text:
Terminology of atomic spectroscopy: Difference Among Term, States and Level
In A Primer on
Quantum Numbers and Spectroscopic Notation
Contents, the concept of a term and level is described as

a) The level is the set of 2J+1 states with specific values of L, S,
  and J. The difference in the energy between two levels gives the
  wavelength or frequency of an atomic transition.
b) The term is the set of levels characterized by a specific S and L.

1) Is there a better way to understand the subtle difference among the concept of a level, a state, and a term. A example would be helpful.
2) A side question is what are the German equivalents of term, state and level. What do they call it in atomic spectroscopy literature?
Thanks.

The figure below shows the situation between configuration for a $p^2$ configuration, terms, levels and states. The word 'state' tends to be used colloquially to mean any of  Term, Level or State.
The Configuration such as  $(1s)^2$ or ...$(2p)^2$ etc. tell us which orbitals are occupied.
These are split with electrostatic (Coulomb) coupling to form Terms.
The Term Symbol informs about the angular momentum. The number of levels in each Term is $(2J+1)$, the number of States is $(2L+1)(2S+1)$ .
The term symbol is
$$\large ^{2S+1}L_J$$
where $S$ is the total spin and $L$ the orbital angular momentum.  $J$ represents the angular momentum as quantum numbers $(J=L+S)$ and is determined by adding individual spin numbers according to Glebsch-Gordon series, i.e. if $q_1,\,q_2$ are angular momentum quantum numbers  then the series is all terms (separated by unit steps) with values $q_1+q_2 , q_1+q_2-1,\cdots \to |q_1-q_2|$. (There may be only be term).
The Terms are split with a magnetic interaction such as spin-orbit coupling, i.e coupling of spin angular momentum with that of the orbital. It is not present for S orbitals because here $L=0$.
Levels are distinguished by total angular momentum $J$, and are $2J+1$ degenerate.
An external magnetic field will remove all remaining degeneracy to form States. These are distinguished by $m_J$ and there are $2J+1$ of them.

Excellent. Could you share the reference for this figure? So in a rigorous sense, when we talk about electronic transitions in atomic spectra, the transitions are in levels. And transition among states can be seen in Zeeman effect.

I used the figure in lectures years ago, and I don't recall where it came from  so perhaps drew it. And yes Zeeman effect (or Stark) splits levels to make states so this spectroscopy identifies states. I think this was how these energy levels were discovered in the first place; anomalous Zeeman effect and all that history.

The concept of terms is really old, and the origin is most likely from the series concept, where each member of series is called a term.

To answer the side question about German language terminology, I'll pick up terms from @porphyrin's answer.
But first of all, term = Term (same as in maths)

The word 'state' tends to be used colloquially to mean any of Term, Level or State.

All those colloquial states are Zustände (pl., sg. Zustand) in German or you can use Term, Niveau (level) and Zustand (state).
I'd say that in German, Zustand is an Oberbegriff (umbrella term) that includes specific types of Zustände such as Terme (Zustand with fixed L and S) and Niveaus (Zustand with fixed L, S, and J).
angular momentum = Drehimpuls (drehen = to turn, Impuls = momentum)
But Eigendrehimpuls ("intrinsic angular momentum") is usually just Spin also in German.
Thus, the Anzahl der Zustände (2S + 1) is the Spinmultiplizität.
spin-orbit-coupling = Spin-Bahn-Kopplung or Spin-Bahn-Wechselwirkung
degenerate = entartet
See e.g. The German wiki page https://de.wikipedia.org/wiki/Spin-Bahn-Kopplung#LS-Kopplung_bei_mehreren_Elektronen for all those terms in their native environment.


text:
How to know dichromate is being reduced from Cr⁶⁺ to Cr³⁺?
When potassium dichromate is being used as an oxidising agent, it gets reduced from chromium(VI) to chromium(III). Is there any way to know that or calculate that other than being familiar with general oxidation states?

Long story short, no.

Similar to other redox reactions (e.g., permanganate [purple] to either manganate [greenish, at a basic pH] or $\ce{Mn^{2+}}$ [colorless, at an acidic pH]), the reduction of chromate to $\ce{Cr^{3+}}$ in aqueous solutions occurrs with a significant change of color of this solution.  Even more in acidic solutions if you have dichromate being reduced to $\ce{Cr^{3+}}$.  And you could infer «being there» since there are both stable compounds with chromate / dichromate, as well as $\ce{Cr^{3+}}$ you can dissolve in water as a visual reference to check with.
But:

it is not so easy to see if your other reagent equally dyes your solution
the color change does not tell you the oxidation state as a number.  UV-Vis spectroscopy might hint you to which electronic transitions now become more probable to access, though.

I would add that magnetic properties of the solution (if you know the concentration and all other ions are diamagnetic) can also give an indication about the oxidation state.

In my question understanding, the OP is not interested in how to experimentally determine it, but rather if there is possibility to predict it without being familiar with chromium chemistry, just from basic principles. Unless one is a quantrum chemistry magician, I would refer to the Ivan's comment. I guess it is a Holy Graal of chemistry students - How to avoid a lot of empirical knowledge ?

That is the colour change. There are several reactions of colour change as you are probably more familiar with the orange dichromate(VI) ion ($\ce{Cr2O7^{2-}}$) than the yellow chromate(VI) ion ($\ce{CrO4^{2-}}$) . Changing between them is easy. If dilute sulfuric acid is added to the yellow solution, it turns orange. If you add sodium hydroxide solution to the orange solution it turns yellow. There are more reaction of this you can find on internet and this reaction itself is elaborated in the following manner:
\begin{align}
\ce{H2SO4 + 2Na2CrO4 &-> Na2Cr2O7 + Na2SO4 + H2O}\\
\ce{Na2Cr2O7 + 2NaOH &-> 2Na2CrO4 + H2O}
\end{align}

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text:
Determining solubility of a gas in a molten salt
If pressure and temperature are known constants, how can I determine the solubility of a known amount of gas in a mixture of molten salts?
Assume the mixture of salts is homogeneous and the composition (as well as amount of said compounds) is known.

Long story short, you can't.

Arbitrary gas and salt mixture? Why even think there's non-negligable solubility?

@IvanNeretin of course it is possible. Geologists and volcanologists have spent decades experimentally measuring the solubility of various gases in various molten liquids (some of which are not too different from molten salts).

@Mithoron once you increase pressure above atmospheric, you can dissolve pretty much any gas in any liquid.

Pumice illustrates this solubility of gases in lava.

All right, "determine" is an awfully broad term. Can you calculate it with a few simple rules? No. Can you measure it experimentally? Sure, why not.

If you know the phase diagram of the gas, and can calculate the exact pressure the gas should have given the volume outside of the molten salt, you can use the measured pressure/temperature to determine any difference in pressure. Then you can calculate the number of moles of gas that must be absorbed to generate the pressure difference.
If the gas is relatively inert (which it probably needs to be if it wont react with molten salt), you can probably use the ideal gas equation:
$$P_0V = NRT$$
If you compare to the actual pressure, you can determine the number of moles absorbed by the salt:
$$\delta N = (P_0-P_\mathrm{exp})V/(RT)$$

This answer is incorrect. The solubility does not depend (at least directly) on volumes. The solubility can be parameterised experimentally for any liquid-has pair, and then calculated using that.

I don't think you understood why I used the ideal gas law here. The solubility is directly related to how much gas is absorbed at a specific pressure/temperature. You can use the ideal gas law to calculate an estimate on how much gas has been absorbed by the molten salt. If the amount of salt is known, then the solubility can be calculated trivially.

"You can use the ideal gas law to calculate an estimate on how much gas has been absorbed by the molten salt." - No, you cannot.

Can you explain your logic? Why would a pressure loss not be due to absorbance? It is not helpful to simply say "you're wrong".

Because once you're a bit higher than atmospheric pressure, things do not act as ideal gases any more. For example, oxygen fugacity can be an order of magnitude higher than oxygen partial pressure. As solubility is determined by fugacity, not partial pressure. Furthermore, the composition of the molten salt has a very strong influence on the solubility of gases in it. Furthermore, at temperatures high enough to melt salts, you will get salt vapour as well and your atmosphere is not the pure gas any more. And let's remember that you have an unknown molar volume of the gas component in the salt.

As temperature increases, intermolecular attractive forces decrease as the kinetic terms dominate. If you look at the van der Waals equation of state, a high temperature (or low pressure) results in behavior that is approximately equal to an ideal gas. You are correct about salt vapor pressure coming into play, however it is negligible. NaCl for example has pressure <100kPa at 1700K (well beyond the melting point). One can know how many moles are in the salt using experimental data for the gas P-T (or equation of state if data unavailable) and finding the difference between vapor pressure.

The salt vapour pressure is negligible compared to what? To the solubility of gases in the salt melt? It is also going to be negligible at high temperature. So you are comparing one negligible thing to another. And let's not forget that salt melts are extremely corrosive and will react with the container to some degree, leading to another source of uncertainty. Doing this kind of stuff is literally my job. If you find a way to do it according to your method, let me know. We probably have a Nature Geoscience or Nature Chemistry paper here.

Of course my suggestion isn't supposed to be perfectly accurate, but according to this chem rev paper, solubility of several gasses in molten salts obey Henry's law: https://pubs.acs.org/doi/abs/10.1021/cr60242a003
The proportionality constants may be determined from existing data to get an estimation.

Noble gas solubilities in molten salts in atmospheric pressures are irrelevant. OP specifically asked for pressure to be a variable, and this will deviate strongly from ideal behaviour once you have things that are not noble gases.

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text:
Do K+ and F- ions have the same ionic radii?
Since fluorine and potassium form ions to gain the electronic configuration of neon and argon respectively, then shouldn’t $\ce{K+}$ have a greater radius, as argon is larger than neon?
Also which would have a bigger atomic radius, potassium or fluorine?
The original question is
The ionic radii of $\ce{K+}$ and $\ce{F-}$ are nearly the same (1.34), then the atomic radii of $\ce{K}$ and $\ce{F}$ will be
A) 1.34, 1.34
B) 0.72, 1.96
C) 1.96, 0.72
D) 1.96, 1.34
All units in Angstrom ($\pu{10^{-10} m}$).

"All units in Armstrong." No, those look more like Ångström. And the question is impossible to answer without eliminating the listed choices (as both radii are typically determined experimentally), which makes it ill-posed IMO.

Relax people! It was the auto-correct. I failed to notice it. Sorry for bother you with a typo. I assume it was obvious that I meant Angstrom. No need to get so triggered over it.

@andselisk it’s really not advanced chemistry. I think it can found at easily, since the one with the smaller size will have lesser radius. It being an MCQ just makes it painfully obvious. All I have a problem is why K+ is smaller than F-

"…since the one with the smaller size will have lesser radius" — this statement is both tautologic and sophistic. Nothing really is easy if you look deeper, and potassium cation is not smaller than fluorine anion (in terms of ionic radii). $\ce{K+}$ with C.N. 4 is comparable in size with $\ce{F-},$ and then the values of ionic radius for higher C.N. skyrocket to 1.64 Å for C.N. 12.

@andselisk I don't think the examiner is really concerned about the real values (pfft, haha, who is? Those weirdos . . . chemists). OP This question does strike me as pretty trivial though. What you're inquiring about is already addressed when a reference point is given: The ionic radii. It's really obvious and if you think about it, you just missed it. There's really nothing to be learned here.

Why do K+ and F− ions have the same ionic radii?

They don't according to wikipedia:

Source: https://commons.wikimedia.org/wiki/File:Atomic_%26_ionic_radii.svg
They do according to "Data taken from John Emsley, The Elements, 3rd edition.  Oxford:  Clarendon Press, 1998."

Source: https://www.angelo.edu/faculty/kboudrea/periodic/trends_atomicradius.htm
As @andselisk wisely mentions in his comments, radii depend on coordination number (and on how distance between anion and cation are turned into a single radius - you have to set one radius somewhat arbitrarily).

The original question

Irrespective of the actual numbers, one can see nicely in the first figure that monovalent cations are much smaller than the corresponding atom (one less shell filled), and monovalent anions are somewhat larger than the corresponding atom (one last electron placed into the outer shell). Answer c) is the only one that follows the expected pattern.

Question has been edited, please have a look again, Since the context was missing

It's a common misconception to assign a single ionic radii to the ion. In general it works OK, but for some ions (such as potassium, actually), ionic radius varies a lot with coordination number. From 97th edition of CRC Handbook of Chemistry and Physics, p. 12-12 $R_\mathrm{i}(\ce{F-}, \mathrm{C.N.}~6) = \pu{1.33 Å}$ and $R_\mathrm{i}(\ce{K+}, \mathrm{C.N.}~4) = \pu{1.37 Å}$ which is indeed pretty close.

Answer has been edited, please have a look again (@andselisk).

I would like to know the reason. Why are monovalent cations smaller than the their orignal atom

@Aditya Look at the electron configuration of the atom, [Ar]4s1, and of the ion, [Ar]. So the atom uses the 4th shell, and the ion has 3rd shell as the highest, with the same number of protons in the nucleus. That explains the big difference.


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What is the binding mode of the ligand in complexes with quadruple metal-metal bonds?
I'm wondering which atomic orbital is left in a complex with quadruple metal-metal bond to bind the steric ligand, since all d orbitals except the x2-y2 are used for the metal-metal bonds?
or is it the x2-y2 atomic orbital that forms the bonding?
and if so: what atomic orbital is used in metal-metal quintuple complexes?
Here is an example complex: https://en.wikipedia.org/wiki/Quintuple_bond

To determine what orbitals do, it is often a good idea to look at the symmetry group as it can provide a rough overview over what can (symmetry-allowed) and cannot (symmetry-forbidden) happen. In the case of a quadruple bond such as $\ce{[Re2Cl4]^2-}$, we can approximate the rhenium(III) centres as being close to $C_\mathrm{4,v}$. A quick glance at the corresponding group theory table (copied below to avoid page loading issues in the large linked page) will show us which irreducible representations we may need to consider.
$$\begin{array}{c|ccccc|cc} \hline
C_\mathrm{4v} & E & 2C_4 & C_2 & 2\sigma_\mathrm{v} & 2\sigma_\mathrm{d} & & \\ \hline
\mathrm{A_1} & 1 & 1 & 1 & 1 & 1 & z & x^2+y^2, z^2 \\
\mathrm{A_2} & 1 & 1 & 1 & -1 & -1 & R_z & \\
\mathrm{B_1} & 1 & -1 & 1 & 1 & -1 & & x^2-y^2 \\
\mathrm{B_2} & 1 & -1 & 1 & -1 & 1 & & xy \\
\mathrm{E} & 2 & 0 & -2 & 0 & 0 & (x,y),(R_x,R_y) & (xz,yz) \\ \hline
\end{array}$$
The final two columns show the relevant orbitals. The s orbital always transforms as $\mathrm{a_1}$, the p orbitals as $\mathrm{a_1+e}$ and the d orbitals as $\mathrm{a_1+b_1+b_2+e}$. The four chlorine atoms will bond using p orbitals (but the orbital type does not matter as we will only be considering σ bonds). They transform as $\mathrm{a_1+b_1+e}$ (calculation not shown). Immediately, we notice that $\mathrm d_{xy}$ does not contribute to bonding to the ligands but only to the $\ce{Re-Re}$ quadruple bond. Furthermore, we notice that $\mathrm d_{x^2-y^2}$ is practically required to bond to the chlorines as there is no other rhenium orbitals of that symmetry.
To determine what is left, it may be easiest to draw out the potential chlorine orbitals. $\mathrm{a_1}$ means that all phases pointing to rhenium are the same. This can bond either with rhenium’s 6s or $\mathrm{5d}_{z^2}$ orbital or a linear combination of both. The $\mathrm p_z$ orbital does not have any contribution in the $xy$ plane and thus cannot participate.
The chlorine’s $\mathrm e$ orbitals correspond to two in which two opposite chlorines have opposite phases and the other two have zero contribution (in one case, the $x$ chlorines have contribution, in the other the $y$ ones). Thus, to bond with these we will use $\mathrm{6p}_x$ and $\mathrm{6p}_y$. Again, this is because the corresponding d orbitals do not have any contribution in this $xy$ plane.
Thus, to sum things up:

$\mathrm{6p}_x,\mathrm{6p}_y$ and $\mathrm{5d}_{x^2-y^2}$ can bond to chlorines
$\mathrm{6p}_z, \mathrm{5d}_{xz}, \mathrm{5d}_{yz}$ and $\mathrm{5d}_{xy}$ can bond to the other rhenium
the $\mathrm{6s}$ and $\mathrm{5d}_{z^2}$ orbitals can bond to either; they probably linear combine with $\mathrm{6p}_z$ to form two bonds and keep an empty orbital.

If the symmetry around a rhenium atom is assumed to be $C_4$ instead of $C_\mathrm{4,v}$, that doesn’t change much of the analysis; only the distinction between $\mathrm d_{x^2-y^2}$ and $\mathrm d_{xy}$ becomes irrelevant.


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What is the largest known zero-point energy for a molecule?
I was surprised to find out that the zero-point energy (ZPE) for the ground state of $\ce{BH3}$ is a few thousand $\pu{cm-1}$, because for molecules I usually work with, it's much smaller. Does anyone know the largest recorded ZPE?
I have recently asked a similar question: What is the biggest known difference between $r_e$ and $r_0$?

Liquid helium, maybe?

@OscarLanzi: Why liquid helium?

My understanding is this is what keeps it liquid to absolute zero unless compressed.

@OscarLanzi: I see. When I wrote "molecule" in the title, I didn't have a condensed phase material (like a liquid) in mind. I suppose I just had gas phase molecules in a vacuum, in mind.

Is the question about vibrational zpe? see eg https://www.nist.gov/sites/default/files/documents/srd/jpcrd362007389p.pdf

@buck of interest in your reference: Mg2 (which may form a weak bond if s-p mixing occurs in its MOs), NaLi (which may be considered a lithide).

I think this question has a trivial answer as written because the ZPE grows with the system size. A potentially more interesting question, and probably the spirit of the question, is what is the largest known zero point energy per atom? I think this may also somewhat trivially be H2 though... This is all assuming OP means vibrational ZPE which aligns with the example about BH3 I believe.

@jheindel: Perhaps H2 has the highest ZPE for a neutral diatomic in its ground electronic state. Yes I meant vibrational ZPE (the energy between v=0 and the minimum of the potential). We would guess this because H2 is light, so the levels are far apart. However the shape of the potential also matters right? He$_2$ has only one v-level and its outer turning point is all the way out at 20 Angstroms right? So could He2 have a bigger ZPE?

So, two things. Generally the ZPE in the v=0 level will be largest on the ground electronic state as higher electronically excited states tend to have larger anharmonicity. Also, the vibrational ZPE of a diatomic is greatly affected by the masses of the atoms involved, so H2 certainly has a higher ZPE than He2. Also H2 is much more strongly bound, so it will have a larger force constant. But as I said, the ZPE can be arbitrarily large, so I'm not sure exactly what you're interested in knowing about.


text:
Significant figures and uncertainty
A narrow-necked volumetric flask with a capacity of 200mL has an uncertainty of ±0.150mL so should the number of significant figures after the decimal point in my final answer be three or two?

I assume that you have a narrow-necked flask (Class A) with a nominal capacity of 200 ml or 250 ml since these are the only ones with a maximum permitted error of 0.150 ml. Could you please clarify? And don't omit the units!

Giving to the uncertainty more than 2 valid digits would be a kind of statistical ignorance.
If you searched for the fundamental constants, measured with many orders of precision, the uncertainty is never published with more than 2 valid digits.
Even if there is done 100 repeated measurements, the SD uncertainty CI95% is 0.88*SD to 1.16*SD, so for 0.150 it is 0.132..0.174.
So $\pu{\pm 0.15 ml}$ would be much better.
See also https://stats.stackexchange.com/q/631/138724

Another possibility is the 0.150 (whatever unit ) is the maximal allowed deviation from the nominal value, given by the quality control. That can be achieved easily.
But then, you have no idea about the statistical distribution of the actual error, as you do not know the probability it would not pass the check.
It is quite possible the s belongs somewhere to interval 0.05..0.06 ( whatever unit ).
0.05 would mean about 0.3% would not pass.( $3\sigma$)
0.075 would mean about 5% would pass. ( $2\sigma$ )

I am curious what process would lead to triangular distribution, unless it is just  arbitrary approximation of normal distribution by triangular one as this one with base at cca 2.3 sigma.: https://math.stackexchange.com/questions/1374785/triangular-vs-normal-distribution

But it would return us to uncertainty of sigma. 0.150 with 3 digits must be a hard cut, not statistical one.

From QUAM:2012.P1 (EURACHEM/CITAC Guide, “Quantifying Uncertainty in Analytical Measurement”, 3rd Ed., 2012.), Example A1 on pp 38-39: "The value of the uncertainty is given without a confidence level or distribution function, so an assumption is necessary. Here, the standard uncertainty is calculated assuming a triangular distribution.  (skip little eqn).  NOTE: A triangular distribution was chosen, because in an effective production process, the nominal value is more likely than extremes. The resulting distribution is better represented by a triangular distribution than a rectangular one."

Sure, triangle one is definitely better than rectangular one :-) So sigma for the OP could be estimated as 0.15/2.3=0.065.

I think the triangular distribution just serves as a minimally informed estimate of a normal distribution: it has the centrality characteristic, but imposes nothing else.

I suppose so. But to be consistent with  error propagation rules, it is better to convert it to estimations of sigma or CI.

Actually, the little equation I skipped showed the uncertainty as the magnitude of the uncertainty divided by the square root of 6. Thus, 0.15/SQRT(6) = 0.15/2.45 = 0.061.

I suppose it is analysis of the triangular distribution variance.

You are exactly right! By the way, that QUAM Example A1 is for "Preparation of a Calibration Standard" (cadmium standard for AAS) and it is pretty good. They go through all the measurement error sources and show how to combine them into an overall measurement uncertainty.

61 Vs 65, it is still good , considering the SD of SD.


text:
Isotopes and elements
this is a gold bar. If we look at its structure, will we see that there are atoms that have a different number of neutrons (aka isotopes) (as far as I'm concerned, an isotope is an atom with a different amount of neutrons in it)
or has it to be another bar that consists of this isotopes?

Gold has only one naturally occuring isotope: 197-Au.

Ed V, what about others, do people make them? (and thank you, a lot) And 197-Au consists of atoms that have the same number of neutrons, right?

Sure! Wikipedia (check it out!) shows 4 major synthetic gold isotopes and my guess is that there are more. The 197-Au has 79 protons and 118 neutrons.

I just checked Brookhaven National Laboratory's National Nuclear Data Center "Nuclides Map" and it shows gold isotopes from 169-Au up to 205-Au. You should search for that map: it is very interesting and amazing! Sorry I do not have a link: I downloaded it years ago.

The URL is http://www.nndc.bnl.gov/nudat2/ . It is completely different than the old giant PDF and it turns out more gold isotopes exist: 169-Au up to 210-Au.

@EdV Maybe you know it, maybe of interest for you / colleagues of you / users of ChemSE: the Karlsruhe nuclide map's review about its current 10th edition / 2018 (open access here: https://doi.org/10.1051/epjn/2019004) and an elder review, including context to other maps, here: www.nucleonica1.com/wiki/images/9/96/NuclearData_FTU_2013.pdf).

@Buttonwood  Wow, really nice! Many thanks! Looks like the question will be closed, but the links are worth their weight in gold! :-)

@EdV Then take these as little «thank you» for hinting in one of your answers or comments somewhere in the second quarter of 2019 -- I don't recall the exact spot -- to the work by Joel Tellinghuisen around how to process and apply statistics on experimental data in chemistry and calibration / regression curves.  It wasn't taught this way in the classes I attended, and often, a point to reflect how to do better.

@Buttonwood I asked Joel for his publication list and he sent it to me last August. If you want it, or any of the 20 or so papers of his that I have, please send me an e-mail! I don’t want to post my e-mail address, but it is pretty easy to find. I would have liked to have had Joel as a professor or as a colleague later.

Gold has only one natural isotope: 197. There are artificial gold isotopes, but they do not end up in gold bars. Even if they did, their half life is measured in days so don’t expect much to remain after a year or so.
The “structure” of the metal bar, or atomic arrangement, has nothing to do with its isotopic composition.
Even if you were to measure somehow trace amounts of radioactive gold in your bar (for example gold-195) using methods such as mass spectrometry, it would be hard to distinguish from platinum-195, a very common impurity in gold. To complicate it even more, you will not know whether the Pt-195 was there to begin with, or formed by the decay of Au-195.
If you had a gold bar made of pure gold-195, you would be dead soon from radiation sickness unless you took the appropriate safety measures.

A 1 kg gold bar of pure Au-195 in 0.5 m working distance gives a dose rate of about 24 Sv/h. So first symptoms of sickness would occur after about half an hour. If you then stop working with it, you would die in about one week. Anyway, the bar would be red hot since it's releasing about 2.7 kW of isotopic power.


text:
What is the physical meaning of the statements like "weight percentage of FeO in Fe"?
I have always had the notion of calculating weight percentage of Fe in FeO. It always seems that we are to calculate at times the weight percentage of FeO in Fe.Particularly when one weight percentage is involved.However "weight percentage of FeO in Fe" is something that seems extremely ridiculous on my part.I would be extremely grateful if someone can correct my wrong notion/give me an intuition to what I am asking for.

Please clarify the question and ideally post the exact question in the textbook. It makes little sense as written.

@M.Farooq My question is how do I calculate weight of Fe from the weight of FeO if I have a mixture of Fe and FeO?

I have read this three times and it does not make sense. Please do exactly what @M.Farooq suggested.

Still makes no sense, so I will take a guess. You know that the mass % of Fe in FeO is 77.73%. Therefore, the reciprocal of 0.7773 is 1.2865. Hence, 1 unit (g, ton, mol, whatever) of Fe can be oxidixed to 1.2865 units (same as used for Fe). If this is not what you are attempting to ask about, then I have no clue.

@EdV Your prediction of my question is absolutely correct.That is exactly what I am asking for.Like % mass of Fe in FeO is something that I understand but what does the reciprocal of that percentage give us?That is my question.

It just tells you how much you could, in principle, make. Pure iron metal does not contain any of the several iron oxides, so there is no physical meaning to "weight percent of iron oxide in iron". Hemoglobin also contains Fe, so there is physical meaning to mass (or weight) % Fe in hemoglobin, but not to % hemoglobin in Fe.

"It just tells you how much you could, in principle, make". Please elaborate on this.I don't get this completely.

Let us continue this discussion in chat.

Based on the comments and extended chat, this is the kind of question the OP actually wanted to ask: Consider a mixture of Fe and FeO. The total percentage (by weight) of iron is 86.638%. What is the percentage of iron metal in the mixture?
Answer: For simplicity, assume exactly 100 g of the mixture. Hence the total iron is 86.638g and the remaining 13.362g is oxygen. Therefore, the number of moles of oxygen atoms is 0.83516 moles, i.e., 13.362g divided by 15.9994g/mole. This is also the number of moles of Fe in FeO, so the mass of Fe in FeO is 46.639g, i.e., 0.83516 moles times 55.845 g/mole. Subtracting this from 86.638g yields 39.999g metallic iron. So the 100 g of mixture was 39.999g of Fe and 60.001g of FeO, i.e., 40% Fe and 60% FeO.


text:
Where will the oxygen atom be inserted on the Baeyer-Villiger esterification?
On our exam in organic chemistry, there was a question that I failed to answer correctly. The question is about the Baeyer-Villiger esterification of propiophenone or ethyl phenyl ketone:

On the exam, I answered with this structure:

But when the exam paper came back, I learned that the answer was where the oxygen is inserted adjacent to the aromatic ring:

How can I resolve this dilemma of mine so I can prevent this mistake from happening?

Google for "migratory aptitude Baeyer-Villiger", there will surely be one site (maybe even on Stack Exchange) explaining it. There's no point memorising the list without first understanding the reaction mechanism, so make sure you know that first (your question doesn't give any indication of what you know about the reaction, beyond that it inserts oxygen atoms next to carbonyl groups).

@Kent de los Reyes:  In addion to the Baeyer-Villiger reaction, there a few rearrangement reactions taught at about the same level of organic chemistry 101 with residues moving with different ease.  Have a look for example on pinacolon rearrangement of diols, or Beckmann rearrangement of oximes (e.g., to access precursors of nylon).


text:
Are there any datasets containing molecules with more than 38 heavy atoms?
I have been testing a machine learning approach for molecular energy prediction. The current dataset that I have is QM9, which is consist of molecules with up to 9 heavy atoms.
I was wondering if anyone knows of available datasets that contain molecules with large numbers of heavy atoms. I will be testing ZINC, which has up to 38 atoms. Anyone knows of a larger dataset available?

In retrospect -- after giving providing an answer (molecules encoded as SMILES), while «user1271772» provided an other answer about molecules keeping geometries, does your question for a larger data set refer to a dataset containing molecules larger than 38 atoms, or a dataset with more molecules than ZINC?

Larger than 38 atoms. I was gonna say, but I appreciated your detail answer very much

Considering your additional input, I added a second part to my initial answer.  It is about an example which retains molecular geometries (similar to .xyz files) rather than a simplified representation as SMILES / InChi string.

The ISOL24 database (http://www.thch.uni-bonn.de/tc.old/downloads/GMTKN/GMTKN55/ISOL24.html) contains molecules with up to 81 atoms!
The other answer says that there's a database called "OE" with molecules that have up to 174 atoms, but it is "not yet publicly available".

Given the initial check in the work by Stuke et al. shown below, do you know about a similar survey about the entries in the ISOL24 database?

@Buttonwood: I do not understand your question.

The question was if there equally is a plot like figure 2 extracted from doi/10.1063/1.5086105 which a) depicts which elements are present or/and b) the the distribution of number of atoms per molecule for the ISOL24 database.  From its associated publiation in PCCP (Phys. Chem. Chem. Phys.,
2017, 19, 32184, DOI: 10.1039/c7cp04913g) I retain it was setup to benchmark functionals and it were nice to compare it this way with the three sets below, too.

@Buttonwood: I agree that it would be nice to compare such a graphic representation. I do not know any such plot for ISOL24 though.

This sounds like you were exploring work at least related to the work by the Lilienfeld group equally hosting a dedicated site here about data sets already used in their earlier and ongoing exploration of chemical space, programs used to work with the data, and publications.
To go considerably higher in molecule count than QM9, you could either go for

GDB-11 about small organic molecules up to 11 atoms of C, N, O and F which «contains 26.4 million molecules (110.9 million stereoisomers), including three- and four-membered rings and triple bonds», described in J. Chem. Inf. Model. 2007, 47, 342-353 (doi.org/10.1021/ci600423u), or
GDB-13, about «small organic molecules up to 13 atoms of C, N, O, S and Cl following simple chemical stability and synthetic feasibility rules. With 977 468 314 structures, GDB-13 is the largest publicly available small organic molecule database to date».  This one was described in J. Am. Chem. Soc. 2009, 131, 8732-8733 (doi.org/10.1021/ja902302h)

Convienently, you can download both -- including sub-sets like «containing only carbon and nitrogen», or «chlorine and sulfur», or «fragrance like» in case you don't want to fetch 2GB of already compressed data -- from the Reymond group. To quote: «All the molecules are stored in dearomatized, canonized SMILES format.»
The even larger GDB-17 («of up to 17 atoms of C, N, O, S, and halogens» with an universe of 166 billion entries, described in J. Chem. Inf. Model. 2012, 52, 2864-2875, [doi.org/10.1021/ci300415d, open access]) is accessible to the public on this site as a 50 million random subset only, partly because the gzipped archive is about 400GByte.  Among the publications citing this work is for example the Lilienfeld group again for machine learning (J. Chem. Phys. 143, 084111 (2015), doi.org/10.1063/1.4928757).

Initially, I misinterpreted the question but think the answer may be more rounded by addition of the following complementary publication:  «Chemical diversity in molecular orbital energy predictions with kernel ridge regression» (J. Chem. Phys. 150, 204121 (2019), doi.org/10.1063/1.5086105, preprint available here).  Aiming for a machine-learning analysis, the authors first compared QM9, 44k conformers of proteinogenic amino acids (AA), and a 64k set of organic molecules extracted from the CCDC potentially suitable for organic electronics (OE) for the content of atoms per molecule and found the following distribution:

To shed some light on them:

QM9 represents 133,814 small organic molecules with up to 9 heavy atoms (C, N, O and F)
AA is about «44,004 isolated and cation-coordinated conformers of 20 proteinogenic amino acids and their amino-methylated and acetylated (capped)
dipeptides. The molecular structures are made of up to 39 atoms including H, C, N, O, S, Ca, Sr, Cd, Ba, Hg and Pb.»
OE is about «64,710 large organic molecules with up to 174 atoms extracted from organic crystals in the Cambridge Structural Database (CSD). [...] The OE dataset is not yet publicly available. OE offers the largest chemical diversity among the sets in this work both in terms of size as well as number of different elements (Fig. 2). It contains the 16 different element types H, Li, B, C, N, O,  F, Si, P, S, Cl, As, Se, Br, Te and I.»

(The mentioned restriction sharing the original data relates to the user agreement with the CCDC.)
Further DFT-based property computations with these OE extracted molecular geometries lead to an ensemble of equilibrium molecular structures, and these derived geometries are accessible within a public Jupyter notebook.  Shared with the public here, the deposit comes with a guiding tutorial.ipynb, including an example how to retrieve these optimized geometries and display them with Jmol.

I don't think any of these have more than 81 atoms!

Ok I see the confusion. @Buttonwood perhaps you can answer this question: https://chemistry.stackexchange.com/questions/119804/what-is-the-dataset-with-the-largest-number-of-molecules

@user1271772 I think the updated answer offers both the avenue toward how to access the chemical space of many more (small) molecules (first half) as well as about molecules still considered as small yet more atoms per molecule.  Equally I perceive the diversity in the OE dataset regarding «types of molecules, which atoms are connected with each other» larger than for proteïns.  For a small-molecule chemist's view, the complexity of the later group emerges more from how theses amino acids are arranged in space ($\beta$-sheets, hairpins, helices, to mention a few complications).

Ok, so now you've shown two databases with fewer than 81 atoms, and one that has more than 81 atoms but is "not yet publicly available". The first half of this answer is completely unrelated to the question so I think it doesn't belong here at all.

The first part was based on a misunderstanding, clarified later.  The second   includes the derived coordinates e.g. in the df_62k.json for which both Jupyter notebook tutorial.ipynb (and equally tutorial.html) document section #4 / working with molecular structures.  The optimized molecular structures are saved in xyz_pbe_relaxed. These entries contain molecular coordinates in the standard xyz-format.  See on line In [52] as example # We will write the molecules contained in df_subset2 to an xyz-trajectory (stacked xyz files) and these xyz strings are processed further, too.

The first part was based on a misunderstanding that was clarified later. So why do you still have it in the answer to the question "what is the dataset containing the largest molecules?"

The first version of the question was titled by «Large molecule dataset» as about accessing many molecules which is orthogonal to the title now, «Dataset of Large Molecules».  The target moved, and so did the answer -- reproducible.  To me, the two parts don't contradict, they complement each other.  If QM9 contains up to nine non-H atoms per molecule, extraction of all DFT geometries from the 64k json yields a 128 MByte .xyz still including 471 entries with more than 100 atoms per molecule (up to the one with 174 atoms [75 of them non-H]).

I appreciate that you struggled to understand the original intention of the question, but to me it was very clear from the sentence: "I was wondering if anyone knows of available datasets that contain molecules with large numbers of heavy atoms. I will be testing ZINC, which has up to 38 atoms."  It is not at all true that the "target moved" it is that your understanding of the target moved (the asker never changed their mind about what they wanted). Anyway, I think your first effort was still valuable so I gave you a chance to let it be helpful as an answer to a new question.

Beyond other answers, I'd suggest the original PubChemQC project, which offers ~3 million molecules from PubChem optimized using DFT (B3LYP/6-31G*). Molecules include a wide variety of elements as long as the molecular mass is less than 500 Da. (Roughly speaking that should still handle ~38 carbon atoms.)
"PubChemQC Project: A Large-Scale First-Principles Electronic Structure Database for Data-Driven Chemistry"
J. Chem. Inf. Model. 2017 57(6) pp. 1300-1308
You mention the number of heavy atoms, but keep in mind that QM9 only contains a small subset of elements and ZINC has many more.

It's somewhat of a pain to download the PubChemQC data files but they've demonstrated the use in ML methods both for ground-state energies and excitation energies using TD-DFT.


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Alcohol Nonmeclature - Organic Chemistry
I'm new to organic chemistry and having some trouble naming alcohols. I came across this molecule, I believe it's name is 3-ethyl 2-pentanol, but my online checker system is saying that I am wrong. Am I wrong, or is the online checker looking for 3-ethyl pentan-2-ol?

Thanks for your help!

No space between substituents and parent name. 3-ethylpentan-2-ol

3-Ethyl-2-pentanol is a accepted IUPAC name. However, preffered IUPAC name (PIN) is 3-ethylpentan-2-ol (https://en.wikipedia.org/wiki/Preferred_IUPAC_name).


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Why is only the carbonate anion used to precipitate Barium?

Q: Why is the $\ce{CO3^2-}$ anion used to precipitate Barium from $\ce{BaBr2}$ and not another anion, such as $\ce{PO4^3-}$, or $\ce{SO4^2-}$?

I am studying high school Chemistry, and the current unit is precipitations. The worked example states that $\ce{Na2CO3}$ can be used to precipitate $\ce{Ba^2+}$ from a solution of $\ce{BaBr2}$, which makes sense.
However, it does not explain why the $\ce{CO3^2-}$ anion should be selected, when according to the solubility table $\ce{SO4^2-}$ and $\ce{PO4^3-}$, would both be able to precipitate $\ce{Ba^2+}$ from $\ce{BaBr2}$. Am I missing something obvious? Why is it that the $\ce{CO3^2-}$ anion is favoured over the other two anions?
`

Do you have solubility product constants?  That may tell you something.

Who said that they’re not? Sulfate is commonly used to precipitate barium. Barium sulfate is an extremely insoluble compound.

I don't have solubility product constants, but it seems that the choice of CO3 was just an oversimplification on the part of the publisher. Thank you both for your help!

Actually, the textbook should remove this question. You should discuss this with your teacher. It will be a service to future students. The heavier elements of group II elements form insoluble sulfates, phosphates and carbonates.
Now carbonates display another interesting phenomenon. If you keep bubbling carbon dioxide in into a group II carbonate solution (or better say suspension), the group II ions form water soluble bicarbonates. It was a classical experiment of "old school" chemistry. Bubble carbon dioxide into a clear solution of lime, it would become cloudy, if you continue that, it becomes clear again.
Another reason, which may be valid is that you may wish to re-dissolve precipitated barium carbonate after filtration. That would be easy to do so by adding a small amount of strong acid. However re-dissolving barium sulfate or phosphate is very difficult.


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What is the dataset with the largest number of molecules?
In computational chemistry we often apply techniques to all molecules of a "dataset". Recently it was asked which dataset contains the largest molecules, and a user accidentally  misinterpreted this as asking for the dataset with the largest number of molecules: Datasets of Large Molecules
This question now gives that user an opportunity to give their answer here.

Despite the link to the related question, I think it won't hurt to add some more details what kind of dataset it's all about. The first thing that came to my mind was molecular structures from CCDC database, which I realized is not what is being asked for only after clicking the URL (or maybe I'm just not that smart:) ).

My answer would be the same for both questions: the PubChem database contains millions of structures ans is free.

@andselisk The original answer given, where I thought the question were accessing more and more molecules, got a second part; now pointing toward (retained) geometries of small molecules with more and more atoms.  From my perspective, this provides a more round perspective.  Equally, since you mention CCDC passing this summer the 1 millionth entry, I think COD deserves to be mentioned as well.  Which equally found its way into cheminformatics, e.g. as searchable supplementary file in DataWarrior.

I'm voting to close this question as off-topic because one answer given to the initial question now equally addresses this perspective, too.  And I think it is worth to keep both perspectives, complementary in character, deposit in one spot in common.


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Neutralization between Ca(OH)2 + H2SO4 in a 30% hydrogen peroxide solution
I have performed an experiment where I added an excess of $\ce{Ca(OH)2}$ base to a solution consisting of 5 mL of 30 % hydrogen peroxide (buffered at pH 5) and a very small amount of sulfuric acid (such that the pH of the original solution of 30 % $\ce{H2O2}$ and acid was 1.7).
I am having trouble understanding what reactions maybe occurring between these three substances to give a final pH of 10.5.
I know that $\ce{Ca(OH)2}$ and $\ce{H2O2}$ form $\ce{CaO2}$ when reacted, but shouldn't the final pH be equal to that of $\ce{Ca(OH)2}$ considering it is in excess? What reactions could be occurring here and why are they not allowing the pH to reach that of $\ce{Ca(OH)2}$?
Any help at all would be very much appreciated. Thank you in advance.

There is no CaO2.

@Ivan Neretin https://en.wikipedia.org/wiki/Calcium_peroxide Calcium peroxide is produced by combining calcium salts and hydrogen peroxide: Ca(OH)2 + H2O2 → CaO2 + 2 H2O The octahydrate precipitates upon the reaction of calcium hydroxide with dilute hydrogen peroxide. Upon heating it dehydrates.

All the same, I don't believe it will form in the said conditions.

It is not the same as "there is no CaO2".

Let me add that, if pH is 10.5, it's clearly not "neutralized pH". As for the final question, you should read about buffers.

Nobody express pH with 4 decimal digits, considering effects of activity coefficients. More than 2 digits usually do not make sense to use.

Is the calcium hydroxide in excess wrt sulphuric acid, or wrt hydrogen peroxide ? As there is neutralisation of the peroxide as the weak acid and precipitation of the salt.

@Poutnik in excess with respect to sulphuric acid

pKa of hydrogen peroxide is 12.5  pH 10.5 means the most of it is in its acidic form. That is supported by CaO2 precipitation.

What method is used to determine pH?

@JamesGaidis A glass electrode pH meter

@Poutnik So, the pKa of Hydrogen Peroxide being higher means that the acidic form of the reaction is favored, and so that prevents the solution from reaching the pH of Calcium Hydroxide?

in your question formulation, you have forgotten to take into account $\ce{H2O2}$ is a weak acid.
The title should rather be:
Neutralisation between calcium hydroxide and 30% hydrogen peroxide"
Unless $\ce{Ca(OH)2}$ was in excess over $\ce{H2O2}$ - and it was said it was not - $\mathrm{pH}$ would be always significantly lower than pH of the hydroxide.
$$\mathrm{pH}=\mathrm{p}K_  \mathrm{a,\ce{H2O2}} + \log \frac{[\ce{HO2-}]}{[\ce{H2O2}]}$$
where $\mathrm{p}K_  \mathrm{a,\ce{H2O2}}=11.75$$ by Wikipedia, but see the links below.
If we consider reaction
$$\ce{Ca(OH)2 +  H2O2 ->  H2O + Ca(OH)(HO2)}$$
we need to neutralize 50% of $\ce{H2O2}$ to reach $\mathrm{pH}=\mathrm{p}K_  \mathrm{a,\ce{H2O2}}$
The hydroxide forms from $\ce{H2O2}$  the $\mathrm{pH}$ buffer solution of a weak acid and it's salt.
$$\begin{align}
\ce{Ca(OH)2  &<=>> CaOH+  + OH-}\\
\ce{CaOH+ &<=>> Ca^2+  + OH- }\\
\ce{H2O2 &<<=> H+ + HO2-}\\
\ce{H+ + OH- &<=>> H2O}\\
\end{align}$$
$\ce{Ca(OH)2}$: $\mathrm{p}K_\mathrm{b1} =1.37$, $\mathrm{p}K_\mathrm{b2} =2.43$
( Wikipedia )
Additionally, $\ce{HO2-}$ is partially eliminated by precipitation, therefore ratio $ \frac{[\ce{HO2-}]}{[\ce{H2O2}]}$ is kept low and so does $\mathrm{pH}$.
$$\ce{CaOH+ + HO2- + 7 H2O <=>> CaO2 \cdot 8 H2O v}$$
Note also the hydrogen peroxide is weakly acidic even without addition of sulphuric acid and that it's $ \mathrm{p}K_  \mathrm{a}$ depends on $\ce{H2O2}$ concentration.
H2O2 pH-and-Ionization-Constant
The solubility constant of calcium peroxide octahydrate in relation to temperature; its influence on radiolysis in cement-based materials

Does the pKa in the pH equation refer to H2O2 or Ca(OH)2? Thank you.

H2O2. for Ca(OH)2, there would calcium compounds instead.

Okay. Thank you very much for all of your help.


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Is bonds in CO is all 3 coordinate bonds or 2 covalent and 1 coordinate bonds?
Below image is as available in India's Class 11 NCERT Chemistry book vol 1 Page 104. I am confused as it states all 3 bonds are kind of coordinate bonds. Should it not start with 2 covalent bonds and 3rd as coordinate bond from oxygen to satisfy carbon? There are several sites confirming my expectation. This book is in a place to force students to study and repeat in exams to get marks. So wondering does it get this right? or teaching wrong?

You cannot distinguish these kinds of bonds. The total bonding between the partner atoms can be seen on a spectra from covalent to ionic, but interpreting that is not without bias. In any case, the final structure in your screenshot is wrong, it is missing the formal charges.

This^ is just some drawing dot formulas for beginners and doesn't imply anything like that.


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Why does period 3 of the periodic table contain 8 elements instead of 18?
Period 1 of the periodic table contains 2 elements ($1s^1$ and $1s^2$).
Period 2 contains 8 elements ($2s^1$, $2s^2$, $2p^1$, $2p^3$, ..., $2p^6$).
By the same argument, period 3 might contain 18 elements ($3s^1$, $3s^2$, $3p^1$, $3p^3$, ..., $3p^6$, $3d^1$, $3d^2$, $3d^{10}$). Why does period 3 of the periodic table contain 8 elements instead of 18?
I think that it could be explained by means of atomic stability, but I need help to understand what happens.
Thank you in advance.

The elements Sc through Zn have 3d electrons, but they also have a pair of 4s electrons, and Sc has one more proton than Ca, so it follows Ca in the periodic table. Therefore, Sc through Zn fit between Ca and Ga. The periodic table is ordered by atomic number. As for why the 4s orbitals get filled before the 3d orbitals, there is likely already an answer posted here, so I suggest searching for it.

The energetics is nicely discussed here: https://chemistry.stackexchange.com/a/8426/79678

You could say that the period number tells you about the largest value of principal quantum number in which a electron is present.
Electrons are filled according to $n+l$ rule. It states as $n+l$ increases, the energy of the orbital increases. If $n+l$ is the same, then the bigger $n$ has larger energy.
So $\mathrm{4s}$ would be filled before $\mathrm{3d}$. Therefore elements containing $\mathrm{3d}$ electrons would have their $\mathrm{4s}$ orbital filled, and therefore, come in 4th period. There 3rd period contains elements having $\mathrm{3s}$ and $\mathrm{3p}$ electron as their last electron which are equal to 8.


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Is there any mathematical or logical proof that carbon with 4 different groups would be chiral?
In chemistry; chirality is generally defined in 2 ways.

Lord Kelvin's definition: "I call any geometrical figure, or group of points, chiral, and say it has chirality, if its image in a plane mirror, ideally realized, cannot be brought to coincide with itself" (source: 1. chirality.org , 2. Biochemistry book by Lubert Stryer.)

"If a tetrahedral (sp3) carbon atom contains a different chemical group on each of its 4 hands, the central carbon atom is called chiral. Though there are some exceptions where we do not found any chiral carbon still some compounds can be chiral. Chiral molecules cannot be superposed with their 3-dimensional mirror image" (common classroom-definition for chirality).

But it seems the first definition is the basic definition; and the second definition seems to be a corollary or application of the first definition.
Now I want to know; is it possible to 'prove' or 'derive' the second definition (better to say second statement or second description) from the first (Lord Kelvin's) definition mathematically or geometrically or logically? and if possible then how to prove the second definition from the first definition?
This question is related but different from this question

Build two enantiomeric models out of matchsticks and chewing gum, and try to superimpose them. If that isn't enough of a proof to you, then what is?

The two definitions are not trying to define the same thing, to begin with. (1) refers to the chirality of a molecule, and (2) refers to chiral carbons, and is not directly related to the chirality of the entire molecule (that's why there is a disclaimer saying that there are some exceptions). So you need to be clear about what you're asking: you cannot prove the definition of A using the definition of B.

@IvanNeretin i think that is more of trial-and-testing (than a proof) that if there are 4 different groups the carbon would be chiral.

@orthocresol please consider chiral carbon.

If you want mathematic proof, take a seat and ingest two semesters of linear algebra. The keyword is point groups. Unfortunately most chemistry faculties dont care. ;)

@Karl lovely. it answers my question

Also in biology university level classes the math was taught to me in a repulsive and non-visual way.

I dont have the age and score-credential to get a scope to "take a seat" to learn linear algebra in any academia. Maybe online will help.

If you find a good online course that teaches group theory, please post the link. I would love to get a refresher.

@AlwaysConfused — Geometrically speaking, the two rules are equivalent. It is demonstrable that if a point (i.e. the (a)chiral center) two identical groups, there always is a type of symetry (point, plane or rotation symetry). As a consequence, any chiral tetrahedral point MUST have 4 different groups.

Exhaustive trial and testing is a totally valid mathematical method to prove sth. The possibilities are very limited, after all.

Related : How can I prove that the angles of the tetrahedral structure is 109.5∘ with calculus. I could do it with geometry.

@SteffX No, the two rules re notequivalent. There are far more types of chirality than those generated by tetrahedral carbons.

@matt_black I am totally agree. The carbon tetrahedron's is only a small subtype of broader definition by Lord Kelvin

Yes, with a couple of simplifying conditions, you can easily show that a tetrahedral carbon with 4 different chemical groups (let's call them 1, 2, 3, 4) is chiral. Let's call the atoms directly bound to the carbon L1, L2, L3 and L4.
So the first condition I will impose is that none of the groups have another tetrahedral carbon with 4 different chemical groups attached.
Chemistry
In a tetrahedral carbon, the groups can't swap places. If that happens, we will call it a chemical reaction, and we are not allowing any chemical reactions.
Geometry
Tetrahdral should be taken loosely: For the plane defined by L1, L2 and the carbon atom, L3 is on one side of the plane and L4 on the other. The exact angles don't matter.
Showing that the carbon is chiral
If we put a mirror in the plane defined by L1, L2 and the carbon atom, the mirror operation swaps L3 and L4 (L3' on the side of L4, and L4' on the side of L3 - not sure we need a formal proof for that). I cannot superimpose that mirror image with the original (L1, L2 and the central carbon are already in the same position as the mirror image. I would have to rotate the molecule to get L3 and L4 superposed with L3' and L4', but then L1 would move away from L1', and L2 away from L2' - not sure we need a formal proof for that). Thus, the carbon atom changes configuration when taking the mirror image.
3D illustration
see https://proteopedia.org/wiki/index.php/R/S_nomenclature (you have to turn the molecules a bit to have L1 and L2 point in the same direction as L1' and L2')
Is this trivial?
No, not at all. For trigonal bipyramids there is sometimes pseudorotation, and for ammonia derivatives the trigonal pyramid can "umbrella flip", so these are typically not chiral. If there are four ligands on a central atom and the geometry is square planar, there is no chirality. So a couple of things have to come together to get a chiral center.

Another way to show this is to first define improper torsion angles, and then show that they switch signs upon reflection in a mirror.

I don't think there is any mathematical way to prove it.It's only experimental.
See Chirality is defined for the compounds whose mirror image doesn't superimpose on each other.

-source : LG Wade Orgainc chemistry.
if you want more explanation then you can take help of LG Wade Or Solomons Organic Chemistry(both are available free on net)

Yes empirical observations demonstrate it, but the proof is nothing to do with that: it is pure geometry. A rigid tetrahedron with 4 differently labelled vertices cannot be the same as its mirror image.


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Friedel-Crafts Alkylation of Pyrrole
I thought that I already know how the Electrophilic Aromatic Substitution (EArs) for Aromatic rings work but then this reaction was shown to me and I was really sure that my professors didn't bothered teaching or even mentioning to us this reaction:

In my defense, at my University, our Org Chem professors always use benzene or benzene derivative rings for demonstration of EArS so I never knew how an acylation of a five-membered aromatic ring works.
Can someone please explain to me how this reaction works?

Dimethyl acetamide is not a substrate for F-C acylation, are you certain this is a F-C process?

@Waylander thank you for noticing, I have edited my question.

Pretty sure pyrrole isn't amenable towards FC alkylation. It's notoriously prone to polymerisation in the presence of Lewis or Bronsted acid.

If you're interested in why it goes to C-2 rather than C-3, then this is relevant: Regioselectivity in electrophilic substitution of pyrrole

To my knowledge, Friedel-Crafts alkylation of pyrrole with aluminum chloride ($\ce{AlCl3}$) is unattainable. However, recent literature suggests F-C alkylation of pyrrole with other catalysts is possible. For example, an asymmetric Friedel–Crafts alkylation of pyrrole with nitroalkenes catalyzed by a copper complex of a bisphenol A-derived Schiff base is reported (Ref.1):

Abstract: The asymmetric Friedel–Crafts (FC) alkylation of pyrrole with nitroalkenes was mediated by $\ce{CuBr2}$ and a novel bisphenol A-derived chiral catalyst at room temperature. The catalyst was found to be applicable for the asymmetric FC alkylation of pyrrole with a wide range of nitroalkenes, affording optically active alkylated pyrroles with enantioselectivities up to 94%. Furthermore, enantiomerically pure 3-nitro-2-arylpropanamides were prepared by the oxidative cleavage of the pyrrole rings in the FC products with $\ce{NaIO4/RuCl3}$ to demonstrate the synthetic application of the products.

For other example, see Ref.2 & 3. For a review, see Ref.4.
References:

Halide Sedef Özdemir, Engin Şahin, Murat Çakıcı, Hamdullah Kılıç, “Asymmetric Friedel–Crafts alkylation of pyrrole with nitroalkenes catalyzed by a copper complex of a bisphenol A-derived Schiff base,” Tetrahedron 2015, 71(19), 2882-2890 (https://doi.org/10.1016/j.tet.2015.03.059).
Yuan-Zhao Hua, Xing-Wang Han, Xiao-Chao Yang, Xixi Song, Min-Can Wang, Jun-Biao Chang, “Enantioselective Friedel–Crafts Alkylation of Pyrrole with Chalcones Catalyzed by a Dinuclear Zinc Catalyst,” J. Org. Chem. 2014, 79(23), 11690-11699 (https://doi.org/10.1021/jo5023712).
Hugh J. Anderson, C. W. Huang, “Pyrrole chemistry. X. Friedel–Crafts alkylation of some pyrrole and furan derivatives,” Canadian Journal of Chemistry 1970, 48(10), 1550-1553 (https://doi.org/10.1139/v70-252).
Hai-Hua Lu, Fen Tan, Wen-Jing Xiao, “Enantioselective Organocatalytic Friedel-Crafts Alkylations,” Current Organic Chemistry 2011, 15(24), 4022-4045 (https://doi.org/10.2174/138527211798109196).


text:
How a plane of symmetry makes compounds lose their optical activity
I understand that if a compound has a plane of symmetry it loses its optical activity, it is very easy to understand this when we have four different substituents on a single carbon atom but when we have larger more complicated structures for example rings with many substituents which are themselves stereogenic how can we be so sure it wont have any enantiomers if we only find one fortunate plane which produces a perfect mirror image?

And why you think there's any difference?

That's really simple. Imagine two different vessels which are precise mirror images of each other; so are all the molecules inside. Imagine that you shine polarized light on them. Being the mirror images of each other, they have to rotate light in opposite directions. But as the compound itself is symmetric, they are in fact identical, so they have to rotate light in the same direction. How can these conditions be reconciled? Just like that: don't rotate light at all.
So it goes.

I have never seen it put that way thanks and on another note i really loved your answer on the question why absolutely zero is unattainable it was hilarious and informatic


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Simple every-day-item hydrometer for distinguishing regular soda and diet sodas
When ordering a diet soda (no sugar soda) in a cafe or restaurant I would like to be sure I am  getting a diet soda and not a regular sugar-infested soda. Because a regular soda, is through all the sugar, far more dense than a diet soda, my idea is to use a simple hydrometer. Ideally this hydrometer would be a simple everyday item that sinks in diet soda but floats in regular soda. I already tested this this idea with some simple items such as coins or small toys. However all tested items either sink in both cases or float in both cases. Anybody has a good idea for a suitable item that sinks in diet soda but floats in regular soda ?
UPDATE although it didn't really answer my question, the experiment suggested by matt_black was too simple not to try out. However what happened when I performed the experiment was that both the can of regular coke and zero coke sank. I discovered there are a lot of video's on youtube showing a can of diet coke floating and only the can of regular coke sinking. So I am quite surprised by my own findings.

How about a capsule containing the sugary stuff - the enclosing material should ideally have a density in the range of the two solutions' densities, and the capsule would have to be free of air. It's an interesting idea but sounds like a mess. Aren't your taste buds good enough?

Saccharometer. Also, if all you want is to quantitatively distinguish between two glasses of coke, something primitive like a plastic straw with a small blob of chewing gum on one end would also do. Just make sure it freely floats in your sugar juice. Alternatively, you can borrow a set of areometers from a friend who does some home brewing:)

Won't the dissolved CO2 affect the density?

Unopened cans of Diet Coke float in water. Sugary cans sink. You don't need anything more complex.

@matt_black, a simple test indeed and I will definitely try it out. However I can't use it to test a served drink in a glass.

Better than a hydrometer, use an Abbe refractometer to measure the index of refraction, which varies with sugar concentration. It requires just a drop of liquid, and that small sample is unlikely to have bubbles, reducing density. You can make a refractometer to measure the sugar concentration, or you can buy one for ~$US20.
That said, diet sodas have been linked to a number of health issues, such as diabetes and diabetic retinopathy. You might be better off with unsweetened products such as seltzer, club soda, tonic water or two cents plain.
On the other hand, if you want to explore brewing or candy making, that refractometer could be useful...
À votre santé!

Those supposed health risks from diet drinks are poor quality nonsense, especially so since they usually don't compare risks to the consumption of very sugary drinks. You really should not promote them until someone does a s study with decent scientific respectability and doesn't exaggerate the implications.

@matt_black, seltzer and other bubbly drinks mentioned above are just carbonated water, without any "health benefits" or artificial sweeteners. They could cause gas pains, if consumed in excess, though ;-)

The refractometer is an elegant solution, indeed.
Other option is to use a simple, home made, small container with water tight cap, with adjusted mass of e.g. fine sand or salt content to have container average density between both colas.
You would adjust it to sink in diet cola and float in regular cola.
Note that I am not a cola drinker,but I guess there is a reduced sugar cola as well. So perhaps reduced sugar cola should be used instead of regular cola. Or you can have 2 such devices to distinguish all 3 kinds of cola.


text:
Magnetic mercury
Well, since I'm not sure, quite simply: Is there a possible way (apart from running a current through in the presence of a perpendicular $\mathbf{B}$ field) to make a small mixture of mercury magnetic so that it would respond to an applied magnetic field, like from a permanent magnet?

Now I know why this rang a bell: https://physics.stackexchange.com/q/490107 .

Mercury, being a liquid metal - is magnetic. But, since it is a liquid, it cannot hold a static magnetization. So, you can certainly perturb it - but it will relax. You must use an alternating magnetic field, if set up correctly, you will have a permanent displacement.

Don't use a mixture, just use pure Hg. Chill to 4.2 K, conveniently reached with liquid He, and all magnetic fields are excluded. Levitate on a magnet, sans yogi.
Mercury in a static magnetic field, with direct current, forms a simple homopolar electric motor, creating a vortex. A safer way (sans neurotoxic Hg) to demonstrate a homopolar motor requires just some copper wire, a magnet and a cell. Or make a linear motor. A nice explanation of the demonstration is also available. [Sorry, this discursion might be better in Electrical Engineering StackExchang].
But if you insist on using Hg at standard temperature and pressure, just use a varying magnetic field and induced current - no direct electrical connection required.
N.B. Mercury is volatile, and its vapor is extremely toxic and hard to remove completely. A building in Massachusetts, USA, was recently condemned and a number of animals died due to Hg contamination. Use gallium, instead, currently available for ~US$15 for 50 grams, and you can also demonstrate gallium attacking the crystal structure of aluminum, or make a Ga mirror. Gallium is also a superconductor, though it needs to be colder than Hg. To liquefy, just warm to ~30 °C in your hand or in warm water.

well pardon my language , I meant using pure Hg not a mixture with other metals or substances. In my apparatus I already will have a convection type mhd pump where current will pass through the mercury in the presence of an applied perpendicular B field, I was just thinking that maybe that will not be enough to keep it steady given that the channel into which the mercury will be itself will be rotating. Like a reverse mhd pump where the liquid metal needs to stay steady in a single position with respect to the pump walls moving.

PS. what was in the video link the other person is referring to?

The last link is region-restricted.

@PeterTaylor Unsurprisingly, many companies outside Europe take the attitude of "Eff GDPR; we'll just block access from the EU instead of complying with it."

@DrMoishe Pippik Could you please repost the YouTube link you originally had? As per the Physics.SE post by the OP (see the link in my comment to the OP), direct electrical contact with the mercury was desired, though the question here is about using a permanent magnet, no direct electrical Hg contact, etc. Thanks!

Well since i'm using a conduction method/type pump I could use DC current through mercury with a permanent magnet providing the perpendicular B field or AC current with an AC magnet.

Would that make much of a difference? I thought the mercury only responded differently if the B field was only applied from outside the mercury like in an induction type pump

Not sure who is being addressed, but I have no quantitative understanding of how mhd (magnetohydrodynamics?) works. This is really a physics problem. In ant event, good luck and stay safe with liquid mercury experiments. The galinstan is far safer, but it wets glass and silica, so wall coating with tin oxide (if I remember right) is necessary. Gallium itself also sticks to silica and glass, so the tin oxide coating may also be necessary. But it is fairly harmless.


text:
10 g of NaCl is dissolved in 250 g of water. The mass fraction and molality of the solution is

10 g of NaCl is dissolved in 250 g of water. The mass fraction and molality of the solution is

For calculating mass fraction, simply divided $10/250$ to get $0.04$. My book says it’s wrong. Is mine the right answer?
Another one,
To calculate molality, no. moles of $\ce{NaCl}$ is $10/58.5$
So
$$\begin{align}
m&=\frac{10}{58.5\times250}\\[3pt]
&=0.684\end{align}$$
Again, my book says it’s wrong, and I am starting to trust is even less, but I still need to get it checked. Am I right, or is the book the winner here?

No, your first answer isn't correct as 250 g isn't the total mass of the solution, and the second one is wrong because you for some reason dropped all units whereas the author of the textbook probably didn't.

Generally, it is preferable to write down the complete quantity equation before plugging in the numbers. If necessary, perform all mathematic operations on quantities symbolically. Do not plug in the numbers until you have only one equation for the desired result.
Always write the values with the correct units and carry the units through the calculation. Do not omit the units while performing intermediate steps and do not just reintroduce units at the end of the calculation.
When a number is given without any further information, it is generally interpreted so that the last digit is rounded with a rounding range equal to 1 in the last digit. Make sure that your result is given with the correct number of significant digits, accordingly. However, it may be necessary to retain additional digits for intermediate results in order to avoid round-off errors in subsequent calculations.

Mass fraction is defined as
$$w_\mathrm B=\frac{m_\mathrm B}m$$
where
$w_\mathrm B$ is the mass fraction of substance B,
$m_\mathrm B$ is the mass of substance B, and
$m$ is the total mass of the mixture.
And you know that, in this case, the total mass of the mixture is given as
$$m=m_\mathrm A+m_\mathrm B$$
where
$m$ is the total mass of the mixture,
$m_\mathrm A$ is the mass of the solvent A, and
$m_\mathrm B$ is the mass of substance B.
Thus,
$$\begin{align}
w_\mathrm B&=\frac{m_\mathrm B}m\\[3pt]
&=\frac{m_\mathrm B}{m_\mathrm A+m_\mathrm B}\\[3pt]
&=\frac{10\ \mathrm g}{250\ \mathrm g+10\ \mathrm g}\\[3pt]
&=0.038
\end{align}$$

Molality is defined as
$$b_\mathrm B=\frac{n_\mathrm B}{m_\mathrm A}$$
where
$b_\mathrm B$ is the molality of solute B,
$n_\mathrm B$ is the amount of substance of solute B, and
$m_\mathrm A$ is the mass of the solvent A.
You also know that molar mass is defined as
$$M=\frac mn$$
where
$M$ is molar mass,
$m$ is mass, and
$n$ is amount of substance.
And you know that, for $\ce{NaCl}$, $M=58.44277\ \mathrm{g\ mol^{-1}}$.
Thus,
$$\begin{align}
b_\mathrm B&=\frac{n_\mathrm B}{m_\mathrm A}\\[3pt]
&=\frac{m_\mathrm B}{M_\mathrm B\cdot m_\mathrm A}\\[3pt]
&=\frac{10\ \mathrm g}{58.44277\ \mathrm{g\ mol^{-1}}\times 250\ \mathrm g}\\[3pt]
&=0.00068\ \mathrm{mol\ g^{-1}}\\[3pt]
&=0.68\ \mathrm{mol\ kg^{-1}}
\end{align}$$

That's probably a weird question, but why does the multiplication sign in the last equation for $b_\ce{B}$ undergo a change from a centered dot $(\cdot)$ to a times sign $(\times)?$

@andselisk According to ISO: "If the point is used as the decimal sign, the cross and not the half-high dot should be used as the multiplication sign between numbers expressed with digits."

Reading this the next day, I am wondering how I even made that mistake. I was really tired and wasn’t thinking straight, and perhaps out of frustration I asked if over here.

@FadedGiant Although it's off-topic: How do you distinguish multiplication from the cross product in this case? For example $1.2 \times\vec{v} \times \vec{B}$, where the first $\times$ stands for a multiplication and the second one for the cross product of the two vectors $\vec{v}$ and $\vec{B}$.


text:
How can the free amino acid B be enantioslectively synthesitzed?
I would like to know how this reaction actually works. I think its an ring metathesis, espececially with the Grubbs catalysator II. But I dont know how to this reaction examole really works especially with respect to enantioselectivity.
May anyone help?

Which bit do you not understand. The alkylations or the metathesis?

@Waylander seems to me that it's the diastereoselectivity of the alkylation(s).

As far as I can tell, the stereoselectivity is completely unimportant as you're doing a double allylation at the same position.

I'm guessing this is a homework type question. Yet, I have some concern of this being a exam question, probably one already taken. So I give a step by step answer depicted in following diagram (courtesy of https://web.chemdoodle.com/demos/sketcher/):

I don't actually agree: in step (b) the reaction proceeds via an enolate, which destroys the preexisting stereochemistry, and the alkylation will again occur anti to the t-butyl group. So it is the allyl group and not the homoallyl group which ends up on the top face.

@ orthocresol: I agree with you 100%. However, OP didn't give final structure so I gave him what possible logical answer for him/her. I might have to use ChemDraw to do proper explanation, which I don't have an access during the weekend. :-)


text:
How to calculate the standard Gibbs free energy of formation of bromine vapor at room temperature?

The normal boiling point of liquid bromine is $\pu{58.2 ^\circ C}$. At $\pu{9.3 ^\circ C}$ the equilibrium vapor pressure of liquid bromine is $\pu{100 torr}$.
  From this data, calculate the standard state Gibbs energy of formation of bromine vapor at room temperature, $\Delta G ^\circ_\mathrm{f}$, $\pu{298 K}$.

I tried using Clausius Clapeyron equation and get $\Delta H$ but I can't manage to solve this problem.
I just used Clapeyron equation to get around $\pu{32 kJ/mol}$ but I can't really manage to get Gibbs free energy at standard state.

If you calculate the equilibrium vapor pressure at 298K (25 C), you can calculate the change in Gibbs free energy of the vapor in going from this pressure to the hypothetical state of 1 bar and 298K using $dG=VdP=RTd\ln P$.  Neglecting the Poynting correction, the free energy of the liquid at 298 and at the equilibrium vapor pressure is 0.  And this is also the free energy of the vapor at 298 and at the equilibrium vapor pressure.

Standard Gibbs energy at normal boiling point
At the normal boiling point, all species are at standard state (pure liquid, gas at normal pressure). The system is at equilibrium, so the Gibbs energy of the process is zero. Because all species are at standard state, the standard Gibbs energy is zero as well.
Standard Gibbs energy at room temperature
The OP used the Clausius Clapeyron equation to get the enthalpy of the process from two Gibbs energies at different temperatures. Now, we are asking for the Gibbs energy at a third temperature, which can be determined by using Clausius Clapeyron in the other direction.


text:
Can I use Ascorbic acid powder as a preservative to extend the shelf life in pickled brine recipe?
Can I use Ascorbic acid powder as a preservative to extend the shelf life  in pickled brine recipe made with vinegar?

Ascorbic acid may serve as antioxidant preservative, but not as antimicrobial preservative.
Unless it is present in comparable amount as added acetic acid and  lactic acid formed by fermentation, but it would keep it redundant.


text:
Why does boiled water have a different taste as compared to cool water?
I have noticed it frequently and decided to check with some of my peers and friends about this and they too have felt such a thing.
The taste comparison occurs after the boiled water is cooled down to room temperature.
My question is: does water even have a taste?
And whether heating it may change the taste.

Well, taste of everything is temperature dependent for 2 reasons:: volatility and  sensory response dependency on temperature.

Does that warm-water taste you're after remain once the heated water is cool again?

You taste senses feel your standby mouth taste, teeth last brushed five hours ago, some remains of a recent lunch, a coffee twenty minutes ago, and that one teeth that should see a dentist. ;) If you now sip some fresh water, the mouth chemistry suddenly changes, and you notice a difference.

Actually feeling of hot and cold in mouth are parts of the taste - that's how capsaicin and menthol work.

@cbeleites yes. The taste seems to remain after being cooled too. It's easy to identify water that has been boiled and cooled by taste

Better suited for Biology SE ?

Slightly related: it is well-known to veteran drink mixers that freezing recently boiled water results in clearer-looking ice cubes; certainly the impurities lost to boiling make all the difference whether in taste or ice cube clarity. Here is an experiment for your consideration: will this difference in taste remain if you had used distilled or deionized water?

The changed taste that stays after cooling the heated water
Hot (boiling) water has a much lower capacity for dissolving gases. When boiling water, $\ce{CO2}$ is removed and through the equilibria behind ($\ce{CO2 + H2O <=> H2CO3}$ and $\ce{H2CO3 <=> H+ + HCO3-}$) so is $\ce{HCO3-}$.
Basically, this removes  $\ce{CO2}$, overall carbonic acid species and also acid equivalents: the resulting water will be more basic (less acidic).
$\ce{CO2}$, $\ce{H2CO3}$ and $\ce{HCO3-}$ are not only present in carbonated (sparkling, fizzy) water but also in tap water. For example, this analysis sheet of the Munich municipal water supply cites 300 mg/l $\ce{HCO3-}$, 15 mg/l $\ce{CO2}$ and 0.5 mg/l $\ce{CO3^2-}$ and a slightly basic pH.
It is known that mammals have $\ce{CO2}$ receptors in their acid taste cells, and this is thought to contribute to the water taste. (I don't have the full text, but Zocchi et al: The cellular mechanism for water detection in the mammalian taste system, Nature Neuroscience volume 20, pages 927–933 (2017) looks promising.)
BTW, two further candidates for that bland/flat/stale taste are
deionized water and reverse osmosis drinking water.
Warm vs. cold water taste
There will be further changes with temperature, including acid base equilibria and pH, partial pressure of volatile molecules and so on between cold and warm water. And this also applies to our sensing in the taste cells. Put this together with taste/smell sensory input being subject to further processing in our brain to produce the taste/smell impression, and I think it is very hard to compare taste/smell of cool and hot things.

in the question I meant the taste of boiled water, that is left to cool down. The taste is compared between the unboiled water with the boiled, then cooled one


text:
Which DFT functional is best for predicting macroscopic properties of molecular liquids?
Which DFT functional (KS-DFT, to be more specific) performs best at predicting macroscopic properties (density, surface tension, boiling point, relative permittivity etc.) of molecular liquids (molecular as opposed to ionic liquids or molten metals)? I'm interested in both water and commonly used organic solvents.
Or, alternatively, if there is no definitive answer: Is there at least a set of functionals which are known to perform poorly in that respect, functionals that I should not use?

Generally, molecular dynamics is used to predict macroscopic properties. Are you thinking about quantum dynamics or do you want to predict those properties from individual quantum calculations?

DFA cannot provide macroscopic properties. Without corrections it doesn't even get the interactions between two molecules right (dispersion, hydrogen-bonding, ...). Maybe COSMO-RS is an option for you, it is based on DFA. Otherwise you'll need to go the molecular dynamics route as @Raphaël suggested.

Thanks for the comments. To be a bit more specific: I'm looking for something along the lines of this paper, but more comprehensive.
https://www.researchgate.net/profile/Mark_Tuckerman/publication/230593872_Ab_initio_molecular_dynamics_study_of_water_at_constant_pressure_using_converged_basis_sets_and_empirical_dispersion_corrections/links/02e7e5307e09ce901b000000/Ab-initio-molecular-dynamics-study-of-water-at-constant-pressure-using-converged-basis-sets-and-empirical-dispersion-corrections.pdf


text:
How ethylbenzene and 1,2‐xylene are chain isomers?
How the following two structures are chain isomers?

According to what I have studied in nomenclature when a compound has both cyclic and straight chain, we consider that the chain as parent chain according to the following order:

functional group;
multiple bonds;
maximum number of carbons.

Following the above order, I came to the conclusion that both structures have benzene as their parent chain. Since both have benzene, then how they come to be chain isomers?
Please explain to me in simple terms because I am just a beginner.

Forget benzene. Constitutional (structural) isomers have the same molecular formula with different atom connectivity. Remove ch3 and an ortho H from ethylbenzene and switch them. Now you have ortho-xylene.

This reminds me of a recent question https://chemistry.stackexchange.com/questions/119523/how-many-ether-metamers-are-possible-for-c4h10o?noredirect=1&lq=1 Same problem, "chain isomers" have very little in common but their identic sum formula.

The substituents (chains of carbon other than parent) should also be same
For example 2-methylpentane and 3-methylpentane are positional isomer-


text:
Haworth's research of glucose
Firstly haworth used CH3I/Ag2O with glucose which replaced 5 H by CH3 forming a acetal which locked the anomer and stopped the equilibirium of ring chain tautomerism. Under dilute acidic conditions acetal group hydrolyze and ethers do not.
So my text (Organic chemistry by L.G. Wade 8th edition) says through these reactions haworth suggested that C5  contains free hydroxyl group and hence proved the pyranose(6 membered cyclic ring) form of glucose. But I fail to understand how did he find that the hydroxyl group is present on the 5th carbon in the chain after hydrolyzation by dilute acid.

You might want to look here: http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/carbohydrates44.pdf. About half way through "Methylation of Pyranoses"

It lists all the reactions I have listed, ok get the reaction but I want to know how did he tell the OH group on 5th after completing(it's not that he knew the product before)

I cannot guarantee that the following explanation for the proof of the ring size of the principal cyclic form of glucose is due to Haworth. The link I provided in the comment above explains one method. In "Methylation of Pyranoses" methods are illustrated for the permethylation of the hemiacetal form of glucose and its subsequent hydrolysis to the tetramethoxy hemiacetal. Although the 6-membered ring is shown, a priori the ring size is unknown. This is the issue to be determined. continued below
In "Ring Size of Pyranosides" mild HNO3 oxidation of the tetramethoxy hemiacetal affords a keto carboxylic acid (upper left). Continued oxidation of either enol of the keto group leads to oxidative degradation to an  optically-active dimethoxysuccinic acid (red path) and a meso trimethoxyglutaric acid (green path). These two dicarboxylic acids are only derived from a 6-membered ring of glucose.

Source: http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/carbohydrates44.pdf


text:
Increase of peak intensity in case of hydrogen bonding
Typical motif in case of hydrogen bonds is charge transfer from electron rich acceptor Y to (usually) anti-bonding X-H σ* orbital. If so, the X-H bonding distance increases, bond weakens, stretching frequency decreases. In IR spectrum, it is visible as red shift of X-H peak and that particular peak intensifies.
Usually the peak intensity has to do with the amount of bonds it represents in the studied material.
Why does peak intensify in this case since it represents the same X-H bond, and the number of bonds is unchanged?
I would've guessed it would, on contrary, decrease, since proton exchange between X and Y may occur and sometimes X-H...Y would become X...H-Y.
What am I missunderstanding here?

Rather than intensifies it broadens, isn't? However, they are no more the same bond, ie the oscillator strength is different

@Alchimista iupac recommendation for hydrogen bond says "increases", as multiple other sources. I actually never read a source that says broadens. "The length of the X–H bond usually increases on hydrogen bond formation leading to a red shiftin the infrared X–H stretching frequency and an increase in the infrared absorption cross-section for the X–H stretching vibration." Could you make "they are no more the same bond, ie the oscillator strength is different" a bit more clear?

The point is that isn't anymore the same bond as you put it using iupac. Cross section increases and intensity does but then you shouldn't take the peak height - intensity being the integral under the curve. A broad bell like band can be overall more intense that a taller sharp one. The broadening is due to the fact that you may have a collection of slightly different hydrogen bonds. Concentration of a specific bond and its own oscillator strength both count on determining the spectrum appearance, ie H-bond is not the best for applying a linear Abs vs conc relationship


text:
Why can't calorimetry be used to measure fast reactions?
As my previous post was a bit unclear and unspecific, I am reformulating it. First of all this is a question that I had while reading some concepts my book introduces, so this isn't a homework question in anyway.
The context is: I have started reading the thermochemistry chapter of my book, it starts explaining the concept of heat and how a calorimeter works. After that, it shows the story behind Hess' Law and says that one of the reasons it was created is because a calorimeter can't be used to measure the enthalpy variation of too slow reactions, explosive reactions and too fast reactions. As a result, there was a need to create an indirect way of calculating the enthalpy variation, and this problem could be solved with Hess Law.
Now, I know why we can't use the calorimeter for too slow reactions and explosive reactions, but I would like to understand what is the problem with too fast reactions. Too fast reactions releases/absorbs heat on a small time period, but this shouldn't be a problem, because the calorimeter will get at equilibrium as other reactions.
As suggested, I am adding the paragraph of the textbook:

As we saw, the heat involved in a transformation can be determined using a calorimeter. However, for many reasons, the calorimeter can't be used to determine the heat for every chemical reaction. There are situations where the reaction can be too slow or fast, can be explosive, it can even occur simultaneously with others undesirable reaction, what makes it impossible to measure the enthalpy variation for a specific reaction. How can we determine the heat involved in these type of transformations?

( I am brazilizan and the book is called Química: Cotidiano e Transformações)
So, what is the problem with too fast reactions?

Feel free to introduce a reference to the textbook and maybe even quote the excerpt. It still sounds rather strange and you haven't added more information to the question, so the answer is liable to be same as for your previous question.

Possible duplicate of Question About Calorimeter and Fast Reactions

Followed your recommendations.

I think your question is very reasonable and hopefully someone here can provide the insight you need. However, could you now please delete the other question you posted. Ideally you would edit a question put on hold so as to remove the hold.

Too fast reactions will start (and probably end) before you close the lid, which is bad for the precision.

@IvanNeretin Exactly! That is why I said this in my comment to the OP’s other same question: “So, provided a fast reaction does not start unless and until you trigger it in the calorimeter, I do not see a problem. The calorimeter simply makes a heat measurement. Your intuition on this seems correct: the water temperature will change accordingly.”

Suppose the “fast” reaction was that of vinegar and baking soda. It would be quite comical and futile to mix them in the calorimeter, get the top properly screwed on, etc. But, change the procedure to add a starting time delay: put the measured quantity of vinegar in the calorimeter, drop in the baking soda in gelatin capsules, properly assemble the calorimeter, etc., and then the reaction will start when enough gelatin dissolves. So, reaction speed is not necessarily a problem and the book’s advice was not definitive.

So my book is wrong? I have seen other books that don't say anything about fast reactions, they just point out for too slow and explosive reactions.

Many reactions are so rapid they they occur far faster than the reactants can be mixed together. In this case it is usual to use a laser which produces pulse of  short duration to start the reaction by exciting an initially inert  substrate. This  excited molecule  then starts the reaction. This is called the pump-probe method. Fast reactions, but ones that are slower than the mixing time, can be studied by the 'stopped flow' method.

So, a calorimeter itself is not enough to calculate the enthalpy variation of too fast reactions, so there Hess Law enters.

No, look again at what I said about the baking soda in gelatin capsules. You simply need a way to either delay the start of the fast reaction, until everything is safely assembled or you need to trigger the start of the reaction in the calorimeter. Usually, this is done electrically. But a modified calorimeter, with a thick sapphire window, would allow laser triggering, for example. So it is the combination of the calorimeter and the protocol that determines feasibility. Short answer: your book is not definitive. Never assume that textbooks are always correct or complete!


text:
Are there chemicals that absorb and release upon heating oxygen more efficiently than potassium oxide?
$\ce{K2O}$ absorbs oxygen forming $\ce{K2O2},$ which decomposes to $\ce{K2O}$ and $\ce{O2}$ at 500 °C. This means that $\ce{K2O}$ can absorb up to 17% its mass of oxygen. Are there other substances that exceed this ratio and release oxygen below 1000 °C?

If you are looking for something that can be made using elemental  oxygen as an oxidizer and reversibly release the excess oxygen below 1000 °C, sodium peroxide beats potassium peroxide by using lighter metal atoms.  It can be prepared by reacting sodium with oxygen in two stages (forming sodium oxide in the intermediate stage) at 130–200 °C and releases the excess oxygen at 657 °C (Wikipedia).
Lithium, magnesium and of course hydrogen form peroxides with even higher excess oxygen percentages, but these are not made via reversible reactions with elemental oxygen.


text:
Calculating thermodynamic quantities for hydrogenation of iron nitride

HW#6.3 If the reaction
$$\ce{Fe2N(s) + 3/2 H2(g) <=> 2 Fe(s) + NH3(g)}$$
comes to an equilibrium at a total pressure of $\pu{1 bar},$ analysis of the gas shows that at $\pu{700 K}$ and $\pu{800 K},$ $p_\ce{NH3}/p_\ce{H2}$ are $2.165$ and $1.083,$ respectively, if only $\ce{H2(g)}$ was initially in the gas phase and $\ce{Fe2N(s)}$ was in excess:
a. Calcualte $K_p$ at $\pu{700 K}$ and $\pu{800 K}.$
b. Calculate $Δ_\mathrm{r}S^\circ$ at $\pu{700 K}$ and $\pu{800 K}$ and $Δ_\mathrm{r}H^\circ$ assuming that it is independent of temperature.
c. Calculate $Δ_\mathrm{r}G^\circ$ for this reaction at $\pu{298 K}.$

At first I ignored the excess data, then I assumed that the pressure exerted by the solids is null and adding partial pressures I got an answer (sum of pressure of gases equals 1).
The problem is that part a in the book gives answer $K_p = 1.777$ at $\pu{700 K}$ which I think is wrong. Please help.

I get 3.85 at 700K

I got that too but the book says 1.777 guess I can't trust the back of the book now ,using Castellan Physical Chemistry


text:
Potential energy of electron density at the nucleus
Consider the hydrogen atom for simplicity.
The electronic density at the nucleus is not null.
The attractive potential between a small volume of electronic density $\mathrm dV$ (at position $\vec r$) and the nucleus (at position $\vec R$) is:
$$\frac{-Z\rho(\vec r)}{|\vec R - \vec r|}\mathrm dV$$
How is the energy not going to (negative) infinity near the nucleus then? How is this accounted for to give the traditional $\pu{-0.5 Ha}$ energy value of the hydrogen atom?

It is accounted for by solving the Schroedinger equation. In a very small region the attraction is quite high. In other regions not so much. And it all volume averages out.

I'd say it is accounted for by integration. Yes, an integral of an unbounded function can pretty well have a meaningful, finite value, and those familiar with calculus can even find it.

You are taking into account that the nucleus, even a hydrogen atom, has a finite size?

I think it's useful to expand slightly on the comments because I think the language here should be as clear as possible.
You're concerned about the convergence of the integral of a function when some piece of the function is unbounded, i.e., for $f(x)$ on domain $[a, b]$, $\exists c \in [a,b]$ such that $f(c) = \infty$. Frequently, integrals of this type do not converge.
Your function certainly looks like that but the resolution of this problem is actually hidden in the $dV$. Typically, we treat the integration of the wave function as a separable problem, i.e, we integrate the radial and spherical components separately. When you look at the integral over the distance from the nucleus, you get a factor $4\pi r^{2}$ because you are integrating shells and $dV = 4\pi r^{2} dr$. So even though $\frac{1}{r}$ diverges at zero, $\frac{r^{2}}{r}$, while not defined at zero, does converge.
$$\lim_{r\rightarrow 0}\frac{r^{2}}{r} = 0$$
From this perspective, no piece of the integral is infinite, and the integral converges.

Actually, some integrals of unbounded functions converge all right.

@IvanNeretin Argh. You're absolutely right. I will clean up the language here.

Good explanation, the volume shell element dV goes to zero at the origin. But I don't get why you should write that $r^2/r$ is not defined at the origin? It doesn't diverge, it is zero (because the volume of the shell is zero).

$\frac{x^{2}}{x}$ is effectly just $x$ except at $x = 0$ where it is not defined, since it is $\frac{0}{0}$.

Hmmm, while I see what you are doing, I wonder if you aren't allowed to use L'Hopitals rule.

To be fair, that's not the function you're actually integrating...


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Electrolytic cell - Potentials

I'm an electronic engineer student and I'm trying to understand the basics of electrolytic cells and practical values' meaning. Based on my figure and electrode potencial values (Vce, Vre and Vwe), can you help me how to describe the following concepts:

Standard cell potencial
Equilibrium potential
Overpotential
Open circuit potential
Reference voltage (electrochemical perspective)

EDITED:
Actually I'm asking about direct relations between this definitions and circuit voltages if there are any
For example, is OCP = Vre when it's not applied external potential?
Thank you for your time!

I didn't get your point, I know how the potentiostat works. I'm asking about electrochemical concepts and relation between them and potentiostat voltages

Also, it's unclear what the question is. "Can you help me how to describe the following concepts" is a vague request suggesting certain amount of tutoring, which is a poor fit for a Q&A site. @EdV made a good point that preliminary literature/internet research might give you a better overview of the topics you've listed. Refer to the Help Center, especially How do I ask a good question?

Ok, maybe I didn't explain it right. Actually I'm asking about direct relations between this definitions and circuit voltages if there are any. For example, I think OCP is Vre when its not applied external potencials, is it right?

Ok, I need to review my question I get it. Btw I'm not asking about electronics. Thank you anyway


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Why does indolizine undergo EAS on C3 and not C5 of the pyrrole ring?
In Clayden, Greeves, & Warren: Organic Chemistry Second Edition, it is said that indolizine (pyrrocoline) reacts in electrophilic aromatic substitution at the C3 of the pyrrole ring, as shown.
I understand that it reacts at the pyrrole ring because the pyrrole ring is more nucleophilic than the pyridine-like ring, but I don't understand why it doesn't react at the C5 position of the pyrrole ring (like pyrrole does); the sigma complex at the C5 position has more extensive delocalization, whereas the sigma complex at C3 is less stable cross-conjugation.

Meh, it's not like things like that can be reliably explained like this, in general.

@Mithoron Then what would you say are the concepts which explain this?


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Why is lead's first ionisation enthalpy greater than that of tin?
Using Slater's rule, I found that both lead and tin's $Z_\mathrm{eff}$ for 6p and 5p electrons equal. However, the ionisation energy is different. Is comparing ionisation enthalpy on the basis of $Z_\mathrm{eff}$ not correct?
Also, I found Clementi-Raimondi Effective Nuclear Charge which probably explains the trend, but why is this different to Slater's rule?

Perhaps Slater's rule is more of a suggestion.

This is an example of a relativistic effect, so Clementi-Raimondi effective nuclear charges (non-relativistic in nature) likely also cannot describe the behaviour adequately. You may find this question tangentially useful.