
This question originates on a forum dedicated to the art of "Makin' Bacon", in which "Cure #2" a.k.a. "Prague Powder #2", is used.  It contains 6.25% sodium nitrite, 4% sodium nitrate, and 89.75% table salt.  It is typically combined with additional table salt and sugar, and applied as a dry cure to pork belly.  The question is posed as: will the addition of Black Pepper (in "significant" quantity) affect the efficacy of the nitrite in accomplishing its intended "curing" effect?  For example, by virtue of a (potential) reaction between the pepper and the nitrite (or nitrate), a) could the desired "cure" be compromised, or b) could there be undesirable byproducts of any such reaction.  As it is impossible to discern whether meat has been "adequately" or "properly" cured by inspection, discerning any negative effects of adding pepper to the process is not easily done (discounting catastrophic cases, of which none (to my knowledge) are reported.
The structure formula of piperine, as shown by wikipedia:

Black pepper has zero capsaicin - it's not chilli.

@Mithoron - My bad! I assumed all peppers' heat was caused by the same compound!  So, it's apparently "piperine" that I'm wondering about then.

Dunno how English can call chilli a pepper and not a paprika, it's dumb, but whatever... You should edit the q. if you don't want it closed.

Well, there are concerns about using nitrites in food. I think it's best to not use them at all, or at least not much. I doubt anyone analysed it vs piperine specifically and it would be missing the point, but it can nitrosate some stuff, perhaps piperine too, as it is an amide.

Depending on local law, I assume food inspection is not only a visual check (as in by naked eye), but equally one related to biochemistry/microbiology. After all, preservation / inhibiting the growth of unwanted bacteria is the main goal of curing where NaCl already lifts much (not all) of the weight by dehydration. Because wikipedia's article points to Nitrosyl-heme, I speculate the nitrites' action there is more likely as the heme e.g., from blood (name) probably is in higher concentration than piperine. A nitration of piperine by the nitrites likely requires elevated temperatures, too.


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Safely dilute this 35% hydrogen peroxide solution to 3% and store after opening (no stabilizer)?
The problem: eating yellow curry with naan and no spoon for the first time resulted in a wonderful dinner but some permanent yellow stain on my cotton shirt. I assume it's coming at least in part from the curcumin from turmeric.
I've had some success removing bleach-resistant stains with household ~3% hydrogen peroxide. However that's surprisingly difficult to find where I live.
So I stopped by a chemistry supply shop, and they handed me a plastic bottle of 35% $\ce{H2O2}$ solution, specific gravity 1.196.
The Chinese writing on the label are common sense safety instructions (store away from light, not safe for food preparation, etc.).
Unlike household peroxide, and like some kinds of oxidizes in spacecraft (Does the Soyuz spacecraft use chemically stabilized hydrogen peroxide (H₂O₂) to remain ISS-dockable for 200+ days?) the does not appear that there are any stabilizers in the solution.
Question: How can I safely dilute to 3%, and best store after opening this 35% hydrogen peroxide solution?
I plan to dilute enough for my present needs, and try to just screw the cap back on the rest of the bottle, but maybe there's more things to consider?
Click images for full size:

The usual precautions are enough. Gloves are the must to prevent immediate whitening of the skin. If available also protecting glasses. Use clean jars or vessels to prevent catalytic violent decomposition.

@Poutnik and the usual precautions for diluting hydrogen peroxide would be...? If this is something I do usually I wouldn't need to post the question. The concentrated peroxide won't liberate oxygen over time and pressurize the bottle?

To the close voter, how might "How can I safely dilute to 3%, and best store after opening this 35% hydrogen peroxide solution?" require clarity? Why not add a comment and highlight what part isn't clear?

@uhoh, Oxygen bleaches only work in alkaline medium. Curry stain is a two pronged problem. You have oil and oil has dissolved curcuminoids from turmeric. First get rid oil by letting only dishwashing liquid on the stain overnight. After that rub that stain with lemon and keep washing the stain under warm water. Let sunlight work on it and it should be gone. Getting rid of oil is more important.

I have meant general usual precautions for manipulation with potentially harmful chemicals. Yes, it can pressurize the bottle. But it IS usually stabilized, AFAIK.

@Poutnik yes I'd never seen just pure H2O2 sold over the counter, albeit a chemistry supply counter. Maybe it's something I don't actually want to keep around for any long period of time at all, rather than worrying about how to store un-stabilized peroxide.

@AChem Thanks for that! I did hit it with water as soon as I'd noticed and worked in a lot of liquid soap soon after that, it's still in its dried-liquid-soap stage now. I'll continue with that procedure and luckily I have a friend with an all-day Sun-exposed roof who can give it a good dose. Hmm, but now I've got all this peroxide, maybe "How to have fun with 35% peroxide?" is my next question (humor).

@AChem your stain removal tutorial worked like a charm! :-)

@uhoh, Glad to know my proposed stain removing chemistry worked :-) I used to do these sun "bleaching" experiments during school days.


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What happens if you carbonate ethanol?
The electronics youtuber bigclivedotcom has an on-and-off-again series where he carbonates various types of alcohol and comments on the taste. One thing he's noticed is that the stronger the alcohol, the more carbon dioxide it will absorb; so, he recently tried carbonating 'moonshine' (70% lab ethanol, 30% water).
This absorbed a staggering 22g of CO₂ per litre (and ended up completely undrinkable, or at least more so than it was previously).
I know that when you carbonate water you end up with a weak carbonic acid solution based on the H₂O + CO₂ ←→ H₂CO₃ equilibrium. But the references I've found says that carbon dioxide is much less soluble in ethanol than in water because ethanol is less polar. So, I'd expect that the stronger the ethanol solution, the less carbon dioxide would be absorbed. This is the complete opposite of what he actually observed.
So, what's happening here?
Reference: https://www.youtube.com/watch?v=yArcH80PiP4

The thought of drinking this mixture gave me rather intense mixed feelings. What can be more noble than experimenting on oneself?

Well, that's one way to stay sober.

@IvanNeretin - my personal favorite is http://www.jstor.org/stable/1660704 "Concerning the taste of heavy water" by Urey...

@JonCuster I'd note the more recent people who have tasted heavy water and disagreed with Urey's conclusion.

Carbon dioxide is in fact roughly ten times as soluble in ethanol as in water [1](https://doi.org/10.1016/j.fluid.2006.04.017) (meaning you need more dissolved under pressure to get that bubbling effect; alcoholic beverages that bubble or sparkle are mostly water). The figure below from the reference illustrates the increasing solubility of carbon dioxide under pressure as we increase the ethanol content in an ethanol-water mixture.

Carbon dioxide is actually non-dipolar, and while the strong quadrupole enhances its solubility in water, it still fits better with less polar solvents such as ethanol.
Reference

I. Dalmolin, E. Skovroinski, A. Biasi, M.L. Corazza, C. Dariva, J. Vladimir Oliveira (2005). "Solubility of carbon dioxide in binary and ternary mixtures with ethanol and water". Fluid Phase Equilibria
245, 2, 193-200,
ISSN 0378-3812,
https://doi.org/10.1016/j.fluid.2006.04.017.

Do you know if diethylcarbonate formation is at all a factor with high ethanol?

The authors do not seem to say. In water carbonic acid formation is not the main driver of carbon dioxide solubility, so I would guess no.

Huh. Google clearly lied to me (although it's possible I saw 'carbon dioxide is only weakly soluble in ethanol' and forgot to ask whether that was more or less soluble than in water). I can't see the article content, sadly --- is there anything interesting about the chemistry caused by the carbon dioxide reacting with the ethanol, in a way analogous to the production of carbonic acid?

@DavidGiven That seems unlikely. The ethyl analogue of carbonic acid would be ethyl hydrogen carbonate, a.k.a. "etabonic acid"; it's apparently unstable, if it exists as a bound molecule at all, although salts of the corresponding etabonate anion $\ce{C2H5OCO2-}$ are apparently stable enough to be synthesized.

@IlmariKaronen I agree. Carbonic acid is only a minor apecies in the water solution and I would expect nothing different in alcohols.


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Which hydrogen in 3-ethynyltoluene is the most acidic?

Arrange the three types of hydrogens in 1‐ethynyl‐3‐methylbenzene in the order from the most acidic to the least.

From my understanding, the hydrogen on methyl would be the most acidic as the negative charge after removing the H would be in resonance.
The next one would be the the ethyne hydrogen because the negative charge after removing it would stay on a sp-carbon.
The least acidic H would be the one from the benzene ring as the negative charge would fall on a sp2-carbon.
But the answer says the correct order is: ethyne's H > methyl's H > benzene ring's H. Where am I going wrong?

Your approach is fine. It is just that resonance is a thing, and $sp$ acidity is a thing too, and you can't easily tell which of these things is bigger.

People who pose these question seriously should go to school and learn about chemistry.

Well, obvious conclusion from the answer is that sp-carbon effect is bigger than single ring effect, and it's indeed true.

@Mithoron True. I just used to think that the stabilizing effect of resonance is multitudes greater than negative charge on an sp Carbon.

@Satya One approach could be to 1) compute the heat of formation of the structure above, 2) compute/look up the heat of formation for a proton, 3) compute the heat of formation of the deprotonated structure of the above so that one can apply Hess' law. It would be an iteration for all positions with a proton attached. But solvation of the molecules and ions has to be considered, too. Though there still is a need to predict pKa, C-H acidities are not always considered in models based on empiric data like MolGpKa.

@Buttonwood I've seen more pKa values in my life than any reasonable person should ;p In a question like that you either know ethynyl is more acidic than benzyl, or not - the difference in pKa is huge.

Consider how easily acetylene forms salts, compared to aromatic or aliphatic C-H bonds.

Based on the papers I've found, the answer is correct.
According to Bouchoux,[1] ethynylbenzene – an analogue to the compound in question – the $\Delta G$ of deprotonation for terminal $\ce{H}$ of ethyne group is $\pu{1522.2 kJ mol-1}$ while that for ring $\ce{H}$s of benzene at o-, m-, p-positions are $1612, 1614,$ and $\pu{1614.1 kJ mol-1}$, respectively.
According to Koppel,[2] $\Delta G$ of methyl $\ce{H}$ for toluene is $\pu{373.7 kcal mol-1}$ (equivalent to $\pu{1564.23 kJ mol-1}$).
According to Hansch,[3] the Hammett constant for methyl substitution is $-0.07$ and $-0.17$ when it is at m- and p-position, respectively, which means that being substituted by a methyl of phenyl ring would almost not affect its proton acidity.
So, the sequence is ethyne's $\ce{H} \gt$ methyl's $\ce{H} \gt$  benzene ring's $\ce{H}$.

References:

Bouchoux, G. Gas phase acidity of substituted benzenes. Chem. Phys. Lett. 2011, 506, 167-174. DOI 10.1016/j.cplett.2011.03.032
Koppel, I. A.; Koppel, J.; Pihl, V.; Leito, I.; Mishima, M.; Vlasov, V. M; Yagupolskiid, L. M.; Taft, R. W. Comparison of Brønsted acidities of neutral CH acids in gas phase and dimethyl sulfoxide. J. Chem. Soc., Perkin Trans. 2 2000, 6, 1125-1133. DOI: 10.1039/B001792M
Hansch, C.; Leo, A.; Taft, R. W. A survey of Hammett substituent constants and resonance and field parameters. Chem. Rev. 1991, 91, 165-195. DOI 10.1021/cr00002a004


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Will this reaction apparatus work (refluxing under inert atmosphere)

The nitrogen flow comes from the right side, directly into the flask. Then will go up the condenser and through the bubbler to release pressure. I am worried that my solvent that evaporates when refluxing could be dragged up by the nitrogen and I will lose solvent over time... or will it still condense when it hits the cold water from the condenser and come back down and be able to undergo reflux?

Looks OK with a gentle flow of nitrogen

What you really want on top of your condenser in one of these https://www.asynt.com/product/gas-bubblers/


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Methanol Evaporation/Degradation From Surfaces
I've seen resources indicate how quickly methanol in small amounts evaporates from a dry surface in open air. Does this apply to methanol mixed with other substances, such as that in wiper fluid? I'm curious if some residual methanol would just remain on a surface with a small amount of fluid residue (such as a countertop, car interior, etc.) indefinitely until washed off.

Good point! Since methanol and water are miscible:

The remaining liquid water would tend to "hold onto" methanol.
As the water molecules evaporate, that removes heat energy and slows evaporation of methanol.

It would take the methanol longer to evaporate in toto. However, even so, water evaporates readily, too, so there's no need to wash off the remaining mixture, at least under normal use outdoors. If you were to spill a 4 liter bottle of windshield (AKA windscreen in the UK) wiper fluid in a garage, that would be a more serious problem. An ordinary wet/dry shop vac could cause fire or explosion, so repeated wet mopping and wringing of the mop would take a while.

Fortunately nothing like that! Just got splashed with some wiper fluid while standing next to a car a couple of weeks ago and touched a few surfaces in the process of getting cleaned up that I wasn't sure were adequately cleaned after, some of which were in the car itself. Thanks for the reply!

This is not a site for health info, but, generally speaking, though methanol is absorbed through skin, absorption is not rapid, so I'd not worry about a bit splashed on open skin and rinsed. If it saturated clothing, holding it in contact with skin, or if I had to deal with large amounts on a regular basis, there would be reason to take action. Trivia: ethanol was used to reduce toxicity of methanol: https://mdpoison.com/media/SOP/mdpoisoncom/ToxTidbits/2013/June%202013%20ToxTidbits.pdf

Makes sense! I talked to poison control immediately after for the medical advice, was just curious about persistence on surfaces.


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Explanation for the low melting point of zinc nitrate in terms of its lattice energy and electrostatic interactions?
I understand that ionic compounds tend to have higher melting points, but I was surprised to see zinc nitrate having an unusually low melting point. I tried understanding it in terms of lattice energy and the electrostatic interactions between the ions but I don't have a conclusive answer.

Your melting point is not a melting point. Rather it represents the loss of water from a hydrate salt, which commonly occurs at much lower temperature than what would be required to actually liquefy the ions.
Nor does zinc nitrate truly melt at higher temperature; instead it decomposes to an oxide like most other nitrate salts. Thermogravinetric analysis, for instance from Ref. 1, indicates that the process of dehydration and decomposition is complete at 300-400°C.

Gregorio, Flores & Carrillo-López, J & López, J. & Martinez, Rafael & Espinosa, Nestor & Rabanal, Maria. (2014). "Structural and Morphological Properties of Nanostructured ZnO Particles Grown by Ultrasonic Spray Pyrolysis Method with Horizontal Furnace." Advances in Materials Science and Engineering 6. 10.1155/2014/780206.

Thanks for the reply, this leads to another question, so when we observe a "colourless liquid" upon heating initially, what is this substance that we are observing, is it simply aqueous zinc nitrate?

Possibly. Hydrated salts sometimes dissolve into their own water of hydration as well. I see no reference fir this, so I did not mention it in tye answer, but for instance ferric chloride hexahydrate is well known to do this.


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How to find the percentage of chloroderivatives after monochlorination of a compound
There are series of examples in my textbook to decipher the chlorination selectivity.

The example was to find percentage of monohlorderviatives of n-pentane after free radical halogenation giving products 1,2,and 3-chloropentane, (assuming relative reactivity of all H-atoms to be same).

The amount of each isomers formed depends on its rate the formation, therefore, the ratio of amounts (i.e., relative amounts) is equal to the ratio of rates.

Relative amount of products=(Total number of equivalent Hydrogen)*(Relative Reactivity)

In 1-Chloropentane, there are two 1°Carbon so six 1°H atoms.
After this I cannot understand how to get 4 and 2 H-equivlents for followup products.

First, I tried to do it the same way I did for n-butane (six 1°H and four 2°H equivalents for chlorbutane and 2-chlorobutane, respectively.) But my answer was too wrong, I searched some sites to find how to find what I was missing?
It seems to me, at least my initial approach was right, H-equivalents means 1°/2°/3° H-atoms and the rest is to go by formula to get the answer? I tried again, but I am still no able to understand how are 4 and 2 equivalents for the rest of product?
Do H-equivalents means something else and what key component am I missing?
Here is one of the solution (without assuming same reactivity for H equivalents):

C1 and C5 have 6H, C2 and C4 have 4H, C3 has 2H. With simplifying assumption of the same reactivity, these numbers are ratio of products. Sure, relative reactivity of primary and secondary C has to be considered. The picture says secondary carbons are 3.8 times more reactive.

There are 12 H-atoms of three different types in pentane. They are equally reactive, so the distribution of products will be the same as the proportion of each type of H-atom in the molecule. (You are not chlorinating 1 chloropentane). So what is the problem? The first calculation looks right.

One thing to point out that you don't have tertiary carbon in pentane. Other than methyl groups at the ends, other three carbons are secondery. That's why they were considered extra 3.4 times reactive (reactivity of $\ce{R2CH^.}$ vs $\ce{RCH2^.}$).

A normal alkane $\ce{C_nH_{2n+2}}$ has two sorts of Carbon atoms : $1$) terminal $\ce{C}$ atoms, which hold $3$ $\ce{H}$ atoms, and $2$) internal $\ce{C}$ atoms, holding only $2$ $\ce{H}$ atoms. Removing one $\ce{H}$ atom from anywhere in the molecule produces n-alkyl radicals $\ce{C_nH_{2n+1}}$. And these radicals are primary $1$-alkyl if the $\ce{H}$ atom was attached to the terminal position, and secondary alkyl if the $\ce{H}$ atom comes from internal $\ce{C}$ atoms.
The first alkane with both terminal and internal $\ce{C}$ atoms is propane $\ce{C3H8}$ with $n=3$.
In propane $\ce{C3H8}$, the first and the last $\ce{C}$ atoms are terminal atoms. Removing one of the $3$ $\ce{H}$ atoms par terminal $\ce{C}$ atoms produce $1$-propyl radicals. Each terminal $\ce{C}$ atom can produce $1$-propyl radicals.  Removing one of the two internal $\ce{H}$ atoms produce $2$-propyl radicals. In total removing one $\ce{H}$ of the eight $\ce{H}$ atoms in propane can produce six different $1$-propyl radicals and two different $2$-propyl radicals.
For butane $\ce{C_4H_{10}}$, there are two terminal $\ce{C}$ atoms and two internal $\ce{C}$atoms. Removing one $\ce{H}$ atom among the $10$ available can produce $2· 3 = 6$ different $1$-butyl radicals (the same number of $1$-alkyl radicals as for propane). But removing one of the $4$ $\ce{H}$ atom attached to one of the two internal $\ce{C}$ atoms can produce $4$ different $2$-butyl radicals. In total, removing one $\ce{H}$ of the ten $\ce{H}$ atoms in butane can produce six different $1$-butyl radicals and four different $2$-butyl radicals.
Now I think you can discover yourself why pentane $\ce{C_5H_{12}}$ can be chlorinated to produce $12$ possibilities of chloropentane : six $1$-chloropentane, four $2$-chloropentane, and two $3$-chloropentane

Thank you! 6,4,2 possibilities for respective chlorinated products. Chlorination selectivity has more to with possibility of that derivative existing than 1°/2°/3° of hydrogen. That approach worked in n-butane because of more straighforward structure but not in n-pentane or more complicated compounds. The book did not clarify in details. Your explanation was very helpful.


text:
Why is the dipole moment of Anisole less than that of Phenol?
According to Wikidata, anisole has a dipole moment of 1.38D and according to this page on Researchgate, phenol has a dipole moment of 1.70D.
Resonance in anisole:

Resonance in phenol:
According to my reasoning, $\ce{-CH3}$ can show a good +I effect and hence the positive charge on oxygen is more stabilized in the canonical forms and so the negative charge must be more delocalized and hence anisole should have a higher dipole moment. My concern also extends to the reaction of both anisole and phenol with Lewis acids like $\ce{AlCl3}$ in case of the Friedel-Crafts reaction. This paper states that phenol forms a complex with $\ce{AlCl3}$ and consequently its activity is decreased. But this states that anisole prefers C-acylation. Now, with some hints from this answer, I have come to believe that the reason for the more easy Friedel-Crafts reaction by anisole is due to its less dipole moment causing it to react less with the Lewis Acids and activate the ring more. But it still doesn't make sense to me, and I cannot explain to myself how even the dipole moment of anisole is less than that of phenol. Please explain in regard to Friedel-Crafts also. Thanks!

Short answer is phenol have pKa of about 10 and so that negative charge on oxygen is much prominent than that in anisole. Hence the difference in dipole moment.

related https://chemistry.stackexchange.com/questions/139822/estimation-of-dipole-moments-in-aromatic-compounds

The answer lies in the fact that OH group is more activating than OR group that is, it increases the electron density inside the benzene ring to a greater extent.
This can be explained by Bent's rule. Since carbon is more electronegative than hydrogen, the lone pair bearing orbital of Oxygen in OR gets enriched in the percentage s character, which decreases its tendency to donate electrons via mesomeric effect (as the s character in a hybrid orbital increases, the electrons in that orbital become more tightly bound with the atom).
Indeed it is true as you pointed out, that the alkyl group exerts a +I effect, although it is not as significant as compared to the mesomeric effect.
Now once we have established this, the dipole moments can be easily compared. As the donating tendency of the group attached to the ring increases, the electron density of the ring increases and that on the attached group decreases, thus increasing the dipole moment.
The explanation about the Friedel Crafts acylation can be given by the same reasoning. The Oxygen in Anisole does not donate its electrons as effectively and hence does not form an effective complex.
I hope that cleared your doubt.

Based on Pauling's assigned electronegativites (EN), oxygen is more electron demanding (EN: 3.44), than carbon (EN: 2.55), or hydrogen (EN: 2.20).  Hence the bonds $\ce{O-C}$ and $\ce{O-H}$ are polarized towards oxygen.
Conversely, hydrogen (in phenol), or the methyl group (in anisole) are electron donating toward oxygen though this inductive effect by the methyl group exceeds the one by a hydrogen alone.  Hence – already more satisfied by the electron donation by the methyl group than by the hydrogen – the polarization between the phenyl group and oxygen is attenuated. This yields a less pronounced molecular dipole moment for anisole, and retains a higher (remaining) pi-electron density in the arene which can react as a Lewis base, e.g. in the course of a Friedel-Crafts reaction.


text:
What parameters do you use to find the order in which elements will leave a solution during electrolysis?
I am looking to perform electrolysis on a molten solution of many elements, specifically a mix of iron, sodium, potassium, silicon, aluminum, and titanium oxides and chlorides. What I'm trying to figure out is how to determine what order the elements will be reduced in, and how this is determined. Some ideas I have are to base it off of the electronegativities or the Gibbs energies, but I'm not sure how feasible calculating the Gibbs energies even is in such a messy solution.
What parameters or formulas would be used to determine this?

The published potentials are valid for aqueous solutions. They are not valid for molten and anhydrous solutions like here. I am afraid you would have to establish this order yourself by trails and errors. .

You might research what others have found in similar experiments, e.g., http://dspace.mit.edu/handle/1721.1/98589 or https://www.researchgate.net/publication/268400452_Direct_Electrolysis_of_Molten_Lunar_Regolith_for_the_Production_of_Oxygen_and_Metals_on_the_Moon , or even contact authors directly.


text:
DIY Turbidity Standards?
I am trying to help my high school chemistry class make their own turbidity standards to calibrate the arduino turbidimeters they are making. Last year we tried calcium carbonate as it makes a cloudy solution but I don’t think that the concentrations were very accurate since there was so much precipitate sitting on the bottom. It was recommended to me to use milk, but the idea of their standards being perishable is non ideal (and I hate to think of what would happen if a kid spilled milk all over their circuit board!) Does anyone know of any relatively safe, relatively available compounds that would produce a consistently cloudy solution for turbidity standards?

So, why would milk not be good, again? You need actual colloid, not suspension like CaCO3. I guess there's some other milk-like stuff around, but I doubt spilling it would be any better.


text:
What range of relative humidity is present in the combustion byproducts of a natural gas stove, immediately after combustion?
If I have a traditional gas range-top, burning natural gas (instead of LPG or propane), what is the typical range of humidity present in the combustion byproducts immediately above the flame? The combustion process adds water to the air, and the natural gas has an inherent humidity as well. However, the combustion process also dramatically raises the temperature, thus significantly drying the air.

Why would there be any relative humidity immediately after combustion? Temperature is higher then boiling point of water.

@Mithoron No problem as long as there still exists a saturation pressure at that temperature. For example, pure steam at 120 °C and 1 bar has a relative humidity of about 50 %. Whether this is useful information, is a different question.

Whoa, generally in SE we don't downvote without leaving explanations, esp. for new contributors. If the question is dumb, please explain why. FWIW, it looks like there is an answer, which is that it is undefined.

Well, you just shouldn't ask about relative, but absolute or specific humidity. Relative one is highly dependent on temperature, even if it was below critical temperature. We don't really know what you want to find out here, I'm afraid.

@Mithoron thanks for the clarification. I'm curious if the combustion process makes the air generally wetter or generally drier. I've taken classes in thermodynamics related to powerplants, so I've never worked with temperatures so high (or low) that we've discussed critical points as something other than curiosities.

Well, then I guess it's mostly about your misconception in the last sentence. I don't get why you'd think heating dries air.

@Mithoron relative humidity changes with temperature. I guess it comes down to whether we consider air is dry because of RH or molar mass. I'm happy to learn my usage of the word "dry" is wrong, and subsequently reword my question.

So again this relative humidity... the air isn't getting any drier, it's just that the saturation vapor pressure increases. That's again a reason not to use RH here.

Burning 0.1 m3 of NG provides (after mixing with air and cooling) formally 0.2 m3 of pure vapor. If the room has 50 m3, it is 0.004 m3/m3. Saturated vapor pressure at 25 deg C is 3130 Pa, what is approximately 0.03 m3 vapor/ 1 m3 air. 0.004/0.03 .100 = approx 13% of the room relative humidity increase. For lower T it would be more. Note also that relative humidity falls to by rule of thumb to 1/2 by air warming up by 10 deg C. So the actual change if relative humidity would very much depend on the rate of burning and of room cooling down.

Assuming ideal conditions starting from dry air, the theoretical mixture immediately after complete combustion contains $19\ \%$ $\ce{H2O}$ (molar fraction) since the oxygen from the air is replaced according to
$$\ce{CH4 + 2 O2 -> CO2 + 2 H2O}$$
However, the adiabatic flame temperature in air (not oxygen) is $2236\ \mathrm K$, which is well above the critical point temperature of $647\ \mathrm K$. Therefore, the relative humidity (i.e. the  ratio of the partial pressure of water vapor to the saturation vapor pressure) is not defined at such temperatures immediately after combustion.

"relative humidity  is not defined at such temperatures immediately after combustion." Thanks for this, I had no idea. I'm going to have to read about that to get a better sense of what it even means.


text:
What does the χ (chi) symbol represent in a Slater determinant?
Context: Computational Chemistry
So, I am looking at the molecular orbital approximation, which allows the assumption that the overall wavefunction can be written as individual molecular orbitals
Ψ(r1,r2..,rn) = ψ1(r1) x ψ2(r2) x .... ψn(rn)  (the numbers and n are subscripted)
However, the above equation is incorrect because it does not take into account spin. Therefore, when spin is taken into account, the molecular orbital ψ(r) is replaced by molecular spin orbitals χ(r,ω). The r and ω are often combined and denoted as xi (the i is subscripted). Does this mean that χ = ψ? If that is true, I don't understand why the symbol was changed from ψ to χ? If someone could explain. Thank you.

"the numbers and n are subscripted" — please take a look at FAQ: How can I format math/chemistry expressions on Chemistry Stack Exchange?. That said, it's quite hard, perhaps even impossible, to answer your question. Where did you get this equation from? What is the context within which it was written? It is very possible that the author is just making a simplification (albeit an incorrect one) where spin is ignored. But we can't really read their mind; and certainly not when we only have one equation of their writing to work with.

(Like, sure, you said it's about comp chem and the MO approximation; but that aspect of it is already quite clear from the equation you quoted, at least to anybody who's studied it before. The real question is: what did the author write before and after the equation? In their article or book, have they consistently ignored the spin degrees of freedom, using only $\psi$'s (spatial orbitals) instead of $\chi$'s (spin orbitals)? Or did they actually cover spin, and are just using a different notation from what you're expecting? These are all possible answers; but it's impossible to say which.)

Hello, thank you for linking the FAQ. I'll use it in the not too distant future.

Hello, thank you for linking the FAQ. I'll use it in the not too distant future. The book I am reading, Computational Chemistry by Jeremy Harvey (Oxford Primer), is where I found this equation. The section of the book this equation is from is introducing the Hartree-Fock theory. In the build up to the equation, spin is ignored. Shortly after the introduction of the equation, the author states the equation is not quite right as it ignores spin. Then the spin coordinate, ω, and spin part is introduced, with spin up and down. Then ($r_i$, $ω_i$) = $x_i$

($r_i$, $ω_i$) = $x_i$   Is the combination of the spin and spatial variable for a given electron. A few statements are then made. The first is about how electrons are need to be treated as identical particles. The second is about swapping electrons being antisymmetric. As a result of these statements and the spin terminology introduced, the equation in the question becomes ψ($x_1$,$x_2$,$x_3$,$x_4$)=-ψ($x_1$,$x_3$,$x_2$,$x_4$) for a 4 electron system.

I think maybe there is some confusion in terminology. Maybe these notes will help you:

When we take spin into account, we combine the spatial orbital of the ith electron $ \psi_i(\mathbf{r}) $ that depends on  the position with the spin function (of course we have two of them), $ \alpha(\omega) $ or $ \beta(\omega)$ that depend on the spin coordinate. Formally we don't combine $ \mathbf{r} $ and $ \omega $, we take the product of $ \psi_i(\mathbf{r}) $ and $ \alpha(\omega) $ or $ \beta(\omega) $. However, we combine these two coordinates in $ \mathbf{x} = (\mathbf{r},\omega) $.
The product of this new function is the spin orbital $ \chi(\mathbf{x}) = \psi(\mathbf{r})\alpha(\omega) $ or $ \chi(\mathbf{x}) = \psi(\mathbf{r})\beta(\omega) $ .
What we want to calculate is the N-electron wave function, this is the guy you wrote there in the left-hand side $ \Phi(\mathbf{x}_1, ...,\mathbf{x}_N) $. However, when I read the title of the question you talk about the Slater determinant, maybe you wanted to write the following
\begin{equation}
  \Phi(\mathbf{x}_1, ...,\mathbf{x}_N) = \frac{1}{\sqrt{N!}} \det
  \begin{bmatrix}{}
   \chi_i(\mathbf{x}_1) & \chi_j(\mathbf{x}_1) & \cdots & \chi_k(\mathbf{x}_1) \\
   \chi_i(\mathbf{x}_2) & \chi_j(\mathbf{x}_2) & \cdots & \chi_k(\mathbf{x}_2) \\
   \vdots & \vdots & \vdots & \vdots \\
   \chi_i(\mathbf{x}_N) & \chi_j(\mathbf{x}_N) & \cdots & \chi_k(\mathbf{x}_N) \\
  \end{bmatrix}
\end{equation}

So, in summary:

When we take spin into account, we change $ \psi $ for $ \chi $.
What goes in the Slater determinant is always $ \chi $.
The N-electron wave function is the capital $ \Phi $.

Hello, thank you very much. The second point answered my question. Also, thanks for the additional information on the Slater determinant.

My pleasure! I assume you are studying the Hartree-Fock method, it is tough, somewhat heavy math, but at the end is very, very rewarding...


text:
How to explain the acidity difference between these β-diketones?
In DMSO, pentane‐2,4‐dione (1) and 1,3‐diphenylpropane‐1,3‐dione (2) possess nearly identical $\mathrm{p}K_\mathrm{a},$ whereas 1‐phenylbutane‐1,3‐dione (3) is almost 1 $\mathrm{p}K_\mathrm{a}$ unit weaker than the former two:

#
Name
Structure
Solvent
$\mathrm{p}K_\mathrm{a}$
Ref

1
Pentane‐2,4‐dione

DMSO
13.33±0.05
[1]

2
1,3‐Diphenylpropane‐1,3‐dione

DMSO
13.36±0.06
[1]

3
Phenylbutane‐1,3‐dione

DMSO
14.2
[2]

It seems that we needn't take solvent effect into consideration, as DMSO is the solvent. It can be easily explained why (2) is more acidic than (3), as the extra benzene ring could provide more stabilization of the anion through conjugation.
How can (1) be so acidic even though the two methyls cannot provide any significant stabilization?
References

Olmstead, W. N.; Bordwell, F. G. Ion-Pair Association Constants in Dimethyl Sulfoxide. J. Org. Chem. 1980, 45 (16), 3299–3305. DOI: 10.1021/jo01304a033.
Bordwell, F. G.; Harrelson Jr, J. A. Acidities and Homolytic Bond Dissociation Energies of the αC—H Bonds in Ketones in DMSO. Can. J. Chem. 1990, 68 (10), 1714–1718. DOI: 10.1139/v90-266.

You're getting into minuscule things here. Rings here are cross conjugated with anionic part - don't delocalise the charge mesomerically. One can get into solvation effects etc. but I think the difference is likely an artefact of method - they didn't get the data the same way, or did they?

The OP's question is:

How can (1) be so acidic even though the two methyls cannot provide any significant stablization?

Answer to that is simple. In Table II of Ref.1 shows in the entries 1-3 that $\mathrm{p}K_\mathrm{a}$ values would effected less by the groups attached to other sidede of the carbonyl groups other than those decreasing the negative charge density of carbonyl carbon such as ethoxy (entry 4) or amino groups or keeping the fixed geometry (entries 5 & 6) of enol once deprotonated. Entry 1 is pent-2,4-dione, which has shown $\mathrm{p}K_\mathrm{a}$ of $13.33 \pm 0.05$. Entry 3 is 1,3-diphynylprop-1,3-dione that has shown $\mathrm{p}K_\mathrm{a}$ of $13.36 \pm 0.06$. The $\mathrm{p}K_\mathrm{a}$s of entries 1 & 3 have shown not much different values even though end methyl groups in entry 1 are replaced by phenyl groups in entry 3, meaning electron withdrawing inductive effect of phenyl ring is overcome by its electron donating mesomeric effect (evidently). However, sigificant increase in $\mathrm{p}K_\mathrm{a}$ has been shown when the change has made to the $\alpha$-carbon of the entry 1. When a methyl group is added to $\alpha$-carbon of the Entry 1, which would become 3-methylpent-2,4-dione, has shown $\mathrm{p}K_\mathrm{a}$ of $15.07 \pm 0.06$ (Entry 2), a 1.74 $\mathrm{p}K_\mathrm{a}$ units increment. Thus, it is safe to say that changes ar reaction center is much more prominently effect the $\mathrm{p}K_\mathrm{a}$ of substrate than changes to any other places.
Followings are some other insights to drastic changes in $\mathrm{p}K_\mathrm{a}$ values of some concerned organic acids. According to the Ref.1, it seems the value of apperent $\mathrm{p}K_\mathrm{a}$ of an organic acid in DMSO is depended in so many conditions such as cation of the base used, its concentration, etc. Eventhough Ref.1 described the condition in details, Ref.2 doesn't even if the principal author in both papers is same. As Ref.1 pointed out the the geometry of the resultant enol plays a role as well.
The Ref.3 provided another dimension affecting $\mathrm{p}K_\mathrm{a}$, which is the steric bulkiness.  For example, the effect of bulkiness is evident in the values given in Ref.2. When one hydrogen in each methyl of acetone is changed to methyl group (pent-3-one), the $\mathrm{p}K_\mathrm{a}$ has been changed from 26.5 to 27.1 (0.6 $\mathrm{p}K_\mathrm{a}$ units). The change could be purely electronic and/or mix with steric. But, change of bulkiness is to isopropyl groups (2,4-dimethylpent-3-one) the change of $\mathrm{p}K_\mathrm{a}$ is by 1.7 units, showing clear steric effect.
Then again, When one hydrogen in the methyl of acetophenone is changed to a methyl group (1-phenylprp-1-one), the $\mathrm{p}K_\mathrm{a}$ has actually been decreased from 24.7 to 24.4 (-0.3 $\mathrm{p}K_\mathrm{a}$ units). The change could be purely electronic influenced by presence of phenyl group. However, the change is inceased to an isopropyl group (2-methyl-1-phenylprp-1-one), the $\mathrm{p}K_\mathrm{a}$ is increased by 1.6 units (from 24.7 to 26.3), again showing clear steric effect.
References:

William N. Olmstead and Frederick G. Bordwell, "Ion-pair association constants in dimethyl sulfoxide," J. Org. Chem. 1980, 45(16), 3299–3305 (DOI: https://doi.org/10.1021/jo01304a033).
Frederick G. Bordwell and John A. Harrelson, Jr., "Acidities and homolytic bond dissociation energies of the $\alpha \ \ce{C-H}$ bonds in ketones in DMSO," Can. J. Chem. 1990, 68(10), 1714–1718 (DOI: https://doi.org/10.1139/v90-266).
Michelle H. Dunn, Nicholas Konstandaras, Marcus L. Cole, and Jason B. Harper, "Targeted and Systematic Approach to the Study of $\mathrm{p}K_\mathrm{a}$ Values of Imidazolium Salts in Dimethyl Sulfoxide," J. Org. Chem. 2017, 82(14), 7324–7331 (DOI: https://doi.org/10.1021/acs.joc.7b00716).

I have not (and cannot) read these papers. Did they consider anything in regards to the keto-enol equilibrium affecting the values? Should that even be considered? I know that the equilibrium is heavily dependent on the solvent, but I cannot remember where it lies with DMSO.

@Martin-マーチン:  None of these papers have considered equilibrium constent of keto-enol. But REF.1 HAS SAID (it's also in the abstract) three enol geometries exist and affected the ion pair in three different ways, But pKa is calculated for infinite dilution.


text:
Stability of different tautomers of vitamin C

Above all four tautomers represent Vitamin C (Ascorbic acid)
So here are some basic rules that I learnt.
1.Usually keto form is more stable due to bond enthalpy considerations
2.Enol form is stable if :
  (1)It is aromatic (diamagnetic ring current stabilization)
  (2)It is highly conjugated
  (3)it has 5/6 membered intramolecular hydrogen bonding
  (4)keto has highly acidic alpha-hydrogen
  (5)keto is cyclic 1,2 diketo
  (6)keto is 1,3 dicarbonyl

Using these rules I was able to conclude :

4th structure is least stable
2nd structure is 2nd least stable

But for the 1st and 3rd structures both satisfy 2(3) point given above.
What is the final stability order?

Only two of structures are tautomers. Other two are oxidized forms.

@mat not really, oxidised form has three carbonyls. Check how double bond goes from C=C to C=O.

@OP You're wrong about 3rd structure, enols of carboxyls are much worse, even than ketones.

@Mithoron why is that?

one theoretical explanation for comparing and 1st and 3rd structures could be that the carbonyl group of the cylic ester (in structure 1) has an oxygen with higher negative charger density due to inductive and electronic effects, as compared to the oxygen of the carbonyl group in structure 3. This results in stronger hydrogen bonding?


text:
Why are orbits shown out of plane in popular cartoons of atomic structure?
The (former) U.S. Atomic Energy Commission had a seal showing a symbolic view of the atom that is unrealistic but instantly recognizable:

Source: https://blog.nuclearsecrecy.com/2012/04/13/friday-image-the-atomic-energy-commission-seal-1949/
This symbol was part of the visual language of the atomic age, used by towns, Disney, General Electric, and unicode (U+269B) starting in 1948 or earlier.
Taking the depicted image literally, this would represent the atom beryllium (because it shows 4 electrons). The blog author in the source points out that the most important part for nuclear energy (the nucleus) might have been shown with more detail. In fact, the current agency uses the word "nuclear" instead of "atomic".
While we know today that electrons behave very differently, I am wondering why the electrons are shown in orbits that are at an angle to each other. Many cartoons of atomic structure do this (the most common one has three orthogonal orbits), and I am sure many pre-date this particular example.

If the inspiration for Bohr's model was the solar system, wouldn't you show the orbits all in the same plane, with different radii like in the solar system, like the image above? Was there something known at the time Bohr formulated the model (or before 1949, when the seal was created) that would motivate out-of-plane orbits (maybe angular momentum) while sticking with the inaccurate Bohr model? Or is this a purely artistic choice, helping to make the cartoon more compelling and three-dimensional?
UPDATE: Bohr's drawing do not look like today's textbook drawings of it. Here is a diagram of Xenon (1921-1923).

Source: https://physicstoday.scitation.org/doi/10.1063/PT.3.1978
The atomic whirl symbol for an atom is often associated with the Rutherford model, even if Rutherford said very little about the position or states of the electrons (perhaps because classical physics did not have a model without contradictions, and a satisfying model required the combination of experimental data with quantum theory). In Rutherford's 1911 book "Radioactive Substances and their Radiations", he states:

So he leaves open whether electrons are spherically distributed or present in concentric paths in a plane, and acknowledges the earlier hypothesis by Nagaoka of a positively charged center with electrons surrounding it in some way (Saturnian or otherwise). At this time, the neutron had not been discovered, and it was not clear where the high-energy electrons of  beta radiation came from (from the nucleus or elsewhere).

Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on [meta], or in [chat]. Comments continuing discussion may be removed.

this has bothered me for over a half-century; I wonder if migration to, or a related question posted in HSM SE might be productive? Something like "What was the first..." or Was this ever justified/defended/explained scientifically?" or something along those lines.

@uhoh This post states that the Rutherford model is fuzzy on electronic structure. It does not, however, say when the "atom whirl" was first used. I found sources all the way back to 1948. During WWII, nuclear physics was top secret. Later, there was an effort to make the public comfortable with nuclear power (while also training them how to use fall-out shelters in the cold war). This was probably the time when the Rutherford model (with balls for electrons rather than some fuzzy quantum representation) was popular to popularize nuclear physics.

@uhoh I posted another question here: https://hsm.stackexchange.com/q/15248/16431

indeed you did, looking forward to its answers!

Similar to coat of arms and traditional principles of vexillology, a good logo balances the provision of enough detail to be recognizable – regardless if it is mounted to a lectern, or on a business card – while avoiding to much detail of information. This is what Tufte describes as aim for a high data-ink ratio
$$ \frac{ \text{data ink} } { \text{total ink} } =
\frac{ \text{elements conveying data information} }
{\text{all elements of the chart}}$$
with intentional remove of detail within reason (ref).  See, for example, the flags of the states, D.C., and territories of the United States of America:

(image source)
– at this scale of representation, many are perceived with some content in their centre.  The ones of Texas, New Mexico, Colorado are among those which stand out.
From perspective of chemistry/physics, the depiction of the beryllium atom is an improvement in comparison of the seal of AEC because the finer structure of the atom nucleus.  At the same time the model is wrong for that diameter of atomic nuclei are orders of magnitude smaller than atomic diameter – in case of helium $\pu{1 fm}$ (or $\pu{10^{-15} m}$) of the nucleus, and $\pu{1 Å}$ (or $\pu{10^{-10} m}$) for the atom.  Hence, a depiction of atom and atom  nucleus in a scale in common often was and remains impractical – see e.g. the logo of the IAEA including an off-centered atom nucleus:

(source)
the Polish Nuclear Research Centre:

(source)
or the logo of The Big Bang Theory

(source)

Tufte focused on representation of data, if memory serves me right, not on creation of logotypes.

Not to mention that electrons don't have a size comparable to that of a nucleus, as depicted in the logos?

I perceive both the representation of data, as well as logotypes as transmission of  information, where intentional white space not used can help in recognition and discern (similar to some fonts of same x-height easier to read at low size/resolution, than others), and hence, related.


text:
Confusion in carbocation rearrangement

In the above dehydration reaction (asked in a competitive examination) I have shown two paths, obviously only one of them is correct.
According to test correct answer is product formed through A but I think other one should be the correct major product :
After the first carbocation forms, it would rearrange; after rearrangement to A by methyl shift, number of alpha hydrogen in A is only 6 (carbocation being 3°) while in B, number of alpha hydrogen is 7 (Hydride shift). B also being 3° carbocation. So yielding more stable carbocation it should rearrange to B and further corresponding Saytzeff product should form.
Also we can't apply migratory aptitude here since there is no scope of back bonding.
So who is right and why ? Please explain.

You might consider the reversibility of the reaction which would favor path A.

So who is right and why?

The person who made the question is correct (OP said according to the test, correct answer is the product formed through inter mediate $\bf{A}$, because that major product, 1,2,3-trimethylcyclohexene is the thermodinamically most stable product (assuming reaction is set under correct condition for thermodynamic control). Since the whole process is reversible except for the last step to give product, there are other posibilities to obtain several minor products mainly based on the conditions applied. Energy barrier to remove those protones should be harder that that to remove tertiary protons in intermediate
The migratory aptitude matters if the intermediate stability increased by the migration. From initial carbo cation, there are two migratory choices: hydride shift or alkide shift, both of which give most stable tertiary carbocations $\bf{B}$ and $\bf{A}$ intermediates, respectively.
Although hydride is superior to alkide in migratory aptitude, methide shift is dominate here because it gives the thermodinamically most stable product with tetrasubstithted double bond. There is also possibility that $\bf{A}$ intermediate can rearrange again by shifting another hydride to give a third tertiary carbocation where positive charge is in the Carbon-2 (of 1,2,3-trimethylcyclohexane frame). That gives an additional tertiary proton (one on C1 and the other on C3) to leave giving the identical product.
The disadvantage on $\bf{B}$ intermediate is it has two secondery protons to leave giving the two different products, 2,4,4-trimethylcyclohexene (as shown in OP's scheme) and 1,3,3-trimethylcyclohexene $\bf{A}$. Thus, those products must be produced as minor products.
Note: There is also posibility to have two cyclopetene and cyclopenten products by ring consruction rearrangement based on condition used. I leaved it to OP to envision.


text:
Why is the melting point of 4-bromoacetanilide so high? Does it have any reason to do with the chemical structure, or the para aspect?
We used melting point to determine the type of bromoacetanilide, and it was closest to 4-bromoacetanilide. How can I explain why the melting point was what it is?

You can't.$,!$

@Ivan There, there, it's probably just supposed to be about some "explanation" why it's m.p. is higher than of other isomers. Some packing effect should be OK I think.

Melting points of solid organic compounds can be affected by several factors. Symmetry is one of these factors. Acetanilide, 2-bromoacetanilide, and 3-bromoacetanilide have melting points in similar range, respectively about $\pu{114}$, $96$-$\pu{101}$, and $87$-$\pu{89 ^\circ C}$. Yet, melting points of 2-bromoacetanilide and 3-bromoacetanilide are significantly lower than that of acetanilide, even without extra bromine atom. Apart from this, 4-bromoacetanilide has the highest melting point among them, $177$-$\pu{181 ^\circ C}$. Thus, you can safely say bromine have nothing to do with their melting point differences.
Both acetanilide and 4-bromoacetanilide have a symmetry element (plane of symmetry) that allows increased $\pi-\pi$ stacking among individual molecules during their crystallization (see the correct educational guess of Mithoron in his comment above). That may be the reason for higher melting points for them compared to compounds lacking that symmetry element. For example, if you replace the bromine with chlorine or iodine, the resulting compounds shows the same trend:
$$
\begin{array}{lr}
\text{Compound} & \text{Melting poing} \\\hline
\text{4-Chloroacetanilide} & \pu{178-179 ^\circ C} \\
\text{3-Chloroacetanilide} & \pu{77-78 ^\circ C} \\
\text{2-Chloroacetanilide} & \pu{87-88 ^\circ C} \\
\text{4-Iodoacetanilide} & \pu{184-185 ^\circ C} \\
\text{3-Iodoacetanilide} & \pu{119 ^\circ C} \\
\text{2-Iodoacetanilide} & \pu{109-111 ^\circ C} \\\hline
\end{array}
$$
Even if the substitution pattern, not the substitution group, affected the melting point of above compounds, there is a peer review paper (Ref.1) showing both substitution pattern and the substitution group have affected on resulting compounds' melting points (interestingly). In an useful table, the result describe th decreasing melting points in 1,4-dihalogenated phenyl derivatives when a halogen or an electrondonating group is introduced as the third substitution (Ref.1). This is actually supporting our caurse here. The parent compound is 1,4-dihalogenated phenyl derivative, which have plane of symmetry element. The introduction of third substitution distrupt this symmetry element and hence lower the melting point. For instance, the melting point of 1,4-dichlorobenzene is listed as $\pu{53.5 ^\circ C}$. The melting point of 2,5-dichloroaniline (when amine is introduced to ortho-position of one chloro group) has been decreased to $47$-$\pu{50 ^\circ C}$.
References:

G. B. Heisig, "Anomalous Effect of Ortho-Para Orienting Groups on the Melting Points of Dihalogenated Benzene Derivatives," J. Am. Chem. Soc. 1928, 50(1), 139–145 (DOI: https://doi.org/10.1021/ja01388a017).


text:
Why do diastereomers have different physical and chemical properties?
While going through my book (Organic Chemistry, Paula Bruice) I came across the following line:

We have seen that enantiomers have identical physical properties. They
also have  identical chemical properties—that is, they react with a
given achiral reagent at the same  rate. Diastereomers, on the other
hand, have different physical properties (meaning different melting
points, boiling points, solubilities, specific rotations, and so on)
and different  chemical properties —that is, they react with a given
achiral reagent at different rates.

There seems to be no further explanation about why this is so, and I am unable to figure out the reason behind the difference in physical and chemical properties of diastereomers. I do understand why it would be so in the case of cis-trans isomerism, but not in the case of optical isomers.
What would be a plausible explanation for this?
Any help would be greatly appreciated.

Your title and your text say different things. Title "Why do diastereomers have different physical and chemical properties?". Text "I am unable to figure out the reason behind the difference in physical and chemical properties of enantiomers. [...] optical isomers".

In any case, the answer to the question you are asking in the title is "asymmetry". If you could not tell left from right, you would not be able to distinguish L-tartaric acid from D-tartaric acid. On the other hand, you would be able to distinguish L or D from meso-tartaric acid, e.g. by the fact that in a particular standard drawing of the molecule the two OH groups are on the same side vs on opposite sides. See https://en.wikipedia.org/wiki/Tartaric_acid. An achiral reagent is like a person who cannot tell left from right.

@user6376297 thanks, I have corrected the typo in the question, I intended to ask about diastereomers. Your answer was quite helpful, thanks a lot


text:
The extinction coefficients of oxymyoglobin and ferryl myoglobin
I want to know what the extinction coefficients of oxymyoglobin and ferryl myoglobin are. I was able to find the one for the oxymyoglobin in this pretty old article (it says that it equals $121 \text{mM}^{-1}\text{cm}^{-1}$), but I couldn't locate any other reference supporting this claim. Furthermore, I couldn't find any reference in the case of ferryl myoglobin.
I am also curious if, in general, there are tables containing the extinction coeffcients of various compounds (at different wavelengths).

Which organism are you interested in? They will be very similar among mammals, but not identical. Also, for which wavelength do you want to know it?

@Karsten I am interested in horse heart Mb and I want to know it for the absorption maximum, so somewhere around 410-425 nm for the Soret band.

Which pH are you interested in?

@Karsten I am interested in pH 9.


text:
Why change of gibbs energy per mole is zero at equilibrium?
I was reviewing this concept and thought of this: $\frac{dG}{dt} = \frac{\partial G}{\partial \xi}\frac{d \xi}{d t}$, for a reaction to be at equilibrium  $\frac{d G}{d t}$ must be zero but I have seen in the formula $\Delta G_r=\Delta G^o_r+RT\ln(Q)$ , the condition for equilibrium is $\Delta G_r=\frac{d G}{d \xi}=0$. My question is, why is necessary $\Delta G_r=0$ to be in equilibrium if it's enough $\frac{d \xi}{d t}$ is zero and that make $\frac{d G}{d t}=0$ ? Is there an explanation of this phenomenon?

See What exactly is minimized when we say that the Gibbs Free Energy of Reaction is minimized at equilibrium? on Physics.SE. Note that Gibbs is the scientist's name, and math operators (\ln) as well as text labels (\Delta_\mathrm{r}G^\circ) should be upright.

What is $t$ here, if it is time it does not enter into classical thermodynamics, which only considers states at equilibrium.  $\xi$ is the extent of reaction so when the change of this with $G$ is zero (i.e. gradient zero) we are at equilibrium and $Q$ is replaced by $K_e$ the equilibrium constant. The extent of reaction is zero initially and $1$ when one mole is converted to product. At eqm $\Delta G_r^\text{o}=-RT\ln(K_e)$.

@porphyrin Thanks for replying, but you cannot describe the extent of reaction as a function of time and their initial reactants? then because a chemical reaction is not quasistatic it cannot be described with classical thermodynamics? I understood this well?

I'm not sure I quite understand your comment, but using a rate equation of course allows [C] vs. t to be measured and so, eventually, an equilibrium constant. But classical thermodynamics does not involve time and only informs us about equilibrium states and changes in energy/entropy etc. in moving from one equilibrium state to another and not about how it happens.

I have never seen, in the context of chemical reaction equilibria regarding classical thermodynamics, that the Gibbs free energy is a function of time.
For a single phase, open system, the total Gibbs energy is a function of the pressure, temperature, and moles of all the species $ G = G(P, T, n_1, n_2, ... n_N)$. Therefore, the total differential of the Gibbs energy is
$$ dG = VdP - SdT + \sum_j \mu_jdn_j \tag{1} $$
By stoichiometry, the moles of chemical species $ j $ can be written as $ n_j = n_{j,0} + \nu_j\varepsilon $, with $ \varepsilon $ defined as the extent of reaction. Thus, at constant pressure and temperature, the condition of chemical equilibrium states that Eq. (1) takes the form
\begin{align}
  dG &= \sum_j \nu_j \mu_j d\varepsilon  = 0 \\
  \dfrac{dG}{d\varepsilon} &= \sum_j \nu_j \mu_j = 0 \\
  \dfrac{dG}{d\varepsilon} &= \Delta_r G = 0 \\
\end{align}
And if you plug $ \mu_j = \mu_j^0 + RT \ln(\hat{f}_j/f_j^0) $ you end up with that equation you wrote
$$ \Delta_r G = \Delta_r G^0 + RT\ln \prod_j\bigg(\dfrac{\hat{f}_j}{f_j^0}\bigg)^{\nu_j} $$
$$ \Delta_r G = \Delta_r G^0 + RT\ln Q$$
So if you want to be systematic:

For a single chemical reaction at constant $P$ and $T$, chemical equilibrium reduces to $ dG/d\varepsilon = 0 $.
Some unknown author (at least to me) decided to call $ \sum_j \nu_j\mu_j = \Delta_r G $.
Solving $ \Delta_r G = 0$ means that you guarantee $ dG/d\varepsilon = 0 $.

The chain rule that you made there is mathematically fine, but in the context of classical thermodynamics, time is not a variable. If you want to know how things change in time, you resort to kinetics. Time may appear when solving transient balances. However, it is always explained that those jumps in time are between equilibrium states, that theoretically take an infinite amount of time. This is to guarantee that you can apply the thermodynamic relations, if not, you can't use them.
Thermodynamics will give you no information of the state of the system at a certain extent of reaction $ \varepsilon $. However, it will tell you that the extent of reaction will not vary when $ t\rightarrow\infty$, and will also give you that value $ \varepsilon=\varepsilon^\text{eq} $.

Thank you for answering my question, I edited my question to fixed those partial derivatives and thanks for giving me the insight of the theorical amount of time needed for a reaction to be in equilibrium. However, that doesn't answer the question of why $\Delta_r G$ needs to be zero at equilibrium. I hope if you could clarify it. Thank you

At equilibrium, the Gibbs free energy has a minimum value, so in the context of a chemical reaction at a constant $P$ and $T$ this means that its derivative with respect to the extent of reaction is zero, i.e., $ dG/d\varepsilon = 0 $, or $ \Delta_r G = 0 $. It seems this is not the question. You want to know why at equilibrium the Gibbs free energy has a minimum value? Why we put a zero there?

Yes, that sounds a better way of phrasing, why Gibbs free energy has a minimum? Is there a principle behind this? Specifically why at equilbrium? Because the rate of extent of reaction approach to zero at equilibrium so I don't see why $\Delta_r G$ needs to be zero if it's enough with the reaction in forward and backward at equilibrium and the sum  of both individuals $\Delta G$ is zero because the rate of reaction of both reactions are the same and their respective $\Delta G$ cancelled each other.


text:
Why is solubility affected by temperature?
Why when we raise temperature  generally  a bigger concentration of a  soluble gets dissolved in a inorganic solvent?Is there a quantum mechanical reason why this is happening?

Quantum mechanical? That's just basic thermodynamics.

@Mithoron explain please.I never really understood it.

Just analyse how Gibbs energy of dissolving changes with temp.

https://chemistry.stackexchange.com/questions/21855/why-do-some-solid-chemicals-dissolve-better-in-colder-water

Dissolution requires energy changes, which is temperature related.

Firstly that is strictly not true. Most solutes that are gaseous at the temperature will be less solvated the higher the temperature of the solvent liquid, for instance. There are other relevant examples as well.
Secondly, if you do not understand this, there is no reason to go digging into quantum chemistry. That is a game best left alone untill you are good and ready.
The most reasonable way to understand this is through understanding energy and thermodynamics. Most substances solvation are favorable to increasing temperature in terms of enthalpy. Energy is required to break the bonds of the pure sample, energy is required to displace solvent-solvent bonds and energy is gained from forming solvent-solute bonds. With temperature all these enthalpies change in magnitude. There is also entropy, the entropy of mixing - which is a strong contributor to increased solvation at elevated temperatures.


text:
Copper plating surface color shifts from light to dark orange to black in less than 12 hours
We have been doing barrel-type copper plating on small parts, and all of a sudden there is an increase of copper plating discoloration after copper plating. The color shifts from bright orange (good) to reddish purple and brown, and finally to black in less than 12 hours. I know that oxides of copper are causing this change, but I don't know yet the cause of abrupt discoloration of the surface. Copper oxides naturally, but not this rapid. What parameters should I check? I already checked the process parameters (plating current, current density, loading ratio) and all are the same as before the discoloration had happened. In terms of plating solution, we are using copper pyrophosphate.

You are correct that the copper is oxidizing. Electrolyte left over on the surface would promote that reaction just as it is well known to do with iron rusting, so check your rinsing and drying.


text:
Colored inorganic liquids: why is SCl2 red, and S2Cl2 orange?
Those do not involve d orbitals, so there is no crystal field theory or ligand theory. Do we have a name of study that only involves p orbitals? Since I think molecular theory is for all s p d f orbitals and transitions.

According to ChatGPT, is due to leaving 2 unpaired electrons in an antibonding orbital, the interaction between these unpaired electrons causes a splitting of the energy levels, resulting in the absorption and reflection of light.

If you feel you have been personally affronted on the site you can use flags or send a message to a community or SE mod so that we can intervene as necessary. It is not usually useful to start a confrontation.

I am deleting your comments on this post as they are not about chemistry. You may start a chat with Poutnik about the meaning of the term "anti-Stokes", but please avoid unnecessary personal arguments.

Also, please note that ChatGPT has a very bad reputation as an unreliable source of information.

Maybe a reader invested in quantum chemistry can perform a computation to estimate which electronic transitions were plausible (and their corresponding gap).

@Buttonwood damn, I thought this was a question for inorganic experts. Thank you for letting me know that this is really a question for quantum chemists.

@Buttonwood when you said "Maybe a reader invested in quantum chemistry can perform a computation to estimate which electronic transitions were plausible" does that mean this question was not answerable before the age of computers? How would a computer program be able to solve this question?

@NealConroy In H2S2, the S-S bond is 2.06 A, in S2 the S=S bond 1.89 A.  For S2Cl2, the S-S distance is 1.95 A, as if there were a little bit of double bond character. Perhaps this is a contribution. For elements of the second period (after Ne), the empiric octet rule does not apply as strictly as on elements up to and including Ne, see e.g. DMSO. Others with said background likely know better.

@Buttonwood oh, so the computational chemistry is to calculate the bond-distance part?

@NealConroy Rather the other way around. The distance Wikipedia reports can be a result of a X-ray / neutron diffraction experiment, microwave spectroscopy, or ... what else. This distance to be shorter than the S-S single bond could be indicative for a small contribution by d orbitals. It is more common for folks in quantum chem than org chem to recognize if such an extension to describe the bonding in S2Cl2 with this additional contribution is significantly better; they can perform the relevant computations (at different levels of theory) and compare the results with data from experiments.

@Buttonwood so you said the S-S single bond is more closer in length to the S=S double bond, indicating a small contribution of d orbitals. Okay now look at my very 1st comment-reply at the top, where I pasted what ChatGPT said, you think what it said could be plausible? ;0

@NealConroy As stated, this enters territory beyond my knowledge, where referral to others who actually are in the field / possibly in the know is the best I can do for now. And I mark the question as «follow» to be notified if there is any advance in future by an answer by an other subscriber.


text:
Confused about underlying reasoning behind reaction quotient/equilibrium constant
I am a high school student taking AP Chemistry. In the unit on chemical equilibrium, there is much emphasis placed on calculating reaction quotients and equilibrium constants using this method: $$aA + bB \rightleftharpoons cC + dD$$
$$Q_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}$$
It has been explained that equilibrium constants much larger than 1 indicate that the reaction favors the products, and equilibrium constants much smaller than 1 indicate that the reaction favors the reactants. I am curious why this is.
I am familiar with the notion that the expressions in the numerator and the denominator are proportional to the probability of particle collisions for both respective sides of the reaction. However, I do not wholly understand the reason behind this, or how this relates to the idea that the equilibrium constant can quantify the degree to which the reactants or products are favored.
I also understand that the reaction quotient is another way of expressing the ratio of the rate constants, $\frac{k_f}{k_r}$, but I do not understand how this, too, relates to the reaction constant's ability to quantify reaction favorability.
Edit: Just to clarify, I understand that when the quotient is greater than 1, the numerator is greater, and vice versa. My confusion is about why the concentrations raised to their stoichiometric coefficients quantify reaction favorability.

The fact that "concentration raised to a stochiometric coefficient" is the right form was first observed experimentally, and proposed as a law. Nowadays, we can also derive this from first principle thermodynamical laws. You can read more about it on Wikipedia, but there are also several questions on this StackExchange, like Is there a reason for the mathematical form of the equilibrium constant?.

@Domen Thanks for the links. These seem very helpful.

Q is a fraction which, like every fraction, could be defined as A/B.
If A > B, Q = A/B is > 1 and the numerator, depending on the final products, is bigger than the denominator, depending on the reactants. The products are more important than the reactants, at the equilibrium.
If A < B, this is the contrary : Q < 1. The reactants are more important than the products.

Thank you for your answer. I understand this much about Q. I am confused why the numerator and denominator accurately quantify the "importance" of the reactants or products. Why has the form concentration raised to a stoichiometric coefficient been used to denote "importance"?

In my comment, the word "Importance" is equivalent to the "product of concentrations raised to a stoichiometric coefficient".

The quotient can take on many values, initially suppose that no product is present then its value is zero. Only at equilibrium do we normally know its value, which is the equilibrium constant, unless we solve the differential equations in time and get the changing concentration ratios that way. At equilibrium the concentrations obviously have their equilibrium value and then the quotient becomes the equilibrium constant which is also the ratio of rate constants $K_e = k_f/k_r$ because only here is the rate of forwards and reverse reactions equal.
If $K_e >1$ there are clearly going to be more products that reactants and so the reaction 'favours the products' but in a more general sense when this is the case the expression $-RT\ln(K_e)$ give the 'free energy' and when this is negative the reaction is called 'spontaneous', although that absolutely does not mean that it will occur at all rapidly. Texts on thermodynamics go into this in detail.

Thank you for your response. It has been very helpful to think about the fact that the rate law is the derivative of time vs. concentration.


text:
What's the exact mechanism of spin coupling in NMR?
The most common one that's found nearly in every textbook and site is that splitting occurs due to the alignment of neighboring proton either with or against the external magnetic field so as it's B effective is increased or decreased, However in Pavia's "Introduction To Spectroscopy" a different mechanism of coupling is described, in chapter 7.1 -Although the first mechanism is mentioned in chapter 5.1- where Dirac vector model is used with pauli exclusion principle to explain coupling, It doesn't state or mention the first mechanism, and actually I'm pretty confused how come both of two mechanisms can be used to explain the same phenomenon, They seem to be contraindicated with the first mechanism stating to be a matter of alignment with external field and second one stating to be a matter of internal electron and protons alignment, I understand both but I'm just not sure I can use them in the right place.

In a system with only one spin, the energy levels available to that spin depend on the strength of the external magnetic field (there is a direct proportionality).
In more complex systems, spin–spin coupling happens when the energy levels of spin 1 are affected by the spin state of spin 2, i.e. whether spin 2 is pointing up or down. This can happen in (at least) two ways:

through-bond coupling, also known as scalar coupling and referred to with the symbol $J$. This is primarily (though not exclusively) mediated by the Fermi contact interaction, which is all the electrons-in-bonds and Pauli exclusion principle stuff that you may have seen.
This only works if the two spins are 'connected' via a series of bonds that isn't too long.

through-space coupling, also known as dipolar coupling and referred to with the symbol $D$. This is because spin 2 generates its own magnetic field, which can either oppose or reinforce the external magnetic field.
This doesn't require the two spins to be linked by covalent bonds. However, in solution-state NMR where molecules rotate freely, the dipolar coupling is averaged to zero* and is therefore not observed in the spectrum.

These two interactions are not mutually exclusive, and the observed coupling can be a sum of both of these. However, in solution-state spectra you only really see scalar coupling because the dipolar coupling is averaged out, and people tend to loosely use the more general term 'coupling' to refer to scalar coupling.
Chances are, what you've been reading is a mixture of both. Practically speaking, for an introductory NMR course on organic molecules, you will only ever really encounter scalar coupling. Dipolar coupling is important in more advanced NMR, as well as solid-state NMR, but most people don't really deal with that at the undergraduate level.

* Higher-order effects of dipolar coupling, such as relaxation and the NOE, do still persist in solution.

I'm very thankful for this explanation, However I don't understand why if it's almost the scalar coupling we're encountering most textbooks and common websites uses the explanation of dipolar coupling, The literature doesn't seem consistent to me at this point, Our doctor talked only about the dipolar mechanism referring to - and using - diagrams of scalar coupling, Is it just a huge misconception?

I can't really comment without looking at the specific source you're using, unfortunately.

Pavia's "introduction to spectroscopy" Chapter 5.1, also if there's a way i can upload Our doctor's PowerPoint. However I think this well enough for my question, Just one last question, If coupling depends on carbons which are non chemically equivalent then why there's a coupling constant for benzene ortho and para protons even though they're the same?

Yup - I don't have that book on hand, though, sorry. Anyway, you're probably asking about magnetically equivalent spins, which is not exactly the same thing as chemically equivalent. Magnetically equivalent spins still have a coupling constant between them, it's just that we don't see it in the spectrum, because of quantum mechanics (see e.g. https://chemistry.stackexchange.com/q/53478/16683). It's a bit weird, but that's how it is.

Can't describe how much your answer helped me, it took me a whole day searching and trying to understand how both the mechanisms relate, Also that coupling constant puzzled me and I couldn't move further, Thx so much!


text:
Haloalkanes gives acid base reactions or SN2
Consider the following reaction:

I am confused whether $SN_2$ reaction will take place or acid - base reaction.
According to me:
This should be an acid - base reaction since the carbonanion(On carbon where chlorine is attached) formed will be stabilized by $p_\pi - d_\pi $ backbonding.
But in many questions, I have seen that they proceed through $SN_2$ mechanism.

They do both, this will give a mixture of alkene and alcohol

Which would be the major?

depends on the conditions

Like I can decide if a polar protic solvent is given to me?

Solvent will be one factor, the choice of counterion of the hydroxide will also be important


text:
Why doesn't oxygen form cyclic pentagonal compounds with itself?
Like we see in benzene, the bond angle between the oxygens of 104.5 degrees is very similar to a pentagon's interior angle of 108 degrees. The formal charges on the oxygens would all be zero and there's no dipoles, weak bonds or areas of high electron density so surely it would be somewhat stable and unreactive? So why don't they form?

https://chemistry.stackexchange.com/questions/51719/do-cyclic-oxygen-molecules-exist-and-their-theoretical-stability

See also: https://chemistry.stackexchange.com/q/35574 and https://chemistry.stackexchange.com/q/431 - I should have checked for previous similar questions

Sulfur does form ring compounds, so why not oxygen?
You're not wrong to wonder, but I think people are reacting to the implication that this species has:

[no] weak bonds

Consider hydrogen peroxide H-O-O-H which has a central O-O bond. $\ce{H2O2}$ is fairly well known to decompose easily and generate radicals.
In fact, there are a variety of peroxides with that R-O-O-R motif. The peroxide refers to the O-O single bond.
In general, if you look up bond dissociation energies for R-O-O-R bonds, they're ~150-160 kJ/mol, about half the dissociation energy for $\ce{H3C}-\ce{CH3}$ and far less than the 498 kJ/mol for the $\ce{O2}$ double bond.
So it's perhaps possible to form such a ring, but with five O-O single bonds, it's likely to decompose readily, presumably into $\ce{O2}$ and $\ce{O3}$ first, which would both have greater bond strengths.


text:
What does the 'S' stand for in base-centered cubic crystals and Bravais lattices?
The 'I' used for body-centered systems is for the German word Innenzentriert and the 'F' for face-centered is for Flächenzentriert, also German but what about the 'S' for a, b, or c base-centered crystals?

S also has German origins: seitenflächenzentriert (side face centered).
Reference: International Tables for Crystallography, Volume A, Space Group Symmetry, Fifth Edition, Springer 2005, pg. 748.
One can trace a 1922 article: Niggli, P. "XII. Die Kristallstruktur einiger Oxyde I." Zeitschrift für Kristallographie-Crystalline Materials 57, no. 1-6 (1922): 253-299
.
A German speaker can identify if this Fraktur S or C. Hard to assess for me.
In general, beware of symbolic stories. The above reference is authentic. Sometime, people tend to invent stories about symbols. My recent discovery was that of a sinc function. The whole web and plenty of books call it sinus cardinalis. It turned out to be a fictitious story. Similarly, most of Schoenflies symbols are not his inventions, and neither are Mulliken's symbols his.

That is a C :) You can check it, e.g., using the Unifraktur Font on Google Fonts: https://fonts.google.com/specimen/UnifrakturMaguntia?preview.text=C2h%20and%20S2h&preview.text_type=custom&query=unifrak


text:
Why is a C4 ring impossible?
I am admittedly not a chemist. I was playing with a molecule set recently and I created a ring molecule where I double-bonded four carbon molecules to each other in a ring.

I was in my university's chemistry department, and a passing biochemist said that such a molecule was impossible, but they couldn't explain why.

There is a thing called "ring strain".

Rings containing 4 carbons do exist but not with all double bonds. Cyclobutane is commercially available and many other compounds contain a cyclobutane unit. But it has strained bonds compared to butane. And adding double bonds makes the strain worse. Cyclobutadiene (two double bonds) is unstable above temperatures of 37 K! though some derivatives are known. Trying to add another two double bonds will increase strain further pushing it into "impossible" territory.

Related: Hydrocarbons with only 4 carbon atoms

Recently asked for oxygen: https://chemistry.stackexchange.com/questions/172272/why-doesnt-oxygen-form-cyclic-pentagonal-compounds-with-itself

I think the important question is why you'd think it was an OK molecule?

@and yeah, particularly this comment https://chemistry.stackexchange.com/questions/19475/hydrocarbons-with-only-4-carbon-atoms#comment133633_19485 :D But Dave's answer probably the most, even if less funny.

@Mithoron Because 0K < 37K? :)

@Mithoron: I thought it was OK too until I saw everybody else was talking about double bonds. I didn't see a problem with a single-bonded version of it except for it's going to be really unstable in an oxygen atmosphere.

@david Cyclobutadiene is hardly thermally unstable, just extremely reactive.  On the other hand square C4 would be at best some transition state. It's not a matter of temperature.

@Mithoron I was making a joke about OK (Okay) vs. 0K (zero K).

A four C ring does exist in many compounds, but it is not possible for double bonds.
Cumulated double bonds $\ce{C=C=C}$ must be linear, in 1 line. They can be bent a little, but it would cause the strain in the bonds, causing instability.
They definitely cannot be bent in right angles. The right angle geometry excludes possibility of overlap of orbitals to form such double bonds.

Knowing ~nothing of chemistry, would a Carbon with four single bonds have the least strain with them ~120 degrees apart?  So you could make a low-strain loop of single bonds with.... 6 Carbons?

Knowing ~nothing of chemistry, your answer makes me think that it can occur, but would simply be "unstable" and ~immediately disconnect? And therefore that maybe a 5-Carbon ring of single-bonds would be less unstable, and therefore maybe "possible"?

With single bonds, it would not be carbon only, but other atoms would be there, typically, but not limited to, hydrogen. Like (CH2)4.

@MooingDuck 6 carbon loops with single bonds or with up to 3 double bonds are very common - for example in graphene and benzene.

@MooingDuck, it's possible with four single bonds (plus hydrogens or other substituents).  It's also possible with two single and two double (plus hydrogens, etc) -- the name of the base molecule of this form is cyclobutadiene.

Right, yes, I did assume the other bonds were connected to other things, probably hydrogen. Is something like (CH2)5 possible, with it's 108 degree bend?  or (CH2)7, which is really close with it's 128 degree bend?

Ring C4 does exist, but it is usually called "rhombic C4".  It is not square, having the shape of a rhombus.  The length of each side is about 1.52 Angstroms and the shorter diagonal is 1.45 Angstroms.
There are two lone pairs, 5 sigma bonds, and one pair of delocalized electrons.

(drawing from Isomers and excitation energies of C4 J. Chem. Phys. 84, 3284 (1986) )
See the following references for more information:
Electronic structure calculations on the C4 cluster J. Chem. Phys. 124, 234304 (2006)
Conversion of Linear to Rhombic C4 in the Gas Phase:  A Joint Experimental and Theoretical Study J. Am. Chem. Soc. 2000, 122, 29, 7105–7113
Observation of laser excitation of rhombic C4 using the coulomb explosion method Zeitschrift für Physik D Atoms, Molecules and Clusters volume 26, pages 340–342 (1993)
Cn(n=2−4): current status Phil. Trans. R. Soc. A. volume 376 (2018)

Ring strain and principles of aromaticity render such a molecule unstable.
Each double bond contributes 2 π electrons, and with four double bonds in one cycle in common, there is a total of 8.  Compounds with (4n + 2) π electrons (where n is an integer number) are considered as Hueckel aromatic with an additional energetic stabilization e.g., in benzene compared to a hypothetical cyclo-1,3,6-cyclohexatriene.  Compounds with 4n π electrons (e.g., 8) however are considered as anti aromatic – not only that they do not benefit the stabilization, they are extra unstable.
It does not say that there is not such a molecule (chemistry in stars / in mass spectroscopy allows for molecules and ions you can't store in bottles on the shelf), but – because of lack of overlap of atomic orbitals to establish bonds and thermodynamics (enthalpy of formation) – very unlikely.
Consider yourself lucky that the molecule set's bonds did not break constructing a cyclobuta-1,2,3,4-tetraene.

I'm not sure I can follow the aromaticity argument. Why would the electrons be confined to the same pi system, when there are plenty of other orbitals to use? I'm quite confident that apparent instability has more profound reasons.

Yeah, it if you go from cyclobutadiene, then two additional "bonds" are perpendicular to the original ones. At such angles they would be very tenuous (lack of overlap, indeed). Thing is while single cyclobutadiene molecule is perfectly stable by itself, just reactive, this most likely would ring open faster than it was made.

Cyclo-carbon with 4 atoms is potentially more likely to exist as cyclobutadiyne, which has two triple bonds instead of four double bonds.
It has been mentioned at least once before in the literature in an article with theoretical computations of hypothetical forms of carbon.
Lopanov, A. N., and E. A. Fanina. "Enthalpies of the transitional forms of carbon." Solid Fuel Chemistry 49.2 (2015): 99-108.
A cyclo-carbon with 18 atoms is known to exist and has been synthesized.

The cyclo(18)carbon is an example where discrete single and triple bonds are conjugated, a pattern like $\ce{-C#C-C#C}$.  (Cyclooctratetraene is an example that this pattern of saturated/unsaturated C-C bonds equally is known for alkenes, $\ce{-C=C-C=C}$. In the presented structure of cyclobutatetraene however, double bonds would be as consecutive / adjacent as in allenes or cumulenes.


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What is the vapor pressure when the vapor and some inert gas is in a sealed steel container?
Suppose that there are two chemical species in a sealed container(1L). X(l) is in phase equilibrium with X(g). The another one is Helium gas, which is 2 atm. If the vapor pressure of X at the given temperature is 0.5atm, what is the partial pressure of X in the sealed container? Should I take helium into account while calculating the pressure of X?

Your problem is not clear. The partial pressure of helium is 2 atm, and the partial pressure of X is 0.5 atm. Does it mean that the total gas pressure is 2.0 atm. Or is it 2.5 atm ?

Actually, I was asking the same question as you. And I think the answer is 2.5atm, since the partial pressure of X is 0.5atm.

Vapor pressure of X = the partial pressure of X at equilibrium conditions.
The total pressure has to be considered only in context as it very slightly increases vapor pressure of X by slightly increasing the chemical potential of the liquid. (You need more energy to add some volume of liquid to a system with higher pressure.)

Thank you for your kind answer! Now I understand that vapor pressure is the partial pressure that the volatile one has. Did I understand well? Also,  Can I have some more explanation about chemical potential increase? I don’t understand it very well.