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Show the acid nature of hydrogen sulfide
We are studying the chalcogens at school. I should prove (show) the acid nature of $\ce{H2S}$ by writing two chemical interactions.
I am really sorry if the terms I am using aren't the right ones. It's the first time I've ever used this part of the language. Which reactions should I write, and why? I think the first one must be neutralization:
$$\ce{H2S + 2 NaOH -> Na2S + 2 H2O}$$

How about the reaction with sodium carbonate?

@Waylander, thank you for the response! I haven't seen this reaction in school, but it might be the reaction I am searching for. What are the reasons you wrote it? I mean what does it show?

Do you know how an (Arrhenius) acid is defined? That may also suggest an equation which can illustrate the acidic behaviour of H2S.

It will show that H2S will give a proton to carbonate, a reaction that is consistent with an acidic nature

@orthocresol, no. Should I read about it?

I suggest you do. It is difficult to describe something if you don't know what it is.

As orthocresol suggested, to illustrate "classic" acidity as it's understood on a school level you unquestionably want to start with Arrhenius definition, which attributes acidic properties to  any substance which increase concentration of hydronium $\ce{H3O+}$ ions in water.
That's said, the very first reaction you probably want to write is dissociation of $\ce{H2S}$ in water and demonstrating that it is a weak acid by showing the corresponding acid dissociation constants $K_\mathrm{a}$ (the larger the $\mathrm{p}K_\mathrm{a},$ the weaker the acid; data from [1, p. 5-87]):
$$
\begin{align}
\ce{H2S + H2O &<=> HS- + H3O+} &\quad \mathrm{p}K_\mathrm{a1} &= 7.05\\
\ce{HS- + H2O &<=> S^2- + H3O+} &\quad \mathrm{p}K_\mathrm{a2} &= 19
\end{align}
$$
Some reactions illustrating acidic properties of $\ce{H2S}$:
Reaction with sodium hydroxide
Note that the reaction
$$\ce{H2S(aq) + 2 NaOH(aq) -> Na2S(aq) + 2 H2O(l)}$$
is for complete neutralization (concentrated $\ce{NaOH}$ solution).
If the reactants are taken in 1:1 ratio (diluted $\ce{NaOH}$ solution), sodium hydrosulfide is formed:
$$\ce{H2S(aq) + NaOH(aq) -> NaHS(aq) + H2O(l)}$$
Reaction with sodium carbonate
Sodium hydrosulfide is also formed when saturated $\ce{H2S}$ solution reacts with $\ce{Na2CO3}$:
$$\ce{H2S(aq) + Na2CO3(aq) -> NaHS(aq) + NaHCO3(aq)}$$
Reaction with ammonia
Reaction with aqueous ammonia solution yields exclusively in ammonium hydrogen sulfide $\ce{NH4HS},$ even when concentrated ammonia solution is used:
$$\ce{H2S(aq) + NH4OH(aq) -> NH4HS(aq) + H2O(l)}$$
Ammonium sulfide $\ce{(NH4)2S}$ doesn't form as ammonium $\ce{NH4+}$ $(\mathrm{p}K_\mathrm{a} = 9.25)$ is a much stronger acid than hydrosulfide $\ce{HS-}$ by a factor of about $10^{19 - 9.25}\approx\pu{6E9}.$
So far the reactions between well-soluble compounds that were carried in aqueous solutions where Arrhenius definition works fine.
However, it cannot be used to explain acidity in non-aqueous media, where a Brønsted–Lowry definition of an acid comes in handy: the acid is a proton donor.
This can be illustrated by the reaction with liquid ammonia:
$$\ce{H2S(g) + 2 NH3(l) ->[\pu{-40 °C}] (NH4)2S(s)}$$
References

Haynes, W. M.; Lide, D. R.; Bruno, T. J., Eds, CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data; 97th Edn.; Taylor & Francis Group (CRC Press): Boca Raton, FL, 2016-2017. ISBN 978-1-4987-5429-3.


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How is antioxidant activity of carotenoids determined?
I am doing some research regarding carotenoids and beta-carotene. After extracting the sample, I have not been able to find a way to measure antioxidant activity of beta-carotene. I have $\pu{0.1mM}$ of DPPH (2,2-diphenyl-1-picrylhydrazyl) solution dissolved in methanol, but I just read that you cannot use DPPH to measure beta-carotene free radical inhibition. Is this true? How should I go about doing this?

You may want to read this: https://www.sciencedirect.com/science/article/pii/S0308814611006108

Also should read: http://dx.doi.org/10.1155/2013/351671.

As Buck Thorn and I commented elsewhere, the antioxidant activity of carotenoids has been determined using $\ce{DPPH}$ radical ($\ce{DPPH^.}$)(Refs 1 & 2). According to Ref.1, antioxidant activities of the isolated α- and β-carotene have been analyzed in five different concentrations of 0.001, 0.005, 0.01, 0.05, and 0.1% ($w/v$) using several assays: The metal chelation, $\ce{DPPH}$ radical scavenging, and superoxide scavenging activities. All of these assays have been showed almost similar results in terms of high activity at lowest concentrations. Thus, it is not true that $\ce{DPPH}$ radical assay cannot be use to measure β-carotene free radical inhibition.
For latest information on $\ce{DPPH}$ radical assay, read Ref.3. In general, following is a typical procedure for the assay:

The $\ce{DPPH}$ assaying of $\beta$-carotene can be performed using known antioxident (e.g., Trolox, ascorbic acid, etc.) as reference compound. The assay can be carried out in either alcoholic media (e.g., methanol or ethanol) or in an aprotic solvent (e.g., acetonitrile). To the degassed solutions ($\pu{2 mL}$) of antioxidant (with a final concentration from $0$ to $\pu{60 μM}$), were added $\pu{2 mL}$ of $\pu{140 μM}$ $\ce{DPPH^.}$ stock solution (70 μM $\ce{DPPH^.}$ final concentration). The solution was stirred at room temperature in the absence of light, and, after $\pu{60 min}$, the absorbance was measured at $\pu{517 nm}$ using an UV-vis spectrometer. The working $\ce{DPPH^.}$ concentration was calculated from the absorbance response at $\pu{517 nm}$ of the initial solution in the absence of antioxidant. The $\ce{DPPH^.}$ scavenging percentage or inhibition ($I\%$) was calculated from the measured absorbance using the following expression:
$$I\% = \frac{A-A_0}{A_0}\times 100$$
  Where $A$ and $A_0$ represent the absorbance at a given antioxidant concentration and the absorbance in the absence of given antioxidant, respectively.

References:

Lars Müller, Kati Fröhlich, Volker Böhm, “Comparative antioxidant activities of carotenoids measured by ferric reducing antioxidant power (FRAP), ABTS bleaching assay (αTEAC), DPPH assay and peroxyl radical scavenging assay,” Food Chem. 2011, 129(1), 139−148 (https://doi.org/10.1016/j.foodchem.2011.04.045).
Surashree Sen Gupta1, Mahua Ghosh, “In Vitro Antioxidative Evaluation of $\alpha$- and $\beta$-Carotene, Isolated from Crude Palm Oil,” Journal of Analytical Methods in Chemistry 2013, Article ID 351671 (10 pages) (http://dx.doi.org/10.1155/2013/351671).
Om P. Sharma, Tej K. Bhat, “DPPH antioxidant assay revisited,” Food Chem. 2009, 113(4), 1202−1205 (https://doi.org/10.1016/j.foodchem.2008.08.008).
The most recent reference for $\ce{DPPH}$ assay: Magda M. Becker, Gilvanda S. Nunes, Danilo B. Ribeiro, Francisco E. P. S. Silva, Gaëlle Catanante, Jean-Louis Marty, “Determination of the Antioxidant Capacity of Red Fruits by Miniaturized Spectrophotometry Assays,” Journal of the Brazilian Chemical Society 2019, 30(5), 1108-1114 (http://dx.doi.org/10.21577/0103-5053.20190003).

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Chemical equilibrium
When calculating equilibrium constant with partial pressures if there is a solid involved should we consider it in the total number of moles in mole fraction?

The phrase 'number of moles' is incorrect, it should be 'amount of substance'.

The answer is No ! Solid have no vapor pressures. And if they still have, this pressure is a constant at the equilibrium conditions. So it does not change between the beginning and the end of the reaction. So it is included in the equilibrium constant.

This answer is wrong. You should look up sublimation.

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How to calculate the number of protons in a sphere given the volume and the pH?
The spherical radius of a nano-droplet of a solution ($\mathrm{pH}$ of which is $6$) is given as $\pu{20nm}$.
I can find the molar concentration of $\ce{H+}$ and subsequently the number of $\ce{H+}$ ions, but I am getting a fractional number ($\approx 2 \times 10^{-2}$). This is obviously wrong as there should be at least $1$ proton in a compartment which has an acidic $\mathrm{pH}$. What am I doing wrong?

It's not wrong - it's just that there would be a 2% chance of presence of a single  H3O+ in one of such droplets if you, say, sprayed such solution.

@Mithoron : But how can the pH of all such droplets be 6 then? Or is it the average pH which is 6?

20 nm is so little that pH loses sense, it isn't even a phase, but some nanodroplet. One could say it's average pH if you had mist made of them.

Chemical equilibriums, similarly as temperature, are statistical thermodynamic properties of large scale systems. They are gradually losing sense for small enough systems. Imagine there is attribute of people with probability 0.01. If there is 10 people and none of them has such an attribute, does it mean it's probability is zero ? Of course not.

As pointed out in the comments, the concept of pH as a thermodynamic property of nanodroplets does not quite make sense unless the nanodroplets are in equilibrium (stable). If they are in equilibrium you can still apply the equation
$$\textrm{pH}=-\log_{10}(a_{\ce{H+}})$$ but now you have to be aware that the activity will be influenced by surface (and other) effects and the approximation $$\textrm{pH}=-\log_{10}(c_{\ce{H+}})$$ may not hold. But assume it does. Then as you point out, the average number of hydronium ions per nanodroplet is
$$\begin{align} n_{\ce{H+}} &=\pu{10^{-6} M}\cdot(\frac{4}{3}\pi\times\pu{20^3 nm^3}\times\pu{10^{-27} {(m/nm)^3}}\times\pu{10^3 L/m^3})\times \pu{6.022\times10^{23} mol^{-1}}\\&=0.020 \end{align}$$
Which is ok. That just means that on average only 1 out of 50 nanodroplets contains a solvated proton. Note by the way that one nanodroplet also contains about a million water molecules.
This is a real "rabbit hole" of a question, given all that is not specified, which is why I used sleight-of-hand and wrote that I am ignoring surface effects and assuming the particles are at equilibrium. It doesn't address how to prepare nanodroplets with the desired average proton concentration or what equilibrium means (are nanodroplets exchanging matter or are they isolated?). As noted in comments, composition (buffering molecules if any, salts and other ions including hydroxyl), droplet charge etc, are being ignored and you can expect statistical fluctuations in concentrations between different droplets. However if I stick to the assumption of equilibrium then I can say (for what it's worth) that ~2% of droplets have ~pH 4.3, and most of the rest are at neutral pH.

"... on average only 1 out of 66 nanodroplets contains a solvated proton" at a given time. If the solution is buffered, the pH will still be 6 as a time-average concentration (or activity) of protons. If the solution is not buffered, a pH of 6 is kind of meaningless even for larger volumes.

I'd flip that last sentence around. It means that $0.015%$ of the protons have a very low pH and the other $0.9985%$ are just plain water

@Poutnik I understand the argument, it is not simple. My point is that if you have a thermodynamicaly stable aerosol you can certainly define a temperature, as well  as the average concentration of anything in the aerosol. You can have fluctuations between the droplets, of course, but you can (ignoring the strict definition of pH in term of activities) simply express the H+ concentration as pH (as an algebraic manipulation).

I would rather say pH, similarly as temperature or chemical equilibrium are statistical phenomena of large scale. For small enough scales, all of them do not have sense anymore.For pH, it turns into probability of H3O+ occurence.

I agree it seems absurd to define a concentration when you only have 1 of something in a volume. Note you have fluctuations within segments in a bulk phase, too. And there is the ergodic principle. The catch is that in the bulk you have material transfer between segments, and surface effects disappear.

I mean both pH and T have some fuzzy border, below which they do not make sense.  I'd a ddople is so small a H3O+ occurence has probability reaching zero, pH stops making sense in this scale. Similarly, at much smaller scale,temperature stops making sense.

It all depends on what you mean with "stops making sense". You can still use pH 6 as shorthand for "average activity of H+ in the droplets is $1 \times 10^{-6} M$ - but yes this is not the strict definition of pH for a bulk sample. And yet, would you deny that in a droplet with 1 million water molecules you can define an activity for one H+ ion? All you need to consider is an ensemble of droplets?

On a nitpicking side, spherical radius isn't the same as radius of a sphere. The volume of the droplet is going to be even smaller.

@andselisk Ok, could you clarify that? You mean $V \ne \frac{4 \pi}{3} r^3$?

@BuckThorn Droplet is not a sphere, it can be envisioned as a cap of the sphere of the radius $r$; so volume of the droplet $V < \frac{4 \pi}{3}r^3$ and depends on the contact angle.

@andselisk ah, the assumption is that we are dealing with a sphere, but droplets are not usually spheres. Reminds me of spherical cow

@BuckThorn …in vacuum, yes, but that is exactly what spherical radius means. If you ever did an engineering drawing with a round-cap stick, its dimension is described with spherical radius.

@andselisk are you sure you don't mean "circular radius"? I've not seen "spherical radius" used in the context you refer to, or certainly not exclusively (memory might fail).

@BuckThorn OMG I forgot how you like to play ping-pong in the comments. No, circular radius is what you used (considering projection of the sphere a circle), and it's not the same as spherical radius. Seriously, have a look at Figure 4 in this paper or just search Google Images with "spherical radius" droplet.

@andselisk yes, I find chat rooms constricting  :-) ping-pong is more for me, allows me freedom to wander. But I don't get it, Figure 4 of the article you linked indicates that spherical radius is.... the radius of a sphere (as I would have expected)? Sure, the sessile water droplet is a slice of that sphere.

Anyway, I see what you mean, good point. Add that to the list of somewhat ambiguous aspects in the question.


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Software for generating a lattice from a unit cell for molecular crystals?
I am looking for a software package which can take a unit cell (and lattice parameters, etc.) and from that generate a crystal lattice. Ideally it will have functionality for orthorhombic, hexagonal, and other types of unit cells which are slightly less trivial than a cubic lattice.
Finally, it should work for molecular crystals.
I have seen the ASE lattice class (https://wiki.fysik.dtu.dk/ase/ase/lattice.html), but it isn't so clear to me how to define a lattice for a system when the unit cell is quite large.
As an example, I have a unit cell for a clathrate hydrate which contains 68 water molecules and I want to generate a lattice from this unit cell. How would one usually go about doing this?
My guess is that one way around this is to use some kind of MD package and just give the unit cell along with the lattice parameters and the type of lattice. Then, after applying periodic boundary conditions, somehow get the package to print out the coordinates of atoms beyond just the original unit cell. This feels a bit hacky to me, but if this is normal and someone knows a relatively simple way to do this, then let me know. Answers involving either open-source or purchasable software are acceptable, although open-source is always preferred :)

As an aside, it seems to me that it may not be too hard to just write code to do this for my specific cases, so if anyone has a good reference for generating the lattice given the type of unit cell, lattice parameters, and a set of Cartesian coordinates, that would also be helpful.

Simply work out the lattice vectors, and then any integer multiple of them when added to a atomic position will be the position of the an equivalent atom.  Forget cell angles and symmetry of the cell, for this all you need is the lattice vectors. https://chemistry.stackexchange.com/questions/94400/counting-body-centre-as-lattice-point-when-the-basis-consists-of-2-atoms/94473#94473 might be of help as well

If I understand the question correctly, VESTA (Free, available for Windows, Linux, macOS) does that. You can create a new structure from scratch by applying the crystallographic system and space group to the set of atomic coordinates, and then "grow" obtained unit cell as much as you want, and finally export to, say, XYZ format. I barely touched this workflow in one of my answers.

@IanBush I could be wrong, but I don't think this is true. The lattice constants are basis vectors for the unit cell and a different transformation matrix is required depending on what symmetry the unit cell is meant to have (e.g. cubic, orthorhombic, hexagonal, etc.). If you are aware of a reference where someone has posted the appropriate transformation matrix for each type of unit cell, then that would be very useful. Maybe these are matrices easy to derive, but I want a reference just to be confident I do this right.

Yes it is true, as long as you pick the appropriate unit cell. 3 non-coplanar vectors span a 3 dimensional space. Crystal unit cells span a 3 dimensional space. Make the lattice vectors the edge of the unit cell and you are sorted. Hexagonal might be the only confusing case, https://en.wikipedia.org/wiki/Hexagonal_crystal_family has a picture to show how it can be done.

In addition to the presentations about CCDC Mercury and Vesta by @andselisk here and later mine about Avogadro could be extended.  As described earlier, Avogadro's capability to read .cif files about crystallographic models allows you to extend these  via the slab builder to any extent you need (GUI entry Crystallography -> Build -> Slab) and to export them either as normal .xyz file, or (via the extension tab, template assisted) in one of multiple formats recognized by quantum chemical programs (e.g., Gaussian, GAMESS-UK, MOPAC).
It equally is possible to start from scratch with a cubic unit cell of 3 Å.  Subsequently, you would alter the lattice parameters ($a, b, c; \alpha, \beta, \gamma$) and enter atoms one, after the other, either in cartesian coordinates (either Angstrom, Bohr, nm or pm), or in fractional atomic coordinates as typically in .cif files. You may wrap the atoms into the cell, and of course impose one of the space group symmetries to fill your cell, too.  You may symmetrize your constructed unit cell and reset (within tolerances you determine) into a standard setting, too. Equally than just reading a .cif,.pdb, or .xyz, these constructed unit cells may be extended with the slab builder, and equally may be exported in one of the computer chemistry formats mentioned earlier, too.

I ended up downloading Avogadro, and it is extremely easy to duplicate the unit cell as needed. This can be done through the build tab as well. Slab seems to do something similar. I will accept this answer after a little bit.

You might also look into Force Field X 'Crystal' module (https://ffx.biochem.uiowa.edu/modules/crystal/index.html), unit conventions can be found here (https://ffx.biochem.uiowa.edu/properties-spacegroup.html).
Also, I feel it's important to mention that the code is open source and freely available, useful if you're interested in exactly how the calculations are performed: the procedure outlined by https://chemistry.stackexchange.com/users/1782/buttonwood is very similar, if not identical, to the implementation, IIRC.
Cheers!

Briefly, I thought about the OP's question to be a little related to how powder-XRD models may be solved, and the shake-and-bake approach for single crystal specimen (https://secure.hwi.buffalo.edu/SnB/) seen to describe larger molecules, too.  Since the program you point to seems to grow among crystallographers for biophysics (e.g., DNA, proteïns), and the concept of SnB entered, for example ShelxD, do you know if there is a similar cell / structure optimization present / planned in Force Field X Crystal, too?

There exists a really useful tool, already implemented, to optimize crystal structures, specifically proteins: https://ffx.biochem.uiowa.edu/commands/realspace.ManyBody.html
The publication can be found here: https://www.cell.com/biophysj/fulltext/S0006-3495(15)00676-1
There also exists a module to further refine structures (including the above-mentioned rotamer optimization): https://ffx.biochem.uiowa.edu/commands-realspace.html
Hope this answers your question, at least to some degree.


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Interpretting relative probabilities based on Boltzmann's probability distribution?
I understand that Boltzmann's probability distribution gives the relative probability of one microscopic state in comparison to the other (P1/P2) but I don't know how exactly to understand this.
For example when looking at Hydrogen's ground state and Hydrogen's first excited state at some temperature T, let's say that P1/P2 is 0.3 what does this mean in words? Does this mean that at whatever temperature T, P1 happesn 30% of the time? Or am I interpreting this incorrectly?

For the example you gave, a ratio of (P1/P0) equal to .3 would suggest that you have more hydrogen molecule in the ground state and specifically would have 3 in the excited state for every 10 in the ground state.


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Charactaristics of Certain Hydroxyl Protecting Groups
I have been reading through Philip J. Kocienski's book Protecting Groups to try to better understand some of the characteristics of hydroxy protecting groups. I am looking to understand the compatibility of silyl ethers, alkyl ethers, alkoxymethyl ethers, tetrahydropyranyl, methylthiomethyl ethers, esters, and carbonates with acidic conditions; however, my book is very vague on this. Which of these classes of protecting groups will be compatible with strongly acidic conditions (specifically, I will need it to remain intact in HBr for an addition reaction). I am just looking to know which categories are best for these conditions so I can narrow my search. So far, based on what I have read, the Troc protecting group would best suit strongly acidic conditions; however, it is highly electron withdrawling, which will likely affect an addition reaction I am looking to perform.

At what temperature will the HBr reaction be run?

I will be running the reaction at -10 C in a 30% HBr-acetic acid solution in toluene (an adaption of the following protocol: https://pubs.acs.org/doi/pdf/10.1021/ed072p848). All future steps are also performed at -10 C.

Plain old O-benzyl ether will probably stand that pretty well

Well "protective groups" are not an absolute way of protecting a moiety. It always depends on the reaction conditions you will use later.

I cannot use a benzyl ether since the product I have to deprotect (I've included an image above - the aliphatic -OH group will have the protecting group-) contains a benzyl ether moiety

Well you need to make a difference between "strongly acidic" and "strongly acidic with water".

It would be strongly acidic without water; however, it would involve a protic acid, acetic acid (anhydrous of course), and an alcohol, phloroglucinol.

@ELIJONES  The molecule you have drawn is a phenyl ether, not a benzyl ether. They react very differently.

Would a tert-butyl diphenyl silyl ether work in this case? The TBDPS ether has a half-life of about 244 minutes in 1% HCl. Would running a hydrobromination involving a 30% HBr solution in acetic acid diluted to half the initial volume using hexanes for 20 minutes at -10 C likely cause a drastic loss of yield? I understand there may be better protecting groups to withstanding these conditions; however, I don't want to have to worry about using harsh conditions to remove these protecting groups.


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Thermochemical data for thiocyanatoiron(III) ion
I am currently writing a lab report about the formation constant of $\ce{FeSCN^2+}.$ All I could find online was a formation constant at $\pu{25 °C}$ $(890),$ which was 70% higher than the experimentally measured value at $\pu{21.7 °C}$ $(292).$
I was hoping to use van't Hoff's equation to calculate the accepted value at $\pu{21.7 °C},$ and I found enthalpies and entropies of formation for $\ce{Fe^3+(aq)}$ and $\ce{SCN-(aq)},$ but nothing for $\ce{FeSCN^2+},$ so I can't calculate the enthalpy or entropy of reaction.
Does anyone have a credible source for this data?


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Kohlrausch Law and conductivity
Im writing up a lab report at the minute and part of it concerns ions in solution and how they affect conductivity. I've been doing lots of reading and I think ive got the grasp of most of it, but its left me with a few questions.

Symbols used:
Resistance ($R$), Resistivity  ($\rho$), Conductance ($G$), Conductivity ($\kappa$), Area of cross-section of the electrode ($A$), distance between the electrodes ($L$), Molar conductivity ($\mathrm{Λ_{m}}$)\, The molar conductivity of the electrolyte at infinite dilution ($\mathrm{Λ_{M}}^0$), The molar conductivities of the component ions at infinite dilution
  ($\lambda{_{M+}}^0+\lambda{_{A+}}^0$)\, Concentration (c)

$$\;
[1]\;\; R = \rho \frac{L}{A}\\$$
$$[2]\;\;R = \frac1G\\
$$
$$\;[3]\;G = \frac{{\kappa}A}{L}$$
$$[4]\;\;\kappa = \frac{GL}{A} = G\kappa_{cell}$$
$$[5]\;\;\kappa_{cell} = \frac{L}{A}$$
$$[6]\;\;\mathrm{Λ_{m}} = \frac{k}{concentration}$$
$$[7]\;\;\mathrm{Λ_{m}} = \mathrm{Λ_{M}}^0 - k\sqrt{c}$$
$$[8]\;\;\mathrm{Λ_{M}}^0 = \lambda{_{M+}}^0+\lambda{_{A+}}^0$$
So as I understand it,
Conductance is the ability of a component to conduct electric current.
Conductivity is the ability of a material to conduct electric current, regardless of its dimensions.
Molar Conductivity is the conductivity of an electrolyte solution divided by the molar concentration of the electrolyte, and so measures the efficiency with which a given electrolyte conducts electricity in solution.
Conductivity is related to the amount of ions in solution, the more ions, the higher the conductivity.
Molar conductivity varies with concentration, perhaps counter intuitively, as concentration increases the molar conductivity decreases. The molar conductivity also varies depending on the amount of dissociation (strong electrolyte/weak electrolyte).
The relationship between molar conductivity and concentration is caused by the asymmetric/relaxation effect and the electrophoretic effect.
Here are my questions:
1) For what condition do the equations [6] and [7] apply? I think that [7] is for strong electrolytes at low concentration though im not sure. So what about at high concentrations of strong electrolytes? What about weak electrolytes?
2) When would you use equation [6], when would you use equation [7]? Are they equivalent?
3) In what situation should you use equation [8]?
4) One of my questions is to calculate the concentration of impurities in tap water, using the conductivity measurement. How can I do that when I have more than one unknown in the equation?
Ive not been given a lecture on this topic so im struggling a bit here. Any help is greatly appreciated

For what condition do the equations 1 and 2 apply? I think that 2 is for strong electrolytes at low concentration though im not sure. So what about at high concentrations of strong electrolytes? What about weak electrolytes? 2) When would you use equation 1, when would you use equation 2? Are they equivalent?

Equation 1 defines the molar conductivity:
$$[6]\;\;\mathrm{Λ_{m}} = \frac{\kappa}{c}$$
It is generally applicable. You use it to compute the molar conductivity from experimental data (conductivity and molar concentration of electrolyte). Here $\kappa$ is the conductivity. For an ideal measurement cell and electrolyte it is defined as
$$\kappa = \frac{l}{RA}$$
where l and A describe the geometry of the cell. A more general definition is possible for an arbitrary geometry or sample composition.
Equation 2 is called Kohlrausch's law and sure enough was derived by Kohlrausch based on the behaviour of strong electrolytes, to which it commonly applies at low concentrations:
$$[7]\;\;\mathrm{Λ_{m}} = \mathrm{Λ_{M}^\circ} - k\sqrt{c}$$
You might fit this equation to concentration versus molar conductivity data to determine limiting molar conductivities $\mathrm{Λ_{m}^\circ}$, and from that obtain the limiting molar conductivities of the component ions. The parameter k (also obtained from the fit) may be modeled by applying the Debye-Huckel-Onsager theory.
For weak electrolytes one might invoke Ostwald's dilution law.

In what situation should you use equation [8]?

Equation [8] is the law of independent migration of ions:
$$[8]\;\;\mathrm{Λ_{M}}^0 = \lambda{_{M+}}^0+\lambda{_{A-}}^0$$
where the $\lambda$ are limiting ionic conductivities, properties of individual ionic species in a specific solvent at infinite dilution. As written the equation applies to a solution containing a single 1:1 electrolyte. More generally $\Lambda_{M}^0 = \sum_i \nu_i\lambda_i^0$, where the $\nu$ are stoichiometric coefficient of individual ions in the electrolyte solution. The equation allows prediction of the limiting conductivity from the electrolyte composition provided limiting conductivities have been compiled. You can also use the equation to parse limiting conductivities of individual components from data for different simple electrolytes.

One of my questions is to calculate the concentration of impurities in tap water, using the conductivity measurement. How can I do that when I have more than one unknown in the equation?

Conductivity measurements can be used to estimate the concentration of total dissolved solids in tap water. Typically standard conversions factors are used to convert from conductivity to TDS, the conversion factor depending on the water source, as explained in the wikipedia:

The relationship of TDS and specific conductance of groundwater can be approximated by the following equation:

TDS = keEC

where TDS is expressed in mg/L and EC is the electrical conductivity in microsiemens per centimeter at 25 °C. The correlation factor ke varies between 0.55 and 0.8.2.

There are also online calculators (such as from lenntech) that convert between TDS and conductivity. However if you have performed a water composition analysis you can combine this with tabulated conductivities for different ions, as the lenntech website explains:

The EC determination can be done in different ways. One possibility is the usage of an ion specific conductivity coefficient. This coefficient is listed in charts though for their usage it is necessary to have an exact water analysis because every single ion affects the conductivity. An example for such a chart can be found in the „Handbook of Chemistry and physics” 76th edition, S. 5-90“.

See also ISO document ISO 7888:1985 and links from the EPA.

Thank you that is greatly appreciated. There is more thing which I am struggling to explain as I have been writing about it, perhaps you could help me once again. I think I understand why molar conductivity increases with dilution, as I stated above it is due to ions having less mobility on account of the asymmetric effect and the electrophoretic effect. Now what im trying to understand is why this doesnt happen with conductivity? I think it is related to the amount of free ions per unit volume but i cant put my finger on it. Why does conductivity not decrease with dilution as well?


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How to convert pepsin into powder form?
I have pepsin powder which is now hydrated due to humidity. I want to convert it back to the powder form. How to do it?
I don't have access to a freeze-dryer .
I prefer any cheap procedure other than freeze-dryer .

What is "cheap"? A lab scale vacuum tray dryer is probably ~1,000 USD cheaper than a lab scale freeze dryer.

An old paper: https://doi.org/10.1021/ac60329a047 describes the cheap way of dehydrating serum. I guess that the low temperature, achieved with dry ice in acetone, prevents denaturation as the solvent is removed. The rest can apparently be done in a vacuum desiccator. As your sample is not solvated, but just wet, I would try the method on a small amount and check if reactivity is preserved

Try using a desiccator with a desiccant such as CaCl2 (cheap) or another -- see https://en.wikipedia.org/wiki/List_of_desiccants.

I accidently add some water in it... 20ml water in 100gm pepsin powder...

Do you need it for its enzymatic property or as nutrient?

as enzymatic property...


text:
Definition of NMR peak
In NMR a spectrum consists of noise and peaks. While steps, such as repeated scans or smoothing of the spectra can reduce noise some still remains. As such one step usually taken is to bin the data so that only peaks are chosen, usually with a threshold and finding maxima/minima.
This raises the question however, which is how do you define a peak. If you use a threshold a poor baseline can result in this failing, for instance if the baseline is above the threshold set.
Is there a widely used method/rule of thumb which is used to decide if a region contains a peak, or is just noise?

What helps in NMR is that for signals originating from a given molecule, the number of hydrogen atoms is an integer, and so signal strengths (integrals) should be approximately ratios of (fairly small) integers. If you have a couple of signals well above the signal to noise level, chances are that all signals originating from that molecule should be above the noise level (unless some signals are broadened significantly).

If your standard 1D spectrum still has a distorted baseline after zeroth and first order phase correction, something was wrong with the measurement. Unfortunately your "experts" at hand may very well not be aware of the problem, which can be any of a dozen.

A general term used in all analytical chemistry is the lower limit of detection (LoD), which is generally defined as 3 times the standard deviation of the baseline/blank. So if you calculate the $\sigma$ standard deviation of the baseline in question, and you have a signal that is over $3 \sigma$, you can generally accept that as a peak. This comes down to the so-called 68-95-99.7 rule, according to which there is only a 0.3% probability that a peak higher than the LoD is an artifact.

How do you work out the sd of your baseline?

To calculate the standard deviation of the baseline, you just have to extract a portion of your NMR spectrum corresponding to the baseline and calculate the sample average and standard deviation values.


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Order of reaction in adsorption process
(1) $\ce{PH3(g) <=>[W]PH3(adsorption)->[W]P(s) + 3/2 H2(g)}$
The decomposition of phosphine on tungsten surface at low pressure is a pseudo first order reaction. This is because the rate of the reaction depends on the partial pressure of phosphine and which depends on the surface area of tungsten covered by phosphine.
$\ce{r= k.(P_{PH3})= k'*(covered surface area of tungsten)}$
But I am not able to understand the pseudo term used for the order of this reaction.
At high pressure, this reaction follows zero order kinetics due to complete coverage of surface area of tungsten.
Rate at high pressure= k
(2) The formation of any gas at low pressure on the surface of tungsten due to adsorption is zero order reaction.
At low pressure, according to Freundlich adsorption isotherm, the extent of adsorption varies linearly with external pressure as:
$\ce{x/m = k*p^{1/n}}$, where $\ce{x}$ and $\ce{m}$ are the masses of adsorbate and adsorbent.
Thus, the extent of adsorption is independent of the concentration of gas. Which means this is a zero order reaction.
Do in anyway (1) and (2) contradict?
(3) Is chemisorption a pseudo second order reaction because of involvement of chemical reaction in the rate determining step?
as mentioned here


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Obtaining mercury from mercury(II) nitrate
I was thinking about obtaining mercury from mercury(II) nitrate. Is it possible?
I've tried putting chunks of Fe into it, I've got the logic from the voltaic table (where Fe is on the left side of Hg, so I think it would get it out of the compound). Can anyone please correct me and show me the right way? I've tried searching on Google, but found nothing.
I'm just a rookie in chemistry, and was interested in it when my teacher announced about a talent show which I'll perform some cool chemistry there (I'm going to perform using fires, I've tried the colored fires and the alcohol carboy). I'm also planning to make a mirror, if anybody has some ideas please comment below.

I personally don't think this question deserves downvotes. There was some sloppy terminology, but it's more or less fixed now. Besides, OP demonstrated knowledge of reactivity series and decided to consult with a broader audience before conducting a potentially dangerous experiment. This IMO deserves some respect.

I think it would be way safer for you to to make a mirror using the silver mirror test for aldehydes.

The process you described would be more appropriately called "reduction of mercury(II) to elemental mercury".
Unfortunately, the trick with iron likely won't work (something more inert like copper would be a better choice though).
Mercury(II) oxide is weakly basic, so mercury salts in general would easily undergo hydrolysis and form basic oxosalts in aqueous solution unless acidified.
Mercury(II) nitrate quickly hydrolyses to poorly soluble yellow mercury(II) oxide upon dilution or addition of alkali:
$$\ce{Hg(NO3)2(s) + H2O(l) -> HgO(s) + 2 HNO3(aq)}\label{rxn:R1}\tag{R1}$$
which can be converted back to nitrate by adding excessive amount of nitric acid, which in turn won't leave any chance for iron not to be oxidized, so you end up with both metals in solution.
I didn't want to dive deep into the discussion of how mercury(II) salts hydrolyze and leave it simple, but after receiving critique from Maurice claiming $\ce{Hg(OH)NO3}$ to be the “real” product, I think I allow myself to add a paragraph or two.
Raman spectroscopy and x-ray scattering studies in late 1960 demonstrated that hydrolysis of mercury(II) salts produces series of polynuclear oxo-bridged species of types $\ce{Hg2OH(H2O)2^3+},$ $\ce{Hg3O(H2O)3^4+}$ or $\ce{Hg4O(OH)(H2O)3^5+}$ [1, 2].
Formation of mercury(II) hydroxide nitrate $\ce{Hg(OH)NO3}$ as a hydrolysis product was taught in 1940–1950s era, and it stuck in several textbooks published later probably because it's listed in all editions of Pauling's General chemistry up to 1988.
However, not only it's oversimplified (granted, reaction \eqref{rxn:R1} is also an oversimplification in a sense it's a boundary case), but is also an incorrect concept.
The most up-to-date summary of what is really happening when $\ce{Hg(NO3)2}$ is dissolved in water can be found in Mercury handbook [3, p. 115]:

$\ce{Hg(NO3)2}$ solutions are only stable in the presence of a certain amount of nitric acid, which prevents hydrolysis.
  $\ce{Hg(NO3)2}$ quickly hydrolyzes in excess water and produces a precipitate of
  $\ce{Hg3O2(NO3)2 · H2O}$ or, when boiled in dilute solutions, forms mercury(II) oxide $(\ce{HgO}).$

Regarding the reduction of mercury(II), there are two pathways: $\ce{Hg^0}$ may be obtained from mercury(II) nitrate using a dry or a wet method.
Speaking of a dry method, the most straightforward way to obtain metallic mercury from nitrate is to heat $\ce{Hg(NO3)}$ in a distillation apparatus (b.p. of mercury is 357 °C).
Above $\pu{400 °C}$ it readily decomposes:
$$\ce{Hg(NO3)2(l) ->[\pu{400 °C}] Hg(g) + 2 NO2(g) + O2(g)}\label{rxn:R2}\tag{R2}$$
\eqref{rxn:R2} is a brutto-reaction; the nitrate first decomposes to the red mercury(II) oxide at lower temperatures (which, in turn, decomposes to the elements):
$$\ce{2 Hg(NO3)2(l) ->[\pu{350 °C}] 2 HgO(s) + 4 NO2(g) + O2(g)}\tag{R3}$$
Wet method suggest mild conditions and a reaction in solution.
For example, formic acid (also used in silver refining) being a strong reducing agent would cause a precipitate from ammoniacal mercury(II) nitrate solution.
Since nitrogen dioxide, mercury vapors as well as mercury salts and oxides are highly toxic, the reactions must be carried in a fume hood which makes it a poor fit for a talent show.
Considering the possible risks and your level or preparation (no offence), I'd strongly advise to take extreme caution doing mercury chemistry and avoid public demonstrations until you become more experienced.
Note: chemical reactions are adopted from [4, p. 310–312]
References

Cooney, R.; Hall, J. Raman Spectra of Mercury(II) Nitrate in Aqueous Solution and as the Crystalline Hydrate. Aust. J. Chem. 1969, 22 (2), 337. https://doi.org/10/b6t3h2.
Johansson, G.; Haugsten, K.; Rasmussen, S. E.; Svensson, S.; Koskikallio, J.; Kachi, S. An X-Ray Investigation of the Hydrolysis Products of Mercury(II) in Solution. Acta Chem. Scand. 1971, 25, 2787–2798. https://doi.org/10/bn5j2g. (PDF)
Kozin, L. F.; Hansen, S.; Kit, M. Mercury Handbook: Chemistry, Applications and Environmental Impact; RSC Publ: Cambridge, 2013. ISBN 978-1-84973-409-7.
R. A. Lidin, V. A. Molochko, and L. L. Andreeva, Reactivity of Inorganic Substances, 3rd ed.; Khimia: Moscow, 2000. (in Russian)

I'd not do it directly. If I remember correctly (and reading wikipedia + basic inorg. chem) the mercuric nitrate can be turned into mercurous nitrate treating it with dilute nitric acid. The mercurous nitrate can then be exposed to UV light and yield elemental mercury and mercuric nitrate. In either case: Making a "mirror" with mercury would require very low temperatures, since it just melts and runs off or it will amalgamate. Then there is as the facts on the ground - the compounds involved are very toxic, or hazardous or both... OP: Find a silver solution instead...

@StianYttervik Yes, sort of, but both conversion to mercury(I) nitrate and photodecomposition are reversible reactions, so someone should take care of eliminating mercury metal from the system. I agree that pyrolysis of mercury(II) nitrate is an unpleasant and dangerous experiment, I just listed it as the most straightforward one (I added a note on that, too). And I completely agree about your suggestion for OP to look at silver, thank you!

Thanks everyone, and also thanks to andselisk. Thanks for warning me about that, I am making the mirror with silver. But when I came across the mercury nitrate, I was so curious on obtaining the mercury. Since we've just learned about electrochemistry, I'm just wondering if the voltaic table stuff really works. But it turns out no. Thank you so much andselisk and everyone, not only that I know I was wrong, I learnt a whole lot of new stuff! I'll reconsider heating the mercury nitrate if it's dangerous, once again thanks!

Here is a possibly interesting safe path to try for a Zinc (or Aluminum) Hg amalgam, which can be used in organic synthesis (e.g., for the Clemmensen reduction) courtesy of the field of Hydrometallurgy. It starts by first placing a sheet of, say, pure Zinc in HCl to create some surface absorbed H atoms (or Al in NaOH, where purity is important to avoid toxic hydrogenated gases) in a fume hood. Remove and rinse with distilled water. Add the treated sheet of metal to a dilute solution of $\ce{Hg(NO3)2}$. This is expected to result in a product of Mercury(I) nitrate in dilute nitric acid. Reaction mechanics:
$\ce{H•(surface)⇌e-(aq) + H+(aq)}$
$\ce{Hg(2+)(aq) +e-(aq)->Hg(+)}$
$\ce{H+ + NO3- = HNO3}$
Apparently, the H• radical functional behaves (per its seemingly reversible formation reaction: e- + H+ = H• ) as apparently an (e-, H+) operating pair. See this source,  'Hydrometallurgy 2008: Proceedings of the Sixth International Symposium', p. 818, which cites a commercial leaching process, to quote Equation (5):
" PbS + 2 H• = Pb + H2S (5) "
which one could view functionally as:
$$
Pb(+2)S(2-) + 2 (e-, H+) = Pb + H2S (g)
$$
Further, per Wikipedia on Mercury(I) nitrate the action of light (or less safe, boiling) results in the liberation of mercury, to quote:
"If the solution is boiled or exposed to light, mercury(I) nitrate undergoes a disproportionation reaction yielding elemental mercury and mercury(II) nitrate:[2]
$$
Hg2(NO3)2 ⇌ Hg + Hg(NO3)2 "
$$
where the Hg(NO3)2 can be further attacked by hydrogen atoms.
Note, if an amalgam is not formed, one just has successfully removed Mercury from the nitrate.

Contrary to Andselisk's opinion, mercury nitrate $\ce{Hg(NO3)2}$ is not decomposed into $\ce{HgO}$ in water. Mercury nitrate is hydrolyzed into a basic salt, whose formula is $\ce{Hg(OH)NO3}$, and the equation is:
$$\ce{Hg(NO3)2 + H2O -> HNO3 + Hg(OH)NO3}$$
To get pure mercury from mercuric nitrate, Andselisk recommends heating their substance to 400 °C. This technique will produce huge clouds of nitrogen dioxide which is brown, toxic and corrosive. An easy way of avoiding this drawback is to mix thoroughly mercury nitrate $\ce{Hg(NO3)2}$ with an equal mass of sodium carbonate $\ce{Na2CO3}$, and heat this mixture. It will produce carbon dioxide $\ce{CO2}$ instead of nitrogen dioxide $\ce{NO2}$. The equation of the reaction is:
$$\ce{2 Hg(NO3)2 + 2 Na2CO3 -> 4 NaNO3 + 2 Hg + 2 CO2 + O2}$$
It should be mentioned that, above 380 °C, $\ce{NaNO3}$ will be slowly decomposed. So it is not recommended to heat to such a high temperature. The reaction starts at 150 °C.
This operation should be done in a retort. The mercury will condense in the spout of the retort, and may be recovered. But mind! Mercury is toxic. And all operations with mercuric compounds are forbidden by the law.

Regarding the products of hydrolysis, see my updated answer. As for the reaction of nitrate with sodium carbonate, it would be nice to see a reference. I'm also wondering why a retort is required as I have never been in an operating chem lab with a retort would be used. "all operations with mercuric compounds are forbidden by the law" – are you sure about that?


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Why does radiocarbon dating only work in nonliving creatures?
I understand how carbon dating works, though I do not understand why it doesn't  happen while a creature is living. Because while we are alive we still have carbon 14 in us, so shouldn't it work?

It should and does. It will correctly show your age as 0.

Even though this is a good question, I'm tempted to close this as a duplicate of How does radiocarbon dating work? Knowing the principles of the analysis would ultimately answer this question too, and that answer indeed does.

An error rate of about +/- 100 years is not all that useful in determining the age of a living creature.

@vsz, not everything that lives is a creature. Some plants live for thousands of years, but... Radio carbon dating will not tell you the age a living tree. Not even if you could read it with absolute precision. The radio carbon clock does not start ticking until the organism dies. (see Poutnik's answer, below.)

A general comment to integrate the answers below. The ratio between C isotopes is constant within the atmosphere. Decaying C14 is constantly replaced.

@andselisk Hm, It seems to me some answers like Buck Thorn's are better then in the original question. Some duplicate closures may cause damage, unless all answers are considered and eventually integrated.

There are plenty of good sources online explaining the principle behind radiocarbon dating. For instance, the wikipedia explains:

During its life, a plant or animal is in equilibrium with its surroundings by exchanging carbon either with the atmosphere, or through its diet. It will therefore have the same proportion of $\ce{^14C}$ as the atmosphere, or in the case of marine animals or plants, with the ocean. Once it dies, it ceases to acquire $\ce{^14C}$, but the $\ce{^14C}$ within its biological material at that time will continue to decay, and so the ratio of $\ce{^14C}$ to $\ce{^12C}$ in its remains will gradually decrease. Because $\ce{^14C}$ decays at a known rate, the proportion of radiocarbon can be used to determine how long it has been since a given sample stopped exchanging carbon – the older the sample, the less $\ce{^14C}$ will be left. [16]

Your question is somewhat subtle: why doesn't decay begin immediately once the isotopes have been incorporated into an organism? Can't measurements of part of an organism tell us when that particular organ or body part was formed?
The key limitation is explained in a document from a company (Beta Analytic) that provides biobased content / renewable carbon measurements:

Although carbon-14 is radioactively decaying away in the body, it is constantly being replaced by new photosynthesis or the ingestion of food, leaving the amount relatively constant.

But if a body part no longer cycles carbon (is in effect dead), then radiocarbon dating should be useful. One application, for instance, is in dendrochronology, the analysis of tree growth rings to assess age and location:

Dendrochronology is useful for determining the precise age of samples, especially those that are too recent for radiocarbon dating, which always produces a range rather than an exact date, to be very accurate. However, for a precise date of the death of the tree a full sample to the edge is needed, which most trimmed timber will not provide. It also gives data on the timing of events and rates of change in the environment (most prominently climate) and also in wood found in archaeology or works of art and architecture, such as old panel paintings. It is also used as a check in radiocarbon dating to calibrate radiocarbon ages.1

Combining tree samples of known age (thanks to dendrochronology) and other sources with radiocarbon ratios provides a calibration method.

The metabolism of living creatures keeps the dynamic equilibrium of their $\ce{^{14}C/^{12}C}$ ratio with the enviromental $\ce{^{14}C/^{12}C}$ ratio via photosynthesis, breath,  food and excrements.
It is not just about one time building, but also about continuous recycling of the body content.
Such an equilibrium means continuous resetting of the $t=0$ on the radiocarbon dating timescale.
After death, this dynamic equilibrium freezes and the $\ce{^{14}C}$ slowly decays according the known exponential curve.
Due minor possible fluctuations of the initial ratio, the method is not usable for short time periods without the initial ratio context.
Additionally, measurent errors affect the short time periods the most.
See more at
Wikipedia - Radiocarbon dating
Wikipedia - Radiocarbon dating considerations

While we are alive, we are constantly replacing our molecules. With every breath cycle, you get rid of some of your carbon (CO2). With every meal, you take in new carbon (carbohydrates, proteins, lipids). The carbon in your body gets cycled as your cells die and are replaced.  This replacement process stops when you die.
After this, whatever carbon remains will decay and can be used to estimate age... or really, time since the organism stopped exchanging carbon.

this was really helpful!


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What happens when Ar and HCl are mixed at 77 K?
So our teacher asked us a simple question in the classroom:

What happens when Ar and HCl are mixed at 77 K?

and no other information was given.
We replied various possible answers for that such as Debye bond will form between Ar−HCl, force of attraction is proportional to the inverse sixth power of Ar−HCl intermolecular distance, Ar has its octet filled, it separates into a different layer.
Though our teacher said all of them are wrong, but I am facing some trouble understanding how the other students get there. Could anyone help me with some hints?

What do you suppose the boiling points and melting points of HCl and Argon are?

Well, we asked if we could take 159k as the melting point, but he said that we don't need to go for the melting or boiling point

https://pubs.acs.org/doi/10.1021/j100382a027

@NilayGhosh The article isn't free, could you help me with something else

What’s the pressure?

Check e.g. https://en.wikipedia.org/wiki/Argon_compounds#Diatomic_van_der_Waals_molecules

At 77 K, both HCl and Ar are solids. So if you cool those gases separately and mix the two solids, you obtain a simple mixture of two powders.
If you mix Ar and HCl at room temperature, you obtain a mixture of gases. By cooling this mixture slowly, HCl will first liquefy at -83°C (density 1.194), then solidify at -112°C. Later Argon will liquefy at -185°C in a liquid heavier than HCL crystallized, with density 1.4, then solidify at -189°C in a solid (density 1.65). You will get two solid layers : HCl above Ar. But there is no chemical reaction between HCl and Ar

What if the pressure is very low so that they are each below their equilibrium vapor pressures?


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Can Michaelis-Menten constant be zero in the Michaelis-Menten Equation, or is this a non-zero constant?
Can Michaelis-Menten constant be zero in the Michaelis-Menten Equation, or is this a non-zero constant?
I am applying the equation in epidemiology so I wanted to assume that it is zero but I am not sure if it is possible for this constant to be zero?

This would kinda defeat the purpose of the model. Short of that, sure, why not.

This is analogous to asking: Can the rate constant for a reaction be zero?

It does not make sense for the constant to be zero. The constant is usually expressed as follows:
$$ K_M = \frac{k_{\mathrm{reverse}} + k_{\mathrm{catalytic}}}{k_{\mathrm{forward}}} $$
Since the $k$ values are strictly non-negative, the constant would only be zero if $k_{\mathrm{reverse}} = k_{\mathrm{catalytic}} = 0$. In theory, $k_{\mathrm{reverse}}$ could just be $0$ (but that would not be called a Michaelis-Menten model if I want to nitpick), but if $k_{\mathrm{catalytic}}=0$ then you have no overall reaction at all.
Another way to see the same thing is to recall that the Michaelis-Menten constant has the same numeric value as the substrate concentration at the half of the maximum reaction rate. Can this be zero? It would not make any sense to me.
In any case, if you would extend your question to contain information about the epidemic model you would like to create, I am sure a lot of people (including me) would love to help finding a good reaction kinetics analogue for it.


text:
What is the scientific term to describe the operation of a bong?
What is the scientific term for the separation of dust particles from air by suction through a specialized flask in a way similar to the operation of a bong?
The thing that led me to this was a vacuum cleaner that uses water to separate dust and other particles from the air by vacuuming the air through a water bowl, and the only thing that is similar to that operation is the bong. It is like sparging, but the gas coming in will contain particles and those particles are the ones that remain in the liquid, and that same gas is the one coming out, filtered to some degree from those unwanted particles.

I think the term describing the initial process of separation of the solid phase (dust particles) from the gas phase via bubbling is a wet scrubbing process.
On an industrial scale, it's more efficient to spray liquid phase, whereas on a laboratory scale bubbling in a compact glass gas scrubber or a gas washing bottle is more convenient.
In order to increase efficiency of the lab apparatus, it's preferred to

keep the surface between gas bubbles and liquid phase as large as possible;
increase time of contact.

This is usually achieved by producing many smaller bubbles by using a spiral tubing (larger particles), frit of long columns filled with Raschig rings.

Figure 1. Left: Büchi miniScrubber, a glass gas scrubber (Source: Büchiglas — Gas scrubber). Right: Corning PYREX® 125mL Gas Washing Bottle with Extra Coarse Fritted Disc (Source: Thomas Scientific)
Eventually collected insoluble dust particles tend to settle out from the solution; this a basis for sedimentation, another separation technique.


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Do colligative properties depend on the number of solute molecules?

Colligative properties are properties that are affected by the number of solute molecules dissolved in the solution. True or false?

I think the statement is false, and the correct one should use "number of solute particles". I asked my TA and he said the molecule version is true, although I show him an example that $\ce{H2CO3}$ can dissolve water in many ways to create different sets of miles for $(\ce{CO3^2-},$ $\ce{HCO3-},$ $\ce{H+}).$ In this case, one mole of molecules $(\ce{H2CO3})$ can result in different total numbers of particles.
However, my TA said that $\ce{CO3^2-},$ $\ce{HCO3-},$ and $\ce{H+}$ can be considered as different molecules. I still don't satisfy with his answer though.
Could someone explain it for me, whether I am right or wrong.

It's not good notation on the TAs part. IUPAC defines a molecule as having to be electrically neutral. I see what they are intending, but solute particles would be more correct. This is why, as you noted, compounds that dissolve into ions produce a greater effect than the number of original molecules would suggest and molecules that aggregate in solution have a smaller effect. Equations for colligative properties depend on the van't Hoff factor, which essentially converts from the original number of molecules to the number of distinct particles.

True. Colligative properties depend only on the number of solute particles, not on the nature of the solute particles. This results in vapor pressure lowering, boiling point elevation, freezing point depression and change in the osmotic pressure.


text:
Is PCl5 really trigonal bipyramidal in shape?
We know that the structure of $\ce{CH4}$ is tetrahedral, and not planar tetragonal, as a tetrahedron enables maximum distribution of $\ce{H}$-atoms in space, as they repulse from each other. This is proved by the fact that each $\ce{H-C-H}$ bond angle is equal in magnitude.
In the case of $\ce{PCl5}$, why should the shape be trigonal bipyramidal? Obviously the top and bottom $\ce{Cl}$-atoms, above and below the plane subtend $90^\circ$ with each of the $3$ planar $\ce{Cl}$-atoms; whilst planar ones subtend $120^\circ$ in between each other. This means all bond angles aren't equal. Why so? Shouldn't there be regular distribution of $\ce{Cl}$-atoms in space?
To my opinion, the XY-plane distribution of $\ce{Cl}$-atoms should be such that each $\ce{Cl-P}$ is $72^\circ$ apart from the other $\ce{Cl-P}$ bond (atoms would not be in the same plane at all. But we should see that the orthogonal projection of each on XY-plane subtends equal angles, i.e. $72^\circ$). Same for YZ-plane. Hence there is regular distribution of atoms along all planes. Now it is to be calculated what the ultimate structure results in.
Sorry, I don't have the necessary tools/knowledge to represent my hypothesis. Else could've just modelled this figure and shown a representative model.

The figure you described is simply not there. People with necessary tools have thought of this before.

See, there are two ways to get anything done: either do it yourself, or have someone do it for you. Those who are qualified to do it for you are telling you that it can't be done. Can you do it yourself?

https://math.stackexchange.com/questions/185441/5-points-uniformly-placed-on-a-sphere

Is your proposal in accordance with experimental results?

The trigonal bipyramidal shape of $\ce{PCl5}$ is indeed the lowest energy conformation.
I have had a very hard time figuring out what your description of the structure meant and I am still not convinced I have actually understood correctly. However I was trying to model it in my head I either arrived at a variety of different angles, or at least two of them were equal to $72^\circ$, or both of these cases.
The only structure $\ce{ML5}$, where all $\ce{L-M-L}$ bond angles are equal is a regular pentagon, i.e. all ligands are in one plane, with $\angle(\ce{L-M-L})=72^\circ$, e.g. $\ce{[XeF5]-}$.
Another possibility for (almost all) equal bond angles is a square pyramid, where $\angle(\ce{L-M-L})=90^\circ$, e.g. $\ce{ClF5}$. Note however, that there are also other angles, so not all bond angles are equal, $\angle(\ce{L-M-L})=180^\circ$.
There is no better solution to arrange 5 point 'uniformly' on a sphere than the trigonal bipyramid. This is also known as the Thompson Problem, which is addressed in the linked question on Mathematics.se, provided by orthocresol: 5 Points uniformly placed on a sphere.
Another point of view would be to ask the question whether there is a way to put 5 points on the surface of the sphere so that they are indistinguishable, which is also answered there. Spoiler warning: there isn't.
You can dive deeper into the matter. For example for $\mathbb{R}^n$ there can only be $n+1$ points equally equidistant to each other. In our perceived reality $n=3$, which makes the tetrahedron the most complex molecule with this property. See on mathematics.se: Proof maximum number of equidistant points for a given dimension (duplicate), Prove that $n+2$ points in $\mathbb{R}^n$ cannot all be at a unit distance from each other (duplicate), $n$ points can be equidistant from each other only in dimensions $\ge n-1$? (Which are all duplicates, but still have answers attached to them.)
And then there are only five Platonic solids, none of which has five vertices.
Furthermore, I would like to add to all of that another word of caution: Molecules are flexible. The trigonal bipyramidal arrangement of ligands just happens to be the low energy conformation for a variety of reasons (considering the necessary and useful approximations, e.g. Born-Oppenheimer approximation).
The molecule (and other analogues) are special though, as the undergo pseudorotation according to the Berry Mechanism (see Wikipedia).

Side note:

We know that the structure of $\ce{CH4}$ is tetrahedral, and not planar tetragonal, as a tetrahedron enables maximum distribution of $\ce{H}$-atoms in space, as they repulse from each other. This is proved by the fact that each $\ce{H-C-H}$ bond angle is equal in magnitude.

This is only half the truth. It is certainly true that protons repel each other, more importantly though is that electrons repel each other, too. The most important part of which is also built into the very simple VSEPR theory. While this theory has its uses, there is a lot of criticism attached to it and one should be aware of that. I shall stress again at this point that VSEPR theory is unrelated to orbital hybridisation; there are quite a few textbooks which do make this post-hoc rationalisation, which can be misleading and dangerous.

The shape of any molecule may be obtained by applying the VSEPR model (Valence Shell Electron Pair Repulsion). In this theory the angle between the bonds must be as large as possible. With your idea, the angles between P-Cl bonds are equal to 72°. It is not very much. In the bipyramidal model, the angles are 90° and 120°. This is much better than with your geometry.


text:
What determines the atom or molecule that receives electrons during oxidation?
I'm no chemist, so bear with me.
From my understanding, the reason metal oxidation occurs is because the metal is losing electrons to the oxygen ions in the air; and that an ion is an atom or molecule which either has more or less electrons than the "standard" of that atom/molecule; which is why it's "taking" electrons from something else.
What prompts my question has to do with soldering. I'm learning that one reason for the use of flux is to prevent this oxidation and protect the soldering tip. So according to what I said before, a soldering iron becomes oxidized because it has lost electrons to the air, and the high temperatures hasten this process.
Now to get to the detailed question: I understand the very basic oxidation process. But, once the metal has lost electrons, doesn't that now make it composed of ions? If so, Why does the metal continue to corrode? Why isn't the metal now taking electrons and virtually "restoring" itself? What is it about oxygen ions that make them the ones that take electrons from metals such as copper? Why isn't copper the one taking electrons from the air? Does it have to do with valence electrons or electron shells or something?

When a metal like iron is oxidized, each iron atom loses 3 electrons. These electrons are kept by oxygen atoms from the air. The iron atom becomes an ion charged 3+, and the oxygen atom becomes an ion charged 2-. The positive ions still attract negative ions, and they will never separate. Two iron ions and three oxygen ions are necessary to make a new compound which must be not charged. This assembly is a new substance called rust, which has a brown color. Once such a transformation has happened, the next iron atoms can repeat this operation : the amount of rust will slowly increase.
All metals can be oxidized by more or less the same way. But it could be quicker or longer.  For copper, this reaction is not so frequent. Copper makes a copper oxide which is black, but you may wait for months to see it.

Patina isn't black and rust isn't Fe2O3. BTW treating such compounds as bare ions was what get OP confused in the first place.


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AX Spin system ordering
I would like to ask a question about the ordering of the $\ce{X}$ spins on the right hand side of an $\ce{AX2}$ energy diagram.
I was given the diagram here below:

and asked to evaluate why this diagram results in a triplet. This is simple, since there are 4 possible energy transitions, but one of them is equivalent to the singlet transition, hence only 3 transitions would appear in the NMR spectrum.
Prior to this though, I was confused because I thought the ordering for the bottom 3 spins was incorrect. Taking $\ce{\alpha}$ as the lower energy spin, I thought the bottom 3 spins should read as:
$$\begin{align}&\beta \beta \\
 \beta \alpha &~~~~ \alpha \beta \\
 &\alpha \alpha \end{align}$$
and the energy levels would be drawn more appropriately.
I attempted but failed to mention this to the lecturer during the presentation.
Am I correct in this instance?

Neither of those diagrams make sense afai can tell. How can you have a transition that doesnt flip any spins?

@BuckThorn I questioned whether the order of the bottom 3 spins was incorrect but I did think about the selection rule and why that was not obeyed in this slide

@BuckThorn it is flipping the A spin. The A spin state is on the left. The X spin states (which don’t flip) are on the right.

@orthocresol Gotcha, I misread the diagram.

@BuckThorn that was a minor point, though. You are quite correct that the diagram is wrong in many ways. It is not a physically accurate description of coupling. The diagram implies that the splitting between the peaks in the triplet has a frequency on the order of an X spin flip, which is completely... well... off.

There are a few things that seem to be wrong with the diagram you present, but since your focus is on the relative energies of the $\alpha$ and $\beta$ spin states I'll address only this. If you check a reliable source such as Cavanagh et al.s NMR textbook [ref 1] it is explained that

the state with $m=+\frac{1}{2}$ is referred to as the $\alpha$ state, and the state with $m=-\frac{1}{2}$ is referred to as the $\beta$ state.  If $\gamma$ [the gyromagnetic ratio] is positive, then the $\alpha$ state has lower energy than the $\beta$ state.

Therefore for protons $\alpha$ is lower energy (since $\gamma>0$). If the nucleus is $\ce{^{15}N}$ ($\gamma<0$), $\beta$ becomes the lower energy state.
So while your concern is valid, it is a relatively minor flaw (in my opinion) with the diagram. Good for you for noticing it.
Reference

John Cavanagh, Nicholas Skelton, Wayne Fairbrother, Mark Rance, Arthur Palmer, III. Protein NMR Spectroscopy, Principles and Practice, 2nd Edition. Academic Press, 2006, ISBN: 9780121644918.


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The "nose" of the periodic table
My teacher said that on the periodic table there is a "nose" formed by Al, Zn, Ag, and Cd. She said that they are all fixed charged (+3, +2, +1, and +2 respectively), and said that if I write them in ionic equations, I just say Silver Nitrate instead of Silver (I) Nitrate. She also said to put all Al as +3 charge in all cases, etc. But I did some research and found out that you do say Silver (I) Nitrate (actually both are fine, but my teacher specifically said never put the (I) in such cases). So is my teacher wrong? And are there cases where the charges of the elements are not what my teacher says?

Saying "silver (I) nitrate" is certainly not wrong, just redundant. Yes there are cases where Ag has charge other than +1, but you probably won't ever encounter them.

So when does Ag have a charge of something other than 1?

Because I put the (I) on the test and the teacher marked it wrong

@BadAtChemistry "Silver(I) nitrate" is definitely not wrong and all of the elements have at least one more oxidation state, zero. As for the noses, tails and other mnemonics used to remember stuff I find them counterproductive and sometimes even harmful.

For regular chemistry, there are not such cases as simplification. For advanced chemistry, there are such cases, like silver (II) or silver(III). Search for advanced resources of element chemistry.

See e.g Ag4O4,( Silver(I,III) oxide), AgF2 or KAgF4.

In probably ten minutes of SciFindering, I could find an exception for each of those elements. But as you have a teacher you’re probably at school so two very old rules apply: 1) you are learning very simplified concepts; and 2) the teacher is always right.

@Jan 1) is true, but should never lead to teching supersimplifications, and 2) is, sorry, the doctrine of idiots trying to fabricate more idiots. Despicable. Sorry if I overlooked some irony tags. ;)

@Karl 2) The fun thing is, it’s both! But essentially, if OP wants top grades in something where scientific consensus does not clearly overrule the teacher’s opinion, or where OP cannot back up their answer by relying on other teachers in the same school, it’s the teacher’s discretion what they accept as correct answers and so asking a random lot of people on the internet whether the teacher is wrong won’t work. (Also, there’s an old German joke of teachers being right in the morning and free in the afternoon xD)

@Jan But as you say, that`s a joke, not a rule. My school friends and I caught our teachers ever so often telling nonsense, and they usually accepted our cold logic (and the facts that we had looked up in next years textbook) in the end. ;)   (Btw. i know that saying originally as "the referee is always right".)

All of these elements can form compounds in other oxidation states. Aluminium forms some compounds in the +1 state (e.g. see the section in https://en.wikipedia.org/wiki/Aluminium_iodide), as does Zinc (see the section in https://en.wikipedia.org/wiki/Compounds_of_zinc) and Cadmium (e.g. https://en.wikipedia.org/wiki/Cadmium(I)_tetrachloroaluminate). But Silver is the element that shows this most often in the list given - for instance it forms fluorides in the +1, +2 and +3 oxidation states (https://en.wikipedia.org/wiki/Silver_fluoride). As such I wouldn't say including the oxidation states in the formula is wrong, but on the other hand if you omit it everybody who knows there is an ambiguity will understand what is implied.


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Dissolving plastic resin without damaging aluminium surface
I have an aluminium surface covered with some plastic, which I intend to dissolve in order to clean the surface. On the note for the plastic I have applied, it says that it contains the following:

2-butanone Naphtha
Hexane
Toluene
Resins

What solvent would you recommend to dissolve this plastic, without damaging the aluminium underneath? Could I simply reapply the solvents that were used initially (naphta, hexane, toluene and the ketone)? Regular petrol should contain up to 15% of the toluene, would it be safe to clean it with petrol?

This isn't what plastic is made of, more like a composition for casting, well, a resin. All other components are solvents and stabilizers, so first I'd try literally one the components you've listed. And VM&P and Naphtha should be together: it's one of the trade names for petroleum ether.


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What is the pH and pKa relationship of LaCl3 in water and its precipitant La(OH)3?
In light of this question and its answers, I hope to get some insights into the $\mathrm{pH}$ of $\ce{LaCl3}$ in water and its $\mathrm{pH}$ relationship to $\ce{La(OH)3}$ precipitant's $\mathrm{p}K_\mathrm{a}$?
What can be done in order to prevent $\ce{LaCl3}$ from forming $\ce{La(OH)3}$ and maintaining the $\mathrm{pH}$ $\pu{24^\circ C} - \pu{25^\circ C}$?
When creating salt solutions of $\ce{AlCl3}$, $\ce{LaCl3}$ and $\ce{FeCl3}$; $\ce{AlCl3}$ seem to have a pretty low $\mathrm{pH}$ in water while $\ce{LaCl3}$ rapidly precipitate into $\ce{La(OH)3}$. With the rapid formation of $\ce{La(OH)3}$, $\ce{LaCl3}$ seems to show extremely low solubility in water.
Does $\ce{LaCl3}$ also acts as a Lewis acid?

$\ce{LaCl3}$ doesn't hydrolyze to $\ce{La(OH)3}$, you need to really make solution alkaline for that. Even if you boil the solution for a long time, you'll likely only end up with hydroxochloride $\ce{La(OH)2Cl}$. In distilled water lanthane (III) should be fine. The aquacation participates in equilibrium $$\ce{[La(H2O)9]^3+ + H2O<=> [La(H2O)8(OH)]^2+ + H3O+}$$ which can be easily shifted to the left by acidifying the solution ($\mathrm{p}K_\mathrm{a}$ of hydroxoaquacation is approx. 11). And yes, it's a Lewis acid. Also, what is "percipient" in this context? Do you mean "precipitate"?

@andselisk great reply, well "precipitate", yes. I am using LaCl3·7H2O. That's why I suspect La(OH)3 is the formed precipitate. What happens is, I am trying to create a LaCl3 in pH 7-7.4. It is not only impossible (due to the starting pH of the solution being low pH, there's no way or not make sense to use HCl), but solubility is so low, a precipitate is formed rapidly. Are you saying the precipitate is [La(H2O)8(OH)]2+ ?

Formation of $\ce{La(H2O)8(OH)]^2+}$ is just a first step in olation process which I used for illustration, it's not a precipitate. I'm wondering how you get pH 7.0–7.4. By dissolving the salt in pure distilled water the pH should be well below 7. Do you add a base? If so, then yes, you do get $\ce{La(OH)3}$ all right. And what do you mean by "I am trying to create a LaCl3 in pH 7-7.4"? JIC, $\ce{LaCl3}$ crystallizes as heptahydrate from aqueous solutions, and if you try to dehydrate it by heating up, you end up with oxosalt $\ce{LaOCl}.$

@andselisk, for all my salts, I want to maintain a pH 7-7.4 range. Of course in this case for LaCl3.7H2O, it's impossible to reach even above pH 5. No adding of base. In distilled water the (aq) of LaCl3 is well below 7, yes. My question and perplexity is that, why there's a white precipitate, that too rapidly forming, when dissolving the salt in water at temperature (24'c-27'c)? LaCl3 might be crystallising as heptahydrate from aqueous solutions. Is there a reference that can be used as evidence for this reaction of forming crystals or/and, what occasion LaCl3 in water may form La(OH)3?

@andselisk is it possible to have a chat with you in the chemistry blackboard Periodic table?


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Rate of reaction- Arrhenius equation
Can the Arrhenius equation be used if the reactants are in the solid state and if I can use it for any state, how do I do so?

Clear distinction has to be made between Arrhenius-equation and transition state theory. The formulas for TST can be derived, as opposed to Arrhenius equation, which by itself, is not based on a solid physical derivation. Basically any process can be modeled by an Arrhenius equation that has the formula:
$$ r \propto \exp{(-E_a/RT)} $$
Any molecular process can be reasonably approximated by this formula if it is a statistical transition between states with different energy. The reason for this is that the statistical probability of a state if proportional to the Boltzmann factor $g \exp{(-E/RT)} $, which is readily Arrhenius-like.
So what are some examples for Arrhenius-type behaviors? For example, if you look up the term Arrhenius-type equation on ScienceDirect you can see that research papers have been published on solid strain rates, membrane permeabilities, solubility rates and corrosion.
So, as you can see, an Arrhenius-type equation can describe a huge number of physical/chemical processes, so it is quite reasonable to use it for solid state reactions. The question is only how to find the energy of activation and the pre-exponent factor. For this, there is no general recipe. The best thing to do is to use them as empirical model parameters to be measured by experiments.


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pKa of phenols using UV-Vis spectroscopy
I have done an experiment where I dissolved four different substituted phenols in acidic, basic, and buffer solutions, and recorded the UV-Vis spectra for each. I'm supposed to pick a wavelength where the absorbances for the protonated, normal and dissociated forms are very different. I then use these equations given to me to calculate the $\mathrm{p}K_\mathrm{a}$ of that phenol:
$$x = \frac{A_\text{buffer} - A_\text{acid}}{A_\text{base} - A_\text{acid}}\tag{1}$$
where $x$ is the mole fraction; $A$ is the absorbance in corresponding medium, and
$$\mathrm{p}K_\mathrm{a} = \mathrm{pH} - \log\left(\frac{1-x}{x}\right)\tag{2}$$
I've calculated the $\mathrm{p}K_\mathrm{a}$ values of each phenol, but none of them are close to literature values. I'm not sure what I've done wrong. Can anyone help explain these equations and any ways I could have messed up?

To quote the key points of an easy to read publication by Salgado and Vargas-Hernández (doi 10.4236/ajac.2014.517135, open access):
All starts with the dissociation equilibrium between the acid $\ce{HA}$ and its anion $\ce{A-}$ for which you write
$$\ce{HA + H2O <=> H3O+ + A-}$$
By consequence of this, the recorded total absorbance $A_\mathrm{t}$ for system, assuming only the acid and its anion absorb light in the region inspected, equates to
$$A_\mathrm{t} = d(c_{\ce{HA}} \cdot \varepsilon_\ce{HA} + c_\ce{A-} \cdot \varepsilon_\ce{A-}) = d \cdot c_\mathrm{t} \cdot \varepsilon_\mathrm{t}$$
with $c_\mathrm{t}$ as total concentration; $\varepsilon$, the absorption coefficient; and $d$, the optical path length.
For one, you record a spectrum at low pH to convert all of your sample into the acid form ($c_\mathrm{t} = c_\mathrm{HA}$, and $c_\ce{A^-} = 0$) where all absorption is due to the acid.  Equally, you record a spectrum at a pH this high that all or your sample is deprotonated (then  $c_\mathrm{t} = c_\mathrm{A^-}$).  You need then spectra between the two extrema, with some of $\ce{HA}$ in presence of some of $\ce{A-}$, because the proportion of the concentrations of acid to its anion may be expressed by
$$\frac{c_\ce{A-}}{c_\ce{HA}} = \frac{[\ce{A-}]}{[\ce{HA}]} = \frac{A_\mathrm{t} - A_\ce{HA}}{A_\ce{A-} - A_\mathrm{t}}$$
This is why you search for the isobestic point, where the absorption spectra recorded do not vary while varying the pH value of your analyte, and where $\varepsilon_\ce{HA} = \varepsilon_\ce{A^-}$.

(credit: doi 10.4236/ajac.2014.517135)
Because the equilibrium between the acid and its anion may be described by
$$\log{K_\mathrm{a}} = \log[\ce{H3O+}] + \log{\frac{[\ce{A^-}]} {[\ce{HA}]}} $$
and the definition of $pK_\mathrm{a} = - \log{K_\mathrm{a}}$ you may substitute absorptions by concentrations to yield
$$\log{\left(\frac{A_\mathrm{t} - A_\ce{HA}}  {A_\ce{A-} - A_\mathrm{t}}\right)} = \mathrm{pH} - pK_\mathrm{a} $$
Hence, plotting $\log{\left(\frac{A_\mathrm{t} - A_\ce{HA}}  {A_\ce{A-} - A_\mathrm{t}}\right)}$ in function of $\mathrm{pH}$ yields a slope with an intercept of $y = - pK_\mathrm{a}$.

(credit: loc. cit.)

Addition:
With WebPlotDigitizer it was possible to  convert above mentioned figure 8 into numbers, separate for the basic (b), the acidic (a) and the intermediate / about neuter run at pH = 7.7 (n).  (The result, a .csv file, is deposit here.)  For a wavelength of about $550\,\mathrm{nm}$ the readout of Abs equals to 1.9935 (b), 0.6502 (n) and 0.2241 (a), and the computation
$$ pK_\mathrm{a} = 7.7 - \log_{10}{\frac {0.4534 - 0.2984} {1.2121 - 0.4534}} $$
$$ pK_\mathrm{a} = 7.7 - \log_{10}{\frac {0.1550} {0.7587}} $$
$$ pK_\mathrm{a} (550\,\mathrm{nm}) = 8.39 $$
By similar token, $pK_\mathrm{a} (604\,\mathrm{nm}) = 8.20$ (with Abs of 1.9935 (b), 0.6502 (n), and 0.2241 (a)); or $pK_\mathrm{a} (432\,\mathrm{nm}) = 8.51$ (with Abs of 0.3411 (b), 0.8198 (n), and 0.8940 (a)).
To put this into perspectivive, the arithmetical mean of the four methods compared in Salgado's paper equals to $pK_\mathrm{a} = 8.277$.  Possibly, the difference between their result and the «recuperation»  here were smaller if either a) WebPlotDigitizer were used with a lesser data increment, b) using an interpolation / smoothing function among the data extracted by the program, c) figure 7 were reconstructed from their figure 4.

Thank you for the clarification. I'm still wondering about the second equation I mentioned in my question. Is pH just the pH of my buffer solution? The article you included has a similar equation where it is the pH of the intermediate solution, but that equation was used over all wavelengths. I'm required to look only at the absorbances at a specific wavelength though, so I'm a little bit confused still.

As far as I remember, you measure the pure acid and the pure base and then the buffered solutions. Those buffered solutions have a defined pH value and these are the ones you take.

We do this practical with our students. Either they use the nominal value of the pH of the buffered stock solutions or they use a pH-meter to determine the values of the pH. With rare exceptions both methods yields statistically compatible results.

@cjperkie I equally think that you use the pH value of your buffer solution including your phenoles.  Of course, be sure to provide enough buffer capacity, otherwise the addition of the phenols may alter significantly the pH value of this solution in common.  (Which may be checked by a pH electrode.)


text:
How many times should I run a solvent extration experiment through?
For advanced higher chemistry project, I am extracting cinnamaldehyde from cinnamon via first steam distillation and then solvent extraction using hexane. I'll be adding sodium chloride to the water/cinnamaldehyde mixture. I'm not sure how many times I should run the solvent extraction or if there is a calculation for it.

Do three extractions, evaporate the hexane, weigh residue. Do 3 more extractions of the aqueous layer, evaporate hexane and weigh. Compare the weights to get an idea of the efficiency of the extraction. If first much greater than the second, then the extraction was efficient. Use the same volume of hexane in each extraction.


text:
Can a per-acetylated sugar anomerize?
So I have a question concerning a reaction mechanism for the reaction below.

I know for certain that I get the alfa anomer, but I can't explain it. When I draw out the mechanism it doesn't make sense to me.

I think what throws me off is the NGP from the Ac- group since it would 'force' the beta configuration. I was then considering anomerization of the beta anomer to the alpha anomer, but when I look at mechanisms for this I can't see how it would be possible when I have a bromide on C-1.

I was then wondering if the endocyclic oxygen gets protonated in the first step rather than the exocyclic oxygen in the mechanism I drew, but that would lead to NGP again because we get an oxocarbenium ion.


text:
Can there be a compound which stops the pesticide from entering into the fruit or vegetable?
Can there be a compound which stops the pesticide from entering into the fruit or vegetable? For example, a solution which is sprayed before pesticides and somehow blocks pesticides from entering into the fruit?

You mean a substance which itself stays completely on the surface, does not get washed off and pollutes the surroundings, is totally harmless to birds and insects, and completely decays into water and carbon dioxide within a few weeks?

In addition to all the qualities @Karl stated (harmless, edible, overspray decomposes, etc.), such a coating would also need to stretch as the produce grows. Such a protective covering might even obviate the need for pesticides, if it were to be created.
That said, there are edible wax and protein coatings to preserve produce after it's picked. Unfortunately, these are applied after pesticides may have been sprayed on the food, so one hopes that the produce is washed before spraying seals in toxins.


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Why does dipole must lie parallel to a rotation symmetry axis?
While studying group theory, I've got confused by the following statement:

For a molecule to have a permanent dipole moment, it must have an asymmetric charge distribution. The point group of the molecule not only determines whether the molecule may have a dipole moment but also in which direction(s) it may point. If a molecule has a $C_n$ axis with $n > 1,$ it cannot have a dipole moment perpendicular to the axis of rotation (for example, a $C_2$ rotation would interchange the ends of such a dipole moment and reverse the polarity, which is not allowed — rotations with higher values of $n$ would also change the direction in which the dipole points). Any dipole must lie parallel to a $C_n$ axis.

why does the dipole need to be parallel to the $C_n$ axis? If I put a fluorine atom in just one of the light gray atoms of the following molecule, wouldn't it have a molecular dipole that isn't parallel to the main axis?

In that case you would no longer have a main axis at all.

Perhaps you are thinking about this in the wrong way. Changing to different types of atoms produces the dipole and at the same time changes the point group, for example octahedral MX$_6$ to MX$_5$Y.

No, it would not
A dipole means that there is a difference (in charge) from one "side" to another "side".
If we have determined that something is symmetric, it means the "sides" are equal to each other — and that as you rotate $\left(\frac{1}{n}\right)$ around the symmetry axis, everything stays the same. If there is anything that shifts out of position, that symmetry is not there (it is broken).
Now, your image is a bit hard for me to read, is it supposed to be linear or is it on purpose not? I am going to assume it is not linear, but that the light grey are symmetrically placed… So in your image, there is only a $C_2$ rotational axis, in the approximate "up/down" direction. By placing a fluorine in one end  (and something else at the other end) you would destroy the $C_2$ symmetry. If you rotate it $180$ degrees around (if you could see it through the backside of your monitor…) the fluorine would be on the other side, clearly not identical to $E$, the original. No symmetry.
If it was linear, the light grey and dark gray arranged as a line, then it would still retain a $C_\infty$ rotational axis along the dark grey atoms. Rotating around this in any fraction or direction would still keep everything looking the same.

That the dipole moment must be parallel to (and aligned with) the axis (all axes) of rotational symmetry of a molecule is a direct consequence of the observed symmetry.
Essentially, you can assume each bond between two atoms to carry an individual bond polarity moment. This is a vector that will be aligned with and parallel to the bond axis. To arrive at an overall dipole moment for the entire molecule, these individual bond polarity moments must be added, but because they are vectors they follow the rules of vector addition.
If there is an axis of rotational symmetry in the molecule, then one $\ce{A-B}$ bond will necessarily be transformed onto another $\ce{A^*-B^*}$ bond in the molecule (either A and A* or B and B* can be identical). The underlying mathematics show that for all sets of $\ce{A-B}$ bonds that are symmetry equivalent, the ultimate resulting vecor will be parallel to the axis of rotation. Thus, you end up with a number of intermediate result vectors all of which are parallel to the axis of rotation; the resulting dipole moment must also be. (They can, however, cancel out; this is typically the case if there are orthogonal axes of symmetry that must be considered. The resulting dipole moment vector turns out to be $\vec 0$.)
If, however, you change your molecule (as you suggest in your example replacing one hydrogen with a fluorine atom) you likely have just destroyed the axis of rotation as suddenly an $\ce{A-B}$ bond will be transformed onto an $\ce{A^*-C}$ bond. Since $\ce B \ne \ce C$, the axis of symmetry no longer exists and the dipole moment can be anywhere.


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Difference between Ziegler–Natta method and Phillips method
I have to compare the similarities and differences between the Zieglerr–Natta method and Phillips method of producing polyethylene. However, the only difference I can find is the difference in catalyst used.
Any help with this is greatly appreciated!


text:
Why is the rate of effusion of a gas inversely proportional to (absolute) temperature?

(iv) Rate of effusion of a gas: The rate of passing of a gas through an orifice may be given as:
$$r = \frac{PA}{\sqrt{2\pi RTm}}$$
where
$P = $ Partial pressure of the gas
$A = $ Area of cross-section of the orifice
$R = $ Gas constant
$T = $ Absolute temperature
$m = $ Molar mass of gas

I am unable to get two things straight:

Shouldn't higher temperatures mean higher energy for the molecules and therefore a greater rate of effusion?
Even if this is so, isn't absolute temperature the lowest temperature at which a gas can exist (zero volume) with a constant value corresponding to $\pu{-273.15 K}$?

Edit: I have resolved the first query, $P$ can be rewritten as $RT/V$ bringing the $\sqrt{T}$ term to the numerator.
I am however still confused with what is meant by absolute temperature.

Wrt question 1, this gives an answer. Wrt q2: you misunderstand the meaning of absolute temperature.

But doesn't absolute temperature refer to absolute zero?

The absolute temperature scale defines an absolute zero which cannot be surpassed (unlike say Fahrenheit or Celsius, where negative temperatures are possible).

So absolute temperature just means temperature in the   Kelvin scale?

@JohnTony Thermodynamic temperature.

@JohnTony Yes, that's correct.

The equation can be rewritten by substituting P = RT/V from the Ideal Gas Equation.
This gives us (A√RT/(V√2πm) showing that increase in temperature leads to increase in rate of effusion.
Absolute Temperature simply refers to the temperature on the Kelvin scale of absolute temperature.

✓ is a check mark symbol that has nothing to do with a square root notation; this makes hard to read and parse your formulas. Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown, or just have a closer look at the corrections done to your question and use the markup as an example.


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Oxidation of certain acids by potassium permanganate
Why does potassium permanganate oxidize formic acid and oxalic acid, but has no reaction with benzoic acid and acetic acid?
The reactions were heated in a water bath (~70-80 °C). Oxalic acid had turned colorless:
$$\ce{C2H2O4 + KMnO4 + 6 H+ -> CO2 + H2O + Mn^2+ + KOH}$$
The same reaction could be said for formic, but with the formation of a brownish precipitate:
$$\ce{HCOOH + KMnO4 + 6 H+ -> CO2 + H2O + MnO2 + KOH}$$
For both the benzoic and acetic acid, they remained purple. I do not know whether the structure could be the reason for not reacting at that certain temperature, or something. Could it be possible that the temperature alone is responsible for not allowing the redox reaction to occur for both acetic acid and benzoic acid?
I looked around online and read that alkanes and aromatic compounds don't react with $\ce{KMnO4}$ from one source (Experiment 5 – Reactions of Hydrocarbons, laney.edu). I was not aware of the fact that potassium permanganate could oxidize anything under certain conditions, thank you for that fact.

We have a policy which states that you should show your thoughts, effort and attempts to answer your question yourself. It'll make us certain that we aren't doing your homework for you, and that the Q/A is beneficial for broad audience.   As "homework-like questions" are considered literal homeworks, self-study questions, puzzles, worked examples etc.  Please [edit] in your full reasoning or thoughts on this. See Homework.  Otherwise, the question may get closed as "low OP effort homework-like question."

Under the right conditions, KMnO4 will oxidize $any$ organic matter. You have to specify conditions and details. Also clarify the context of your question.

Your equations are basically wrong for two reasons. 1) It is simply impossible to produce a base like KOH from the reaction of an acid like HCOOH or H2C2O4. An acid cannot produce a base. If this would happen, the first molecule of the base will be immediately destroyed by the next molecule of acid. 2) The charges cannot disappear in chemistry. In your second equation, you have a total of 6 positive charges at the left-hand side, and no charges at all at the right-hand side. The same error occurs in your first equation !

Ah my apologies, I assumed this was right as my teacher had written these. But disregarding my equations, what could be the reason for the non-reaction of the two substances? Or could I not have gotten any conclusive results because of my equations?


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How to dispose of NaOH solution?
I have a range of sodium hydroxide solutions with different concentrations. (from 1.0 mol dm^-3 to 0.5 mol dm^-3). The max volume of each is 100ml. How can you easily dispose of these? Can I pour them down the sink?

NaOH is the main component of drain cleaners, so…

It should not be an issue, just use plenty of water.

Range for sinkable Acids/Bases is between 5.5 and 12 (has to do with the piping, so that may vary a bit depending on the city). So if you're in that range, you're fine to drain it.

Depends where you are and what waste water permits you have. At my company, pouring any chemical solutions down the drain is forbidden.

We cannot guess what your local drainage laws or company/university regulations are. Thus, we cannot answer. You better ask whoever is responsible for lab safety/waste disposal. In Munich, pouring acids and bases down the sink was generally fine if large amounts of water were poured after because of a neutralising device in the cellar. But your mileage may vary.

You may pour them down the sink. NaOH will be soon neutralized and destroyed by the CO2 from the atmosphere, or by the bicarbonate ions present in all drinkable calcareous waters.

Or, add a little NaHCO3 to the NaOH solutions. That gets the process of neutralization started earlier. It takes the edge off, and you can't add too much!


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How can I find the Van der Waals constants (a and b) without any critical values?

I'm given a container of gas that is kept at constant temperature, and whose volume is decreased by applying pressure, from A to E, as the graph shows. I am given the temperature, the total moles in the container, and the following data:
At B the molar volume is 1.153, and the pressure is 17.76.
At C the molar volume is .8167, and the pressure is 17.76.
At D the molar volume is .032, and the pressure is 17.76.
Given this data, and I emphasize, only this data (no critical P, T, or Vmolar), find the van der Waals constants, a and b, for the gas.
I've been working on it for 3 hours and so far all I can come up with is a single equation in terms of a and b at point B: 17.76=RT/(Vmolar-b) - a/Vmolar^2
At point C, the equation would be the same as what I came up with for B, since the temperature and pressure remains the same as B, and we're in the region of co-existence, so Vmolar is at its maximum for the gaseous phase and therefore also is the same as in B (any further increase in Vmolar causes precipitation).
Any and all help is greatly appreciated.

Point C is nothing but smoke and mirrors. It contains no new information. Of course the pressure in it is the same as in B and D; we could have guessed that much ourselves. Use B and D. Write down the equation, plug in all the known values...

I don't quite understand why this question is getting closed as homework because OP clearly demonstrated their attempt and explained the problem in details. And if the question were unclear, there probably weren't two answers already…

@MishaelHibshoosh It would be nice if you could add some information regarding the source of the image (e.g. a textbook reference/website URL).

I would approach this a little differently.  I would integrate the pressure between $V_D$ and $V_B$, such that the average pressure is 17.76 bars:  $$\frac{\int_{V_D}^{V_B}{PdV}}{(V_B-V_D)}=17.76$$or,
$$17.76(V_B-V_D)=RT\ln{\frac{(V_B-b)}{(V_D-b)}}-a\left[\frac{(V_B-V_D)}{(V_B-b)(V_D-b)}\right]$$
This would provide an equation for expressing a in terms of b.  So the pressure could then be expressed exclusively in terms of b.  Then, to determine the value of the constant b, I would set $P(V_B)$ equal to $P(V_D)$ and solve the resulting equation for b.

Took me a second, but this is very elegant and clever. Me gusta.

What this ultimately boils down to is that this is a system of equations which we're trying to solve. Since we have two unknowns, we need two equations. Luckily, we have 3.
$$17.76 atm = \frac{RT}{V_m - b} - \frac{a}{V_m^2}$$
where our three equations come from having three different known values of the molar volume.
As is typical with systems of equations, easiest way to deal with them is to reduce a first equation to a single unknown variable, and then substitute that variable into a second equation to explicitly solve for the other variable. Then, turn around and use that known second variable to explicitly calculate the first.
Van der Waals equation is kinda ugly to work with, but we can do that first step by solving for a with an equation 1 (I'm going to use subscripts on our molar volumes to allow us to keep track easily)
$$a = -V_{m,1}^2\Bigg[P-\frac{RT}{V_{m,1}^2 - b}\Bigg]$$
we'll take this piece and stick it into this equation where a is
$$17.76 atm = \frac{RT}{V_{m,2} - b} - \frac{a}{V_{m,2}^2}$$
and now since we have a single equation with a single unknown, just rearrange all the variables to solve for b. Use b back in equation 1 to solve for a.


text:
What should be IUPAC name of given compound (see image)?
I think the name should be 1-carbamoylpentane-3-sulphonic acid, but the answer sheet mentions it as 2-carbamoylpentane-3-sulphonic acid:

Can anyone please explain why the chain starting with methyl group is considered parent chain instead of one starting with amide?

The methyl group is part of the pentane chain. Therefore, the carbamoyl group is at carbon 2.

hint: your answer has neglected the methyl group entirely.