text:
For a 1st order reaction, how do you derive the integrated rate law for a product versus time?
For a chemical reaction in which some stoichiometric quantity of arbitrary chemical $A$ decomposes into products $B$ and $C$,
$$a \ce{A} \ce{->} b \ce{B} + c \ce{C}$$
or more simply:
$$a\ce{A} \ce{->} \text{Products}$$
The depletion of substrate A at a concentration [A] can be modeled:
$$\frac{{\rm d}[\ce{A}]}{{\rm d} t} = - k[\ce{A}]^n$$
If this is a first order reaction then n = 1:
d[A] / dt = -k * [A]
This is a separable differential equation. When integrated from time 0 to time t and initial concentration [A]0 to [A]t which is the concentration at time t, the integrated form is this:
ln([A]t) = ln([A]0) - k * t
or in terms of [A]:
[A]t = [A]0 * e-k * t
This is useful if you want to know how the concentration of A as [A] changes with respect to time. But if you want the concentration of products, I think you could consider the total concentration:
[A]0 = [A]t + [P]t
Substitution into the the last equation yields a form in terms of products:
[P]t = [A]0 * (1 - e -k * t)
This kind of makes sense to me. The substrate is being exponentially depleted and the products are being exponentially produced to a maximum concentration which is [A]0. The following assumptions were made that the reverse reaction is negligible and that the reaction proceeds to completion which isn't always true. In fact, the time coordinate where these two curves intersect is actually the half-life!
If [P]t = [B]t + [C]t
Solving the equation for [B]t yields:
[B]t = [A]0 * (1 - e -k * t) - [C]t
This doesn't make sense because at the start of the experiment (time = 0),
negative concentration of [B] would be observed. Am I doing something wrong?
I have never seen integrated rate laws modeling the concentration of a single product versus time, only the depletion of reagent concentration versus time.

What is wrong with
$$[\ce{B}]=[\ce{B}]_0 + \frac ba \cdot  [\ce{A}]_0 \cdot (1 - \exp{(-kt)} )$$
$$[\ce{C}]=[\ce{C}]_0 + \frac ca \cdot  [\ce{A}]_0 \cdot (1 - \exp{(-kt)} )$$

The flaw in your reasoning is [A]$_0$ = [A]$_t$ + [P]$_t$. Look at your reaction scheme. If a=b=c=1, for example, [P]$_t$=2[A]$_0$.

If you are doing this on a molar basis, you have to be careful about assumptions like: $$[A_o] = [A]_t + [P]_t$$
In the same way that you have defined a component balance for the reactant system, you can also do so for the product system. Let's look at the concentration of B in your reaction equation $\ce{aA -> bB + cC}$.
Defining the component balance,
$$\frac{d[B]}{dt}=k[A]$$
$$[B]_o = 0, t = 0$$
This is essentially saying that the generation of $B$ is proportional to the inverse depletion of $A$ (as defined in your component balance). We know that from the solution of the A component balance that, $$[A](t) = [A]_oe^{-kt}$$ Plugging in this expression to the differential equation for $[B](t)$ yields,
$$\frac{d[B]}{dt} = k[A]_oe^{-kt}$$ Solving this differential equation will yield the following expression for $[B]$,
$$[B](t)=-[A]_oe^{-kt}+C_1$$
Using our intial condition that $[B]=0, t = 0$ we solve for our constant $C_1$ and end up with,
$$[B](t)=[A]_o(1-e^{-kt})$$
Note that this is similar to the form you had already derived, however... this is only for $B$. Since the kinetic expression will be the same assuming no initial mols of $B$ or $C$ and same stoiciometric coefficients (i.e $b = c$) then we can reasonably assume that $$[P]_t = 2[A]_o(1-e^{-kt})$$


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Do we really have technology with the resolution to distinguish between layers of single atoms of materials?
My textbook, Solid-State Physics, Fluidics, and Analytical Techniques in Micro- and Nanotechnology, by Madou, says the following:

X-ray analysis reveals the symmetries of crystals (lattice type), distances between atomic planes (lattice parameter), the positions of atoms in crystals, the types of atoms from the intensities of diffracted x-rays, and the degree of crystallinity (ordering).

I found this quite fascinating. Do we really have technology with the resolution to distinguish between layers of single atoms of materials? In other words, we can tell when one layer of single atoms ends and the next layer of single atoms begins, and analyse them as such?

Yes, it does exist, but the meaning of the quotation was not that.

@Greg oh? My apologies. Can you please explain what the author meant?

If I wanted to nitpick, I would say that this short text piece already contains two errors. Is the rest of the book written as sloppily? ;)

@Karl I don’t know what the errors are, so I can’t say. There aren’t any obvious errors; although, some of the explanations could be better.

Well, you cannot identify the elementar composition of a sample by xray diffraction. That is utterly ridiculous. XPS is a spectroscopic technique, hence the "S", not diffraction. And strictly speaking, XRD works on the distribution of electron density in your sample, not on "atoms". But that is indeed nitpicking. ;)

@Karl Hmm, can you please rephrase what you said? You were referring to x-ray diffraction, but then jump to saying that XPS is a spectroscopic technique (what is XPS?), so it is a bit difficult to follow what you're saying.

XPS does identify elementar composition. https://en.wikipedia.org/wiki/X-ray_photoelectron_spectroscopy

@Karl, the point about electron density is more significant than just nitpicking imo; as I understand it, this is exactly why XRD can not identify exactly what atom it is looking at. Although I am happy to be corrected or nitpicked at if I am wrong. :)

@orthocresol Of course, Xrays only interact with electron density. But for such an introductory sentence, its probably OK. ;)

You might be interested on this https://www.google.com/amp/s/www.upi.com/amp/Science_News/2020/01/17/Scientists-film-chemical-bond-making-breaking/1861579295504/

X-rays do not allow you to look  at the material just like an microscope. You get the spacing information from the diffraction pattern of only crystalline materials and a lot of mathematics. On the other hand, atomic force microscopy may actually be able to tell you where one atomic layer starts and where it ends on a surface. For example, imagine a gold surface, onto which you deposit a layer of molecules (say on local spot by some means). AFM would allow you to create an image (again by indirect means, as it is not an optical technique) of that molecular layer.
See google images generated by AFM and you will see how fancy they are.
To the OP: A good way to studying is to rely on multiple references of the subject. Do not take everything literally. The authors are human beings after all with limited knowledge and of course they are not the experts of $every$ topic covered in the textbook. When I was a student, I used to spend a lot of time trying to decipher what the author was saying in a particular sentence. With maturity you learn that the better way (read: more scholarly way) is to look at the same topic in another book and see what the other author is saying.

Thanks for the answer. Hmm, those atomic force microscopes look almost identical to optical microscopes, no?

Sort of. Optical microscopes have objective lens and a telescope (eye piece), there is nothing like that. I have never used AFM but seen the images. People have imaged single molecules on surfaces.

Do you happen to know what @Greg meant in his comment? Was I misunderstanding something?

The quote in the books explains the capabilities of X-rays, but X-rays cannot be used for scanning at a resolution of atomic scales. So it is $not$ a direct observing technique. It is better to use electrons. Read about transmission electron microscope.

But the point it's making is that X-rays are used as an indirect observation technique for layers of single atoms, right?

Here's an excerpt from later in the same textbook: "In Chapter 5 we will see that the formation of an image, according to Abbe’s theory, is a two-stage, double-diffraction process: an image is the diffraction pattern of the diffraction pattern of an object. In x-ray diffraction there is no lens to focus the x-rays, so we have to use a computer to reassemble the image: the x-ray diffraction patterns from a crystal are related to the object diffracting the waves through a Fourier transform."

Yes, that was my point. X-rays are an indirect way of observing distances and positions. I am not sure if you can study single layers of atoms by X-rays. TEM is good for extremely thin layers.

@ThePointer An AFM is (very basically) scanning a very sharp probe tip across a surface in 2 dimensions and measuring the surface hight at each point. This is nothing like how an optical microscope works and doesnt provide the same sort of information that an optical microscope does.

@Matt you’re right; I was just commenting based on the images I saw on Google images, which seemed to look very similar to optical microscopes.

That excerpt ("In Chapter 5...) is also a bit weird. Optical image formation is scattering (at your sample), followed by diffraction (at your lens). While strongly related, scattering and diffraction are not the same thing. And the last part is two correct statements linked by a piece of mumbo jumbo. There is no image to be reassembled.

@Karl, seems like "Optical image formation is scattering (at your sample), followed by diffraction (at your lens)."  is pure non-sense. The textbook writer is confused himself.

@M.Farooq That´s what I wrote. Now Im confused too. ;)

@Karl, I thought the textbook writer wrote that. The proper word would be refraction at the lenses. Diffraction requires interference. I think you just mistranslated some German terminology but meant refraction.

@M.Farooq You are right. The German terminology is equally clear. diffraction (Beugung) happens only at the edge of your lense. It´s coherent scattering.

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Indicator for H2S
In there an indicator for sulfur compounds like H2S (also organic)?
Or some straight forward way to measure if the H2S concentration in the air is higher than usual? Or is the concentration too low anyway?

Your nose is an excellent and extremely sensitive detector. ;)

(do not overtax it however. 200ppm of hydrogen sulfide are quite toxic, and your nose is already in overload then, you hardly smell it any more.) The lower human detection limit is somewhere around .5 ppb according to wikipedia.

There are simple test tubes available for the measurement of $\ce{H2S}$ and similar compounds in air. For example this one by Dräger (no affiliation), which uses oxidation by iodine.
$$\ce{H2S + I2 -> 2 HI + S}$$
The (discoloration of the) brown color of the absorbed iodine serves as indicator.
In order to get a quantitative result for the concentration in air, you would need to pump a defined volume of air through the test tube.

"Standard Measuring Range: 0.2 to 7 vol. %" The lower end of that range is lethal  to humans within minutes.

On the premises of an oil refinery, if you are not staying completely indoors, everybody gets one of these:
They give you an alarm if the H2S concentration goes above a few ppms, and the sensor inside typically can (quantitatively) measure up to one or two hundred ppm before it goes into saturation.

Oil production areas use a personal badge; lead acetate , it darkens if H2S is present ( Draeger was a brand for detection equipment).

To OP: this was and probably still is the standard H2S indicator also in analitical labs, and in school and university labs.

@Alchimista I think the use of H2S bottles in school or undergrad labs has cheased in the meantime (TAA was standard at my university twenty years ago already), and if you wanted one for your research, I´m sure your lab safety people will equip you with a modern electronic sensor that beeps. ;)

@Karl TAA was in use already at my times. Still, lead acetate paper was part of the kit for every one. I guess that for safety the electronic device should be the choice because of its alerting beep. If a lab might have dozens of them to be used atop dozens of beckers almost simultaneously I don't know. pHmetres don't replace litmus paper, at the end. Ps: H2S bottles probably means Kipp's apparatuses.

@Alchimista I never worked with gaseous H2S, but yea, according to my old firstyear textbook, it would have come from a Kipp, not a bottle. ;) It also mentions lead acetate paper, but I guess our profs thought that the chances of anyone who couldn´t smell H2S any more noticing the blackening of a piece of paper were just too small to be worth caring about.

@Karl I just have said that lead acetate paper is the common indicator for sulphides and sulphydric acid. I hope that the Lassaigne test is still performed, and with a stripe not an electronic detector :)

@Alchimista Now you´ve lost me. ;) Cyanides can be part of my analysis, but why and how should I use this Lassaigne?

@Karl Sulphur of course

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Reference for Hammett acidities of dilute H2SO4?
When $\ce{H2SO4}$ is dissolved in water in little amounts ($<\pu{1M}$), the $\mathrm{pH}$ scale can be used to measure the acidity of the solution. However, in larger concentrations, this breaks down and we have to use the Hammett acidity instead.
I have found this thesis, which has a table of $\mathrm{H_0}$ (Hammett acidity function) values for 95% and higher $\ce{H2SO4}$.
However, I would like a table with Hammett acidites of more dilute systems, like 70% $\ce{H2SO4}$ and 50% $\ce{H2SO4}$. Is there a reference for this?

Figure 1 on p. 8 in your PDF already contains a plot of $H_0$ function for the concentration range from 0 to 100%. What else do you want to know?

Didn't notice that, thanks. But a table with precise values would be better.

Check the Typical Values section in Wikipedia.


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Why does methyl not shift in the mechanism of isoborneol to camphene?

After the formation of carbocation, I was confused why would the reaction not go in favor of forming bornylene. I am still not sure why but I think that $\mathrm{sp^2}$ hybridized carbon will cause a lot of strain in the ring. Another thing confuses me, why does methyl not shift in place of the ring itself breaking? Is this anywhere similar to Bredt's rule? Please help me with these two doubts.

There is no alkyl shift here; the carbocation is inherently non-classical in nature. Note how the “shift” is drawn using resonance arrows <->, not a reaction arrow ->. Google ‘norbornyl cation’ for more info.

In addition, the bond that does migrate has good overlap with the initially formed carbocation while the methyl C-C sigma bond is essentially orthogonal to the carbocation p-orbital.

This is a class of carbocation 1,2-rearrangement reactions, and historically named Wagner–Meerwein rearrangement after two chemists who discovered the reaction (see following diagram):

Although it can be explained considering as an rearrangement of a classical carbocation, there are a tremendous amount of work has been done considering 2-norbornene cation as non-classical carbocation (see comment of  orthocresol elsewhere). However, my interned task is to explain why it is not possible to have 1,2-methide shift instead of alkyl shift (Wagner–Meerwein rearrangement) considering classical carbocation.
First, why the reaction not go in favor of forming bornylene? Since the reaction is in acid medium, fast equlibrium of protonation (to give initial 2-norbonyl carbocation derivative) and deprotonation (to give bornylene derivative) could be expected until most thermodynamically stable product is formed. This can be expected from a $2^\circ$-carbocation, if there is a possibility of rearrangement to a more stable $3^\circ$-carbocation.
Now, the initial 2-norbonyl carbocation derivative is $2^\circ$-carbocation with a possibility of rearranging to a more stable $3^\circ$-carbocation by either 1,2-methide shift or alkyl shift (which is the one eventually happens). If you look at initial carbocation without the bridge-head, the bottom cyclohexyl part is in boat confirmation, which is locked in by the rigid bridge-head carbon. Now that restricts the 1,2-methide migration, which would made energetically unfavored triangularpyramidal $3^\circ$-carbocation. Thus, the readily available alkyl migration is energetically favored and fast. The rearanged crbocation would readily dehyrogenate to give camphene (2,2-dimethyl-3-methylidenebicyclo[2.2.1]heptane).
Also keep in mind that not only it is not favored by the steric resons, but also 1,2-hydride shift would generate a $3^\circ$-carbocation, which would have violate Bredt's rule.
Note: Nonetheless, a recent publication has discussed this scenario and 1,3-hydride migration to give 7,7-dimethyl-2-methylidenebicyclo[2.2.1]heptane as elimination product (Ref.1).

References:
Huidong  Zheng, Yi Lin, Mengchan Wang, Jie Liu, Dan Wu, Jingjing Chen, Guanghua Yin, S. Ted Oyama, Suying Zhao, “The influence of solvent polarity on the dehydrogenation of isoborneol over a $\ce{Cu/ZnO/Al2O3}$  catalyst,” Catalysis Today 2019, 323, 44-53 (https://doi.org/10.1016/j.cattod.2018.10.021).


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What are the assumption of Gibbs free energy equation referring to the system and the surrounding?
The system for
$\mathrm dG=\mathrm dH-T\,\mathrm dS$
is a constant temperature and pressure throughout the process, only volume work is possible and it is in thermal equilibrium with the surrounding. Is that right? Are there any other constraints?

The only restriction on your equation is that the system is at constant temperature.  The system can be open or closed, the pressure does not have to be constant, and any type of work is allowed.
Let's start with the basic definition of G in terms of H, which is completely unrestricted:
$$G \equiv H-TS \Rightarrow dG = dH - d(TS) = dH -TdS -SdT$$
Then, to get your equation, the only restriction we need to add is constant $T$:
$$dG= dH -TdS$$
Why, then, might you be associating this equation with the more extensive restrictions of constant $T$ and $p$, and no non-$pV$ work?  I suspect it's because, to use $dG = dH -TdS = 0$ as a condition for equilibrium, these other restrictions are required. For more on this, see the end of my answer at What is wrong in this argument that dG must always be zero?


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Does ice increase the humidity of its surroundings?
Does ice increase the humidity of its surroundings? I'm trying to use the BME280 for an Arduino project regarding locating ice, but I don't know whether the atmosphere is more humid around ice. I'm trying to see whether humidity increases near ice.

It can do both very much depending on the air around your machine

What is a BME280 ? It might help your answer if you were able to tell us what it is. One option for humidity sensing at low temperatures would be a quartz micro-balance combined with a peilter cooler.

Why not sensing for T? One should really know the setup and the scenario in which it will operate.

It depends on the current humidity.
Water vapor certainly does evaporate from ice, as you can see in a vapor pressure chart. If the humidity is 100%, and the air temperature is the same as that of the ice, then the evaporation and condensation are in equilibrium; i.e. water evaporates as fast as it condenses, so there is no net change in humidity.
However, if the ice is warmer than the air, or the air has less than 100% moisture content at that temperature, then ice will evaporate more than it condenses, raising the humidity.
Conversely, if the ice is colder than the atmosphere, it can lower the humidity.

No. It is the contrary. Water from the atmospheric humidity is condensed on objects whose temperature is 0°C.

In order for condensation to occur, local relative humidity must reach 100%, so actually, yes, the ice does increase the relative humidity of its immediate surroundings by cooling the air. . .

Not to mention that if the air initially has no humidity, then the ice will humidify the air.

Atmospheric humidity condenses on surfaces with temperature below its dew point. it may be above or below 0 °C. Stratospheric humidity would not condense even on surface with temperature -50 °C, if its dew point is -80 °C ( what is common ).


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Why n-Butane has high B. p than iso-Pentane although both have equal no. of carbon atoms along the chain length?
n-Butane =B. p =-1 oC
iso pentane =b. p= 28 oC

Pentanes are heavier than butanes, whatever their geometry. So the boiling point of pentanes are higher. This is a general tendency in the series of hydrocarbons. Pentanes and butanes are good examples to show that the chain length has not the most important effect on the boiling point.
n-Pentane boils at $\pu{36°C}$, 2,2-dimethylpropane at $\pu{9.5°C}$. So all pentanes boil between $\pu{9.5°C}$ and $\pu{36°C}$.
In contrast, all butanes boil under $\pu{0°C}$. n-Butane boils at $\pu{-1°C}$ and isobutane at $\pu{-10°C}$.


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Transposition in vicinal diols dehydration

In this reaction, I don't know how I could place a methyl in the benzyl carbon. My first idea was to treat the compound with OsO4/H2O2 to form the vicinal diol, and then dehydrate in acidic medium and heat, but in this way I assume that carbocation is formed in the most substituted carbon, so I suppose that it should form in the other carbon and transpose a methyl, then with an oxidation get the carbonyl in the desired position, but this is It contradicts my idea of the formation of carbocation in the most substituted carbon. Any other idea?

You are on the right track. With acid it is a pinacol rearrangement initiated by forming the benzyl carbocation. THere will be no need for an oxidation after the methyl migration. Just lose a proton from the oxygen.

You're absolutely right, I hadn't taken into account the great stability of the carbocation in a benzyl position. Thank you


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Definition of a Homologous Series
My textbook states that members of a homologous series have the same general formula (ex. $\ce{C_{n}H_{2n}}$ for alkenes) as they only differ from one to another by additional $\ce{CH_2}$ (methylene) groups. However, alkenes can have more than 1 carbon-carbon double bond, which should alter their general formula right?

Luke here for good definition and description.

@MathewMahindaratne Are you suggesting that members of a homologous series need not have the general formula, only the same functional group?

Read carefully and look at examples given.

There is the series for dienes and whatever number of double bounds you may wish to count and classify. Also , if the number of double bounds gets higher, usually trends and comparison is interesting with respect of the latter different n. This leads to a series, in a way, except that is of analogous rather than homologous - oligo- and polyenes.

If a molecule has more than one double bond C=C, it is not an alkene. It is an alkadiene, or an alkapolyene.


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extracting parameters from .SKF files ,
I have a question about interpreting .SKF files, I am using these parameters to model defective graphene,
when I use parameter set given by dftbBaby/C-C (I think this is from Hotbit)
or the parameter set given by the following paper:
Porezag, D., Frauenheim, T., Köhler, T., Seifert, G., & Kaschner, R. (1995). Construction of tight-binding-like potentials on the basis of density-functional theory: Application to carbon. Physical Review B, 51(19), 12947.
(both of which give approximately $H_{pp\pi} ~ -3 eV$ at ~1.4 Angstrom) then after performing slater-koster transformations and embedding the H0 and S values into the secular equation, I get the correct band structure.
however, for other parameterizations, for example say dftb/mio, the value I get for $H_{pp\pi}$ from interpolation of hopping integrals is much smaller ~ 2eV and gives wrong band structure:

here for sure it is me who is doing sth wrong because using the same parameter set DFTB+ gives the correct band structure,
I am not sure whether I am not interpreting the SKF files properly (in a naive attempt the band structure can be corrected by increasing the value of r0 given in the SKF file which is equal to shifting the curves downwards) or is there any other step I need to perform before embedding the H and S values into the secular equation ?
Thank you for your help

I figured that I was using the wrong format, the detailed format of .SKF files is given here
The correct extracted parameterization profile is given below:


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Magnesium oxide dissolving in distilled water
Magnesium oxide is a salt, and salts dissolve in distilled water, due to water being polarized. Therefore I would expect that you would find free ions, which is confirmed by the water conducting electricity. But most sources I find states that magnesium hydroxide is created through a reaction with the water. I would appreciate if someone could tell me what is correct and why the other one is false.

The first step is to write up the equations.

All of that is true, these facts do not contradict each other.
The missed part is quantification. Salts and generally ion compounds are soluble, but solubility comes in range of many orders of magnitude.
Also, some compounds, even if largely ionic, react with water.
So magnesium oxide reacts with water forming magnesium hydroxide, both being almost insoluble in water.
Oxide anion  as a very strong Lewis base immediatelly reacts with water $\ce{O^2- + H2O -> 2 OH-}$. So the only chance for metal oxides to stay being oxides is to be insoluble, with their lattice energy too high to be broken by water hydration of ions.

One question, how come the distilled water with the magnesium hydroxide dissolved in it can conduct electricity? Does the magnesium hydroxide split up into free ions? (Considering dissolving something in distilled water and seeing if it conducts electricity often is a good way to see if it is a salt)

Yes, magnesium hydroxide dissociates. But in this case, ion concentration is very low. Sodium hydroxide and oxalic acid conduct electricity very well, but are not salts.

@poutnik how is sodium hydroxide not a salt? IUPAC's definition allows ionic hydroxides along with other ionic compounds to be salts.

@OscarLanzi Sure, depends on wideness of salt definition. It is the sodium salt of the weak acid water. OTOH, practical chemists rarely consider it as a salt.

When Melvin says that "salts dissolve in distilled water, due to water being polarized", it is by far not a general law. Some salts are soluble and some are not. Nobody is really able to predict the solubility of a given salt. There are plenty of rules describing this subject, using polarity, lattice energy, or other parameters. They are valid for plenty of compounds. But there are always strange exceptions.
A good example is the solubilities of calcium halogenides. These compounds must have similar structures. $\ce{CaCl2}$, $\ce{CaBr2}$ and $\ce{CaI2}$ are all extremely soluble in water, being dissolved in less that their weight of water. But $\ce{CaF2}$ is extremely insoluble in water, as it is one of the most important minerals on Earth to produce fluorine compounds. If $\ce{CaF2}$ was a tiny little bit soluble in water, this mineral would have been already washed out by the rains, and drained to the seas. It is not the case, as $\ce{CaF2}$ is a common mineral that can be found in many different countries.
The general theory for predicting the solubility of any compound has yet to be discovered. From time to time articles are published in papers like the Journal of Chemical Education, for explaining the solubility with new parameters like the electronegativity, the polarizability, the atomic dimensions, the hydrogen bonds,  etc. They are able to explain quite a lot of compounds. But there are always exceptions.


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What is the correct term of a chemical formula for very large molecules if not just "chemical formula"?
I know that many molecules, wihch I as a non chemist, might define as "small" if their two dimensional diagram can be sketched on an A4 page, are often formulated;
That is → by one or more processes given a chemical formula;
Maybe after being isolated from a natural deposit or organism.
For example, here is such formula for vitaminB12 of the type hydroxocobalamin:
C62H89CoN13O15P

My problem
Whenever I examined a wiki article about a protein molecule, I understood it as too big to be represented in a "chemical formula", let alone in a two dimensional diagram on an A4 page, such as those I have described.
I though for example:

As these are isolated molecules after all, they do have a "chemical formula" and should be standardly represented somehow.
Such large molecules doesn't have to be proteins → maybe proteins and other molecules in similar size have different ways to be represented.

My question
What is the correct term of a chemical formula for very large molecules (such as proteins), if not just "chemical formula"?
From prior research I understand a "protein sequence" is the standard parallel for "chemical formula" in the context of proteins but I don't know about a standard parallel to such large molecules in general.

Chemical formula is chemical formula, not much to say about it — whether it's one of the largest proteins, titin $(\ce{C169719H270466N45688O52238S911})$, polyoxomolybdate $(\ce{(NH4)12[Mo36(NO)4O108(H2O)16] · 33 H2O})$ or ammonia $(\ce{NH3}).$ I suspect your confusion arises from mixing up molecular formula with its abbreviated forms and various graphical representations of structural formulas.

There can't be and isn't such thing in general. There are, however, certain local approaches in the areas which are important to us for some reason or another: protein sequence, DNA sequence, etc.

What does JohnDoea want to change ? Why is he not satisfied about the word "Chemical formula" ? Chemical formula can be applied to very large molecules. Where is the problem ?

@Maurice there is no problem and I am totally satisfied with the word - I used my own words then to find the parallel or shortcuts for very large materials where a chemical formula line is very long. Anyone is welcome to communally suggest an edit to try to make the question more reachable and shorter.

Chemical formulas of the type you describe are just as valid for large molecules as for small ones. The reason they are not commonly used is that the number of possible molecules described by a given formula goes up exponentially with the number of atoms.
Thus, it is much more common to describe larger molecules with more precise terms. In the case of polymers like proteins or nucleic acids, the primary sequence of the constituents (amino acids or bases) is often used if space allows. Alternatively, a protein or nucleic acid  might be referred to by a common name with which a sequence is associated.
Even for small molecules, IUPAC formal names or common names are used in preference to molecular formula in many cases, because the names provide information about the atom connectivity that is absent in the molecular formula. Other representations that include connectivity information are forms such as SMILES, which is a standardized system for representing connectivity in text.
In the scientific literature, names of molecules are often accompanied by pictorial representations that provide atom connectivity information in an easily understood format. For very large molecules like proteins whose connectivity cannot be easily represented in a small image, pictorial representations are more often used to show secondary and tertiary structural features.

Nice answer as the question could attract various and reasonable comments that do not remove the doubt of the asker. This goes to the points.


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Leaving capacity of halogens
Why is F a bad leaving group? Say in HF and HI, F is more electronegative than I, then shouldn’t HF be a better acid ?

Consider the strength of the C-F bond

When HF loses proton the conjugate base formed is F- Fluoride anion,Since fluorine is small and compact atom having lower affinity to keep electrons because of lack of d orbitals thus fluoride ion is unstable conjugate base compared to iodide ion which is stable.Happy conjugate base(stable), more reactive acid. Hence HI is more acidic than HF.
Even though fluorine is highly electronegative(tendency to drag shared electrons in a covalant bond),it has lower affinity to keep this electrons(lower electron affinity) due to its smaller size .


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Electric double layer formation
What textbook states about the formation of electric double layer:

When silver nitrate solution is added to potassium iodide solution the precipitated silver iodide adsorbs iodide ion from dispersion medium and negatively charged colloidal solution results. Having acquired a negative charge by selective adsorption on the surface of colloidal particles, this layer attracts counterion from the medium forming second layer. The combination of these two layers of opposite charge is called Helmholtz electrical double layer. The first layer of ions is firmly held and termed as fixed layer while second layer is mobile and termed as diffused layer.
    Reference : chapter 5; page :141
    http://ncertbooks.prashanthellina.com/class_12.Chemistry.ChemistryI/index.html

What I understand about electrical double layer by browsing various internet sources is:
An electric double layer consists:

Surface charge: charged ions adsorbed on the surface of the colloidal particles or intrinsic surface charge or through loss or gain of electrons as in case of electrodes.
Stern layer (fixed layer): counterion attracted to particle surface and closely attached to it by electrostatic forces.
Diffusion layer: a film of dispersions medium adjacent to the particle containing higher concentration of counterion due to the residual electrostatic forces from the charged colloidal particles.

Is my understanding correct? Does the textbook statement means something more or do I miss something? I like to have a almost clarified understanding about the electric double layer.

Your second and third explanation are good and not basically different from one another. On the other end the first sentence is not as good, as there is no gain or loss of electrons in or through the double layer.

Please add a citation to the quote you are using.

NCERT textbook really is no proper citation; there should be an author (or editor), a title, a publisher, a year, an edition, and the chapter/page where the quote can be found on. This is to help identify which actual book we are talking about, and to see if changes to future editions make this question be perceived differently.

If you are confused about the electrical double layer, it is completely fine because electrochemists spent several decades to understand it and yet the concept remains "grey" in the minds of most electrochemists. You will find different stories and versions on the web and of course different terminologies.
Your understanding is correct but I will call the surface charge not only due to permanently adsorbed ions but actually there are negatively or positively charged functional groups of the molecules making up the the material. For example, if we have a polystyrene bead, we sulfonate it to $covalently$ attach $\ce{-SO3H^-}$ groups. In that case, one should say the surface charge of the beads is negative. Surface charge is related to the zeta potential (an experimentally measurable quantity). The sign of the zeta potential tells us the sign of the surface charge. If we measure the zeta potential of the colloids particles, say, sulfonate polystrene, we see a negative zeta potential in water.
This diagram is basically a depiction of your explanation and a minor correction of surface charge.


text:
How does potassium hexfluoronickelate(IV) exist in a stable state?
Nickel generally does not exist in a $+4$ oxidation state. Wikipedia has the following to say regarding nickel(IV):

Ni(IV) is present in the mixed oxide $\ce{BaNiO3}$. ....... Ni(IV) remains a rare oxidation state of nickel and very few compounds are known to date.

This article directly contradicts the first statement, as it says:

the photoelectron spectroscopy data suggest the description of $\ce{BaNiO3}$ in terms of trivalent Ni and oxygen anion holes.In terms of a local ionic model the idealized description of this high-valent oxometalate would correspond to $\ce{Ba^2+Ni^3+(O2)^2O^•-}$

Now, we end up coming to the curious case of potassium hexfluoronickelate(IV), which decomposes at 350°C, meaning that it is relatively stable. Not much literature is available on this compound online, except this article, which is unfortunately behind a paywall.
Could someone explain how nickel exists in a (IV) state in this compound instead of degenerating to $\ce{K4[NiF6]^2-}$? Also, are there any more nickel compounds exhibiting the (IV) state? Explanations regarding the formation and stability of this compound are also welcome!

The same way barium ferrate(VI) does, maybe?  You can get things into an amazing variety of oxidation states given the right environment.

https://en.wikipedia.org/wiki/Fluoronickelate gives you a few. https://pubs.acs.org/doi/pdf/10.1021/ja00145a013 describes the synthesis of NiF4. Low spin octahedral d6 is very well stabilised. See Gold V Fluoride

"are there any more Nickel compounds exhibiting the (IV) state?" Yes, there are some Ni(IV) complexes: https://onlinelibrary.wiley.com/doi/abs/10.1002/anie.201002181 ... https://pubs.acs.org/doi/abs/10.1021/ja00978a065 ... https://science.sciencemag.org/content/347/6227/1218 ... https://pubs.acs.org/doi/abs/10.1021/ja00753a022

To quote from the comments: you can get things into an amazing variety of oxidation state given the right environment.
In the case of maximal oxidation states, particularly above $+2$ or $+3$ in the $3d$ transition series, the right environment (at least in simple species) involves complexation with ligands that are highly electronegative, capable of pulling electrons even from a strongly positive central atom, and resistant to oxidation themselves in an electron-poor setting.  Meaning, in practice, oxide or fluoride ligands.
We can therefore roughly classify high oxidation state complexes into oxo and fluoro complexes.  Oxo complexes are more likely to be seen with early transition metals, where the central metal ion contributes few $d$ electrons and there are a lot of vacant orbitals with which the strongly $\pi$-donating oxygen can form back-bonds.  The best known examples being chromate/dichromate ($\ce{Cr(VI)}$), manganate and permanganate ($\ce{Mn(VI), Mn(VII)}$), and ferrate (most commonly, $\ce{Fe(VI)}$).  Even cobalt reaches its apparent maximum $+5$ oxidation state with an oxide-ligand species, $\ce{[CoO4]^{3-}}$.
Later transition metals, which have lower maximum oxidation states due to increased electronegativity/ionization energy, have more $d$ electrons coming from the cationic shells and thus are no longer so stabilized by $\pi$ back-donation.  Then the fluoro complexes, with a relatively $\pi$-inactive ligand, take over.  In the $3d$ series cobalt, cited above as forming an oxide complex with the formula $\ce{[CoO4]^{3-}}$, also forms a +4 fluoride complex $\ce{[CoF6]^{2-}}$(1).  Fluoride complexes are also seen with both nickel and copper in oxidation states above $+2$, including the $\ce{[NiF6]^{2-}}$ ion mentioned in this question and the copper(III) species $\ce{[CuF6]^{3-}}$.  Zinc in the $3d$ series is limited to oxidation states of $+2$ or less, but if mercury has a tetrafluoride $\ce{HgF4}$ (disputed), it's no coincidence that this too is a fluoride.
Reference

J.W. Quail and G.A. Rivett, "Fluoride Complexes of Tetravalent
Cobalt",  Can J Chem 50:2447 (1972).

Interesting. A lot of these fluoro complexes also adopt the $\ce{K2[PtCl6]}$ lattice structure and also seem to be similar in their properties. I'm assuming that all these fluoro complexes are low spin, given the high charge on the central metal atom. Could you please confirm?

At least for the $\ce{NiF6^{2-}}$ ion it is low-spin.


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CO2 and Aeration
I am an aquarist, and I am looking into adding CO2 into my aquarium in order to help plant growth.
This is a pretty standard practice, but some of the science spread within the community is contradictory.
My main question is: will adding surface agitation, or aeration to an aquarium with high levels of CO2, cause the CO2 to dissipate from the water?
My understanding is that, surface agitation with cause the water to ‘equalize’. Meaning the fluid with less CO2 (ambient air or water), will ‘steal’ CO2 from the liquid that has more until they are equal.

At equilibrium, the concentration of CO2 in the water will not equal the concentration of CO2 in the air.  Instead, at equilibrium, their chemical potentials will be equal.   At room temperature, CO2 likes being in a gaseous state more than in a aqueous state.  Consequently, when their chemical potentials are equal, the concentration of CO2 in the water will be quite a bit less than that in the air.
Specifically:
According to https://climate.nasa.gov/news/2915/the-atmosphere-getting-a-handle-on-carbon-dioxide/ , the current concentration of CO2 in the air is 412 ppm.  Extrapolating from this table (https://sites.chem.colostate.edu/diverdi/all_courses/CRC%20reference%20data/solubility%20of%20carbon%20dioxide%20in%20water.pdf) yields the following equilibrium concentrations of CO2 in water:
At 20 C: 29 ppm
At 25 C: 26 ppm
Extensive aeration, with room air, will help to maintain these equilibrium concentrations (you might want to consider a bubble wand in addition to air stones).  Surface agitation would also help, but I would think the increase in surface area you would get with extensive aeration would be greater than that you would get with surface agitation.   Plus, with surface agitation, you might end up with a concentration gradient of CO2.  By contrast, aeration will help ensure the aquarium is equilibrated with the air at all depths.
Note also that the 412 ppm value is for outdoor air.  Indoor air can be twice that or more, especially in a poorly-ventilated room with people.
The concentration of CO2 in water is approximately linearly proportional to its concentration in air, so if the indoor air were ~800 ppm, then the corresponding equilibrium concentrations in the water would be about:
At 20 C: 60 ppm
At 25 C: 50 ppm

This is awesome! Thanks for all the information. Just so I can make sure I understand. At room temperature, 25C agitated water will absorb CO2 until it’s reached 26ppm (assuming the CO2 levels are the same as outside). If I added CO2 and brought the level up to 30ppm, the CO2 would drop to 26ppm?

@JarredAwesome  That's right.   If you bubble in CO2 whose partial pressure is greater than that of the atmospheric CO2 (easily achievable with, say, pure CO2, since there its partial pressure will be ~ 1 atm), you will increase the concentration of CO2 in the water.  If you stop bubbling in the CO2, it will drop back to the equilibrium concentration (in this case, 26 ppm).  [Of course, you don't truly have an equil. sys, since the plants&fish perturb the conc of CO2; heavy aeration w/ ambient air should mostly overcome this, giving a steady-state conc. of CO2 about equal to the equil. value.]

Some gases can displace dissolved air by a simple process of bubbling. The formal wording for it is "sparging" For example, it was a standard practice to bubble helium in organic solvents to remove dissolve air. Similarly, if you sparge $\ce{N2}$ for a long time in water, you will displace oxygen and dissolved air from water. I feel that exposing water to high levels of carbon dioxide may not be a good idea for the fishes because you will deplete dissolved oxygen.

The co2 is added in very low amounts. There is equipment we use to make sure we don’t add too much. So in this case, would surface agitation remove dissolved co2 from the water?

I don't think so because water has an affinity for carbon dioxide (it is not completely inert like oxygen, nitrogen etc towards water). But let us go back to nature. Search the amount dissolved carbon dioxide in seawater and in rivers and see what are the levels in air. They are highly agitated systems.

I just took a look. I’m not a chemist, so my understanding is pretty base level. According to to the attache link, co2 in rivers wants to reach  equilibrium but doesn’t due to other factors. Ideal ppm in an aquarium is 30ppm co2. And the co2 indoors is more then 10 times that. So I would assume the surface agitation adds co2 to water... do you guys feel that’s accurate? https://www.researchgate.net/publication/289707721_Dissolved_CO2_in_Freshwater_Systems fresh water systems

When you use a typical air pump in aquariums, it is also pumping carbon dioxide from the air. I feel this is good enough. Surface agitation will certainly help to dissolve more CO2 and air of course, if CO2 level in air is way higher.

I am going the other way: I have a 125 gal with only 3 average size tropicals and numerous plants. I use aeration to get CO2 from the atmosphere into the water . As you said ; aeration promotes equilibrium between gases in the atmosphere and gases dissolved in the water.

I was thinking of going that way too. I was considering doing a lily pipe. Apparently that adds co2 also.
However, I decided to go pressurized co2. I’m just trying to figure out how much, of any co’2 my air pump is going to off gas

To add CO2, the answer is buffering as occurs naturally in nature (with carbonate rocks, for example). To quote a source:

An important breakthrough came when the researchers examined their results from the perspective of carbon dioxide in the oceans. In ocean waters with high buffering, that is, a greater capacity to neutralize acid, carbon dioxide and oxygen behave very differently than they do in waters with low buffering. For one, much more excess carbon dioxide lingers in highly buffered ocean water.
Researchers examined their samples and found that the degree of carbonate buffering controls the amount of excess carbon dioxide in surface freshwater. Alkalinity is a good marker for the degree of buffering and is an important consideration when looking at carbon dioxide in rivers.

A precise answer to your question is : Yes. Adding surface agitation, or aeration to an aquarium with high levels of $CO_2$, causes the $CO_2$ to dissipate from the water.

Does the level of co2 in the water have to be higher in the water than in the air?

OP skip this answer as the result is history related. Of course shaking soda cans content lowers the CO2 concentration... See theorist and Farooq answer.


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Reaction of alc. KOH with alkyl halide
The following question was given in FIITJEE study material:

At first, I thought it should be elimination. But, soon I realized that there is no heating, plus a primary alkyl halide is used. So, "no elimination".
Then, I thought it would proceed through $\ce{S_N2}$, and hence (B) should be correct, as it involves concerted mech.. But, again I was wrong. Why is it so?

The correct answer given is:

 (D)

My hint would be "enolate".

Just as @Karl mentioned, use the base to abstract the $\alpha$ $\ce{H}$, and form an enolate and subsequently use it as a nucleophile to substitute the $\ce{Cl}$. thus forming the 3 membered ring.

Ok, now I got it. Thank you, Karl and SirArthur7 :)

@RahulVerma How do you have to change the reaction to get (C)?

@Karl: it is a good one. I guess I would need a carbanion which would then do SN2Ar?

Wait... I think I should make a ketal first, then do EAS and then deprotect the ketal. Isn't it?

@SirArthur7: can I call this SNi mechanism?

If you're asking about the actual question, then I would better attribute this reaction to NGP (neighboring group participation). SNi is a bit rare and specific to chlorination of alcohols with thionyl chloride, or the decomposition of alkyl chloroformates, though basically both involve the same concepts. But NGP includes other cases too, so I feel its safe to use it. I'd like someone else to contradict my belief if there are any.

Did you learn about Friedel–Crafts reaction yet? If so, do you see the relation between Karl's question and Friedel–Crafts reaction?

@MathewMahindaratne 1) I've completed school last year, so you can expect that ;) 2) yes, I was trying acetal + friedel crafts + hydrolysis. But, I am not sure till now.

Correction: This is not acyl (acetal is wrong terms to use here). It's alkyl. Acyl part is already attached to aromatic ring.

@MathewMahindaratne: I'm sorry to use keywords, but I meant to say "acetal formation" from that carbonyl part, which will activate the ring towards EAS. I think.

The solvent used is a polar protic solvent which heavily solvates the OH- radical thereby decreasing it's nucleophilicity as the solvent forms a cage around the radical anion .Even though the basic strength of the anion is also decreased by solvation still the decrease in nucleophilicity is dominant as it is more difficult for a large solvated anion to attack at a carbon atom than to sneakily abstract an acidic hydrogen and hence the enolate is formed which then , proceeding through the SN NGP (neighbouring group participation) mechanism eliminates the Cl ion forming a cyclopropane.
Hope that this helps.


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Is dialkyldimethylammonium chloride suitable for sterilization in mycology?
In hobby-mycology 70% rubbing alcohol is commonly used for sterilizing work surfaces, bags, syringes, still air boxes, basically everything.
Alcohol however, seems to have some downsides, when it comes to mycology as it needs a long soak time to actually kill anything and it doesn't kill mold and mold spores at all, which is one of the most common contaminants.
So I am looking for something, which does not create toxic fumes, does kill bacteria and fungi spores including mold and acts quickly.
Is dialkyldimethylammonium chloride the right solution for me? If yes, what dilution should I use? Would 2% be sufficient? Does it become denatured after 24h mixed with water like bleach? And most importantly, is it safe for use at home? Will it cause residue buildup, like bleach would?

I am not aware of any ammonium chloride desinfectant abilities. Wikipedia article does not mention it either.

What about dialkyldimethylammonium chloride? Or is that something completely different?

Yes, it is something completely different. With long alkyls, it is cation active tenside, that could serve as desinfectant.

I edited my question.


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How can Ellingham diagrams be linked to Gibbs energies of formation?
My textbook1 quotes the following as the limitation of Ellingham diagram:

The interpretation of $\Delta G_\mathrm{f}^\circ$(standard Gibbs free energy of formation) is based on $K$ ($\Delta G_\mathrm{f}^\circ = -RT \ln K$).Thus it is presumed that the reactants and products are in equilibrium.

Definition of Standard Gibbs free energy of formation:

The standard Gibbs free energy of formation of a compound is the change of Gibbs free energy that accompanies the formation of $\pu{1 mol}$ of a substance in its standard state from its constituent elements in their standard states (the most stable form of the element at $\pu{1 bar}$ of pressure and the specified temperature, usually $\pu{298.15 K}$ or $\pu{25 ^\circ C}$).

But in Ellingham diagram  reactants are normalized to consume one mole of oxygen having partial pressure of $\pu{1 atm}$. Thus products formed are not one mole. So how can it be a graph of Standard Gibbs free energy of formation and how this limitation even applies?

1. NCERT Chemistry Textbook for Class XII - Part I, Chapter 6, page 152

The Ellingham diagram doesn't actually use molar Gibbs energies of formation $\Delta G_\mathrm{f}^\circ$ per se; it is more accurate to say that it uses molar Gibbs energies of reaction $\Delta G_\mathrm{r}^\circ$. The difference is that the formation energy is only relevant to one specific chemical equation, for example:
$$\ce{Ca + 1/2O2 -> CaO} \qquad \qquad \Delta G_\mathrm{r}^\circ = \Delta G_\mathrm{f}^\circ(\ce{CaO})$$
in which the stoichiometric coefficient of CaO is equal to 1.** On the other hand, for any (balanced) equation with any stoichiometric coefficients, it is valid to define a Gibbs energy of reaction:
$$\ce{2Ca + O2 -> 2CaO} \qquad \qquad \Delta G_\mathrm{r}^\circ = 2\times \Delta G_\mathrm{f}^\circ(\ce{CaO})$$
which is related to the energy of formation, but is not the same thing, as evidenced by the factor of 2.
In the Ellingham diagram, every reaction has the same stoichiometric coefficient for $\ce{O2}$, which is typically 1. This is needed to make sure that different reactions are comparable. Let's say, for example, you want to see whether the reaction
$$\ce{C + 2CaO -> CO2 + 2Ca}$$
is feasible. This is done by checking the sign of $\Delta G_\mathrm{r}^\circ$: if it is negative, then the reaction is feasible, and vice versa. The point is that this $\Delta G_\mathrm{r}^\circ$ can be calculated by subtracting two reactions together:
$$\begin{align}
\ce{C + O2 &-> CO2} & \Delta G_\mathrm{r}^\circ &= c_1 = \Delta G_\mathrm{f}^\circ(\ce{CO2}) \\
\ce{2Ca + O2 &-> 2CaO} & \Delta G_\mathrm{r}^\circ &= c_2 = 2 \times \Delta G_\mathrm{f}^\circ(\ce{CaO}) \\ \hline
\ce{C + 2CaO &-> 2Ca + CO2} & \Delta G_\mathrm{r}^\circ &= c_1 - c_2 \\
\end{align}$$
but these two equations add up nicely only if the coefficients of $\ce{O2}$ in both equations are the same. What the Ellingham diagram does is to plot the Gibbs energies of reaction, $c_1$ and $c_2$: if $c_1 < c_2$, then the reaction is feasible. It doesn't plot the Gibbs energies of formation, because comparing those wouldn't tell us anything about the sign of $c_1 - c_2$.
As a final remark, note also that the equation
$$\Delta G_\mathrm{r}^\circ = -RT \ln K$$
holds true for any reaction, whether or not it actually corresponds to a formation reaction.

** Having a stoichiometric coefficient equal to $x$ does not mean the same thing as $x$ moles of the compound are produced in the reaction. The coefficient is purely a mathematical expression which tells us the stoichiometric relationship between different species in the reaction. It does not correspond to a real-life reaction, where a defined quantity of reactant is added to a defined quantity of product. To illustrate this, let's say you go to a lab and mix 0.4 mol of HCl to 0.4 mol of NaOH. You're asked to write a balanced equation for this. You can write
$$\ce{0.4 HCl + 0.4 NaOH -> 0.4 NaCl + 0.4 H2O,}$$
and that would be correct, but it is hardly the only correct possibility: the more conventional
$$\ce{HCl + NaOH -> NaCl + H2O}$$
is equally correct, even though the stoichiometric coefficients (1 in all cases) do not match up with the actual amount of substance used in the reaction (0.4 mol). Note also that the units are different: stoichiometric coefficients are dimensionless, but amount of substance is measured in moles. The difference is subtle, but one well worth pondering about, as mixing these two up can lead to a lot of misconceptions in thermodynamics.

We may interpret the energy variable as free energy of formation per mole of $\ce{O2}$.  Thus, for instance, silicon would be favored to react with a limited amount of $\ce{O2}$ versus iron, because silica has a more negative free energy of formation per mole of $\ce{O2}$ than iron oxides; even though $\ce{Fe3O4}$ might be more negative per mole of compound because $\ce{Fe3O4}$ uses two moles of $\ce{O2}$ per mole of complound versus $\ce{SiO2}$ using one.  This, of course, is one of the main drivers behind silicon going into slag when we smelt and purify iron.

For the Ellingham diagram:
(1) We write down the reaction/reactions that we are interested in. The $\Delta G_f$ (formation) per mole of product is calculated using the individual free energies of any intermediate reaction, and using Hess' law to sum them up.
(2) This energy is then normalised per mole of $O_2$, using stoichiometry.
(3) The normalised energy is plotted on the diagram.
Coming to your question - how is the Ellingham diagram representative if the products are not necessarily 1 mole?
There are two ways to resolve this:
(1) The easier, slightly less accurate : You consider $O_2$ to be the main product, such that you are writing a reduction reaction. Then just reverse the reaction to make it an oxidation reaction (this can be done for redox reactions). Consider this just a convention. You could equally make a diagram of the reduction reactions; it would convey the same information if you interpret it correctly.
(2) The more abstract, more accurate reason: As shown by @orthocresol, conversion from individual free energies of formation to final free energy of the reaction requires a combination of several reactions. Stoichiometry demands that different reactions be normalised. Since we are dealing with redox reactions that necessarily contain oxygen here, it makes the most sense to normalise the net reaction with respect to oxygen.


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Why can't C-C double bonds rotate?

The alkyl chains of saturated fat molecules can readily rotate about their carbon-carbon single bonds, giving them flexibility and viscosity.

What is different about the carbon-carbon double bonds of unsaturated fat molecules, preventing inter-conversion of cis and trans forms?

A double bond between 2 carbon atoms means there is a $\pi$ bond with the side orbital overlaps in 1 plane, in parallel to the axial $\sigma$  bond.
The consequence is, free rotation  would need to break this $\pi$ bond, making the double bond just a single bond, what would require a strong torque and a lot of energy.
By other words, particular rotation positions around a single bond have ( almost ) the same energies. But for double  bonds, the molecule energy strongly depends on on this orientation. The $\pi$ bond can exist, only if the bystander groups on both carbon atoms   are about in the same plane. If they wete on about perpendicular planes, the  orbitals of both atoms cannot overlap to form the bond.

Poutnik we must agree that somehow is hard to see these things out of balls & sticks solid body representation.... OP please don't be confused by this comment of mine which is not adverse to the answer but it is a kind of discussion by an experienced chemist that somehow and sometimes has the same pondering of you.

@Alchimista We need not agree on that. :-) People, for whom is less confusing quantum chemical explanation, need not such an analogy.   But there are people, who can easier grasp the idea from things they can imagine. It is an analogy, so it does not say things are exactly like that. The truth is, some people have problem with analogies in general, wanting analogy not to have part of things different.

I am down voting this for the potato part. This is so far from helpful and does convey a picture that is by far too simple for it to ever be useful. Imagine anyone actually trying to learn more about this m will always be burdened with this picture. For a triple bond is even worse, because it suggests it cannot rotate, which is wrong.

It does not suggest it cannot rotate. But OK, I will remove it.

The removal of the potato wasn't my intent at first. Perhaps I wasn't clear that my comment wasn't aimed at rectification but pointing to the fact that a rigid anchorage of bonds and nucleus must be taken for grant as the tip of a stick cannot move inside the potato. I was fine with the analogy, contrary to Martin.


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Can eggs boil indefinitely?
At breakfast someone asked if the eggs were boiling. While obviously they meant the water for the eggs, but it lead to a discussion regarding if you can make an egg boil without it coagulating or even if you can heat a coagulated egg to the point that it starts boiling under special circumstances such as in an oxygen free environment to prevent combustion.
So would it be possible to make an egg boil without coagulation or otherwise destroying it? The shell would break either way so we can remove that for the experiment.

Boiling means dynamic evaporation of liquid.
Infinite boiling would mean to have infinite amount of that liquid and therefore the egg would have infinite mass.
Such an egg would collapse into a black hole and we would not be able to continue in our observation.
In reality, in inert atmosphere, with overlapping stages:

protein coagulation
water boiling
organic matter thermal dehydration
general organic matter decomposion.
carbon stage

If we kept temperature just above $\pu{100 ^\circ C}$ to keep boiling,  stage 3-4 would be very slow and we would not reach carbon.

That is a great scientific answer. In hindsight I should probably have worded the question a bit differently such as boiling away without coagulation, charring, etc. But you touched upon that as well. I think this is probably one of the better answers to a breakfast discussion I've had. I will leave the question unmarked for a few hours in case someone else wants to take a shot at producing a better answer.

Would it go through some nuclear burning stage or it would collapse directly into a black hole?

@Alchimista As it would already be inside in event horizon, together with the observer, both would already form the black hole, together with the rest of Universe. But I am definitely not an expert in GR and cosmology. :-D

I took it as infinite in mass but an egg :))

Being an egg and having infinite mass excludes each other. It may bypass a nuclear fusion and go directly to neutron star collapse and whatever is beyond that.

While I agree some chemistry is involved in this operatory laboratorium experiment, I believe it may be a mere physics quest. Taking the theoretic existence of a blackhole into account, one can infer that this is a very hypothetical question. Scientists have discovered a way to test this finding in laboratoriums relating to physics.
Now.. it is essential to find a physics laboratory and ask questions for the laboratorium supervisors. An infinite mass of egg, such a computational integer exists not. Perhaps finding the forma for your hypothesis and simply taking the limit to infinity. The time and progress of an egg to boil indefinitely will be of answer to the equation. This will be an educated guess and a quite remarkable and extraordinary result.


text:
Shortcuts for finding geometrical isomers
Our teacher provides us extra questions for pratice, so one of the questions was :-

Find number of products if following are reacted with excess $\ce{NH2OH}$ (hydroxyl amine):
Cyclohex -
(i) 1,4-dione
(ii) 1,3-dione
(iii) 1,2-dione
(iv) 1,2,3 trione
(v) 1,2,4-trione
(vi) 1,3,5-trione  and
(vii) cyclobut-1,2,3,4-tetraone

To solve this, our teacher told us replace each $\ce{=O}$ group by 1 $\ce{-CH3}$ and 1 $\ce{-H}$ , then find the number of stereoisomers (including enantiomers) of the resultant compound, which is the answer.(if you actually count it's tough to visualize the lone pair and $\ce{-OH}$ bond and it gets long).
I have cross-checked , the answers pretty much match. So my doubt is, is this trick always right and if yes then please prove it. I would also be content if you can provide any links, if the question has already been answered.

I had another doubt, about another trick my teacher told us about. Let's say there's a compound with $n$ carbon atoms forming a closed ring of all $n$ carbons. At each carbon, out of 2 $\ce{-H}$ , 1 $\ce{-H}$ is replaced by $\ce{-Cl}$ (chlorine). The trick if $n$ is odd ,the number of geometrical isomers of the compound is $2^{(\frac{n-1}{2})}$ and if $n$ is even then the number of geometrical isomers is $2^{(\frac{n}{2})}$. Please prove this also.
Thank you in advance.

In the last sentence starting by "The trick is", the word "odd" has to be replaced by "even"

@SirArthur7 You have the Suggested Edits queue red and full with your enthusiasm! Go on by all means(our site is getting better!), but I would urge you to make more complete edits. Deal with all problems in the same suggestion, don't just handle formatting/markdown and leave out grammar.

Thanks @WilliamR.Ebenezer currently on another edit. Could I come back later?

I am pretty much sorry for that, actually when you have to delve into such huge articles and you're rushing with time, you might miss out on a few things. But I will try to look after that from now on, thanks for the suggestion. Happy to be of help :)


text:
Is it possible to hold and rotate molecules to relativistic speeds around axis in observable way?
Also, is it possible to see frame-shortening effects on electrostatic forces/van der waals at the edges, or are molecules too weak or necessarily neutral to be manipulated this way?

Rotation speed cannot be made arbitrarily close to the speed of light. Molecular rotation is quantized, and every molecule has a finite number of rotational levels, just like it has a finite number of vibrational levels based on the shape of its potential energy surface (the same can be said of vibrations: the molecule cannot vibrate arbitrarily quickly without dissociating).
However forces like the van der Waals force do get noticeably affected at the highest supported molecular energy levels. Specifically the $C_6/r^6$ van der Waals force becomes $C_7/r^7$ at certain distances, due to the Casimir-Polder effect in quantum electrodynamics (an intrinsically relativistic theory of the quantum world). The reason for this is that the derivation of the $C_6/r^6$ van der Waals force does not take into account the fact that information cannot travel at infinite speed, so it actually takes a finite amount of time (based on the finite speed of light) for one atom to "feel" the force of a distant atom. This "slowing down" of the energy transfer, is called "relativistic retardation" because "retardation" means slowing down. This causes the force to become weaker for larger distances ($1/r^7$ is much smaller than $1/r^6$ for larger $r$).
This is usually not taught at the undergraduate level, or even graduate level courses, so there is no easy "textbook" example to demonstrate it, so unfortunately we have to resort to publications in scientific journals, where the audience is far more advanced: But in the image below, I hope you can at least see that "with retardation" the experimental values here are much closer to the theoretical values of Tang (2009), versus the experimental values derived from a model "without retardation":

This shows evidence that the theoretical idea of "relativistic retardation" is in fact a much better description of the experiment than the non-relativistic theory, so it is indeed fair to say that the effect of light (electromagnetic radiation) having finite (rather than infinite) speed is observed in a molecule. The molecule here was Li$_2$ and the above image is from Figure 6 of this paper which deals with deriving information from experimental spectra of a molecule:
http://scienide2.uwaterloo.ca/~rleroy/Pubn/09JCP-Li2AX.pdf

Comments are not for extended discussion; this conversation has been moved to chat.


text:
Accurate method to calculate the pH of a salt from a weak acid and weak base
Can someone help me with this?
The problem is as follows:

Find the $\mathrm{pH}$ of $\pu{1M}$ solution of $\ce{NH4CN}$. Their respective
  hydrolysis constants are:
$K_h = 5.6\times 10^{-10}$
$K_h = 2.5 \times 10^{-5}$

The alternatives given are as follows:
$\begin{array}{ll}
1.&2.8\\
2.&3.5\\
3.&6.4\\
4.&11.7\\
5.&12.2\\
\end{array}$
What I've attempted to do was to establish the following reactions:
$\begin{array}{lllll}
\ce{NH4+ + H2O <=> NH3 + H3O+} &k_{h}=\frac{K_\mathrm{w}}{K_\mathrm{b}}\\
\ce{CN- + H2O <=> HCN + OH-} &k_{h}=\frac{K_\mathrm{w}}{K_\mathrm{a}}\\
\ce{H3O+ + OH- <=> 2H2O}&k_{h}=\frac{1}{K_\mathrm{w}}\\
\end{array}$
This meant that:
$K=\frac{K_\mathrm{w}}{K_\mathrm{a}\times K_\mathrm{b}}$
For the given information it can be calculated:
$K=\frac{10^{-14}}{\frac{10^{-14}}{5.6\times 10^{-10}}\times\frac{10^{-14}}{2.5 \times 10^{-5}}}=1.4$
All of this is translated into:
$\ce{NH4+ + CN- <=> NH3 + HCN}$
From this equation it can be established the equilibrium equation:
$K=\frac{\ce{[NH3][HCN]}}{\ce{[NH4+][CN-]}}$
Returning to the previous equation I'm getting:
$\begin{array}{cccccc}
&\ce{NH4+ + CN-  & <=> &  NH3 + HCN }\\
i&1&1 & & &\\
c&-x&-x &  &+x &+x\\ \hline
e&(1-x)&(1-x) &  &x &x\\
\end{array}$
Using this relation with what was established earlier it becomes into:
$K=\frac{\ce{[NH3][HCN]}}{\ce{[NH4+][CN-]}}$
$1.4=\frac{\ce{[NH3][HCN]}}{\ce{[NH4+][CN-]}}$
$1.4=\frac{x^2}{(1-x)^{2}}$
Solving this for $x$ results into:
$x=0.54196$
and
$x=6.45804$
The thing here comes on which to choose for the concentration. I decided that the first seems more logical as it will not produce negative concentration in the denominator.
Therefore the concentration for $x=0.54196$
This means:
$\ce{[NH3]=[HCN]}=0.54196$
$\ce{[CN-]=[NH4+]}=1-0.54196= 0.45804$
Then to get the $\mathrm{pH}$ can be established from the hydrolyzation of $\ce{HCN}$:
$\ce{HCN + H2O <=> CN- + H3O+}$
Then:
$K=\frac{10^{-14}}{2.5\times 10^{-5}}$
$K=\frac{\ce{[CN-][H3O+]}}{\ce{[HCN]}}$
$4.0\times 10^{-10}=\frac{\ce{[H3O+]}(0.45804)}{(0.54196)}$
$[\ce{H3O+}]=4.73286\times 10^{-10}$
$\mathrm{pH}=-\log(4.73286\times 10^{-10})=9.3248$
Which seems kind of logical that the $\mathrm{pH}$ is within the range of an alkaline solution as $K_\mathrm{b}>K_\mathrm{a}$.
But since this option does not appear within the alternatives. I'm not convinced whether my solution is right or not.
I've found this equation which relates the $\mathrm{pH}$ with the $\mathrm{p}K_\mathrm{w}$ and $\mathrm{p}K_\mathrm{a}$ and $\mathrm{p}K_\mathrm{b}$ and the concentration:
It states:
$\mathrm{pH}=\frac{1}{2}[\mathrm{p}K_\mathrm{w} - \mathrm{p}K_\mathrm{b} + \mathrm{p}K_\mathrm{a} + \log c]$
where $c = \text{concentration}$ of the salt.
Inserting the given values in the above equation yields:
$\mathrm{pH}=(0.5)\left[-\log(10^{-14})+\log\left(\frac{10^{-14}}{5.6\times 10^{-10}}\right)-\log\left(\frac{10^{-14}}{2.5\times 10^{-5}}\right)+\log(1)\right]$
$\mathrm{pH}=9.3248$
which also checks with what was obtained previously.
But again none of these seem to check with any of the alternatives given. Could it be that my procedure was wrong?
The thing which it bothers me here is if this could be accepted as a general procedure to calculate the $\mathrm{pH}$ of a salt which has a weak acid and weak base ions. Can somebody help me here? And more importantly, how to prove that equation for $\mathrm{pH}$? I'm not sure if the hydrolyzation constants given in the problem are right or accurate. Does somebody has access to cross reference and check if those values are correct?
Can somebody help me with that part too?

You might find some help here https://chemistry.stackexchange.com/questions/60068/how-to-set-up-equation-for-buffer-reaction .and in the link to a related answer in a comment.

Does this answer your question? How to set up equation for buffer reaction?

@SirArthur7 The existing answer doesn't delve into the specific of hydrolisis of both ion and is more about establishing the equation for a buffer coming from acid and its conjugate salt which is well explained in many textbooks. It doesn't clear out my confusion on whether is it alright to multiply the constants and considering an interaction of acidic and basic ions for make water in the end.

@porphyrin That's and interesting entry and I've checked into it but it was not exactly what I was looking for.

@ChrisSteinbeckBell there are currently two votes to close as duplicate, it takes five. If you can edit your question and in the beginning add a link to the ducplicate and explain exactly why your question is not answered there, detailing how an answer to this question needs to differ from what's in those answers, it may help to keep this question open.

Ignoring activities, activity coefficients, ionic strength effects, and so on, I get a pH of around 9.3. Given the 1 M concentration, I would not bet a lot of money on this, though. My solution, assuming a more typical solution concentration, is given in the two figures below. Sorry these are figures rather than proper formatting: I have been away for 3 months and am beyond rusty at this point. If the mods wish to delete this answer, they should feel totally free to do so.

There are obviously other ways to do this problem and what I have posted is simply an approximation: any pH problem solution, of this type, that does not explicitly involve the formal concentration of the salt, i.e., "F", always gives me pause. But maybe this helps a bit.

So in general terms (it is a relief), does this validates the way how I proceeded?. I'm wondering if you take into an account the other aspects (ionic strength, and so on) would it change the value of pH which you have obtained?.

The approximation which you mentioned at the end of your answer, what's its justification?. Can you explain this part please?. Out of curiosity what software or tool did you used to draw your cursive sketch?

In general terms, I think your approach is fine! If all the fancy stuff is included, the pH will probably not be 9.3, but it is a good enough zero order approximation. It really all comes down to setting up all the equilibria expressions, in their full glory, and then solving them. Which is why it is useful to have a good piece of software that can do that number crunching. I do not have such a program, which would be useful, at the very least, to check manually calculated results. The trouble with pH problems is that concentrations need to be in the Goldilocks regime: not too concentrated,

not too dilute, but ‘just right’. Otherwise, you have slightly dirty water or need activities, etc. The software was an iPad app called Noteshelf. What I posted was screenshots from that app. They were from the last time (2015) that I taught the second semester general chemistry course. Finally, the simple approximation just recognizes that there can only be one pH in a solution. Nothing deep about it.


text:
Is this step justified?
This is a question from JEE Advanced  2019 (a highly valued engineering entrance exam in India), but the fact that we are ignoring basic rules of electrophilic substitution is disturbing. Correct me if I am wrong or missing something really important, but according to synergic electrophilic substitution, it should be an ortho attack from methoxy group as we know that an oxygen atom's conjugation would be more contributing than carbon's hyperconjugation and inductive effects combined.
Overall, attack from 2-position (to alkyl group) instead of 3, as done in the question is what I mean to say.
But I also know that it (attack of 3-position) will make the compound to have a bridgehead, will it be stable then? And for overall compound, the stability of the compound, do we have to do a meta attack even in an activated ring?

Could you crop your image to remove all irrelevant stuff and resize it for ease of readability? Also could you frame your question better, it took me sometime to make out what you were actually claiming.

Please rephrase your question. Are you suggesting that the cyclisation should go into a different position?

Yes ,cyclisation should go from 2 position and not 3

Is the editing helpful now? @waylander?

If this ? is from an engineering exam, it is inefficient because the ketoacid is more readily available (FC of anisole and succinic anhydride) than the ketoaldehyde. My lab partner did the prep of 7-methoxytetralone. Works fine.

If I guessed it right then maybe what you are trying to say is that anisole and succinic anhydride FC( idk what FC stands for) is a major product but the product given by me can also be prepared but in special circumstances I.e is minor.

I'm a little confused what OP's argument is. If you say "it should be a meta attack for methoxide group," then the question is showing an electrophilic substitution at meta-position to methoxy group. But, I understand you are wondering why substitution is at the meta-position to methoxy group rather than on ortho-position to methyl group. If that's what you meant, this is the answer to your curiosity:
The step in question is an example of Friedel-Crafts reaction. More precisely Friedel-Crafts acylation reaction. Here though, the substrate (aromatic nucleus) and the reagent (acyl chloride) are in the same molecule unlike any other typical inter-molecular reactions. In any inter-molecular Friedel-Crafts reaction involving 4-alkylanisole, substitution of acyl group mainly occurs on ortho-position to methoxy group, as you pointed out that "an oxygen atom's conjugation would be more contributing than carbon's hyperconjugation and inductive effects combined." I said here mainly because there would be some minor product from the substitution at ortho-position to alkyl group as well. For example, according to my experience, when Friedel-Crafts acylation reaction is performed in toluene and chlorobenzene, there were mixtures of products including o- and p-chloroacylbenzenes.
Now, we look at the problem in hand. This reaction come down to inter- versus intra-molecular acylation reactions. It's always intra-molecular reaction is much more faster than same type of inter-molecular reactions (here is acylation reaction of both kind) due to proximity. Thus, this phenomenon overcomes the deficiency of activation by methyl group versus methoxy group here, and gives most thermodynamically stable product, 7-methoxytetralone as the major product.

Edit: As OP's request, I include here an example for intra- vs inter-molecular acylation reaction (Ref.1):

Reference:

Yingju Xu, Mark McLaughlin, Cheng-yi Chen, Robert A. Reamer, Peter G. Dormer, Ian W. Davies, "A General Method for the Synthesis of 3,5-Diarylcyclopentenones via Friedel−Crafts Acylation of Vinyl Chlorides," J. Org. Chem. 2009, 74(14), 5100-5103 (https://doi.org/10.1021/jo900696k).

So it's a minor but thermodynamically most stable product?

Also can you provide a link for the fact stating the inter vs intra molecular acylation theory, it would be really helpful.

The given product is the major product, if it is not the only product. My guess is it is the only product.

You can easily goggle intra- vs inter-molecular reactions.

You'd find the same example here.

It is worth remembering that the MeO group activates all positions on the ring, meta less than ortho but it is still activated.


text:
How diphenylcarbazide is used to detect Mercury II ions?
Diphenylcarbazide is used in test strips for heavy metals detection in drinking water but I can't find much info on this reagent, apart from it's MSDS and a preparation for detecting Chromium VI ions, in which diphenylcarbazide 250 mg is dissolved in 50ml acetone and few drops added to test sol which is then mixed with sulfuric acid. Also, "Vogel's qualitative inorganic analysis" references this reagent but it is very vague explanation.
If you can elaborate on how the reagent works to detect mercury, and possibly how it is stored and synthesized, that would be really appreciated.
Also, how are the test strips for heavy metals in drinking water made from this compound, I'm curious?

The Diphenylcarbazide Test for the presence of mercury was used as early as 1920s. However, since the test is very delicate (the test can detect $\pu{250 \mu g/L}$ mercury in solution), it has been used predominantly in neutral solutions, until it have been adapted to Quantitative Analysis Scheme by Scott in 1929 (Ref.1) to ditect in acidic medium:

The mercury is precipitated in Group I is dissolved in aqua regia, that of Group II is dissoved in hydrochloric acid and sodium chlorate. In either case, the solution is evaporate to a volume of about $\pu{1 cc.}$, placed in a test tube, and diluted with $5$ to $\pu{6 cc.}$ of water. Four to eight drops of a saturated alcoholic solution of diphenylcarbazide is added and a large excess of solid sodium carbonate is gradually dropped into the solution. When the mercury is present, the form produced on neutralization assumes a blue tinge, and after the addition of an excess of carbonate the entire solution turns blue.

The author stated at the end that neither sodium acetate nor sodium bicarbonate can be successfully substituted for the sodium carbonate, indicating the sensitivity of the test for $\mathrm{pH}$. It is worth noting that the test outlined here (vide supra) has been successfully used in Quantitative Analysis Courses in University of Georgia, Athens, GA for two years before publication.
The major drawback of the test is the deterioration of stock solution of diphenylcarbazide in different solvents. Thus, it needs to be freshly prepared. To avoid this problem, a solution has been found in the publication by Urone by 1955 (Ref.2). Accordingly, non-aqueous ethyl acetate and acetone were the better solvents, diphenylcarbazide solutions of which are stable for months. Diphenylcarbazide solutions of Methyl ethyl ketone, methyl cellosolve (2-methoxyethanol), and 2-propanol are usable for 1-2 weeks. Aqueous solutions and solvents tending to be basic such as methanol and ethanol, and those containing traces of water and basic impurities, do not make good solvents for stock solutions of the colorimetric reagent (Ref.2).
Notes: (1) All solutions tested (diphenylcarbazide solutions in seven different solvents) in Ref.2 are $0.25\%$ in diphenylcarbazide. (2) Also, you don't have to synthesize 1,5-diphenylcarbazide; it is commercially available in Millipore-Sigma. (3) I don't think anybody can answer your question about "how are the test strips for heavy metals in drinking water made from this compound," because it should be a corporate secret, similar to $\ce{KI}$ papers for oxidant check.
References:

Alfred W. Scott, “Adaptation of the Diphenylcarbazide Test for Mercury to the Scheme of Qualitative Analysis,” J. Am. Chem. Soc. 1929, 51(11), 3351-3352 (https://doi.org/10.1021/ja01386a503).
P. F. Urone, “Stability of Colorimetric Reagent for Chromium, s-Diphenylcarbazide, in Various Solvents,” Anal. Chem. 1955, 27(8), 1354-1355 (https://doi.org/10.1021/ac60104a048).

In the beginning of the $20th$ century, the following procedure was developed to prove the presence of mercury in solution. One drop of the solution to be tested is deposited on a filter paper which had been dipped into a freshly prepared 1% alcoholic solution of diphenylcarbazide. Mercury salts produce a purple spot, even in a very diluted solution. Chromates and molybdates produce the same reaction. See more detailed reports :
W. Böttiger, Bestimmung kleiner Mengen Quechsilbersalz in stärker Verdünnung. Z. Elektrochem. 22, 69 (1916)
A. Stock, E. Rohland, Kolorimetrische Bestimmung sehr kleiner Quecksilbermengen, Z. Angew. Chem. 39. 791 (1926)
A. Stock, W. Zimmermann, Zur Bestimmung kleinster Quecksilbermengen, Z. Angew. Chem. 41, 546 (1928)
A. W. Scott, Adaptation of the Diphenylcarbazide Test for Mercury to the Scheme of Qualitative Analysis, J. Am. Ch. Soc. 51, 3351, (1929)

Synthesis routes
From molbase[dot]com :
Phenylhydrazine + urea -> 1,5-Diphenylcarbazide                                         ~96% yield
Pasha; Madhusudana Reddy Synthetic Communications, 2009 , vol. 39, # 16 p. 2928 - 2934
Phenylhydrazine + urethane -> 1,5-Diphenylcarbazide + 1-Phenylsemicarbazide + Ammonia
Journal of the Chemical Society, vol. 53, p. 551
Phenylhydrazine + phosgene + Et2O + Toluene -> 1,5-Diphenylcarbazide
Justus Liebigs Annalen der Chemie, vol. 263, p. 277
Phenylhydrazine + 1-Methyl-1-phenylure -> 1,5-Diphenylcarbazide
Gazzetta Chimica Italiana, , vol. 68, p. 516,520
Phenylhydrazine + 1,1-Diphenylurea -> 1,5-Diphenylcarbazide
Gazzetta Chimica Italiana, , vol. 68, p. 516,520
Phenylhydrazine + 1-Phenylsemicarbazide -> 1,5-Diphenylcarbazide
Journal of the Chemical Society, , vol. 53, p. 556
1-benzhydrylidene-5-phenyl carbonohydrazide -> 1,5-Diphenylcarbazide + 1,3-bis(benzhydrylideneamino)urea
Journal of the Chemical Society, , p. 2376 Journal of the Chemical Society, , p. 2114 Anm.
Phenylhydrazine + urethane -> 1,5-Diphenylcarbazide + 1-Phenylsemicarbazide
Journal of the Chemical Society, , vol. 53, p. 556 Chemische Berichte, , vol. 23, p. 2837
Phenylhydrazine + urethane -> 1,5-Diphenylcarbazide
Fresenius' Zeitschrift fuer Analytische Chemie, , vol. 72, p. 119
Phenylhydrazine + ? -> 1,5-Diphenylcarbazide
Zhurnal Organicheskoi Khimii, , vol. 4, p. 837 - 844,814 - 820
Phenylhydrazine + 1,3-diphenylurea -> 1,5-Diphenylcarbazide
Gazzetta Chimica Italiana, , vol. 68, p. 516,520
Phenylhydrazine + phosgene -> 1,5-Diphenylcarbazide
Justus Liebigs Annalen der Chemie, , vol. 263, p. 278
? -> 1,5-Diphenylcarbazide
Chemische Berichte, , vol. 44, p. 3744,3748
1,5-diphenyl-3-nitroformazan -> 1,5-Diphenylcarbazide
Chemische Berichte, , vol. 44, p. 3744,3748
Phenylhydrazine + diphenyl carbonate -> 1,5-Diphenylcarbazide
Comptes Rendus Hebdomadaires des Seances de l'Academie des Sciences, vol. 129, p. 1256
Phenylhydrazine + Guaiacol carbonate -> 1,5-Diphenylcarbazide
Justus Liebigs Annalen der Chemie, , vol. 446, p. 263 Journal of the American Chemical Society, , vol. 52, p. 1132 Comptes Rendus Hebdomadaires des Seances de l'Academie des Sciences, , vol. 129, p. 1256 Bulletin de la Societe Chimique de France, , vol. <3> 23, p. 52


text:
salts affecting the salinity of water
I know there are different types of salts like Himalayan salts table salts etc. I was wonder do different types of salts have a different level in salinity when dissolved in water.

Table salt is obtained from various sources from mines and seas. The world's second largest salt mine (Khewara salt mine) is in Pakistan and this is where this so-called Himalayan pink salt originates. All of it is mostly sodium chloride however it appears pinkish because of iron impurities. If you dissolve it in water, a very small amount of brown precipitate can be seen which is mostly iron oxides.
There is no dedicated study which shows which salt is better for health or not. Himalayan pink salt has recently become a craze in gourmet (cooking) world and among the pseudo-medical sciences which claim that this pink salt contains 99 minerals and it can cure everything.

wonder do different types of salts have a different level in salinity when dissolved in water.

You cannot quantify your personal taste for salinity. As chemists, we only care how much sodium chloride is there by mass percentage. Whatever table salt ends up in grocery stores is highly refined and purified. Grocery store salts contain small amounts of additives such as potassium iodide, potassium iodate, silicates, prussiates, thiosulfate, and so on. So the taste may be slightly different from brand to brand.


text:
What substances do humans consume that are caloric but neither protein, carb, nor fat?
On the Medical Sciences StackExchange site I asked Can a calorie be neither protein, carb, nor fat?
 and got a very helpful answer, which was that ethanol (in alcoholic drinks) is caloric but neither a protein, nor a carb, nor a fat.  I then asked a follow-up question as a comment: Where can I find a list of all such substances?  That is, substances that are caloric but neither protein, nor carb, nor fat.  The person who answered my question recommended I post the question to this SE site.

If you go to https://eur-lex.europa.eu/legal-content/EN/TXT/PDF/?uri=CELEX:01990L0496-20081211&qid=1580028914722&from=EN you will see a fairly extensive list of caloric compounds (screenshot pasted below).  This list is probably not comprehensive -- likely there are many compounds we (which includes our gut bacteria) can metabolize for calories.  Rather, it's limited to componds that are caloric and approved for use in food.  [It also doesn't include compounds that are naturally found in foods, and may be caloric, but are only present in very small quantities, e.g., nucleic acids.]
You can see that, in addition to fat, proteins, carbohydrates, and ethanol, the table lists polyols, organic acids, salatrims, and fiber.
Some explanation of the table may be helpful:
Polyols are artifical sweeteners. They include lactitol, maltitol, mannitol, sorbitol, xylitol, erythritol, glycerol, hydrogenated starch hydrolysates,  and isomalt. [Though note that while erythritol is a polyol, it is listed separately as non-caloric.  And while glycerol, aka glycerin or glycerine, is a sweetener, it is also used in food for other purposes.]
Organic acids include acetic acid (the main component of vinegar, other than water), as well as citric acid, ascorbic acid, and malic acid (the latter three are found in citrus fruits).
Fiber refers to dietary fiber, which is the type of fiber we can digest completely.
Salatrims are "short and long chain acyl triglyceride molecules"; they are a type of low-calorie fat substitute.
Source:  Consolidated text: Council Directive of 24 September 1990 on nutrition labelling for foodstuffs (90/496/EEC) Select: 6 CELEX number: 01990L0496-20081211 Author: Council of the European Union Date of document: 11/12/2008

Minor point: Digestible fiber structurally is a carbohydrate. A more general answer is that we can metabolize an enormous range of compounds, as long as they are composed primarily of carbon, hydrogen, oxygen and nitrogen. A little bit of sulfur or phosphorus is okay too. Some of these compounds will not be net sources of calories, but you have to address that case by case.

@Andrew As the OP was interested in the types of compounds that could be caloric, I thought it made sense to use dietary categories, which distinguish dietary carbohydrates (sugars and starches) from dietary fiber, in part because, on average, they have significantly different caloric values.  Having said that, it is a good point that,  structurally, dietary fibers are carbohydrates.  At the same time, it is worth noting (also a minor point) that not all dietary fibers are purely carbohydrates.  Carageenan contains sulfur, and chitin contains nitrogen.

@BuckThorn Thanks for the fix, and the bibliographic info.  I incorporated both into my post.

The chemical answer is that everything that is not in its thermodynamically most stable state (for many elements that means maximally oxidised or reduced; depending on the element in question) can be transformed into something else under the release of energy. This excludes:

water
cations and many anions e.g. in salt
the carbon dioxide in fizzy drinks
gold, if you eat something like gold-plated sausage for lols and giggles or to feel rich

There are probably one or two more but as you can see this is a very short list compared to all the things that one might possibly eat or drink.
A good rule of thumb is that if it is in food, it probably derives from nature. A second one is that if it was generated by a living thing and it is not in its thermodynamically most stable state, it can probably be broken down by another living thing. The only question that remains is whether humans can break it down and use it as an energy source. One very simplified way of looking at it is: if you eat it, it passes through your digestive tract and you do not recognise it when it comes out, then either you or your gut microbiome were able to metabolise it; it should probably count as caloric.

What about a cup of hot water?

@BuckThorn Interesting one! I think (but this is based on logical reasoning, not actual experimental evidence): As hot water is adding thermic energy to your system, this will cause one of two things: if you’re cold, you will use less caloric energy to keep yourself warm; or if you’re hot you will need to expend more resources to cool you down. So, drinking lots of hot water should mean that you use less caloric energy of your food for keeping your body temperature in most parts of the world and it thus has an indirect effect as if hot water were caloric. But I’m open to corrections.


text:
What happens to concrete if you add lye to the mix?
What happens if you’re mixing concrete and you add some lye to it? Does it make it weaker, stronger or does nothing at all? I’m guessing concrete is basic so adding lye to it might expedite the reaction, causing it harden faster?  Alternatively, it may react to with the water and starve the mixture?

It makes the cement harder faster as shown in this test https://www.sciencedirect.com/science/article/abs/pii/000888468990046X

Cement is basically made of impure $Ca_2SiO_4$, which is made of independent $Ca^{2+}$ and $SiO_4^{4-}$ ions. When mixed with water, the following reaction happens :
$$Ca_2SiO_4  +  H_2O \unicode{x2192}Ca(OH)_2  +  CaSiO_3$$
The useful part of this final mixture is $CaSiO_3$ which has the structure of a polymer made of a long chain $H-O-(-Si(O^-)_2-O-Si(O^-)_2 -O-)_n-H$. And of course there are $Ca^{2+}$ ions near each Si atom to compensate the charges. This makes a long chain of covalences $-Si-O-Si-O-$  and this is why the concrete is so hard and so resistant.
However to be correct, one should not write $CaSiO_3$ but rather $Ca_n(SiO_3)_n$ and the previous equation should be written : $$n Ca_2SiO_4  +  nH_2O  \unicode{x2192} nCa(OH)_2  +  Ca_n(SiO_3)_n$$
To go back to the question of User148298, one sees that $Ca(OH)_2$ is a by-product in making cement. So it no use adding it to the original cement. It does not help.

Isn’t lye Na(OH)2?  Your point still stands.

The article referenced in the comment above (Ref 1) is not relevant to concrete. It analyzes solutions with a water to solids (cement plus $NaOH$) ratio of 1, whereas ordinary concrete seldom goes above 0.5. This is a paste, not a structural material. The $NaOH$ speeded up early hydration at 1 M, and retarded it at 2 M. It was apparently affecting the aluminates more than the silicates.
The composition of typical portland cements varies, but is usually considered to be mostly $C_3S$, a faster reactor and $C_2S$, a slower, but eventually stronger component. Ref 2

Many admixtures are used in portland cement concrete (Ref 3), but sodium hydroxide is not one of them. $NaOH$ is used as a promoter with silica-rich additives like flyash in highly modified cements, but portland cement is already silica-poor. When portland cement hydrates, $Ca(OH)_2$ is formed, which is not a really good strengthener, except that over long periods of time, it will become incorporated into the silicate gel, strengthening the concrete somewhat.
However, there are adverse reactions noted with silica-rich aggregates and the high alkalinity of sodium hydroxide (Ref 4), so it is always advisable to keep the sodium level low, whether from an alkali, an admixture, or external applications, like salt.
Ref 1: https://www.sciencedirect.com/science/article/abs/pii/000888468990046X
Ref 2: https://en.wikipedia.org/wiki/Portland_cement
Ref 3: https://www.cement.org/cement-concrete-applications/concrete-materials/chemical-admixtures
Ref 4: https://www.concreteconstruction.net/how-to/repair/chemical-attack-on-hardened-concrete_o


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Why doesn't nitrogen react with oxygen at room temperature?
I already know that nitrogen's triple bond requires a lot of energy to be broken, so the activation energy is very high, but the book also lists the endothermic energy change as a reason.

Does the positive enthalpy change have something to do with the reaction not being spontaneous i.e. due to positive free energy?
Is there a link between $E_a$ and free energy when deciding whether a reaction will occur?

Yes, you are right, the reason primarily looking at it from a theoretical point of view is the high bond dissociation energy of $\ce{N#N}$
Let's look at this aspect from every point of view

Theoretical- High bond dissociation energy of $\ce{N#N}$
Kinetics- We have the Arrhenius equation given below, which states the relationship between the rate constant and activation energy of the reaction.

$$k = A \mathrm e^{-E_\mathrm a/(R T)}$$
So essentially, this relation shows that if activation energy is higher, then the rate is slower. However, if we increase the temperature sufficiently, then the reaction does start to happen significantly. (Hence, oxygen and nitrogen do react when lightning strikes)

Thermodynamics- We have the very famous relation of thermodynamics

$$\Delta G= \Delta H-T \Delta S$$ So now lets again look at each term here

$H$- Enthalpy of the reaction, it is very high, as we discussed before, the bond dissociation energy of $\ce{N#N}$ is very high.
$S$- Entropy change won't be any significant as the reactants and products both are gases, so this is not an entropy-driven reaction.

Thus
, in every way possible we can conclude that nitrogen and oxygen do not normally react at room temperature.

Thanks a lot! You clearer a lot of things up. One last thing. From what I understand, even if the reaction was exothermic somehow and was thus thermodynamically favourable, it still wouldn’t occur due to the high activation energy barrier. So should the enthalpy change be listed at all as a reason for the reaction not occurring?

https://chemistry.stackexchange.com/questions/105580/why-is-the-combustion-of-nitrogen-endothermic

It is endothermic not exothermic

I know it is. That was not my question

Ohh I got your question, we'll see what you are talking about is two different things. Activation is a kinetically factor, and enthalpy is thermodynamic factor. They are related. But the realtion between kinetics and thermodynamics depends on other factors too. Hence, in every reaction you would hear phrases like it is kinetically favourable but not thermodynamically so, although activatiom and enthalpy are related, don't use only one of them to conclud your reaction

@HahaHahaha Not related to this question, but to https://chemistry.stackexchange.com/questions/118011/how-does-sodium-carbonate-reduce-ph, an old question you attributed a bounty. Please add a comment below this question (the one about pH value) why you attributed a bounty; what is the point requiring further clarification?