text:
Bubbling air through hydrochloric acid to generate HCl gas
I am trying to build a system to generate dry HCl gas.  I was wondering if you bubble air through diluted HCl will it produce a gas with mostly water vapor or HCL vapor, can we vary the proportions of each by varying the temp of the inlet air and the acid, like hot air bubbling through cold acid or vice versa giving more acid or water vapor.
I am aware of tried and tested techniques to generate dry HCl gas, but this question came to my mind exploring techniques if one does not have access to conc. sulfuric acid. Surely something similar has been studied and researched before where the rate of evaporation has been measured.

Basically you are trying to find something that can dry the air without reacting with the acid. This will rule out the common ones such as $\ce{CaCl2, KOH}$.

https://patents.google.com/patent/US5958356A/en - Check out this patent...!

And bubbling air through $\ce{HCl}$ would not be advisable, rather heating it to form the vapours and passing the vapours through $\ce{MgCl2}$ supported on an activated carbon or silica gel substrate activated by heating to a temperature between 150° C. (302° F.) and 300° C. (572° F.) under vacuum will remove water at partial pressures of below 0.5 torr. (according to the patent above)

The problem you face is that HCl is very soluble in water and blowing air through dilute HCl will not be very effective. At the very least you will have to spend a huge effort to dry the resulting gas stream which is going to be very annoying and time consuming versus almost any other method of generating dry HCl gas.

@HarikrishnanM I'm not sure that patent helps at all since it would require access to both relatively sophisticated apparatus and chemicals. If you can't get even sulfuric acid, those are not going to be readily available.

Whats wrong with calcium chloride, it generates relatively dry HCl for gassing salts.

Purchase anhydrous HCl gas bottles.

HCl forms a high boiling azeotrope at 110 Celsius and 20% HCl. This azeotrope has been well studied as a gravitational standard for acidimetric titrations. Low concentrations will concentrate on distillation until the azeotrope is reached. More concentrated will lose HCl. Distilling neither will give dry HCl gas but concentrated 37%HCl should come close. The recommended drying agent is concentrated H2SO4 so a better method would be to react NaCl with sulfuric acid directly. Molecular sieves might work; check with the manufacturers. Other reactions such as PCl3 or PCl5 with water are also complicated involving hard to obtain and nasty chemicals These reactions could possibly be engineered to give dry HCl. Tetra chlorides of tin, titanium or silicon or AlCl3 might also work. Look them up for details.

For the first question:

will it produce a gas with mostly water vapor or HCL vapor

The gas would be mostly water vapour because dilute HCl is present as ions so it's concentration as whole (unbroken) HCl would be low.
For second question:

can we vary the proportions of each by varying the temp

If you bubbled gas that is 200°C then entire water along with HCl would get evaporated. This is the only way to achieve large amount of HCl by bubbling but mole fraction of HCl in our gas would be still very low.


text:
Activated charcoal for water filtration: Granules vs Pellets
NOTE:
After further investigation into the producer I found out that granules are made of 100% coconut shell whereas pellets of 100% hard coal.
This renders this question meaningless.
Apologies for the inconvenience and thank you for the knowledgeable and enriching answers.
I'm looking for activated charcoal for my water distiller and I found a producer offering both granules and pellets made of steam activated carbon from hard coal and coconut shells.
Considering the similar grain size / diameter, I'm surprised to find out that the two have opposite properties.
The following is a table provided by the retailer:
                                  Granules   Pellets
Wastewater - water purification   ✓          ✓
Inorganic chemicals                          ✓
aquarium water                    ✓          ✓
chlorine / ozone                  ✓
Discoloration / Toxins            ✓
Electroplating                               ✓
Organic Chemicals                            ✓
Drinking water treatment          ✓          ✓

Summary, as I understand it:
Granules  let go  inorganic and organic chemicals  stop    chlorine and ozone
Pellets   stop    inorganic and organic chemicals  let go  chlorine and ozone

This is a trade off that I'm obviously not willing to settle with, I just would like to drink CLEAN water.
I'm considering mixing granules and pellets in order to avail of both properties, if that could possibly work.
I would like to know why granules and pellets made of the same material have such differences.
I'd also be glad to have an advice on which one may be best for water distiller.
Here are the specifications of both products:
                              Granules          Pellets
Grain size / diameter         0,6 - 2,36mm      Ø1,5mm
Bulk density                  500 +/- 30kg/m³   530 +/- 30kg/m³
Water content at packing      <5%               <5%
Iodine value                  >1050 mg/g        >900 mg/g
CCI4 - Adsorption             n.V.              >60%
Specific surface area (BET)   ca. 1100m²/g      ca. 1050m²/g
Ash content                   <5%               n.V.

Unless you ask the vendor or producer what is the difference and why, all would be just speculation. // The data the vendor provided are rather irrelevant for the difference. The statement claiming capturing organic and inorganic chemicals is a misleading statement of century. And even if it were true, what are chlorine and ozone if not inorganic chemicals? // What purity is guaranted for saltys added to distilled water? Are they certified for food industry usage? // Using distillation for producing drinkable water is rather foolish, unless there is no other way.

Pellets are obtained by pyrolizing vegetal material which is of course made of living cells. When heating this material to high temperature, water is eliminated, the inner parts of the cells are destroyed and mainly evaporated; the cells explode, then the membranes are pyrolized and transformed into impure charcoal, which usually stays unmodified in place. The obtained material has lost water and a large proportion of its mass. It is rather light.
Under the microscope, it looks like of a huge amount of broken and empty more or less spheric bubbles. Organic substances may enter the rest of these spheres where they are adsorbed, but they have more difficulty to get out. That is why this pyrolized substance can stop big  chemicals present in air around. But this material cannot fix small molecules like chlorine or ozone. To adsorb these small molecules, granules having much smaller cages are needed, which are produced by much smaller living cells, after pyrolysis. And probably another surface adsorption mechanism is needed to explain this effect.
You can use a mixture of both charcoals to purify drinkable water.

Both granules and pellets are made of the same activated charcoal.
Why only granules are advertised as chlorine and ozone adsorbent?
Why only pellets are advertised as inorganic and organic chemicals adsorbent?

@Claudio. Granules and pellets are not obtained form the same sort of living cells. They are not made of the same activated charcoal.

from the producer:"We obtain our activated carbon from hard coal and coconut shells. The production process is technically complex. We use the following method: Steam activation".
from Wikipedia:"Extruded activated carbon (EAC) combines powdered activated carbon with a binder, which are fused together and extruded into a cylindrical shaped activated carbon block"
@Maurice are you saying that in order to use a binder a different base material must be used? Is the producer false advertising?

@Claudio. The producer must be right, because my knowledge of this process is from many years ago. Apparently there has been changes in the process used during recent times. I did not know that activated carbon and a binder can be fused together and extruded into cylindrical carbon black.

source: Stack Exchange

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

sort: votes


text:
Small-Angle X-Ray Scattering of Bulk Polymers
Does anyone have any recommendations on how to get useful information from SAXS of polymers when the system is not dilute?
This is the type of data we're working with:

My understanding is that since there's no plateau at the beginning we can't use the correlation function to get parameters like Lc.
I'd really appreciate if anyone has any thoughts on how to alternatively calculate this!

What are the axes on your figure $A(q)$ vs $q$ ?

The more relavant details your provide, the more relevant answers arrive. and vice versa.

So sorry, the axes are momentum transfer (q) and Intensity. So it's a plot of I(q). This sample is a semicrystalline polymer film of ~20% degree of crystallinity by WAXS

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


text:
A first order gaseous reaction - getting the rate law
There is a specific example cited in our NCERT Class 12 textbook, to find out the rate law equation for a general gaseous reaction, $$\ce{A->B + C}$$
It describes the entire procedure as shown:

Let us consider a typical first order gas phase reaction
$\ce{A(g) -> B(g) + C(g)}$.
Let $\pu{p_i}$ be the initial pressure of A and $\pu{p_t}$, the total pressure at
time 'ť'. Integrated rate equation for such a reaction can be derived as $$\mathrm{Total \space pressure}\space\pu{p_t = p_A + p_B + p_C}\space \mathrm{(pressure\space units)}$$
If $\pu{x atm}$ be the decrease in pressure of A at time $t$ and one mole each of B and C is being formed, the increase in pressure of B and C will also be $\pu{x atm}$ each.

\begin{array}{|c|c|c|c|}
    \hline
    & \ce{A} & \ce{B} & \ce{C} \\
    \hline
    \text{At } t = 0 & \pu{p_i} & 0 & 0 \\
    \hline
    \text{At time } t & \pu{p_i - x} & x & x \\
    \hline
\end{array}
where, $\pu{p_i}$ is the initial pressure at $t= 0$.
$$\pu{p_t=(p_i-x)+x+x=p_i+x}$$
$$\pu{x=p_t-p_i}$$
where,
$$\pu{p_A=p_i-x=p_i-(p_t-p_i)= 2p_i-p_t}$$
Finally,
$$\pu{k=\frac{2.303}{t}log(\frac{p_i}{2p_i-p_t})}$$

Now I am trying to derive something for a general equation. The final formula would look like:
$$\pu{\frac{2.303}{t}log(\frac{(z-1)p_i}{zp_i-p_t})}$$ where z is number of gaseous products from one mole of gaseous reactant. Fine! But one thing: I want it with respect to difference in gas moles.
Considering the reaction $\ce{aA->bB + cC}$, I did try it, and got something like this.

Does anyone know any alternative method?

$\Delta n_g$ is the difference in gas moles, to clarify.

There seems to be an error in the LCM taking. You finally get $$\pu{k=\frac{2.303}{t}log(\frac{zp_i}{(z+1)p_i-p_t})}$$ but again that does not match the expression given.

The above comments are for V-1 of the question.

Note that using photos/screenshots of text instead of typing text itself is highly discouraged.  The image text content cannot be indexed nor searched for, nor can be reused in answers.  Specifically handwritten scripts can be difficult to decipher.  Consider  copy/pasting or rewriting of essential parts.
// Optional: Formatting guides for texts and formulas/equations/expressions.

@Poutnik the Mathjax of the same is quite cumbersome and the user is new. Also I feel that the crux of the question is covered in the typing itself.

@HktalksCasually your derivation should be done with the equation $$\ce{A-> bB + cC}$$ you can redo and post it. I'll edit your $\LaTeX$.

My formula given above is the final equation, IMO.


text:
why is the suffix of alcohols -ol but not -al?
According to the IUPAC nomenclature, we put the suffix “-ol” to alcohols while we put the suffix “-al” to aldehydes.
This will be related to some etymology issues. I presume “-ol” and “-al” are derived from the ending of “alcohol” and the beginning of “aldehyde” respectively.
I get very confused about this. Both “alcohols” and “aldehydes”, I mean the English words, start with “al”. Then, why alcohols cannot have the “al” suffix?

If alkanols R-OH became alkanals, what would happen with alkanals R-CHO ? // By other words,generally, if having two different groups suggests having the same name, only one can keep it. The other must have assigned some other name.

Chemical nomenclature did not emerge systematically. Some names have been borrowed from other languages centuries ago including German, French and Arabic. Alcohol is derived from Arabic, from the the time of Middle Ages. The word al-kali (alkali) are remnants of those times. There is no way current nomenclature can be changed.
From the Elsevier Dictionary of Chemetymology- The Whies and Whences of Chemical Nomenclature and Terminology, we can see:

Alcohol derived from al-kuhul (Arabic: powder) referring to the
alchemistic procedure of reducing samples, through several
purification steps, to a powder, the supposedly pure essence of the
substance in question; thus alcohol is the essence (or spirit) of
wine; the same word, spelled kohl, is used for powdered antimony as a
cosmetic preparation

Aldehyde is rather recent

Aldehyde coined by contraction of alcohol dehydrogenatus (New Latin: aldehyde  literally dehydrogenated alcohol)

Hope that solves the mystery behind -ol and -al. There is a need to distinguish -ol and -al groups and most chemists have accepted this name to represent an alcohol and an aldehyde, at least in the English language.
There will be certain molecules when both functional groups are present and the suffixes will be determined by IUPAC's priority list.

To complement the answer: Wikipedia equally has sections about etymology regarding the names e.g., of alcohols, and aldehydes, the later including a footnote to Liebig's definition in a publication in 1835.


text:
Question on electron density
I was working with ammonia ($\ce{NH3}$) and phosphine($\ce{PH3}$).  Both have lone pair electron.  But $\ce{NH3}$  is stronger base (Lewis base) than $\ce{PH3}$.
I was told by my instructor that it is because of size and polarity of ($\ce{NH3}$). He said  for the polarity of nitrogen, it draws electron density toward itself and for this excess electron density it donates partial electron density via coordinate covalent bond.
But how much electron density  is too much and when a molecule would donate partial electron density?( to be exact) [ would better if mention other cases or general case also]
And also can't phosphine do the same?


text:
Common Ion effect with sulphuric acid
Sulphuric acid dissociates completely (H2SO4 -> HSO4⁻ + H⁺).
Its product HSO4⁻ has a pKa of 1.92 which should then dissociate partially into SO4²⁻ and H⁺.
I was trying to find out the pH of a 0.05M H2SO4 solution and used the Common Ion Effect to shift the reaction towards the reactants, getting a pH of 1.23. The professor's solution does not take into account the hydrogen ions produced during the first dissociation and calculates a pH of 1.16. The CIE only applies to weak acids which according to the definition I have should be anything with a pKa > 0 (approx).
Could you tell me which solution is correct and maybe why I shouldn't be taking into account the H⁺ produced during the first step?

Any solution is correct if it respects 1/ mass and charge balances, 2/ equations of equilibrium 3/ proper usage of justified strong inequalities for  equation simplification.

@Poutnik And am I being too bold if I ask which if not both (or non) of the solutions satisfies your propositions?

Why do not you check it? BTW they are not propositions, but mandatory requirements  // Take the equilibrium equation for K_a2 and try to express [HSO4-] and [SO4^2-] using [H+] variable.

@Poutnik Yes, you are right they aren't propositions. My bad. Yes, they satisfy those requirements the only difference is how I define [H⁺] 1) = [SO4²⁻] + 0.05 (mine) 2) = [SO4²⁻] (professor's). My TA said that the professor was right from a thermo point of view but no clue why

H+ from the 1st dissociation cannot be ignored, forming majority of H+. // $K_\text{a2} = \dfrac{[\ce{H+}][\ce{SO4^2-}]}{[\ce{HSO4-}]} \approx \dfrac{[\ce{H+}]([\ce{H+}] - \pu{0.05 M})}{\pu{0.1 M} - [\ce{H+}] }$, with neglection of water autodissociation.

@Poutnik Yes that's what I did. Still have to convince the professor :')

Let us continue this discussion in chat.


text:
Bancroft's point
In the last problem (Problem 10.43) within the problem section of Chapter 10 in "Introduction to Chemical Engineering Thermodynamics" (J.M. Smith, H.C. Van Ness, M.M. Abbott, McGraw-Hill, 7th edition, 2004), the last exercise states the following:

Vapour/liquid equilibrium azeotropy is impossible for binary systems rigorously descripted by Raoult's law (Pb. 10.5). For real systems (those with $\gamma_i \neq 1$), azeotropy is inevitable at temperatures where the $P_i^\mathrm{sat}$ are equal. Such a temperature is called a Bancroft point. Not all binary systems exhibit such a point. (...)"

In other words, if the vapour-liquid equilibrium curves of two different compounds intersect, they will inevitably present an azeotrope once mixed. This claim seems to be very general and unexpected. I tried it on many compounds, just for curiosity, and surprisingly it holds. I exemplify with water+2-butanol below:

This system has a Bancroft point at $\approx \pu{87.7 °C}$, and indeed shows an azeotrope. A reference is found here. I am unable to find any information about this topic.
Question
Is there any way to prove this very general statement? Is it a mistake? Any hint?
Edit
When the authors claim that "not all binary systems exhibit such a point", they refer to the next idea. Suppose $\ce{A}$ presentes VLE equilibrium in the range $\pu{50-150 °C}$, and compound $\ce{B}$ in $\pu{160-250 °C}$, then both curves can't intersect and thus the Bancroft's point can't exist. However, if the ranges overlap and the VLE curves do intersect, then azeotropy is a fact.


text:
Why do reduced variables for gases use their critical constants?
I understand why it's useful to have reduced variables, especially as it is the basis for the theorem of corresponding states. However, I'm having a hard time understanding why the critical constants work best as the scale for reduced variables.
Unlike standard values for things like $\Delta G$, they clearly cannot be some arbitrary value since a substance is not necessarily a gas at any given state (hence the entire concept of a critical temperature). I'm more curious about why the critical temperature works so well with the law of corresponding states in the first place.

You got any better idea? Critical parameters do characterise p/T properties of substances in quite practical way.

What else to use?

The reason to use the critical properties come from the two conditions that happen for pure fluids:
\begin{align}
  \left(\frac{\partial p}{\partial V}\right)_T = 0 \hspace{1 cm}
  \left(\frac{\partial^2 p}{\partial V^2}\right)_T = 0 \tag{1,2}
\end{align}
i.e., the slope and the curvature of the pressure $p$ as a function of the molar volume $V$ (at constant temperature $T$) are zero. When you submit the vdW equation of state to these conditions, $a$ and $b$ are put in function of those critical properties. The math leads to
\begin{equation}
  a = \frac{9RV_\mathrm{c}T_\mathrm{c}}{8} \hspace{2 cm}
  b = \frac{V_\mathrm{c}}{3} \tag{3,4}
\end{equation}
Books do the proof in the natural or straightforward way. First they define the reduced variables and show that they are consistent with the theorem of the corresponding states.
But, we can do the opposite. We replace Eqs. (3) and (4) in the vdW equation of state and see what transformatitions we should choose in order to arrive at the theorem. We begin
\begin{align}
  p &= \frac{RT}{V - b} - \frac{a}{V^2} \\
  p &= \frac{RT}{V - V_\mathrm{c}/3} -
  \left(\frac{9RV_\mathrm{c}T_\mathrm{c}}{8}\right) \frac{1}{V^2} \\
  p &= \left(\frac{RT}{V_\mathrm{c}}\right)\frac{1}{(V/V_\mathrm{c}) - 1/3} -
  \left(\frac{9RV_\mathrm{c}^2T_\mathrm{c}}{8V_\mathrm{c}}\right)
  \frac{1}{V^2} \\
  p &= \left(\frac{RT}{V_\mathrm{c}}\right)\frac{1}{(V/V_\mathrm{c}) - 1/3} -
  \left(\frac{9RT_\mathrm{c}}{8V_\mathrm{c}}\right)
  \frac{1}{(V/V_\mathrm{c})^2} \\
  p &= \left(\frac{RT_\mathrm{c}}{V_\mathrm{c}}\right) \left[
  \frac{T/T_\mathrm{c}}{(V/V_\mathrm{c}) - 1/3} -
  \left(\frac{9}{8}\right) \frac{1}{(V/V_\mathrm{c})^2} \right] \\
  \frac{pV_\mathrm{c}}{RT_\mathrm{c}} &=
  \frac{T/T_\mathrm{c}}{(V/V_\mathrm{c}) - 1/3} -
  \left(\frac{9}{8}\right) \frac{1}{(V/V_\mathrm{c})^2} \\
  \frac{p_\mathrm{c}}{p_\mathrm{c}}
  \left(\frac{pV_\mathrm{c}}{RT_\mathrm{c}}\right) &=
  \frac{T/T_\mathrm{c}}{(V/V_\mathrm{c}) - 1/3} -
  \left(\frac{9}{8}\right) \frac{1}{(V/V_\mathrm{c})^2}\\
  \frac{p}{p_\mathrm{c}}
  \underbrace{\left(\frac{p_\mathrm{c}V_\mathrm{c}}{RT_\mathrm{c}}\right)}_
  {Z_\mathrm{c}} &=
  \frac{T/T_\mathrm{c}}{(V/V_\mathrm{c}) - 1/3} -
  \left(\frac{9}{8}\right) \frac{1}{(V/V_\mathrm{c})^2} \\
  \left(\frac{p}{p_\mathrm{c}}\right)Z_\mathrm{c} &=
  \frac{T/T_\mathrm{c}}{(V/V_\mathrm{c}) - 1/3} -
  \left(\frac{9}{8}\right) \frac{1}{(V/V_\mathrm{c})^2} \\
  \frac{p}{p_\mathrm{c}} &= \frac{1}{Z_\mathrm{c}} \left[
  \frac{T/T_\mathrm{c}}{(V/V_\mathrm{c}) - 1/3} -
  \left(\frac{9}{8}\right) \frac{1}{(V/V_\mathrm{c})^2} \right] \tag{5}
\end{align}
If you remember, the vdW equation of state predicts a universal critical compressibility factor of $3/8$. Therefore, Eq. (5) yields
\begin{align}
  \frac{p}{p_\mathrm{c}} &= \frac{8}{3} \left[
  \frac{T/T_\mathrm{c}}{(V/V_\mathrm{c}) - 1/3} -
  \left(\frac{9}{8}\right) \frac{1}{(V/V_\mathrm{c})^2} \right] \\
  \frac{p}{p_\mathrm{c}} &=
  \frac{8(T/T_\mathrm{c})}{3(V/V_\mathrm{c}) - 1} -
   \frac{3}{(V/V_\mathrm{c})^2} \tag{6} \\
\end{align}
Eq. (6) suggests that the reduced variables must be in the form $x_\mathrm{r} \equiv x/x_\mathrm{c}$. This is the only way that we can arrive to a universal function, for which all fluids at the same reduced volume and temperature will lead to the same deviation from ideality.
If you want to be creative, such as $T_\mathrm{r} \equiv T / \sqrt{T_\mathrm{c}T_\mathrm{t}}$, where $T_\mathrm{t}$ is the triple temperature, this won't work. $T_\mathrm{t}$ will be in some part of the equation, and we will have to say that "all fluids at the same reduced temperature, reduced volume, and triple temperature will lead to the same deviation from ideality". This is not the statement of the theorem, because it adds one more thing. You are left with no other choice.

I'm more curious about why the critical temperature works so well with the law of corresponding states in the first place.

Unfortunately, this principle works well with very simple fluids. I have used much more complex equations of states and is unreliable for most organic compounds that are moderately-highly polar such as ammonia, all alcohols, aromatic compounds with some groups, amines, etc. When in the last paragraph I sad 'add one more thing', this is exactly what some other people did. You can add another parameter to this theorem, Kenneth Pitzer did it in the 1955 with the accentric factor and created a 'three-parameter theorem of corresponding states'. Still, for polar fluids, it sometimes fails pretty bad.

Do not you have the derivatives (3,4) reciprocal ? there should be dp/dV and d2p/dV^2 and not dV/dp and d2V/dp^2

@Poutnik Yeah, you are right.... How come I didn't notice that? Thank you very much for the observation...

Perhaps we are inclined to consider p as independent and V as dependent variable, so volume derivative per pressure seems more natural.

Yes, that's how it goes. Equations of states are explicit in $V$ and $T$, and when you see the $p-V$ graph, you see the maximum righ there. Actually, we can think of $V$ (independent variable) as a function of $p$ and $T$, but you may need like 5 papers to write those three roots. No one does that.

In context of the van der Waals equation, it has the clear mathematical analysis reasoning.
The function $p=\text{f}(V)_{T=\text{const}}$ must have at $T_\mathrm{c}$ and $p_\mathrm{c}$ an inflection point and zero slope, so the first and second derivative of pressure per volume must be zero. It also means the infinite derivative of volume per pressure.
It allows splitting the unique value of volume for given temperature and pressure to two values ( the 3rd without physical meaning) below the critical temperature.
It then comes naturally as advantageous to express the equation in terms of the reduced parameters wrt the critical ones.


text:
Why would 1-ethenyl-2,6-dimethylbicyclo[2.2.0]hexane give off 8 H1 signals?

I am trying to figure out this problem, and can only find seven signals, two from the double bond, one from the single bond attached to the double, there is symmetry, so one signal on the methyl, one on the carbon in the ring, and attached to the methyl, one on the bottom left carbon, and one on the middle bottom.

Think about the 3D structure of this molecule. Are the protons on the [4] rings all in the same environment?

The molecule contains six protons from $2$ $\ce{CH3}$, the $2$ times $2$ protons form $2$ secondary atoms (in cycles), then three tertiary carbon atoms (in cycles), and three $\ce{H}$ atoms on the double bond. Total $6 + 4 + 3 + 3 = 16$ H atoms.

The structure as drawn does not represent a single unique 3D structure but several structures. The bicyclic ring is likely to be fairly rigid so both the hydrogens and methyls on the ring can be locked in position above or below the ring. Whichever structure is intended, this matters for the NMR shift of the hydrogens. Try building a 3D structure and counting the distinct hydrogens.


text:
Give a molecule, how do we determine the serial number of each atom in it to put in a pdb file?
Example:
For molecule of 1-chloro-4-nitrobenzene, using RDKit's MolToPDBFile() function, I can convert this molecule to the pdb file shown below.
MODEL        1
HETATM    1  C1  UNL     1      -0.155   1.178  -0.139  1.00  0.00           C
HETATM    2  C2  UNL     1       1.202   1.142  -0.326  1.00  0.00           C
HETATM    3  C3  UNL     1       1.836  -0.081  -0.343  1.00  0.00           C
HETATM    4  C4  UNL     1       1.109  -1.240  -0.174  1.00  0.00           C
HETATM    5  C5  UNL     1      -0.255  -1.180   0.011  1.00  0.00           C
HETATM    6  C6  UNL     1      -0.900   0.033   0.031  1.00  0.00           C
HETATM    7  N1  UNL     1      -2.307   0.122   0.221  1.00  0.00           N1+
HETATM    8  O1  UNL     1      -2.865   1.224   0.234  1.00  0.00           O
HETATM    9  O2  UNL     1      -3.070  -1.007   0.392  1.00  0.00           O1-
HETATM   10 CL1  UNL     1       3.569  -0.109  -0.582  1.00  0.00          CL
HETATM   11  H1  UNL     1      -0.657   2.138  -0.125  1.00  0.00           H
HETATM   12  H2  UNL     1       1.780   2.067  -0.459  1.00  0.00           H
HETATM   13  H3  UNL     1       1.588  -2.220  -0.184  1.00  0.00           H
HETATM   14  H4  UNL     1      -0.875  -2.066   0.149  1.00  0.00           H
CONECT    1    2    2    6   11
CONECT    2    3   12
CONECT    3    4    4   10
CONECT    4    5   13
CONECT    5    6    6   14
CONECT    6    7
CONECT    7    8    8    9
END
ENDMDL

However, I do not understand how does it determine which carbon to start with. I thought there is an algorithm or rule behind, but I have not yet found any clue after 1hr of searching on google.
Any idea or suggestion about where I may find answer?

There are IUPAC naming rules. Then again, why bother? A PDB file with incorrect numbering is as good as any other.

I want to analyze a group of molecules(more than 50). Each of them has an atom of interest that seemingly always listed at the same order across all pdb files related. Now i want to run an analysis software using script and this idea can only work if I can confirm that those atoms would always have the same serial numbers across all my molecules, or at least I know how to calculate the serieal numbers they would belong to.

Then they probably adhere to the said rules. Not sure if it is strictly enforced, though.

yeah, that's where I get confused about, I have read a couple of articles discussing about format of pdb file, but none of them mentioned how exactly do they determine the serial number for each atom....

@Yuuya It is not yet evident why the atom of interest must show up on the same line of each file for your analysis. For instance in a Kabsch test to compare conformations of one molecule, the relative position of an atom surrounded by other atoms is more important. By default, the implementation by Charnley et al. addresses the issue of atoms' sequence in a xyz file by the Hungarian algorithm; i.e. to retain atom type, but disregard an atom label.

it is not a conclusion but rather an observation, that's why I would like to bring this up for discussion. what i noticed was that atoms with similar relative structure/position seems to have same serial number as other atoms from other molecules, but i have no clue/reference to assert this is true.. @Buttonwood

@Yuuya Internally, RDKit's bookkeeping of atoms might (to some degree) rely on atom indices, integers starting by 0.  You may visualize them (see for instance RDKit's entry in the cookbook).  One may argue once assigned, the report to pdb first states C, N, O, Cl (increasing atomic number) and eventually H (the report of [explicit] hydrogen atoms in a pdb is not mandatory).

@Buttonwood can you confirm if pdb file would inherit this rule when being converted from a mol object using rdkit?

You know that "p" in pdb is protein? This format isn't designed for some random stuff. And the number is largely arbitrary, it can be even negative.

@Yuuya The sequence of C-N-O-Cl is specific to your file. Try from rdkit import Chem; smi = "c1ccncc1"; mol = Chem.MolFromSmiles(smi); Chem.MolToPDBFile(mol, 'test.pdb') about pyridine, and move N (e.g. describe py as c1ccccn1).  Vary now with explicit hydrogens in from rdkit import Chem; smi = "c1ccccn1"; mol = Chem.MolFromSmiles(smi); mol = Chem.AddHs(mol); Chem.MolToPDBFile(mol, 'test.pdb'); hydrogens now included trail the non-H atoms. But rdkit and obabel report non-H in pdb in blocks, e.g. c1ccccn1.c1cocc1 (a benzene and a furane), trailed by H. The sequence depends on the input.

@Yuuya And in similar tune to the comment by Mithoron: if you describe a small molecule, .mol/.sdf, or .mol2 are more suitable (if you don't need explicit coordinates, consider SMILES/SMARTS) than .pdb.  The later can be a container for small molecules, but pdb structures often contain a large protein + a couple of water molecules + one or a few small molecule ligand(s) + some cations / anions of the buffer used to isolate/crystallize the sample and hence the file format wants to retain these information and perhaps some comments, too.

@Yuuya To name C1, C2, C3 serial number would be a bit unusual, though.  It is more common to describe them as atom label, and for convenience the element symbol typically is used as a part of the string.  Which differs from atom type, really about what element is meant (and in some file types, amended with additional information as if it is an alkyl carbon, an aromatic carbon, etc).

[OP] I thought there is an algorithm or rule behind, but I have not yet found any clue after 1hr of searching on google.

There is no rule. You have to rely on the "CONECT" records and the coordinates to find corresponding atoms.
In the example case, 1-chloro-4-nitrobenzene, the presence of symmetry makes matching pairs of atoms in two molecules ambiguous. The way the carbon atoms are named, C3 is connected to the chloride and C6 is connected to the nitro groups (this is unexpected given the numbering in the name). C1 and C2 can be swapped for C5 and C4 without changing the connectivity. Also, naming of the two oxygen atoms is ambiguous.
If you want to find the best superposition of two 1-chloro-4-nitrobenzene molecules, you have to consider these ambiguities and not rely on atoms with the same name matching.
If you are lucky and the files were generated by the same software, there might be some method in the naming madness. However, it is best not to rely on it but rather pair up atoms based on "CONECT" records and pairwise distances.


text:
Does pH affect the activation energy of decomposition of hydrogen peroxide using iron (III) nitrate solution?
I know that pH can affect the activation energy of the decomposition of hydrogen peroxide when catalase is used, as catalase is an enzyme. However, if I use some nonenzymatic catalysts like iron (III) ions or manganese (II) oxide, is controlling pH important anymore if I want to measure the activation energy of the reaction? Any response will be appreciated.

As the reaction rates would strongly depend on pH, controlling pH is essential if the Ea is determined e.g. from the Arrhenius equation and temperature dependence of the reation rate. As the rate=f(T, pH).   As pH affects availability of respective forms of iron and manganese in solution or behaviour of surface functional groups.

@Poutnik would you mind explaining more about "pH affects availability of respective forms of iron and manganese in solution or behaviour of surface functional groups"? I haven't been introduced to this concept yet (still a secondary school student). Thank you!


text:
How to get 1-ethyl-3-isopropylbenzene fron benzene

I have to synthesize the compound shown above in a multi-step synthesis. My biggest issue is that I have no idea how to obtain one of the alkyl group in the meta-position. Getting the first alkyl group is fairly easy, but I don't know where to go from there.

Using a meta-directing group is a great idea. For instance, I could introduce a nitro group in the meta position. But where do I proceed from here? How can I transform the nitro group into an alkyl group.

FC Acylation to make acetophenone 2)Alkylation 3) Clemmensen or Wolff-Kishner reduction to remove the carbonyl

Hello! Thank you for the answers, everyone! Could you please elaborate on what you mean by 'Alkylation'? Specifically, what are you referring to?
@Harikrishnan M your method with FC after Nitration will sadly not work. The Nitro group deactivates the ring, causing FC reaction to no occur.

What specific conditions would those be? If I were to write this in an exam, which reagents and conditions should I include

Step 1 - Friedel-Crafts acetylation of benzene with acetyl chloride/AlCl3 to give acetophenone as described here
Step 2 - Alkylation of acetophenone with 2-chloropropane/AlCl3 as documented in section 11.3 here. The exact isopropylation reaction is described here
Step 3 - reduction of the carbonyl by Clemmensen details here/Wolff-Kishner details here or sodium borohydride followed by acid catalysed hydrogenation

Hello, thank you for your response! I have one question: Does the ketone introduced in your initial step deactivate the ring, thereby preventing further Friedel-Crafts reactions

Any electron-withdrawing group deactivates the ring, but depending on the conditions the reaction can still be done - I have added another reference to my answer.


text:
Purpose of sulphuric acid in archaic matchboxes
The first attempts at instantaneous matchboxes usually included some sort of oxidizer, usually potassium chlorate and an initiating fuel (first sugar was used then they moved on to $\ce{Sb2S3}$ etc.) However the stick had to be dipped into a sulphuric acid solution and withdrawn for it to catch fire.
My question is:- How come the matchstick tip had to be dipped in $\ce{H2SO4}$ in order for it to catch fire? What was the acid doing? Was the active oxidant $\ce{HClO3}$?

Well, you gotta do something to set a match on fire - contact oxidiser with reducer. If one uses  a solution for that, I doubt you could think of a better one.

I don't think I get you, isn't sulphuric acid also an oxidant?

H2SO4 is not much of an oxidant, certainly not enough to start a fire by itself.

https://chemistry.stackexchange.com/questions/4158/how-does-sulfuric-acid-catalyze-the-reaction-of-potassium-chlorate-and-sucrose?rq=1 It's the dehydrating nature of sulphuric acid i overlooked. I think this answers my question. Do i delete/close this?

Well, if you wanna get an automatic question ban, then go ahead and delete ;>

https://chemistry.meta.stackexchange.com/a/4398/9961

@GauravSaiMaddipati Instead of delete, you can self close the question by linking that question.

Good to know, however it seems my answer-hunting may have led to an incorrect conclusion as the answer below shows.

You are right. $\ce{HClO3}$ is at the origin of the flame, and the dehydrating power of $\ce{H2SO4}$ has nothing to do here.
Anhydrous $\ce{H2SO4}$ reacts with $\ce{KClO3}$ to produce $\ce{HClO3}$ (and $\ce{KHSO4}$). When anhydrous, this $\ce{HClO3}$ is the active oxidant, but it is also an unstable compound, which is quickly and spontaneously decomposed. As soon as it is in contact with a reducing agent, like sugar of $\ce{Sb2S3}$, $\ce{HClO3}$ reacts strongly, producing $\ce{KCl}$ and oxygen, and these oxygen atoms create a flame with the sugar or $\ce{Sb2S3}$. In the $19$th century, the first matches were made with a mixture of molten $\ce{KClO3}$ and sugar (or another flammable substance) which were deposited on bits of wood. To get a flame, the matches had to be wetted with a tiny amount of concentrated sulfuric acid.
Today the acid has been suppressed. Matches contain only $\ce{KClO3}$ on wooden sticks. To start a fire, the matches are rubbed on a thin layer of red phosphorous $\ce{P_\mathrm{n}}$ : the friction heats a bit, enough to produce a flame, due to the reaction of $\ce{KClO3}$ on $\ce{P_\mathrm{n}}$ (and $\ce{P_\mathrm{n}}$ is probably partially transformed by friction into white phosphorous $\ce{P4}$, which is much more flammable).

Didn't we dip it into solution? How would it be anhydrous?

The last paragraph reminds me on Walter White (Breaking Bad) and the scene where he explains the modern design (link to a snippet on youtube).  But how would one carry safely anhydrous $\ce{H2SO4}$ around?  Fuming $\ce{H2SO4}$ is the closest one to your description I came to use so far, and I'm glad it was stored in a solid glass bottle.

These old fire-kits certainly werent carried around, that's for sure.


text:
How is nitrogen the most powerful electron donor in aromatic rings?
In Clayden, page 735, Aromatic Heterocycles 1:Reactions:

The other simple five-membered heterocycles are furan, with an oxygen atom instead of nitrogen, and thiophene, with a sulfur atom. They also undergo electrophilic aromatic substitution very readily, although not so readily as pyrrole. Nitrogen is the most powerful electron donor of the three, oxygen the next, and sulfur the least. Thiophene is very similar to benzene
in reactivity.

This is in the context of electrophilic aromatic substitution.
My question is, why is nitrogen the most powerful electron donor? Sulphur is the least electronegative, so will it not be less reluctant to donate electrons?
Or are there some resonance/aromatic or other effects I'm failing to see in pyrrole, furan, and thiophene?

Sulphur doesn't like bond order higher than one. One could think Se would be even better, but it's likely the opposite.

I remember this on heterocyclic chemistry class. It is obvious that oxygen would donate less than nitrogen due to electronegativity. Sulphur however, are on the 3rd period. This allow it to contained more than 8 electrons. Basically, because you're allowed to draw thiophene with $=S=$ bond in the resonance structure, making Sulphur the least donating of the three.

I once heard an explanation related to larger atoms having weaker resonance because they are geometrically less able to. I wonder if this is correct

Sulfur is not a good donor to carbon pi systems due to its larger size. Sulfur's larger 3p orbital (you can imagine a lone pair there) does not overlap efficiently with the pi system of the ring.

It can't be simplified to comment length, but essentially, most of the times longer bonds mean weaker bonds. This is usually due to more nodes in the eave function and less efficient overlap. Most importantly though, since it came up, there are no d orbitals in binding for sulfur. There's not more than 8 electrons allowed. And I have no clue what geometrically stable could mean in this context.

Nitrogen is able to donate its electrons to the aromatic pi-system more readily than oxygen and sulfur because its p-orbital (which contains the lone pair) is energetically closer to carbon (i.e., only slightly lower in energy than carbon's p-orbitals). Oxygen's p-orbitals are lower in energy still, meaning that overlap is weakened, and sulfur's p-orbitals are much higher in energy and larger in size making overlap much weaker.
You can think about it in terms of which atom's p-orbitals are most similar to those on carbon, since the aromatic system is mostly made of carbon p-orbitals.


text:
How to separate rhodium sulfate from zirconium sulfate?
I dissolved a zirconia ceramic containing rhodium with sulfuric acid, i tought this process will dissolve only zirconia ZrO2 at 100 °C and lower but it turned out that rhodium in form of powder can dissolve also.
I Proceeded in 3 way's the last one turned brown (no water were added) compared to the first and second one's  (red; note here i diluted with water from the begining).

Are you sure rhodium is not present as a fine suspension in the solution ? Have you tried to filtrate it? After Greenwood and Earnshaw, Chemistry of the Elements,p. 1292, Rhodium is not soluble in aqua regia. The only way of dissolving it is to fuse it with $\ce{NaHSO4}$. The behavior of rhodium in presence of acids is qualified as "extreme inertness".

@Maurice https://chemistry-europe.onlinelibrary.wiley.com/doi/10.1002/ejic.201601247

@DavePhD But there's 400 °C not 100.

@Mithoron this https://pubs.acs.org/doi/10.1021/ja01624a009 (cited in the above article) says 0 °C and dilute sulfuric acid to form a yellow sulfate from the trihydroxide and heating that to 100 °C yields a red sulfate.

@DavePhD But from hydroxide, not metal.

@Tintin. The video shows that rhodium is attacked by concentrated sulfuric acid if the reaction is maintained $80$ hours (more than $3$ days) at $435$°C. This is more or less equivalent to dissolve it in fused $\ce{NaHSO4}$, as I stated it $20$ hours ago. .

@ªTintin. According to the reference, rhodium is dissolved in sulfuric acid at 435°C in a sealed ampule. The inside pressure is not given. If sulfuric acid gets decomposed à 320°C at atmospheric pressure, its decomposition is probably prevented at higher pressure.Now metallic rhodium may be separated by filtration.

Please explain how you have obtained the seven figures. Fig. 1 looks an emulsion of a black stuff. How was it obtained ? Fig.2 is a colorless solution with a "mirror" on the bottom of the flask. Fig. 3 is an orange-brown solution. How was it obtained ? Fig .4 shows two plastic bottles with colorless but poorly defined content. Fig. 5 is a colorless precipitate in a colorless solution. How was it obtained ? Fig. 6  is a dark precipitate. Fig. 7 is a colorless precipitate in a pale yellow solution. How were all these solutions and precipitates obtained? Explain !

@Tintin. You aren't clear. You state that Fig. $1$ is "where you did not dilute" the acid. What is this "where" ? Is it the result of the action of sulfuric acid on zirconia ? It should be mentioned. You state that Fig. 7 is the remain slurry of it". So the remain slurry is white !! How come if the liquid is black ? You state that Fig.2 is the same solution. Well. Which same solution ? Is it the black liquid from Fig.1 ? What is this mirror ? All the other figures are equally unclear. I will not describe my trouble for all. It would take too much time.


text:
How to coat stainless steel?
I am trying to coat stainless steel with carbon black. I am using Polyvinylidene fluoride PVDF 2% as a binder. I dip and I heat at 100C.
The coating that comes out is very fragile where you can blow on it and it would break.
What am I doing wrong? or is there a better method?
(Note: I am trying to maintain electrical conductivity between the steel and the carbon)

You might consult the authors of https://www.sciencedirect.com/science/article/abs/pii/S1359836819343483. Also, are you dipping into PVDF powder, or a solution?

solution with DMSO as a solvent.


text:
Disproportionation reactions
Are there any examples of a disproportionation reaction being harnessed to do useful electrical work? It is of course possible in a comproportionation reactions, such as a lead-acid cell.

How would it know which way to work?

Electrochemical reactions from which we can harness electrical work must involve the trasfer of electrons through an electrolyte/electrode interface. Afterwards, they can travel externally and we can use them. In lead-acid batteries, you may "say" that you we comproportionation because lead ($0$) and lead(IV) oxide ($+4$) react and give lead(II) sulphate ($+2$). However, this is the sum of individual reactions that do take place in the cell. The global reaction is just an addition to help us, but strictly speaking, it doesn't take place anywhere in the cell.


text:
Isoelectric Point
How do we derive the formula of isoelectric point?
pI=(pKa1 + pKa2)/2
I read that isoelectric point is defined as the pH at which degree of protonation in amino group is same as degree of deprotonation at carboxylic group, but I still could not derive the relation.

Wikipedia — Isoelectric point literally points you to the derivation: "For derivation of this expression see acid dissociation constant". And your relation (and the "definition") is for the special case only, and does not hold true for the isoelectric point as defined by IUPAC.

The isoelectric point is the pH at which the amino acid is in its zwitterionic form (neutral, typically with a deprotonated carboxylic group and a protonated amino group).
The titration of an amino acid is similar as the titration of a diprotic weak acid, with 2 buffer regions at pH=3 and pH=9 (which correspond to the pKa of the alpha carboxylic and amino groups)

The amino acid is neutral when the carboxylic group is fully deprotonated, and the amino group is still fully protonated. This occurs at the first equivalence. The pH at the first equivalence is given by pH = pI = (pKa1 + pKa2)/2
I'm pretty sure there are more complicated ways and more accurate formulas to find the pH at the first equivalence by considering the 2 equilibria of protonation and deprotonation of the zwitterion.
I hope this helps


text:
Isolating the zinc core from the copper shell of a U.S. penny
I am a grad student who has a friend with a pretty nice personal lab, one thing I have always sought out to do as a chemist is to remove only the copper (and also obviously the copper oxide) from the outside of a penny. I know that a common experiment is dissolving the zinc core as I have done that many times for demonstrations but wanted to turn the tables. I have seem some things about using Potassium Cyanide but I feel like there is potentially a safer way that i’m just not thinking about. Any alternatives y’all can think of?

Good luck with that :/  Zn is a lot more reactive than Cu.

Cheat: electroplate Zn onto the copper cladding -- it will react as Zn (until it dissolves off). ;-)


text:
Stereoselectivity of the homogeneous catalytic hydrogenation of 2-Methylenebicyclo[2.2.1]heptane using wilkinson's catalyst
For the following reaction, 2 products are available: in one, the methyl group is axial, and in the other, it is equatorial.

The addition of hydrogen to the double bond is a syn addition to the less sterically hindered side of the bond, but how can I know which side is more sterically hindered in this case? Is the "bottom" face with 2 methylene groups more hindered than the "top" face with only one methylene group? or are there other factors to consider (such as the stability of an equatorial methyl group compared to an axial one)?
Thanks in advance for the help.

It'd be better if you put the complete reaction in the attachment.


text:
solubility of glutamic acid in water
As we have learned in school, generally speaking polar molecules can dissolve in solutions with other polar molecules. However in the case of glutamic acid, it appears that it is unable to dissolve in water (proven by experiment, which we have conducted in class). What is not clear to me is why it is unable to dissolve within water, as the glutamic acid in a solution with pH 7 should be in fact an anion with a formal charge of -1.
After we have added sodium hydroxide, eventually the glutamic acid dissolved. This process does make sense to me, as the hydroxide deprotonates the glutamic acid which then becomes an anion with a formal charge of -2. But it still does not explain why the anion with an formal charge of -1 did not already dissolve in water, as glutamic acid also contains several partially polar functional groups.
Thanks in advance, and I hope I was able articulate my question in an understandable way :)

It is incorrect to say it is insoluble. The measured solubility of Glutamic Acid in water is 8.57g/L

Ah, yes. Sorry for the misunderstanding. Indeed it dissolves to a certain degree, but $8.57g/L$ really is very less.

You said the pH was 7. Did you adjust it to 7 after adding the glutamic acid or was that the pH of the water before you added the acid? If you did not adjust it to 7 after adding the acid, then it will be much lower and the charge will not be -1.

@Andrew Oh, that might indeed by the reason. As you assumed correctly I did not adjust the pH, it was the pH before adding the acid. Thanks for pointing that out.

Glutamic acid is a weak acid. It does not totally protonate (give H+ or hydronium ion H3O+), to form all glutamate ion with formal charge -1. (Only few of them.) An acid which totally protonates and dissolves in water should be a strong acid such as hydrochloric acid. Low solubility of glutamic acid is also because of it is an organic compound, with not high enough polarity.

Ah, I see. So, basically the molecules that are present in water only consists partially of ions and most of them are just "normal" glutamic acid molecules?
Also I don't fully get why being an organic compound leads to the low solubility?

Organic compounds tend to have a lot of C-C and C-H bonds which create regions of hydrophobicity - the more of them the less likely to be water soluble

Ah, I see. I was just unaware that this is the meaning of the term (I am not an English student). I have taken this to consideration as well however what still does not make sense to me is that all the partially polar groups that should be able to create hydrogen bonds with water are not outweighing the "small" organic compound?


text:
Ideal Chi-squared for X-ray pattern Le Bail fit
I have made a potentially novel solid state phase. I know the general parameters it should have, and have been trying to Le Bail fit XRD data about this phase prior to a synchrotron run in May. I have achieved a chi-square of 11.1 using a Bruker D2 Phaser and FullProf refinement software.
Does this mean I have my compound, but perhaps there are small impurities? Or does it mean that the fit is worthless? I'm afraid I do not know what the "non-ideal but still ok" range of fit is.

First off, what is this mystery matetial?

(If I am correct), it is a monoclinic Wadsley-Roth phase. The lattice parameters and space groups of these materials can be easily predicted as a function of the stoichiometry and oxygen deficiency. When using these predicted parameters, I get a relatively decent looking fit, but it has a chisquare of 11.1. It is very possible that there are small impurities in here, one of the precursors is very expensive to acquire better than 99% purity of, so I want to make sure the phase is correct before buying the more expensive precursor. I just don't know if 11.1 means "fit is wrong" or "impurities".


text:
Standard formation enthalpy or entropy. What is temperature is standard conditions?
I am reading Atkins's Physical Chemistry.
In section Thermochemistry, in subsection 2.7 Standard enthalpy changes
It states that

As implied by the examples, standard enthalpies may be reported for any temperature.
However, the conventional temperature for reporting thermodynamic data is
298.15 K (corresponding to 25.00°C). Unless otherwise mentioned, all thermodynamic
data in this text will refer to this conventional temperature.

I found many examples in the book that give data at 298 K. It is omitting the decimals or is it a different criterion?
If you check wikipedia page of Standard enthalpy of formation, both numbers are used? Are there many criteria?


text:
What happens to pressure when a liquid goes supercritical?
Fill a pressure vessel with water, no airspace. Heat it past 374 Degrees C, so that the water goes supercritical but can't expand. What happens to the pressure at that point? Does it continue to increase linearly as the temperature continues to increase, or would the graph form a hockey stick there, as the rate of increase (pressure) suddenly increased, or does it increase exponentially (curved graph) or????
Put another way, is the pressure at 700 Kelvin twice that at 350 K, or???

The vessel would crack due liquid water thermal dilation long before reaching the critical T.

Huh, one would need a line of constant density drawn on P-V phase diagram to tell, but I don't think you'd notice anything at this temp. Why not? Because there would be no vapor phase and pressure would get over crit. value long before crit. temp.

At such high pressures supercrit. fluids are liquid-like and there is no phase transition at crossing crit. temp. The pressure would increase in fashion that is neither linear nor exponential, or any simple function afaict.

Over my head, guys. I can't copy and paste the chart on page 2, here https://www-pub.iaea.org/MTCD/publications/PDF/P1500_CD_Web/htm/pdf/topic5/5S03_I.%20Pioro.pdf

continued: --d C,  so they never get too close to 374 C.  I know that zirconium combines with the oxygen in steam and supercritical water, releasing hydrogen; don't care; I need to know what happens to the pressure if that plant overheats and its coolant goes supercritical. Again, if your plant is running at 325 C, and it overheats to 650 C, does the pressure double, or what?

Guys, guys, all I want to know is what happens to pressure when a fluid goes supercritical. The rest of this is irrelevant and over my head. Does pressure go through the roof when the fluid in a closed, filled vessel goes supercritical, or does double the temperature produce double the pressure regardless of crossing the supercritical point? Does the pressure inside a nuclear power plant double if the temperature (in Kelvins) doubles, of what?

If your plant is running at 325 C, and it overheats to 650 C, it's gonna blow up.

Typically, isochores in a $p-T$ diagram are straight lines. This is evident in the ideal gas law, but also a prediction from the simple van der Waals' equation of state.
We put this to test using a more sophisticated equation of state, the Peng Robinson's equation of state
\begin{equation}
  p = \frac{RT}{V - b} - \frac{a}{[V + (1 - \sqrt{2})b)][V + (1 + \sqrt{2})b]}
  \tag{1}
\end{equation}
Even though it still has two parameters, the one related to the attraction forces has a dependency with temperature
\begin{equation}
  a = 0.45724\,\frac{\alpha(T_\mathrm{r},\omega)R^2T_c^2}{p_\mathrm{c}}
  \tag{2}
\end{equation}
in which the accentric factor is taken into account via the correlation established by Peng and Robinson
\begin{equation}
  \alpha(T_\mathrm{r},\omega) =
  [1 + (0.37464 + 1.54226\omega - 0.26992\omega^2)(1 - \sqrt{T_r})]^2 \tag{3}
\end{equation}
The results are shown below:

This is a $p - T$ diagram but in reduced coordinates. There are ten isochores which are also displayed in reduced form. The supercritical region that interests you lies in the rectangle for which $p_\mathrm{r} > 1$ and $T_\mathrm{r} > 1 $. When $T_\mathrm{r}$ is rather low, there seems to be some curvature in all the isochores due to that both terms are important in the equation. However, for the supercritical region we have essentially straight lines. This agrees with the explanation given by Oscar Lanzi.

References

The parameters for water used in the equation of state were extracted from the DIPPR Data Compilation of Pure Compound Properties, ASCII FIles, National Institute of Science and Technology, Standard Reference Data, Gaithersburg, MD, 1458 chemicals, extant 1995.

\begin{array}{|c|c|c|c|}
  \hline
  T_\mathrm{c} \; \pu{(K)} & p_\mathrm{c} \; \pu{(bar)} &
  V_\mathrm{c} \; \pu{(dm3/mol)} & \text{Accentric factor} \\ \hline
  647.1 & 220.55 & 0.0559 & 0.345 \\
  \hline
\end{array}

Assuming that the vessel does not blow up, the "gas" pressure ultimately becomes proportional to absolute temperature, but the proportionality constant may be greater than that derived from the ideal gas law.
Illustrating with the Redlich-Kwong equation:
$P=\dfrac{RT}{V-b}-\dfrac{a}{\sqrt{T}V(V+b)},$
where P is pressure, V is molar volume, T is absolute temperature and $a,b$ are deemed constants for a given gas. Under the conditions assumed in the problem $V$ is also constant as the temperature is raised and therefore the first term becomes mathematically dominant. Thus
$P\approx\dfrac{RT}{V-b},$
which exceeds the ideal gas law through the "excluded volume" from the rigid structure in the gas molecules. The excluded volume effect is also present in ordinary gases, but the relative effect would be much larger in the pressure-vessel situation considered here.

Well, that's illustrative, but for details one would could rater use some modified Tait_equation, I think.

Mostly way over my head--it would really help if you would define your variables--but I gather that pressure is proportional to temperature, and that crossing the supercritical point doesn't change that, so that twice the temperature (in Kelvins) means twice the pressure. Interesting that twice the pressure means half again the velocity in firearms, and twice the temperature seems to equal about half again the efficiency in Rankine-cycle power plants. Must be another law of thermodynamics in there ... . I'm thinking of the differences between pressurized water nuclear power plants and molten salt reactors, and if I got this right I got what I needed, so thanx, guyz.


text:
Soluble Ion separation
Is it possible to separate ions dissolved in solution? I know that the charges have to be balanced, so thinking that maybe it is possible to artificially give that charge? Take for example, sodium hydroxide solution, would there be a way to put all the sodium cations in one area separate from the hydroxide anions?

Ever heard about Coulomb's law? Electric forces are really, really strong, about as strong as the strongest thing you can imagine, only stronger.

There is a very well established ion separation technique, it is called ion-chromatography. You can "isolate" a given ion of interest from solution. Keep in mind that even in that technique, the charge balance is always maintained. Wikipedia has a nice detailed article.

Applying Coulomb's law, the force necessary to separate $1$ micromole $\ce{Na+}$ from $1$ micromole $\ce{Cl-}$ at a distance of $1$ meter is about $\pu{10^{12}}$ Newton. It is so huge that it cannot be represented by any reasonable way. And it would be even much greater if the separation distance would be smaller. If the separation distance is $1$ micron ($\ce{10^{-6}}$ meter) instead of $1$ meter, Coulomb's law states that this force would be about $\pu{10^{24}}$ Newton. It is not possible to find a reasonable way for describing such a huge force.

Ions can be separated by a strong electric field of even by mechanical action. Examples are static electricity, lightning, both are recombination of separated ions recombining by a cascade mechanism. Mass spectrometers and particle accelerators separate ions and use the ions for analysis and nuclear reactions.  alpha and beta nuclear decay initially have separated ions that eventually, or at least possibly, become neutral atoms. Electrical neutrality means that somewhere, sometime there will be a +charge for every -charge. In solution things happen faster but ions migrate anions to anode and cations to cathode and each type congregates.


text:
Why O-H bond and not C-O bond breaks in Acid Catalyzed Hydration of Alkene?
Recently, while studying Hydrocarbons I came across this reaction (Acid catalyzed hydration of an alkene)
What I don't get here is that the breaking of O-H bond (in Step 3) and not the C-O bond, even though the C-O bond is weaker than O-H bond.

If the C-O breaks then you have gone back to the product of step 1. Note that these steps are all reversible, but you only get the product by the O-H bond breaking, and experiments show that you do get the product

To reiterate: the C-O bond does break all the time, and form again, and break again, and form again.

@Waylander The 3rd step is mentioned to be fast so i think it's almost irreversible ?

What I don't get is the reason why O-H bond breaks when C-O bond can break more readily ig

Fast does not equal irreversible. The product alcohol can and does protonate under the reaction conditions

You are equating the RO-H bond strength with the strength of the bond of ROH2+ they are not the same

Also note that "bond strength" values typically refer to homolytic bond cleavage, whereas all of the steps shown involve heterolytic cleavage, so your premise is incorrect.

As I can see in the 3rd step you’re referring to its a reversible process thus major and minor products are also formed so I think the breakage of the C-O bond would fall under one of the minor product, and the breakage O-H bond will fall under major product.
I also think the oxygen's octet conditions would not be allowing it to do so.

I think the bond energy of         C-O(358kJ/mol) is less than  O-H(463kJ/mol)

I can see your concern about the Bond Dissociation Energy of the O-H bond being higher than that of the C-O bond.
A key point to consider here would be that the oxygen in question carries a positive formal charge, making it more electronegative and exhibiting a stronger -I effect.
This effect is experienced by both the adjacent carbon and the hydrogens; However, the carbon in question is a tertiary carbon and the three methyl groups stabilise the 'electron deficiency' of the carbon attached to oxygen by compensating with +I effects.
Hence, the hydrogen attached is electron deficient, and carries a partial positive; making it susceptible to attack by the lone pair on oxygen.
Note that either Hydrogens could be attacked, and we get the desired product.
Hope this helps :)


text:
Why does the P-V graph for a cyclic process make a loop?
Assuming the same path is not followed when the gas performs expansion and compression irreversibly in a cyclic process, why is a loop formed on the P-V graph?
Isn't a rectangular hyperbola made when we plot a graph for an inverse ratio?
If considered individually (that is possible, right?), wouldn't it look something like this:

is this 'loop' correct too?
Am I correct in my approach or am I missing out on something important here?

@Poutnik https://mathworld.wolfram.com/RectangularHyperbola.html

See the answer update. I have also deleted my obsolete comments.

@Mel  I know the last graph is drawn for illustrative purposes, but the "orange" curve doesn't satisfy the requirements of a mathematical function. Therefore, I don't know if that is possible to achieve...

@MetalStorm Is that the reason the apparently 'correct' one (the one which we've been taught) is inverted to form the loop the other way?

@Mel If by the correct one you refer to the one we can see in the second image, then yes. The lower curve that returns to the starting point at least mathematically makes sense.

A hyperbolic process state path happens for isothermal processes.
The other curve describes very peculiar process, when the gas is during expansion being heated up by a specific way and than being cooled down to the initial temperature.
If endpoints of both paths have the same temperature and pressure, it leads inevitably to a closed cyclic path.

A pure rectangular hyperbole means an isothermal process (for ideal gases).
If there is a deviation from a hyperbole toward the origin, the gas is temporarily cooled down on the process path. Or the amount of gas has temporarily decreased, like in temporarily open systems in the Otto cycle for gasoline combustion engines.
If there is a deviation from a hyperbole away of the origin, the gas is temporarily warmed up on the process path. Or the amount of gas has temporarily increased.
The individual expansion part shape suggests the gas temporarily obtain absolute temperature about twice as high, compared to the endpoint values.

I added another image, could you confirm if it's correct?

I assume it is. I have not been used to the "rectangular hyperbola" term, but it was after while clear what it means.

Ah, I see. It does make sense now. Can't upvote this (I don't have the reputation), but thanks! As for the figure though, another user pointed out it's not mathematically correct, which probably is the case, or else it would have been there on my textbook.

@Mel You can still click the "tick" sign on the left to accept the answer. This should be below the arrow that is pointing down.


text:
Is methylenecyclopropane a cycloalkene?
Are all compounds with a ring and double bonds considered cycloalkenes? Can the double bond be outside the ring in a cycloalkene (example: methylenecyclopropane)

No, methylidenecyclopropane is not a cycloalkene.
According to the Glossary of class names of organic compounds and reactivity intermediates based on structure (IUPAC Recommendations 1995), the double bond of a cycloalkene has to be part of the ring:

Unsaturated monocyclic hydrocarbons having one endocyclic double or one triple bond are called cycloalkenes and cycloalkynes, respectively.

until someone proves me wrong...
I think this is a matter of perspective. In terms of nomenclature, I would consider  methylenecyclopropane a cycloalkene I defer to IUPAC, as pointed out by Loong's answer. In terms of reactivity, one usually thinks of double bonds outside of the ring as more reactive than when part of the ring (but there can be ring strain effects, too). But I cannot think of a reaction that would distinguish between methylenecyclopropane and methylcyclopropene.
For aromaticity, we have something: Addition of bromine to the double bonds of 5-Methylene-1,3-cyclohexadiene (i.e. toluene with one H moved from the methyl group to a ring C) would proceed, but for toluene, radical substitution on the methyl group or electrophilic aromatic substitution are the only sensible possibilities.
So my argument is this: it somewhat does not matter if methylenecyclopropane is a cycloalkene or not because the reactivity is not fundamentally different.
... and you can reach a different conclusion.


text:
Calculate the volume (in mL) of 0.200 M CaCl₂
Trying to calculate the volume (in mL) of 0.200 M CaCl₂ needed to produce 2.00 g of CaCO₃ (s).  There is an excess of Na₂CO₃.  Molar Mass of Carbon Carbonate = 100.09 g/mol.  I am getting the answer as Volume (in mL)≈99.9mL  Is this right?
To calculate: I got a 1:1 CaCl₂ to CaCO₃ in the stoichiometry of the reaction. Moles of CaCO₃
≈0.01998mol so it was the same for CaCl₂.
Moles of CaCO₃≈0.01998mol
volume in L = Moles/molarity
V = 0.0999L×1000mL/L
Volume (in mL)≈99.9mL  (is this correct?)

The calculation is correct. The only trouble is the precision of the final result. First: the concentration of the $\ce{CaCl2}$ solution is $0.200$ M, so that its uncertainty is ±$0.001$ M; this corresponds to a relative uncertainty ±$0.5$%. Second the mass of $\ce{CaCO3}$ is $2.00$ g, known with a precision ± $0.01$ g, and this corresponds to a relative uncertainty ±$0.5$ %. As a consequence, the relative uncertainty of the final volume is $0.5$ % + $0.5$ % = $1$%. The absolute uncertainty on the final volume is ± $0.01 · 99.9$ mL = $0.999$ mL. As an uncertainty has only one significant number, the uncertainty on the volume is ± $1$ mL, and not $0.999$ mL. The final volume is therefore :  V = $100$ mL ± $1$ mL


text:
Are there any yellow ferrous (ideally stainless) alloys? Perhaps after heating?
I recently learned about purple gold, which is roughly 80%-20% Au - Al by weight. (It's beautiful.) I also know that Titanium will turn bright purple if heated sufficiently (probably an oxide layer). The 300 series stainless will also turn into various colors after welding. I am curious if there are any stainless steels (ideally) or other ferrous alloys that are YELLOW, (and if by a heat-induced coating, a uniform yellow).

Perhaps titanium nitride coated steel? Commonly used for drill bits: https://homesteady.com/12504204/what-are-titanium-drills-used-for, https://en.wikipedia.org/wiki/Titanium_nitride


text:
Why does balancing the reduction of dichromate(VI) to chromium(III) require four water molecules among the products?
Problem

Balance the following chemical equation using oxidation numbers:
$$\ce{Cr2O7^2–(aq) + HNO2(aq) –> Cr3+(aq) + NO3–(aq)}$$

Solution (in Swedish)

$$\ce{\overset{+VI}{Cr}_2O7^2–(aq) + H\overset{+III}{N}O2(aq) –> \overset{+III}{Cr}^3+(aq) + \overset{+V}{N}O3–(aq)}$$
\begin{align}
\text{Elektronövergång:}\quad
\ce{\overset{+VI}{Cr}_2 + 6 e- &-> 2 \overset{+III}{Cr}} &\quad &|\times 1 \\
\ce{\overset{+III}{N} &-> \overset{+V}{N} + 2 e-} &\quad &|\times 3
\end{align}
$$\ce{Cr2O7^2–(aq) + 3 HNO2(aq) –> 2 Cr^3+(aq) + 3 NO3–(aq)}$$
\begin{align}
\text{Laddningar:}\quad
&\text{vänsterled} &\quad &= -2 &\quad &|\text{lägg till}~\ce{5 H+} \\
&\text{högerled} &\quad 2\cdot(+3) + 3\cdot(-1) &= +3 &\quad &|\text{lägg till}~\ce{4 H2O} \\
\hline
&\text{skillnad} &\quad &= +5
\end{align}
$$\ce{Cr2O7^2–(aq) + 3 HNO2(aq) + 5 H+(aq) –> 2 Cr3+(aq) + 3 NO3–(aq) + 4 H2O}$$

Question
I can understand everything about the solution, except for the added $\ce{4 H2O}$ on the right side of the equation. Why is it that $\ce{4 H2O}$ is added to the right side and not $\ce{2.5 H2O}?$ How come the equation is balanced despite the oxygen atoms not being equal on both sides?

This question is getting traction in Meta.

$$\ce{Cr2O7^2–(aq) + 3 HNO2(aq) + 5 H+(aq) -> 2 Cr^3+(aq) + 3 NO3–(aq) + 4 H2O} $$
The left side has $1\cdot7 + 3\cdot2 = 13$ oxygen atoms and $3\cdot1 + 5\cdot1 = 8$ hydrogen atoms. The right side has $3\cdot3 + 4\cdot1 = 13$ oxygen atoms and $4\cdot2 = 8$ hydrogen atoms. It is not clear to my why you would expect 2.5 water molecules instead of 4. You can't just count the $\ce{5 H+}$, you have to consider that $\ce{HNO2}$ contains hydrogen as well.

Thank you for the reply. So, when the coefficient in front of the H2O molecule was chosen it was to balance the amount of O atoms and H atoms. It was a miscalculation on my part thinking the O atoms weren't balanced. And the coefficient 2.5 I got the idea from my textbook. I must've interpreted that passage wrong. Thank you once again.


text:
Can the relative reactivity of metals be explained by ionization energy and/or electronegativity?
I'm a 9th grade physical science teacher, and we're currently learning periodic trends. This is the first chemistry my students have learned. One of our labs for this unit has to do with the relative reactivity of Al, Mg, and Ca in water and 1M HCl.
In past years, the lab has been fun, but not very educational, because students aren't connecting what they see (bubbles of hydrogen gas forming) with what they know about periodic trends. I'm trying to find a way to explain this that isn't overly-complicated, but also isn't terribly inaccurate.
My current plan is to explain that acids contain H+ ions, and during a reaction with a metal, electrons from the metal are given to the H+ ions, which then become H2 gas, which bubbles up out of the test tube. The lower the ionization energy and electronegativity, the easier or faster it is for the H+ to take the ions from the metal, which is why we see a faster and more violent reaction with the Ca, for example, than with the Mg.
Is this an accurate understanding? (Or at least accurate-enough understanding in a "lies-to-children" sort of way?) Is there any information that I'm missing or that would clarify the connection between the reaction and the properties of the metals? Or is there a better lab to demonstrate this concept altogether?

In this case it might be, but just talking about redox potential would be more accurate. Or just showing reactivity somewhat correlates with place in Periodic Table would maybe be less likely to bite them later on. BTW we're rather against lies-to-student, misconceptions, etc. around here.

So you're saying that in this particular case, reactivity is caused by electronegativity, but that isn't always the case?

By lies-to-children, I guess I meant things like teaching using Bohr models, because even though it's not actually accurate, it is a useful model up to a point. I always try to tell the students when we're simplifying things.

Even with acids passivation can bring surprises, on the other hand sodium reacts especially violently.

What age is 9th grade? Not everybody has the same education system

9th grade here is 14-15.


text:
Why does polar solvents favor Ketonization and non-polar solvents favor Enolization?
I was just studying isomerism, when I came across this statement for keto-enol tautomerism:

A polar solvent favors ketonization, while a non-polar solvent favors enolization.

And I just can't get my head around that why is it so.
Can someone explain the mechanism by which polar solvents favor keto product formation?
PS: I also found this statement in the same text. I don't know if this would help in understanding the above question:

Polar solvents are known to stablize anions by solvating the negative charge.

Related: https://chemistry.stackexchange.com/q/83824/135348


text:
Transition of electrons between t2g and eg to give colour of coordination compounds
In the case of a hexaaquairon (III) complex, the iron (III) ion has 5 electrons, each singly occupying one d orbital. As the d orbitals are occupied, the vacant 4s, 4p and 4d orbitals are hybridised to give 6 sp3d2 orbitals to be used to overlap with the atomic orbitals of the ligands to do coordinative covalent bond. In this case since the t2g and eg orbitals are not completely filled transition can take place to give colour.
But in case of [Cr(H2O)6]+3. there will be 3 single electrons in 3 orbitals t2g. and all the other orbitals will be hybridized d2sp3. And according to what I understand, there will be coordinate covalent bonds between the lone pair of electrons of the ligands and these hybridized orbitals, meaning that these orbitals will be filled. So, how the electrons will transfer from the t2g orbitals to any higher filled orbital? thus, How could this compound have a color?

You are mixing CFT and VBT

Related: https://chemistry.stackexchange.com/a/180366/135348

Don't use hybridization for transition metal complexes. See https://chemistry.stackexchange.com/questions/76726/why-is-it-wrong-to-use-the-concept-of-hybridization-for-transition-metal-complex for an explanation as to why it is just wrong.

So, how the electrons will transfer from the t2g orbitals to any higher filled orbital?

This concept of t2g and eg comes in crystal field theory.
According to NCERT chemistry class 12 it is written that:

The crystal field theory (CFT) is an electrostatic model which considers
the metal-ligand bond to be ionic arising purely from electrostatic
interactions between the metal ion and the ligand.

Ligands would cause splitting of d orbitals but they do not transfer there electrons to these t2g and eg orbitals and hence electrons are free to jump in them.
Apart from CFT, other theories like LFT also tell the same thing that: t2g and eg orbitals are not occupied by ligands.


text:
What is the number of different compounds formed from the monochlorination of cis-1,2-dimethyl cyclopentane?
I encountered this question in the Black Book of Organic Chemistry by Ashish Mishra while preparing for a college entrance exam.
According to the answer key the answer is 12. I have only managed to draw 11 possible structures.
If someone could help me find out the 12th structure it would be greatly appreciated.

 In this the 1st, 2nd, 5th, and 6th structures are optically active, giving a total of $7+4 =11$ compounds.
PS: I used this to draw these structures. If there is more preferable way to upload structures let me know and I will update them.

The 7th structure is also optically active. (1S,2R)-1-(Chloromethyl)-2-methylcyclopentane is different from (1R,2S)-1-(chloromethyl)-2-methylcyclopentane.

The molecule (1S,2R) exists here. But the molecule (1R,2S) is not produced here.

@Maurice I believe both will form depending on which methyl group is chlorinated, is the answer not correct?

Yes, Grimm. Sorry ! It is correct. I had not thought about this possibility. Thank you.


text:
Difference between HOMO and LUMO and corresponding spectral gap in Hamiltonian
Coming from a Quantum Computing background, I am familiar with the singular value definition of spectral gap, I am currently trying to run Quantum Chemistry simulations and I would like to know if $|\text{HOMO} - \text{LUMO}|$ is same as the singular value definition.
The singular value definition of the spectral gap is defined over Hamiltonians. It is the difference between the ground-state energy and the first excited energy, i.e. the difference between the smallest and the second smallest singular value of the Hamiltonian.
I am referring to the Pitt Quantum Repository (link) that provides me with details about molecules such as Methane (link) which contains molecular orbital energy levels along with HOMO and LUMO. I would like to know if I can use the difference in their value as spectral gap in Hartree units.
For the Hamiltonian representation as an operator, I plan on using Pennylane's repository (link) that performs HF transformation and STO-3G basis.

With what you have said above I think the best we can say is "maybe". Firstly quite to what approximation to the Schrödinger equation do the Hamiltonians you use in Quantum Computing correspond? If they are some kind of of post Hartee-Fock or post DFT there may well be no orbitals at all, so the HOMO-LUMO gap can't be talked about. in such a case it might be hoped that the HOMO-LUMO gap is a good approximation to the first excitation energy, this is essentially Koopman's theorem, but there's not guarantee it will be the case.

Secondly if in your calculations you do have orbitals quite how do you chose the zero of energy when calculating your Hamiltonians? If you choose it so the Hamiltonian is positive semi-definite then the Singular Value and Eigen Decompositions are the same, and hence the HOMO-LUMO gap and the first excitation energy as you define it above will be the same.  However is the Hamiltonian is not positive demi-definite I can't see why this correspondence should occur, but I haven't sat down and  done the maths.


text:
Hydrolyse a fluorophosphonic compound / Which protective group strategy can I use for phosphonate esters to access the free acid after a reaction?
I have a synthetic problem in my PhD thesis, in which I synthesize a phosphonate ester.
To be more precise, it is a phosphonofluoridic compound. In my specific case, the fluorine attached to the phosphorus is the result of the acidic reaction conditions successfully applied in the previous step. I do not want to have a fluorine atom at this position in the final molecule and want to replace it by oxygen. Until now, I was not able to remove this fluorine. I tried to use base, but until now, to no avail.

In the literature I found during the similarity research, that compounds with a methyl group (like the toxic organofluor esters) attached to the phosphorus are in contrast hydrolyzed quite easyly. Here, only water is utilized to remove the fluorine.

In my case in contrast to that behaviour, the removal by base seems to be inhibited by the partly deprotonation of the oxygen in proximity to the fluorine. Seemingly, the compound has a small pka-value (strong acid), therefore the proton in neighboring position is likely to be fully deprotonated and therefore the negative charge will stabilize the fluorine in this position. My idea was now, when I can't get the hydrolysis to work to attach a protective group so that the discussed fluorine is not introduced in the first place, perform all the reactions that form "R" and than as the last step perform a deprotection reaction to release the free acid/ the salt. In the Greens book on protective groups in organic chemistry unfortunately, most strategies show only monodeprotection when not too acidic conditions are utilized:

I am wondering, what could be a valid approach, to achieve full deprotection to receive the free acid? The protective group should be stable under acidic conditions and in the presence of fluoride ions.
Currently I would say the following groups would be most interesting: Benzyl and 1,2-Phenylene

However, the protective chemistry for phosphate offers many different options. Maybe you would have a better possible route to solve this problem and recommend a different group? Do you have another idea how to kick out that fluorine?

How does the compound, and the base, really look?

For the compound, currently I am not at liberty to disclose, but for the base I utilized ammonia up to pH 12 because its a polar structure and utilizing a non volatile base like NaOH would bring challages in terms of purification. I did additional research recently, and it is maybe possible to remove excess base by HILIC chromatography, so I might try this again with stronger base up to pH 14.

sigh Is it phosphonic acid or ester that gets hydrolysed? Your post is unclear.

OK. I want to hydrolyze a substituted phosphonofluoridic acid to a free phosphonic acid. Until now I was not able to achieve this, therefore I had the idea to acess the free acid by a deprotection reaction that does not rely on acid or base hydrolosis but for example on hydrogenation. More clear?

O-Allyl is another possibility. Can you silylate the P-OH group then run the hydrolysis? The fluoride released by hydrolysis may then cleave the silyl group.

I was considering O allyl, but after reading about it, it came clear that it is prone to nucleophilic attack and not as stable as O-Benzyl. I was trying to get results with an O-Benzyl precursor, but the analytics are until now inconclusive.

@Waylander Something like TMSBr will not work, it's to acidic, the compound will die. But maybe there are other alternatives that are not as harsh? The compound tolerates TMSOMe. Do you think TMSDEA or HMDS might work? P-OH is more like P-O- in reality. When the negative charge of the oxygen is eliminated, this should indeed help to enable hydrolysis. I observed this, when P-O was activated with (COCl)2 and than subjected to alcoholysis. After the alcohol was coupled pH 11-12 was enough to remove the fluorine.

I am wondering however, if the flouride would hydrolyse before the siliyl group?

TMSDEA might work, but consider Bis(trimethylsilyl)acetamide (BSA) for silylating the P-OH

I was thinking about the problem and I did not have the opportunity to test silyation yet, but maybe just coupling benzylalkohol with oxalylchloride/DMF, DMAP, than removing the fluorine by slightly basic conditions, than pd/c benzyl deprotection maybe is a way to proceed.


text:
Acid base titration molar ratio
Consider the following neutralisation reaction:
$$ \ce{2HNO3 + Ba(OH)2 -> Ba(NO3)2 + 2H2O} \tag{1} $$
I would be glad for help in how to solve this problem with $c_1V_1 = c_2 V_2$, because the molar ratio here is not $1:1$.
What volume of a $\pu{0.15M}$ solution of $\ce{Ba(OH)2}$ is required to neutralise $\pu{45 mL}$ of a $\ce{HNO3}$ with concentration of 0.29M solution?

"45 Millilitres HNO3 solution..." At what concentration? 2. They an only be naturalised if they contain foreign substances. 3. Show work.

If you keep using C1V1=C2V2 formula, you will never learn how to solve titration problems.

You can use the equation for H+ and OH-. Or you can go via equivalent substance amounts. Note that nobody would use barium hydroxide to titrate HNO3, it is a made up task.

This is why NaOH is usually used to titrate acids it is not necessary to multiply the molarity by 2 to get equivalents of hydroxide.


text:
Why do the carbonyl oxygens in EDTA not participate in coordination?
I recently had two, very brief lectures on coordination compounds. In those lectures, my teacher mentioned EDTA because of its importance in analytical chemistry and its very stable complexes and mentioned it is a hexadentate ligand after the removal of the 4 hydrogens from the carboxylic acid groups, the other two bonds coming from the nitrogen bonds.
My question is why do the carbonyl oxygens not participate in coordination even though they have two lone pairs of electrons. I suppose it is because of the closeness of the carbonylic oxygen atoms and the nitrogen atoms, because if the two of them bonded via electron pair donation, the bonds would be too close to each other and would create instability in the structure. Is that correct? If yes, why does the nitrogen atom donate instead of the oxygen atom?

Having two oxygen atoms attach to both the same carbon and the same metal atom is a four-membered ring, not best.

Now, Oscar answered question "Why not both oxygens of carboxylic group bind?" Which is a pretty good one, and sometimes they do. If you'd ask why "carbonylic" one doesn't bind instead of the other, it's not a good one - both oxygens in uncoordinated carboxylate group are the same.


text:
How do I find the amount of moles from the amount of units?
The problem:
Find out the amount of moles of a certain substance knowing that the amount of units of said substance is $\pu{1.8E23}$. I have to find how many moles of carbon, hydrogen, and oxygen that amounts to. The number $\pu{1.8E23}$ was calculated in a previous question where I was asked to find the amount of units in $n(\ce{C6H12O6}) = \pu{0.3 mol}$.
The answer key:
\begin{align}
  &\ce{C}: \pu{1.1E24}  \\
  &\ce{H}: \pu{2.2E24} \\
  &\ce{O}: \pu{1.1E24}
\end{align}
Try 1:
\begin{align}
  m(\ce{C6H12O6}) &= M(\ce{C6H12O6}) \cdot n(\ce{C6H12O6}) \\
  &= (\pu{180.156 g mol-1})(\pu{0.3 mol}) = \pu{54.0468 g}
\end{align}
so that
\begin{equation}
  n(\ce{C}) = \frac{\pu{54.0468 g}}{(\pu{12.01 Da})(\pu{1.6603E-24 g/Da})}
  \approx \pu{2.71E24}
\end{equation}
Which is not the right answer. The right answer for carbon is $\ce{C}: \pu{1.1E24}$.
Try 2:
\begin{equation}
  n(\ce{C}) = \left(\frac{\pu{1.8E23}}{\pu{1.6603E-24 g/Da}}\right)
  (\pu{12.01 Da}) \approx \pu{1.30E48}
\end{equation}
Which is not the right answer either.
How do I go about this problem to get to the right answer?

Mere minutes after having spent a lot of time fiddling with MathJax to make this post I figure out how to get the right answer. I need to do this: multiply 1.8 times 10^23 with 6 (the amount of coal atoms in one glucose molecule). I'm unsure why that is, so an explanation would still be greatly appreciated...

Like many school chemistry problems, this one is so simple as to give the students a hard time for not believing the right answer could be as simple as it is. Don't be deceived: yes it is that simple, simpler than you think, and simpler than that, too. See: mass is not a part of the problem and not needed at all, as much so as the words like trousers, serendipity, or rhinoceros. Moles is all that matter. You have moles of glucose, you need moles of carbon. How many carbons are in one glucose? You'll never find out, unless you look at its formula.

Each molecule of glucose has 6 carbon atoms. Hence a given number of molecules of glucose would have 6 times the carbon atoms.
Since you have $\pu{1.8E23}$ atoms of glucose, the number of atoms of carbon would be this number multiplied by 6 which gives us $\pu{1.08E24}$ atoms.
Similarly proceed for Hydrogen and Oxygen.


text:
Bodenstein Hydrogen Iodide equilibirum experiment query

This text says that the same amount of HI was formed during the bulbs' time spent in the thermostat baths, regardless of the varying initial amount of HI each bulb contained.
Is this true? I always thought equilibrium meant only that, at a specific temperature, the forward reaction rate = reverse reaction rate. Does anyone know where to find Bodenstein's original published article on this experiment?

Type "Bodenstain works" on Google, or "Bodenstein Hydrogen Iodide". You will get all the information you need.

You seem to forget that you are describing a chemical reaction and at the start of the experiment there was no chemical equilibrium, you have the pure compounds.

You could check any textbook of Physical Chemistry and the kinetics of reversible reaction


text:
Making an UV inactive salt UV active during purification of charged anion with HILIC purification?
In the context of a purification setup I am asking myself the question: Can a UV inactive charged substance made be UV active by a counter ion that absorbs UV light during a column run?
If so, when no other UV active material is present, than would it be a legitimate statement for me to say in a publication, that this UV signal of the counter ion is directly proportional to my product during the elution run and every product fraction would absorb (Or any anion that might present as an impurity?) Because a sole anion could not travel on a normal or hilic phase, right?
For example imagine a mixture of two components impurity $A^-$ and product $B^-$ which are very polar and do not show absorption because the cation $C^+$  (same for both) is inactive. I would assume that you see nothing on the UV/Vis detector. However, I was able to exchange recently the counter ion to one that absorbs at 254 nm: $C^+_{UV}$, resulting in overall stoichiometry of the impurity $AC_{UV}$ and product $BC_{2,UV}$. However, if I would add buffer salts to the mobile phase, the stoichiometry of the different components might change during the run, and product could elute which is not paired with the UV active counter ion? To avoid this, can I run the column completely without any additional buffer salts or is this not advisable?

Clarify your question, please? Normal phase and HILIC are two different worlds. Write the column, mobile phase, UV-Vis absorption properties of A and B.

@Achem: 1) If I pair my UV inactive anion with a UV active cation, can I assume, that all my compound will be visible during a LC run and, not  only portion? 2) During a prep run, with a buffered mobile phase, I would assume  that the stoichiometry of a binary salt will change due mixing with the mobile phase. Can this be avoided by working with a pure aqueous phase?

You did not answer my question: What is the column (name/chemistry), what is the mobile phase composition and what are the UV-Vis properties of A and B?

I was not yet finished ;). I plan to use conventional normal phase silica in a purification run to remove impurities. I understand that normal phase is not equal hilic, but hilic can be used on normal phase with an increasing water gradient and ACN, methanol or isopropanol as an organic modifier as far as I know. The column I want to use is Biotage Sfär Silica – 60 μm or just bulk material. A and B are completely UV inactive because they do not contain any chromophores. But C contains an aromactic system and absorbs strongly at 254 nm.

In addition, I find the normal phase approach promising, because I still want to keep hard ionic species like salts from previous reaction steps to be completely traped at the top of the column where on special functionalized silica these species may come of with increasing water gradient.

You case study is an interesting scenario but in the question there are implicit misconceptions associated with hydrophilic interaction liquid chromatography (HILIC). The late inventor of this technique was a good acquaintance and we often discussed HILIC phases in conferences. The punch line is that HILIC mode does not work without buffers. As you have studied about HILIC, the role of buffer is to promote the formation of water layer on the surface of the stationary phase. The "thicker" the water layer, higher is the retention of a polar molecule. He used to emphasize that you must have high concentration of buffer in the mobile phase in order to use the HILIC mode (even on bare silica). The typical concentration should be 25 mM salts; usually ammonium salts adjusted to desired pH.  Otherwise, one is basically using mixed-mode retention and ion-exchange properties of the stationary phase/silica surface.
You are right, in the presence of buffers one cannot guarantee that the ion-pair AB or BC are still intact.
Suppose you do not add any buffer to ACN and water, and use bare silica as a stationary phase. Your analytes, A and B are anions, silica at neutral pH has a negative zeta potential due to ionized silanols. Both ions will be repelled, at times, and they may elute even before the dead time and C may show some retention. Note that there are ions from everywhere, including ambient dissolved CO2, so there might not be a very strict charge balance condition for AC and BC to travel together. Ions can also come from glass and metals in the chromatograph. If your molecules are bigger peptides etc., they will show a horrible peak shape on bare silica. In short, salts are needed, with the right pH to ensure a good peak shape of analyte in HILIC mode. I do not recall seeing any HILIC separation without salts. Also, bare silica is not a good HILIC phase for biological compounds.

Yes, I was referring to another technique which you may call pure aqueous liquid chromatography (PALC). It was mentioned here: https://www.chromatographyonline.com/view/hilic-pros-and-cons In that case In the past I made advantage of IEX properties of silica than. For my scenario, I did not recognize a difference with or without the addition of ammonia as buffer additive for my specific use case, the compounds showed similar retention on TLC and in booth cases very sharp peaks, indeed much sharper than on normal phase TLC.

Regarding the contact ion pairs, it would a nice trick if they could stay intact under specific conditions. I would like to believe, that here it could make a difference if you combine a soft anion with a soft cation and have a good match. I would need check the fractions each if there is good overlap or not. Nevertheless when I see, that I need additional buffer components, I would use volatiles whose excess can be fully removed and than as a final step perform an ion exchange to get a defined stoichiometry. My compounds are all small and highly charged,  peptides I would purify differently.


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Oxygen bleach discolored white tablecloth
My white polyester tablecloth got stained with red wine. Stains were about the size of a quarter.I proceeded to launder it several hours later.  I used Chlorx2 ColorSafe Bleach and as  per manufacturer's directions, poured some of the product directly on stains to pre-treat. The stains immediately turned a dark blackish purple and spread over the tablecloth. Without allowing any time to elapse, I washed the tablecloth separately in cool water with a dye and perfume free liquid detergent. The stains remained.  Further washing in cool water and detergent only did not budge the stains. Why did this happen? I would have expected that either the stains were released or that they would have remained the original color and size.  How can I avoid having this happen again when I use the new tablecloth I had to replace the ruined one with? Thanks.

Bleaches are often alkaline. The coloring material of wine, a pH-sensitive anthocyanin, like those in red cabbage, blueberries and hydrangeas, is an indicator of pH (acidity or alkalinity), and in this case it changed from reddish to bluish. some of these anthocyanins are quite stable to oxygen, so oxygen (or even chlorine) bleaches might not affect them.
To revert to red wine stain, soak in vinegar or lemon juice. Interesting demonstration, but not what you want.
There are two things holding that stain to the cloth:

The original anthocyanins were dissolved in ethyl alcohol (part of wine), a relatively nonpolar solvent, unlike more-polar water.
To some extent, the anthocyanins may adhere to the fibers of the fabric.

To remove the stain, one might try a few things:

First, wash the spot with alcohol, rather than water. One could use white wine or vodka, but ordinary "rubbing alcohol", 70% or stronger, ethyl or isopropyl, should work even better. Put an absorbent pad (e.g., an old terry-cloth towel) under the stain, and pour the alcohol and blot repeatedly over the stain with another absorbent pad.

Second, to help break the anthocyanin bond to the fabric, launder after applying a stain-removal product, such as these in CNN, or in Better Homes & Gardens, to the spot, rather than a bleach.

Oxygen bleaches for laundry rarely work in cold water. The color change of the stain was simply due to pH change since they are usually sodium percarbonate or sodium perborate salts. Simply wash the table cloth at the highest allowed temperature.
Alternatively, Clorox website has a detailed guide about bleaching polyester fabrics.
Clorox guidelines for polyester

HGTV.com gives tips for various materials, which have some differences. In general, they favor using organic materials; bleach is generally not favored. For your case eith polyester they recommebt rreatment with detergent plus white vinegar, followed if necessary by rubbing alcohol:

Stains can cling to synthetic materials. Presoaking the stain can help minimize the damage.
Blot as much of the wet wine off the garment with a dry towel or cloth.
Fill a bowl with 4 cups of warm water, a few drops of liquid detergent and a tablespoon of white vinegar.
Soak the stain for 15 minutes and then rinse it in clean water to remove detergent.
Lay the garment flat. If there is still a stain, blot it with a clean sponge dampened with rubbing alcohol.