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Why ethyl cyanide and propanol have the same boiling point?
As I understand, the intermolecular forces of ethyl cyanide are of van der Waals nature (dipole-dipole) but in the propanol molecule there are hydrogen bonds which are in principle much stronger. Why then their boiling points are equal (97 ºC)? Their molar mass is very similar too.
Obviously, the triple bond of the cyanide group must be the cause, but I do not know how it can affect the boiling point, as the C-N bond is less polar than the O-H bond.

Van der Waals bonds usually describe a different class of (weaker) interactions than dipole-dipole interactions (ultimately induced dipole interactions between molecules that don't have strong dipolar features). Whether this is a factor here where both molecules have strong dipoles is less clear.

@matt_black  The term van der Waals force ...... is sometimes used loosely for all intermolecular forces. The term always includes the London dispersion force between instantaneously induced dipoles. It is sometimes applied to the Debye force between a permanent dipole and a corresponding induced dipole or to the Keesom force between permanent molecular dipoles.

It goes in the direction of being a coincidence. You can ask the same question for a huge number of pairs of substances. All you need is picking up values from tables or literature. That is because you have a number of factors (even interconnected) which contributes to the phase diagram. From molecular mass to molecular shape, from packing to polarity, etc. In the specific case, cyanide molecules are quite polar. CN is an electron withdrawing group, you should look at its effect on the whole molecule.

Use the term propiononitrile rather than ethyl cyanide lest someone gets the idea it is toxic in the sense of cyanide.

@Alchimista it is correct that this kind of similarities can be found searching through data tables but even so, it is interesting to know why this happens. Some days ago I posted a question asking why hydrogen bonds are stronger than VW forces. After, explaining this to my high school students, I used randomly a polar molecule like ethyl cyanide to illustrate the VW forces and ethanol for hydrogen bond and I got the surprise. Now I know that I need to be more careful regarding this general statements about intermolecular forces like hydrogen bonds are always stronger than VW forces.

I mean that you cannot really know why it happens without perhaps modelling and calculating a lot for each couple. I did mention tables not as an activity, but to say that one can find similar boiling points or melting points for a multitude of molecules, even belonging different classes. As for there is nothing that alone determine such properties.

And, as you are a teacher, in the case you proposed and for those purpose, it is not important how much che CN bond are polar, but how much the molecule is.

Yes, I just found that the dipolar moment of the ethyl cyanide (propiononitrile) is 4,04 Debye, almos 2,5 times higher than propanol's... this may be the most important factor, I think, to explain its high boiling point.

@fich good. Glad it helped.


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In metal complex electron counting, is the neutral or ionic formalism more correct?
I have been practicing finding total valence electron counts by both methods, and a few points have arisen that have confused me.
1) Which method gives the true d-electron count?
From what I understand, the ionic method assumes that both electrons from a bond go to the more electronegative atom, and the bond is 100% ionic. This lines up with formal oxidation state which uses the same premise. As a result, d-electrons are usually calculated as completely lost from the metal, giving a positive formal oxidation state.
In the neutral method, the d-electron count doesn't change as all calculations assume that M-L bonds are cleaved such that both species are neutral, and metal oxidation state stays at 0.
This results in 2 different values for a d-electron count, and I was wondering if the value actually means anything. I guess it doesn't matter in the end as the same number of electrons are put into the overall MO scheme, but I was wondering if being asked the d-electron count has any point to it. This can be seen in the below example from here:

2) If the neutral formalism is used, where does the electron deficiency in positively charged complexes come from?
In the ionic method, we assume everything is charged and just take electrons out of the metal to account for the overall charge (even this seems strange, why do we take electrons from the metal when we could take them from anywhere else to give the same MO scheme?). But in the neutral formalism, if everything is assumed neutral, where are we supposed take the electrons from?
3) If the electrons don't have to be taken from the metal, what is the point of assigning metal oxidation state?
I always hear statements like "d8 metals such as Pt(II) commonly form square planar complexes", but both the assignment of d8 and Pt(II) differ when using a different formalism. So is the use of the ionic method and oxidation of the metal specifically simply a convention that allows us to compare complexes quickly by their total electron count?
It seems to me like the correct answer is somewhere in between the 2 methods (between 100% ionic and 100% covalent), and the use of one or the other is arbitrary, but any corrections would be appreciated.
Also, if oxidation state is arbitrary, is the argument that "metals with lower ionisation energies reach higher oxidation states so can form more bonds" the same as saying "metals that have lower ionisation energies generally have larger radii so can fit more covalent bonds around them"?

You have good questions, and I don't have the time to address all in detail, but the gist is that the electron counting method is arbitrary (that's why they give the same answer, and you should be worried if they don't). Those merely tell you the total number of electrons in the MO diagram, they are not meant to reflect some physical reality about their distribution. Oxidation state and d-electron count aren't arbitrary. Those tell you how many of those electrons are in the metal d-orbitals (or t2g/eg orbitals, etc. for different geometry). That has a great influence on chemical behaviour.

Well, technically the concepts of oxidation state and d-electron count can be stretched to a point where they don't make much sense (e.g. when there is heavy delocalisation), but as long as you don't go there...

@orthocresol Looking at an MO scheme, it doesn't seem to matter where you decide to take the electrons from, it will still give the same number of electrons in t2g/eg orbitals. Does this mean using the formal oxidation state of the metal and taking electrons from the metal to account for charge gives you a not so arbitrary quantification of eg/t2g electron count that we just CALL a d-electron count, ignoring the ligand character?

The eg/t2g orbitals are mainly metal d-orbitals (you can see that from an MO scheme), so yes, the oxidation state or "d-electron count" pretty much ignore the ligand contribution to these orbitals. Of course, these are not 100% metal d-orbital character, which is why the "truth" is somewhere in between. That starts to really break down when... well, I already mentioned it. :-)

I think, the answer to number 1 will remove the need to answer the subsequent questions:

1) Which method gives the true d-electron count?

Neither!
You cannot, and I want to stress that, derive the oxidation state of a metal in a complex solely by looking at its ligands. You can make educated guesses and in some cases only one oxidation state is plausible but there are many cases—most notably those denoted $\ce{\{FeNO\}^n}$ where such an assingment is meaningless to impossible. Without consulting spectroscopic data, it is absolutely impossible to tell whether a $\ce{\{FeNO\}^7}$ compound is $\ce{[\overset{+II}{Fe}(NO^.)]}$ or a $\ce{[\overset{+III}{Fe}(NO^-)]}$. In fact, entire papers have been published arguing one way or the other.
This essentially renders your follow-up questions entirely moot, but I do wish to take the time to point out that the metal’s oxidation state along with the corresponding oxidation state of the ligands is theoretically important information that can predict the electrochemical behaviour of a complex. The only problem is that it cannot always be neatly predicted as shown above. It is worth noting, though, that certain complexes have been assigned metal oxidation states beyond reasonable doubts (e.g. $\ce{[Cu(NH3)4(H2O)2]}$), and in such cases the predictions of reactivity are absolutely adequate.

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Chemical to dissolve rubber/latex glove
I seem to have a drain pipe in my flat blocked by what seems to be a latex/rubber glove. I am unable to pull it out — is there a chemical that can reliably dissolve the glove and at the same time not damage the PVC pipe itself?

Look for PVC compatibility to turpentine. The latter should soften the latex gloves. It will require drying the drain, in case it does not affect the pipe.

There are a number of issues with the question: 1. How do you keep the solvent in contact with the glove until it completely dissolves?
2. Such solvents do not belong in a sewage system. Use a snake.

Thanks. I tried snake already, also endoscope camera. The pipe is completely blocked and not possible to take the glove out, that’s why I’m looking for something to dissolve it. I already tried all other alternatives.

Any chance of a plumber?

Usually it's M.Farooq, but now it's my turn to get surprised for the negative reception of a question. This appears to be a decently-written practical problem which is on-topic for Chemistry.SE, and which I personally encountered several times in a teaching lab. Fortunately, mechanical cleaning did the job for me, but a chemistry-related solution (sorry for the tautology) could also be interesting.

This answer by @andselisk may be relevant. I think disulfur dichloride would dissolve latex, maybe slowly, but it is a chemical weapons precursor, and not something to flush down a drain anyway. An ordinary sodium hydroxide/Al/sodium nitrate drain cleaner, e.g., “Drano” in the USA, generates lots of heat, is extremely caustic, and presumably would not destroy the PVC pipes. But experimenting on your flat’s plumbing could get expensive: plumbers make big bucks because DIY plumbers often just make things worse. (No downvote from me, BTW.)

Thanks for all the comments and suggestions. Naturally I don't want to use anything too aggressive / dangerous but I've exhausted all other options (including plumbing snake, qualified plumber etc.) other than dissolving it and destroying half of my kitchen to reach the pipe. I am assuming that turpentine or sodium hydroxide/Al/sodium nitrate (Drano) are the best options, unless someone has better suggestion?

This is really a plumbing issue much more than a chemical dissolution issue. The problem is uncertain: "seems" to be a latex rubber glove; complete blockage.
I'm trying to envision how a rubber glove completely blocks a drain. Perhaps by partially blocking the drain, then catching hair and other particles - but on what? Perhaps on something else sticking into the drain - a root? Some other partial blockage?
Whatever; let's imagine pouring some solvent into the drain. For now, let's forget the PVC pipe: it is much thicker than the glove and will not dissolve before the glove does. As soon as the glove dissolves, the built-up head of solvent should flush away and prolonged rinsing with water will pretty much clean off the PVC pipe.
Unless the glove is caught on some projection and merely softens, while the 5 gallons of solvent drips away over a few minutes, leaving the glove softened and intermixed with the other captured trash. Now the problem has intensified. As solvent dissipates, the glove will become the adhesive that binds the trash together and to the PVC. But then the solution will be obvious: discover where the blockage is (with your snake), then remove that portion of the pipe and replace it. And don't do that ever again.
But if you insist on a chemical solution, I suggest air: pump up the pipe with a tire pump or one of those air blasters for toilet blockages and push the blockage down to larger pipe. It would also be helpful to load the blocked pipe up to your drain opening with a solution of dish detergent for lubrication and to decrease the volume that you have to pressurize.
If you must get more aggressive, there are tables of glove resistance (https://www.augusta.edu/services/ehs/chemsafe/PDF%20files/gloveselechart.pdf) online, but it would be easier to do your own experimenting. Since you have suspected it is a rubber glove, you might have other rubber gloves available that you could put into various solutions for a few hours or days to see how the glove behaves. But I think the glove is only part of the problem, and a good commercial drain cleaner would attack what I suspect is the helper-blocker, and dissolve that, and alkali isn't too gentle with latex rubber anyway (although the chart suggests natural rubber withstands alkali fairly well - except for ammonia). Also bleach is tough on natural rubber. Of course, YOU MUST NOT EVEN THINK OF MIXING AMMONIA WITH BLEACH!!!

(+1) Nice answer, though calling the plumber is still likely the best option. And the “some projection”, if there is one, may very possibly be an errant toothbrush.

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Where can I find the released combustion energy in NASA CEA rocket problem?
I am using NASA CEA program for a study project.
I have read the CEA Nasa user's manual over and over, yet I haven't found an answer to my question which is :
where can I found in the output of NASA CEA the energy released during a combustion in a rocket problem ?
For instance, using CEA I have simulated the following problem :

Rocket problem with infinite area-combustor and assuming equilibrium in the combustion chamber
pressure chamber at 1 bar
combustion of H2 with O2

And in the output I have these data, but I am not sure where or how I can find the energy released by the combustion of H2 and O2 : Is it the yellow-highlighted one ? Or is it the red-circled one ? Or it's not given, I have to calculate using the output data ?
Thank you

A little subtraction should do the trick.

Do you mean subtracting Hthroat - Hchamber = - 1293.48 - 6.6529 ?  I thought that formation enthalpies should be involved somehow


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Any idea why tap water goes stale overnight?
We have great tap water here in Hungary. My only problem that it goes stale overnight sometimes in some glasses or sometimes even the 2L water boiler has this problem too. But microbes cannot resist 100°C or 20% vinegar, so I thought it might be rather a chemical compound what causes it. It is really hard to get rid of it. When I use boiling water or vinegar it does not remove it. When I use soap and water several times, it removes it most of the times, but only for a few days. The only working solution was not using the affected glass for a month. It is hard to descibe the taste, but it is something like how new Chinese clothes or paint thinner smell so it is sort of strong plastic taste. The longer I use a glass without washing it every day, the bigger the risk it develops this feature (2 days are enough for certain glasses, while others never develop it), so maybe microbes are producing it after all, but the compound is long lasting even if they die off. Any idea what causes it and more importantly how to get rid of it permanently without having to wash my glasses, boiler, water jug each day?

You should get your stale water tested from a laboratory. Normal water will not change its taste overnight in normal containers nor should it develop plastic taste. Are your glasses food grade glasses? How do you store it? In fridge, water can absorb food odors.

@M.Farooq Nope it is in the living room at room temp.

Just so I get this right, you don't want to wash your dishes?

@Karl Haha, lol. I don't want to wash them this often and washing them does not seem to solve it. Even after washing them with boiling water and soap I can taste it a few hours later. So far I tried it with glass and stainless steel. I'll try it with ceramic too.

You're contradicting yourself. You said the taste occurs sometimes if you leave a glass with water out overnight. And that it only occurs if the glass hasn't been washed for 2+ days.

There is a nozzle on most faucets, whose purpose most people don't realise is to aerate the water as it comes out. Without any other information or testing I would have assumed this is the important effect.

Do you use a lot of fragrances (perfumes/colognes/air fresheners)? As a person that is highly sensitive to most perfumes, I have found that a glass of water left out tastes off when exposed to a heavily scented environment. You may find this article informative: How to Know If Your Perfume Is Poisoning You.

"This [well] water [which is where 75% of Hungary's water comes from], when brought to the surface from great depths, is rich in various mineral solutions and may be very hot." – Hydrology of Hungary - I live in Chicago; "good" tap water doesn't go stale.

Exclude reasons one by one
Surface biological contamination: use two vessels, wipe the one of them with H2O2, hydrogen peroxide solution, leave another untreated. Check if time till smell appears is different by a few days. Bacterias can survive 20% vinegar and 100*C for a short time, some of them, if they are in a sleeping stage, it is not quite reliable method i would say.
Tap water biological contamination: use two vessels with water, add a few tablespoons H2O2 with water in one, and just water in other. Check if time till smell appears is different by a few days.
Tap water chemical contamination: buy some bottled water, use two vessels, one with tap water, another with bottled water. Check if it will be different in a few days.
Bulk material of the vessel releasing chemicals: leave water in different material vessels, one is glass, another one is your preferred one. Check if glass vessel has less smell adter a few days. Glass releases the least amount of chemicals into the water.
Dust accumulation: use two vessels, cover one and leave another one open, check if smell will appear faster in one.
Psychological factor: use two identical vessels, ask your friend to give you one or another, randomly, blindfold yourself, do a test 10 times, check if your score is above 8 correct guesses out of 10 tries.
After completing this tests you will narrow down your search significantly to give you some idea about what is going on.

Thanks! I'll do the experiments. I'll rather use KMnO4, I find it safer for drinking water not sure how much contaminants the H2O2 would contain. It is hard to get decent chemicals in my country.

KMnO4 also works, just as good, you can use it too. I offered H2O2 because its usually easier to get, used in medicine for small wounds treatment. And H2O2 is exceptionally sensitive to contaminants, it decomposes if any dirt is present, unlike KMnO4. And KMnO4 is prohibited in some countries because something related to drugs. And H2O2 decomposes into safe products, water and oxygen.

It was prohibited here for a few years to have big quantities of KMnO4, but currently it isn't. Wineries use it a lot afaik. I have a little, maybe 5-10g crystals from an organic chemistry research lab, so I am pretty sure it is pure asf. I really liked to work in labs, but I ended up in IT somehow. I still miss it, I would do both if I could.

KMnO4 is fine from toxicology point of view, but maybe not for this experiment since it has a distinct taste.

@cbeleitesunhappywithSX In theory it is sweet. I don't think it will be hard to distinguish it from the other taste.

A few more thoughts:

fresh water taste is AFAIK associated with come $\ce{CO_2}$ being dissolved in the water. This can get lost, and make the water taste stale.
Here's an easy experiment to check this: compare the taste of:

fresh tap water,
water gone stale "your way" (cooled to the temp of the fresh tap water), and
de-gassed fresh tap water: take water, boil it to remove dissolved $\ce{CO_2}$, cool to same temp as the other 2.

If the boiled water tastes stale like the stale water, loss of $\ce{CO_2}$ is likely the cause.
I wouldn't describe this as anything like paint-thinner or plastic or the like. But for me it is the textbook meaning of "stale" for water.

Sidethought about "chlorine" smell. What we usually describe as chlorine smell in water are chloramines (mono-, di-, tri-). In first approximation, we can say that they form from e.g. from chlorine used for water treatment with amino groups from organic matter. AFAIK, our nose is more sensitive to them than to chlorine itself. I.e. the smell indicates that some organic contamination occured, but also that chlorine was there and did work - but it does not indicate whether the water is now safe from a microbiological point of view, the amount of chlorine may not have been sufficient.
So if your tap water arrives fine and with some residual chlorine, but you then get organic matter into the water, chloramine smell may develop later on.

In my supported opinion, it’s first chemistry (actually Fenton-type chemistry and contributing photocatalytic sunlight ). I am assuming one has copper (or iron) metal pipes bringing in the water containing dissolved oxygen into their home. Further, it has been treated with say Chlorine (creating residual Hypochlorous acid, HOCl) or even Chlorine dioxide (where .ClO2 is stable free radical) and left sitting out in some light.
The second piece of the puzzle is the introduction, at some point, of organisms that will flourish upon removal of said chlorinating agent or Chlorine dioxide.
The chemistry-based explanation, in words, I would describe as, for example, a Fenton (or Fenton-type and photo-Fenton) reaction occurring with transition metal ions in the presence of dissolved oxygen creating the so-called superoxide radical anion.  The later radical can directly react with Chlorine dioxide creating the Chlorite anion (ClO2-),  which is unstable and far less effective in disinfecting water than Chlorine dioxide.  The superoxide can also react with Hypochlorous acid in a decomposition reaction forming powerful, but unfortunately, very transient radicals.
Some background material:
Ozone treatment while effective produces no residual continuing disinfecting action, but can reduce the subsequent amount of chlorine, for example, required, so no further mention of ozone is needed.
$\ce{ClO_2}$ is a gas that is highly soluble in water (some 8 g/L at 20 °C). However, per Wikipedia , to quote:

It does not hydrolyze when it enters water, and is usually handled as a dissolved gas in solution in water.

Also, from  a government source makes the statement:

In air, chlorine dioxide readily dissociates both thermally and photochemically and may form chlorine, oxygen, hydrogen chloride, HClO3, HClO4.ClO, chlorine peroxide, and/or chlorine trioxide, dependent on temperature and humidity. Chlorine dioxide dissociates in water into chlorite and chloride, and to a lesser extent into chlorate (Budavari et al. 1996).

Further, a key supporting  Dutch study: Chlorine Dioxide as a Post-Disinfectant for Dutch Drinking Water, to quote:

In this paper, results are presented of experiments into the consumption and reaction kinetics of chlorine dioxide in a number of (drinking) waters in The Netherlands. It was found that chlorine dioxide consumption is related to the dissolved oxygen content (DOC) of the water and the reaction time.

Now, relating to the superoxide radical anion, it can be created by a so-called metal auto-oxidation reaction with say ferrous (as in $\ce{Fe(HCO3)_2}$) or cuprous presence:
$\ce{Fe(II)/Cu(I) + O2 <=> Fe(III)/Cu(II) + .O2-}$  (See also, Figure 1 here)
Further, an implied radical reaction to account for the loss of Chlorine dioxide:
$\ce{.O2- +  .ClO2 ->  O2 +  ClO2-}$
In summary, the presence of generated superoxide from various sources in natural oxygen-rich waters (from sunlight and transition metals,..., interacting with O2), which has been treated with say chlorine dioxide, where superoxide concentration is connected to dissolved oxygen presence, is one of the more likely pathways to account for the reported removal of $\ce{ClO_2}$.
On references citing superoxide presence in natural waters, I refer those interested to Measurement of Antioxidant Activity toward Superoxide in Natural Waters, which mentions a photochemical pathway for superoxide, which is likely the largest source for sunlit waters.  Also,  ferrous concentration in natural waters is discussed in this paper, where, it should be noted, that absence light, transition metals and dissolved oxygen could be sources of superoxide.
Bottom line,  oxygenated water, transition metal ions (newly introduced from household plumbing), and associated water disinfecting agents can interact resulting in the eventual removal of said agent. This leaves the water vulnerable to colonization by microbes, algae, or whatever opportunistic organism finds the aqueous environment favorable.
Note, my proposed explanation accounts for how the treated tap water after entering the home and exposed to light, could soon lose its disinfecting power, and in time, can become readily contaminated, therein altering the water's taste.
[EDIT] As a side comment, the mechanism at work here (radical chemistry assisted by UV light exposure and biological) may similarly be at play in the observed off-taste that develops in previously open water bottles left in cars subject to UV light, added air and bacteria (from partial consumption) and/or warming. My personal experience strongly suggests, at times, a UV-assisted hydroxyl radical attack on the plastic vessel itself introducing problematic organic breakdown products.

This is good. I mean we can add another factor to the experiment "Surprised Seagull" suggested: sunlight or artificial light. I am not sure about the exact material of the pipes, but they are made of metal.

Tap water in Germany is by law absolutely free of any disinfectants. I have never heard about anybody having a "paint thinner" taste in the morning.

@Karl: also in Germany, drinking water is treated routinely, and chlorine treatment is one way. We may have two advantages over Hungary, though: I know from some water engineer friend that many German regions have soil temperatures that are just sufficiently low to make drinking water microbiological safety easier than it would be at only a few degrees warmer. Hungary may need a bit more chlorine (and it may be unable to go for e.g. UV disinfection only for the same reason) in order to have micobiological safety all the way to the tap. The second aspect is the state of the supply network:...

... how likely is microbiological/organic contamination to happen after the treatment point of the local drinking water supply. Not sure how Germany vs. Hungary compare in that respect, but I'd anyways expect substantial regional differences. All this is not paint-thinner like, though.

@cbeleitesunhappywithSX There are big differences between regions here. I live in a mountain region, where water quality is better than in lowlands.

My daughters all kept their fresh drinking water in the fridge and never complained about the taste of water that had been kept there in a covered plastic jug even after several days. When I was a child living in Malawi in the 1950s, all our water was drawn from the river. My mother kept our boiled filtered drinking water in a charcoal cooler. That primitive device would never have reached a temperature anywhere near that of any modern fridge, but I still found that a glass of water straight from the cooler was very refreshing and tasted sweet, even if it had been kept there for a day or so. Cool water always tastes fresher than water at room temperature. I suggest you do a simple before and after comparison. Take the temperature of your fresh tap water and check its temperature again before you taste it the next morning.


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Combustion of NaCl(aq) when testing conductivity
In a experiment that dealt with electric conduction of many concentrated solutions, NaCl(s) created an unknown, to me, substance after adding a drop or two of water. Furthermore, it isn't the reaction of water and NaCl(s) that I don't understand, but the reaction between the anode and cathode when they came in contact with NaCl(aq), sparks were present and it seemed to have "burned" NaCl(s). The question is: what is the correct name for the part that "burnt"?

Nothing was burnt. You have just had a dangerous experiment. Apparently you have connected the two electrodes to a generator before dipping them into the solution. If an electric spark happened when touching the solution, it means that, before touching the liquid, there was a tension between the electrodes and no current. This is dangerous. You also may have been crossed yourself by an electric current. With wet hands, you may have been dead. You should never do this. Please put the electrodes into the solution before engaging the generator.

What kind of voltage did you apply here?

Not sure what is wrong with this question! Have you never seen sparking of closely placed electrodes in electrolytes?

To answer both questions at once: the voltage was your normal household voltage, well at least I think it was; the professor asked us particularly about what the "burnt" part of the salt was and she also added, in the question, why were there sparks, that was my reasoning.

@Mate, Your question was initially voted to be closed. My comment was for the folks voting to close it. Your query was okay.

There is not "burning", one might see a spark. It is a well known phenomenon. Do you see lightening in the sky? Air is an insulator, yet when very high voltage is present it becomes a conductor. It is called dielectric barrier discharge. In the same way, discharge between two electrodes can occur in pure water (almost an insulator). Adding NaCl to water makes it a good conductor but still it is mostly water, and if the electrodes are close enough, electrical breakdown can occur. Instead of ions carrying the current as it should happen in an aqueous conducting solution, the current becomes electronic even in the presence of water. Apparently, the local heating between the electrodes is so high, that a gaseous plasma is formed between the electrodes and plasma are electrically conducting.
As suggested in the comment, always exercise cause caution and all experiments should be done under proper supervision by a trained person/teacher.


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Gibbs free energy as a difference of irreversible and reversible heats
I read in a book that ∆G=∆H-T∆S and then ∆H and T∆S were substituted with Q and Qrev respectively(where Q and Qrev are heats added to the system in any process and reversible process respectively) and then there were two cases,

Case 1:irreversible process
Q was replaced with Qirr(heat added to system irreversibly)so ∆G=Qirr-Qrev
Case 2:reversible process
Q was replaced with Qrev and so ∆G=0

What I don't understand is that replacing T∆S with Qrev in both the cases implies that the state change in both cases is same(∆H,∆S and ∆G are state functions) but the values of ∆H and ∆G imply otherwise.Is the book wrong or am I thinking wrong?

Please provide a reference for your book.

@ChetMiller R.C. Mukherjee's modern approach to physical chemistry(1) applications of the concept of equilibrium


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Effect of dilution on weak acids pH
My textbook has a question which says what would happen to the pH of a weak acid if it is diluted with water. The answer is its pH would decrease.
My teacher’s argument is that it would indeed decrease because water ionizes to hydronium ions which would increase the concentration of H+ ions and therefore, decrease the pH.
My argument is that it would increase because when the acid is diluted, the concentration of H+ ions will decrease and hence, the pH will increase. Further details in the attached photo.
My father, who is a chemist, argues that it would decrease then increase. He says that it varies upon the amount of water added and that it is not a direct relation. Instead, it would form a graph similar to the letter V.
I am honestly lost. What would exactly happen?

Simplified formula is pH = 0.5 ( pKa - log c )

Things get weird when there is very little water, as in glacial acetic acid. See e.g. https://www.sciencedirect.com/science/article/abs/pii/S0167732216329154

This would be crystal clear with a derivation of the appropriate pH equation from scratch, with no approximations. This question is closely related. The first graph in the top answer directly shows the shape of the curve you seek - the pH of a weak acid solution rises smoothly and monotonically until it becomes so dilute that the pH is pegged by the autoionisation of water at pH ~ 7. You are correct.

The pH will increase because if you dilute enough you will have a solution with acid concentration and its $\ce{[H]^+} \lt 10^{-7} $ and then the dissociation of water will keep the pH close to 7.

Whatever the nature of the acids, weak or strong, dilution of an acidic solution produces an increase of pH. The dissociation coefficient may increase, but the pH will always increase. The expression "like a Y letter" is pure fantasy.
Your calculations about acetic acid are correct. Just a small remark : in the second part, you say :"if  you double the volume". It should be stated that this operation is made by adding pure water. In this case, the total concentration of acid is divided by 2. Surprisingly, the concentration of $\ce{H+}$ is not divided by $2$ : $\ce{[H^+]}$ = $2.99 ·10^{-3} mol L^{-1}$. So it has been divided by $√2$ (and not by $2$).
Now, let's speak about the number of moles. During the dilution, the total number of moles of acetic acid has not changed, but the total number of moles of $\ce{H^+}$ has slightly increased during the dilution, which means that the acid is more dissociated in the diluted solution.

The degree of dissociation will increase in this case, so more ionization will take place,but it wont b able to counter the decreased concentration.
Actually, if you write the equilibrium equation, taking c as the initial concentration of acetic acid, and x as the degree of dissociation, then the acid dissociation constant, Kc will approximately be equal to $cx^2$, and the concentration of H+ will be cx.
Now, Kc is constant for a given  acid at a given temperature,
therefore, $cx^2= cx*x$ is constant.
So if x increases on dilution, the cx will decrease, hence the concentration of H+ will decrease.
So, the Ph has to increase in this case.


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Permeability of containers to poop smells
I do a lot of hiking trips that involve driving for an hour or two to the trailhead in the predawn hours. I'm a morning pooper, so I need to poop during the drive. Before COVID, I would usually plan a stop along the way at a Starbucks, get coffee, and use their bathroom. Then if necessary I'd also use a bathroom at the trailhead. Right now, during COVID, these bathrooms are all closed, so my new normal is that I pull over somewhere in the countryside, poop, collect my poop using one of my dog's poop bags, and bring it with me to a town or trailhead where I can dispose of it.
Sometimes this was stinking up my car, even though the bag was tied shut, so I tried putting the poop bag inside a large plastic (PETE) can that some nuts came in, and sealing it with the screw-on lid. Even then, there have been times when the smell seeped out within 30 minutes. Sometimes the smell has been like sour milk, sometimes like sulfur or eggs.
If I want to make this work better, what should I be focusing on? Is ~0.5 mm of PETE likely to be permeable to the smelly volatile compounds involved? Or is the issue more likely to be that there's a poor seal at the screw threads? Would I be better off trying something like the glass jars with clamping lids that people use for their coffee? A tupperware-style container with an o-ring seal? What kind of material should I be looking for in a gasket or o-ring?
(The other option I've tried successfully is to clamp the poop bag under a windshield wiper, so it's outside. But I'm worried this would, uh, catastrophically fail if I then drove at freeway speeds.)

Wow, three downvotes and a close vote, none of them with any explanation whatsoever. Are people just that offended by the subject matter? This is just a question about permeability of plastics to organic chemicals. Please think before having this type of knee-jerk reaction of hostility.

Yeah, Chemistry SE is one of the least welcoming communities I've seen. They downvote/close everything that moves. What doesn't move - they push and then downvote/close.

I am against this down voting phenomenon which is only observed here in chem. The problem is chemistry gets a lot of lazy homework or medical questions. On the other hand, many good genuine and innocent questions  get a habitual down vote by some stubborn offenders. I rarely down vote but this question because it was too vividly described with unsolicited details. It could have been asked in better and formal terms.

The downvotes are one thing, and I imagine that’s related to the subject matter, although I can’t guess for sure. But I don’t understand the close votes, to be honest. How / why is this “opinion-based”?

On the other hand, though, Buttonwood’s answer does suggest that you might find more practical answers elsewhere on the SE network, too. There is not really all that much... chemistry in the answer. Perhaps somebody has studied the relevant permeabilities......

I chose the "opinion based" option on VTC, largely because Professor Crowell has not provided us with details about the composition of the feces in question, which certainly is required information to begin to address this question. Also, it's not focused: there are at least 5 sub-questions contained herein. Finally, the superfluous information concerning the morning stooling and unfortunate closure of the nearby Starbucks detract from the question.

@orthocresol True, there were no headspace GC-MS data provided, as sometimes used by Firmenich and Givaudan to offer new components for perfumes.  Despite some of the ingredients have «some» history (e.g., 3-methyl indole, by the way not liquid at ambient temperature), the revision aimed for something «ready to be used», with less sorrows than the chemical toilets seen on construction sites / festivals.

@ToddMinehardt, opinion-based question would be "which producer of goggles do you like the most" - there're no correct answers to such questions. If a question is missing details - you can ask for them. The question doesn't become invalid simply because you need more details to answer it (moreover some people are asking for extra details which aren't actually needed for answers, I see this all the time. This isn't specific to ChemSE though). As for "not focused" - that's when a post contains multiple unrelated questions. The question above is perfectly focused on one problem.

@StanislavBashkyrtsev Thank you for sharing your opinion.

Out of curiosity, is it against the law to dispose of poop bags in public garbage bins (as in a park) in your area? Does the same apply to human feces?

@BuckThorn This (obviously) depends on where you are.  In Switzerland, for example, you find dedicated garbage bins, robidog by Robidog AG: bins, bags, and even phone apps to localize the next bin. The local municipality decides if dog poop (only in a bag) may enter the normal public bin. And no, thankfully never seen to be used for human feces, there are public toilets (https://pee.place/en).

@Buttonwood I have been in similar situations to that described by the OP and have never had this problem using standard poop bags (double bag and tie a tight knot!). Maybe he's special. I guess my thoughts are that this is both logistical problem regarding access of a disposal site as much as of odor containment.

@BuckThorn Possibly yes, «logistical problem regarding access of a disposal site». I read about his in the newspapers, e.g. comparing the climb by Norgay and Hillary back in 1953 to today's situation in some of the camps today (https://en.wikipedia.org/wiki/Mount_Everest#Waste_management).  Esthetics aside, poop, especially in high altitudes, may affect a local ecosystem e.g., as an unwanted fertilizer.  But this is a different story.

@Buttonwood Reminds me of one of the ultimate poop bag/logistical disposal problems ever: http://www.spacesafetymagazine.com/aerospace-engineering/nuclear-propulsion/will-anyone-recover-apollo-13s-plutonium/

The problem likely is not (so much) the plastic material of your container, either solid, or bag.*)  Assuming there is no contamination on the outer side of the container, most likely the smells are noticed because of closing the container by either knot or lid is not tight enough.
After some search on sibling sites LifehacksSE and OutdoorsSE I learned hikers found multiple solutions to this problem (example).  Assuming you stay on ground and don't climb, a) a poop tube may be suitable for you. In this tube, you carry the waste with you until you find back to a place connected to a waste management / waste water facility.

(credit, source describes both construction and use)
Step #3 of this guide includes an important clue how to reduce the smell.  To quote:

«Before you go out camping, hiking, or climbing, preload the airtight bags with 2-3 small scoops of kitty litter or powder detergent to remove moisture. This will also keep obnoxious odors from leaking out of your poop tube.»

b) The idea to catch liquids and thus smell in a bag filled partially with some of (advanced?) cat litter is commercialized elsewhere, too (example).
Maybe this avenue is working well enough that additional sealing of the lid by an O-rings of suitable diameter and material (viton is a material resistant to many chemicals, but I don't dismantle a vacuum deposition to check this field of application) is not necessary.
*) Although permeabilities of liquids and gases across plastics may be determined experimentally.


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Why is gasoline consisting of heavily branched alkanes better fuel than gasoline consisting of more linear alkanes?
As far as I know, the reason isooctane is better than octane in gasoline engines is because the octane is more prone to ignition by pressure. This makes it ignite when it is not supposed to, which damages the engine.
Isooctane, however, is somehow less prone to this. Why is that? Why is isooctane "more resistant to the potential of ignition incited by pressure"? How are the higher number of ligands involved in this?

Who says they are, by what quantitative measure?

@Karl That's what I've learned. Is this wrong?

@Karl Also, more heavily branched isomers have lower boiling points than their less branched isomer counterparts. I think that may be related.

Yes, branched isomers have a lower Bp., and that is of course one factor. The rest is debateable. Who taught you that, what was their reasoning? Do you know how the Diesel engine works?

@Karl Not quite sure who taught me that, I think it is a mash of stuff from different places. So, if the branched isomers are not more combustible than those lesser branched isomers, does this mean they are less combustible? Or is the amount of branches not the most defining factor behind combustion? I believe the diesel engine works by compressing the diesel hydrocarbons, which makes the combust?

Something tells me this is an extrapolation of the octane rating of a fuel, where 2,2,4-trimethylpentane (isooctane) is seen as a good fuel, whereas n-heptane is not. Whether this extrapolation is correct, and whether it has an explanation, I do not know.

Yes, diesel oil (very much linear alkanes) autoignites on compression. You don't want that in a regular gasoline engine, that's why gasoline has a high content of branched hydrocarbons.

@NicolauSakerNeto "Good" for a gasonline engine means no autoignition even at high compression. For a Diesel, you want autoignition without explosion. There is the cetane number for that.

@Karl Okay, so having branched hydrocarbons is good for making it "pressure-resistant", in the sense that it wont ignite before it is supposed to (before the fuse ignites it), and as such, won't cause knocking in the engine? How come linear alkanes are more prone to ignition due to compression than non-linear alkanes?

That's a tough one. You should ask a new question. To this one, there is no linear scale for "combustibility". There is the flash point, ignition temperature, vapour pressure, heat of combustion, ....

@NilayGhosh Yeah,  I checked out that question and it did lead me to some good information about radical-based reactions, but the research I did also made me believe more in this notion that more heavily branched hydrocarbons are more combustible. This is because I thought that the propagation step would be more abundant in radicals due to the higher amount of ligands to cut off, made more available due to the higher surface area.

https://chemistry.stackexchange.com/questions/123093/deciding-the-order-of-heat-of-combustion-of-isomeric-alkanes/130811#130811

the real problem here is not that this is a bad question but that it is (essentially) a duplicate of https://chemistry.stackexchange.com/q/99333/81

"higher number of ligands" seems to be the wrong terms, ligand implies there is a metal complex. Do you mean "more branched" or something similar?

@matt_black - you can use the "close" button at the bottom of the question to flag it as a duplicate of the question you linked.


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Find the molecular formula of a gas
The problem is:

A quantity of $\pu{35.2 g}$ if a certain hydrocarbon gas, occupies $\pu{13.2 L}$, measured at $\pu{1 atm}$ and at $\pu{323 K}$. Knowing that 85.5% of is C, find the molecular formula for the hydrocarbon.

Attempt. So I tried with the first step doing $0.85\cdot \pu{35.2 g} = \pu{29.92 g}$ of C and I'm pretty sure that's the wrong step but I'm not fully sure, so anyways, I continue. After finding the mass of C on the hydrocarbure, I subtracted $35.2 - 29.92 = \pu{5.28 g}$ of $\ce{H}$ to get the mass of $\ce{C}$ on the hydrocarbon. Now basically:
$$\pu{29.92 g} \text{ of C}\cdot \frac{\pu{1 mol}~\ce{C}}{\pu{12 g}~\ce{C}} = 2.5$$
$$\pu{5.28 g} \text{ of H}\cdot \frac{\pu{1 mol}~\ce{H}}{\pu{1 g}~\ce{H}} = 5.28$$
and this is why I think I'm wrong, I can't get a relationship of natural numbers for the empirical formula. May anyone give some input on this?
Continuation. So bravely approximating to the empirical formula by diving by the smaller one, i get $\ce{(C1H2)_n}$ so the empirical molar mass call it, $M_\text{empirical} = 12 + 2 = \pu{14g/mol}$. Getting the $n$:
$$n = \frac{M_\text{real}}{M_\text{empirical}} = \frac{M_\text{real}}{\pu{14g/mol}}$$
and to get the $M_\text{real}$, we go to
$$PV = nRT \implies PV= \frac{m}{M_\text{real}}RT \implies M_\text{real} = \frac{mRT}{PV}$$
hence
$$n=\frac{\frac{mRT}{PV}}{\pu{14g/mol}} = \frac{\frac{\pu{35.2 g}\cdot \pu{0.082 L atm//K mol}\cdot \pu{323 K}}{\pu{1 atm}\cdot \pu{13.2 L}}}{\pu{14g/mol}} = 5$$
then I get $\ce{(C1H2)_5}$ hence $F_\text{molecular}=\ce{C5H10}$. Yes, right?

You don't need to approximate it to $1:2$ so early. You could always leave it as $\ce{C_{2.5n}H_{5.28n}}$, which means a molar mass of $(2.5n \cdot \pu{12 g/mol}) + (5.28n \cdot \pu{1 g/mol})$.


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Is reforming and isomerization the same thing?
Okay, just to give an idea of my current understanding. Reforming is a step one can do after one has cracked some hydrocarbons. As such, the process of reforming is dependent on a separate process, cracking. What it involves is cracking the hydrocarbons, then putting them together in new ways, whilst still having the same chemical formula. In a sense, one is reversing the work done by the cracking, since that is a process that changes the chemical formula, however, when one puts the molecules back together, one changes them structurally. Basically, one is making isomers.
However, I do wonder if reformation also includes the saturation of hydrocarbons; as in, making alkenes and alkynes into alkanes. This obviously requires cracking, since double and triple bonds must be broken down into single bonds. If this is a part of reformation, that would mean the term applies to a wider process than just making isomers, since an alkane cannot be an isomer of an alkene/alkyne.
Perhaps isomerization is the creation of isomers, but within any field of chemistry, whereas reforming applies specifically to petrochemistry?

Generally speaking, reforming is not that exact in that you will produce a range of compounds. Since these compounds do not have the same chemical formulas as the original, it should not be considered an isomerization in the strictest sense.

Okay so would for example the conversion of an alkene to an alkane fall under the category of reforming @Zhe? Also, is reforming basically just taking mostly linear hydrocarbons and turning them into non-linear hydrocarbons, that may differ in chemical formula from their original chemical formulas?

The general conversion of an alkene to alkane is not reforming. An example would be cracking the alkene into smaller pieces, which reform into a different compound, an alkene. Note that this specific example is not an isomerization in any step, since compounds of different chemical formulas are generated at each step, including the overall process. The exception in the overall transformation would be a monoalkene to a cycloalkane of the same formula.

TL;DR: Isomerization is a subpart of reforming process. There are four major reforming reactions. Isomerization is one of them.

Reforming is a processing technique by which the molecular structure of a hydrocarbon is rearranged to alter its properties. The process is frequently applied to low-quality gasoline stocks to improve their octane number and thereby improving combustion.
There are four major types of reforming process:

Dehydrogenation of naphthenes

Dehydrogenation and aromatization of paraffins to aromatics (dehydrocyclization)

Hydrocracking of paraffins

Isomerization of normal paraffins to isoparaffins

You can see that during all the process except isomerization, there is a loss of atoms either carbon or hydrogen. However, there is no change of atoms during isomerization process.

So cracking is a part of reforming? And why is it only dehydrogenation that is a part of reforming? Why isn't hydrogenation a part of reforming? As in, making alkenes and alkynes into alkanes?

@A.Kvåle 1. Yes 2. No 3. Hydrogenation comes under hydrocracking. This process is made up of two reaction a) hydrogenation b) cracking. In cracking, the heavier hydrocarbons are broken into smaller hydrocarbons and in hydrogenation, the aromatics and olefins(if any present) are converted into alkanes and cycloalkanes. This two reaction goes hand in hand. The example shown is actually cracking.

Isomerization is a process in which any compound converts itself into its isometric forms. The chemical composition remains the same, but the structure & configuration of the molecule of the compound changes.
Coming to reforming. Reforming is a technique through which molecular structure of a hydrocarbon can be rearranged to alter its properties. This is basic definition of reforming.
So, you can now conclude the differences between the two.

You should give an example to showcase the difference.

Sorry. I currently don't have any examples with me.


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ethanol/water %vol/vol and freezing temperature
Main Question: Where can I find a table (or realistic formula) for isopropanol/water %vol/vol mixes and corresponding freezing temperature with public access?
Background:
Trying to get the freezing point of my car windshield wash fluid.
What I have tried so far:
Anti-freeze is may be ethanol or isopropanol. I got tables for ethanol/water %vol/vol and temperature, but could not find them for isopropanol. I learned that for ethanol the calculated (using simple formula) freezing temperature does not match the real one if using high concentrations, so this is likely true also for isopropanol, and why I am asking for this table.
Additional bonus question: is there a (household-practical) way to determine what was filled (and possibly even mixed during garage service): ethanol/isopropanal-based anti-freeze...?
(PS: wasn't sure about which forum, chemistry, physics... - please guide me if this question belongs somewhere else)

Googling ‘ isopropyl alcohol water phase diagram’ yields what you want.

Thanks, but actually I am having a hard time to find proper results: they refer to the vapor pressure or other alcohols...  Can find multiple dataon ethanol, even just inthe Wikipedia datapages, but not for isopropanol/isopropyl alcohol (tried several variants) That's why I was asking.

Ethanol/water: https://en.wikipedia.org/wiki/Ethanol_(data_page); isopropanol/water e.g. on https://www.engineeringtoolbox.com/ispropanol-water-d_988.html  The bonus question: no, as far as I know my tools in the kitchen, not with household items.

Thanks - may sound simple to you but exactly what I was looking for and could not find. Can you post as answer so that I can acept it?


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Aluminum Battery with PAC setup : Rapid voltage drop problem after a charge
I am a retired man and seek for activity. I dont have background in chemistry but I want to learn an aluminum battery..the known problem is my aluminum metal get eaten and also corrosion even if using sodium electrolite and worse if using Aluminum chloride.
So I just try PAC for water treatment and it look like the reduction is minimum. But as usual the usual voltage drop always problem in any set up of galvanic.
What happen and how to prevent that? my experiment setup is a square aluminum that act as a case also as cathode. Fill with carbon powder separated by paper. in the center is a graphyte rod pour with PAC in water...this gives a 2,5V
please review what kind of chemical formula during equilibrium and during recharge? why voltage is droping or how to make this setup more better and stable as battery?

What PAC stands for  ?

It's no use trying with aluminum. As soon as a piece of aluminum is in  contact with air, its surface  gets oxidized and covered by a thin and continuous layer of aluminum oxide, which is insulating and stops all electrochemical reaction. If you prevent this formation for example by amalgamation, the metal reacts quickly with air and with water, according to : $\ce{2 Al + 6 H2O -> 2Al(OH)3 + 3 H2}$. So it is destroyed and cannot be reduced to Al metal later on. The only way for Al to generate electrons in a battery is to use a non-aqueous solvant like toluene in contact with Al and AlCl3.

Aluminum works in a strongly alkaline cell, where the alkali can erode through the surface oxide film. Such a cell can provide temporary service, but the aluminum quickly reacts with the alkali and water, forming an aluminate.

PAC is poly alumunium Chloride

Poly Aluminium Chloride (PAC), is a water-soluble inorganic polymer between AlCI3 and Al(OH)3, its general formula is [Al2(OH)nCl6-nLm]. I already tried submerge alumunium foil in PAC solutuon and its not corroding...what is your comment  on this? and the voltage is at least 1V wirh 1,80 mA of current on load 5k ohmi

Thx maurice, I look up toulene didnt find many info on how to do it. Could you pls elaborate for me on how to do the experiment...? the setup


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Green stuff from natural crystal
I continuously soak this natural quartz (it is originally white) in basin of water and salt, it has a metal on it too.
What caused the green stuff?
How to remove it?

Looks like algae to me.

yes, it may be algae, how do you remove the algae?

What is the logic behind soaking it in water and salt.? Sound more like $Alchemy$ work rather than chemistry.

I'm studying the Raman spectra of quartz soaked in water salt to see if there is any changes without liquid immersion, the raman spectrum i got is this https://imagizer.imageshack.com/img922/8730/5FQzZB.jpg

What on earth gave you the idea that quartz could be modified by soaking it in cold water?

The green stuff in algae is chlorophyll. If you are able to get resonance Raman spectra of chlorophyll, you can be reasonably sure you have algae (going confocal on the surface would certainly help as well). As for getting rid of them, how about soap, water + elbow grease?

How did chlorophyll end up there? My Raman is 532nm laser. Can i do resonance Raman with it? How?

Do you know the origin of the sample?  Prior to deposit into tab water, was the sample already this green (e.g., a photo taken with the same color temperature of the light source [use of a color checker is seldom seen, even in institutions dealing with optics]) back then as now or / and UV-Vis reflectance spectrum?  For me, it is unlikely such a hard material will pick up and incorporate easily dye at ambient conditions (temperature, pressure, reaction time of some days or even months).  Think about glass ware and optic lenses, for example.
If the green does not go away despite washing with soap and a brush (quartz easier scratches than is being scratched if in contact with many materials) and you have experimental evidence that it is quartz, surprisingly*) -- according to Wikipedia -- it could be prasiolithe, a rare variety of green quartz e.g., from Brasil, or Lower Silesia in Poland:

Worth noting the Swedish edition extending sources of origin and showing, e.g. this

If still passing the tests till here, maybe there is something worth a spectroscopic check. E.g., Caltech moderates a spectroscopic directory about minerals and references to quartzes of color.  Since quartzes may be grown in the lab (green hinting for  $\ce{Fe^{2+}}$?), perhaps the Günter group at ETHZ previously collecting expertise to discern diamonds of natural and artificial origin could be of help.
*) No, I didn't study geology.

It is originally milky clear. Im sure of it.

Here is proof it's not within the crystal. This is a medallion also soaked in the water with salt for about 10 years. How do I remove the green? https://imagizer.imageshack.com/img924/1773/ZoS3tT.jpg

No. It was not green originally. I bought a lot of it before to investigate piezoelectricity. I sold them and retained it. It was soaked in water with salt for 20 years (I remembered now). It may just be algae which also got stuck on the metal. How can I remove all the algae? is it possible?

@Jtl The medaillon, assuming it were wet metal, reminds a little to malachite and some copper plates exposed to weather (e.g. Bartholdi's Statue of Liberty) but it equally could be plated copper with only a thin surface of copper on a magnetic metal, brass, or bronze.  Thus, because a remote diagnosis via a photo is guessing only, I refrain from a recommendation.

@Jtl The condensers within the rotary evaporators catch algae on the inner side (where tap water flows), too to yield a greenish bio film. So we drain the water, dismount them, fill them half with diluted (about 5 to 10 vol%) of aqueous HCl and gentle shake them.  Because quartz will resist to diluted HCl, too, you may try this too (put on gloves and googles, fill a plastic jar with diluted HCl, gently rub the wetted quartz with a synthetic brush.  Since the material of the medaillon is not ascertained, I would NOT use the acid there.


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Can I extract solid sugar from any fruit juice at home?
The goal here is to isolate the sugar from all other compounds in the juice. In this case, I'm using apple juice and grape juice. The sugar does not need to be edible, as it is being used to make r-candy. Boiling off the water from the juice doesn't work in this particular application, since the syrup is still contaminated with other compounds. I've tried mixing juice in denatured alcohol, and I didn't see any sugar precipitate (felt like a kid staying up all night waiting for Santa to come). Do y'all have any ideas?

Have you checked how a sugar factory works? Far as I remember, the most important tricks are Ca(OH)2, coagulants, filtration, and recrystalllisation. Possibly a bit of crystalline sugar as nucleation agent.

Hard to see why you are concerned with stuff from the fruits and not of denatured alcohol. My idea is to buy a pack of sugar and do a better use of the fruits.

@Karl I just did haha, I guess I'll need to do more research into that process. The only issue I can see myself running into is the centrifugal filtration process that's used to separate sugar crystals from molasses (I don't have a centrifuge, is there an alternative process I could use?). Also, is the Ca(OH)2 be added to balance the pH of the juice?

The road to scientific success is paved with literature studies. :-))

Let the sugar crystals grow until they are mm-sized, then use a sieve? Slow cooling might help, i.e. let the sugar crystallise at elevated temperature. Calcium hydroxide firstly inhibits the "inversion" (check wikipedia) of sucrose, that might not be relevant for you. The high pH also precipitates polyvalent ions, that's a first semester chemistry lab classic. The molasse is in the factory later neutralised again by CO2 (which comes from the inhouse production of Ca(OH)2 ), precipitating calcium carbonate again.

What the hell is "r-candy"?

I'll have to try those methods^

R-candy, short for "rocket candy", is a type of rocket fuel made with sucrose and potassium nitrate. For some reason, every resource says to use a 13g:7g potassium nitrate:sucrose ratio, even though it doesn't balance out stoichiometrically.

Sorry if I insist. But why not to get the required ingredients and prepare salad fruits instead? Is for the pleasure of making all DIY?

@Alchimista It's a proof-of-concept/I'm-bored-and-want-to-do-something kind of project

Fine. It didn't look so judging from the question.


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Which mechanism would be followed by a certain reaction with good leaving group?
In my class test, I was asked this question:

Which mechanism would be followed by a certain reaction with good
leaving group?

E1 reaction

E2 reaction

No reaction

both E1 and E2

I answered 4 because they both occur simultaneously the main difference comes in the amount of product formed by them. But, I was awarded with negative marks.
Can anyone please tell if I was wrong on my part or my teacher?

One point is: Your comparison of E1, E2 and E1cb should equally consider if there is synchronicity (e.g., the approach of a base and leaving of an other group across a transition state) or one after the other (passing intermediates, e.g. a base deprotonates the substrate, the carbanion then expels an other group) to yield e.g., an alkene.  Which is not the same as letting $\ce{X-}$ leave to generate an intermediate carbocation eventually loosing H to yield an uncharged alkene.

the reaction can follow any pathway just the conitions make one case or the other more probable

Yes there are reactions which possibly could proceed by more than one pathway, already because on occasion it may be easier to reject a suggested mechanism than to collect supporting experimental evidence.  Thus the suggest that you add to your question the reaction in question, to set it in specific context to your question.

Take note also of the temperature at which the reaction is allowed to move— E1 reactions usually perform under high temeperatures (based from my Org Chem class in first year).

IMO this question is nonsensical because you can't generally make a strong statement on ideas in organic chemistry. There are no hard and fast rules.

@Buraian but then the options were also nonsensical. Either provide appropriate(non debatable options)

Well of course, nonsensical questions can only have nonsensical options :-)


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Judging orders in a series of compounds
My teacher frequently gives us questions where we have to arrange the given binary compounds on basis of their melting points or solubilities. He did give us some instructions in order to judge this order. But they don't seem to work for any questions. I asked a senior student as to how to tackle this problem, and he simply said inorganic predictions are tough to make, wo just stick to the data and make appropriate relations between the given compounds. But the problem is we don't get data in tests and have to judge the compounds solely on the basis of logic. But I have no clue as to how to figure out this logic. So is it really true that all the orders are simply unpredictable without their actual data? Is every order just not following the standard rules to figure out these orders? It would really help if someone is able to guide me how to approach such questions without knowing the related data and maybe even share some appropriate further readings. I tried looking on Wikipedia and google, but everywhere I go, they quote the experimental data. I even tried looking at some answers here on stackexchange, but the people in comments tend to confuse me even more.(some have even written the obvious answer-"stick to experimental data").

Don't worry too much about it, this educational exercise will be over when the exam is over. In real life nobody can predict the melting point or solubility in water just by looking at the structure of a given molecule. There are software like ACD, ChemDraw and SciFinder database which have predicted melting points and boiling points of most known compounds but those values are often off. Sometimes the values are correct. This is why only the experimental values of are considered reliable. Maybe in future, our understanding will become so much better that melting/boiling points and solubilities in water would be accurately predictable.
All your teacher wants is to remember some general trends and periodic properties, which should be in your mind (in other words memorize, just like 2+2 is always 4). Take the example of halogens (diatomic molecules). As you move down the group the melting point increases. Fluorine has the lowest melting point and iodine has the highest.
In the alkali metals, the melting point decreases down the group. All you have to keep in mind the melting point trends down in the groups as well as across the periodic table for simple molecules. Similarly, look for the trends in alkali halides, hydrogen halides, other hydrides of the oxygen group.
Read this article on Is there a trend in melting points on the periodic table? https://antoine.frostburg.edu/chem/senese/101/periodic/faq/melting-point-trend.shtml


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Can Fe and Ca react and become unusable in some foods?
In Turkish cuisine there is a popular food called "tahin pekmez" (tahini with grape molasses).
According to a 2017 article on some Turkish news site, mixing these is supposedly counter-productive due to the calcium in the tahini bonding with the iron in the molasses and becoming unusable to the human body.[1]  From what I can tell, the article doesn't cite any study for this claim.
Obviously I'm inclined to believe that the claim is click-bait, but I thought let's give it a chance and ask people who know their chemistry.
I hope this isn't too basic or otherwise off-topic for Chemistry StackExchange.
[1] https://www.yenisafak.com/gundem/tahin-ve-pekmezi-karistirmayin-2804300 (in Turkish)

Calcium itself does not react with iron. What makes iron unavailable is oxidation to iron(III), not enough acidic pH leading to precipitation and forming stable complexes with organic acids.

Molasses is not a very good source of iron and sensible dieticians do not recommend it as a source of iron not least because it is high in sugar and that is usually not good as a significant part of your diet.

Studies have indicated that calcium inhibits iron absorption in humans. However, numerous dietary factors influence iron absorption, and the effect of calcium may not be as pronounced when calcium is served as part of a whole diet. Calcium does not have any significant long-term effect on iron absorption. In addition, milk and milk products have not been found to affect iron absorption.
References:

Lönnerdal B. Calcium and iron absorption--mechanisms and public health relevance. Int J Vitam Nutr Res. 2010 Oct; 80(4-5):293-9. doi: 10.1024/0300-9831/a000036.
Grinder-Pedersen et.al., Calcium from milk or calcium-fortified foods does not inhibit nonheme-iron absorption from a whole diet consumed over a 4-d period, The American Journal of Clinical Nutrition, Volume 80, Issue 2, August 2004, Pages 404–409, doi: 10.1093/ajcn/80.2.404
https://www.dairynutrition.ca/nutrients-in-milk-products/calcium/calcium-and-iron-absorption-is-there-an-interaction


text:
Origin of Virtual Molecular Orbitals in Hartree-Fock equations
Let's look at scheme
Do I understand correctly, the origin of Virtual Orbitals is a solution of (for, exampre, restricted) HF equation eigenfuncions?

Yes, they result from the HF equations just like the occupied orbitals. The diagram should also be show them as present in the initial set of equations, unless the basis  (used for the LCAO) was changed between the two states shown. The number of occupied orbitals plus the number of virtual orbitals is equal to the basis set size because they are the eigensolutions of
$$
 \mathbf{F}\mathbf{C} = \mathbf{\epsilon}\mathbf{S}\mathbf{C},
$$
where $\mathbf{F}$ is the Fock matrix, $\mathbf{C}$ is the matrix of orbital coefficients, $\mathbf{\epsilon}$ are the eigenvalues which are the orbital energies, and $\mathbf{S}$ is the atomic orbital overlap matrix (because the basis set used is typically not orthogonal). The dimension of all matrices involved is determined by the size of the basis set, which will be finite because otherwise the computational effort becomes infinite.

Can you, please, clarify (may be, it is need some math) why the number of occupied orbitals plus the number of virtual orbitals is equal to the basis set size, because I thaught the sum is infinite, as any problem for eigenvalues and eigenfunctions.

@Sergio I have extended my answer.

It seems, I understend.The virtual orbital in H2 molecule is a unbonding orbital. Am I right?

That depends on the basis set. Assuming the minimal basis set of two hydrogen-like orbitals, you are correct.

Yes, I mean minimal basis set.  Are there virtual orbitals if the HF method is applied to atoms? Or this is a feature of HF with MO LCAO?

The number of atoms makes no difference, only the size of the basis set vs. number of electrons. So typically, yes, there are virtual orbitals in single-atom calculations.

Let's start with HF equations (we omit the procedure for their derivation), and  obtain the pseudo eigenwertproblem:
\begin{equation}\label{Fock}\tag{Fock}
    \hat{F}_i\phi_i' = \varepsilon_i\phi_i'.
\end{equation}
The number of such equations is equal to the number of electrons $N$ (we do not consider such methods as RHF, ROHF in general), so $i = 1 \ldots N$ and $\phi' = \{\phi_1, \phi_2, \ldots, \phi_N \}$ This equation is actually only well defined for the orbitals that give the lowest energy.
The next step to find an approximation to actually solve these still pretty complicated systems.
In this approach we map the solutions $\phi_i$ in the form of linear combination of some basis functions:
\begin{equation}\label{fLCAO}\tag{fLCAO}
        \phi_i'(\mathbf{x}) = \sum_{a=1}^{\infty} c_{a,i}\chi_a(\mathbf{x}).
\end{equation}
The number of such basis functions is infinite.
But in reality, taking into account the savings in computer computing power, the number of basis functions should be reduced to $M$, thus, we already approximate:
\begin{equation}\label{LCAO}\tag{LCAO}
        \phi_i'(\mathbf{x}) = \sum_{a=1}^{M} c_{a,i}\chi_a(\mathbf{x}).
\end{equation}
The basis functions are known to us, but only the coefficients in the expansion remain unknown. The problem of solving equations is reduced to a purely algebraic one. How are these coefficients found?
Substitute \ref{LCAO} into \ref{Fock} and multiply left by $ \left\langle\chi_a\right| $:
\begin{equation}\label{}
    \sum_{b = 1}^M  c_{ib} \langle\chi_a|\hat{F}_i|\chi_b\rangle = \varepsilon_i \sum_{b=1}^M c_{ib} \langle\chi_a|\chi_b\rangle.
\end{equation}
or in a more expanded form
\begin{equation}\label{sys}\tag{sys}
    \begin{cases}
        c_{i,1} (F_{11} - \varepsilon_i S_{11}) + c_{i,2} (F_{12} - \varepsilon_i S_{12}) + \ldots + c_{i,M} (F_{1M} - \varepsilon_i S_{1M}) = 0, \\
        c_{i,1} (F_{21} - \varepsilon_i S_{21}) + c_{i,2} (F_{22} - \varepsilon_i S_{22}) + \ldots + c_{i,M} (F_{2M} - \varepsilon_i S_{2M}) = 0, \\
        \vdots \\
        c_{i,1} (F_{M1} - \varepsilon_i S_{M1}) + c_{i,2} (F_{M2} - \varepsilon_i S_{M2})  +\ldots + c_{i,M} (F_{MM} - \varepsilon_i S_{MM}) = 0.
    \end{cases}
\end{equation}
The coefficients $ c_{ib} $, the number of which is $ M $, will now be unknown. But we still have $ \varepsilon_i $ unknown. To find them, we must use the fact that non-trivial roots must meet the condition:
\begin{equation}\label{}
    \left|
    \begin{matrix}
        F_{11} - \varepsilon_i S_{11} & F_{12} - \varepsilon_i S_{12} & \cdots & F_{1M} - \varepsilon_i S_{1M} \\
        F_{21} - \varepsilon_i S_{21} & F_{22} - \varepsilon_i S_{22} & \cdots & F_{2M} - \varepsilon_i S_{2M} \\
        \vdots               & \vdots               & \ddots & \vdots               \\
        F_{M1} - \varepsilon_i S_{M1} & F_{M2} - \varepsilon_i S_{M2} & \cdots & F_{MM} - \varepsilon_i S_{MM}
    \end{matrix} \right| = 0.
\end{equation}
This equation is $ M $ -th degree with respect to $ \varepsilon_i $, so there must be solutions to $ M $ of$ \varepsilon = \{\varepsilon_1, \varepsilon_2, \ldots, \varepsilon_M \} $. Substituting one of these values into the equation \eqref{sys} (for example $ \varepsilon_2 $), we obtain as a solution the coefficients $ \{c_{21}, c_{22}, \ldots, c_{2M} \} $ .
And knowing these coefficients, from the formula \eqref{LCAO} we get the orbital $ \phi_2 $. Similarly, we get all $ M $ orbitals
$$\phi' = \{\phi_1, \phi_2, \ldots, \phi_N, \phi_{N + 1}, \ldots, \phi_M \}.$$
Next is the self-consisting procedure, which improves coefitients.
That is, if we had $ N $ electrons, and at the same time had to have $ N $ orbitals, then by introducing $ M $ pieces of basic functions, we expanded the number of orbitals to $ M $, and hence some of these orbitals ($ M-N $ ) will not be occupied. These orbitals are called virtual.
From these orbitals, one can construct a Slater determinant by discarding the $ M-N $ virtual orbitals, thus we have
$$\phi' = \{\phi_1, \phi_2, \ldots, \phi_N \}$$
When approximating a bounded basis by $M$ functions, it is difficult to give a physical meaning to virtual orbitals $\phi'_{virtual} = \{\phi_{N + 1}, \phi_{N + 2}, \ldots, \phi_M \}$. But if we choose the expansion in the form of \eqref{fLCAO}, then theoretically, they would mean the real excited states of the system (see Configuration interaction).


text:
Retro Diels–Alder reaction in synthesis of corallopyronin A
I was looking at the synthesis of corallopyronin A in the following paper by Rentsch and Kalesse: Angew. Chem. Int. Ed. 2012, 51 (45), 11381–11384.
There is one reaction which is described as

oxidation and a retro Diels–Alder reaction generated pyrone

I understand that the DMP oxidises the secondary alcohol to a ketone, and at the end there is an intramolecular esterification to produce the lactone, but I don't quite understand how the retro Diels–Alder reaction works. Could someone explain the reaction steps?

Here is my assessment of the mechanism. You are correct about the Dess-Martin periodinane (DMP) oxidation of 1 to 2. The resulting ketone 2 can also exist as the enols (E)-3 and (Z)-4. Any one of the three can undergo a retro-Diels-Alder-like thermolysis, but I have chosen (Z)-4 for the convenience of cyclizing ketene 5, which has lost acetone in the thermolysis. Finally, the ketene 5 affords pyranone 6.
ADDENDUM: Citation for the "retro-Diels-Alder". A. R. Katritzky, Z. Wang, M. Wang, C. D. Hall and K. Suzuki, J. Org. Chem. 2005, 70, 4854. https://doi.org/10.1021/jo050307m


text:
For what chemical purpose did soldiers in WW2 carry small ion exchange resin columns in their survival kits?
My textbook [Chemistry Live, Declan Kennedy] states that ''During WW2 small ion exchange resin columns were part of a soldier's survival kit.''. I can't find any further info about this online or elsewhere. The chapter is on water purification and hard/soft water.
My thoughts are to either 1) to aid purification of urine for consumption 2) to remove ions from water that has been poisoned 3) to purify water that may be polluted due to all the metal waste (bullets, shrapnel, etc.)
Can anyone confirm the actual reason that soldiers carried these kits ?

Sites such as https://www.wearethemighty.com/mighty-history/this-is-the-average-gear-a-soldier-in-ww2-carried/ do not support this. Or, if they were handed out they may not have gotten carried for very long.

I believe all your guesses are way off. Ion exchangers are used to make sea water potable. Not sure they were around as early as WWII, though.

Why a down vote for this perfectly valid question with a reference? Ion exchangers played a big role in WWII and a subject of intense study in the 1940s-70s.

Still, the unwanted foreseen use what to make water pounds or whatever more drinkable. It was a survival kit, at the end.


text:
Best pH indicator for titration

Which substance is best suited as a pH indicator for titration of $\pu{15 mL}$ of $\pu{0.1 M}$ ascorbic acid $(K_\mathrm{a} = \pu{7.9E-5})$ with $\pu{1.0 M}$ sodium hydroxide solution?
Using the image below to find the best suited substance.

My progress:
$$\mathrm{p}K_\mathrm{a} = -\log (\pu{7.9E-5}) = 4.102,$$
then
$$\mathrm{pH} = 4.102 + \log\frac{\pu{0.1 M}}{\pu{1.0 M}} = 3.102$$
I am quite sure this is wrong, but don't know what I should do.

Your calculation does not make sense. If you want to titrate a solution of a weak acid with a strong base like NaOH, you should choose an indicateur whose pKa is approximately half-way between the pKa of your weak acid and the pH of you basic solution. For your ascorbic acid solution, whose pKa is $4.1$, titrated by NaOH whose pH is  $14$, the average is pH = $9$. Phenolphtalein is the best choice, because its color changes between $8.2$ and $9.5$

thanks so much for the answer, what about if we would titrate a solution of a strong acid with a weak base, how would I choose the best suited indicator?

Usually this titration is made in the other way. The basic solution is in a beaker where the strong acid is added drop by drop from a burette. Anyhow, you choose the indicator by calculating the average of the pH of the strong acid and of the pKa of the weak base (14 - pKb of this base).

@Maurice Usually? You cannot titrate by adding the weak base/acid. You'd need to know the concentrations beforehand, for startes, and the curve would be so shallow you cannot make out anything. Not to mention the complications by huge concentration change.

@Karl. So you would replace "usually" by "always" in my last comment. I agree with you. I simply did'nt want to criticize the author too abruptly.

@Maurice Yea, hard to find a balance there. What I don't get is how teachers and books can teach chemistry so badly that students or pupils are left with such crass misunderstandings.

Find pH of the point of equivalence, what can be taken as pH of solution of base with pKb=14-pKa.  pOH=0.5(pKb - log c) -> pH = 14 - 0.5(pKb - log c )=14 - 0.5(14 - pKa - log c )=9.05 + 0.5 log c) // Then choose the indicator with the range the equivalence point about falls into.

@Poutnik Can you covert this comment to answer? :)

For titration of weak acids by strong bases, usable are generally indicators with transition from less to more intense colour in mild pH range. Typical, available and frequently used one is phenolphthalein with range 8.2-10.0.
To take it more analytically, you can find pH of the point of equivalence  as pH of solution of weak base with $ \mathrm{p}K_\mathrm{b}=14-\mathrm{p}K_\mathrm{a}$.
We can justify two simplifying assumptions $c \gg [\ce{OH-}] \gg [\ce{H+}]$
$\mathrm{pOH} =0.5(\mathrm{p}K_\mathrm{b} - \log {c})$
$\mathrm{pH} =14 - 0.5(14 - \mathrm{p}K_\mathrm{a} - \log {c})$
$\mathrm{pH} =7 + 0.5(\mathrm{p}K_\mathrm{a} + \log {c})$
$\mathrm{pH} =9.05 + 0.5 \log{0.1}=8.65$
Then choose the indicator with the range the equivalence point about falls into. Phenolphthalein will be ideal.


text:
How would I make a basic protein folding algorithm?
On the face of it it doesn't seem especially difficult.
Model the amino acids as solid objects linked together. Approximate the 3D force fields around them using a coarse approximation, then set the simulation going in a computer and see what happens.
Why is this such a difficult problem that it has taken decades to crack? Are there subtle difficulties not apparent in the above model?

The pdb has thousands of structures you can download, but I suggest reading a bit of liters before you start. There’s lots of info already out there about different approaches and how they work.

It is certainly easy for the protein to find its folded conformation (in most cases, it takes less than a second). It is difficult for us, however, and for molecular dynamics or other simulations (most attempts get it wrong). Deep neural networks, fed with ten thousands of crystal structures showing how proteins actually fold, now do a decent job with proteins of unknown structure (not sure about the computational cost).

One issue is that md steps have to be tiny, far, far less than a nanosecond to be accurate but it takes billions and billions of steps to fold to the proper structure as there are so many options that have to be tried to  find the true structure and time just runs out to complete the task. Just recently,  training with neural networks appears to have worked v. well, but it may not tell us just how the real protein did this, which may, or may not matter, depending on your viewpoint.

First, don't try from scratch: there are many attempts at this over decades and, until recently, few have worked well. Learn from them to save yourself a great deal of effort. Second, don't start your attempt until you can afford a supercomputer or equivalent. That is what the professional teams have.

@matt_black True, but I would imagine if someone were to find a way to solve the problem using desktop computing that would be quite useful. (If a little unlikely!)

@zooby I took out the request for datasets (as that separately if you want to keep questions on a single topic). I hope this gets it reopened as it isn't a bad question. But re-edit if I have corrupted your intent.

It is far, far more difficult that a naive approach suggests
The reason why simulating protein folding is extremely hard is the combinatorial complexity of the options you have to explore to get the "right" answer. This exhausts your computers memory or computing capacity far faster than it will for simple molecules.
It is possible to see why this is so from very simple considerations. The backbone of a protein consists of peptide bonds (which can rotate fairly freely) plus two rotating bonds in each amino acid. even ignoring the amino acid side chains (which often have multiple rotatable bonds) even a dipeptide has 5 bonds where you have to explore the possible rotations to find the optimal configuration of the extremely simple molecule. Since proteins consist of between 50 and 2000 amino acids that leaves a gargantuan number of options to consider, even if we can ignore all the other things that matter.
And those other things do matter. A simple example is that the configuration at a specific bond in the backbone may depend on the spatial orientation of other bonds dozens of bonds away so every choice at one site depends on all the other choices on all the other bonds. Optimising one bond at a time just doesn't work. On top of that is the complexity of including the side chains. That makes the problem harger by another gargantuan degree even if you don't consider the specific possible interactions of those side chains with each other (which are know to be important in many real proteins in stabilising specific spatial structures.
On top of those complexities, real proteins fold themselves in the environment of the inside of cells where interactions with water matter. So you have to consider those interactions as well. And, on top of that, the machinery that makes proteins consists of other pieces of molecular machinery designed to force specific shapes to emerge (heat shock proteins or chaperones are known to do this job).
One way to simplify the task so its difficulty is apparent is to realise that a simple protein chain has an extraordinary number of possible ways to fold only one of which is the "right" way in a living system. The problem is not finding a possible structure, but the right structure. Given the almost-impossible-to-count number of options, it is beyond the capacity of computation to find the right one by any known search algorithm.
This is why large teams of professionals have failed over decades to find a good general approach even when using vast amounts of specific biological knowledge from known structures. It is simply not a problem that can be tackled by amateurs.

When you talk about "choices" are you talking about quantum choices, where the system can jump into different states? Otherwise wouldn't the protein just wiggle around in a predictable classical manner until it gets stuck in a local minima?

@zooby By "choices" I mean simple things like the angle at each rotatable bond. So quantum effects are not important and the complexity will be extremely high whether we worry about quantum effect or not.

But my understanding is a protein doesn't make choices then, it just wiggles around according to the forces acting on it and the electric charges. In fact, worse than that, the protein itself is like a little machine that can move about into several configurations so it is not even a solid object.

@zooby I'm using the word "choices" looseley. To model the protein you have to explore all the possible bond angles for every rotatable bond in the protein. I'm calling each possibility a "choice". For every combination of all the possible choices, you have to calculate the energy state and not just the local state around each bond, but the global state of the whole protein. And even that isn't enough as the global minimal energy might not be the configuration found in nature.

But you still haven't told me how the protein itself explores all the possible bond angles.

@zooby If anyone knew how proteins achieve their structures we would not need to write any algorithms to guess the mechanism or the structure. The machinery of cells is complex and proteins are most certainly not made by random exploration.

In short: Joining amino acids to build proteins does not just assemble rigid bodies together.
Maybe you think it is joining a carboxylic acid and an amine to form an amide and to identify a conformation corresponding to the global energetic minimum around this unit.  It surly was not this easy to enter successfully CASP14.  Without intent to list all possible complications on the way, you may encounter additional problems:

many of the essential $\alpha$-amino acids contain flexible chains.  Their orientation, which may interfere with other chains of other amino acids nearby, equally contributes to the overall energy of the protein.  Thus, it isn't only the angle $\omega$ within the amide group which has to find a / the optimum, equally $\psi$ in the back of the amide's NH.

(credit)
A contribution to simplify the problem however is experimental evidence what is most likely for the amino acids accessible as illustrated by the Ramachandran plot

(credit loc. cit.)

a force field literally universal (UFF) covering all atoms may be too coarse to identify the globally best spatial arrangement.  Some intramolecular angles and distances will be estimated to small, and some to big compared to the true values.  Even more advanced ones which computationally are more costly, e.g. MMFF94, are based on experimentally determined data about small molecules, rather than proteins.  Thus you likely start to parameterize / develop a new model to describe experimental evidence good enough without paying too much on computation and include results e.g., from dedicated protein NMR spectroscopy,  protein crystallography and modeling.

natural proteins not only are large (number of atoms, number of amino acids, conformational flexible side chains), their higher structure (e.g., hairpin loops, $\beta$-sheets) equally is affected by intra- and intermolecular interactions of the protein itself, but may enclose (metal) ions and water molecules.  Contrary to an amide bond, their presence is not due to covalent bonds

One element preventing AlphaFold2 (this years winner of CASP) to identify a reasonable solution just by brute force is a layer of machine learning / deep learning.  For example, if the unknown protein structure contains a sequence of Ala-Cys-Arg, data bases are queried for other (known) protein structures with this sequence.  Their local structure for this very sequence is then used for the unknown protein structure as a start structure to be tested.  If you have about an hour, Yannic Kilcher presents some of the theory used by AlphaFold2's predecessor in a video.
There are numerous projects related to protein folding e.g., on GitHub which both publish their code and provide access to data for training and testing, e.g., ProteinNet equally linking to the original CASP7 (2006) data set.  The text based version is a tar.gz archive of 3.2 GB.  The consultation of ChemSE's sibling site MatterModellingSE may be of help, too.

What confuses me is that the question is posed in terms of the protein finding the minimum energy. Why would the protein (in real life) not just get stuck in a local mimnum? (Or perhaps a lot of proteins do but only the ones that work get stuck in the right local minimum).  When a protein wiggles around it tries lots of combinations but at what point does it fix itself. How does it "know" it is in the right position?

Does the protein fold in a classical way, or does it jump into diffrent configurations in a quantum way?

I suppose the surrounding water molecules could knock it out of local minima until it found a lower energy state?

@zooby Of course a protein structure may be stuck in a local minimum, the identification of a global minimum would require activation to leave this «pit» first.  Assuming linear (instead of convergent) protein synthesis, the protein chain is exposed to other molecules (e.g., water) and will start to fold right there, without waiting for the whole sequence to be synthesized in fist place.  This synchronicity may narrow the folding options, possibly excluding access to the global minimum.  About water around and water trapped in a protein, see e.g. https://www.pnas.org/content/101/10/3325 .

@zooby or this review publication: https://pubs.acs.org/doi/10.1021/acs.chemrev.5b00664

Interesting. Yes, I would assume the protein would start to fold itself as it is being made... and also be obstructed from folding by the very structures that are creating the protein! I'm starting to see why this is a more difficult problem than first appears.


text:
Why is the gradient of the curve of a strong base titrated with strong acid small up until equivalence?
I understand the cause of the sharp decrease in pH upon reaching the equivalence point, but I don't understand why addition of relatively large amounts of the strong acid would be accompanied with small changes in pH before that.
When looking at how pH changes when adding a strong acid to water there is a very fast drop in pH, as to be expected due to the logarithmic nature of the pH scale. The pH scale (I believe) also explains the flat region at the start of the curve for titrating a strong acid with a strong base.

But I don't understand it for the titration in question. I presume it must be something to do with the neutralisation but I'm not sure.

Below pH=7, you have H+ ions because you physically added them. Above pH=7, you have H+ ions because of the autoprotolysis of water.

Let's calculate the $p$H by changing the volume of $\ce{HCl}$ added. I will use the unit millimole ($1 mmol = 0.001 mol$), which is more practical than the mole.
In the very beginning the solution contains $50 mL~ \ce{NaOH}~ 0.1 M$, or $5 mmol$ $\ce{NaOH}$. Let's take, for example, the point when you have added $30  mL~ \ce{HCl} ~0.1 M$ (or $3.0 mmol$$ ˙\ce{HCl}$). After this addition, the solution contains $5 - 3.0 = 2.0 mmol$ NaOH in a volume $50 mL + 30 mL = 80 mL$. The concentration of $\ce{NaOH}$ is $\frac{2.0 mmol}{80 mL} = 0.025 M$. The concentration $\ce{[H+] = 10^{-14}/0.025 = 4 10^{-13}M}$. So $p$H = $12.4$. You see that the pH has fallen from $13.0$ to $12.4$ for more than half the trip to the equivalencce point. A change from $13.0$ to $12.4$ is rather small. The point of the curve is still nearly at its initial value. The gradient of pH between the beginning and here is small.
If you take the same calculation with an addition of $45 mL~ \ce{ NaOH}~ 0.1 M$, or $4.5 mmol$, the solution contains still $ 5 mmol - 4.5 mmol = 0.5 mmol ~\ce{NaOH}$ in $95 mL$. The concentration in $\ce{NaOH}$ is $\frac{0.5 mmol}{95 mL} = 5.26·10^{-3} M$. $\ce{[H^+] =  1,9·10^{-12}}$, and $p$H = $11.7$ We are near the equivalence point but the pH has not decreased a lot. It is still $11.7$. As you see the gradient (or the slope) of the curve is not important in the part of the curve between $0 mL$ and $45 mL$.
And suddenly between $45 mL$ and $50 mL$, the $p$H drops from $11.7$ to $7$. This is a huge change in just $5 mL$.
If you want you can calculate one or two other points along this titration. Try for example $49 mL$. You will be surprised ! It still gives a rather high $p$H value.

The titration curve is a deception of mathematics to our eyes. Look at the y-axis carefully. What do you have on the y-axis? A log function operating on hydrogen ion concentration. A unit change in pH means that hydrogen ions concentration has changed by a factor of 10. As more and more hydroxide ions are removed from the solution during the titration with HCl it takes lesser and lesser volume of acid to change the pH, hence a steep drop in your titration curve.
The concentration of hydrogen ions decreases proportionally with the addition of base . You add x-moles of hydroxide ions, x-moles of hydrogen ions are consumed.


text:
Can copper sulfate react with terpenes (such as limonene and terpinenes) at room temperature?
I am just wondering, can I make a metalorganic or organometallic complex of copper and limonene/terpinene/phellandrene in the form of lime oil? If so, what is the formula of the complex formed? I tried the reaction with lime oil in iPrOH, and a copper sulfate solution. The result was a blue crystalline solid. What are the crystals? I'd like to know.

I have a video of the reaction that I performed: https://www.youtube.com/watch?v=PeJhFKj1nuw

By the way, the compound is an opaque blue solid, and I used iPrOH to dissolve and extract the terpenes, as well as to wash the crystals.

The terpenes in the oil, I mean. It's poorly soluble in water, and insoluble in iPrOH.

Most of the evidence you present, together with solubility data from Wikipedia, seems consistent with just precipitating the copper sulfate.  Note that because of spectrochemical effects, we would expect the complex to change color if organic ligands were to displace water.

Yes there is nothing in terpenes to coordinate with copper ion. You got copper sulfate again.

For the record, after M. Farook's agreement I am making my comment an answer.
Most of the evidence you present, together with solubility data from Wikipedia, seems consistent with just precipitating the copper sulfate. Note that because of spectrochemical effects, we would expect the complex to change color if organic ligands were to displace water.


text:
(Probabilistic) entropy associated with many equally probable microstates
This is a pretty naive question. The probabilistic definition of entropy is (for a mole of particles)
$$-R\sum_i p_i\ln p_i $$ in which $p_i$ is the probability of a given microstate and $\sum p_i =1, p_i\geq 0.$
If we have a large system whose microstates are equally probable the terms of $-R\sum p\ln p$ will trace a curve like that below (image on Wolfram Alpha), the area under which is (up to some multiplicative constant, omitting R) $ A=0.25.$

My question is whether there is any physical intuition or significance to the shape of the curve. In particular, why would the entropy associated with identically probable  microstates not be the same?
In case it sharpens the question, I would expect that as we add up entropy associated with equal-probability microstates the entropy would accumulate in a linear way. That is evidently not the case.
Thanks for any insights.

The conversation has been moved to chat.

Time does not appear here so the idea of adding bits of entropy seriatim is artificial. Maybe better to think of it as an instantaneous picture of an existing system. Does the picture make physical sense in that case?


text:
Is (1R,3aR,4S,6aS)‐1,4‐dibromo‐octahydropentalene chiral or achiral?
Is this molecule on the left chiral or achiral?

I believe it is achiral because its mirror image (on the right) is essentially the same molecule. But I was told that it is chiral because it doesn't have an internal plane of symmetry. Could someone tell me which explanation is correct?

For the relationship of symmetry to chiral vs. achiral, see https://chemistry.stackexchange.com/a/123129/72973

This compound is no different from the figure 8 knot, which is achiral. http://ursula.chem.yale.edu/~chem220/chem220js/PROBSETS/PS13spr/PS4-S13-ans/fig8knot.htm

It indeed is achiral, because the red dot is the center of symmetry, hence it is superimposible with its mirror image.
Having an internal plane of symmetry is not the only criterion, the presence of a centre of symmetry also leads to the compound being achiral.

If you rotate the mirror image on the right 180 degrees about the axis shown by the blue line, you arrive at the original molecule in the original orientation shown on the left.

Yeah thats why i said it is achiral.

Yes, but the OP did not only ask for the answer but also wanted to know which explanation is correct.