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Stability of geometrical isomers in cycloalkanes
Among the following, which should be the most stable compound?
1)Cis-cyclohexane-1,2-diol
2)Trans-cyclohexane-1,2-diol
3)Cis-cyclohexane-1,3-diol
4)Trans-cyclohexane-1,3-diol
My thought process is-cis isomers with adjacent OH groups lead to Hydrogen bonding so 1) should be the most stable, I'm not sure if repulsion will be the dominating factor and change the answer- what is the correct way to think about this?

This question seems a little bit invalid since 1,2-diol cannot be directly compared to 1,3-diol counterpart. From hydrogen bonding you can tell that cis- counterpart wins in stability but both are not comparable without actually knowing the values.

To answer this question, you needs to know  what are the values for each acting forces on the cyclohexane frame work. Without that information, nobody would know if the repulsion forces (steric) would or would not be the dominating factor over hyrdogen bonding (electronic). Once we know the answer for that, you'd be able to calculate and find out the most stable structure. For your convenience, I attached a diagram with the chair conformations of all four structures and their Newman projections to make you understand the hydrogen bonding capabilities:

As you see, if the substitutions are not capable of hydrogen bonding, it is obvious that 1,3-cis structure is the most stable considering only stecic factors (eq-eq conformer). 1,3-trans structure (eq-eq conformer) is also stable relative to others but has an additional gauche interaction of vicinal bulky groups (1,2-substitutions) compared to its 1,3-counterpart.
In your case, however, all four molecules contain hydroxyl groups, so that three of them could be able to have structure stabilizing hydrogen bonding (the exception is trans-cyclohexane-1,3-diol; see Newman projection of IV-eq-ax). Both cis- and trans-cyclohexane-1,2-diols are structurally similar to 1,2-ethanediol, which favors a gauche conformation instead of an anti conformation, because only the gauche conformation allows hydrogen bonding between the two vicinal hydroxyl groups. Thus, both cis- and trans-cyclohexane-1,2-diols would adopt gauche conformations (see Newman projections of I-eq-ax and II-eq-eq) Ref.1. Keep in mind that, for cis-cyclohexane-1,2-diol (I), the anti conformation is impossible because of the restriction of ring structure. In trans-1,2-cyclohexanediol (II), however, the anti conformation is physically possible, but it would require the hydroxyl groups to be in diaxial positions, which is much less less stable than diequatorial due to 1,3-interactions imposed by axial 3,5-hydrogens (ax-ax chair confirmation).  According to the literature (Ref.1), cis-cyclohexane-1,2-diol is more acidic than trans-cyclohexane-1,2-diol by about $\pu{1.5 kcal/mol}$, hence 1,2-cis-compound would impose stronger hydrogen bonding than its trans-counterpart.
According to the same study (Ref.1), cis-cyclohexane-1,3-diol is $\pu{10.9 kcal/mol}$ more acidic than trans-cyclohexane-1,3-diol, thus cis-compound is highly stable than its trans-counterpart (also remember, trans-cyclohexane-1,3-diol cannot have hydrogen bonding (vide spupra)). However, between conformers, the diaxial and diequatorial conformations in cis-cyclohexane-1,3-diol are relatively close in energy, so it is not immediately clear which conformation is dominant (Ref.1). Yet, a solution NMR study of this molecule showed that the diequatorial conformer (III-eq-eq) is more stable than diaxial conformar (III-eq-eq) Ref.2, although  III-eq-eq is impossible to make hydrogen bonding (see Newman projections of III-eq-eq and III-ax-ax). This shows that hydrogen bonding itself cannot overcome the steric repulsions induced by the 1,3-ax-ax-interactions. My conclusion is without knowing values of these attraction and repulsion interactions, it is impossible to say which of the three structures, I, II, and III), is the most stable.
References

X. Chen, D. A. Walthall, and J. I. Brauman, Acidities in Cyclohexanediols Enhanced by Intramolecular Hydrogen Bonds, J. Am. Chem. Soc., 2004, 126(39), pp. 12614–12620.
R. J. Abraham, E. J. Chambers, W. A. Thomas, Conformational analysis. Part 21. Conformational isomerism in cis-cyclohexane-1,3-diol, J. Chem. Soc., Perkin Trans. 2, 1993, (6), pp. 1061–1066.

The most stable compound should be cis-cyclohexane-1,3-diol because the formation of the hydrogen bond and less crowding.
the less stable  is the tran-cyclohexane-1,3-diol there is no way of making hydrogen bond.
the second more stable is the cis-cyclohezane-1,2-diol it also has a hydrogen bond and then with little difference the trans-cyclohexane-1,2-diolñ

How can I tell whether 1,3 is too far to form a hydrogen bond or not? and whether crowding dominates or H-bonding in cis-cyclohexane-1,2-diol?

if you use any 3D program to write the molecules you can find the data.

I'm preparing for an exam where no relevant data, or even calculators are provided.  So I can't take the help of models while learning to answer these questions.


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Is sugar necessary for the sweetening effect of food?
A lot of food packages, nowadays, are mentioning "0 g sugar" or "sugar free". Then how do those still taste sweet?

Lead acetate is sweet and isn't sugar.  Though I'm sincerely hoping that's not what is in your sugar-free snacks.

Interesting. This is new info. Thanks @AlaskaRon. No, it's not in my diet. Haha! As long as I don't plan to kill myself!!

@AlaskaRon Who tasted that?

https://chemistry.stackexchange.com/questions/27901/what-chemicals-help-define-taste

@Abcd Everyone who put it into mouth ;) Check out the link^

The protein Thaumatin is found in some tropical fruits and is many, many thousands of times sweeter than sugar.

There are plenty of sweet things that are not sugar(s)
People have a sweet tooth (or, to put it another way humans like sweet things). This is an evolutionary adaptation because fresh fruits are both good food (both nutritionally and in providing lots of energy) and rare in pre-modern food gathering cultures. So we are adapted to like things that are sweet as eating them is good for us (in moderation).
In the modern world we can have as much sweet stuff as we want but this is no longer a survival advantage as too much sugar provides more energy than we need to survive or thrive and eating it just makes us fat.
So the modern food industry has adapted to give us sweet-tasting things that don't overload us with unneeded energy. Many of those food ingredients mimic the sweetness of sugar without containing many calories, so allow us to eat sweet food without the same risk of consuming too many empty calories and getting fat.
There are many chemical that taste sweet but are not sugars. The successful ones tend to tase a great deal sweeter than the typical sugars found in fruits and related plants (glucose, sucrose, fructose and others). For example, aspartame is hundreds of time sweeter than sucrose but is a peptide not a sugar; Saccharin is an aromatic suphimide and is also hundreds of times sweeter than sucrose; acesulfame K is similar to saccharin.
There are also sugar-like sweetness that the body cannot digest but which taste sweet. Steria is a complex sugar with sweetness but few calories; Sucralose is a modified sugar the body can't digest so tastes sweet without containing the calories of sucrose.
The point of the claims on "sugar-free" foods is that they are sweet but non-calorific. Plenty of compounds can mimic the sweetness of real sugars like sucrose but are calorie free either because they are far sweeter (so much less stuff is required for the same level of sweetness) or because the body can't digest the compound and turn it into unneeded calories.

Thanks Matt. This is good info. But I would like to ask you the same question which asked Todd. "This 'sugar' that people with diabetes mellitus are advised not to consume in large quantities, does that include only sucrose sugar or all these kinds of sugar?"

@TheDude Diabetics are not affected by sugars you can't digest. They are affected by most simple natural sugars (fructose, glucose, sucrose etc.). If a product says it is "sugar free" the sweeteners it contains won't affect them. even the sugar-derived artificial sweetness are not a problem for them.

Wow! I guess, I now have a holistic, confusion-free idea about this whole 'zero sugar' issue. Thanks again Matt for enlightening.

Most of the sweetening is done by using sugar alcohols such as xylitol, which allows producers to legitimately claim that there is "no sugar" in their products, as long as "sugar" is considered in the traditional sense (sucrose).
Stevia - another sugar substitute - is quite widely used. The Wikipedia entry states that as far back as 2006, stevia accounted for 40% of the sweetener market in Japan (for instance).

These are rather rare in comparison with aspartame and few others.

Wow! Sugar alcohols now! Ok, so this 'sugar' that people with diabetes mellitus are advised not to consume in large quantities, does that include only sucrose sugar or all these kinds of sugar?

@TheDude - Many of the sugar alcohols (and some of the other substitutes) have non-zero glycemic index values and do affect one's insulin response, so the answer is "yes." For example, see this UCSF page on sugar alcohols with respect to diabetes.

@Todd Thanks a lot. Not only was this reply helpful in better understanding this issue, but also the link you posted is resourceful and I've bookmarked it for future reference. So, thanks again!

It is wrong to group stevia with sugar alcohols as it is a very different type of sweeter and has no nutritional calories at all. And sugar alcohols do not account for "most" sweetening agents in foods.

As matt_black said, polyols (sugar alcohols) are not the predominant non-sugar sweetening agent.  Low-calorie sweeteners are far more common. Per unit of sweetness, low-calorie sweeteners are much less expensive, on average, than polyols.  Yet, in spite of this, the global market for low-calorie sweeteners is (in dollar value) about 5 - 10 times that for polyols.  That should give you a picture of how much more of the sweetening is done by low-calorie sweeteners than by poloyols

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

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Mix quaternary ammonium compounds with washing powder
So to get rid of bad smells from clothes that have stayed wet for long, I am wondering if I can just add a disinfecting product to my wash in addition to regular washing powder.
I have a product called Plano Odorclean. The active ingredients are as far as I could find 5-10% alkyldimethylbenzylammonium chloride, 5-10% didecyldimethylammonium chloride, 1-5% ethoxylated isotridecanol. It has a pH of 6.5.
My question is, will any of this react and somehow neutralize with a washing powder, for example Ariel Color?
I am thinking of putting the powder in and as the water carries it away, add 20 ml of the disinfecting product into the running water. So then the clothes would be washed with a mixture of washing powder and the disinfectant.
Will the disinfecting ingredients be neutralized by the washing powder or will they still work fine?

Not sure what is in Ariel, but I assume that washing powder is based on anionic surfactants, whereas the quaternary ammonium compounds are cationic. Probably they would interfere with each other. Why not do a normal wash, rinse, and then wash the clothes a second time with the quaternary ammonium product?

Thanks for the answer! From now on I will make sure that they are used separately, ie with a prewash or by washing a second time. Just so far I had been doing it like this and I wondered if it was useless after reading on the Wikipedia page "Quaternary ammonium cation" that common soap deactivates it.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


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What metals have been used in photography?
I'm interested in knowing about the various processes that have been used in chemical photography. As I understand it, most are based on silver. Are there any major ones which have not used silver? What other metals have been used? Are there any processes which don't use metals at all?

TEM film is/was barium based.

Thanks but can you clarify that or provide a reference? I'm assuming TEM means "Transmission Electron Microscopy"? And the barium was part of a chemical reaction, not an inert substrate layer?

I wanted to contribute what I know so far:

Most photography processes are based on silver halides, which are light sensitive, decomposing into metallic silver and halogens. Here is a claim that silver salts are the most light-sensitive, but it is not well-supported. Silver processes include black and white, as well as more complex color processes like E-6 and C-41 which use dye couplers and color-sensitive layers. I don't know what makes the layers color sensitive, but the reaction is like $$\ce{AgCl + photon -> Ag(s) + {1\over 2}Cl_2(g)}$$ Apparently, after a photon is received, freeing a silver atom, this atom must combine with another free silver atom through diffusion over the lattice of the silver halide crystal before it can become part of the image. There may be a better reference, but this fact is mentioned in the photographic developer Wikipedia article.
Platinum and palladium photography exists. These are also based on metal halides, but they usually include ferric oxalate as a sensitizer. Also, they appear to be based on alkali salts like sodium tetrachloropalladate. Palladium is above platinum on the periodic table, and was used as a cheaper alternative. I saw some old black-and-white photographs in a museum recently, which were labelled as platinum or platinum-palladium prints.
Other metals with light-sensitive salts include mercury and gold, but I am not aware of these being used in photographic paper (although mercury saw early use as a developer in the Daguerreotype process). These undergo a different reaction than silver halides, at least in the case of mercury - undergoing disproportionation rather than releasing free halogen.
The Cyanotype process is based on iron, using both ferric ammonium citrate and potassium ferricyanide. It is the source of the name "blueprint". The light-sensitive compound is ferric ammonium citrate; the citrate reduces the ferric ion under exposure to UV light. The resulting ferrous ion then reacts with potassium ferricyanide to form insoluble ferric ferrocyanide (Prussian blue).

Light-sensitive compounds outside of chemical photography:

In biological systems, chlorophyll is mostly organic, but contains magnesium in a key role. This contrasts with the opsins such as rhodopsin which are responsible for vision. They are proteins bound to cofactors, I think the cofactor is always retinal which is organic - so no metals in vision as far as I know.
Light-sensitive synthetic organic molecules exist, such as aryl phthalate esters. However, I am aware of no photographic process based on organic molecules.
Glow-in-the-dark toys are usually based on metallic phosphors such as zinc sulfide and strontium aluminate.
Wikipedia has a category of light-sensitive chemicals which seems under-populated, but includes many inorganic compounds, as well as hydrogen peroxide and LSD.
Digital photography of course uses metals, with CCDs containing capacitors formed from metal-oxide-semiconductors. Whether this can be usefully related to the chemistry of chemical photography I do not know.

I'm not sure what the overall conclusion is. Why metals are important but not necessary; why silver seems to be uniquely sensitive to light, and how to classify and understand this uniqueness...


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Is cobalt(II) in the hexaamminecobalt(II) complex high spin or low spin?
I want to determine the  electronic configuration of $\ce{Co}$ in $\ce{[Co(NH3)_6]Cl2}$ with Crystal Field Theory.
I know that since the complex is octahedral, there would be degeneracy breaking and the d-orbitals would undergo splitting to form a $\mathrm{t_{2g}}$ level having $3$ orbitals and an $\mathrm{e_g}$ level above that with $2$ orbitals.
As cobalt is in the $+2$ oxidation state, it would have a $\mathrm{d^7}$ configuration.
Thus, if the complex is high spin, the electronic configuration is $\mathrm{(t_{2g})^5}\ \mathrm{(e_g)^2}$ and $\mathrm{(t_{2g})^6}\ \mathrm{(e_g)^1}$ if it is low spin.
This is where I ran into a problem, what type of complex is this?
Ammonia is generally considered strong field ligand, but it is weaker than say, $\ce{CO}$ as it is a σ only donor.
Wikipedia does not give anything definitive and I get inconclusive/contradictory information elsewhere on the Internet, so can someone please clarify whether the electronic configuration of cobalt in $\ce{[Co(NH3)_6]Cl2}$ is high spin or low spin?

Please also note that crystal field theory has been superseded by ligand field theory for a better description of bonding.

Theoretically, you cannot predict a priori whether a compound is high- or low-spin. For some reason, a lot of people seem to think that it depends only on the ligand and that it is possible to unambiguously use the position of the ligand in the spectrochemical series to figure out whether a complex is high- or low-spin. This is a very narrow viewpoint and leads to lots of mistakes: for example $\ce{[Co(H2O)6]^3+}$ is low-spin although $\ce{H2O}$ is fairly low on the spectrochemical series. In truth it depends on (at least) the ligand, the metal, as well as the oxidation state, and there is no magic formula or rule that allows you to combine all three factors.
However, as a general rule of thumb, most 3d metal complexes are high-spin. If you return to the fundamental criterion for high- vs low-spin, i.e. the relative magnitudes of $\Delta_\mathrm o$ and the pairing energy, you will find that for many 3d metals, $\Delta_\mathrm o$ is small due to the poor overlap of the 3d orbitals with ligand orbitals. $\ce{[Co(NH3)6]^2+}$ falls squarely into this category.
The main exceptions occur when you have ligands that are very high on the spectrochemical series. These are nearly always π-acceptors, such as $\ce{CN-}$, bpy, phen, etc. Even so, it should be noted that there are some 3d π-acceptor complexes that are still high-spin, such as $\ce{[Co(bpy)3]^2+}$, so this shouldn't be taken as a rule but rather a rough generalisation. The other big exception is when you have high oxidation states, mainly +3 or higher. So, for example, Co(III) is nearly always low-spin except in $\ce{[CoF6]^3-}$. But there are plenty of high-spin Mn(III) and Fe(III) complexes, so even there it is not a clear-cut rule.


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Is it incorrect to say that oil is insoluble in water?
I'm a biology teacher and therefore not a chemistry specialist.
When I was a student, my teachers used a mixture of water and olive oil to illustrate the concept of soluble and insoluble substances in water. Later, in high school and at university, I learned that the correct terms are miscible and immiscible.
Is it incorrect to use sugar and olive oil to illustrate the concepts of solubility and insolubility to 5th grade students?

You can comment on answers  to your own question. And it's obvious thing it's correct, it's "miscibility" that's additional. Solubility is universal for all states of matter (any exceptions would be obscure).

It's perfectly fine to say that oil is not soluble in water.

That's what i thought and believed but someone here , from chemistry (!), placed that doubt in me. Many thanks

Those are just different ways of looking at the same thing.
A solution is a homogeneous mixture where the principle component is a liquid. Miscible implies that the mixing of two liquids is homogeneous.
However, miscibility can only be applied to two liquids, so if you're talking about a brick and water, you cannot say that they are not miscible because that makes no sense.

However, solid solutions do exit, such as many metal alloys, presumably then one solid is miscible in another.

@porphyrin You are totally correct. According to the IUPAC gold book, miscible is fine as long as a single phase, so it's not just for liquids... https://goldbook.iupac.org/html/M/MT07230.html


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Can 32% hydrochloric acid be stored in a milk jug (HDPE)?
Can $32\%$ ($20$-degree Baume) hydrochloric acid (typically found as 'muriatic acid' in hardware stores) be stored in a milk jug?

I have read that acids can be safely stored in polyethylene bottles. The milk jug I have has the HDPE resin ID code on it. However looking at Wikipedia, it is clear that HDPE is used to make a wide range of products with different "strengths". Given the "squishiness" of milk jugs, I would assume it is on the lower end of the "strength" scale. Does this make it any less suitable for holding diluted hydrochloric acid?
Context: I plan on etching some circuit boards as per this tutorial. And I am looking for places to store the resulting copper (II) chloride solution.
$$
\ce{Cu + 2HCl + H2O2 -> CuCl_2 + 2H2O}
$$
I am assuming that if the container can hold hydrochloric acid, it can hold the resulting copper(II) chloride solution.

I guess there'd be no problem, but you can easily get more solid HDPE cannisters. Just google it. Anyway, please label you bottle properly. Spray it in yellow with a pink skull on it, so everybody get's it that this is not to be openend, spilled, drunk, etc.

Is everyone etching circuits for some weird reason? https://chemistry.stackexchange.com/questions/662/handling-and-storing-sodium-hydroxide-hydrochloric-acid-hydrogen-peroxide-and https://chemistry.stackexchange.com/questions/35185/how-do-i-store-hcl-safely  https://chemistry.stackexchange.com/questions/83845/how-to-properly-store-acids-at-home

https://chemistry.stackexchange.com/questions/60946/is-it-safe-to-store-muriatic-acid-hydrogen-peroxide-and-acetone

Can't you store it in the vessel it's stored in when you purchase it?

@ToddMinehardt I would except, I won't use all of it... and I don't want to mix the "pure" with the "used"

@Mithoron ah that's funny. I had come across two of those articles but was curious about milk jugs. The rust shown in one of those questions is alarming! Is the milk jug more "permeable" than a "stronger" container? Is the copper (II) chloride that potent, or is HCl a special case?

@Karl, the prices I am seeing on amazon are astronomical... where do people typically buy these canisters?

Hm. I find a dozen online sources for 5€ within a second. You local hardware store? Agricultural supplies? There you should get them for 2€. I can get one filled with 5 liters of new wine for ~10 at every other corner at this time of the year.

There is more than just material composition to container selection. Yes HDPE is compatible with hydrochloric acid, but you have to consider rigidity and failure probability. Think: would you want to store Concentrated Hydrochloric acid in a 2 mil bag? Probably not because the probability of puncture and resulting loss/cost from the spill would be sufficiently high. Though not as extreme your milk jug is along those lines as it is softer and less rigid as the original container.
There is also the issue of legal considerations. In the US I'm sure that milk jug would not meet the DOT requirements for transport, and OSHA prohibits using containers typically used for food as being used for chemical or waste storage.
That said for home use only, you will probably be okay to store it, certainly not recommended, but this is the real world. I would at least recommend that if you do use a milk jug, you at least use a basin that can hold all of the contents for secondary containment in case of a leak which milk jugs occasionally do  AND certainly label the contents.

I like the reminder about secondary containment. You could also put the milk jug in a plastic bag and tie the bag top to help contain vapors of HCl. Milk bottle tops are not really snug, especially after use, and HCl tends to corrode things before dissipating.

Another important consideration is the effects of wall thickness on HCL escaping from the container, as it's happening even with thick-walled HDPE containers. It's certainly ok for short-term storage, but without knowing the actual impact I would definitely consider getting a better container for long-term storage, and also ensure that it's not stored on or near materials that would be quickly corroded by HCL.

I prepared the very same etching solution you mentioned to make a custom shield for an Arduino project (it worked great by the way) and still have the etching solution sitting in my garage. It's in a small HDPE tote and has been for over 9 months without any problems. However, the tote has thicker sidewalls than a milk jug. I wouldn't worry about the chemical compatibility and would worry mostly about structural integrity of the container. It might be wise to buy a thicker walled HDPE or Polypropylene (PP) container at the dollar store.

Thanks for the reply! Does your tote have a cover (liquid and air tight)? Doing a google image search for "HDPE tote" and can't quite picture what you have. Also did you experience any problems with the muriatic acid and rust? It seems to be a common problem with storing the acid... and people resort to storing it outdoors =/? Glad to hear the etching process works!

I looked for a photo of the kind of box/tote I bought and couldn't find one that matches well. If I find a good image of the box/tote I used, I'll post it. It does have a lid, but I wouldn't trust it to be waterproof. Now that I think more about it, the Muriatic acid I used is this one.
https://www.osh.com/Osh-Categories/Outdoor/Outdoor-Living/Pool/Hot-Tub-%26-Pool-Chemicals/Kem-Tek-2X1-Gal-Pool-Muriatic-Acid/p/2821262
It comes in two one-gallon milk jug like containers. The walls are thicker than you would find on a normal milk jug.

It looks like a tiny version of this.  http://zamza.info/plastic-boxes-with-lids/plastic-boxes-with-lids-plastic-boxes-with-hinged-lids-uk-small-plastic-container-with-lid-and-handle/

According to this chart from a manufacturer of plastics, HDPE is resistant to concentrated $\ce{HCl}$ acid as well as to copper salts (other manufacturers report the same - Google if needed). As mentioned in the comment thread, do be sure to label and store the containers properly.


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When reacting a strong acid with a weak base, would all of the strong acid react with the weak base
If I do the following reaction:
$$\ce{Na3PO4 + 3HNO3 -> 3NaNO3 + H3PO4}$$
Does nitric acid completely react with the Trisodium phosphate to form a weak acid and a salt, or does it form a small amount of the weak acid or nothing? If it does form the weak acid, will this reaction create heat?
I have only really dealt with neutralization reaction so I don't really understand what will happen, any help would be greatly appreciated.

This would be a simpler question to answer for a salt of a monobasic acid, like sodium acetate. But, for the specific reaction you ask about:
First, we must consider the conjugate acid of trisodium phosphate: phosphoric acid, which will be formed when nitric acid is added. It has three $\mathrm{p}K_\mathrm{a}$ values for the three decreasingly acidic hydrogens: $2.15,$ $7.20,$ and $12.32.$ These values are measured in dilute water solutions, so they do not apply exactly to a reaction between pure nitric acid and pure trisodium phosphate. But these values do indicate that, by combining pure trisodium phosphate with three equivalents of a strong acid like nitric acid $(\mathrm{p}K_\mathrm{a} < 0)$ you would effectively have a complete conversion to phosphoric acid and sodium nitrate. And, yes, the reaction will be exothermic.
That was the simple answer. But if the reaction is carried out in water solution, which would help to keep the reaction from overheating and allow control of the rate of combining the reactants, we note that the first $\mathrm{p}K_\mathrm{a}$ of phosphoric acid is only $2.15$ units higher than that of nitric acid in water (effectively, hydronium ion with a $\mathrm{p}K_\mathrm{a}$ of $0)$ — in other words, phosphoric acid is $10^{-2.15}$ or about $1\%$ as strong an acid as hydronium ion.
So, after the reaction is complete, we would have about $99\%$ of the phosphate in the form of phosphoric acid and about $1\%$ in the form of the dihydrogen phosphate anion. There would also be small amounts present of monohydrogen phosphate dianion and phosphate trianion, whose  concentrations can calculated as $10^{-\mathrm{p}K_\mathrm{a}}$ for the other two $\mathrm{p}K_\mathrm{a}$ values. So, technically the reaction will be about $99\%$ "complete".
Meanwhile, keep in mind that the sodium ions are "observer" species that do not participate in the acid-base reaction.


text:
Convert Free base to 2HCl form
I've been using this drug SID 347829124 - PubChem in my experiments from a vendor that supplied it as a $\ce{. 2HCl}$ version, and I switched to another supplier that has a free base instead. The two compounds have different solubility  in water and it would be desirable for me to obtain the $\ce{. 2HCl}$ form to avoid using $20\%$ dimethylformamide to solubilize the free base powder. I thought that using water with a stochiometric quantity of $\ce{HCl}$ would do the trick, but if somebody from the community could confirm this it would be very helpful.

Yeah, it should work.

If you want the dihydrochloride you need to use 2 molar equivalents of HCl.

It is not clear to me why 2 equivalent of HCl are needed though, there seems to be just one OH group highlighted in red capable of dissociation, why is there another oxygen atom highlighted?

It should work but it is not guaranteed and will depend on application. Drug efficacy is sensitive to the polymorph of the drug and for that reason, a slight change in manufacturing requires a complete redo of any certification process. I assume you are not in the manufacturing business, but it is something to consider and be aware of that your results may vary just by this simple change.

Thank you for the answer. The efficacy is the same and I use the drug for in vivo research on mouse models.


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Comaparing melting points of group 14 elements
I found that melting point of $\ce{Sn(232°C)}$ is less than $\ce {Pb (327.5°C)} $ but i also saw that the bond enthalpy of $\ce {Sn-Sn(187.1 ±0.3 kJ mol^{-1})}$ is more than $\ce{Pb-Pb  86.6 ±0.8 kJ mol^{-1}}$ in the gaseous state.

I do not have any data for their bond enthalpies in solid state. And since melting involves breaking of bonds in solid state to convert it to liquid state should bond enthalpy of $\ce {Sn-Sn}$ should be less than $\ce{Pb-Pb}$ in solid state? Why or why not? I would like to know if here melting points depend on some other factor. Source -https://www.webelements.com/lead/compound_properties.html I want to know about Bond enthalpy comparison of $\ce {Sn-Sn} $ and $\ce {Pb-Pb} $

The matter is qualitatively discussed here: https://www.chemguide.co.uk/inorganic/group4/properties.html
And indeed, there is a correlation between the trend of ionization energies and that of melting points (see: https://en.wikipedia.org/wiki/Ionization_energies_of_the_elements_(data_page).

@The_Vinz the wiki page has the data but i dont understand how ionization energy would effect melting point. The significance of bond enthalpy is more I feel. However the first link I am not able to view

I guess that ionization energy might correlate with the strength of the metal bond (think of Fermi's metal bonding model, of atoms in a sea of electrons). Probably the ease with which a metal leaves its electrons might contribute to the strength of the metal bond

@The_Vinz what are the bond enthalpies of Sn-Sn and Pb-Pb in solid state is the question I am more interested in


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Which of these chemicals share a likeness in structure with benzodiazepines causing false-positives in urine tests?
Below are potential culprits:
Quetiapine, Oxcarbazepine, and Valproate feature in the current medicine regime.
Olanzapine was discontinued quite recently.
Quetiapine XR (increased to 400mg) and an increased dose of Oxcarbazepine (900mg) were quite recently added to the medicine regime.
Following that, a 5-panel urine test came out positive for benzodiazepines along with lab results confirming: 88 nanograms.
Please assume that the individual in question is honest about not ingesting any benzodiazepines.
Thank you!

Let me say that the 5-panel drug test is a preliminary immunoassay, suitable for qualitative testing. If you are sure that the result is a false-positive, you should request a better test (HPLC, GC-MS...)

Rather depends on what structural features the assay is picking up but Quetiapine (Seroquel) certainly has similarities to benzodiazepines. Oxcarbazepine does to a lesser extent, valproate is not really a candidate.


text:
Nucleophilic comparison of epoxides and ketones

Why does $\ce{H+}$ attack on epoxide and not on the double bonded oxygen?

The answer is that H+ does protonate the carbonyl group in equilibrium with the protonated epoxide. However this is of no consequence because the protonated carbonyl form is a dead end, all it can do it swap with the protonated epoxide; whereas the protonated epoxide has this forward route open to it.


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Why do the 4f and 5d orbitals have similar energies?
I've come to know this phenomena when learning about the exceptions to the aufbau principle. But there is no explanation of why 4f and 5d have similar energies (as I've read in some websites).

The energy of a hydrogenic orbital with principal quantum number $n$ is proportional to $-(Z/n)^2$, so as $n$ increases the orbitals become more and more bunched together.
In a multi-electron atom the degeneracy of orbitals with different orbital angular momentum $l$ is lifted, i.e. s/p/d/f orbitals are no longer equal in energy. Since the gap between $n = 2$ and $n = 3$ is fairly large, one would not expect the 2p and 3s orbitals to be particularly similar in energy.
But the gap between $n = 4$ and $n = 5$ is smaller, so it is inevitable and not all that surprising that some of those constituent subshells have similar energies when the degeneracy is lifted. This is how it looks like for lithium (diagram from Modern Spectroscopy 4th ed., Hollas, p 214). The diagram will not be the same for the 4f/5d elements – in particular, I would expect the 4p/4d/4f orbitals to have a more significant difference in energy – but the idea is the same.


text:
Would then base react with water?
Imagine you fill a tank with undecayed neutrons and water .Now neutrons will decay into proton and electrons. Assume that the wall of the container is so thick that it doesn't allow the protons to pass while allow beta rays to pass. The water will now have h + ions and they will make h3o + ions in water that is water will become acidic. Will now
1. A metal having normal reactivity will give H2 gas
2. NAOH react with solution to give water

Your proton-charged water will not last long enough to react selectivy with any electropositive metal or a base that you add.  Electrostatic forces are powerful.  The excess protons will grab electrons, generating ions that themselves grab other electrons, until eventually the beta particle electrons that were emitted to begin with get consumed to rebalance the charges.  You might get some unpredicted chemical reactions but no lingering excess charge.


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Maximum Concentration of Benzene possible
A question for my PhysOrg course states:

Assume you are performing a reaction in which benzene is a reagent. What is the maximum concentration of benzene possible? Hint: what facts do you need to know to determine concentration?

I understand concentration is amount per volume in mol/L, but I'm not sure how to proceed from there.

Sorry, I figured it out- just substitute mols for mass/MW and density  for volume and rearrange until you get density/MW

You mean divide 876g/l by 78 g/mol, which gives you 11.2 mol/l. Jup. Water, for comparison, is 55.4 mol/l.


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Will reaction of hydrogen sulfide + copper chloride/potassium chloride occur and why?
I'm pretty new to chemistry and came across a last year's chemical olympiad in our country and I got stuck on this (I tried googling this problem but after a half of hour I gave up):
Will these reactions occur? If they do, what will be the products?
$$\ce{H2S + CuCl2 -> ???}$$
$$\ce{H2S + KCl -> ???}$$
In my opinion, the copper will replace the hydrogen, but I'm pretty sure the potassium won't. Am I even correct? And what's the reason behind it?
Your help is really appreciated!

Here some hints

Copper and sulphur will react to form $\ce{CuS}$ and $\ce{HCl}$. The equation needs to be balanced!

I'll let you check both copper (II) sulphide and copper (II) chloride solubility in water. It might lead you to a justification for the products.

Second one, no reaction.

You can make analogous reasoning for the second equation.

Some extra thoughts
In both cases: Can we understand this behaviour microscopically? Why $\ce{CuS}$ has such a low solubility compared to $\ce{CuCl2}$
Copper and sulphur ions have similar size and sulphur higher electric charge than chloride ion. So we can undestand microscopically why $\ce{CuS}$ is less soluble than $\ce{CuCl2}$.


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Compensating for hardness in aqueous NaOH solution?
I'm trying to strip the seasoning (polymerized oil) off a carbon steel skillet.
Following online instructions, I've prepared a lye solution, using tap water, 2.5% concentration by weight, which ought to produce results in a matter of days; two weeks in, it's still far from being done.
It occurred to me we have hard water here, 13 dKH/15 dGH, so I'm guessing the solution just isn't as basic as it was expected to be.
Heating it up isn't an option, so I'm planning to add more crystals of sodium hydroxide to speed the reaction up.
Assuming that my diagnosis is correct, and the instructions were based on tap water hardness that is typical for the US, how much more hydroxide should I be adding to compensate for the higher hardness?

The water hardness wouldn't have any measurable effect.

If you're just trying to clean the cooking surface, try some fine steel wool and a little elbow grease.

Need a stronger solution. I don't know where you are based but here in the UK I can buy KOH in polyethylene glycol gel which cleans metal surface in a couple of hours.

@Waylander How much stronger? While I do wear protection, I don't want to end up with a solution that requires too much extra care, like neutralization before disposal.

@MaxW I am aware the coating can be removed mechanically, but that takes a ton of time and work. I tried and it's too much, which led me to giving up and buying sodium hydroxide, that I would have rather done without.

Try 10% and dilute it x 4 before disposal. If you use hot water this will accelerate the process.

The short answer to your question is that the dGH of your water is negligible compared to the amount of lye you are using, so no adjustment needs to be made on the basis of hardness. 15 dGH corresponds to 150 mg/L or 0.015% as CaO, which is trivial compared to your 2.5% lye solution.
Other possible reasons for the slow cleaning process include the way in which the oil has polymerized (as a result of what has been cooked and under what conditions) and the quantity of polymerized oil present. Regardless, the means by which you can speed up the cleaning using your method are to increase temperature and increase lye concentration. If heating really is not an option for you, then you could try a 10% solution (as noted by Waylander in the comments). Disposal of this solution is not a problem, just dilute the solution first or during disposal by running water and pouring slowly. Many household cleaning products use lye so this really isn't an issue.


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Significant Figures
I recently did my first chemistry lab where we had to find the average mass of our aluminum cylinder. We were required to weigh and record its mass 3 times using an electric balance which gave us a value up to the thousandths place. After recording I got these values: $\pu{7.195 g,}$ $\pu{ ~ 7.198 g,~}$and $\pu {7.197 g}$. My professor asked to find the average of these. Adding them up I got $\pu{21.59 g}$. To get the average I had to divide by $3$ since I have $3$ quantities. My question here is, how many significant figures should I have in my answer?

So would I just simply add a 0 to match 3 decimal places after my decimal since my original results had 3 places after the decimal place?

The 3 in the integer is infinite precision. It has an infinite number of 0's after the decimal place, unless you're not sure you counted properly. Otherwise, it's exactly 3.

Significant figures are a sloppy method for doing error propagation. The point with significant figures is to maintain "reasonable" precision in the answer.

Ideally you'd assume that the measurements followed the normal distribution and you'd take many more measurements to get a better average and standard deviation. From the standard deviation of the individual values and the number of measurements you could calculate the standard deviation of the mean.

For addition or subtraction the least significant digit is the limiting factor.   Since each of your three measurements had 3 significant figures after the decimal, your sum and average should too. So your sum was not 21.59 g, but rather 21.590 g. (In math the trailing zeros are not significant so that 21.59 = 21.590, but when using significant figures such zeros in the sum do matter.) Now consider the estimator for the mean:
$$\dfrac{21.590}{3} = 7.1966666666666666666666666...$$
Now considering significant figures you round the answer to three significant figures after the decimal yielding 7.197. The whole point with significant figures is to keep you from going wild and suddenly turn this average into say 27 significant figures.
In general it is wise to carry at least two extra digits in intermediate calculations and only round the final result to the proper number of significant figures to try to avoid rounding errors. The extra digits are no problem with calculators these days. However when I started chemistry we used log tables or a slide rule. With a slide rule three significant figures were the best that you could do for intermediate calculations. With a good set of log tables you could do 4 significant figures for intermediate calculations.

So for addition if you had values of 17.19 (not 17.190!), 7.198g, and 7.197 g than the average would be rounded to the hundredths place since 17.19 only has two significant figures after the decimal.

For multiplying it is the number of significant figures that matters. So multiplying $17.19\times3.198 = 54.97$ since each of of factors has 4 significant figures. Multiplying $17.19\times3.20 = 55.0$ since $3.20$ only has three significant figures.

As an error is not explicitly given and as your first number is  $7.195$ g this means that the absolute error bound implied is 0.0005, which is to say that  the error (uncertainty)  is in the decimal place after the last decimal place and in your example and is the same for each number. Your first number should be written as is $7.195\pm 0.0005$ and similarly for the others. The average total uncertainty is $(\sqrt{ \sigma_1^2+\sigma_2^2+\cdots })/3$ where $\sigma = 0.0005$ and the error works out as $\pm0.0003$ when rounded up so that 3 decimal places are needed in your average since the uncertainty is the fourth place.

Your calculation of the estimated theoretical error as ±0.0003 g is correct. However, the experimental data show a standard deviation of ±0.0015 g and thus an experimental standard error of ±0.0009 g, which is larger than your estimated theoretical error of ±0.0003 g. Therefore, there is another source of measurement uncertainty and the reported result should not be more precise than the total uncertainty.

@Loong where did the 0.0015 come from ? We know that the data has three decimal places as these are given so are known, thus  the uncertainty has to be beyond this hence $\pm 0.0005$ is the standard assumption.

0.0015 is the experimental stadard deviation estimated from the individual values 7.195, 7.198, and 7.197 as $\sigma=\sqrt{\frac{\sum_{i=1}^N\left(x_i-\overline{x}\right)^2}{N-1}}=0.0015$. Since the resulting standard error of 0.0009 is larger than your correct estimate of the random error 0.0003 coming from the used electrical balance, there might be an additional source of error (e.g. from handling or touching the sample, temperature drift, air pressure, humidity etc.) that is not included in your estimate.

Furthermore, there might be a systematic error of the used electrical balance that cannot be eliminated by repeated measurements with the same electrical balance. We both have not taken such systematic error into account yet.

@Loong, I see; it is of course ok to do this but we now have uncertainty in the 3rd decimal place (and also rounding the error to 0.002) so the result has only 2 decimal places that are significant. This is also ok as it is always better to err on the cautious side with errors.

The discussion of the standard deviation goes beyond significant figures. Also the standard deviation of the mean for the three measurements has a whopping error in itself. You'd of course need to assume the normal distribution for the three measurements, use the Student's T value, and calculate a confidence interval for the standard deviation of the mean.

Assuming Loong's value, 0.0015 for the standard deviation is correct, for 2 degrees of freedom at the 95% confidence level, The Student's T value is 4.303. So the 95% CI about the average for an individual measurement is +/- 0.0065. The 95% CI for the standard deviation of the mean is +/- 0.0037.

The significant figures rule for averaging is to identify the lowest significant place value in each addend and then round the average to the highest of these minimal place values.  Here $7.195, 7.198, 7.197$ are significant to the place values $0.001, 0.001, 0.001$ respectively.  The highest of these place values is $0.001$ so you round the average to the nearest $0.001$. Thus, $(7.195+7.198+7.197)/3=7.197$.


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should charge balance be based on activities?
We have a fluid with $\pu{0.01 M}~\ce{HCl}$ and $\pu{0.02 M}~\ce{K2HPO4}$, so if we assume activity coefficients are $1$ throughout we can find pH as $6.800012$ and confirm that at that pH there is approximately zero balance ($6.104839\!\times\!10^{-13}$) between positive and negative charges calculated as
$$\mathrm{  K^+ +H^+ – OH^- - Cl^- -}\ce{H2PO4-}\mathrm{ - 2\times} \ce{HPO4^2-} \mathrm{\ -\ 3\ \times}\ce{PO4^3-}$$
Now we can find the ionic strength and via Debye-Hückel/Davies find activity coefficients for absolute charge 1-3. This allows correction of the thermodynamic dissociation coefficients for phosphoric acid (and water dissociation) so we can correct the concentrations and find a new pH to minimize charge balance. After $4$ iterations we have pH 6.436864 at a charge balance of $1.429412\!\times\!10^{-15}$. This procedure allows reproduction of the results in a recent paper (Schell et al J Chem Eng Data 2018; 63: 2151-2161) with correlation $>0.9999$.
Now the question: these results requires the charge balance being formulated strictly in concentrations multi-plied by the relevant charges. I simply wonder why we should not for the charge balance equation multiply the found activity coefficients on the charged species so making the balance to minimize a function of $f_3\times 3\times [\ce{PO4}]$ and so forth with $f_3$ being the trivalent activity coefficient from Davies?

It doesn't seem strange at all to me.  You are doing a balance of charges, and the charges don't know what activity is.

Sticking with concentrations certainly gives a perfect fit compared to other methods, but all the literature on single ion activity coefficients for strong ions  made me wonder if the charge should not be discounted. I am happy if that is not the case.

Would there be any online literature Chester Miller could cite to make it obvious that charges are oblivious to activities?

I don't know.  But check out Handbook of Aqueous Electrolyte Thermodynamics

This Handbook of Aqueous Electrolyte Thermodynamics is not available for me  - I would really like a reference if it exists :-) -

I would have corrected the typesetting, but I do not understand the formula at all. 2*K+H – OH- Cl -H2PO4 - 2*HPO4 - 3*PO4 What exactly are you adding or subtracting there? It is certainly not the elements/ions, but some kind of property of them. Please try to be more precise about this. I still don't understand the question and believe it should remain closed until it can be clarified.

@Martin-マーチン I had the same problem. My guess was concentrations, but its definitely not standard notation and since the rest of the question is about activities, I couldn't be sure.

So sorry for the confusion - I had a 2 x K in the charge balance, it was all meant to be concentrations  multiplied with charges. Chester Miller mentioned that charges don't know about activities. If so I have no problem. But if in a concentrated fluid Na has activity coeff of say  0.5  I just thought it reasonable to think its charge was also somewhat discounted. But it disturbs certainly the fit to data to make that assumption so I would like to believe it were wrong.

No, charge is not really a activity based property.  It is a number.  Just as the activity coefficients don't change the number (or moles) of potassium atoms or phosphate anions, they don't change the charge either.
If you told me that one liter of a 1.0 molar potassium chloride solution had activity coefficients of 21 for potassium and 186 for chloride, then the amount of charge in the solution is still one mole of potassium cations and one mole chloride anions.


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Why do *tert*-butyl ethers in diluted sulfuric acid undergo elimation instead of solvolysis?
My organic textbook states the pathway for propyl tert-butyl ether in dilute sulfuric acid will follow E1 with a tertiary carbocation intermediate. It mentions that Sn1 will only occur with good nucleophiles.
It does not explain why however. Why is this any different than a tertiary haloalkane in water? I expect that the alcohol is a good leaving group, similar to water.

Vollhardt, P.; Schore, N. Molecules: Structure and Function, 7th edition; W. H. Freeman: New York, 2014, p350

Tertiary carbocation? Protonate the ether oxygen, followed by splitting off the propanol together with an H-shift, and then elimination? That sounds a bit weird. You might as welldirectly attack the tertiary H-Atom in the protonated ether with H2O, making this an E2.

There is no H-shift as far as I know. I have included the relevant excerpt from the book in the post now.

Undoubtedly E1 over E2.

This is tert-butyl ether, not isobutyl!

It isn't necessarily the same situation as the haloalkane in water. In the sulfuric acid case, there is a large amount of acid present, and as you rightly say the alcohol and water are similar leaving groups, so the SN1 product is easily decomposed back to the tertiary carbocation. As a result, the mechanism tends towards the thermodynamic product, which apparently is the alkene (probably for entropic reasons.

Compare this with the alkyl halide in water. Here, the halide is a much better leaving group than water. Additionally, there is a low concentration of acid in solution (at most 1 equivalent, from the generation of HX). This means that loss of water from the SN1 product is much less favourable.
That said, I'm sure you'd get a mixture of E1 and SN1 products from the haloalkane reaction you describe, I merely wanted to point out that the ether and haloalkane cases aren't as similar as you may have presumed.

Is there a difference in temperature between the top and bottom mechanisms as well? (e.g. is the top one under heated conditions vs the bottom one under cold conditions)


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Why do noble metals like silver not form natural cubic crystals?
Pyrite forms cubic crystals, both in theory and in nature:

However, the noble metals, like silver, that have cubic, face-centered lattices, do not seem to form cubic crystals in nature.
One possible explanation is that silver and other noble metals are invariably contaminated by trace amounts of other elements that prevent perfect crystal formation. For example, salt (NaCl) is also cubic, and when very pure deposits of salt are found, they may have large cubic crystals, but normal salt generally does not form such crystals presumably because of contamination.
Is this the explanation why gold and silver do not form cubes when they solidify? If so, imagine we somehow obtained perfectly pure silver as a gas or liquid, would it crystallize on cooling to a cube?

Native silver crystals can be found in cubic, octahedral, or dodecahedral forms. As you mention, silver often is found with gold, copper and other metals, and with gold, tends to hexagonal stacking, rather than cubic crystals.
Silver may also form pseudomorphs, in which the crystal shape is from its progenitor or its surroundings. For example, dyscrasite, $\ce{Ag3Sb}$, or acanthite, $\ce{Ag2S}$ may be replaced by silver in the form of the antimony or sulfur minerals.


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Dimerisation of HOOC-COOH in non polar solvents
In a textbook I read, the following question is my doubt:

Why is the answer B)? Why do the other molecules not dimerise? How does it work?
(P.S- Why dosen't the B) option only form hydrogen bonds with the same type B)
?)
Edit: Turns out I was looking at the wrong answer key and the answer is actually a) :p

Hint: think about the types of intermolecular bonding that can occur in these molecules

I believe your answer key is wrong on this one. Carboxylic acids like acetic acid (a) are well known to form doubly hydrogen bonded dimers in which each the carbonyl group on one acid is hydrogen bonded to the OH of the other.

For more explanation and structures, see https://www.chemguide.co.uk/organicprops/acids/background.html.
Oxalic acid, being a carboxylic acid, can participate in this kind of intermolecular association, but the result would be a polymeric structure, rather than a simple dimer.
The alcohols can hydrogen bond to each other only singly - too weakly to form stable dimers.
Kudos to you for having the right instincts to challenge your answer key an probe more deeply!


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Detection limit of selective ion electrodes
I've been studying about the selective ion electrodes and I came across a question which I seem to not be able to answer. It has to do with why the detection limit for iodide ions is lower than the one for chloride ions.
I know that the detection limit depends on the type of the electrode, temperature, pH, the interfering ions etc., but how can I understand the difference between the detection limit of I- and Cl- ions?

I'd suggest that the difference is indicated by difference in solubility products for the silver salts. The $K_{sp}$ for silver chloride is $1.77\times10^{−10}$, but for silver iodide the $K_{sp}$ is $8.52 \times10^{ −17}$.


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What side reactions are suppressed by the use of TMSCl in the acyloin condensation?
I was reading about the acyloin condensation and came across a statemant according to which, the use of TMSCl obviates competing reactions. My question is what kind of competing reaction would happen if it was not used and how? I've read about the possiblity of some sort of Claisen condensation but I fail to see how would that be possible.
https://en.wikipedia.org/wiki/Acyloin_condensation

Your acyloin product contains an alcohol under basic conditions this could easily transesterify with starting ester material

The acyloin condensation of diethyl adipate (1) on your Wikipedia citation generates ethoxide that can cause a Dieckmann condensation (intramolecular Claisen condensation, structure 6). TMSCl scavenges ethoxide and silylates the enediolate 3 formed in the acyloin condensation. The process is a 4-electron reduction (see details below). Mild hydrolysis of 4 gives the acyloin 5.

The article by Rühlmann, https://doi.org/10.1055/s-1971-21707 (German), where this was introduced details this.


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How can I recognize aromatic rings in a connection table?
Given a connection table (that is, a list of atoms, their connectivities/bonds, and the type of bonds), how can I detect the presence of an aromatic ring?
For reference, the aromatic rings in the structure will always be depicted in Kekule form (alternating single and double bonds).
It is simple enough for me to use a cycle finding algorithm on the graph representing the bonds to find candidates. The problem is now recognizing which of those rings can be aromatic.
A naive approach for at least six-membered carbocyclic rings would be to check if the ring bonds alternate between single and double bonds. But, the structures below would be problematic:

where the center ring in phenanthrene and the six-membered ring in 2H-indene are decidedly aromatic, even if there is no complete alternation of single and double bonds. If you try to check exocyclic double bonds, the leftmost structure could be a false positive.
Pointers to algorithms would be very much appreciated. I am ignoring heterocyclics like thiophene or imidazole, or cations like tropylium for now, but if you know how to handle them too, it would be a nice bonus.

I haven't gotten an adequate explanation of why people thought the question was "too broad"; thus, to restrict the question, here is a drastically short reformulation:

Given a connection table for a molecule that contains any number of six-membered rings (possibly fused, spiro, etc.) and is represented in Kekule form, can one write an algorithm to determine which rings are aromatic and which are not? Atomic coordinates are available if necessary.

Wouldn't simply checking for hybridization of all ring carbons solve all 3 problems presented here? Do you need to differentiate aromatic/antiaromatic?

This may be me being slow, but all the carbon atoms in the six-membered rings in all three atoms are sp2-hybridized, yes? But, the six-membered ring in the first structure shouldn't be aromatic... and yes, excluding antiaromatics would be immensely useful.

Well, when looking at aromaticity of fulvalenes and similar structures I don't think I can be sure if that exact structure would be aromatic, antiaromatic or non-aromatic. And another question, are you looking into what ring is aromatic or if the structure contains at least one aromatic ring?

I basically want to mark rings as aromatic or non-aromatic, for further analyses I do not want to detail here.

I wouldn't say 2H-indene is aromatic, much like 1,2-benzoquinone also isn't.

Since I'm pretty sure it is done in context of QSAR, I suggest to move from analysis of aromaticity to analysis of conjugated pi-systems in general. I also have to say that connectivity matrix lacks information of 3d structure and it can easily missdetect some cases of aromaticity/conjugation.

@permeakra, I should have mentioned I have coordinates too, sorry. It should be easy enough to use a least-squares method to check if e.g. the six carbon atoms in a ring are planar, along with checking if their hybridization is sp2. Would that suffice, or are there pitfalls I haven't considered?

...after some thinking, that check I proposed might be problematic for things like hexahelicene, where the rings are still aromatic even though coordinates would show the rings to be bent quite a bit.

I will reiterate what I have written in a comment earlier, as of why I find this a bit problematic (especially for a cheminformatics approach).
There still is no clear cut definition of aromaticity, there even is an (at the time of writing) upcoming conference Aromaticity 2018 to get more clarity on the matter.
There are a couple of questions on the site, where I have written an answer for which demonstrate the issue. I don't want to be shamelessly promoting my own posts, but I also don't want to repeat myself. I think one of the crucial examples is How to explain (non-/anti-) aromaticity in fulvene with the help of resonance structures?
From examples like this, it seems clear that from structure and Kekule form alone it appears impossible to determine whether a ring has aromatic character or not. Mithoron stated in the comments something quite categorical, which I find in its extend debatable; but I wouldn't want to dismiss it either:

I wouldn't say 2H-indene is aromatic, much like 1,2-benzoquinone also isn't.

Take the first as an example, it has a high C-H acidity (higher than indene), because the resulting structure would have a fully conjugated system very closely related to naphthalene. This structure is undoubtedly more aromatic (handwavy, but if you could quantify it, it probably would), so any kind of structural diagram, yes/no-parameter would misrepresent that.

Given a connection table for a molecule that contains any number of six-membered rings (possibly fused, spiro, etc.) and is represented in Kekule form, can one write an algorithm to determine which rings are aromatic and which are not? Atomic coordinates are available if necessary.

I'm afraid the answer to this question is no.

What you could do, and where you'll probably be right many times is using the bond lengths as an indicator. The bonds in such rings should all be very, very similar (maybe ±10%), an given Pekka Pyykkö's data (pdf from his homepage) fall somewhere around 140 pm. Couple that with the ring-finder and I would assume it's going to be close many times.


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Question about the definition of an isotopes
My answer was as follows : atoms of the same element with differrnt relative atomic masses.
However in the mark scheme the answer was simply : atoms of the same element with different masses.
Was there something wrong with my answer?
Why does it matter?

Usually the term "relative atomic mass" for an element is a weighted average of the atomic masses of the various isotopes of an element, with the weighting determined by their abundance. So, since the atomic masses of various isotopes enter into this determination, your first answer is not too good.

İ still don't understand. :(

https://en.wikipedia.org/wiki/Relative_atomic_mass#Current_definition

the term "relative atomic mass" already implies a mixture of isotopes, so you can't use that term for the atomic mass of a single isotope

So the relative atomic mass is not the mass the mass of a single isotope then ?

yes $\phantom{}$

You would only use “relative atomic mass” to describe the mass number you see on the periodic table, which is essentially the average mass of all the isotopes of that element, proportional to their abundances.
Isotopes are atoms of an element which have the same number of protons but differing numbers of neutrons, and therefore, different masses.


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Fischer-Speier Esterification: Sulfuric acid removal following creation of ester
I'm a beginning chemistry student, and I was wondering what the best way of removing sulfuric acid was following the creation of benzyl acetate. So far, I have that benzyl alcohol and glacial acetic acid are mixed in a 1:1 ratio to create the benzyl acetate in a beaker filled with molecular sieves (to remove the water and shift the reaction to the right). However, I am unsure as to how much sulfuric acid to add and how best to remove it from the final solution of ester, water, and sulfuric acid.
Thanks!

Welcome to ChemSE.  If you accessed a reference stating the molar ratio of alcohol and acetic acid (why not acetanhydride?), likely the same reference will provide you a hint about the catalytic amount of sulfuric acid necessary.  Give you a mid- and longterm favour, narrow your question, and show evidence you read further the lab protocol provided to you.  Or take bus #69 to either Irchel or Hönggerberg.

In a typical esterification, the sulfuric acid plays a catalytic role, so you don't need a lot, and removing it should not be a concern. Also, to drive the equilibrium reaction forward, you would want a slight excess of one of the reactants, usually the cheaper one of the acid or the alcohol (acetic acid in your particular case). The sulfuric acid is also a good dehydrating agent, so the sieves seem excessive. Finally, the ester is a bit more volatile than the other components (check a handbook for boiling points!), so distillation is standard. Otherwise, look into a Dean-Stark trap.
For other ideas on how one usually conducts an esterification, look up related syntheses in Organic Synthesis; e.g. this one or this one or this one.

Catalytic sulfuric acid is required so just a few drops should do it.
To remove (or as it normally expressed work up) dilute the mixture with a relatively low boiling point  organic solvent - ethyl acetate is good - and wash the solution with dil. sodium hydrogen carbonate solution. Separate off the organic phase (upper phase if you used EtOAc), dry it with a solid drying agent such as anhydrous magnesium sulfate, filter and remove the solvent under reduced pressure.

If the use of acetic acid/H2SO4 isn't mandatory, why not use acetic anhydride?


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Ring Expansion And Final Structure
[If you don't like to read the question just look at the 3 pictures; the first one has the carbocation which has to be stabilised by ring expansions. The other 2 are my predicted structure so please confirm which one is correct or if both are wrong please provide the correct one]
Please help me in expanding the rings in the following structure :

I tried a lot expanding the the rings twice to get this :

But the intemediate in this structure is unstable as it has a bridge carbocation.
Instead we can have a intermediate with sigma-pi resonace stablising the carbocation by expanding the ring once as follow  :

I can't wrap my head around which one is correct?

There are mainly two factors governing this phenomenon of ring expansion and rearrangement of carbocation, mainly Stability factor and Thermodynamic factor. When you have a carbocation which can rearrange within itself, it will definitely prefer a rearranged structure which has the highest stability. Not only that, the carbocation also needs some energy for the rearrangements to occur. If it needs multiple steps of rearrangements to attain a structure which has decent stability, but the loss in the potential energy can't compensate the requirement of energy for rearrangement, it is not also possible for the molecule to undergo that many number of steps thermodynamically.

Based on these two factors, we can at least logically conclude which will be the final rearranged product. In both of your predicted compounds, first there will be a rearrangement by Hydride shift. Now it is highly difficult for a molecule to undergo two rearrangements at once unless it gets  extreme high stability by the second case. In this case, it does by forming a cyclopropyl methyl cation which is unusually highly stable due to non-classical resonance and thus ring-expansion occurs in this case. So, by these two steps you have got your first possible carbocation.

Now , if we look for the third possible rearrangement of ring-expansion, here energy requirement is extremely high as it becomes more and more difficult for a molecule to undergo rearrangement as its number increases as a single reaction intermediates. Also, the newly formed carbocation is tertiary which is obviously less stable than cyclopropyl methyl carbocation. Thus, the last ring-expansion is energetically less favourable.

So, I think the correct one should be the image you have drawn at the end in your question, as further more rearrangements on that system is both thermodynamically and stability wise less feasible.

But carbocation stability isn't the only factor consider - relief of the torsional and bond angle strain may provide the additional driving force for the final ring expansion step?

But that factor isn't strong enough to compensate for the lose of stability of cyclopropyl methyl cation , which is much more dominating.

I'm not sure why you think the final compound is so unstable. It's still tertiary, although the adjacent substituents are maybe marginally less electron-rich. It absolutely is not a bridgehead carbocation, since the compound is a fused rather than bridged bicyclic.


text:
What are some example of the use of argentometric titration?
Like, we use alkalimetric titration to calculate the real concentration of table vinegar. What about argentometric titration?

Methods based on precipitation of silver(I) compounds are collectively called argentometric methods. They are most often used for the determination of chloride ions, but they can be used also for other halides (bromide, iodide) and some pseudohalides (thiocyanate). Titration is in aqueous solution against a solution of silver nitrate of known concentration for silver(I) salts that gives an insoluble precipitate. See Wikipedia overview and titrations.info — Argentometry - Mohr method for the example of a lab procedure.
It is also possible to use indirect argentometric methods for determination of anions, that create insoluble salts with silver(I) (for example, phosphate $\ce{PO4^{3-}}$, arsenate $\ce{AsO4^{3-}}$ and chromate $\ce{CrO4^{2-}}).$ These methods are based on back titration of excess silver with standardized thiocyanate solution. See Wolfram Demonstrations Project — Argentometric Titration by Precipitation of Silver Salts for more details.


text:
Unique carbons on rings

In rings in general, how do I determine how many unique carbons there are? In this example, I am thinking that there are 2 unique carbons (one on top and one with a double bond) since this molecule is symmetrical. Any feedback is appreciated.

Yes, one would try to find points or lines of symmetry in general to check for equivalent ring atoms. It immensely helps to have a model and a lot of practice.
In your specific example, notice that you have two lines of symmetry in your (idealized) diagram: one passing through the sp3 carbons, and one crossing the two double bonds. You can thus "fold" the diagram into "quarters", leaving you with two non-equivalent atoms.

Don't think "fold" is the right word. The symmetry would be better expressed as rotations, reflections and inversion.

indeed that's the formal, mathematical way to put it, as reflections and other symmetry operations. I have a geometric mindset, so I always visualize things being literally sliced by planes, or being folded into equivalent halves or quarters

so based on this, how many unique carbons would benzene have? would it be 1?

yes $\phantom{}$


text:
Arrange the following questions in the order of basicity? Also Tell why?

The main confusion is between the compound (a) and compound (b) because in compound (c) we get aromatic ring(above) upon cleavage of double bond on N so it must be best whereas in compound (d) we get an anti aromatic ring (above ) so it must be least but what about coumpound (a) and (b).

A positively charged species is unlikEly to attract H+

The stability of the conjugate acid in case of a) is lesser than that of b), because of the increased electron withdrawing effects of N atom.

After the cleavage of the double bond between N and Upper ring , the positive charge of the nitrogen will no longer be there so it will the surely attract H+   @Waylander


text:
How many watts would I need to get CO2 from 30 °C to 4000 °C in 1 hour
Scenario: After reading some articles on new sorts of ways to deal with $\ce{CO2}$ in the atmosphere, a question came to mind: If I had 100 kg of $\ce{CO2}$, and I would like to heat it up from 30 °C to 4000 °C in a given period (lets say 1 hour), how much energy would I need? What would that be in watts? And from there, assuming a solar panel with 20 % efficiency for 1 m² (therefore 200 watts/hour), what would be the area of solar panels needed to reach heat the $\ce{CO2}$?
Obs: From my very forgotten knowledge of Chemistry from high school, I found the specific heat for $\ce{CO2}$ should be 1000 to 1800 kJ/(kg K), and from this page http://www.heatweb.com/techtips/useful.html I got a formula and tried calculating it, but my units are probably all wrong, because the value was considerably low. I then multiplied 100 kg * 1500000 * 3970 °C. With the energy value, I divided by time (1 hour) to have the output in watts.
Obs2: I know that I am forgetting or do not know about many other limitations (such as the dissociation above a certain temperate). I also know that it would be inefficient and very costly, but would nonetheless like to find out the values, to try and figure what would be necessary for this procedure.
Question: Is this a viable way to do this calculation? What else should I consider?

I approached this in a similar way utilizing the heat capacity of $\ce{CO2}$ over a temperature range.
An estimation can be made by approximating the change in enthalpy of $\ce{CO2(g)}$ over a given temperature range. This can be accomplished through the integral form for changes in Enthalpy:
$$\Delta{H} = \bar{H_1} - \bar{H_2} = \int_{T_1}^{T_2}C_p(T)\,\mathrm dT$$
Which has units of [energy/mass].
Often times, the function $C_p(T)$ for certain gases and liquids will be tabulated in textbooks or chemical databases (such as NIST). In this case, I will utilize a function for the $C_p(T)$ of $\ce{CO2}$ in two temperature ranges (300–1200 K, 1200–6000 K)$^1$:
$$C_p(T)=A+BT+CT^2+DT^3+\frac{E}{T^2}$$
Range 1 (300–1200 K) coefficients:
$A=24.997,$ $B=55.187,$ $C=-33.691,$ $D=7.948,$ $E=-0.137$
Range 2 (1200–6000 K) coefficients:
$A=58.166,$ $B=2.720,$ $C=-0.492,$ $D=0.039,$ $E=-6.447$
Next, we can obtain the energy required for this temperature change by multiplying $\Delta{H}$ by the mass of $\ce{CO2}$ we're interested in, such that:
$$Q = m_\ce{CO2}\,\Delta{H} =  m_\ce{CO2}\int_{T_\text{initial}}^{T_\text{final}}C_p(T)\,\mathrm dT = m_\ce{CO2}\left[\int_{T_1}^{T_2}C_p(T)\,\mathrm dT + \int_{T_2}^{T_3}C_p(T)\,\mathrm dT\right]$$
Evaluating the integrals (keeping in mind that A, B, C, and D will change based on the temperature range in this case) yields:
$$m_\mathrm{CO2}\times7.266\times10^{12}\ \mathrm{\frac{J}{mol}}$$
$$\frac{100}{0.04401}\ \mathrm{\frac{kg}{\frac{kg}{mol}}}\times7.266\times10^{12}\ \mathrm{\frac{J}{mol}} = 1.651\times10^{16}\ \mathrm J$$
In watts, this value becomes:
$$\left(1.651\times10^{16}\ \mathrm J\right) \left(\frac{1}{3600}\ \mathrm{\frac{h}{s}}\right) = 4.560\times10^{12}\ \mathrm W$$
Calculating the solar-panned area required:
$$A_\text{panels} = \frac{4.560\times10^{12}}{200}\ \mathrm{\frac{W}{\frac{W}{m^2}}} = 2\times10^{10}\ \mathrm{m^2}$$
Source [1]: “Carbon Dioxide.” Carbon Dioxide, National Institute of Standards and Technology, webbook.nist.gov/cgi/cbook.cgi?ID=C124389&Mask=1#Thermo-Gas.


text:
Compound having the highest enol content will be?

The answer written in book is 3 but no solution is given.
After referring to some other question I think that hydrogen bonding can take place in compounds like 3 but am not sure about it. Am I right that it is the only reason why it has the highest enol content?
I would like to ask one more thing that, while comparing the keto and enol content, if in a compound more than one ketone is present, than do we enolize any one of them only to check?

It's a 1,3-dicarbonyl...

3 is the only 1,3 diketone so there is a delocalisation effect between the enol and the adjacent ketone

What do you mean by delocalisation effect .@Waylander . Does it means what is shown here ->  https://chemistry.stackexchange.com/questions/42738/which-of-the-following-has-higher-enol-content?rq=1

If you can form an enol where the carbon-carbon double bond is conjugated to one or more other double bonds, the enol is more stable and the equilubrium mixture will therefore have more.  With that in mind, try seeing what happens to (3) if you tautomerize with the proton on the bridging carbon between the two carbonyl groups.
For a further exercise: render the structure for cyclohexa-2,4-dien-1-one where the carbonyl group is conjugated to two carbon-carbon double bonds in the ring.  Now tautomerize to the enol and see what "enol" you get.  When the conjugation of the enolic double bond forms an aromatic ring, the stabilization effect is so strong that the enol takes over completely.

(This is too elaborate to fit into a comment.)
Although one would, at first blush, immediately think of the β-diketone (3) as quite acidic, a moment's pause of what the putative enol should look like will confront you with a violation of Bredt's rule. Thus, it would prefer to enolize at the non-bridgehead hydrogens, and the acidity would likely not be quite different from the other three choices.

(-1) Bredt's rule is irrelevant here.

Why is it so? @orthocresol

I also thought of your point @derek correa and hence made a guess of (2) because in that case we get a conjugated system also but then i saw this https://chemistry.stackexchange.com/questions/18904/which-is-the-more-stable-enol-form?rq=1 ..please tell me that can hydrogen bonding effect solve this problem?


text:
Why do different gases have the same molar volume?
Molar volume:

The molar volume, symbol $V_\mathrm m$ is the volume occupied by one mole of a substance (chemical element or chemical compound) at a given temperature and pressure.

At S.T.P conditions the molar volume of any ideal gas is $\pu{22.711 dm^3}$. My question is: how can two gases with giant difference in molecular size (say $\ce{He}$ and $\ce{Fe}$) have the same volume, considering that the size difference between them is huge, shouldn't the volume they occupy also be different?
Thanks to @MaxW for pointing out a fault.

Most of the volume in a gas is empty space, not atoms.

You have the wrong molar volume. Since 1982, STP is defined as a temperature of 273.15 K (0 °C, 32 °F) and an absolute pressure of exactly $10^5$ Pa (100 kPa, 1 bar) which gives a molar volume of $22.711 \mathrm{dm}^3\mathrm{/mol}$.

Calculate the average volume containing a single molecule at STP.  It's about 36,000 nm3 (cubic nanometers).  A single simple molecule (like CO2, SF6, etc) is about 0.03 nm3, meaning that at STP a volume of gas is about 99.9999% empty space.  Even if the molecule in question becomes quite large it is still effectively occupying a similarly negligible amount of physical space.

You have the right thought process; if two real gases have different sized molecules, they won't have the same molar volume. The issue is that your statement about gases at STP refers to ideal gases which by assumption are just point particles that only interact via elastic collisions. To get a better sense of the difference that considering the finite size of molecules and their electrostatic interactions can make, you will want to read up on the various real gas models used, such as the Van der Waals, Redlich-Kwong, and Peng-Robinson models.
As a small side note, MaxW is correct that the definition of STP you are using is out of date so the molar volume at STP with the new definition is slightly higher. Its still quite common to see people accidentally continuing to use the old definition, so be on the lookout for that.

So, what you are saying is that they don't have same volume? I wasn't aware of the change regarding STP, will read into it as well.

@Any3nymoususer they would not have the same volume, though in most cases the difference is very small because molecules don't take up a lot of space. The specifics of how much the molecular size matters depends on the model you are using to describe a real gas.

Very small doesn't make them equal. Anyhow so lemme see if I understood correctly: The difference although really small (as atoms themselves are small), is there and in most events they won't be equal?

Can you possibly give a reference read. It'd be great if I could read up on this? Thanks for taking the time to answer

@Any3nymoususer for now, I would start with the Chemlibretext page on real gases. It has a table of the molar volume of a few gases at STP (actually the old STP, but like I said a lot of people seem to have missed the memo on the change in definition) and explains the difference in how you treat individual molecule effects that cause nonideality.

Alright, thanks a bunch for the answer. Marked as accepted and thanks for the link

@Any3nymoususer remember that ideal gas law is even previous to the periodic table. It was found experimentally, and explains fairly well behaviour of gases. You can wonder about molecules' size, but take that into account i.e all microscopical interpretations come centuries later.


text:
Protection of Alcohol
I carried out an experiment and it goes like this:
2-bromoethanol + dihydropyran (DHP) ---> 2-(2-bromoethoxy)oxane
Acidic ion exchange resin was added alongside as a source of H+ ions (catalyst).
If, instead of using acidic ion exchange resin, a conventional acid such as p-toluenesulfonic acid was used as a catalyst. How can it be removed from the product?
I thought of using a weak base, but would there be any side products formed?

The catalytic p-TsOH is easily removed at the end of the reaction by washing with aqueous NaHCO3 (bicarb). Failing that, if you plan on clumping the protected product it generally sticks to the silica gel .

The silica gel that you are talking about is not the acidic ion exchange resin right?

I have run this exact reaction with pTSA in CH2Cl2 many times. As @NotEvans says above, wash with saturated  sodium hydrogen carbonate solution, dry over anhydrous sodium sulfate, filter and concentrate under reduced pressure


text:
What is a protein interface account?
I'm not a specialist but I'd like to get into meaning of what is written in some articles. In the sentences:

"The large numbers of salt bridges and hydrogen bonds in the protein
interface account ..."
"Alanine scanning mutagenesis have showed that only some of the amino
acids of the interface account ..."

what is "account" stands for? As long as I understand, an "interface" is a protein part/domain that contacts with another interacting protein. But what is an "interface account"?

I'm voting to close this question as off-topic because it's a not about chemical nomenclature, but common word.

Account seems to be used as a verb here, but I can't tell without more context.

"Account" in this case is used as a verb.
The original paper for the second quote appears to be a PLoS One article, and the exact quote is

Alanine scanning mutagenesis have showed that only some of the amino
  acids of the interface account for the binding free energy.

We could equivalent rewrite this as

Alanine scanning mutagenesis have showed that only some of the amino
  acids of the interface are responsible for the binding free energy

or

Alanine scanning mutagenesis have showed that only some of the amino
  acids of the interface contribute significantly to the binding free energy

I haven't looked at the first quote, but I suspect it's in a similar boat.

It does make sense. Thank you for the answer.


text:
Molecular electronic state symmetry
I am trying to work out the ground state symmetry of electronic states. The paper states the GS symmetry is $^6A_1$ for the right hand side electron configuration and $^4B_1$ for the LHS in the D$_{2\mathrm{d}}$ point group. I understand that the symmetry is determined from direct products of the molecular orbital symmetries of the singly occupied orbitals. However I cannot get to the two symmetries listed. For the left electronic configuration it is $b_1 \times a_1 \times b_2 \times e \times e = [a_1+a_2+b_1+b_2]$ so I'm unsure of how to get $A_1$. Likewise the product e x e for the right electronic configuration produces the same results and again I'm unsure of how to arrive at $B_1$.

Each of the terms you generate has a different energy, and may have different spin multiplicity. The spin multiplicity $(2S+1)$ is given by the left superscript (6 and 4) . The quartet (right) it seems has the lowest energy with spin of a triplet (sym e) x doublet ($b_2) \to$  triplet + quartet.  To do this add spin vectors triplet $S_T=1/2+1/2$ , doublet $S_D= 1/2$ and  make series $S_T+S_D \to |S_T-S_D|$ in unit steps, (Clebsh-Gordon series) this gives total spin 3/2 & 1/2 or quartet($\uparrow\uparrow\uparrow$) + doublet($\uparrow\uparrow\downarrow$). The symmetry has next to be associated with the spin multiplicity and this is complicated. It is necessary to find the character ($\chi$) under, say, quartet multiplicity for each symmetry operation $R$ for each symmetry species $A_1,A_2 ..$  etc. $R^2$  means square the operation. In this case this is
$$\chi(quartet) = (1/6)\left([\chi(R)]^3-3\chi(R)\chi(R^2) +2\chi(R^3)\right)$$
The result has to be reduced into symmetric $\chi^+$ and antisymmetric components $\chi^-$  as $\chi^+=\left( [\chi(R)]^2+\chi(R^2)\right)/2 $ and   as $\chi^-=\left( [\chi(R)]^2-\chi(R^2)\right)/2 $ and the set of characters produced reduced to find the symmetry species.
Chapter 5 of 'Symmetry & Spectroscopy' by Harris & Bertolucci gives a detailed account of how to do this.

Thanks for the reply. I must add that the two electron configurations are for different metals. I'm reading Symmetry and Spectroscopy however is all this analysis necessary to derive the term symbol 4B1 for the given electron configuration on the left?

unfortunately it seems so , the direct product gives you many possibilities, finding which ones have a given spin is the difficult part as you realise.

Ok thanks lot for the recommending "Symmetry & spectroscopy"..it's an absolute gold mine


text:
Unique carbons on long carbon chains
If we have a long carbon chain like octane for example, do we determine if a carbon is unique by just looking at the immediate surrounding atoms of each  carbon (in this case 3 unique carbons), or do we need to take into account the symmetry of the whole molecule (many more unique carbons)? Edit: This is for NMR purposes

It depends on why do you need that in the first place. For the purposes of most chemical properties, all CH2's are pretty much the same; from the NMR point of view, they are different.

oh yeah forgot to add that this is for NMR purposes

In general, the problem is a mathematical one. What you are asking for is the 'equivalence classes' of the atoms under the symmetry group. As pointed out in the comments, this may or may not be chemically important.
One simple way to divide the atoms into equivalence classes is to use Morgan numbers. The algorithm is particularly simple:

Label the atoms with their degree (so terminal carbons are "1")
Update each atom's label by adding the labels of the neighbouring atoms
Repeat until the number of different labels is constant

So, for a linear hydrocarbon, the labels start out as [1, 2, 2, ..., 2, 1] and update to [3, 5, 6, 6, ..., 6, 5, 3]. After a couple more rounds, you get a pattern of labels like [a, b, c, d, d, c, b, a].
This works reasonably well, although there are more complex algorithms for determining non-equivalent atoms. As you have worked out, the atoms beyond the immediate neighbourhood are necessary.

Thanks for the feedback, also I forgot to note that this is for NMR purposes

would what you  said still hold true for NMR purposes? Thanks

If you don't look just at the closest neighbors, but see what the whole groups at each atom are, you see that in the case of octane, there are 2 carbons with one heptyl, 2 carbons with one methyl and one hexyl, 2 carbons one ethyl and one pentyl, and 2 carbons with one propyl and one butyl; i.e. 4 different carbon atoms (each one twice).

Thanks for the feedback, also I forgot to note that this is for NMR purposes


text:
Effect of oxidation aluminium reflectivity
Aluminium is a fairly good reflector, but oxidation will reduce it's optical reflective properties.
Aluminium is used for reflectors either being deposited on a substrate, or polished and protected by a protective layer to avoid oxidation.
How will the oxide layer of a several year exposed polished aluminium (indoor) will impact the reflectivity.
Will it be considerable or minimal ?

Likely drastic. Telescope mirrors are made with a protective layer of SiO2.

@DrMoishePippik I haven't been able to find any sort of data of reflectivity of anodized aluminium unfortunately

The lifetime of aluminum depends on the environment. If it's near an ocean, where air is humid and often contains salt particles, it may become pitted or look a milk white, rather than shiny, in just months. Anodizing can protect it for years, though.
Because the surface of anodized aluminum is porous (until sealed), it may have a somewhat more or less matte finish depending on treatment. Bright dip anodizing, depending on the aluminum alloy, offers a more shiny finish.
The Alzak process is used for lighting reflectors, and there is information on its specular reflectivity.


text:
Why does vinegar and baking soda yield a far smaller volume of carbon dioxide gas than expected from calculation?
I experimented $\ce{CO2}$ generation by adding $\pu{4 g}$ of $25\%$ acetic acid water solution ($= \pu{1 g}$ of acetic acid) to $\pu{1.4 g}$ of pure baking soda. But this mixture yielded only $50\%$ less than the calculated amount $\pu{373 mL}$ of $\ce{CO2}$.
$$
M(\ce{CH3COOH}) = \pu{60g/mol}, M(\ce{NaHCO3}) = \pu{84 g/mol}\\
  \implies \ce{CH3COOH}~\pu{1g} : \ce{NaHCO3} \pu{1.4 g}. \\
\ce{CH3COOH + NaHCO3 -> CH3COONa + H2O + CO2}\\
\text{mol to mol} = \pu{1 mol} ~\ce{CO2}, M(\ce{CO2}) = \pu{44 g/mol} \\
  \implies \ce{CO2}~\pu{0.733g} : \ce{CH3COOH}~\pu{1g} : \ce{NaHCO3}~\pu{1.4g}\\
\text{Gas} = \pu{22.4 L/mol} \implies \pu{22.4 L}/60 = \pu{373.3 mL} (= \pu{0.733 g}) \text{(Theoretical yield)}.\\
\text{Actual yield} = \pu{190 mL} \text{by volume measurement} (50\%)\\
\text{Actual yield} = \pu{0.586 g} (=\pu{298.6 mL}) \text{by mass measurement} (80\%)\\
$$
I set the water filled measuring cup upside down above the water filled plate, and attached hose underneath the skirt of the cup so that any gas generated from the sealed bottle of the mixture to get provided into the measuring cup.
Moreover, I checked the amount of $\ce{CO2}$ yielded by the weight loss method. That is, I measured the mixture above the scale to see how much $\ce{CO2}$ is evaporated during the chemical reaction.
The above 2nd method provided me with $80\%$ yield.
$50\%$ of air volume and $80\%$ of mass are coming with the consistency during the several times of experiments.
What was wrong with my experiment??

Watch this guy. http://youtube.com/watch?v=5pPPFI8E33U . He also has consistant 45% percent yields.

The rest of CO2 is still dissolved in water. CO2 is partly soluble in water.

I have updated (at least I tried) your post with chemistry markup. If you want to know more, please have a look here and here. It would have been easier if you were not using some text equations, please check if I have destroyed anything in the process.

I'm assuming your calculations and materials are correct here. Acetic acid is a weak acid and as such will for a buffer solution with a weak base. So at half your reaction completion you will have generated a sodium acetate buffer with pH ~5. At this stage the acidity of your solution might not be enough to protonate all the carbonate.
Try to add a stronger acid if you can.

Oh, I missed one thing. I already tried 4g of acetic acid + 1.4g of sodium bicarbonate only to have just around 50% yield. In this case, acidity is already high enough and still keeps high even after reacting with 1.4g of sodium bicarbonate, but why the yield doesn't get higher??

How did you test for acidity? What pH did you have?

I didn't. I added 100% acetic acid solution bought from lab material shopping mall to the tap water. Tests were done in order of #1 1 g acetic acid without water, #2 1g acetic + 3g water, #3 2g acetic + 3g water, #4 3g acetic + 3g water, #5 4g acetic + 3g water, #6 2g acetic without water, #7 3g acetic without water, with fixed amount of Arm & Hammer backing soda which is 1.4g.

You can try different amounts of baking soda to make sure it is actually pure bicarbonate.

Watch this guy. https://www.youtube.com/watch?v=5pPPFI8E33U . He also has consistant 45% percent yields.

Other papers insist that they yield at least 90% of calculation.

Okay, could you please update your question to include the math you did. Especually the part where you determine how potent your reagents are.

I added the calculation.

First, use freshly bought $\ce{NaHCO3}$ as on warming, it converts into more alkaline $\ce{Na2CO3}$.
$$\ce{2 NaHCO3 -> Na2CO3 + H2O (g) + CO2 (g)}$$
See this article: 'Vanishing Baking Soda' at the Scientific American.
Next, my experience with generating $\ce{CO2}$ is that you can collect it over water, but depending on the level agitation/bubbling, you likely suffer a loss due to the dissolution of the $\ce{CO2}$! In fact, you can use this to crush a plastic vessel fill with $\ce{CO2}$ and a little water which is tightly sealed. Upon shaking, the $\ce{CO2}$ rapidly goes into solution causing an implosion.
Bottom line, bubbling $\ce{CO2}$ through water may result in seemingly reduced yield from carbon dioxide dissolving.
To test, replace the tap water with freshly boiled (and still hot) $\ce{NaCl}$ rich water, and report results.


text:
Relative strengths of methyl, primary, secondary, and tertiary C–H bonds

Which of the highlighted C–H bonds in the following compounds is weakest? (R = generic alkyl group)

I personally thought that the answer was 1 but it is given as 4 in the answer key. I didn't understand how it is so.

I believe hyperconjugation is a possible cause for this phenomonen. In general, the greater the amount of electron donation to the antibonding MO of a bond, the weaker the bond. Note also that it is widely-acclaimed that $\ce {C-C}$ $\sigma$ bonding MOs are better hyperconjugative donors than $\ce {C-H}$ $\sigma$ bonding MOs.
In the case of methane, there is no hyperconjugation occuring. Thus, naturally, we would predict that it has the strongest of the $\ce {C-H}$ bonds. In the case of the $\ce {C-H}$ bond made to the tertiary carbon, there is the most numerous number of hyperconjugative donations of electron density to the $\ce {C-H}$ $\sigma$ antibonding MO due to there being the most number of adjacent $\ce {\alpha C - \beta C}$ and $\ce {\alpha C - \beta H}$ bonds acting as hyperconjugative donors.

It is probably Cα–Hβ (or Cα–Cβ) σ MOs acting as the donor orbital, analogous to ethane as in the diagram here, not the immediately adjacent C–Cα which have very poor overlap with the C–H σ*

@orthocresol Yes I agree. I should clarify that in my answer.

$\ce{X–H}$ bond dissociation energies usually correspond inversely to $\ce{X^.}$ radical stability, so another way of looking at it is to recall that the stability of radicals increases in the order methyl < primary < secondary < tertiary.
The root cause of this is hyperconjugation (from $\sigma_\ce{C–H}$ to singly filled $\mathrm{p}$ orbital on carbon), analogous to the stabilisation of carbocations. So, it is just the second side of the same coin, really.

My approach of analysing the hyperconjugative interactions to the C-H sigma * MO in the molecule is equally valid right?

Indeed, that’s the same thing in the end.

Consider ethane.
The bond lengths are as follows: $\ce{C-H}$: $\pu{110 pm}$. Bond angle is:
$\ce{H-C-H}$: $109.6^\circ$.
And, consider those for methane:  lengths are $\ce{C-H}$: $\pu{109 pm}$ and bond angle is: $\ce{H-C-H}$: $109.5^\circ$.
Though small, the angle has risen from methane to ethane (0° to 1°). Also, the bond length has risen. These influences are (supposed to have been) caused by steric effects of a methyl group in ethane. And thereby causing a decrease in bond energy. Also, due to the presence of three alkyl groups, the electrophilic character of carbon (though small already) is reduced reflecting in its share of electrons in the $\ce{C-H}$ bond of a 3° alkane causing the bond to weaken (I emphasize, this effect has very very little share).


text:
What is the term used as correction for attractive forces in the Van der Waals Equation?
I have seen a question in my textbook which is little bit confusing,

The term that corrects for attractive forces present in real gas in van der Waal equation is
(i) ${nb}$
(ii) $\frac{an^2}{V^2}$
(iii) $\frac{-an^2}{V^2}$
(iv) ${-nb}$

Since the question is about term  "forces" I do not see any of the four options dimensionally correct as (i)&(iv) are related to volume and (ii)&(iii) are related to pressure.
But in my textbook answer the correct answer given is (ii) so how is it dimensionally correct?

Kudos for thinking of dimensional analysis. It is a very handy tool to have in your toolbox.

Nothing states that the correction is itself a force, only that the correction is for attractive forces. There is thus no need for the correction to have the dimensions of a force.

I think a better way to understand the term is to look at the Van der Waals equation itself. You start with the ideal gas law
$$PV=nRT$$
Then add a fudge factor to the pressure to account for the attractive forces between molecules, $\dfrac{an^2}{V^2}$, and another fudge factor to account for the actual volume of the gas molecules, $nb$, themselves.
$$\left(P+\dfrac{an^2}{V^2}\right)(V-nb) = nRT$$
So doing a dimensional analysis on $\dfrac{an^2}{V^2}$ must yield the same pressure unit as $P$.
Doing a dimensional analysis on $nb$ must yield the same volume unit as $V$.

The term $an^2/V^2$ is not a force but a pressure (force/area) and  is a term that corrects for the effect of attractive forces between molecules. The constant $a$  has units of  $\mathrm{dm^6 bar }$ so $an^2/V^2$ has units of pressure.  When the vdW equation is expanded it becomes $an^2/V$ which has units of pressure $\times$ volume ($pV$) or energy (Joules) which is the unit of $RT$. The $1/V$ is there because at a constant number of moles, when the volume is large, the molecules spend less time close to one another and so the effect of intermolecular forces is reduced as one would expect.


text:
the pattern of losing electron
The question is the electron configuration of $\ce{Fe2+}$.
Why does it lose the electron of $4s^2$ first instead of losing $3d^2$?
I consider the pattern of the losing electrons in the $p$ orbital. The electron in $p_y$ first but not the $p_z$ in the $Mn$.