text:
Fischer Esterification of the Unsaturated 3-Butenoic Acid with Methyl Alcohol
I am attempting to perform a Fisher esterification of 3-butenoic acid (vinyl acetic acid). I believe that this should theoretically work and not disturb the unsaturated region of the molecule, but I'm not certain. Has anyone had any practical experience with this in the lab or seen any literature referencing a Fischer esterification being performed on an unsaturated acid? Theoretically, since the esterification is being performed with anhydrous methanol with a sulfuric acid catalyst the acid-catalyzed hydration should not be possible; however, as the reaction progresses, water present in the reaction mixture may hydrate the alkene. Will this not be an issue since there will still be a large excess of methanol or will I need to find a physical method to remove the water from the reaction as it forms? Given, a Dean-Stark apparatus cannot be used. I'm thinking about using molecular sieves to remove any trace of water as it forms in situ but I don't know if this would work? Has anyone ever attempted this or something similar?

You can add equivalents of methyl orthoformate or orthoacetate to remove water. The byproducts will be methyl formate or acetate, respectively. Distillation should be trivial.

Use BF3.MeOH complex as catalyst instead of Sulfuric

Are you asking us to perform your literature research for you?

I am just wondering if anyone has had any specific experience with this reaction in the lab, as I could not find any literature on this. I may have missed something in my literature review, so I just wanted to include that question in the case, but I'm mostly worried about practical advice, as I am relatively new to synthetic research.

I will edit my question, however, to be more specific

Be careful handling crotonic (butenoic) acid esters. They are vesicants. Alas, I speak from experience. By the way, Google Clinton-Laskowski method for the esterification of acids to methyl esters.


text:
How does liquid mercury conduct electricity?
It is stated that mercury is a liquid at STP because the high relativistic mass of its valence electrons stabilizes their orbits and thus prevents the electrons from being shared. If so, liquid mercury would seem to have no "free" electrons; how, then, does it conduct electricity?

Perhaps wherever stated that is, well, incorrect? Note that Pt, Au, Tl, Pb, Bi, etc. are all solid at room temperature. Various sources try to make a big deal about 'relativistic' electron orbitals without really backing things up.

How is this off topic, or a homework question? This seems like a valid question someone would ask after reading some article or textbook.

Keep in mind that "stabilizing" orbitals does not "prevent" electrons from being shared, only that they would be shared to a smaller extent.  The very fact that mercury is a liquid and not a gas shows that there is some bonding in mercury, even if it is weaker than the bonding in most metals.  So, it is conductive the same way other metals are, just less conductive.  You can look up the conductivity of mercury, and you will find it to be lower than most metals.

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text:
Question about the role of d-orbitals in polarisation
I know that, although many elementary books say that d-orbitals are involved in the bonding of hypervalent main group compounds period 3 or higher such as the sulfate ion but in reality they are only barely, if at all, involved in bonding.
The question is- is it true that "the d-orbitals are not directly involved in bonding but rather in polarisation" and what does that exactly mean? I saw that statement on the comment sections of a post of mine but don't know the proof.

Well, please be more specific about the molecule(s) in question for that quotation.  d-orbitals are certainly involved in some bonding, especially for the transition metals, but you are right that the participation of d-orbitals in the bonding of main group covalent molecules and polyatomic ions has been overstated in the past and is in fact fairly minimal.
As you mentioned, there is very little, if any, d-orbital participation in the bonding of sulfate.  As for d-orbitals being involved in polarization but not bonding, that literally does not make sense for something like sulfate.  Sulfur's d-orbitals are empty, so it is impossible for them to polarize without bonding.  Only when there are occupied d-orbitals (as in transition metals) can non-bonding d-orbitals affect things like the shape and polarity of the compound (as in crystal field theory/LFT).
So, long story short, I think that comment you read is simply wrong, and your original understanding is correct.  See the below references if you want to read more about it.
References

Reed, A. E.; Schleyer, P. v. R. Chemical Bonding in Hypervalent Molecules. The Dominance of Ionic Bonding and Negative Hyperconjugation over d-Orbital Participation. J. Am. Chem. Soc. 1990, 112 (4), 1434–1445. https://doi.org/10.1021/ja00160a022.
Cunningham, T. P.; Cooper, D. L.; Gerratt, J.; Karadakov, P. B.; Raimondi, M. Chemical Bonding in Oxofluorides of Hypercoordinate Sulfur. Faraday Trans. 1997, 93 (13), 2247–2254. https://doi.org/10.1039/a700708f.

I am certain the original comment about d orbitals makes sense in a way. In order to have a p orbital polarised (mathematically) you need d orbital like functions. In approximate solutions of the Schrödinger equation, like HF, MP2, KS-DFT, etc. these would show up as d orbital contributions. These functions (or orbitals) then also do partake in binding.

I concur with Martin. In computational Chemistry, orbitals of higher angular momentum are included in extended basis sets, and these polarisation functions add directionality to the bonding (for example, they allow the p orbitals in the antibonding π* orbital of a diatomic to be polarised away from each other, better representing the actual situation as compared to a minimal basis set).

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Ligand substitution reaction in coordination compounds
I know that the following substitution reaction can take place:
$$\ce{[Cu(H2O)4]^2+ (aq) + 4NH3 (aq) -> [Cu(NH3)4]^2+ (aq) + 4H2O (l)}$$
This is because ammonia is a stronger ligand than water.
What can we say about the product formed from the following reaction?
$$\ce{[Cu(NH3)4]^2+ (aq) + excess HNO3 (aq) ->}$$
Nitric acid is a strong oxidizing agent but $\ce{Cu}$ can not be further oxidised. Is there a possibility of forming $\ce{[Cu(H2O)4]^2+ (aq)}$?

Consider the fact that nitric acid is able to protonate the ammonia molecules, resulting in a change in the position of equilibrium of the ammonia-ammonium equilibrium when it is added. Then consider the effect on the ligand exchange equilibrium.

The reaction you wrote down is wrong on two counts. The reactant is not a hypothetical tetraaqua complex and the product is not a hypothetical tetraammin complex. The correct reaction is as shown below:
$$\ce{[Cu(H2O)6]^2+ (aq) + 4 NH3 (aq) <=> [Cu(NH3)4(H2O)2]^2+ (aq) + 4 H2O (l)}\tag{1}$$
Note that I have used an equilibrium arrow here: the reaction can proceed in both directions depending on the concentration of ammonia in your sample (in reality, there is also the competing precipitation of $\ce{Cu(OH)2}$ at slightly basic pH levels but I will ignore that here).
As an equilibrium, this can be pushed either way. So let’s take a look at what nitric acid can do. It is both an acid and an oxidising agent – but as you noted, copper(II) is already very highly oxidised and it is generally not possible to sustain copper(III) or higher copper oxidation states in aquaeous solution (it does exist as an intermediate in various copper-catalysed organic reactions). On the other hand, $\ce{HNO3}$ can very much act as an acid in your solution as there is a basic compound able to receive protons (see spoiler after figuring out the reaction yourself).

 $$\ce{HNO3 (aq) + NH3 (aq) -> NH4+ (aq) + NO3- (aq)}\tag{2}$$

Note that while $(2)$ is also a reversible reaction, the equilibrium is far to the product side so we can essentially treat it as completed. Obviously, this has an effect on the original reaction – and with a simple application of Le Chatelier’s principle the resulting answer should be obvious.

 Yes, since ammonia will be protonated to give ammonium, equilibrium $(1)$ will be shifted back to the reactant side, regenerating hexaaquacopper(II) $\ce{[Cu(H2O)6]^2+}$.

In equation (1), why ammonia is replacing exactly 4 water molecules?

@Apurvium I’m afraid the best answer I can give is that ‘that’s just how nature works’. Analysing the complex formed in aquaeous solution shows that it contains four ammine ligands and two water ligands. In liquid ammonia, a hexaamminecopper(II) complex can be synthesised but that is unstable in water and will degrade to the tetraammine complex (if the ammonia concentration is sufficiently high) or all the way to the aqua complex.


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Is it possible to put wax or PE or any kind of insoluble coating on soluble material (e.g. edible film)?
I'm working on a project to make a cup that can holds hot/cold water inside but can be dissolved from the outside. I'm just wondering if that would be possible scientifically??

Try it. Make a sugar-glass container and line it with something insoluble.


text:
Question about d-orbitals
I saw somewhere that, "in PF5, the 3d orbitals of phosphorus end up in the anti-bonding molecular orbitals, and thus contribute less to the final molecule" and that "the hypervalency of some hypervalent molecules can be explained more efficiently by canonical forms involving ionic bonds, and some by involving d-orbitals".
I also saw somewhere that "the effect of d-orbitals on bonding is ALWAYS negligible in hypervalent molecules". (I said "negligible" since it cannot mathematically be zero)
My question is, is the first statement right or is the second one right? And, if the first one is right, is there a way to crudely estimate how much the d-orbitals affect bonds, for example a way based on the electronegativities/number of lone pairs(etc) of the subsituents attached to the hypervalent atom?

If the antibonding MOs are not populated, then they don't describe the bonding. Empty orbitals don't exist. So there is no contradiction.

Let me take two parts from the question I can address of the bat

Are d-orbitals essecial to describe bonding in 3+ periods?

If we are talking about practical calculations, than yes, absolutely. d-Type basis function are a vital correction to p-orbitals. For $\ce{PF5}$ population analysis shows considerable projection of electron density on $d$-orbitals... but same applies to $\ce{PF3}$
[1]

Is bonding in $\ce{PF5}$ and the likes better described as extended on full octet.

The theoretical analysis suggests that extended octet isn't needed and the structure is best described as a resonance hybrid of several structures with few having significant charge separation. [2]


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Difference between the terms “analyte” and “analyate” and their applicability
“Analyte” is the term that is usually universally understood and is listed in  IUPAC Gold Book:

analyte The component of a system to be analysed.

However, recently I stumbled upon the word “analyate”.
Initially I thought of it being a typo or an erroneous translation (I'm not a native English speaker), but upon inspection of the literature sources it turned out be used in cited publications in the context similar to the term “analyte” (emphasis mine):

McMaster's LC/MS: a practical user's guide (323 citations) [1, pp. 2, 16]:

The mass analyzer is quickly overwhelmed by the signal from the solvent if the HPLC output is introduced directly into the mass spectrometer. The analyate signal is buried beneath this solvent signal avalanche. […] The HPLC solvent gradient used to resolve closely eluting HPLC peaks and decrease HPLC run times also produces solvent composition changes that further complicate the solvent-masking effect of analyate signal.
[…]
The purpose of an interface is twofold. It must remove as much solvent as possible without losing the analyate, and it must volatilize and ionize its components for submission to the analyzer.

At the same time in the textbook the term “analyte” is utilized on p. 61:

The ionized analyte ions formed in the interface between the HPLC and
  the mass spectrometer cannot survive collision with air molecules in the analyzer.

… and on p. 151:

Gas chromatography. A separation technique in which the volatile analyte is swept by a carrier gas down a column packed with packing coated with an absorbing liquid.

Zhao and Swager (272 citations) [2, p. 9379]:

It is informative to investigate the fluorescence quenching activities of P1-3 in solution. Conducting these studies in chloroform, a good solvent, allows us to evaluate the responses of individual polymer chains and the relative
  analyate affinities to the polymers under these conditions.

Patent US20050137471A1 (182 citations) [3]:

The Signals are processed in a processing unit to calculate the concentration
  of the analyate.

DeLouise et al. (146 citations) [4, p. 3224]:

It also simplifies quantitation of the optical response relative to competitive techniques based on evanescent waves where the signal response decays exponentially from the surface causing sensitivity to depend on the surface-analyate distance.

Hart et al. (130 citations) [5, p. 461]:

It was found that as structural similarity between the imprint molecule and the analyate molecule diminishes, both retention and selectivity decrease.
[…]
Making a relatively small perturbation to the analyate by removing the aromatic hydroxyl group results in a lower, but still significant, separation factor of $α = 2.5.$

… and many more other less cited papers, over 250 in total over 1887–2019 time span according to Google Scholar (I used Harzing's PoP software for query).
So, is there a difference between the terms “analyte” and “analyate”?
If they are not fully interchangeable, then what exactly “analyate” is, or in what context its usage is more appropriate?
References:

McMaster, M. C. LC/MS: A Practical User’s Guide; John Wiley: Hoboken, N.J, 2005. ISBN 978-0-471-65531-2.
Zhao, D.; Swager, T. M. Sensory Responses in Solution vs Solid State:  A Fluorescence Quenching Study of Poly(Iptycenebutadiynylene)s. Macromolecules 2005, 38 (22), 9377–9384. https://doi.org/10.1021/ma051584y.
Haar, H.-P.; List, H.; Baader, F.; Harttig, H.; Meacham, G. Continuous Glucose Monitoring Device. US20050137471A1, June 23, 2005. (PDF)
DeLouise, L. A.; Kou, P. M.; Miller, B. L. Cross-Correlation of Optical Microcavity Biosensor Response with Immobilized Enzyme Activity. Insights into Biosensor Sensitivity. Anal. Chem. 2005, 77 (10), 3222–3230. https://doi.org/10.1021/ac048144+.
Hart, B. R.; Rush, D. J.; Shea, K. J. Discrimination between Enantiomers of Structurally Related Molecules:  Separation of Benzodiazepines by Molecularly Imprinted Polymers. J. Am. Chem. Soc. 2000, 122 (3), 460–465. https://doi.org/10.1021/ja9926313.

In my opinion it is,. although recurrent in your examples, just a typo or a mistake of the writer.

The unabridged Oxford English Dictionary, available by subscription, is the ultimate source to confirm the usages of English words. I checked the word "analyate" and it does not exist in the entire dictionary. The word analyte is also very recent (mid 1950s). I am not surprised that wrong spellings are published in high-end journals. If you study the statistics of wrong references, that is even more shocking. How many times it has happened that you check a citation and it turns out that the citation never mentioned anything like that.
Here is an excerpt from the OED

Origin: Apparently formed within English, by compounding. Etymons: analyse v., -lyte comb. form.
  Etymology: Apparently < ana- (in analyse v.) + -lyte comb. form, perhaps after electrolyte n.
  Chemistry.
  Categories »
A substance that is analysed or determined, esp. quantitatively.
1955   C. M. Crawford in Chem. & Engin. News 5 Dec. 5262/2   There is
  not, as far as I am aware, any short word denoting ‘the substance
  being analysed for’ in use... My proposal is that the word ‘analyte’
  be understood to convey the above meaning.

Yep, I also got a similar idea by looking at the Google Ngram for both terms. "Analyate" isn't a known term, but there is a little change it's been used so rarely Google classifies it as an outlier.


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The reaction rate when the rate-determining step has a back reaction
For some reason, I am reviewing general chemistry and I find that the elementary reaction for the rate-determining step (RDS) is assumed to be irreversible, that is, single arrow, in all questions that I see.
For example, think a reaction consisting of two elementary reactions:
$$\ce{A + B <=>[k_1][k_1^\prime] C} \tag{fast, reversible}$$
$$\ce{C ->[k_2] D} \tag{slow}$$
Then the overall reaction is $\ce{A + B -> D}$, and the rate is then
$$\text{rate} = k_{2}[\mathrm{C}].$$
Now, I'm curious about the rate when the equation becomes like
$$
\ce{A + B <=>[k_{1}][k_{1}^\prime] C \:(\mathrm{fast, reversible)} \\
C \color{red}{<=>[k_{2}][k_{2}^\prime]} D\:(\mathrm{slow)} }
$$
In this case, I think the rate expression should be changed.
From the overall chemical equation ($\ce{A + B -> D}$), the rate itself is defined as
$$
\mathrm{rate} = \frac{\mathrm{d}[\mathrm{D}]}{\mathrm{d}t},
$$
and checking the reactions related to $[\mathrm{D}]$ in the RDS,
$$
\frac{\mathrm{d}[\mathrm{D}]}{\mathrm{d}t} = k_{2}[\mathrm{C}] - k_{2}'[\mathrm{D}].
$$
Thus, the rate is
$$
\mathrm{rate} = k_{2}[\mathrm{C}] - k_{2}'[\mathrm{D}].
$$
Though, I'm not sure if the reasoning is right. Is it correct?

The model of the rate determining step is somewhat ambiguous and in a way not up to date anymore. It really only is a good approximation, if all other steps in the system are so much smaller that they can be ignored; and this is not the whole truth either. The overall kinetics of the system is more complex and a better approximation is the energy span model (https://pubs.acs.org/doi/10.1021/ar1000956). In your case A,B,C,D are in equilibrium, so the kinetics will be determined by the maximum energy between any of these and a TS connecting them.

It would be mathematically easier to consider the reaction $A\leftrightharpoons C\to D$ and solve the rate equations which can be done exactly, messy, but a sum and difference of exponentials. Then add your conditions as to which rate constants are bigger than the others etc. and look at the results or plot as graphs.

Thus, the rate is
  $$
\mathrm{rate} = k_{2}[\mathrm{C}] - k{_2}'[\mathrm{D}].
$$
  Though, I'm not sure if the reasoning is right. Is it correct?

This is correct but does not give you much insight into how [D] changes over time. This system will reach equilibrium eventually. If you call the equilibrium concentrations $[\ce{C}]_{eq}$ and $[\ce{D}]_{eq}$, and $\epsilon$ the difference between current and equilibrium concentration for C, you get the following:
$$[\ce{C}] = [\ce{C}]_{eq} + \epsilon\tag{1}$$
$$[\ce{D}] = [\ce{D}]_{eq} - \epsilon\tag{2}$$
Also, the rates are equal at equilibrium:
$$ k_2 [\ce{C}]_{eq} - k{_2}'[\ce{D}]_{eq} = 0\tag{3}$$
Plugging all of this into the rate equation gives:
$$\frac{d[\ce{D}]}{dt} = -\frac{d\epsilon}{dt} = k_{2}[\mathrm{C}] - k{_2}'[\mathrm{D}]\tag{4}$$
$$= k_{2}([\ce{C}]_{eq} + \epsilon) - k{_2}'([\ce{D}]_{eq} - \epsilon)$$
$$= k_{2} \epsilon - k{_2}'(- \epsilon)\tag{*}$$
$$ = (k_{2} + k{_2}') \cdot \epsilon$$
The simplification yielding the expression labelled '*' is because the net rate at equilibrium is zero (substituting (3) into the previous expression). As a result, you get a first order behavior of approching equilibrium with respect to the "excess concentration" $\epsilon$. When you integrate this, $\epsilon$ decays exponentially, with the sum of the forward and reverse rate constant in the exponent. This is somewhat surprising because we started out with these rate constants appearing with opposite signs in the (correct) rate expression given by the OP. However, this result is correct and well-known from temperature jump kinetics.
What about A and B?
The derivation above is for a two-species system $\ce{C<=>D}$. If A and B are in fast equilibrium with C, equation (1) or (2) has to be modified to account for A and B supplying more C as it reacts to D. One way is to define $[\mathcal{C}] = [\ce{A}] + [\ce{C}]$ and write (instead of (1)):
$$[\mathcal{C}] = [\mathcal{C}]_{eq} + \epsilon\tag{1'}$$
Because the first equilibrium is fast, you can calculate $\ce{A, B and C}$ from $[\mathcal{C}]$ at any point in time, and you would have to adjust equation (4) accordingly. The details depend on whether $\ce{A and B}$ are present at stoichiometric amounts or one of them is in excess. If one of them is in large excess, the result would just be different by a constant $\beta$, i.e.
$$ \mathrm{rate} = (\beta k_{2} + {k_2}') \cdot \epsilon$$
where $\beta$ depends on the equilibrium constant of the first (fast) reaction and the concentration of the reactant in excess.


text:
Antiseptic role of kerosene
When I was kid and got a huge cut on my finger, I remember my grandmother putting lots of kerosene on my wound, as there was no health facility around.
I want to confirm. Do chemists agree on the antiseptic role of kerosene, or was grandma just using useless remedy?

All medical advice is off topic here.

Just as a note to anyone considering it as an antiseptic agent, it is toxic and probably absorbs through the skin.
Any compound that causes a change in osmotic pressure within a pathogen may kill it through plasmolysis, depending on how much of the compound is soluble in water.  There may be other mechanisms by which a hydrocarbon poisons pathogens like the interruption of metabolic pathways.
Anyhow petroleum distillates are dangerous to apply to the skin and nobody should ever do it on purpose.


text:
Can anyone please explain the statement on solubility and precipitation
"It is because of high solubility product of Mg(OH)2 as compared to that of MgCO3 that Mg(OH)2 is precipitated." Can anyone please explain the statement.

As it is written, it is nonsense, as the higher sulubility product generally means higher solubility.
But the direct comparison of solubility, based on solubility products, is possible just for precipitations of the same type ( binary as $\ce{BaSO4}$ , ternary as $\ce{PbI2}$) and for solutions not containing the common ion, ideally just water.
About precipitation conditions,
$\ce{Mg(OH)2}$ precipitates if $$[\ce{Mg^2+}] \gt \frac {K_\mathrm{sp, \ce{Mg(OH)2}}}{{[\ce{OH-}]}^2}$$
$\ce{MgCO3}$ precipitates if $$[\ce{Mg^2+}] \gt \frac {K_\mathrm{sp, \ce{MgCO3}}}{[\ce{CO3^2-}]}$$
The priority of precipitation depends on which of magnesium concentration thresholds above is lower.
For kinetic reasons, both precipitates may form, if both thresholds are crossed. But in long term, the precipitate with the lower threshold takes precedence  for thermodynamic reasons.
$ K_\mathrm{sp, \ce{Mg(OH)2}}=5.61\cdot 10^{-12}$
$ \mathrm{p}K_\mathrm{sp, \ce{Mg(OH)2}}=11.25$
$ \mathrm{pMg}=11.25 - 2 \cdot  \mathrm{pOH} =-16.75 + 2 \cdot  \mathrm{pH}$
It means at $ \mathrm{pH}=10$ the threshold magnesium precipitation concentration is at level of typical magnesium concentration in drink water.
$ \mathrm{p}K_\mathrm{sp, \ce{MgCO3}}=7.8$

Should be $\ce{[CO3^2-]}$ without a square in the denominator.

@Aniruddha Deb Yes, sure, it was a copy/paste/should_edit error. Fixed now. thanks.

what does threshold mean?

Generally, it is a limit value, when something happens, or a different decision is made, if the value is crossed. In this case - precipitation.


text:
Why does a substance such as methane have 4 sp3 orbitals rather than 3 orbitals
I have been learning about SP3 hybridisation and slightly confused why it it results in 4 sp3 orbitals.
My understanding is this...
The S orbital is spherical and uniform whilst the p orbitals are made of 2 lobes.
There are 3 p orbitals Px Py and Pz.
So if each of these p orbitals combines respectively with the s orbital you end up with 3 new ones. Where does the remaining orbital come from?
Thanks very much,
Daniel

It would be quite wrong to say that each sp3 orbital comes from specific p orbital. Instead, they are all mixed together.

The number of atomic orbitals equals the number of (hybrid) molecular orbitals.

Actually one s orbital and three p "mix" together, hybridize to form four sp3 orbitals.
Maybe this could help:

The 2s and 2p orbitals are close to one another in energy.  In the case of a single hydrogen atom, these states are in fact the same in energy, although you would not see a hydrogen atom with an electron in the 2p orbital unless it had been excited by light or a collison.
Because these orbitals are close in energy, the hybridization concept promotes the idea that these orbitals readily mix. It's a good thing that they do, because this causes the tetrahedral shape of fully saturated carbon.

Uhhh... here we go with the toxic downvoting of reasonable responses.


text:
Why homolytic fission of polar molecules is possible at higher temperature?
While studying alkanes I came across the nitration of alkanes using Nitric acid at very high temperature, which apparently follows "free radical" mechanisms, by formation of OH°  °NO2 free radical ! How "homolytic" cleavage of N-OH (a polar bond)  is possible at any temperature instead of heterolytic cleavage i.e formation of ions  ?

That's how molecules work. Use high enough temperature and they break. In this case stability of NO2 eases this.

It is possible for both ions and radicals to be produced and it all depends on the energy. Molecules have many excited states that 'cross' the ground state at high energy and certain bond extension and this means that radical or ions are produced depending on the conditions. NaI can form Na$\cdot$ + I$\cdot$, see A. Zewail Nobel prize lecture for pictures of the energy diagram.


text:
Is activated charcoal an air purifier?
It's well known that activated charcoal is an adsorbent and due to very big surface area per unit of mass it can adsorb a lot of particles.
Let's take as an example for my question this product on Amazon.
My questions are the followings:

How much do this apply to the air purifiers? According to them it is supposed to work as an air purifier but I don't see it working unless the air you want to purify is being pushed through a activated charcoal filter. AFAIK by itself the activated charcoal is not attracting the smelly particles. Or is it?
I have read that activated charcoal adsorbs bad smells(gases from air) and when the activated charcoal is placed in a hot environment it releases the adsorbed gases. What is that temperature at which adsorbed gases are released back in the air? Is this why they advice placing the product in the sun? Taken from product description: "Placing activated bamboo charcoal bag in the sun for 2 hours once a month, naturally rejuvenates the charcoal by removing excess moisture - extending the life of these charcoal room odor absorbers for up to 2 years."

Agreed: unless air is pushed through the activated charcoal bed, few impurities would be adsorbed.
It would depend on how tightly the substance clings to the carbon, i.e. it would vary with impurity, but Trox Technik KS Filters recommends, "saturated activated carbon is reprocessed... with a gradual increase of temperature from 20 to 830 °C and a delay period of approximately 30 minutes."
Finally, how long does the air-stream remain in contact with the filter? It would seem that most air bypasses the pores of the carbon on inexpensive commercial air filters. Activated charcoal might be of more use in reducing foul odors in a closed container, such as bottles of medicines.

The total amount of harmful substances in the use environment and the frequency of desorption. Because the quality of the harmful gas adsorbed by activated carbon can be close to or even reach its own quality, and in the air of ordinary home space, the quality of harmful gas is far less than the amount of activated carbon used. Therefore, activated carbon can be used for a long time as long as it is often placed in the sun and exposed to the sun.


text:
Making Warming hand sanitizer at home
What can I  mix with hand sanitizer to make it heat up as I apply to/rub into my hands?  Something that doesn't affect the sanitizing.

Sanitizer is 95% alcohol: Adding something that can make it heat up would only evaporate more of the alcohol, neutralizing all sanitizing effects.


text:
Is the pyrolysis of tetrachloroaluminate(III) anion possible?
Wikipedia says that lithium tetrafluoroborate decomposes to lithium fluoride and boron trifluoride when heated. Does the same hold for sodium tetrachloroaluminate (all anhydrous) as well?

Unlike lithium tetrafluoroborate $\ce{LiBF4},$ sodium tetrachloroaluminate $\ce{NaAlCl4}$ doesn't decompose upon melting, and doesn't show any signs of decomposition or phase transition up to approx. $\pu{800 °C}$ [1, p. 688]:

FIGURE 3. Calculated $(\ce{NaCl + AlCl3})$ phase diagram at $\pu{0.1 MPa}$ (dotted lines are liquid–liquid miscibility gap boundary and monotectic at $P > \pu{0.1 MPa}).$

Molten sodium tetrachloroaluminate is used as electrolyte in ZEBRA batteries with normal operating temperature range of $\pu{270–350 °C}.$
References

Robelin, C.; Chartrand, P.; Pelton, A. D. Thermodynamic Evaluation and Optimization of the $(\ce{NaCl + KCl + AlCl3})$ System. The Journal of Chemical Thermodynamics 2004, 36 (8), 683–699. https://doi.org/10.1016/j.jct.2004.04.011.


text:
What is the role of hydration enthalpy in oxidation?
My textbook says that the greater oxidising behavior of fluorine than chlorine can be attributed to the high hydration enthalpy of $\ce{F-}$ ions and the low dissociation enthalpy of $\ce{F-F}$ bond.
While the second reason makes perfect sense to me, I can't reason why hydration enthalpy comes into play. I know that these elements act as oxidising agents in aqueous media, but I still can't understand why we need to consider hydration enthalpies.

@NisargBhavsar Oxford dictionary says "oxidize/-zing, UK English also oxidise/-sing".  So "s" is not exclusive even in UK.

Related

The hydration enthalpy is the amount of energy released on dilution of one mole of gaseous ions or we can say the heat energy released when new bonds are made between the ions and the water molecules is called hydration enthalpy.
The magnitude of hydration enthalpy depends on the charge density of the ions. The charge density is more in case of small ions and that's why the smaller ions have high hydration enthalpy. The higher is the charge density the higher will be the force of attraction between ion and the water polar end. So the value of hydration enthalpy is higher for small ions (...)

(...) Both the oxidising power of an element and the hydration enthalpy are related with its charge density. The higher the charge density greater will be the oxidizing power ( or strength of oxidising agent) and also the higher will be hydration enthalpy. So, the hydration enthalpy and strength of oxidizing agent are directly related if one is high then other will also be high (...)

(...)As Fluorine has high charge density due to its very small size therefore, it has high oxidising power and also have high hydration enthalpy. As explained above because of small size and higher charge density the attraction between fluorine ion and water will more so the fluorine generates a large amount of heat when it forms its hydrated ion and hence, the hydration enthalpy of fluorine will also be high.

The above writings were fetched from here. Overall, oxidizing character/ability depends on two more factors other than hydration enthalpy: heat of atomization/dissociation and electron affinity.

Fluorine is a stronger oxidizing agent than chlorine. So fluorine oxidizes chloride ion as per the following reaction:
$$\ce{F2 + 2Cl- -> 2F- + Cl2}$$
It can be observed that low enthalpy of dissociation of $\ce{F-F}$ bond lowers the energy of activation of the reaction mentioned above and hence favours the formation of products. Due to the small size and high electronegativity of fluoride ion, it has a higher hydration enthalpy or in other words it's easily covered by a lot of solvent molecules. This factor greatly reduces the rate of the reverse reaction. This process also shifts the equilibrium towards the products. This is how hydration enthalpy plays an important role in the oxidizing capability of fluorine.

In aqueous solutions, halogens are powerful oxidising agents (especially fluorine). Let us confine our discussion to fluorine and chlorine. Oxidising ability can be demonstrated using the enthalpy change for the following reaction.
$$\ce{1/2 X2(g) + e- -> X-(aq)}$$
Let $\Delta H$ be the enthalpy change for the above mentioned reaction. This reaction can be achieved in three steps as follows:

Dissociation of halogen into gaseous atoms.
$$\ce{ 1/2X2(g) -> X(g)}$$
Let $D_0$ be bond dissociation enthalpy.

Converting gaseous atoms into gaseous ions.
$$\ce{ X(g) + e- -> X-(g)}$$
Let $\Delta_\mathrm{eg}H$ be electron gain enthalpy or electron affinity value.

Hydration of gaseous ions into hydrated ions.
$$\ce{X-(g) -> X-(aq)}$$
Let $\Delta_\mathrm{hyd}H$ be the hydration enthalpy.

Based on Hess law,
$$\Delta H = D_0 + \Delta_\mathrm{eg}H + \Delta_\mathrm{hyd}H$$
What we need now is this net enthalpy $\Delta H$. The more negative the enthalpy, the more the oxidising power. As fluorine has low bond dissociation enthalpy and high hydration enthalpy it act as powerful oxidising agent than chlorine. So, now it can easily be understood why we should consider enthalpy of hydration in comparing oxidising power.

Why are we considering only aqueous solutions? Is fluorine not a stronger oxidizing agent than chlorine outside aqueous solutions?

Fluorine is a stronger oxidising agent than chlorine even outside the aqueous medium because of its high electronegativity. In general, we use water as a medium for many redox reactions that you perform in the lab. Is your doubt how hydration energy comes into picture if we perform reactions outside water ?

You may think that hydration energy is indirectly linked to oxidising power. Because we can't define hydration energy in mediums other than water. So, hydration energy is a consequence for mediums other than water, but it is a factor which decides oxidising power in aqueous solutions.

Mostly, this kind of books we usually use generalises the statements rather than discussing particular situations.

I have awarded the bounty to you, but I have a request. Please include the comments that you just made about hydration energy and oxidizing power in your answer (and possibly expand on those comments!)


text:
Rotational spectroscopy application of different equations
So we were given two problems and I don't understand why one was solved one way, and another was solved another way
Estimate the bond length of 12C16O (pure rotational spectrum) given J''=3 (15.356 cm-1) and J'=9 (38.356)
To solve we used
$12B = (38.356 - 15.365)=1.916$
to solve for B and then used to determine the bond length.
In another problem we are told that the J=3 to J=4 transition for a diatomic occurs at 0.50 cm-1. What is the J=6 to J=7 transition for this molecule assuming rigid rotor.
In this problem We used
$E(rotation) = Bh[J'(J'+1) - J''(J''+1)]$
Which when solved/rearranged becomes
$E(rotation) = 2BhJ'$
Which can be rearranged to solve for B, because we know the energy as a wavenumber. This B is then used to calculate the transition wavenumber for 6 to 7 transition.
I am lost as to why in the first one we can simply assume each transition is 2B apart and so 6 transitions would be 12B and solve whereas in the second we cannot make this assumption.

The transitions themselves aren't 2B, they spaced apart by 2B. In the first case, you had two transitions, so you could determine B by their spacing. In the second case, you only know the location of one, but B is also related to the energy of a single transition by the equation given.

In the second case, once you have B, you can just use the separation to solve for the energy of the 2nd transition. They happened to use a different equation but that doesn't mean you couldn't use the same equation in both cases.


text:
The product of the reaction of chloroiodoalkane and cyanide ion in acetone

What would be the product of the following reaction $\ce{Cl-CH2-CH2-CH2-I + KCN} $ ($\pu{1mol}$ each) in acetone?

I think the  reaction should follow the second order nucleophilic substitution mechanism due to primary halide and strong nucleophile. Thus, iodine atom should be replaced giving $\ce{Cl-CH2-CH2-CH2-CN}$ as product but the book says the product is $\ce{I-CH2-CH2-CH2-CN}$. Where am I wrong?

Iodide ion is recycled in the system. $\ce{KI}$ is quite soluble in acetone $(\pu{1.31 g}/\pu{100 ml}),$ $\ce{KCl}$ is essentially insoluble.
You are quite correct that the first reaction is $\ce{CN-}$ displacing iodide giving $\ce{Cl-CH2-CH2-CH2-CN}$ and $\ce{KI}.$ The soluble $\ce{I-}$ will then do a nucleophilic attack on the $\ce{CH2-Cl}$ to give $\ce{CH2-I}$ and the $\ce{KCl}$ produced precipitates out making the reaction irreversible.

Missing a CH2. A lot depends on the stoichiometry and mode of addition of reagents. Glutaronitrile, NC-CH2-CH2-CH2-CN is also. an option

Fixed. I agree it is likely some glutaronitrile would be formed.


text:
What formula allows adjusting balance of (Ethyl Alcohol / Water / Caffeine powder) to know concentration for the Caffeine solution?
I am experimenting with different levels of concentrated caffeine solution. The solution has a small percentage of ethyl alcohol, in the $\ce{H2O}$.
Problems encountered too much ethyl alcohol solidifies the powder into clumps, adding water dissolves the clumps further into smaller clumps.
What formula allows me adjust the balance of the ethyl alcohol / water / caffeine powder, so that I know what concentration I have for the caffeine for each semi-solidified solution?
My goal is to find a compromise for the highest concentration of caffeine solution for a given semi-solidified solution.
For Example:
If I add $\pu{60 mL}$ of $\ce{H2O}$ to $\pu{31,090 mg}$ of caffeine powder then add $\pu{10 mL}$ of ethyl alcohol, the powder fully absorbs all $\pu{60 mL}$ of the $\ce{H2O}$ so it becomes a solid clump. If I add another $\pu{90 mL}$ of $\ce{H2O}$ the clump turns into a solution. But, I am uncertain what concentration the solution is. I need to experiment with the semi-solidification state and concentration level. To find the highest concentration for a compromise of the semi-solidified solution state.
Reference:
if 1 gm of Caffeine dissolves in 46 mL water at room temp.
if 1 gm of Caffeine dissolves in 5.5 mL water at 80 degree C.
if 1 gm of Caffeine dissolves in 1.5 mL water boiling water.
if 1 gm of Caffeine dissolves in 66 mL alcohol at room temperature.
if 1 gm of Caffeine dissolves in 22 mL of alcohol at 60 degree C.
?1.) What is the formula for adjusting volume H2O / Ethyl Alcohol at a certain temperature, which takes into account the Caffeine grams included is always dissolved at 40 degrees C constant?
?2.) What is formula for finding the minimum amount of $\ce{H2O}$ / ethyl alcohol that fully dissolves a given amount of caffeine powder at a certain temperature, to help find the highest concentration limit for a given temperature?
For example:
If the temperature constant is 40 degree C for both H2O and Ethanol, and there is 1000 mg of Caffeine to dissolve.
What is the maximum concentration per mL of the 1000 mg at 40 degrees C, that fully dissolves caffeine?  Basically what is the minimum amount of liquid volume at 40 degrees C that dissolves the Caffeine, resulting in highest concentration.

You mean 31.1g of coffeine, right?

The solubility of coffeine in water is 20g/l, even a bit less in ethanol (at room temperature). What you have is a suspension, not a solution

There is no straightforward formula. You determine the solubility in a few alcohol/water ratios by drying and weighing the saturated solutions, and then interpolate.

Yes 31.09 grams

Can't I use Enthalpy?

It appears Caffeine liquifies with the lowest mLwhen heat is applied. But, I am restricted to 40 degrees Celsius.

1.5 mL of water seems like the reasonable amount of liquid concentration and concentration I need say for 1 gram of Caffeine. But, I cannot go that high in temperature to dissolve 1 gram in 1.5 mL. I am limited to 40 degrees C. But, I also need to add Ethanol. Do, you know a general approximation formula we can create to obtain some test trials?  I want 1.5 mL suspension of H2O/Ethanol and 1 gram of Caffeine at 40 degrees C, to have almost totally dissolved Caffeine. Need to find the correct ratio of H20/Ethanol.

What analytical method are you using to measure the dissolved caffeine? For info, solvent mixtures frequently demonstrate non-linear effects when it comes to solubility.

A structured approach to this is to use at least five mixtures of the solvents and saturate with caffeine,  stirring for at least four hours.
The five mixtures are not picked arbitrarily but can be set at idealised levels.
If you use 100% ethanol and water as the extremes, then you want 50:50 and two inbetween on either side.
These can them be combined with your process variable if you desire, which seems to be temperature.
Measure the concentrations of each and derive the equation using line of best fit (Excel and Minitab can do this).
If you have JMP or Design Expert for example, you can set this up as a DOE.  In either case, now you can choose your optimum.


text:
Difference between equations for spectroscopy wavenumbers
What is the difference between these two equations and when is appropriate to use which one?
We were told that $F(J)$ in this case is
$F(J)= E/h$ for units in frequency
And
$F(J) = BJ(J+1)$
$\nu = 2B(J+1)$
(where $\nu$ (nu) is meant to signify wave number)
I have seen difference example problems use difference ones but it appears to me as though these two equations give the same result: wavenumber. So how is it possible that it has two completely separate values?

The $\nu=$ equation is obtained for a rotational transition $J\to J+1$ in a diatomic molecule so is not the same as the second equation  $F(J)=BJ(J+1)$. This equation is that for the energy level itself so $\nu =F(J+1)-F(J)$. The first equation has no meaning unless you define $E$ in some way. $B$ is often given in wavenumbers.

@porphyrin What do you mean by the "first equation has no meaning"? Which equation are you referring to? Please clarify

$F(J)=E/h$ unless you specify what $E$ is and in what units.

Ultimately, they are just different expressions for the same fundamental physical quantity: the energies of quantum states and the differences between them. There are different units you can use to express that quantity - energy, frequency, wavelength, wavenumber - but at the end of the day they are all equivalent, and can be converted among each other, as
$$E=h\nu = h\frac{c}{\lambda}=hc\tilde{\nu}.$$
There's no hard and fast rule what to use when, just a lot of conventions that can unfortunately differ between closely related fields. For example, microwave spectroscopists will typically talk about their spectra in units of frequencies, and infrared spectroscopists about theirs in units of wavenumbers. Throw electronic states into the mix and you'll also get electronvolts (a unit of energy). - If you want people from a certain field to understand what you're talking about, it makes sense to stick with the units they are familiar with. Otherwise, pick your favorite, or what most suits your problem and available data at hand.
NB: The $E/h$ you have written yields a frequency, not a wavenumber. Things may sometimes differ depending on how the rotational constant is defined, but again a look at the units will tell you what you need to know.
Also, $\nu$ without the tilde is only ever a symbol for a frequency, not a wavenumber. To get the wavenumber $\tilde{\nu}$, write $\tilde{\nu}$.


text:
Copolymerization regioselectivity
When undergoing copolymerization via free radical mechanism Buna-N forms a certain way.
Actual product and possible products
Can we predict which of the products will be formed based on the substituents?
If so, how?


text:
Calculating the total uncertainty of the final product
I have performed a large number of calculations in order to determine the ethanol content of wine. These included titrations using burettes, mass measurements using analytical balances, and others. I know the uncertainties of the apparatus.
I am given a task to find the ethanol content of wine (done) but also to calculate the total uncertainty by propagation of errors.
As I understand it, the total uncertainty is given by
$$\delta F=\sqrt{\left(\delta q\right)^2+\left(\delta w\right)^2+...}$$
But - does this include the standard error? That is, to calculate the total uncertainty of the final value, will I need to find the standard error for all the sets of titrations (eg - for titration 1, a set of 3 values, find the standard error taking into account all results, even non-concordant ones, and do this for titration 2), and treat those values as fractional uncertainties? Should I add this value to the fractional uncertainties of the burette measurements (ie its own measurement uncertainty, not the standard errors from the readings), and to the measurement uncertainties of the other apparatus?
Finally - does the operation which I apply to the values that the uncertainties effect change the equation for the total uncertainty? Or do I just use the same equation as above?

Can you give the equation which uses all of the calculated values? Uncertainties propogate according to the functions acting on the variables and parameters in them. You can't just take the sum of squared uncertainties. Show me the equation you use to calculate ethanol content please.

General approach
\begin{equation}
    \sigma_{F}^2 = \sum \left( \frac{\partial F}{\partial x_i} \right)^2 \mathrm{d}x_i^2
\end{equation}
Here we neglect all correlation terms formed by squaring what is essentially the first order Taylor expansion with respect to the variables about their uncertainty. $x_i$ is a variable or parameter which contains uncertainty and $F$ is the function which depends on the variables and parameters $x_i$. $\mathrm{d}x_i$ is the uncertainty in $x_i,$ i.e., you measure the mass of your solution 10 times and it is 1 kg ± 0.1 kg. Your $\mathrm{d}x_i$ for mass is 0.1 kg. If you want you can think of $\mathrm{d}x_i$ as $\Delta x_i$ instead.
For a given calculation $f$ you take its derivative $F$ wrt each variable/parameter, hence you know its rate of change, and $\mathrm{d}x$ is the known/measured uncertainty of that variable/parameters, i.e., how much the function might be out for a given error in a specific parameter.
So, in your original expression you only accounted for the uncertainty in the parameters, but not how their error would propagate into each and every function/equation/calculation they are used. If a function is linear in a given $x_i$ then at most it will be out by the uncertainties magnitude multiplies by some constant in the equation. However, it may be that the function contains $x_i^{10}$ in which case the error in $x_i$ will be to the 10th power which could be huge! So, use the above equation, with the appropriate derivatives to "propagate" error.
Some background derivation
The change in a function due to a variable/parameter can follow from its differentiation
\begin{equation}
\mathrm{d}f = \left( \frac{\partial f}{\partial x} \right) \mathrm{d}x + \left( \frac{\partial f}{\partial y} \right) \mathrm{d}y
\end{equation}
It is possible that the error due to variable $x$ is equal but opposite to $y,$ and the errors cancel. However, when we square it, which gives us
\begin{equation}
\sigma_f^2 = \left( \frac{\partial f}{\partial x} \right)^2 \mathrm{d}x^2 + \left( \frac{\partial f}{\partial y} \right)^2 \mathrm{d}y^2 + 2\left( \frac{\partial f}{\partial x} \right)\left( \frac{\partial f}{\partial y} \right) \mathrm{d}xy,
\end{equation}
this way errors don't cancel. Also, we almost always assume the correlation term $\mathrm{d}xy$ is negligible. Neglecting the correlation terms, we find ourselves left with the top equation.
More info can be found here: Wikipedia uncertainty propagation. Also, don't forget to donate to Wikipedia. :)


text:
Keeping Li-Ion battery on charger vs. plug the charger in and out regularly
I have read on numerous occasions a set of recommendations to keep a Li-Ion battery happy: avoid discharging the device at all costs, don't charge it up completely, keep it as much as possible between 40% and 80% and don't charge it too quickly.
I'm wondering, however, which one of those two workflows are least harmful to a Li-Ion battery:

Keeping the device on (slow) charger most of the time the whole day while using the device, the battery level remaining at 100% most of the time, and dropping sometimes to 80%.
Unplug the (slow) charger regularly during the day, and then plug it again a hour or two later, the battery level constantly fluctuating between 50% and 100%.

I'm asking this question because while it is quite unpractical for me to keep the battery level of my phone between 40% and 80%, it is perfectly possible for me to either keep the phone on charger nearly all the time when I'm at home or at work, or to switch regularly between on battery and on charger modes.

What kills a Li-Ion battery is 1.) overtemperature, 2.) overtemperature, 3.) overtemperature.

This is more of an electrical engineering question, nonetheless plugging a battery in 24/7 isn't an issue for modern devices, it's mostly early Li-Ion battery devices that had that problem

One thing worse then overtemperature is overtemperature at charging.
One thing worse then overtemperature at charging is overtemperature at charging with charging >70-9x%.
That speeds up the side reactions of eventual metallic lithium deposits with dialkylcarbonate at high charge.

The reaction of either DMC or DEC with lithium was initiated at 180 K, a temperature much lower than their bulk melting temperatures, producing lithium methyl carbonate, methyllithium and lithium ethyl carbonate, and ethyllithium, respectively. At temperatures greater than 270−300 K, the lithium alkyl carbonates start to decompose with $\ce{Li2O}$, elemental carbon, and alkyllithium as products on the surface.

Those reactions decrease the cell capacity and increase the cell internal resistance.
Things go worse in summer.
The rule number one is never heavily use the phone ( e.g. at gaming ), while charging it at high percentages.
If it must be charged, then when it is just topping,
or better at low charge percentage
or the best not being charged at all.

Charging scheme is also a question of the cell static aging, that applies more to laptops than to short life cycle of mobile phones.
Another thing is that cells prefer rather  more of shallow cycles than less of deep cycles. That may be unfortunately counteracted by durability of charging connectors.
Fully charged cell is aging several times faster than about half charged one. As at full charge, lithium ions may form a metallic lithium deposit instead of lithium graphite intercalate, which very slowly reacts with dialkylcarbonate solvent.

I frequently charge my phone from my powerbank to avoid immobility of the phone during charging, what gives me the option to charge the phone whenever I want. I do not charge the phone overnight nor keeping  it long time being fully charged.  But the truth is, phone life is typically 2-3 years max, what does  not have much chance to show consequences of not optimal cell maintenance.
My Lenovo T400 laptop from 2009 with 40-80% battery maintenance had its battery ( in late 2017) in better condition than T430 from 2013. (70-75% versus 60% capacity). The latter running RHEL Linux was fully charging for several years, until I found the way to achieve 40-80% mode as well. Both were running mostly with AC charger attached on the office table.

So basically Pokémon GO is the death to all lithium batteries out there, especially in summer.

@Jan Ať least it is a challenge for them, close to dying very slow.

Explains why I regularly need a new phone battery O:)

@Jan If you do, that is why. Otherwise, I do not generally say you do. :-) My previous phone Sony Xperia M dual lasted about 4 years , until became progressively unusable for other than battery reasons, having the original battery.


text:
Tight binding model with Extended Huckel Approximation
I've been reading Albright's Orbital Interactions in Chemistry. In the chapter on solids, he provided a general approach to find the band structure of a solid state system

Now if we are to model a one dimensional hydrogen chain, let $|1s(r)\rangle$ denote the $1s$ orbital centered at $r$. The band energy at each $k$ point will just be $e(k)=\frac{H_{11}(k)}{S_{11}(k)}$ according to the secular determinant. I then use a first neighbor interaction simplification, so $H_{11}(k)=\langle 1s(r)|H_{eff}|1s(r)\rangle+2\cdot\exp(ika)\langle 1s(r)|H_{eff}|1s(r-a)\rangle$ and $S_{11}=1+2\cdot\exp(ika)\langle1s(r)|1s(r-a)\rangle$.
It's worth noticing that assuming orthogonality of neighboring orbitals will set $S_{11}=1$, then $\langle 1s(r)|H_{eff}|1s(r-a) \rangle$ and $\langle 1s(r)|H_{eff}|1s(r)\rangle$ are just analogous to $\beta$ and $\alpha$ parameters in the simple Huckel model. But for the sake of realism I’m considering overlaps(the textbooks is also doing that) That is why the plots below show an asymmetric split with regard to hydrogen 1s ground state energy.
So far so good. I can even get a quick numerical estimate of $H_{11}$ using the extended Huckel approximation because then I just need to know the overlap integral between neighboring $1s$ orbitals. However, when I plot the result in Matlab it's different from the plot in the textbook:

Especially, the energy at the edge of the Brillouin zone is too positive. Anybody knows why that might happen? Thanks!

Since it's been a few days and you still haven't received an answer, I thought I might draw your attention to a dedicated computational chemical physics community that you might be interested in. There is a proposal for a Stack Exchange for Materials Modeling: https://area51.stackexchange.com/proposals/122958/materials-modeling?referrer=ODVjZmM2YzJmOTc0MzBkZDVkMTZmMTIzN2EyNmZmNTQyOWYyY2Y0YWFmNGYxNTQ2OTQzNmMwOWIyYjhhOTZjMhwblBomlTMtPu8EgXeAKtp-8dFsxghojkD0vBSe1wZl0. Perhaps you might consider supporting it if you think it would be useful for you!

Already there's some tight-binding questions there!

I don't know offhand, but I'd guess that you're using somewhat different parameters. If you tune $\beta$ the graph will change, so I suspect that's the main difference. Your approach looks fine to first approximation.


text:
Production of Succinic Acid from Crude Glycerol using Fermentation
So I am writing a short paper on how one can produce Succinic acid from glycerol. I have to find a microorganism which can do that. I have found the following microorganism, which are traditionally being used for fermentation: Anaerobiospirillum succiniciproducens, Actinobacillus succinogenes, Mannheimia succiniciproducens and recombinant Escherichia coli strains. But the problem is that whatever article I find, they are written with respect to Glucose.
Does anyone know any article/paper/hints or some information they can share?
I just need to understand which organism can be used to do the fermentation, and how much succinic acid can be made from it.

I'm not an expert but saw this for E. Coli https://www.nature.com/articles/srep36526

Glycerol has 3 carbon atoms and succinic acid has 4. So you need (apart from some redox reactions) a way to add another carbon atom.
Also, succinic acid appears in the citric acid cycle. What you have to research are anaplerotic reactions. When you remove any of the metabolites (like succinic acid) from the cycle, you have to replenish them. One way is to turn pyruvate (3 carbons) into oxaloacetate (4 carbons), as shown in the wikipedia article on anaplerotic reactions:
$$\ce{pyruvate + HCO3− + ATP -> oxaloacetate + ADP + P_i + H2O}$$
Then, you have to find a pathway from glycerol to pyruvate. Here is a schematic of glycerol metabolism:

Source: https://www.researchgate.net/publication/224950286_Effect_of_impurities_in_biodiesel-derived_waste_glycerol_on_the_performance_and_feasibility_of_biotechnological_processes/figures?lo=1
As you can see, it is possible to turn glycerol into pyruvate fairly directly (using reactions of glycolysis).
To put everything together, you just need an organism that secretes succinic acid, and make sure that these pathways are turned on sufficiently that most of the provided glycerol gets metabolized to succinic acid. As this involved oxidation, the organism has to have access to oxygen or some other oxidation agent.
UPDATE
Prompted by Andrew's comment, I found a paper where they optimize succinate production from glucose. As you can see from the scheme above, glycerol can be funneled into glycolysis by two enzymes (using one ATP for phosphorylation and one NAD+ for oxidation). The enzymes used for the remainder of the reaction (and to provide NADPH via the pentosephosphate pathway) are shown in the figure from the paper below. You would need to provide some glucose to feed in the pentosephosphate pathway to have sufficient NADPH, or invoke some other pathways:

Source: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4983090/figure/Fig1/?report=objectonly

Succinic acid can be produced through oxidative or reductive reactions. The high-yield succinic acid-producing bacteria use both, balancing them so that no oxygen is needed. In addition to using substrate-level phosphorylation, some of these organisms can even gain ATP through coupling of fumarate reduction to oxidative phosphorylation.

@Andrew Based on your comment, I found https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4983090/pdf/12934_2016_Article_536.pdf. They start with glucose, though, not glycerol.

As you pointed out in your answer, glycerol is readily converted to the glycolytic intermediate DHAP, so the carbon source is essentially independent of the succinic acid production. Only difference is stoichiometry due to different redox state.

glucose is not required for sufficient NADPH. E. coli can grow on glycerol as a sole carbon source. If NADPH availability limits growth, many strategies are available to increase production. See https://www.frontiersin.org/articles/10.3389/fmicb.2015.00742/full.     2) the scheme lacks the oxidative pathway to succinic acid (via the normal direction of the TCA cycle), which is important for balancing the redox equivalents.

@Andrew Your comments are better than my answer - either your research is more extensive or you have more prior knowledge- I can’t tell which.

I used to work in industrial fermentation and metabolic engineering. So probably a bit more prior knowledge. . .


text:
Is it tennessINE because it has 117 electrons or because it behaves like a halogen?
The NPR News podcast and transcript The Convoluted Story Of How The First Atoms Of Tennessine Were Created includes the following exchange between science reporter Joe Palca and Oak Ridge nuclear engineer Julie Ezold:

PALCA: In other words, in five months, they only got six atoms of tennessine. When I first heard that name, I thought it was a mistake. All the other synthetic elements ended in -ium - think plutonium or americium. Oak Ridge nuclear engineer Julie Ezold understands my confusion.
JULIE EZOLD: We all thought it was going to be that way as well. But remember - the periodic table has rules.
PALCA: Ezold says Element 117 fits in a column of the periodic table filled with compounds called halogens - fluorine, chlorine, bromine.
EZOLD: Therefore, its name had to end in -ine. So instead of tennessium (ph), it's tennessine.

The more I thought about this, the more I questioned how elements discovered in the future might or might not be classified as halogens, especially if an "island of stability" is ever found and even more elements are discovered.

Is there really a rule or at least a firm guideline that ultimately requires the name of element 117 to end in -ine simply because it has 117 electrons?
Or was it somehow necessary to experimentally demonstrate some chemical property associated with halogens?
Or would an atom with 117 electrons certainly and without question behave like a halogen?
Or will all current and future elements that obeys certain numerical rules be called halogens no matter how they behave chemically?

It's worth noting that we see the same thing with element 118: it falls in the noble gas group and, consistent with this, ends in -on instead of -ium: oganesson.  See:  https://en.wikipedia.org/wiki/Systematic_element_name

It comes after Astatine so it is a kind of halogen, in the sense that it has expected valence configuration ns2np5. It is expected to be rather metallic/metalloid.

It is thought that tennessine will have very different chemistry from the lighter halogens. Astatine would already be something of an oddball, if only it could be isolated in reasonable amounts. The discontinuity of the chemical properties between the last common halogen (iodine) and tennessine would be very stark indeed.

Is it tennessINE because it has 117 electrons or because it behaves like a halogen? ... Or will all current and future elements that obeys certain numerical rules be called halogens no matter how they behave chemically?

Yup, that last bit is the most correct.  The math and physics behind how electrons form orbitals meant that scientists already knew that Element-117 was going to be one-electron-shy of filling that outermost shell of valence electrons (and they knew it for decades before they were actually able to synthesize it in a lab).

[Tennessine has a -INE name like all the other halogens. Did they test whether it actually acts like 'a halogen'? Or is it simply assumed to behave like the rest? Or does it get to be an honorary halogen just because its electron configuration shoehorns it into the same column as others?]

The half-life for the Tennesine-293 isotope are only about 25ms (and 51ms for Ts-294).  That is only really enough time to see how the atom continues traveling through the particle accelerator and tracking the pieces after it decays (to make sure all the masses add up correctly).  It is not enough time to do the kind of "real chemistry" of testing whether Tennessine "acts like a proper halogen".
So, is it assumed/calculated to behave like the other halogens? Well that depends what you mean exactly. Fluorine, Chlorine, Bromine and Iodine are all "Non-Metals" whereas Astatine might be a "Metaloid".  Tennessine is likely to be even more-so metallic than Astatine - being either a definite "Metaloid" or maybe even a full-on "Metal".  One could possibly argue that being a metal disqualifies it from being a 'proper' halogen (while still being Group-17)... but Astatine has been skating by on its shaky credentials since the 1940s so it's probably fine to let them both keep the -INE names.  Also, it's not like folks are all that strict about differentiating 'proper' Chalcogens or 'proper' Pnictogens, so why start now with 'proper' Halogens?  Sure, it is a bit of a subjective judgement call, but most of the chemistry world is on board with it.

... I questioned how elements discovered in the future might or might not be classified as halogens, especially if an "island of stability" is ever found and even more elements are discovered.

This might be trickier to answer.  Part of the reason why scientists were so easily able to just fit Tennessine into the modern table was because Period-7 follows the same pattern of orbitals as Period-6.  Period-6 elements mostly have stable isotopes so this essentially gave a mostly-complete template for what the Period-7 row should also look like. (And it worked like a charm as scientists filled out the gaps in the Actinide series and the gaps in the heaviest transitions metals from like 1930 to now.)
We have zero examples of elements in Period-8, so we can only hypothesize that there is a g-block of orbitals that gets added on.  Assuming that is all correct, the expected number of s/p/d/f/g electrons needed to fill a Period-8 shell means that the next 'halogen' element would need to have an atomic number of about 167-171... which is a long way off if only fleeting unstable isotopes are left to discover.
If there is an "Island of Stability" then we might luck out and be able to learn more about the electron configuration of Period-8 enough to either keep the g-block idea or scrap it for something better based on that new evidence... but either way you look at it, another new halogen seems unlikely anytime soon.

I vaguely remember that there is an equation which predicts the maximum atomic number any hypothetical element could have. But I clearly remember that limit is quite lower than the estimated atomic number of the next possible halogen.

@Priyansh2003 I think I saw a $2n^2$ pattern mentioned somewhere while googling all of this that describes the (2, 8, 18, 32...) pattern of the (K,L,M,N...) shells which would mean an O-shell of 50 pins the next Noble Gas at 168.  But https://en.wikipedia.org/wiki/Extended_periodic_table seems to indicate that the latest simulations favor the idea of a Noble Gas occurring at 172 (if I understood correctly, due to relativistic effects letting more electrons squeeze into the shell that would normally be allowed by the $2n^2$ rule).

Excellent answer +1. Iodine already shows incipient metallic character as bonding is becoming delocalized in the solid. For tennessine, Syysziński (pp. 139-146) reports that a compound with nihoinum would have negative charge on the nihonium.

Tennessine was named after the land of Tennessee in USA. The element was created by a joint team of Russian and American researchers. When it was finally confirmed to have recreated by a collaboration of German and American scientists, IUPAC and IUPAP handed over the right to name the element to the former team of discoverers.  They suggested to name it Tennessine after the land of Tennessee in USA.
The element's name ends with -ine instead of -ium probably because it is in group 17 all of whose elements are non-metals as names of most metals preferentially ends with a -ium.
It's not possible as of yet to determine its experimental properties given that only six atoms of Tennessine have been created. This is why we can't establish its chemical similarities or dissimilarities with other halogens.
Future elements can be named in many other ways that we don't yet know but maybe scientists would prefer to keep certain trends because periodic table is intended to show patterns and rules.

Thank you for your answer! I'm still wondering; as discussed in my question, how does one know that element 117 is in group 17? Were any halogenic properties used to isolate the atoms, or is this from some calculation, or from some simple rule?

The main criteria to put it in Group 17 is it's electronic configuration ns²np⁵. It matches with general outer configuration of halogens - 7s²7p⁵.

Got it, I guess this is something that a chemist would know intuitively starting with the number 117.

Probably, because all halogens, including Tennessine ends with a -ine.

I guess I'm still trying to understand how one starts with 117 and arrives at 7s²7p⁵ with sufficient certainty to call it a halogen. To my question "Is it tennessINE because it has 117 electrons or because it behaves like a halogen?" I think your answer is "It's -ine because it has 117 electrons, which means we expect its configuration to be 7s²7p⁵  which would then mean it's a halogen which would then mean we'd use -ine." Is that a fair summary of your answer?

To summarise, element 116 (Livermorium) was discovered before element 117. So it should naturally come after Livermorium. Also it's name has nothing to do with it being 117. Because only temporary names are assigned according to atomic numbers which was Ununseptium for Tennessine. 'Un' means 1 and 'sept' means 7 which make Ununseptium 117.


text:
Why is CO2 more likely to dissolve in a basic solution than an acidic solution?
I'm having trouble thinking through this problem. I know that when $\ce{CO2}$ dissolves in water it can release $\ce{HCO3-}$ and $\ce{H+}$ and cause a drop in pH. But can I use this to justify why $\ce{CO2}$ (g) might dissolve more easily in a basic solution than an acidic one? Is the water acting as the base?

Hydroxide ions keep concentration of dissolved carbon dioxide low, converting it to bicarbonate and eventually carbonate. Therefore while alkalic enough, the solution does not gets saturated.

A simple way to think about this is as follows (assuming we are talking about an aqueous solution).
The dissolution of $\ce{CO2}$ in water is followed by three chemical reactions:
$$
\begin{align}
\ce{CO2 + H2O &<=> H2CO3}\label{rxn:R1}\tag{R1}\\
\ce{H2CO3 &<=> HCO3- + H+}\label{rxn:R2}\tag{R2}\\
\ce{HCO3- &<=> CO3^2- + H+}\label{rxn:R3}\tag{R3}
\end{align}
$$
Now the more basic the solution becomes, the further reactions \eqref{rxn:R2} and \eqref{rxn:R3} (\eqref{rxn:R3} becomes significant at high pH) lie to the right as hydroxide consumes $\ce{H+}$. As reaction \eqref{rxn:R2} depletes $\ce{H2CO3}$, reaction \eqref{rxn:R1}  is also forced to the right. The net result is more $\ce{CO2}$ residing in solution (dissolved) as the solution becomes more basic.

The carbonic acid equilibrium lies massively to the left,  but will be driven to the right  by the hydrogencarbonate formation. However, you also have the hydroxide acting as a direct nucleophile on carbon dioxide which cannot be discounted.

There is need to extend the equilibrium chain by the carbonate formation in alkalic solutions.

The forward part of reaction (1) is slow, with the rate constant 23 s^-1


text:
Safety requirements for the laboratory work with salts of heavy metals
So I am going to start a set of experiments involving the following metal nitrates:

Cobalt nitrate hexahydrate
Barium nitrate
Cerium nitrate hexahydrate
Ammonium nitrate nonahydrate
Iron(II) nitrate nonahydrate

All the products are from Sigma-Aldrich.
As I was going through the SDS data sheets, my gloves and googles matched the safety standard on all these products, except my gas mask.
I have a P2 level gas mask, but the recommended is a P3 level mask.
I will be performing all my experiments in a standard fume hood and it has all the required ventilation.
Do I still need the P3 level mask or the fume hood is enough?
Asking this because I am kinda scared doing these experiments as I will be working with heavy metals for the first time.

SDS are tailored on industrial production and in the vast majority of cases the common-and-good sense of a chemist suffices. LiF has mask or respirator mentioned too, so I suppose this is necessary in a mine or a mill, just to say. I would personally don't worry at all. SDS are essentials but in a small scale research lab shall be read with chemistry in mind. A big plant with workers doing all kind of tasks everyday for a life or that poses hazards is another story. To be even clear: diethyl ether is of common use in every lab. Not so in industry.

Will you be weighing/dispensing in a suitable fume hood though?

Wow, I guess you're really new to this. Just use what teacher tells you. You don't need to know any SDS.

Generally, personal protective equipment (PPE) is the last measure. First the employer has to check whether the risks to safety and health can be avoided (e.g. by replacing the toxic substances with substances that are not toxic). After that, the risks must be limited by technical means or collective protection or by measures, methods and procedures of work organisation (e.g. removal of dust by suction close to the source). Only if risks cannot be eliminated or adequately controlled on a general or collective basis, the employer resorts to providing PPE. (Note, for example, that PPE only protects the wearer.)
The corresponding general rule in the European regulation (89/656/EEC) reads as follows.

Personal protective equipment shall be used when the risks cannot be avoided or sufficiently limited by technical means of collective protection or by measures, methods or procedures of work organization.

In your case, the fume hood comes first, after that you can resort to using a suitable mask with particle filter. Therefore, if the risks are sufficiently limited by technical means (the fume hood), you do not need PPE (the mask with particle filter).


text:
Easy way to make carbonic acid from other chemicals?
I would like to test for the reaction between carbonic acid and copper (to simulate the effect of acid rain on copper). However, I find that many supplies failed to provide carbonic acid (either in powder or liquid form). Some says that carbonic acid is unstable and thus cannot supply it. Therefore, I want to make it myself. Are there any simple ways to make carbonic acid from other chemicals (e.g. sodium carbonate?).
Also, I have found some carbonic acid powder in some commercial product such as this one.
But it is for bathing instead of chemical use. Can they produce carbonic acid?
Any answers and comments are welcome. Thank you very much for helping.

Carbonic acid does not exist

Carbonic acid does not exist either in powder either in liquid form near normal conditions.. ( It can be created and detected at special kryogenic or gaseous phase conditions  not applicable for its usage as "on shelf acid".)
In a pure form, it exists only as its salts: bicarbonates ( of alkali metals ) and carbonates.
As an acid, it exists at ambient conditions only in minor concentration in water solutions of carbon dioxide ( soda water, mineral water ) and/or bicarbonates ( drink water, mineral water, baking soda solution ), with the equilibrium strongly shifted in favour of the oxide.
$$\ce{CO2(aq) + H2O <<=> H2CO3(aq)}$$

Note that the acidity of acid rain is not based on carbonic acid, that gives to natural rain just mild acidity about $\mathrm{pH=5.7}$.
Acid rain contains traces of strong mineral acids as sulphuric and nitric acid, that are formed from oxides of sulphur and nitrogen, present in air.
I suggest for the testing to use very diluted ( 0.1-1 mM ) solution of one of the above acids, or their mixture.

Carbonic acid has been reported in the solid phase, though not in a form you can easily make and put in a bottle - https://www.sciencedirect.com/science/article/abs/pii/0584853991800973?via%3Dihub . See also https://en.wikipedia.org/wiki/Carbonic_acid

Yes, I know, but I tried to keep it simple for the OP. And yes, I have read and provuded the Wikipedia link as well. Anyway, thanks for comment.

@IanBush at any drastic system, you can make chemicals that are not known to exist at room temperature. You can even make compounds of krypton at such systems. So, though Carbonic acid is not known to exist at room temperature theoretically, by changing the conditions drastically like super low temperature, super high pressure, you can confirm its existence.

I think Perrier has a pH about 4.5. Other carbonated mineral waters are probably similar. These would be a good approximation to acid rain. I suggest using a closed vessel to contain the copper, the liquid, and some empty space to contain CO2 and O2 to oxidize the copper.

It is better to follow the real acid rain conditions than simulate it by unnecessarily very different system. E.g the carbonised water has much higher buffering capacity than very diluted mineral acids, brings bubbly conditions and extra support for forming cupper carbonates.


text:
Why upon melting of PEO between crossed polars colours appear?
Crystalline PEO when under crossed polars shows distinct Maltese cross pattern. Being birefringent, when polymer crystal align with the polarizer or analyser it shows extinction regions (effect known as zero amplitude extinction). And as PEO crystallize in spherulites, it does lead to Maltese cross pattern. (As parallel and perpendicular crystal fibrilis lead to extinction). Because the fibrilis twist upon radial growth it also leads to banded extinction regions, known as zero birefringence extinction.
Upon melting, first the crystal fibrilis realign themselves enhancing contrast of crystalline spherulite, and then I can observe some of the colours appearing in the banded structure, that I haven't seen before. Why does it happen?

What is abbreviated by "PEO"?


text:
Nomenclature of ester with cyclic substituents
I have to tell the name of the following compound

The name is cyclopropylmethyl cyclobutanecarboxylate. All of it makes sense but I don't understand why the suffix carboxylate comes? Why are we writing the single carbon left unaccounted for by using carboxyl ? Kindly clarify.

The group -COO- is called "carboxylate" if it is included in a molecule with some organic radicals attached on its left and on its right-hand side. It is a definition, due to the fact that there is a carbon atom bound to an oxygen atom. Joined together, it makes carboxy.
This terminology is used when no simpler way of naming this structure exists.
Examples.
The structure CH3-COO- may be called methylcarboxylate. But usually it is called acetate. It is shorter.
The structure C6H5COO- may be called phenylcarboxylate. But usually it is called benzoate, as it is a shorter name.

The names of esters are usually formed by placing the alcoholic component in front of the name of the acid component as a separate word. The corresponding rules in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) read as follows.

P-65.6.1 General methodology
Neutral salts and esters are both named using the name of the anion derived from the name of the acid. Anion names are formed by changing an ‘-ic acid’ ending of an acid name to ‘-ate’ and an ‘-ous acid’ ending of an acid name to ‘-ite’. Then, salts are named using the names of cations, and esters the names of organyl groups, cited as separate words in front of the name of the anion.

Since the acid component of the ester that is given in the question is cyclobutanecarboxylic acid, the corresponding ester is named as … cyclobutanecarboxylate.
However, that does not completely answer your question why the extra carbon atom is named using the ‘carboxy’ part.
Simple carboxylic acids are named substitutively using the suffix ‘oic acid’ as follows.

P-65.1.2.1 Carboxylic acid groups, $\ce{-COOH}$, that conceptually replace a $\ce{-CH3}$ group of methane or terminate an unbranched hydrocarbon chain are named by replacing the final ‘e’ of the name of the corresponding hydrocarbon by the suffix ‘oic acid’. No locants are necessary to denote the positions of the carboxylic acid groups in a hydrocarbon chain; locants are used when hydrocarbon chains are modified by skeletal replacement, as shown in P-15.4.3.2.3. Except for formic acid, acetic acid, oxalic acid (see P-65.1.1.1), and oxamic acid (see P-65.1.1.1), systematically formed names are preferred IUPAC names; the names given in P-65 .1.1.2 are retained names for use in general nomenclature.

Thus, the ‘carboxy’ part is not used in such simple cases. For example, pentanoic acid is not named ‘butanecarboxylic acid’.
However, the acid component of the ester that is given in the question is not such a simple case since the $\ce{-COOH}$ group does not terminate the chain of the parent structure (the cyclobutene ring). In such cases, the suffix ‘carboxylic acid’ is used instead.

P-65.1.2.2 The suffix ‘carboxylic acid’ is used for all carboxylic acids not covered by P-65.1.2.1, except for benzoic acid, a retained name (see P-65.1.1.1). The carboxy group can be attached to any atom, carbon or heteroatom, of any parent hydride; (…).


text:
College Chemistry - Raised Boiling Point
Determine the molar mass of an unknown if you dissolve enough of the unknown into Benzene to make a 1.55% mass percent mixture of unknown to benzene and find that the boiling point of Benzene was raised by 2.3%
For this question, I just have a problem understanding how to get the molar mass without having grams or moles given. Now I don't know if I have to chose my own grams or if there is another way to find the molar mass of the unknown.

You have mass fraction, that is what you need. Mass alone is not useful. Molar mass of unknown then relates mass fraction and molality.

The question is not clearly defined. Saying that the temperature is increased by 2.3% has no meaning. Which temperature scale ? The absolute temperature in Kelvin ? The usual temperature in centigrade degrees ? in Fahrenheit ? 2.3% of 0°C is zero. 2,3% of 273 K is about 5 K. That is not the same.

This question has a lot of flaws to be a college chemistry question. Therefore, I decided to give some insight even though this is a clearly a homework question.
The question did not have enough data such as the ebullioscopic constant ($K_b$) of benzene and the van't Hoff factor ($i$) of the solute. At least you can find the ebullioscopic constant of benzene as $\pu{2.53 K\:kg\:mol-1}$ from Wikipedia. If we assume the van't Hoff factor ($i$) of the solute is unity, we can solve this problem without much difficulty. However, we also have to assume that the percent temperature elevation (although it is unitless in %) is given in Kelvin scale ($\pu{K}$) because of units in $K_b$. If it was given in Celsius ($\pu{^\circ C}$), this percentage would change.
Suppose molar mass of unknown is $M_u$. If the solution is 1.55% mass percent ($w/w$) mixture of unknown to benzene,  the masses used are $\pu{1.55 g}$ of unknown and $\pu{(100-1.55) g}=\pu{98.45 g}$ of benzene to make $\pu{100 g}$ of solution. Therefore, molality of the solution ($m_b$) is:
$$m_b = \frac{(1.55/M_u)\ \pu{mol}}{\pu{98.45 g}\times \frac{\pu{1 kg}}{\pu{1000 g}}} = \frac{1550}{98.45M_u} \pu{mol/kg}$$
Now we can apply the boiling point elevation: $\Delta T = iK_b m_b$, where $K_b = \pu{2.53 K\:kg\:mol-1}$ and we assumed $i=1$. Numerically:
$$2.53 \times \frac{1550}{98.45M_u} = \Delta T = \left(\frac{\Delta T}{\pu{353.2 K}} \times 100 \right) \times \frac{\pu{353.2 K}}{100} = 2.3 \times \frac{\pu{353.2 K}}{100} $$
$$\therefore \; M_u= 2.53 \times \frac{1550}{98.45} \times \frac{100}{2.3 \times 353.2 }  = \pu{4.90 g/mol}$$
This is a awfully small number for a non-gaseous compound. The only wrong assumption should be the boiling point of benzene in Kelvin so let's make the switch to Celsius scale, which is $\pu{80.1 ^\circ C}$:
$$2.53 \times \frac{1550}{98.45M_u} = \Delta T = \left(\frac{\Delta T}{\pu{80.1 ^\circ C}} \times 100 \right) \times \frac{\pu{80.1 ^\circ C}}{100} = 2.3 \times \frac{\pu{80.1 ^\circ C}}{100} $$
$$\therefore \; M_u= 2.53 \times \frac{1550}{98.45} \times \frac{100}{2.3 \times 80.1 }  = \pu{21.62 g/mol}$$
Which is a better number.


text:
Does myristyl glucoside contain alcohol?
I read that "myristyl glucoside is a product obtained by the condensation of myristyl alcohol with aglucose polymer." Does that mean that alcohol is part of the final product/myristyl glucoside itself? Or is all of the alcohol removed during the making of it?

The  myristyl alcohol becomes part of the final product by chemical reaction. Any excess will be removed by processing after the reaction.


text:
Atomic Orbital and Wave Function
I was reading about the atomic orbital in my chemistry textbook. It says that the atomic orbital (psi) is a mathematical wave function that depends on the coordinates of the electron. It is also mentioned that psi does not carry any physical meaning. That part is understood, however I could not understand what a psi vs r graph represents.(r is the distance from the nucleus) Can someone please clarify ??

You need to roll back your understanding, because it is wrong. $\psi$ does carry a great deal of physical meaning.

Thanks for the response , but then my NCERT textbook says exactly that " The wave function is a mathematical function whose value depends upon the coordinates of the electron in the atom and does not carry any physical meaning." (psi ultimately refers to a wave function) How should I interpret that ?

Maybe the book will later talk about Born's interpretation, and will mention that physical meaning was later attributed to the square power of psi, which represents the orbital's electron (probability) density

Yes that is correct, it does say that psi squared refers to the probability density of the electron. However, I wanted to know how I can understand a psi vs r graph ??

I do understand that a psi squared vs r graph would represent the probability density of the electron in an orbital at a distance r from the nucleus, obviously.

$\psi$ does carry a great amount of physical meaning. The annoying part is, that it doesn't represent anything directly observable. The way (non-relativistic) quantum mechanics is constructed, makes $\psi(x_i, s_i, t)$ ($x_i$ - physical coordinates of $i$-th electron, $s_i$ - spin projection on $z$ of said electron) -  a direct analogue to trajectory, or, in case $\psi$ doesn't depend on $t$ (time),  orbit. *
Orbitals themselves, however, do not have direct physical meaning in that sense that they are nothing more than building blocks we use to construct collective $\psi$ of the electrons in the chemical system after we apply Born-Oppenheimer approximation (that we can ignore nucleus movement when we talk about electron movement). It's a good approximation, so it is almost always in effect in quantum chemistry.
In some special cases (hydrogen-like atoms) the electron wave-function represents exactly one electron of one atom and, ignoring spin contribution, orbital equals wave-function of that one-electron system.
Orbitals of hydrogen-like atoms can be represented ** as $\psi(x)=\psi(r,\theta,\phi)=\psi_r(r)\psi_{angle}(\theta,\phi)$. It is a useful representation, because even if $\psi_r$ changes from level to level, $\psi_{angle}$ doesn't, so you can see parallels between, say, $2p$ and $3p$ orbitals, which differ*** only by $\psi_r$. Again, neither part of the orbital is observable. What IS observable is "electron density" i.e. a probability to observe electron in this particular finite (small) volume. Those can be extracted from various electron or X-ray diffraction experiments. What, however, it is good at is to point out, how good outer electrons shield inner electrons and that chemical properties are mostly guarded by outermost reaches of outer shells.
$\psi_r$ on its own cannot be observed and has no simple interpretation. It has no inherent meaning on its own, and I'm unaware of obvious interpretation. It IS useful when one has to compare different orbitals between each other or to compare various approximations of the same orbital though.
$\psi^2_r$ is fairly useless, imho. It is observable for $s$ orbitals as electron density at given radius, but it isn't very useful. What is useful IMO is density integrated over spherical surface, which for $s$ orbitals is $ |rR|^2 = 4\pi r^2 |\psi_r|^2 $, which shows how much time electron spends at given distance from nucleus. It shows, how diffuse the orbital is, and when plotted against observed total electron density, it shows how good electron is shielded against outer influence.
I hope this helps.
* Note: Here is one of the non-obvious differences between quantum and classical systems lie: you CAN have a quantum system with large kinetic energy and $\psi$ independent on time.)
** Note: it is not the case for all mathematical functions, we are lucky that it is the case for atomic orbitals.
*** Note: orbitals on the same level CAN differ by $\psi_r$. The general rule is that orbitals on the same level have the same number of surfaces where $\psi$ is zero that is equal number of the level minus one. That means that we have only $s$ orbitals on the first level with zero such surfaces, but $2s$ and $2p$ orbitals. For $2p$ orbitals the "zero" surface is one of coordinate planes and for $2s$ orbital$ it is a sphere around the nucleus.

Thanks for the answer. It does give some more clarity. Please do note that my main question was "how I can interpret the psi vs r graph  ? " and it would be great if you can help with that. Also, what would be the difference between a psi squared vs r and a psi vs r graph ?? (basically that's where I got confused when I was reading my textbook)

@AuravSinghTomar I edited in my opinion on the matter.

Thanks for the edit. And the help too !


text:
Hydrolysis of sodium carbonate
This is the equation given by my textbook for hydrolysis of sodium carbonate:

$$\ce{Na2CO3 + 2 H2O -> H2CO3 + 2 Na+ + 2 OH-}$$

and it mentions that sodium ion $(\ce{Na+})$ does not tend to combine with the hydroxide ion $(\ce{OH-})$ and I was wondering what prevents them from combining together to form $\ce{NaOH}.$

What do you mean by "combine", exactly? A precipitation? If so, you might want to look up solubility of NaOH in water and read about solvation.

@andselisk i mean what prevents them from reacting together to form NaOH

You didn't answer the question. There is NaOH all right, it's dissolved in water. Also note that NaOH is a formula unit for ionic solid, so there is no reason to expect NaOH molecules, if that's what you mean.

Additionally, the textbook is wrong, as carbonate hydrolysis to bicarbonate. $$\ce{CO3^2- + H2O <=> HCO3- + OH-}$$ Carbonic acid is formed just in traces $$\ce{HCO3- + H2O <=> H2CO3 + OH-}$$, as the latter reaction is in alkalic environment pushed strongly toward the left side.

Apparently you believe that $\ce{NaOH}$ may be an independent molecule. It is not in this case.
$\ce{NaOH}$ does not exist in a solution. $\ce{NaOH}$ does not exist in the solid state either. $\ce{NaOH}$ does not exist in the solid state as a molecular compound. In the solid state, it is made of a huge pile of $\ce{Na+}$ and $\ce{OH-}$ ions, exactly like a salt grain is made of a pile of $\ce{Na+}$ and $\ce{Cl-}$ ions. On the average, there is the same amount of $\ce{Na+}$ and $\ce{OH-}$ ions. But $\ce{NaOH}$ does not exist as a separate molecule.
When dissolved in water, the water molecules are going to separate the ions, which become independent from one another.


text:
Determination of the iodine value of a fatty oil: Does the concentration of Na2SO3 effect the iodine value?

I want to determine the iodine value of a fatty oil. For this I dissolve $\ce{1.0 g}$ of the oil in chloroform. Iodine bromide is then added and placed in a dark room for at least $\pu{30 min}$. After this, $\ce{KI}$ is added. Then, I am going to titrate with $\pu{0.1 M}$ $\ce{Na2SO3}$. Based on the added volume, I can calculate the iodine value of the unknown fatty oil.

Now I was wondering what the effect should be using $\pu{0.02 M}$ $\ce{Na2SO3}$ instead of $\pu{0.1 M}$? And is there a reason why the solution has to stay $\pu{30 min}$ in a dark room? I think there's no difference in outcome when it will be in a dark room, for example, say $\pu{3 h}$.

The reaction of $\ce{I2}$ or $\ce{IBr}$ with double bonds is slow. You have to wait half an hour for the reaction to be finished. But of course, you may wait a longer time, maybe $\pu{3 h}$ as you suggest. Or one night. Usually we are in a hurry, and we are looking for ways of avoiding to spend too much time for a given operation. That is why the procedure recommends half an hour.
And of course you may carry out the titration with a solution 5 times more diluted than recommended. The only drawback is that you may have to refill many times your burette before going to the end of the titration. Suppose $\pu{42 mL}$ of the $\pu{0.1 M}$ solution is required for performing the titration. This could be done with a usual $50$-$\pu{mL}$ burette. With a $\pu{0.02 M}$ solution, you will use $5 \times \pu{42 mL} = \pu{210 mL}$. In this case you will have to refill four times your burette. Four times! Bad luck!

I have one more question: why is the oil placed in a dark room? I understand that there will occur side reactions if not, but which side reactions exactly? Is it because of the formation of bromine radicals or will the iodide ions be oxidized to iodine?