text:
How Is This Acid Base Map Established?
In the Acid-base map -shown below- each line denoted as an isohydric line shows the pairs of $ PCO_{2}$ & $ [HCO_{3}^{-}] $ that give a certain $ pH $. So each line connects the values of $ PCO_{2}$ & $ [HCO_{3}^{-}] $ that give a value of $ pH $ that is constant across that line..
The 7.4 line shows the isohydric line of physiological pH,the central ellipse shows the normal range.
I understand the Henderson–Hasselbalch equation:
$\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}+\log _{10}\left(\frac{[\mathrm{Base}]}{[\mathrm{Acid}]}\right)$
What I would like to know is how this map established either experimentally or mathematically or both preferably.

The curves are rather obtained clinically, statistically evaluating diagnose, symptoms and results of clinical lab tests.
Metabolic curve follows malfunctioning metabolism keeping too low ( m. acidosis ) or too high ( m. alkalosis ) level of bicarbonate.
Respiratury curves follows states of too intense breathing ( r. alkalosis ) or too shallow breathing ( r. acidosis ), because of too low respectively high level of carbon dioxide partial pressure, caused by abnormal breathing.
Note that the $K_{\ce{H, CO2}}$ Henry constant  or CO2 is $\pu{29.41 atm/(mol/L)}$. the value will be higher because of the body temperature, can be corrected by temperature dependent $\ce{CO2}$ solubility.
$$\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}+\log\left(\frac{[\mathrm{Base}]}{[\mathrm{Acid}]}\right)=\mathrm{p} K_{\mathrm{a}}+\log\left(\frac{[\ce{HCO3-}] \cdot K_{\ce{H, CO2}}}{p_{\ce{CO2}} }\right)$$
Where pKa of carbonic acid is 6.35, if counting the total $\ce{CO2}$ concentration.
There are 3 variables ( $\mathrm{pH}$, $p$, $\ce{[HCO3-]}$ ) and 2 degrees of freedom. $\mathrm{p}$ is controlled by breathing, $\ce{[HCO3-]}$ is controlled by metabolism. $\mathrm{pH}$ is result.

Can the relationship between the 3 variables established mathematically? @Poutnik

If we set a certain pH constant, how do we choose which other variable to make constant? @Poutnik

What clinic state? @Poutnik

Thanks for taking the time to answer, but I think the soluability should be multiplied by the PCO2 in the equation. Otherwise a great answer. @Poutnik

[atm/(atm/(mol/L))]=[mol/L], so pCO2/KH = [CO2]. This KH is the pressure of CO2 needed to form CO2 solution 1 mol/L.


text:
pH scale at various temperatures
I am an electrician apprentice in charge of a large number of pH-sensors/transmitters at a power plant and chemical plant.
I do all the pH-instrumentation related maintenance, calibration, adjustments etc.
By studying pH measurements I have learned about the Nernst equation. The transmitter is able to (via an inbuild Pt1000 sensor) automatically keep the temperature variable in the Nernst equation constantly updated.
However, I have also learned that pH bear no meaning unless a temperature of the measured medium is specified. The transmitter is ALSO able to compensate for this variability, however. This is done by heating up a sample of the process medium and noting down the pH difference at various temperatures. This information can be plotted into a matrix in the transmitter.
HOWEVER, since the process medium composition is ever changing at the plant I work at (waste incineration) this matrix can't be made.
Seeing that many of the proces temperatures at the plant are anywhere between 40 and 80 degree celcius, how large of a discrepancy am I looking at? I know that both chemical composition and temperatures dertermine how much, but realistically, what are we talking about? 0,1 difference in pH? 0,5 difference? More? Worst case? Best case?
Thanks in advance.

Note that one thing is how pH changes with T, another things is, how pH-meter value changes with T. Optimal  would be measuring at 25 deg C. The second best would be probably correlation curve between measured pH at T versus measured pH at 25 deg C. The point is, pH buffers calibrating pH meters assume T=25 deg C.

A Factor is composition may change with temperature . For example ,on a gas well , some components of interest are water, salts, CO2 ,and H2S, The relative proportions in  the water phase change as temperatures change from 250 F to 350 F.

The pH of acidic solutions does not change much with increasing temperature. The pH of basic solutions does change more significantly with increasing temperature.
Ex. : The pH of $0.05$ M Potassium tetroxalate is $1.67$ at $0$°C to $1.81$ at $95$°C
The pH of $0.05$ M potassium hydrogenophtalate is $4.012$ at $0$°C and $4.23$ at $95$°C.
The pH of pure water is $7.47$ at $0$°C and $6.13$ at $100$°C
The pH of $0.1$ M borax is $9.46$ at $0$°C to $8,83$ at $95$°C.
The pH of $0.02$ M calcium hydroxyde is $13.42$ at $0$°C, and $11.45$ at $75$°C
Ref. : D. R. Lide, Handbook of Chemistry and Physics, CRC Press, 43rd Ed. 1961, and 83rd Ed., 2002.

The pH of an acidic solution is not very temperature dependant, neither is the pOH of a basic solution. What's temperature dependant is the pKs.

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text:
How to make SiO₂ hydrophobic aqueous-solution
I bought 100g of SiO₂ hydrophobic of 20 nanometers. I'm trying to make my own "car body wax". This "wax" that you can buy on the internet allows to make water repellent on car bodies : https://www.youtube.com/watch?v=JRtNpGQM53E
I did a quick test by mixing a few grams of SiO₂ with water and then put on my car body. It didn't work, no hydrophobia.
Would you know how to use this powder from SiO₂ to make a hydrophobic liquid ?

Who makes wax by putting hydrophobic components into water ? Do you want hydrophilic wax ? You are supposed to use some hydrophobic matrix for this hydrophobized ( silanized? )  SiO2.

This SiO2-based liquid can be found on the internet and in various car stores.
The final goal is to create a hydrophobic film on the car body so that the water runs off instead of staying on the body.
On the site of a French manufacturer, it is indicated :
"Our products are liquid solutions based on water or ethanol (alcohol) in which the active substance, Silicon Dioxide (SiO2), is immersed". => https://www.nano-protection.fr/content/6-faq

I suppose it is not just water and silanized SiO2. there is  know-how what to add as a thin film of SiO2 binder after water evaporates. "based on"

Maybe, I don't know the solution used to "assemble" them unfortunately.

The claims in the video are surely exaggerated: protection for 6 months? In a garage, perhaps.
The video claims that the hydrophobic silica - $SiO_2$ - is activated by water. The final product is applied to a wet car body with water droplets severely beaded up - i.e., already hydrophobic from the pretreatment washing agent and rub. So, the hydrophobic agent reacts with water on the car to repel water - but is already in an aqueous solution in the bottle?
My guess is that the actual process is 1) the wash which includes a hydrophobizing surfactant, perhaps a silyl ether, 2) a rub with a cloth containing some agent like a high m.w. silicone oil (promotes water beading), followed by 3) the application of fairy dust (informal term for a product that has more imaginary power than chemical activity) in water containing some slightly odoriferous (but safe) solvent. Perhaps it also contains some silyl ether or other silicone oil-type surfactant.
The main character in the video did a great job of showing how to use the manufacturer's product, and it is sure to have some advantage - he said it was a time saver. It also seemed to have the advantage of allowing you to rub your car gently and love it, and beautify it with a wash and rinse. But I was not convinced by Dave, the "chemist". I think he was a mechanic or a salesman. Good graphics, but not really believable. I also liked the spray bottles: very nice misting (not squirting), which implies low surface tension.
A related product called Rain-X is a silyl ether in alcohol. It was wiped onto dry windshields to provide a hydrophobic surface. It had to be applied evenly and very thin, but beaded up water fantastically! Any thick portions also resulted in beading up, but these thicker zones were soft and deformed a tiny bit, increasing the hysteresis and resistance to easy movement. So the beads didn't run, they built up and actually decreased visibility.

I have tested this kind of product and I guarantee you that it is not a "fairy powder".
It works really great. Just put it on the body and rub it until it's dry. The body is also shinier where the product is applied. The solution contains no silicone oil and is really fluid like water. In fact, it is said that you can drink it because the components are harmless.

Experiments are the best proof. I acknowledge your results and have no evidence of the opposite. But I still wouldn't drink it. I'll look around to see if it is available here.

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Will a salt dissociate completely in water to give free ions?
I was reading salt hydrolysis of a salt made from strong acid and weak base. I came across the hydrolysis of $\ce{NH_4Cl}$. These reactions were mentioned

$$\ce{NH_4Cl + H_2O<=>NH_3 + HCl \tag{R1}}$$
$$\ce{NH_4+ + Cl- + H_2O<=>NH3 + H+ + Cl-}\tag{R2}$$
$$\ce{NH_4+ + H_2O<=>NH3 + H3O+\tag{R3}}$$

$\ce{H+}$ and $\ce{Cl-}$ will not combine appreciably and concentration of $\ce{HCl}$ cam be neglected.
But in the second reaction, why did they assume that $\ce{NH_4Cl}$ dissociated completely into ions in water? Is that practically true too?
Edit: I also have another question; when calculating $\text{pH}$ of a salt like $\ce{NH_4Cl}$, is the concentration of hydrogen ions, $\ce{[H+]}, $ inclusive of the contributions from water?

Is this supposed to be in aqueous solution?

A weak acid NH4+(aq) reacts with the strong base OH-, passing to it its proton, forming NH3(aq) and H2O.

NH4OH does not exist. It is a solution of NH3 in water. The formula NH4OH is a wrong  formula invented in the 19th century, and still printed by merchants, because the customers (caretakers and contractors) prefer this name to the name of ammonia solution .

Reactions that are in equilibrium will have both reactants and products present. However, the interesting thing about dissolution reactions is that a solution that is undersaturated (below its solubility limit) will not be in equilibrium (i.e. no solid exists). So unless the example you are talking about specified that the concentration of salt is above the solubility limit and both solid and dissolved salt exist, you can assume full dissolution.
Here is a link to a ChemLibreTexts article that further discusses this topic.
For calculating the pH of a salt solution, an ICE chart can be used which accounts for the concentrations of the ammonium ion, ammonia, and the hydronium ion.
Here is a link that gives an example for calculating the pH of an ammonium chloride solution.

Why dissolution under solubility limit will lead to no equilibrium ?


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How is it possible to draw a radial wave function for the 2p orbital?
Below is a picture of the radial component of the wave function distribution for the 1 through 3s orbitals. It makes sense to me that there are points where the wave function is 0, since by definition, a radial node is where the wave function = 0.

What confuses me is the following graph of the 2p orbital wave function distribution. Since the 2p orbital is shaped like a tear drop, if you went a certain distance r from the nucleus, there would be areas that the tear drop doesn't cover (since a tear drop isn't spherical). Wouldn't the probability of finding an electron in those empty space areas be 0? How can you draw a radial wave function when the 2p orbital depends on angles as well?

Source of images: http://faculty.uml.edu/ndeluca/84.334/topics/topic1.htm

You can certainly plot the radial part of the function, but do need to keep the angular portions in mind.

@JonCuster How can you plot the radial part though? The 2p orbital isn't like a sphere where you can arbitrarily go out a distance r and be within the orbital. With the 2p, if you go out at the wrong angle, won't you be in empty space, outside of the orbital?

Look at the form of the spherical harmonic solutions, and see how $r$ enters into them. Yes, they vary with $\theta$ and $\phi$ as well.

@JonCuster So if we look at only the radial part, doesn't that imply we are fixing some value for theta and phi?


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What is the exact material brand and model number of Lego (ABS)?
Can anyone tell me the exact ABS material (brand and number) used for Lego?
Is it food grade material?
Kind regards

Probably a company secret.

Ask the company. This is not a chemistry question, please delete.

I’m voting to close this question because it is not germane to this site

See another Stachexchange for a good discussion on the topic: https://bricks.stackexchange.com/questions/7471/bpa-in-older-lego This article from the WSJ suggests - food grade perhaps not.
https://www.wsj.com/articles/lego-struggles-to-find-a-plant-based-plastic-that-clicks-11560331800


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Why does the Most Stable State of an Atom Tend to be One with Full s and p Subshells?
I'm new to posting on stack exchange, although I've read a lot of it before. This question seems like it might end up being marked as a duplicate, but I've looked through a lot of the similar questions and I feel like I'm still not getting it (most answers end up just explaining that the octet rule is wrong).
I'm currently taking AP Chemistry and we've been talking about electron configuration. As I understand it, atoms generally are at a lowest energy state when they fill their outermost s and p subshells (resulting in the octet rule, as the 2 + 6 electrons in these shells add up to 8). As far as I can tell, this is mostly due to Coulomb's Law being stronger with a smaller radius and the shielding effect impacting how much protons pull on electrons. But what I don't understand, then, is why an atom would tend to gain electrons in order to fill its valence shell. Wouldn't the lowest energy state be to lose a whole shell and decrease the radius and the shielding effect? It would make the atom very positively charged, but that doesn't seem to be a problem for something like Si4+. For that matter, why have any electrons at all? Wouldn't the most stable state be for an atom to repeatedly lose electrons and just be a nucleus? Presumably this has to do with the ionization energy being too high, but why is that the case? If it's just as simple as because the electrons are attracted to the nucleus, why can any electrons be removed at all? What am I missing or where is the flaw in this reasoning?
P.S. I've found electron configuration really interesting so far, but we haven't been doing much of anything with the math that explains it. I'm taking multivariate calculus right now, and I would really like to learn about the math behind it all. Is it too early for me to start? If not, does anyone have any good introductory sources to learn about that?

The major factor is electrons with the same quantum numbers $\mathrm{n}$ and $\mathrm{l}$ do not mutually shield the nucleus charge well.
Elements approaching the noble gas group in the Periodic table ( $\ce{N, O, F}$ ) get nearly  fully filled respective $\mathrm{p}$ orbitals. The effectively perceived nucleus charge grows for valence electrons. It gets progressively more difficult to ionisate these electrons, and at the same time the energy released by capturing an extra electron grows.
Elements at the opposite table side ( alkaline metals and alkaline earth metals ) have the opposite situation. They start to fill orbitals at the new, higher quantum number $\mathrm{n}$ level. The lower, now fully filled $\mathrm{p}$ orbitals shield the nucleus well. Additionally, the new $\mathrm{s}$ orbital  is farther from nucleus with lower attractive force. Both effects lead to low ionization energy of such atoms and very low affinity to extra electrons.
This leads to the octet rule, which is consequence of the fact, if chemical bonds lead to completing octets, the total electron energy is lower.
It has its limits. Ionization of electrons leads to progressively increasing ionization energy for every next electron. Similarly, accepting too many electrons leads to negative electron affinity, so the electron is released at the nearest convenience. So ions with high positive charge occur only in heavily ionisating environment, respectively the one with highly negative charge need source of electrons. Even in solid matrices there is partially covalent bond.
It may be challenging, but this explains a lot about the screening of the nucleus charge: Slater's rules


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Beer-Lambert law absorption by each species
Light travels length $l$ through a medium with $n$ independently absorbing species with extinction coefficients $\alpha_1,\dots,\alpha_n$. According to Beer-Lambert law, the amount of light absorbed is
$$1-T=1-\exp\left(-l\sum_i \alpha_i\right)=1-\prod_i\exp(-\alpha_i l)$$
How do I find the amount of light absorbed by each species? Is it
$$I_{Ai}=(1-T)\frac{1-\exp{(-\alpha_i l)}}{\sum_i {[1-\exp{(-\alpha_i l)}]}}$$
This is the only self-consistent expression I can think of but I don't know if or why it is correct.

Is it not the fractional absorption, i.e. that for species i divided by that for the whole sample.

@porphyrin I'm not familiar with the concept of fractional absorption. If it is $1-\exp(-\alpha_i l)$, it doesn't add up to the total absorption $1-\prod_i \exp(-\alpha_i l)$

sorry I was not clear, I meant the fraction absorbed by species i so $(1-\exp(-\alpha_i l))/(1-\prod_i \exp(-\alpha_i l))$

@porphyrin that still doesn't add up to $1-T$ though. And on second thought, I feel that ratio between absorption by species $i$ and total absorption should be independent of path length $l$, because incident photons don't know how far they've travelled. Does $(1-T)\alpha_i/\sum_i \alpha_i$ look reasonable? Seems weird.


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How can I determine the size of tropolone molecule?
How can I determine the size of a tropolone molecule?
It's a seven-membered aromatic ring with =O on the first carbon and -OH on the second carbon.
Is there any software available?

You may need to clarify what you mean by size (e.g. length across, volume, collision cross section, some other measure of steric bulk)

Yes, length across.

It's about 3 cm or so on my screen...

I invite you to read comments and answer but it depends on what you are really looking for. Even drawing the molecule in scale while using typical tabulated bond lengths won't drive you much out of reality. Hard to judge how accurate you must be by reading just the question.

You wanted to measure the distance "across tropolone" but you didn't specify from where to where. The distance from a remote hydrogen to each of the oxygens and to the hydroxyl hydrogen are shown below. You can do the measurements you want yourself here: http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/ESPotential/ESPmisc2.html

According to ChemDoodle's ruler tool, this molecule is probably about 5.35 Angstroms across.
I'm a little annoyed that the font doesn't seem to have a proper Angstrom symbol. The symbol is part of the symbol selection widget, but doesn't get used by the ruler tool. Oh well, still cheaper than ChemDraw.
EDIT: The angstrom symbol issue is font related. If I use Arial, it fails to display. But if I switch to Times New Roman, it shows up fine.

This is a cool use of ChemDoodle's ruler tool. The ruler tool in 2D uses the Angstroms/bond length and default bond length styles from the style sheet to calculate distances. It is typically used with imported 3D structures where coordinates are already calculated. You can also use ChemDoodle 3D measurement tools to measure distances after optimizing structures in 3D with force fields.

If you use «Arial Unicode MS» instead of «normal Arial», then you get the Angstrom comme il faut.  And it is not the only font (if working in Windows) supporting this sign (examples in https://chat.stackexchange.com/rooms/3229/the-periodic-table).

This seems to neglect a few hydrogen though.

@Martin: True. Even worse, though, chemically speaking, is that the drawing of the tropolone may be scaled, while drawing the molecule according e.g., policies of a journal / style sheet the program uses implicitly by default, or explicitly to sooth wacky bonds (if one doesn't use a template for a 7-membered ring).  But because the molecule's structure is one graphical object on the canvas of the sketcher, and the ruler an other, the two may be scaled independently from each other, too.  For me, this is like the «poof, cinq» by Hugh Laurie and Helen Mirren, back in 2006 at the Emmy awards.

@Buttonwood (I don't know this skid, but I can imagine it. Will google.) As a rough guesstimate it may still work and it's pretty much a computerised version of drawing it by hand using standard bond lengths. It's rough which you can see, as the other values come in  around 580 - 600 pm. I would suggest a more robust technique for uses other than graphical design.

A quick way to do this would be to enter the molecule in name in the search form on ChemSpider (maintained by the UK Royal Society of Chemistry). Select the 3D representation by hitting the 3D button. In the JSMol window, you can select an atom by double clicking it. To get the largest atom-to-atom distance, I've selected the hydroxyl hydrogen. By moving the mouse to the other atoms, you can see their distances. The largest distance appears to be 0.588 nm.

The atoms themselves have size owing to their electron clouds, so a better measure of the size of the molecule could include the van der Waals radius of the atoms, which you can find a value for on the Wikipedia page. Hydrogen has a radius of 0.11 nm, so the total distance would be 0.558 nm plus the van der Waals radius of the two hydrogen atoms, which would be 0.778 nm.
This is only one way to characterize the size. More complex ways of characterizing the size would be better done in a dedicated program. You can save structure as a .mol file from ChemSpider by clicking the disk icon below the 3D image. Conformers can also be downloaded in different formats from PubChem (maintained by the US National Institutes of Health), which offers many of the same features as ChemSpider.
Commonly used standalone programs for visualizing molecular structures are the programs VMD, PyMol, and Avogadro. They are open source and completely free (Avogadro is GPL'd) or free for educational use (VMD, PyMol). There are many tutorials for using these programs. PyMol will open the .mol file natively, but VMD will not, so you can use OpenBabel to convert the .mol file to a .mol2 file:
obabel tropolone.mol -O tropolone.mol2
Then I open it in VMD:
vmd tropolone.mol2
You can measure distances between pairs of atoms by hitting the "2" key and clicking atoms. Note that these distances are in angstroms (Å).

VMD will give you the atom-to-atom size along the x, y, and z axis using the following commands:
mol new tropolone.mol2
set s [atomselect top all]
set m [measure minmax $s]
puts [vecsub [lindex $m 1] [lindex $m 0]]

The results are 5.42 Å along the x-axis, 5.20 Å along the y-axis, and 0.07 Å along the z-axis.

Since we're comparing: molden has a zmat builder and a force field (amber) implemented. You can start with the obvious choices of getting a structure from a database, or generating a starting structure with Open Babel, or simply click it together. The result will come to something like this (I admit this is even less prettier than VMD, but you can use some custom settings in gmolden to make that look nicer, too):

Obviously, better results (i.e. more accurate estimates) can be achieved with better calculation techniques. A quick and open source option is xtb, which calculates the molecule on the semi-empirical level of theory.
If you want to go full quantum calculation, free options include orca and gamess. I'd say both are reasonably simple to set up and use. There are plenty of more options. For general recommendations I suggest looking through our Mater Modeling partner site's Q&A.

Another nice free (but limited in function) tool for this job is NIH PubChem's viewer:

The computed distance (not clearly visible in the figure) is $\pu{6.10 Å}$
The structure is also from PubChem:
https://pubchem.ncbi.nlm.nih.gov/compound/Tropolone#section=3D-Conformer
You can alternately process the coordinates with a tool such as Matlab/Octave:
data=[-2.1849   -1.2310   -0.2189 16   0  0  0  0  0  0  0  0  0  0  0  0
   -1.8861    1.4126    0.3330 16   0  0  0  0  0  0  0  0  0  0  0  0
   -0.9119   -0.6926   -0.0609 12   0  0  0  0  0  0  0  0  0  0  0  0
   -0.8564    0.7909    0.0658 12   0  0  0  0  0  0  0  0  0  0  0  0
    0.1188   -1.5452   -0.0243 12   0  0  0  0  0  0  0  0  0  0  0  0
    0.3789    1.5596   -0.2139 12   0  0  0  0  0  0  0  0  0  0  0  0
    1.5156   -1.2996    0.2099 12   0  0  0  0  0  0  0  0  0  0  0  0
    1.6492    1.1479   -0.2156 12   0  0  0  0  0  0  0  0  0  0  0  0
    2.1767   -0.1426    0.1249 12   0  0  0  0  0  0  0  0  0  0  0  0
   -0.0939   -2.6096   -0.1302 1   0  0  0  0  0  0  0  0  0  0  0  0
    0.1974    2.6114   -0.4310 1   0  0  0  0  0  0  0  0  0  0  0  0
    2.0899   -2.1870    0.4717 1   0  0  0  0  0  0  0  0  0  0  0  0
    2.4120    1.8907   -0.4446 1   0  0  0  0  0  0  0  0  0  0  0  0
    3.2469   -0.1561    0.3232 1   0  0  0  0  0  0  0  0  0  0  0  0
   -2.1177   -2.1985   -0.2883 1   0  0  0  0  0  0  0  0  0  0  0  0];

xyz = data(:,1:3);

mat=shiftdim(repmat(xyz,1,1,size(xyz,1)),2);
dist2=sum((permute(mat,[2 1 3])-mat).^2,3); % compute distances between all atoms
[d2max,imax]=max(dist2); % find max value (column by column)
[dd2max,jmax]=max(d2max); % find absolute max

disp(['max distance:' sprintf('%10.4f',sqrt(dd2max))])

@andselisk PubChem's viewer also allows visualization of an enclosing box but I haven't figured out what units are used. They are not consistent with the otherwise reasonable estimate of the longest interatomic distance. Unfortunately documentation is difficult to find.

The crystal structure of tropolone has been determined [1] and is freely available via CCDC (CSD Entry: TROPOL10).
One can use JSmol on the web page to find the linear parameters (RMB → Measurements → Click for distance measurement).
I would, however, suggest to use Olex2.
Choosing two atoms of interest, say, H1 and H4, followed a line command, yields in  linear parameter of 5.992(2) Å:

The issue with this approach is that atoms are treated as point particles (other answers use the same approach), which is not entirely correct.
When it comes to the size of a molecule, involving a space-filling model based on van der Waals radii would be more appropriate.
Conveniently, Olex2 has a tool for that too — a wbox command which

Draws a rectangular Wrapping BOX around the chemical moieties on the screen.
The vdW radii employed in the computation are printed.

Like such:

Using the following element radii:
(Default radii source: http://www.ccdc.cam.ac.uk/products/csd/radii)
C   1.7
H   1.09
O   1.52
Wrapping box dimension: 3.453 x 7.348 x 8.156 A
Wrapping box volume: 206.955 A^3

meaning the length across is actually about 8.2 Å.
References

Shimanouchi, H.; Sasada, Y. The Crystal and Molecular Structure of Tropolone. Acta Cryst B Struct Crystallogr Cryst Chem 1973, 29 (1), 81–90. DOI: 10.1107/S0567740873002013.


text:
What does the term 'by mass' mean with respect to different contexts?
I've seen the term 'by mass' being used in various sorts of contexts. Some, while explaining the law of definite proportions say, the 'same proportions by mass', some say 'constant chemical composition by mass'.
What does the term 'by mass' mean? Does it mean in a proportion/ratio, we have mass as both, the denominator and numerator?
When I say 'percentage by mass', what would that mean?

You have the right idea, it's a ratio of masses. You might also see the phrase mass fraction

In $\ce{H2O}$, the proportion of $\ce{H}$ counted by atoms is $\ce{2/3 = 66.7}$ %. In the same $\ce{H2O}$, the mass of $\ce{2 H}$ is $2$ u. As the mass of $\ce{1 O}$ is $16$ u, the proportion of H by mass in $\ce{H2O}$ is $2/18 = 11.1$%. This percentage $11.1$% is rather different from $66.7$%


text:
How do u clean an hp laptop keyboard from nail polish remover
I was cleaning my nails with nail polish remover(isopropyl acetone, methyl ethyl ketone, and n-methyl-pyrrolidone) and I touched my computer and I saw a white stain so I tried to take it off with all-purpose cleaner but that did not work so I look at my computer and I see the whiteness so how can I remove it.

You do not - laptop isn't a nail.

@Mithoron. How could Amineh Elhoodary have discovered that the same question was asked 5 yers ago ?

@Maurice https://chemistry.stackexchange.com/search?q=nail+polish+remover

It is not just a stain that could be cleaned off by dissolving or fine brushing by a wipe. The plastic is already damaged.
I see only few ways, none really good :

Colour the spots by some colour ( markers ? ) to be as least disturbing as possible.

Peel off the damaged spots to the undamaged layers, but the original surface texture would be lost, so the result may not be better..

If it is a free, not functional surface, put there some decoration stick label.


text:
Why is melting point of Vanadium higher than Iron?
The iron is found to have four unpaired electrons while vanadium has 3 of them. we say the binding energy is directly proportional to number of unpaired electrons and hence number of  metallic bonds in the structure. By this logic, iron should have high melting point than vanadium but no. Iron has a m.p of 1535 degree celsius while vanadium has 1917.

Atom-level (chemistry) concepts don’t translate well to crystal (solid state physics) bonding.

@Jon Custer makes a good point in his comment so lets look beyond the electronic shell configuration of V and Fe to help understand the different in melting points.
Consider the following:
Pauling’s Electronegativity Fe > V  1.8 vs 1.6
Vander walls radius Fe < V  0.126 nm vs 0.134 nm
And consistent with Fe having a higher electronegativity the first ionization energy of Fe [761 kJ per mol] is greater than that of V [ 649.1 kJ per mol]
Given that we think of melting as the breaking of lattice forces holding atoms together, Fe being a smaller and more electronegative atom [vs V] is consistent with the relative melting points of these two elements.
Ref:
https://www.lenntech.com/periodic/elements/fe.htm
https://www.lenntech.com/periodic/elements/v.htm


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What's the difference between Zinc Ascorbate and Zinc Gluconate?
I've been looking at a dental product for pets, called MAXI/GUARD Oral Cleansing Gel and it states it uses zinc ascorbate/taurine, however in the ingredients I found elsewhere it instead lists zinc gluconate.

Active Ingredient(s): Deionized water, zinc gluconate, ascorbic acid (vitamin C), methylcellulose, taurine, methylparaben, propylparaben, F.D.&C. blue no. 1. May also contain zinc sulfate.

What is the difference between the two and any idea why they are listed as such?

As I understand your question, “the two” refers to Zinc Gluconate and Zinc Ascorbate.
Zinc gluconate and zinc ascorbate are distinctly different compounds.  They are both adducts of Zn with one Zn and two bidentate oxygen-based ligands of organic acids.  Some basic information from PubChem for both zinc gluconate and zinc ascorbate is given below.
Zinc gluconate: $\ce{C12H22O14Zn}$
Zinc gluconate [Gluconic Acid Zinc (II) Salt] is a zinc salt of gluconic acid comprised of two gluconic acid molecules for each zinc cation ($\ce{Zn^2+}$). Zinc gluconate is a generally recognized as safe (GRAS) substance by FDA.

Of interest, as related to periodontic issues note that a phase 4 clinical trial was completed 2014-12-16 “The effect of Zinc-D-gluconate on bleeding propensity of the periodontium in patients with chronic parodontitis - a randomised, placebo-controlled, double-blind interventional study”.
Zinc ascorbate: $\ce{C12H14O12Zn}$
Zinc ascorbate is a zinc salt of ascorbic acid[Vitamin C] comprised of two ascorbic acid molecules for each zinc cation ($\ce{Zn^2+}$). Zinc ascorbate goes by a variety of different names, L-Ascorbic acid, zinc salt (2:1), L-Ascorbic acid, zinc salt, (-)-Ascorbic acid, zinc salt to name a few.

A ResearchGate reference discusses zinc-gluconate, taurine in reducing halitosis and gingivitis.  A portion of the abstract follows:

Objectives: to evaluate the efficacy in reducing halitosis and gingivitis insurgence in fixed orthodontic patients of a
zinc-gluconate, taurine and cetylpyridinium-chloride (CPC) oral-rinse,
sole or combined with a toothpaste with the same active principles.

References:

https://pubchem.ncbi.nlm.nih.gov/compound/443445
https://pubchem.ncbi.nlm.nih.gov/compound/Zinc-ascorbate#section=Structures
Dalessandri, Michela & Dalessandri, Domenico. (2011). Halitosis And Gingivitis Reduction Using A Zinc-gluconate-taurine-cetylpyridinium-chloride Oral-rinse And Toothpaste. (link)

Thanks very much. Now I see they have listed Zinc Ascorbate as Ascorbic Acid, now it makes sense :)
Do you have a link to the referenced clinical trial mentioned above regarding Zinc Glucanate? I don't believe it was included in your references?

https://www.clinicaltrialsregister.eu/ctr-search/search?query=2013-001866-40  I believe that link will take you to the available information.


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How do I go about solving this question?

So I've studied Redox concepts but in the exam we are asked questions like these.
How am I supposed to know what the reaction is going to be?
How do I go ahead and assume what oxidation state it's going to be and then form the reaction?
Do I need to have knowledge about how the reactants interact to answer these questions or is there something only based on redox concepts that I can use to answer this question?

Think about the products that result from that reaction. Usually, when bubbling chlorine gas over a sodium hydroxide solution, one would get two chlorine based products: chloride and hypochlorite. Here, the conditions are quite different: since the sodium hydroxide solution is both hot and concentrated, it provides a more oxidizing medium. Hence, you might get chloride and a form more oxidized than hypochlorite.

"Do I need to have knowledge about how the reactants interact to answer these questions". Yes, you need to.

How would you like a book title "How would I go about naming this book ?" ? BTW screenshots/photos of easy plain text are highly discouraged for multiple reasons.

Reaction of chlorine with hot hydroxide is easily available in textbooks or online resources.

Yes I understand that. But I wanted to know if students all over the world use anything else - ideas/tricks/mnemonics that'll categorise particular reactions.

For example if we look at organic chemistry, most questions can be easily distinguished and categorised into SN1 & 2. This way it becomes easier to remember bigger chunk of information for competitive exams. Please just don't judge a person's questions without understanding the complete reasons. I couldn't rather googled this instead of posting it here right? There's obviously a reason I want it to be a discussion where I can see how other students deal with questions. For you it might be a simple question because you've explored the subject enough, for a beginner it isn't. @Poutnik

I appreciate the way Maurice has given the answer here.

Questions here are assumed to be well elaborated. If they are not, the OP can blame just himself for wrong assumptions of others, who do not know the background, circumstances and questionable hidden elaboration. Chemistry has a good set on main rules, a lot of empirical evidence to be learnt, and with this knowledge comes recognizing patterns that lay between empiriic knowledge and rules. I prefer teaching fishing to giving fish. Disproportionation of chlorine to -I/+I and -I/+V is such data to learn or pattern already known, depending on knowledge level. Wikipedia is full of great articles.

You are neither the first neither the last who asks here for tricks to learn chemistry. There is no simple trick. The only way not to memorize everything is to understand it and recognize patterns in behaviour. That takes time.

In this case, you have to know that $\ce{Cl2}$ and $\ce{NaOH}$ react at low temperatures according to $$\ce{Cl2 + 2 NaOH -> NaCl + NaClO}$$ and that at high temperature, the hypochlorite $\ce{NaClO}$ gets disproportionated according to $$\ce{3 NaClO -> 2 NaCl + NaClO3}$$ There is no way of predicting it. It must be known. So b) is the good answer.
Note that the preceding reactions may be written with ions instead of sodium compounds.


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Why doesn't separating the layers change the equilibrium point?
I'm following this tutorial for synthesizing esters. I don't understand why adding water to the separation funnel doesn't push the equilibrium reaction point back to the left, or why draining it doesn't push the reaction to the right, given the tutorial's explanation of Le Chatelier's Principle:

When the reaction reaches equilibrium there is still a large amount of reactants left in the mixture resulting in a poor yield of the ester.
The yield of ester can be improved by increasing the concentration of one of the reactants (either the alcohol or the carboxylic acid).
By Le Chatelier's Principle an excess of one reactant will drive the reaction to the right, increasing the production of ester, and therefore increasing the yield of ester.
In our experiment we are using an excess of carboxylic aicd (acetic acid which is also known as ethanoic acid).
From the balanced chemical equation, you will also appreciate that the presence of water in the reaction mixture will drive the equilibrium to the left, favouring the formation of reactants rather than ester.

It is a question of temperature. If water is added at the temperature of the reaction, the back reaction is favored, and the ester will be hydrolyzed. But if water is added to the ester at room temperature, the reaction will be so slow that no visible reaction will happen during the time of the operation.

You are preparing ethyl acetate at elevated temperature in the presence of H2SO4 as a catalyst. The extraction in the presence of water is at room temperature. Huge difference in rate and contact time. BTW the tutorial states that acetic acid is in excess. The volumes of ethanol and acetic acid say otherwise. Ethanol (~0.25 mol); acetic acid (~0.16 mol). Ethyl acetate is partially soluble in water (8.3g/100mL). You should saturate the aqueous layer with salt.

@user55119, why saturate the water with salt? Thanks

You have 2 layers, ethyl acetate and water. But the former is partially soluble in the latter. By adding NaCl and forming a brine solution, the ethyl acetate becomes less soluble in the aqueous phase there by increasing the yield of the ester.

Draining the aqueous phase cannot change anything. If the phases are in equillibrium before (both internally and with each other), they stay that way if they are separated.
Now if you add water to the aqueous phase, the reaction in the unpolar phase (which is saturated with water, as the phases are in equillibrium) is in the first moment not affected.
However the acid, alcohol and ester have different solubilities in the aqueous phase. That means the concentrations in the unpolar phase will change a bit, and so does the reaction equillibrium.
But: This is all irrelevant, because the reaction happens under reflux. While you are standing there with the cooled mixture in the separation funnel, the reaction rate is much lower, you will never notice a change.


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Colligative properties: why does the size of solute not matter?
Colligative properties are defined as the "properties of solutions that depend on the ratio of the number of solute particles to the number of solvent molecules in a solution, and not on the nature of the chemical species present (wikipedia)."
When a solute is dissolved in a solvent, the number of solvent molecules near the surface decreases, and the vapor pressure of the solvent decreases. That's why boiling-point elevation, freezing-point depression etc correspondingly occurs.
If the reason of elevated boiling points is due to decreased surface area of volatile solvent molecules, then why does the size of the solute not matter? Won't larger solutes occupy more space in the solution?

+1 for the question -- Doesn't seem like a homework problem to me....


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Does increasing the number of interferometers increase signal in FT-IR?
I'm trying to understand why most modern FT-IRs have three interferometers. Does this improve the signal in someway or does it have another effect?

Where did you see this? There is only one interferometer in standard FTIR used in chemical labs.

Modern interferometers generally have 3. This is standard in modern instruments

Could you provide a reference which says three interferometers are employed in modern ones? I have seen opened modern FTIRs all one can see is a single interferometer.

This was information in a lecture that I am trying to understand. He did not provide a reference, just stated the fact.

My guess is it somehow improves resolution?

It is not a fact. Most likely there is some misunderstanding. Perhaps if someone is making an astronomical measurement they may employ interferometry. Routine instruments which you will encounter in a university have only one interferometer.

Would increasing the number of interferometers improve resolution?

In interferometry, the resolution is improved by the distance travelled by the mirror. It is a fundamental property of Fourier transforms.

would it increase the signal in some way? Basically we were told that modern instruments have three and we should think about why

Ask your teacher to give a reference. Have you checked your textbook? I am under the assumption you are taking an instrumental analysis course in a chemistry program not astronomy. Each interferometer is extremely expensive, the cost is prohibitive for 3 such system for regular analytical chemistry work.

It is a materials course

Still, check your reference text and see what was the context of the lecture. He must have shown a diagram. Only very fancy custom designed FT-IR systems can employ three or more. I just saw one paper in Nature in 2020. Certainly, this is not common or commercial to begin with. https://www.nature.com/articles/s41378-019-0111-0.

yes thats the same paper i saw

Most FT-IR instruments have one interferometer, as @M.Farooq says. The resolution is inversely proportional to twice the distance (in cm) that the moving mirror moves. But, for a small integrated chip, only an inadequately small mirror movement distance is feasible. So the authors of the 2020 linked paper had the nice idea of using N interferometers, driven by the same actuator, so that the desired travel distance could be divied up into N coherently scanned contiguous shorter distances. Then they process the N interferograms as necessary in order to achieve their goal spectral resolution.

Do have have handouts for the course? A direct quote from the offending slide would be helpful.

I use FTIR in Physics, pretty much top-of the-range commercial instruments (actually more than just IR - ours go down to the UV and would do THz if we had the detectors; this isn't uncommon so I'll mainly refer to FT rather than FTIR).  There's definitely only one interferometer in each of our systems (and the ones used by our friends in Chemistry).  In astronomy FT spectroscopy has been used on the ground, but the best IR work these days is done from space using dispersive instruments (e.g. Spitzer's IRS).
The only common commercial use of three series-connected spectral devices (trying to coin a deliberately vague term) is dispersive instruments, which use three diffraction gratings for high resolution spectra and, for Raman spectroscopy, laser rejection with low stray light. A good benchtop FT instrument (with one interferometer) can give just as good resolution, though slowly and without the dynamic range. I mention this because if you were giving a lecture introducing a range of systems and techniques, and skipped a slide or two, you could jump from FTIR to triple dispersive systems and cause a lot of confusion.


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Is the Hydrogen spectrum only an emission spectrum?
In a statement question it was stated that hydrogen spectrum is an emission spectrum and the scheme considered it to be ture. But my argument is that it can be both emission and abosorption spectrum. Please tell me whether I am correct.

yes both absorption and emission. You need different types of experiment to measure one or the other.

The important thing missing in many questions and in this particular one as well  is the context. In your case, If they discussed a particular spectrum, that is an emission one, they would be true. If they discussed an absorption spectrum, they would be false.

Emission and absorption are inverse processes.
So, naturally, if an atom exhibits some emission spectrum, it will also give an absorption spectrum. So you were right in assuming that.
However, the processes by which these two spectra are generated are completely the opposite of each other.
An atom emits a photon when an electron goes from a higher energy state to a lower energy state, thereby releasing energy in the form of light. When a photon is absorbed, the electron uses this energy to excite to a higher energy state.
What kind of spectrum you have at hand depends on which instrument and detector you are using. When atoms releases photons (light), the detector catches the production of the photon and gives out emission spectra. For absorption spectroscopy, varying wavelengths of photons are incident on the atom and the corresponding photon is absorbed by the atom resulting in no photon of that particular wavelength being detected.
I am attaching the images of the emission and absorption spectra of hydrogen atom. Note how they are complete inverses of each other.

Image source: https://chem.libretexts.org/Courses/Solano_Community_College/Chem_160/Chapter_07%3A_Atomic_Structure_and_Periodicity/7.03_The_Atomic_Spectrum_of_Hydrogen

LibreTexts content is licensed by CC BY-NC-SA 3.0. Please make sure to follow this when using their content.


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How to separate a mixture of sodium chloride and sodium hydroxide?
I recently did a double displacement reaction in which I formed a solution of water with sodium hydroxide and sodium chloride dissolved in it. Now I want to separate both of them. How can I do this?

Full separation would be very difficult. So does partial separation $\ce{NaOH}$ from of $\ce{NaCl}$. Partial separation $\ce{NaCl}$ from of $\ce{NaOH}$ is duable, as @Hans says.

Is there some reason you need to recover the NaOH? I would consider neutralizing with HCl to get a solution that is just NaCl and then distill off the water.

Hans' idea (evaporation + filtration) is not easy to carry out in practice. Concentrated $\ce{NaOH}$ solutions are extremely viscous, and if they are mixed with $\ce{NaC}$l crystals, they cannot be filtered easily. Look ! If the filtration is done on paper, the paper is quickly transformed into a sort of parchment, that prevents any liquid from going through. And if it is done on a glass-sintered filter, the glass will be attacked by $\ce{NaOH}$. As a consequence, I am afraid there is no possibility of separating easily $\ce{NaCl}$ from $\ce{NaOH}$. The only really working process would be centrifugation, which requires an equipment that is not always available in all laboratories.

One solution would be to reduce the amount of solvent (under vacuum for instance), then filter out the precipitated NaCl at high temperature (at higher temperatures NaOH has as much as >10x the water solubility of NaCl).
The common ion effect will play in your favour, as the high water solubility of NaOH at higher temperatures will help you reach NaOH concentrations that strongly disfavours NaCl solubility.
The way to go may be to first reduce the solvent mass under vacuum at acceptably safe temperatures, and once the desired mass is reached redissolving most of the precipitated NaOH by pushing the temperature up before filtration to a level at which most NaOH has redissolved.
Use appropriate protection equipment and contact material to handle very hot NaOH.

Sorry but concentrated NaOH solutions cannot be filtrated. .

@Maurice: You bring up the point that concentrated NaOH solutions are not as easy to filter as any solution, which is a good remark. Two points on that (1) As for the case here we don't know the proportion of hydroxide vs chloride, so there's no indication how basic the solution would be once concentrated. (2) There are ways even for pure NaOH solutions: a) Cotton + Funnel works well, b) Crystalizing slowly and settling works also, and c) Even in the most extreme cases PTFE membranes with a water miscible solvent (to address the materials hydrophobicity), is another possible solution.

I'm writing this only to balance a remark that fully has its place on the topic.

For amateur chemistry, maybe the best bet for separating NaOH from NaCl in solution, is to use low-current electrolysis across a cation exchange membrane, that allows Na+ ions to cross the membrane, while drawing Cl- to the anode. Nurdrage has several YouTubes on electrolytic production of NaOH that might be helpful.

Low current means low ion transport. Also how does getting all the Na+ on one side and Cl- on the other solve the problem?


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Alternative explanation of ethene's pi bond
This tutorial about ethene's bonds has the following paragraph that I hope someone can paraphrase in a better way:

The pi bond dominates the chemistry of ethene. It is very vulnerable to attack - a very negative region of space above and below the plane of the molecule. It is also somewhat distant from the control of the nuclei and so is a weaker bond than the sigma bond joining the two carbons.

What does it mean by:

"vulnerable to attack"?
"negative region of space"?

Just get yourself a good textbook and read about orbitals and electro- and nucleophiles.

Here is the meaning of the words ""vulnerable attack" and "negative region of space".
The nucleus of the six atoms (2 C and 4 H) are situated in the same plane, which is usually defined as horizontal. The electrons of the 4 H atoms and the three first electrons from each Carbon atom are all together included in sigma bonds, and their maximum probability of presence is situated in this plan. These bonds are made of negative charges, that should be able to attract positive charges. But these bonds can hardly attract any arriving positively charged ion, because this attraction would be entirely compensated by the repulsion due to the neighboring nuclei.
The pi electrons are very different. Their maximum probability of presence is not situated in the plan. They make a sort of double long cloud high above and much under the plan of the molecule. They produce a sort of negative cloud (or a negative region of space) surrounding the molecule above and under the plan.
Electrons are negatively charged. So they may attract positive charges. If a positive ion like $\ce{H+}$ or $\ce{Br+}$ approaches an ethene molecule, it will be attracted by this negative $\pi$ cloud. If this approach leads to the formation of a new bond between the ethene molecule and the coming ion, it makes sense to state that the coming ion attacks the ethene. One may also state that the $\pi$ cloud attacks the coming ion. In other words, it also makes sense to say that the $\pi$  bond of ethene is vulnerable to attack, at least more vulnerable than the sigma bonds in the plane.


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"Isotopes" in a molecule?
I just learned that atoms of the same element can have different mass because of they are isotopes. If there are two isotopes in which each reacts to form the same thing. Would the mass differ in each of the two reactions? Is there a term for compounds just like the term isotope is for the atoms?

see:kinetic isotope effect

The "quick questions" without explicitly expressed solving effort are not very welcome. That may lead to hesitation to answer or even to closing a question.  If a question is asked on Chemistry SE site, the site policy expects authors to elaborate the question, e.g. by searching in textbook+online resources, writing what has been found, understood or tried and what is the stumble stone for the answer to be found. SE network aims for a  collection of  high quality answers AND questions.

It means that "I have just learnt that..." is wrong approach, implying you have not even thought about it much yet nor spent time to search for answers.

There is no term for that; we just say "molecules with such-and-such isotopic composition".

@Poutnik Sorry to say, but I actually did some google searches before. I did searched: "isotopes of molecule",
"can the same molecule differs in mass", and many others. Please stop making assumptions.

These molecule are called isotopologues.

@HilbertDJ Show then assumptions were wrong by what you have found. It is a basic topic in textbooks and there is plenty of available articles across Internet. Search rather by keyword set than by expressions that may not be used. For scientific topic, add "Wikipedia" keyword while searching, as it may bring you faster to  the topic. Wikipedia is full of articles, like Isotope or Isotopic labeling.  Isotopologues were  mentioned.

OK, I stand corrected, there is a term, after all. Now, I vaguely recall having heard that those little plastic or metal pieces wrapped around the ends of shoelaces also have a name, but I don't remember that either.

@IvanNeretin I have never heard about isotopologues until today either. :-)

It's kinda sad to see that a legitimate question is being downvoted.

If 2 molecules have different isotopes of some element their masses will differ. In this respect there are 2 different terms:

Molecular mass a.k.a. molecular weight (MW) - is an average mass of a molecule. If we take many molecules (a mole of molecules) of the same substance: some of them will have isotope A, others - isotope B. Their average is going to be MW*. It's used in calculations e.g. when we want to know the mass (in grams) of N moles of some substances. Or vice versa - given the mass (in grams) we'd like to calculate how many moles of there are.
Exact molecular weight - the mass of the molecule with isotope A or isotope B. These are the masses that will differ.

*Note that in atmosphere the isotopes may come in different ratios. E.g. $Cl$ comes as 2 isotopes ($^{35}Cl$ and $^{37}Cl$) - their ratio is 3:1. So MW is not the average of $35$ and $37$, it needs to take the abundance into consideration: $\frac{0.75\times35 + 0.25\times37}{2}$


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How does solutions of two liquids work?
When two liquids mix, and they are not hindered by a polarity mismatch (like oil and water), why don't they always form a homogenous mixture? Due to entropy, a liquid will disperse itself evenly. This even distribution will lead to homogeneity, no? According to an answer to this question (one given when this question looked different), said that they don't always distribute themselves evenly. So, let me lay done some ground rules:

Both the liquids are non-polar or polar
The liquids do not react in a way that creates products

Now, what hinders these liquids from dispersing evenly in each other?
Also, I'm going to clarify a certain conception I have, in case it is wrong:

A homogenous mixture is a mixture where the parts are evenly distributed. A solution is a homogenous mixture, where the solute has been deconstructed into its ionic components.

Ever heard of sugar? Is it a salt?

@IvanNeretin Hmm, I see. We have only been taught about solutions in the context of salts in class. It seemed like it needed to be a salt for a solution to occur.

I know, chemistry is commonly taught in a way that suggests it is completely detached from everyday life. Well, in fact it isn't.

There is no sharp boundary between being polar and nonpolar, so it depends on the quantitative properties like permitivity, dipole moment, ability to form hydrogen bounds or donor-acceptor bonds. Additionally, liquids may mutually react, releasing producs not soluble in them.

"If two nonpolar or two polar liquids are mixed together respectively, will the mixture always be a solution?"  The qualifier of "always" means he answer is NO!  There are oil, [non-polar liquids] when mixed do not form a solution.

What determines whether a mixture of non-polar liquids will be homogeneous or not? @Hal

It has already been pointed out that your text is incorrect. Saccharose (commonly known as sugar) is not an ionic compound yet it dissolves very well in water. It is not alone although I am too lazy to think up a non-carbohydrate example.
I’ll also point out that ionic compounds exist that are insoluble or poorly soluble in water, although some may rightly argue that their bonds tend to be more covalent than ionic.
As for two different liquids mixed together, the answer is: You can’t say a priori. (Or, in a highly logical reading of your question: no, the result will not always be a solution.) Many polar liquids will form a solution with many other polar liquids but in some cases the pairing will not. Conversely, a polar liquid such as ethyl acetate or acetone can form a solution with a practically unpolar liquid such as hexane.

Can you explain (or point to the explanation) why ethyl acetate forms a solution w/ non-polar hexane?

@StanislavBashkyrtsev I cannot. I can run the experiment and see the result but I cannot tell you any underlying calculatory reasons.


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What will a polyurethane resin not adhere to?
For making a mould out of the bicomponent polyurethane resin kit that I have - I am going to need to take out the mould from its form. The mould is what I am actually making from the bicomponent resin.
I have done this once and i know how hard it is to take out the hardened resin out of the form it was poured into.
Is there any material I can apply on the form's walls for a smooth release of the hardened resin ? Maybe some sort of tape or powder ?


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Must a desiccant be made anhydrous before it can be used for drying?
I'm writing about the treatment of methane gas, and in one part of the process, the methane gas is dried in a tank called a scrubber. Here, a desiccant called triethylene glycol is used.
Thinking about another desiccant, magnesium sulfate heptahydrate (Epsom salt), the water must be taken away through evaporation before the salt can be used for drying. The chemical must be anhydrous, so that it'll steal away all the water from the object in need of drying.
However, is this the case with triethylene glycol? Must it be made anhydrous before it can be used for drying, or is its attraction to water molecules great enough to begin with?

answering the general question, no, conc. $\ce{H2SO4}$ is a dessicant that does not need to be made anhydrous for it to have dessicating properties. Not too sure about triethylene glycol though.

What are you going to do with your triethylene glycol once it consumes as much water as it can? Throw it away? No, that's not how things are done...

@IvanNeretin How does my question imply that I'm under the impression the triethylene glycol must be thrown away afterwards? I know that the triethylene glycol is circulated and heated up after it has collected the water. This heating up makes the water molecules evaporate away, and then the triethylene glycol can be used again. However, my question is, before it is even used, must it be heated to the extent that all hydrates in the substance are removed, making it some kind of  anhydrous triethylene glycol? Also, triethylene glycol can become toxic waste if the benzene amount is too high.

So you already knew that it is used for dessication over and over, and dehydrated before each round. As to me, this very much answers your own question.

@IvanNeretin Okay, I'm bad at wording myself: Epsom salt is not a desiccant; not to begin with at least. It must be heated up, so as to let go of the heptahydrate. After doing this, the product is a very reactive ion that will suck out the moisture of something. Triethylene glycol is not an ion, nor does it seem like it is very lacking of hydrogens. So, my thought is, maybe it too needs to be heated up (before it has even been used once), which would create anhydrous triethylene glycol, whatever that would look like. Is this the case? Or, is it already hungry for water from the get-go?

I guess it is good to go as sold.

All desiccants will eventually saturate at some level of water. MAny can be effective desiccants with moderate water loads below the saturation level. Few need to be completely dry to work effectively. So, no, 'anhydrous' is not usually required.

@matt_black So it is basically just a continuum of hydration?

@A.Kvåle Yes. But there will be specific data for specific desiccants so hard to be entirely general about how well one works  at, say, 50% saturation.

@matt_black Oh okay. Also, whenever a desiccant is hydrated, will that amount of hydration be noted as (chemical) * (nhydrate), i.e. magnesium sulfate heptahydrate, where magnesium sulfate and "hepta" is the n-value? If so, saturated triethylene glycol would be written as "tryethylene glycole * nhydrate? The asterisk is supposed to denote the kind of dot that is used when the nature of the bond is unknown or something.

@A.Kvåle Not in general. those hydrates are for compounds that form specific crystalline hydrates of a fixed composition (most of which will be poor desiccants). Glycols don't since they are completely miscible with water. What a commercial glycol would specify is usually the % of water in the product.

In the oil and gas industry, glycols are used in a scrubber to remove water from natural gas [methane] for a number or reasons: water is very soluble in glycols, water is extracted into “dry” glycols due to equilibrium forces, and there is efficient contact between the gases and the liquid.  Another huge part of the stribber is the regeneration portion of the system, the stripper – getting to the point of your question.  Heat is applied to the water/glycol mixture after having been in contact with the “wet” methane, driving off the water and regenerating the sorbent, the glycol.  The glycol can then be recycled.  Although I have described this as a batch process, the actual process is continuous.
In an industrial setting, “throwing something away” is used as a last resort.
Below is a schematic of a glycol dehydrator and regeneration unit.

https://www.sciencedirect.com/science/article/pii/S2352484719301039#fig1

This is a good answer and my question was perhaps badly worded. I know that the triethylene glycol is circulating through the system, going through a heating process so as to evaporate away the water it had attained. Thing is, what I'm wondering about is, does the TEG need to be made anhydrous before it is even used? If I had completely normal TEG on me right now, would it work as a desiccant, or would it need to be treated like Epsom salt?

@A.Kvåle, Does the TEG need to be anhydrous before it ever used?  If you purchase pure TEG with no or minimal water, then It could be use.  It can be purchased from Aldrich as 99% pure.  https://www.sigmaaldrich.com/catalog/product/aldrich/110175?lang=en&region=US

Then again, when you have a factory with a huge scrubbing facility, you don't buy from Aldrich.

Correct, as a scientist in the laboratory Aldrich might be a place to start.  In a refinery TEG by the tank car load might be needed, The point of the Aldrich comment was to show that 99% TEG was available on the market and would not need dehydration to begin with.


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What is the meaning of the slope in a graded dose-response curve?
A graded dose-response relationship is shown below with the drug concentration on the $\log$ scale on the $x$ axis and the response on the $y$ axis :

According to this website:

Slope:  a useful parameter - when it is shallow, it suggests a greater chance of overlap between desired effects and side-effects, whereas a slope which is too steep suggests that it will be difficult to achieve precise control of the effects.

I want to understand how does the slope as $\frac {Response} {concentration}$ have the meaning mentioned above?

Since the drug concentration axis is logarithmic, even a small change in the nominal drug concentration can have a large effect if the slope is steep, so getting the correct dosage could be problematic. Accordingly, this requires care in drug concentration preparation and, in particular, it might make it harder to avoid overdoses.
For a shallow slope, a relatively large change in drug concentration may be needed in order to achieve the desired result. But side effects of the drug do not necessarily follow the same dose response relationship. So increasing dosage by a factor of 10, for example, might pose a much higher than factor of 10 risk of side effects.


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What is the equation for the reduction of water?
I'm a grade 12 student trying to understand redox reactions. What's the equation for the reduction of water?
The back of my textbook says: $\ce{2H2O + 2e- -> H2 + 2OH-}$
I get that this makes sense, but I have no idea how to derive it myself. Any help would be appreciated!

First of all: Reduction is just a fancy way of saying "gain of electrons".
The posed question can thus be reformulated as:
What happens to water, if it gains more electrons?
In order to answer this, we should have a look at the oxidation numbers of water: Since oxygen is more electronegative, it has an oxidation number of -II, while the two hydrogen atoms have an oxidation number of +I (Roman numerals).
As we can see, oxygen is already in its most reduced state. We know this, because oxygen has 6 valence electrons, with an oxidation number of -II this makes eight thus fulfilling the octet rule.
What can be reduced (i.e. gain electrons) here, is thus the hydrogen, going from +I to 0 (electrons are negatively charged, thus reducing the oxidation number).
Hydrogen with an oxidation number of 0 is the hydrogen molecule $\ce{H2}$. In $\ce{OH-}$, both atoms still have the oxidation numbers as in water. Now we just have to balance the reaction.

I see, this makes quite a lot of sense, thank you

Wait a minute. When you said that hydrogen is oxidized, doesn't oxidation mean the oxidation number increases? But when hydogen goes from +1 to 0, it's oxidation number is decreasing. So it it being reduced? Thanks

@SamLiu Yup, I edited the answer. Hydrogen is being reduced and oxygen is already in its most reduced state. Thanks =)

I get that this makes sense, but I have no idea how to derive it myself. Any help would be appreciated!

Good question because such questions were pursued by prominent physical chemists and Nobel laureates (Ostwald/ Nernst and Hoffmann).
The reduction of water can be tested visually and the apparatus to demonstrate it indeed named after Hoffman. Basically, it is an electrolysis of water experiment. What will see at the side connected to the negative terminal of the battery that it produces a gas, which is twice the volume as compared to the side connected to the positive terminal of the battery.
This gas which is very light, burns with a blue flame and it forms water again- so this must be hydrogen. Secondly, another interesting effect is that electrode region which is producing hydrogen becomes highly alkaline. If you put phenolphthalein indicator, the negative terminal becomes pink indicating an excess of OH(-) ions.

So three things can convince us with the following:
$\ce{2H2O + 2e- -> H2 + 2OH-}$
a) The electron rich electrode is the cathode in electrolysis cell. Water can gain electrons here.
b) The cathode becomes highly alkaline during electrolysis
c) The cathode produces hydrogen gas, which is twice in volume as compared to oxygen.
d) You can even calculate the number of electrons used to produce a given volume of hydrogen from the measured current and the time.

Hmm ok, I understand that this reduction reaction can happen in practicality. Is it just a reaction to memorize?

Yes, it is a fact, just like the Sun sets in the West.

In class, we learned that to balance redox reactions, we add water, hydrogen ions, and hydroxide ions in basic solution. This reaction can't really be derived by adding hydrogen ions to both sides, etc. This is just the way it exists?

Balancing redox reactions in acidic/basic medium has nothing to do with water reduction or oxidation. Don't mix them up.

@SamLiu  There is a lot of things needed to be memorized or looked up in chemistry. But if you go after understanding principles, the amount on things to memorize is much less. It may take time when you start to recognize some substance behavioural and other patterns.

Given the image I think you should describe the reactions at the cathode and anode.

As to the balancing of redox reactions with water, protons and hydroxide, look here: https://chemistry.stackexchange.com/questions/140462/balancing-a-reaction-where-dichromate-ion-reacts-with-ethanol-to-give-criii-an/140478#140478

Reduction processes as posed here, “rearranges” the atoms upon taking up two electrons.  One way you may have been able to “derive” or take a good guess is to consider the electrolysis of water, breaking water down into $\ce{H2}$ and $\ce{O2}$.  In this case there has to be both an Oxidation and a Reduction.  In the hydrolysis case, the $\ce{H2}$ is the product from the reduction process.  This is because the $\ce{H}$ in $\ce{H2O}$, formally has a 1+ charge. The $\ce{H}$ [both of them] in $\ce{H2}$ has a formal charge of zero.  The $\ce{H+}$ of water accepted an electron and now has a charge of zero – hence a reduction.  Now, knowing the reduction product, simply balance the reaction.


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Viability of reduction of oil paint binder and solvent flammability upon disposal via addition of water
As an artist, when disposing of a paper towel or cotton rag containing oil paint, turpentine, odorless turpentine substitute or mineral spirits at the end of a painting session, I generally pour water over the paper towel or cotton rag to reduce its flammability. Sometimes I rinse out the paper towel or rag in the sink before disposal to reduce concentration of the solvent. Does mixing water with the oil paint binder and solvent actually reduce its flammability? I understand mixtures can be categorized as suspension, colloidal or solution, and I would be curious to know whether or how this categorization pertains.

It really depends exactly on what you're mixing the water with. Soaking a rag or paper towel will reduce the flammability and is recommended by many safety organizations online (here is one). However, if you are washing off a large amount of solvents or flammables with water and the chemicals are immiscible in water, it can sit on the top of the water and still burn if exposed to some ignition source.
The best thing to do is to look at SDS or MSDS sheets if you are unsure what the best disposal procedure is for some chemical you are using. Additionally, a google search should offer additional information on disposal and hazards of the substances painters used.


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Why is Hydroxide OH negatively charged
I am confused as to why OH has a negative charge, between all of the bonds and the free electrons there are 8 electrons and there are 9 protons (8 from the oxygen and 1 from the hydrogen)
there are 9 protons and 8 electrons in this ion
where did my thinking go wrong

You are forgetting that oxygen also has two inner shell electrons besides 6 valence shell electrons. The charge is for the overall molecule. There are 10 electrons in this ion.

Consider the electron configuration of oxygen. For oxygen to be neutral, it needs 6 valence electrons. So that means that its $1s^2$ orbitals are filled. The valence electrons are in energy level 2: $2s^2$ , $2p^4$. Neutral oxygen has 6 valence electrons but it wants to follow the octet rule. Therefore oxygen wants two more electrons to fill its $p$ orbitals.
In the case of the hydroxide ion, oxygen is going to follow the octet rule. It will have three pairs of electrons surrounding it, and it will participate in a covalent bond with hydrogen, satisfying the octet rule:

If you look at the bond in this picture, you will notice that it changes color half way through. If you draw a circle around oxygen and a circle around hydrogen, splitting the bond in half, you can calculate their formal charges. Remember that a covalent bond is two electrons being shared by two atoms.

Now if we count the electrons inside each circle, we see that oxygen has 7 valence electrons surrounding it, and hydrogen has one.
The oxygen has one more electron than it has protons, so it has a formal charge of $-1$. Hydrogen has $1$ proton and $1$ electron, so its formal charge is $0$. That gives an overall charge of $-1$.

Now if we add another hydrogen it becomes neutral. This is because one of the lone pairs on oxygen can bond with the hydrogen, giving oxygen a formal charge of $0$.
So if we add everything up, there are actually $10$ electrons and $9$ protons in the hydroxide ion.


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Internal pressure, but from other variables
The internal pressure is derived by differentiating internal energy with respect to the natural variable in one term while holding the unnatural variable in the other term constant.
$$\mathrm dE = T\,\mathrm dS - P\,\mathrm dV \tag{1}$$
becomes
$$\left(\frac{\partial E}{\partial V}\right)_T = T\left(\frac{\partial S}{\partial V}\right)_T - P\tag{2}$$
Then, using the relevant Maxwell relation, it becomes
$$\left(\frac{\partial E}{\partial V}\right)_T = T\left(\frac{\partial P}{\partial T}\right)_V - P = \pi_T \tag{3}$$
This procedure is easily modified. For instance,
$$\left(\frac{\partial E}{\partial S}\right)_P = T - P\left(\frac{\partial V}{\partial S}\right)_P \tag{4}$$
Again using a Maxwell relation,
$$\left(\frac{\partial E}{\partial S}\right)_P = T + P\left(\frac{\partial T}{\partial P}\right)_S \tag{5}$$
This procedure can be cycled through the various variables in $H,$ $A,$ and $G$ to produce another six expressions (excluding the ones possible with the mole-chemical potential pair).
Do these expressions also have names like internal pressure? If so, what are they?

What is difference between natural and unnatural variables ? As I have not met such terms yet. Do you mean directly measurable versus computed ?

The only examples I can recall are not on your list and are isothermal compressibility $\displaystyle\alpha \equiv \frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_P$ and bulk modulus $\displaystyle\kappa \equiv-\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_T$

The natural variables of a function are those that, when held constant, maximize or minimize the function. For example, E is minimized at constant T, P, and u (chem potential). A is minimized at constant S, P and u, etc.

Isothermal compressibility and bulk modulus are completely different concepts.

There is an analogous expression for the enthalpy, $\left(\frac{\partial H}{\partial p}\right)_T =V- T\left(\frac{\partial V}{\partial T}\right)p$, but I don't know that it has a name. It is related to the Joule-Thomson coefficient as $-\mu{JT}C_p$

@Buck Thorn Exactly! I'm just seeing if anyone knows if these things have names. I realize they aren't as important as the internal pressure, but they might still have names.


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Identifying a meso compound
I am still a first year student; I was asked in the exam to identify the meso comound out of some choices. The answer key said that the compound below is meso. But I see no plane of symmetry here. Am I wrong or the question? I hope someone clarifies please. Sorry if the question was super silly.

Not all meso compounds have a plane of symmetry.

@Ivan Neretin Don't all achiral molecules have a plane of symmetry? Sometimes the plane is not obvious, but after some changes in conformations it is revealed. But i think this isn't the case here, right?

No. Indeed, many do, but many don't. That's not how chirality is defined.

So how is it defined?

@user716591  the Wikipedia entry is as good a place to start as any https://en.wikipedia.org/wiki/Meso_compound

This molecule has a center of symmetry.

You might want to look at this: https://chemistry.stackexchange.com/questions/48995/is-2r-3s-butane-2-3-diol-chiral/87415#87415

Place your hands palms down with thumb-to-thumb and index fingers touching. Plane of symmetry! Turn over one hand, place thumbs to index fingers. Center of symmetry! You have brought a "racemate of hands" together to form meso "compounds".

As the comments imply, having a center of symmetry makes a molecule achiral.  This is because, geometrically, inverting through the center equals a mirror reflection plus a 180° rotation.  If you start with an atom at point $(x,y,z)$ and you reflect it through the $xy$ plane, the atom is now at $(x,y,-z)$.  Now rotate 180° around the $z$ axis and you're at $(-x,-y,-z)$ just like inverting through the center.
Your molecule has no symmetry with respect to just a mirror reflection or just a 180° rotation.  (It might look like you could rotate 180° within the plane of the paper, but the stereochemistry is different.)  But it does have a symmetry with respect to the combination of the two, which is that center of inversion, so you can match the mirror image to the original by allowing the rotation.  That makes the molecule meso.


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Entropy of reaction from these givens?
Reaction that happens at 380 Kelvin
$\begin{align} \Delta S^\circ _{sys} &=\pu{-130 J/molK}\\
C_p(\ce{C2H4})&=\pu{43.6 J/molK}\\
C_p(\ce{C2H5OH})&=\pu{65.9 J/molK}\\
C_p(\ce{H2O})&=\pu{33.6  J/molK}  \end{align}$
I need to calculate the $\Delta S $ for reaction:
$$\ce{C2H4(g) + H2O(g) <=> C2H5OH(g)}$$
I really have no try, i can calculate it if i was given the entropies for each reactant/product and do products-reactant. but can someone give me the formula needed to solve this?
I'm given the system standard entropy and heat capacities only.
Thanks a lot in advance.

"380k Ds(sys)=-130. Cp(c2h4)=43.6. Cp(c2h5OH)65.9. Cp(h20) 33.6." makes no sense. Please format chemical formulae, physical quantities as well as their units and symbols appropriately. Please visit this page, this page and this one on how to format your posts better with MathJax and Markdown.

The reaction happens at 380K (temperature) and then the Cps were given in J/K*mol of the reactants and products, and im also given the system standart entropy which is -130.
that's all really. @andselisk

What is the meaning of Ds(sys) = -130 ? Unit ? What is the relationship or the difference between dS and Ds ? Why do you have to calculate something where you already have got the answer ?

Thats what i dont understand, they are given me the standart entropy for the system and i need the standard entropy for the specific reaction, i added a picture in the OP to make it clear of whats given @Maurice

For an isobaric process you can integrate the expression $dS = dq/T = C_p dT/T$ to compute the entropy change at another temperature. Since the heat capacities are assumed constant you can proceed as follows:
$$\begin{align}\Delta S_{\pu{380 K}}^\circ &= \Delta S_{\pu{298.15 K}}^\circ + \sum_i \nu_i\int_{298.15 K}^\pu{380 K} \frac{C_{p,i}}{T}dT\\&= \Delta S_{\pu{298.15 K}}^\circ + \int_{298.15 K}^\pu{380 K}\frac{dT}{T} \sum_i \nu_i C_{p,i} \\ &=  \pu{-130 J/molK} + \ln\left( \frac{380}{298.15}\right) \left( 65.9 - 43.6-33.6   \right)\pu{ J/molK}\\ &=  \pu{-133 J/molK}  \end{align}$$ where the stoichiometric coefficients $\nu_i$ are $\pm 1$.


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Why isn't standard Gibbs energy zero at equilibrium?

At equilibrium, everything that $G$ depends on is constant.
The Gibbs' free energy may have been given to you without really emphasizing that it is actually a function of temperature, pressure, and the moles of involved reactants and products:
$$G=f(T, p, n_i)$$ When we are at equilibrium, we really mean all of the following are occurring:

Thermal equilibrium
Pressure equilibrium
Dynamic equilibrium

And hence the  moles of reactants and products is steady/constant (but may react if disturbed).
But when all those things are constant, the Gibbs free energy simply isn't changing at all. If the variables on which $G$ is dependent are constant, $G$ (for a single state) is constant.

Standard Gibbs energy is basically Gibbs energy but at standard states i.e. $\pu{1 bar}$ pressure and substances at their standard states.
So this standard Gibbs energy is also a function of $T$, $p$, and $n$, which at equilibrium are constants. But still $$\Delta G^\circ=-2.303RT\log K$$ at equilibrium which clearly isn't zero.
Why isn't change in standard Gibbs free energy $\Delta G^\circ$ zero at equilibrium while $\Delta G_r$ is?

Standard Gibbs energy at equilibrium is a contradiction in terms.

What is your understanding of the term "change in standard Gibbs free energy $\Delta G^0$?"

@ChetMiller for a reaction $$\ce{aA + bB->cC + dD}$$ $$\Delta G^\circ = cG^\circ_c + dG^\circ_d - aG^\circ_a - bG^\circ_b$$ Basically it is the difference between standard Gibbs energy for products and reactants. As for $G^\circ$ it is the Gibbs Energy of the substances at standard state.

T, p, and n are not constant if the reaction is not at equilibrium, as they will change as the reaction proceeds towards equilibrium (except in the very rare case when the standard conditions happen to represent the equilibrium state)

@Andrew so why is $\Delta G_r$ at equilibrium zero then?

At equilibrium, those parameters are no longer changing. That's what equilibrium means.

So why is $\Delta G^\circ$ non zero at equilibrium?

@ Eyy boss. Apparently you don't understand the difference between $\Delta G$ which is zero at equilibrium and $\Delta G°$ which is the difference between the standard Gibbs formation energies of the products minus those of the reactants. At equilibrium the molar Gibbs energies are very different from the standard Gibbs energies. During the reaction the G values of the reactants are decreasing, and the G of the products are increasing. The equilibrium is reached when the G values of the reactants and those of the products are equal.

There are actually two different methods of arriving at the desired result.  They are very different, and this is the source of your confusion.  I will outline them for you later when I provide my answer.

METHOD 1
Suppose we were to start off with pure reactants in stoichiometric proportions in the standard state of 1 bar, and end up with pure products in corresponding proportions in the standard state of 1 bar.  The change in Gibbs free energy for this process is $\Delta G^0$, as you have answered in your comment to me.  To bring about this change, we use whatever equipment is necessary, typically using a van't Hoff equilibrium box and constant temperature compressions and expansions in cylinders with piston.  If we had assumed different values of the product and reactant pressures in the initial and final states than 1 bar, then the change in free energy would not be $\Delta G^0$, but rather $\Delta G$, with $$\Delta G=\Delta G^0+RT\ln{Q}$$.  If the values of the pressures in this final state happened to be the same as the partial pressures of the reactants and products in an equilibrium reaction mixture, the $\Delta G$ would be 0, and the Q would be equal to the equilibrium constant K.  So we would have $$0=\Delta G^0+RT\ln{K} $$
METHOD 2
Here, we start out with a mixture of various amounts of reactants and products that are not at equilibrium, and we analyze the variation in the free energy of the mixture as the reaction proceeds.  So, let $n_A$, etc. represent the number of moles of reactants and products at any time as the reaction proceeds, let $G^0_A$ etc. represent the free energies of formation of the reactants and products, and let $p_A$ etc. represent the partial pressures of the reactants and products at any time as the reaction proceeds.  So at any time, the Gibbs free energy of the reaction mixture will be:
$$G=n_A(G^0_A+RT\ln{p_A})+n_B(G^0_B+RT\ln{p_B})+n_C(G^0_C+RT\ln{p_C}+n_D(G^0_D+RT\ln{p_D})$$ with $p_A=\frac{n_A}{n_A+n_B+n_C+n_D}P$ , etc., and  $n_A=n_{Ao}-a\xi$, $n_B=n_{Bo}-b\xi$, $n_C=n_{Co}+c\xi$, $n_C=n_{Do}+d\xi$, with $\xi$ representing the extent of the reaction and P representing the total pressure.
If we use this equation to make a plot of G vs $\xi$, the equilibrium of the mixture will be determined when G passes through a minimum.  If we differentiate  this equation with respect to $\xi$ and set the derivative equal to zero, we find algebraically that this happens when $$RT\ln{\left(\frac{p_A^cp_B^d}{p_Ap_B}\right)}=RT\ln{K}=-\Delta G^0$$Work it out yourself and see.
So both very different methods give the same result for the final equilibrium.

So basically at $\Delta G^\circ$ is the difference between Gibbs energy of products and reactants had the pressure been $\pu{1 bar}$. But since the equilibrium is established at a pressure, say $p_t$, at standard state there would be no equilibrium, i.e. $Q\ne K$

Yes, that is correct.


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How to convert Excel comments to ChemDraw .cdx?
Not sure whether this might be a Superuser or StackOverflow question... but I hope some Chemist can help me with this.
I have an Excel table with cells which contain comments "by Cambridge Soft". I assume that these comments somehow contain the ChemDraw data of the structure. It seems to be ASCII-characters only.

How can I get these comments into a .cdx file?
I tried to save these comments as a text file, renamed it to test.cdx and tried to open it with ChemDraw... no success. I don't have ChemOffice available, which possibly might handle this.
So far, I haven't found any useful information about this on the web. The CambridgeSoft pages are not very informative.
Since these are quite a few structures I don't want to do this manually, but in an automated process. Python would be available. Any hints or ideas?

What is the source of the Excel file?

@Andrew, I guess it was an export from some database via Spotfire, but I don't know details.

Because you find these entries in a Excel workbook, it is likely that this spreadsheet was created with Chemdraw's add-in for Excel (company's demo) to embed chemical formulae:

(source).
to provide a bridge between the two worlds to display chemical libraries (promotional example, or manual chapter 13).  To use it, you need both access to Excel as well as ChemDraw.
The (part of the) string shown by you does not look like the typical content of the binary .cdx, nor the human readable .cdxml file which would be understood by openbabel, a program written in the intent to interconvert structure files used in chemistry (it may read .cdx; but read and write .cdxml) on a one-by-one basis (e.g., in a GUI), or programmatically (e.g., C, Python).
Maybe you may identify the author of the spreadsheet.  Or find support by the software's authors (for some years now by PerkinElmer) / add-in maintainers (e.g., on GitHub).  It is not obvious their format were readable by similar products (e.g., ChemAxon's JChem).

Thank you for your answer. I know OpenBabel. Yes, as you say, it doesn't look like .cdx or .cdxml. I don't have ChemOffice and this Excel-add-in. I have Exel and ChemDraw and I can copy&paste ChemDraw OLE-objects from ChemDraw to Excel, but this is different. I guess the origin of the Excel file was an export from some database via Spotfire, but I don't know details. I will try to find out more.

@theozh Some additional ideas were put into the chat, https://chat.stackexchange.com/rooms/3229/the-periodic-table.  Good luck!

thank you for sharing your thoughts and listing different options. My hope and idea was to have SMILES and ChemDraw OLE objects in the same Excel table. Apparently, the database has ChemDraw objects in it but no SMILES and I don't know how to get the ChemDraws out. I guess an unnecessary lengthy workaround would be to export to SDFs, then batch convert these SDFs into SMILES and/or InChI (for this I have a ChemDraw+Python solution :-), further generate CDX, but then the last missing bit would be how to get the CDX in the Excel table again...


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Obtaining potassium silicate from sodium silicate
Is it possible to obtain Potassium silicate ($\ce{K2SiO3}$) from Sodium silicate  ($\ce{Na2SiO3}$) without using expensive lab equipment such as furnaces etc.?
For example, Calcium silicate may be obtained using the following exchange reaction:
$$\ce{CaCl2 + Na2SiO3 ->  CaSiO3 v + 2NaCl}$$
Substituting calcium chloride with potassium chloride probably will not produce a similar result as potassium has different (less) electronegativity than sodium, though I am not sure if the exchange reaction takes place because of electronegativity (only).
(The aim is to be able to not only produce potassium silicate, but also separate it for example, as an aqueous solution without other solutes.)

Feel free to comment as I'm just learning and maybe asking a question that doesn't make sense.

Sodium and potassium silicates have the same solubility in water. You cannot obtain one from the other.

This idea doesn't make much sense, but I guess one could do it, by first precipitating hydrated SiO2 and then reacting it with hydroxide.

I believe aqueous potassium silicate solutions are available commercially if that helps?

Thanks @Hendrix13, but that doesn't help. The question was not about buying potassium silicate.


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Beer Lambert law when complex is tetragonal (Jahn-Teller elongated)
I wondered what modifications there would be to the Beer Lambert law to find the molar extinction coefficients for the solution if the solvent contained the ligands which bind to the metal centre of a metal compound (ie 2 ligands add along the z axis only) and the complex was tetragonal?

As far you can define a concentration, that is as far you can identify a molecular or heavily bond entity nothing change. Stoichiometry  always count of course, if you are following a process.

You must find a wavelength that only the complex absorbs, then its the same as normal


text:
What is the product of ammonia reacting with silicone?
I used a soft silicone hose to carry ammonia at ~70 °C. I was surprised to later find the hose had hardened.
A quick materials compatibility search for silicone had given me a lot of info but an "I" (standing for "Insufficient Information") for ammonia. I wrongly assumed it would be OK.
What happened? What was produced? Can ammonia also react with or leach out elements from glassware? Should I throw the hose away? Will it become brittle?

Do you mean +70 or -70 ?  As if you mean hot ammonia solution, it is quite weird idea what I would avoid at any cost. If you mean freezing cold liquid ammonia, than I bet is just temperature dependent mechanical properties change.  You have included too little context to avoid confusion and wrong assumptions of the readers. You prior weak/sctrong acid question is on the way to be closed as not focused, this may be closed from lack of details or clarity. Try to improveit  in both ways. The full scenario, its  background and detailed observations.

But preliminary guess is NH3 molecule is small, it can diffuse to silicone and chnage its properties.

@Poutnik: Approximately 70C. So it would be NH3 diffusion into the silicone? Is silicone notably inappropriate for this kind of use? Is there any risk of cracking? Do you think the same would happen if the gas had been cooled prior to passing through the tubing?


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Reference Request - bond lengths in PCl2F3
While writing an answer here, I needed references for bond lengths of $\ce{PF3Cl2}$ and $\ce{PF4Cl}$. The latter was easily found in reference (1), whereas I could not find exact references for bond lengths in the former molecule.

References (2) and (3) contain information about the Raman spectra and $\ce{F^{19}}$ NMR spectra, but do not contain information about bond lengths. Reference (2) has information about the bond lengths of $\ce{PCl3F2}$, $\ce{PF5}$ and $\ce{PCl5}$, but not for $\ce{PF3Cl2}$.
A reference for exact bond lengths in $\ce{PF3Cl2}$, similar to the bond lengths shown above, is appreciated.
References:

Leiding, J., Woon, D. E., & Dunning, T. H. (2013). "Bonding in $\ce{PF2Cl}$, $\ce{PF3Cl}$ and $\ce{PF4Cl}$: insight into isomerism and apicophilicity from ab initio calculations and the recoupled pair bonding model". Theoretical Chemistry Accounts, 133(2). doi:10.1007/s00214-013-1428-7
Griffiths, J. E., Carter, R. P., & Holmes, R. R. (1995). "Molecular Structures of $\ce{PCl4F}$, $\ce{PCl3F2}$, $\ce{PCl2F3}$, and $\ce{PF5}$: Infrared and low-temperature Raman Vibrational Spectra". Phosphorus, Sulfur, and Silicon and the Related Elements, 98(1-4), 11–31. doi:10.1080/10426509508036938
Holmes, R. R., Carter, R. P., & Peterson, G. E. (1964). "Molecular Structures of $\ce{PCl4F}$, $\ce{PCl3F2}$, and $\ce{PCl2F3}$: Pure Chlorine Nuclear Quadrupole Resonance and Low Temperature $\ce{F^{19}}$ Nuclear Magnetic Resonance Spectra". Inorganic Chemistry, 3(12), 1748–1754. doi:10.1021/ic50022a021

To complete your literature survey, include this publication cited at least 11 times so far: «Dichlorotrifluorophosphorane (PCl2F3): molecular structure by gas-phase electron diffraction and quadratic force field» by French et al. in Inorg. Chem. 1985, 24, 2774–2777, doi 10.1021/ic00212a014.  Despite the paywall, ACS's «In lieu of an abstract, this is the article's first page.» informs you about some key properties you search, values including ($2\sigma$ estimated) e.s.d.

How did I find this publication?  Because the English Wikipedia has an entry about it (don't overlook the Chinese including complementary information).
But how to know if Wikipedia indexed such a compound?  Way better than trial-and-error is to address this question to the Wikipedia Chemical Structure Explorer allowing to query there by structure or string, which may lead you to the corresponding article (in Wikipedia).  It is backed by a public database (GitHub repository) initiated by Ertl et al.
Source: Ertl, P.; Patiny, L.; Sander, T.; Rufener, C.; Zasso M J. Cheminf., 2015, 7, 10, doi 10.1186/s13321-015-0061-y, (open access).

Great answer, time to bookmark the chemical structure expolorer page :)