text:
Copper (II) Acetate from 5% vinegar + salt, electrochemically
I'm trying to create Copper (II) Acetate crystals, but in these times of Coronavirus it's difficult to come by hydrogen peroxide. I could be patient, but I'm not, so I'm trying to make it electrochemically. I have 5% vinegar and lots of copper scrap, and an adjustable power supply. Unfortunately I can't barely get any current going, so I've thought of adding salt, regular NaCl. I'm curious what effect this will have on the final outcome though. Balancing equations is something I struggle with, but I'd really like to learn the chemistry here. Will the salt interfere with or alter the growth of the copper acetate crystals? Ultimately I'm trying to make calcium copper acetate crystals, I've already made the calcium acetate.

Another ways is to let time and aerial oxygen to oxidate cupper, leading to copper acetate.  $$\ce{4 HAc + O2 + 2 Cu -> 2 H2O + 2 Cu(Ac)2}$$
I once performed vinegar decalcification of aged boiling kettle. Its heating spiral had exposed cupper, with the stainless coating already mostly stripped away. I have shortly boiled vinegar and forgot it to let stay overnight. The solution got pretty blue.

Interesting. I thought about using a small air pump to bubble air through. As it is it's stood on a "hot plate" (coffee maker base wired to 20V DC, gets to about 200°F), with my small copper scrap in the bottom (lots of small pieces of 12 gauge solid wire). It's been about 12 hours and it's very slightly more blue than it was. I am impatient.

With NaCl, there is danger copper would end bound in $\ce{CuCl4^2-}$

Welllll, all I had was a cheap 12v tire pump and some silicone tubing. I closed one end of the tubing and poked a bunch of needle holes in that end, then stuffed the other end into the tire chuck and applied 5v, to make it run a little more slowly. It's chugging along pumping a bunch of small bubbles through the copper. Not quiet, I probably won't leave it running, but it works!

I suppose forcing air must be much more efficient than heating, as heating expels oxygen from solution.

To prepare copper acetate absence H2O2, employ a known method from hydrometallurgy to process copper ore employing aqueous ammonia, air (a source of oxygen) and a small amount of salt (acting as an electrolyte for this, in part, spontaneous electrochemical reaction detailed below). This results in tetra-ammine copper hydroxide. The latter exists only in solution and upon evaporation yields CuO (and possibly some Cu2O also, see this old patent). Add an acid of choice (like vinegar, a source of acetate) to convert CuO into copper acetate.
Related copper chemistry with ammonia and oxygen:
Cited half-reactions:
$\ce{1/2 O2 + H2O + 2 e- -> 2 OH-}$ (cathodic reduction of O2 at surface of the Copper)
And, at the Copper anode, the formation of the complex:
$\ce{Cu + 4 NH3 + 2 H2O -> [Cu(NH3)4(H2O)2](2+) + 2 e-}$ (anodic dissolution of Cu by a complexing agent)
With an overall reaction:
$\ce{Cu + 4 NH3 + 1/2 O2 + 3 H2O -> [Cu(NH3)4(H2O)2](2+) + 2 OH-}$
Also, some interesting cited standard chemical reactions, where copper oscillates between cuprous and cupric states as noted by the source below:
$\ce{2 Cu + 4 NH3 + 1/2 O2 + H2O -> 2 [Cu(NH3)2]OH}$
$\ce{2[Cu(NH3)2]OH + 4 NH3 (aq) + 1/2 O2 + H2O -> 2 [Cu(NH3)4](OH)2}$
$\ce{Cu + [Cu(NH3)4](OH)2 <--> 2 [Cu(NH3)2]OH}$
Reference: "Kinetics and Mechanism of Copper Dissolution In Aqueous Ammonia" and a related work detailing reactions: ‘Copper Dissolution in Ammonia Solutions: Identification of the Mechanism at Low Overpotentials’, which is also a fully available PDF.
Some alternate preparations would be to follow paths to copper patina (see this site, which discusses three paths for the home chemists which includes an embodiment of the ammonia prep, and, with the addition of vinegar, would convert the basic copper carbonate to copper acetate.

Thank you, that's an awesome answer. I'll read more closely when I get home and I'm not on my phone, I think I have some 10% ammonia left from making ammonium sulfate. Thank you for the detailed information!


text:
Question on notation

Here, when they write $v/(\mathrm{mol\ s^{-1}\ (kg\ \text{protein})^{-1}})$ is $0.30$ for the first row first column, do they mean that the value is $0.30\ \mathrm{mol\ s^{-1}\ (kg\ \text{protein})^{-1}}$ or $0.30/(\mathrm{mol\ s^{-1}\ (kg\ \text{protein})^{-1}})$?

$v/x = 0.30$ means that $v = 0.30x$; this applies whether $x$ is a number, a unit, or a collection of units.

The expression
$$v/(\mathrm{mol\ s^{-1}\ kg^{-1}})=30$$
means
$$v=30\ \mathrm{mol\ s^{-1}\ kg^{-1}}$$
This notation is used in the table, so that the entries are all just numbers without any unit symbols.
For an explanation, I copy the following section from my meta answer, which is mainly a collection of rules and examples taken from various standards and guides (see the meta answer for references).

Unit symbols are mathematical entities and not abbreviations. In expressing the value of a quantity as the product of a numerical value and a unit, both the numerical value and the unit may be treated by the ordinary rules of algebra.
(…)
In expressing the value of a quantity as the product of a numerical value and a unit, both the numerical value and the unit may be treated by the ordinary rules of algebra.

the equation $T=293\ \mathrm K$ may equally be written $T/\mathrm K=293$

It is often convenient to write the quotient of a quantity and a unit in this way for the heading of a column in a table, so that the entries in the table are all simply numbers.

$$\begin{array}{ll}
\hline
T/\mathrm K & p/\mathrm{kPa} \\
\hline
300.00&3.5368\\
310.00&6.2311\\
320.00&10.546\\
330.00&17.213\\
\hline
\end{array}$$

The axes of a graph may also be labelled in this way, so that the tick marks are labelled only with numbers.

do not write “mass [kg]” for “mass in kilogram”

Also note that expressions for units shall contain nothing else than unit symbols and mathematical symbols. Any attachment to a unit symbol as a means of giving information about the special nature of the quantity or context of measurement under consideration is not permitted. Therefore the use of the unit “$\mathrm{mol\ s^{-1}\ kg^{-1}\ \text{protein}}$” as written in the question is not permissible. The correct unit is $\mathrm{mol\ s^{-1}\ kg^{-1}}$.

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text:
Does tris(ethylenediamine)iron(III) exhibit optical isomerism?
I'm studying 12th. In my book they mentioned that $\ce{[Co(en)3]^3+}$ exhibits optical isomerism.
But in that same book, there is a question

Which of this doesn't exhibit isomerism?

and the answer is $\ce{[Fe(en)3]^3+}.$ Since both complexes have the same structure, then why they are saying $\ce{[Fe(en)3]^3+}$ doesn't show isomerism?

It would be nice if you could add the citation for your book, including author(s), title, edition, publisher, year, page number(s) and IDs (e.g. ISBN). On Chemistry.SE, ACS citation style is preferred.

It is my textbook!!

Good news: you can cite your textbooks as well.

Okay. They gave these options..

a) [Ni(NH3)4(H2O)2]2+

b) [Pt(NH3)2Cl2] c)[Co(NH3)5SO4]Cl d)[Fe(en)3]3+     Which should I choose?

That's not at all what I'm asking. "To cite" means to provide extensive bibliographic information about your source.

Apparently Vijay refuses to answer Andselisk's question. He refuses to give the author's name of his book,

I think the answer given in your textbook is either incorrect or a misprint. I let you find out the correct answer from given choices.
However, I'd like like to claim that the given answer is wrong because of following facts:

Tris(ethylenediamine) complexes of transition metals have been studied for more than 100 years and $\ce{M(en)3^3+}$ complexes are been considered chiral (e.g., Ref.1 for studies on $\ce{Co(en)3^3+}$).
The enantiomeric nature of these complexes has been verified by X-ray crystallography (e.g., Ref.2 for studies on $\ce{Co(en)3^3+}$).
The crystal structures of the first-row transition-metal series of tris(ethylenediamine) diacetate complexes have been recently studied (Ref.3) to show their chiral nature, and following common structure is given for the complexes with $\ce{M^2+}$ central metal:

The abstract states that:

The crystal structures of the first-row transition-metal series of tris(ethylenediamine-$\kappa^2$N,N′)metal(II) diacetate, $\ce{[M(C2H8N2)3](CH3CO2)2}$, with $\ce{M = Mn, Fe, Co, Ni, Cu,}$ and $\ce{Zn}$, are reported. The complexes are all isostructural, crystallizing in a centrosymmetric triclinic cell and possessing an asymmetric unit composed of one $\ce{[M(en)3]^2+}$ cation and two symmetrically independent acetate anions. In the unit cell, the two complex cations are inversion-generated enantiomers, possessing the energetically favoured Δ(λλλ) and Λ(δδδ) configurations. The complex cations and acetate anions combine through a series of $\ce{N—H⋯O}$ hydrogen bonds to generate a three-dimensional network in the crystals. The other notable feature of the series is a significant Jahn–Teller distortion for the $\mathrm{d^9}$ $\ce{Cu^2+}$ complex.

It is a fact that structures of these same first-row transition-metal series with 3+ oxidation state would give the same octahedral shapes, the cation of which are chiral (e.g., Ref.1 and Ref.2). Examples of few are given below:

Overall, it is true that the $\ce{[Fe(en)3]^3+}$ cation is chiral.
References:

A. Werner, "Zur Kenntnis des asymmetrischen Kobaltatoms. V,"  Berichte der deutschen chemischen Gesellschaft 1912, 45(1), 121-130 (https://doi.org/10.1002/cber.19120450116).
Y. Saito, K. Nakatsu, M. Shiro, H. Kuroya, "The first-row transition-metal series of tris(ethylenediamine) diacetate Determination of the absolute configuration of optically active complex ion, $\ce{[Co(en)3]^3+}$, by means of X-rays,"  Acta Crystallographica 1955, 8, 729-730 (https://doi.org/10.1107/S0365110X55002211).
Duyen N. K. Pham, Mrittika Roy, James A. Golen, David R. Manke, "The first-row transition-metal series of tris(ethylenediamine) diacetate complexes M(en)32 (M is Mn, Fe, Co, Ni, Cu, and Zn),"  Acta Crystallographica Section C 2017, 73(6), 442-446 (https://doi.org/10.1107/S2053229617006738).

this answer misses an important point. The individual ions in the structure are chiral but the overall crystal is not.

@matt_black: The major point of the question is chirality of the complex ion, not the crystal structure. Crystallography is one way to show the absolute stereochemistry. The crystal coordination shows that. Should have read the article.

Take my comment as a suggestion for ow to make your answer clearer. Had you added a simple sentence explaining the point I made, the answer would be totally fine. But that point is buried in useful but unnecessary detail.

@matt_black: Finally, I got your point. I was only try to point out that OP's textbook statement is incorrect. I apologize for my ignorance. The answer has been corrected since.

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text:
Are Hartree-Fock solutions eigenfunctions of the electronic hamiltonian?
$\newcommand{\Ket}[1]{\left|#1\right>}$
$\newcommand{\Bra}[1]{\left<#1\right|}$
I'm currently revising for my qualifiers, and I've come across a point of confusion, one that several places give seemingly contradictory results.
To my understanding:
In Hartree-Fock, the variationally determined ground state, $\Ket\Phi$, is not an eigenfunction of the electronic hamiltonian. It's composed of eigenfunctions of the Fock operator. It's an approximation of the ground state of the Hamiltonian.
(1) Is this all correct so far?
(2) If so, is it justified to use it in numerical derivations as an eigenfunction of the Hamiltonian,
i.e. can we write $\hat{H}\Ket\Phi\overset{?}{=}E_{0}\Ket\Phi$ or are we restricted to $\Bra\Phi\hat{H}\Ket\Phi=E_{0}$

$\newcommand{\ket}[1]{\left|#1\right>}$
$\newcommand{\bra}[1]{\left<#1\right|}$

(1) Is this all correct so far?

Yep.

(2) If so, is it justified to use it in numerical derivations as an
  eigenfunction of the Hamiltonian, i.e. can we write
  $\hat{H}\ket\Phi\overset{?}{=}E_{0}\ket\Phi$ or are we restricted to
  $\bra\Phi\hat{H}\ket\Phi=E_{0}$?

First, you can write $\hat{H}\ket\Phi = E\ket\Phi$ where $\Phi$ is the HF wave function but then $\hat{H}$ has to interpreted as the so-called mean-field Hamiltonian. It is the approximate Hamiltonian in which the exact potential which represents the energy of the Coulomb repulsions between the
electrons ($\hat{V}_\mathrm{ee}$) is replaced with some approximate mean-field potential ($\hat{V}_\mathrm{MF}$) which describes the model system in which electrons do not instantaneously interact with each other, but rather each and every electron interacts with the average, or mean, electric field created by all other electrons. That is basically the definition of the HF wave function, i.e. it is an eigenfunction of the mean-field Hamiltonian:
$$
\hat{H}_\mathrm{MF} \ket\Phi = E_\mathrm{MF} \ket\Phi \, ,
\quad \text{where} \quad
\hat{H}_\mathrm{MF} = \hat{T}_\mathrm{e} + \hat{V}_\mathrm{en} + \hat{V}_\mathrm{MF} \, .
$$
Secondly, despite the fact that the HF wave function is not an eigenfunction of the exact electronic Hamiltonian
$$
\hat{H}_\mathrm{e} \ket\Phi \neq E_\mathrm{e} \ket\Phi \, ,
\quad \text{where} \quad
\hat{H}_\mathrm{e} = \hat{T}_\mathrm{e} + \hat{V}_\mathrm{en} + \hat{V}_\mathrm{ee} \, ,
$$
we could evaluate it energy $\bra\Phi\hat{H}_\mathrm{e}\ket\Phi$ using Slater rules and minimize it to find an upper bound to the ground state energy
$$
\bra\Phi\hat{H}_\mathrm{e}\ket\Phi \geq E_\mathrm{e0}
$$
But the resulting Slater determinant obtained by minimizing the energy is only an approximation to the ground state wave function.
Finally, yes, can we use $\ket\Phi$ as our trial wave function despite the fact that it is not an eigenfunction of $\hat{H}_\mathrm{e}$ since it is not required by the variation method. $\ket\Phi$ is normalizable, it (presumably) satisfies same boundary conditions as the exact wave function, and it has adjustable parameters. That is basically enough to qualify as the trial wave function. And at the end of the day, starting from a known solution for a simpler problem is a very usual practice for variation method.


text:
How do I denote a measurement with a margin of error using ±?
I am trying to express a quantity that is measured to be $\pu{0.75 mL}$, but could be anywhere between $\pu{0.7 mL}$ and $\pu{0.8 mL}$. Is this the correct way to do it?
$$\pu{0.75 mL} \pm \pu{0.05 mL}$$

$\pu{0.75 \pm 0.05 mL}$ typically means a standard deviation of $\pu{0.05 mL}$, not a hard limit of $\pu{\pm 0.05 mL}$.

@MaxW is right on the money: the notation most frequently means sample mean +/- one sample standard deviation, computed from N independent measurements. So degrees of freedom is N-1 and this information should get conveyed. I could add a lot to my answer, but the downvotes on the question mean it probably is not worth the effort. One more thing: the +/- term can reasonably convey hard limits in some cases, e.g., histogram bin center +/- ends of the histogram bin. Other possibilities are +/- half a confidence interval or a standard error. Context is everything.

I'd say it should be written what the interval means, everything else is in practice too ambiguous: I regularly meet situations where it's not clear whether it is the standard deviation across measurements or the standar error of the mean. Not to mention the rather official recommendation in @EdV's answer to report extended uncertainty.

@cbeleitesunhappywithSX I fully agree with you and I was going to reference this paper (D. Coleman, J. Auses, N. Grams, “Regulation – From an industry perspective or Relationships between detection limits, quantitation limits, and significant digits”, Chemom. Intell. Lab. Syst. 37 (1997) 71-80.), where Coleman et al. cogently criticize the inadequacy of the typical x ± y notation. Sadly, we are still having to deal with these needlessly ambiguous matters.

I strongly recommend looking at this: QUAM:2012.P1 (EURACHEM/CITAC Guide, “Quantifying Uncertainty in Analytical Measurement”, 3rd Ed., 2012., 141 pages.) You can easily find it online, for free, and it is all about quantifying uncertainties in measurements, though with an obvious focus on analytical chemistry-related measurements. As an example, the figure below shows the QUAM recommended way to report standard and expanded uncertainties:

9.3. Reporting standard uncertainty
9.3.1. When uncertainty is expressed as the combined standard uncertainty $u_c$ (that is, as a single standard deviation), the following form is recommended:

"(Result): $x$ (units) [with a] standard uncertainty of $u_c$ (units) [where standard uncertainty is as defined in the ISO/IEC Guide to the Expression of Uncertainty in Measurement and corresponds to one standard deviation.]"

NOTE: The use of the symbol $\pm$ is not recommended when using standard uncertainty as the symbol is commonly associated with intervals corresponding to high levels of confidence.
Terms in parentheses [] may be omitted or abbreviated as appropriate.
EXAMPLE:

Total nitrogen: $\pu{3.52 g}/\pu{100 g}$
Standard uncertainty: $\pu{0.07 g}/\pu{100 g}$*
*Standard uncertainty corresponds to one
  standard deviation.

9.4. Reporting expanded uncertainty
9.4.1. Unless otherwise required, the result x should be stated together with the expanded uncertainty $U$ calculated using a coverage factor $k=2$ (or as described in section 8.3.3.). The following form is recommended:

"(Result): $(x \pm U)$ (units)
[where] the reported uncertainty is [an expanded uncertainty as defined in the International Vocabulary of Basic and General terms in metrology, 2nd ed., ISO 1993,] calculated using a coverage factor of 2, [which gives a level of confidence of approximately 95 %]"

Terms in parentheses [] may be omitted or abbreviated as appropriate. The coverage factor should, of course, be adjusted to show the value actually used.
EXAMPLE:

Total nitrogen: $(3.52 \pm 0.14)~\mathrm{g}/100~\mathrm{g}$*
*The reported uncertainty is an expanded uncertainty calculated using a coverage factor of 2 which gives a level of confidence of approximately 95 %.

This is from pages 30-31 of QUAM, referenced above. CITAC is the acronym for Co-Operation on International Traceability In Analytical Chemistry.
All that said, your notation is quite common, so no worries.


text:
Why is the top SN2 reaction faster? (Molecular Orbital Theory)
I'm having trouble understanding why the top SN2 reaction proceeds faster. I know that benzylic and allylic halides react faster in SN2 reactions than non-conjugated halides. I think it's because the transition state is more stable but how can you explain that using MO Theory?

This is what my professor drew for this question but I don't understand it all. Any help would be appreciated :)

Related: https://chemistry.stackexchange.com/questions/28222/why-do-allylic-halides-prefer-sn2-reaction-over-sn1


text:
If 40.0 grams of Aluminum and 25.0 gram of Cl2 reacted together how much AlCl3 can be produced?
The equation goes as follows: Al + Cl2 ----> AlCl3.
The equation is unbalanced so I then balanced the equation.
Heres what I did: 2Al + 3Cl2 ---> 2AlCl3
I then got the Molar mass of 40 g of Aluminum.
40.0g * (1 mol Al/ 26.98)= 1.49 mol Al
I did the same for 25g of Cl2
25.0g * (1 mol Cl2/ 70.9)= .35 mol Cl2
I am unsure as to what to do next. Do I simply multiply the two masses and then times them by two because of the 2AlCl3. I'm struggling with stoichiometry. Next on the list would be identifying the limiting and excess reagents and the % yield. This is my first crack at it.

First, $40$ g of $Al$ is $1.482$ mol $Al$. And $25$ g $Cl_2$ is $0.352$ mol $Cl_2$. You should see that there is too much aluminum. $0.35$ mol $Cl_2$ reacts with $(2/3)·0.352$ mol $Al$ = $0.2347$ mol $Al$. This is less than
$1.428$ mol.
As you want to obtain the mass of $AlCl_3$, you must see that the number of moles of $AlCl_3$ is the same as the amount of the used aluminum, i.e. $0.2347$ mol.
And $0.2347$ mole of $AlCl_3$ weighs $0.2347·133.33$ g = $31.29$ g. This is the theoretical yield.


text:
How to reproduce the Gaussian 09 SDD basis set in Gamess-US?
I'm trying to migrate away from Gaussian 09, using Gamess-US instead for molecular modelling of some heavy alkaline earth metal complexes, using density functional theory (DFT). In some of my previous Gaussian inputs I used the SDD (Stuttgart/Dresden) pseudopotential with the metallic cation ($\ce{Sr^2+}$). Now I intend to reproduce this calculation in Gamess-US.
After a couple of days figuring out how to use mixed basis sets in Gamess, I searched the Basis Set Exchange for this Stuttgart/Dresden ECP specification. Then I got confused, because there's no basis listed with that exact name. The closest I've found available for strontium was Stuttgart RLC and Stuttgart RSC 1997.
Which of these (if any) is the best match to the SDD pseudopotential available in Gaussian?

Have you tried to look on their website? http://www.tc.uni-koeln.de/PP/clickpse.en.html
In your shoes, I would try to compare the pseudo potentials from the Gaussian output file (I hope you still have access to it) and the pseudo potentials from that website.

SDD is a basis set description, which uses Stuttgart/Dresden ECPs. They are readily available for download at the BSE, but also via Univ. Köln. And they are pretty easy to pick out. See the Pseudo keyword for what is Gaussian 16 is using currently, which should be the same for 09. At least in the BSE there hasn't been an update to the S/D ECPs since their initial upload. See the related notes included at BSE and pick the appropriate size. The more problematic part is finding the BS for the orbitals that are actually described. I believe this is the correct choice, i.e. Stuttgart 1997 RSC. Please review the literature specified on the Gaussian homepage about the implementation of SDD (should be about 20 to 30 papers), and then you can cross-reference them to the references saved in the BSE library.
The work is from the earlier days of computational chemistry, so it is quite fragmented. I believe the ECPs are still popular and used in the correlation consistent Dunning basis sets. The orbital parts I wouldn't be too sure about. There are plenty of other choices out there and it would be wise to review the literature to see what is the state of the art for strontium.
Why not use the move away from Gaussian also as a move to a less fragmented, different level of theory? If you want to compare the data to previous calculations, you have to recalculate them anyway, why not choose a method which is a bit more consistent than the SDD basis. And depending on what you are calculating, you might also want to look at different software suits, e.g. Orca, Psi 4, Turbomole. Gamess certainly has its strengths, basis set implementation and handling is unfortunately not among those.


text:
Are there any semiempirical methods that work with heavy alkaline earth metals?
I did some research. Found some semiempirical methods that work even with heavy lanthanide compounds (RM1). But could not find any references to semiempirical methods compatible with the heavier members of the alkali earth group: Sr, Ba and Ra. If there is such method, does any open source / free software implements it? I'm learning to use Gamess-US, but the only semiempirical methods implemented in this package are MNDO, AM1, PM3 and RM1 (refs file).

The implementations of MNDO, AM1, and PM3 along with PM6 and PM7 in MOPAC include Sr and Ba, but not Ra. MOPAC is free for academic use, but modern versions of it are not open source. Most of these models have an open-source implementation in SPARROW (GitHub). An entirely different class of semiempirical models (GFNx) is available in the open-source XTB code (GitHub), which also includes Sr and Ba, but not Ra. The lower portions of the periodic table (lanthanides and beyond) are of less general interest to scientists and have less reference data for fitting purposes, so there is lower demand and higher difficulty in covering this portion of the periodic table. The only parameterization that I know of that covers Ra is the QUASINANO2013 parameter set (J. Chem. Theory Comput. 2013, 9 (9), 4006–4017) in DFTB, which is proprietary and only available in the commercial implementation of DFTB by SCM.
If you only need Ra to behave like a +2 cation with a slightly larger ionic radius than Ba, then you probably could modify the parameter set of one of the aforementioned models and tune the repulsion parameters of Ba to increase its ionic radius in a few reference ionic systems. That would of course take a bit of effort, in both learning what the parameters of the model mean, where they are stored in a particular simulation code, identifying some suitable reference data, and tuning one or more parameters to fit the data.
I'm sorry that there isn't a simple, affirmative answer to your question, but I hope that this helps anyways.

Good answer! For future reference: I fixed your links, they should be terminated with a space. Additionally you can use the [text](url) construct for inline links. There is also a [text][ref] construct. Please see the help pages for more information.


text:
Is it possible to measure light absorbance for analysis of lead concentration without photospectrophotmeter
I’m designing an experiment which requires I measure final and initial lead concentrations for the determination of aqueous lead uptake by different mediums. Because lead concentrations would be measured on the ppm scale, I’d need to use a spectrophotometer to measure light absorbance by the water samplings. I could then use Beer Lambert’s Law to determine concentration.
With recent protocols surrounding COVID-19, it’s been difficult to contact an institution. I was wondering if there was any way I could measure light absorbance of my samples at home. Is there some other apparatus I could look into using (or perhaps even purchase)?
All the best!

Unfortunately no. However small spectrophotometers are commercially available. Check Vernier's website: https://www.vernier.com/product/go-direct-spectrovis-plus-spectrophotometer/

You still need dyes that form a colored complex with lead.

You need a dye, as @M.Farooq said, but there is the long shot possibility of smartphone spectroscopy here. Easy to find other articles by searching, but it would be much better to have an inexpensive real spectrophotometer.

Not an option right now, but one university where I have been working had regular "garage sales" of old-ish equipment. And there's labexchange.com

Another possibility is the grease spot photometer (aka Bunsen photometer) German Wiki page. This can be homemade, you need a piece of paper, some wax or oil, two light sources, a meter stick and for the measurement of solutions also some e.g. cardboard to shield unwanted light. Also the darker the room, the better.
The underlying idea is that light intensity from two different sources can be compared (and found to be equal) with a paper with a grease (waxed) spot: the spot will vanish when as much light is transmitted through the spot as reflected around.
If you move a light source to achieve this condition, you can conclude relative intensities from the distance (intensity goes 1 / d² with d the distance from the respective light source).
Here's a youtube video demonstrating this: https://www.youtube.com/watch?v=sgZdl4qRa48

How to measure photometrically
so schematically we have:
o--U----|----o

+---d---+

with light sources o, cuvette U and the spotted paper |. Measure distance $d$ between equal intensity paper position and the light source with the cuvette.
We have $I_{transmitted} \sim \frac{1}{d^2}$
Next, a calibration series is acquired as usual: blank (which serves as $I_0$) and concentrations over the calibration range. Almost as usual:
$$E = -\lg \frac{I}{I_0} \sim \varepsilon l c$$
$$-\lg \frac{d_0^2}{d^2} = b c$$
$$\lg d = a + b' c$$
$a$ and $b$ are determined by calibration - I use here $a$ instead of $2 \lg d_0$ because several choices are possible and in general the fitted value should be used rather than the theoretical one since it includes some effects in which practice differs from theory (e.g. differences between % transmission in the grease spot from % reflection outside).
Standard addition instead of calibration would be possible as well.

This video is extremely interesting. But I don't see how this apparatus can be used to measure the absorbance of a given solution.

@Maurice: I updated with a rough outline of the procedure.

Interesting but I guess a smartphone might be better

@Alchimista: In what respect?

In the overall procedure. I am sure the Bunsen approach might be surprisingly good if well operated but surely requiring skill. In all cases it depends on the accuracy needed. If this is for a small shop there are relatively cheap solutions on the market.

@Alchimista: yes, the Bunsen approach does have certain requirements. But already the quantitative treatment needed during sample preparation (smart phone or grease spot photometer alike) does require skills which I think should also be a good proxy for being able to operate the grease spot photometer possibly with the one exception that someone with poor vision. All that being said, in real life I'd order the analysis with a lab specialized on environmental analysis: even now I expect at the very least that the preliminary communication should still be possible, and I expect small labs ...

... around here (Germany) to continue working if they have orders (possibly with reduced capacity: owner-only) (At least, I continue to work in my company - this is unproblematic since there's noone showing up here who doesn't live in the same house as I do. A number of other chemistry-related small business owners I know do the same). And the possibility to buy a used proper spectrophotometer has been pointed out already (which would be my 2nd choice)

It is still possible to measure light absorbances of a colored solution without any photometer, simply by comparison.
First you prepare a reference solution with maybe 10 ppm Pb and an excess of a colorless reagent making a colored compound or complex with Pb. You prepare a series of 10 test tubes. You add 1 mL of this reference solution in the first test tube, 2 mL in the second, 3 mL in the third, 4 mL in the fourth, etc. up to 10 mL in the last. Then you add enough solvent to get a total volume of 10 mL in all tubes. You homogenize each test tube.You obtain a series of solutions. They are nearly colorless in the first tubes but the intensity of the color increases gradually from the 1st to the last one.
Then you prepare 10 mL of a new solution with exactly the same procedure used to produce the reference solution, but with the unknown solution instead of the known concentration (10 ppm). By comparison between all these tubes, you can get a good approximation of the concentration of the unknown solution.

Perhaps you can add H2S to create PbS (as a starting step). Then, based on an extract from Wikipedia on Lead Sulfide:

Although of little commercial value, PbS is one of the oldest and most common detection element materials in various infrared detectors.[12] As an infrared detector, PbS functions as a photon detector, responding directly to the photons of radiation, as opposed to thermal detectors, which respond to a change in detector element temperature caused by the radiation. A PbS element can be used to measure radiation in either of two ways: by measuring the tiny photocurrent the photons cause when they hit the PbS material, or by measuring the change in the material's electrical resistance that the photons cause. Measuring the resistance change is the more commonly used method. At room temperature, PbS is sensitive to radiation at wavelengths between approximately 1 and 2.5 μm. This range corresponds to the shorter wavelengths in the infra-red portion of the spectrum, the so-called short-wavelength infrared (SWIR). Only very hot objects emit radiation in these wavelengths.

some ideas. Perhaps examining a comparative study (based on known concentrations of Pb) of the changed in measured electrical resistance of a sample of PbS suspension taken from the solution after treating with the same volume of H2S followed by irradiation may provide a path. This not meant to be a complete answer, but a suggestion on a possible path to be explored.
[EDIT] To be clear, I am assuming that one cannot measure light absorbance, due to lack of equipment, for the purpose of establishing Pb concentration. But, one can perhaps measure electrical resistance. The question is can one construct a path (actually, more likely a graph to which a curve is fitted) to determine Pb concentration from the change in electrical resistance recorded upon applying a fixed amount of infrared radiation to a solution where varying amounts of PbS is introduced. If experimenting with known starting Pb concentrations, converted to PbS, suggests a detectable and measurably path with some degree of goodness-of-fit, then yes, else no (which is the only answer provided so far).
Here is some background on PbS based photodetectors where irradiation exposure varies and the amount of PbS concentration remains fixed.

PbS is a standard SWIR semiconductor detector (1 - 3.3 µm) whereas PbSe is used in the MWIR range (1 - 4.7 µm when uncooled; up to 5.2 µm when cooled). Our lead salt detectors are photoconductive; the detector resistance is reduced during illumination. The crystal structure is polycrystalline and is produced via chemical deposition.

Please note the actual size of the PbS based photodetector.
Note, my proposed study of feasibility avoids the use of a nephelometer (which can measure the concentration of some inert suspended particulates in say a liquid colloid by employing a light 'source' beam and a second light detector at 90° to the source beam) by taking advantage of the properties of irradiated PbS particulates. Also, it should be noted that a spectrophotometer, based on a quantitative measurement of the reflection or transmission properties of a material as a function of wavelength, based again on light, is also presumably not an available option.
[EDIT EDIT] Just found a reference, THE RELATIONSHIP BETWEEN ELECTRICAL RESISTANCE AND DISPERSED PHASE CONCENTRATION IN OIL IN WATER EMULSIONS to quote:

The electrical resistance of a series of oil‐in‐water emulsions has been measured, and used to test the validity of four equations relating resistance with the concentration of dispersed phase. A modification of one of the equations was found to give the best relationship.

In the current context, we have a suspension of irradiated PbS in water (releasing electrons, per this 2013 article 'Photoinduced electron transfer from PbS quantum dots to cobalt(III) Schiff base complexes: light activation of a protein inhibitor'), which I would have guessed is more promising in determining the PbS concentration than say of oil in a water emulsion.
Also found a new method to quote a source:

In summary, we have demonstrated a practical way to measure nanoparticle concentration in a colloidal solution using quartz crystal microgravimetry. Application of a small drop of the nanoparticle colloid in a volatile organic solvent to the crystal surface leaves a dry nanoparticle film after solvent evaporation. Crystal resonant frequency shifts obeyed Sauerbrey’s equation for the dry nanoparticle concentrations up to 1300 µg/mL, as calibrated using a set of serial dilutions of Si and Ag nanopowders in methanol, rhodamine B in methanol, and ferrocene in cyclohexane.

In aqueous solution, adding $\ce{H2S}$ would create a precipitate of $\ce{PbS}$ which would be very fine particulates. If any kind of quantitative analysis could be done on the particulates it would be using a nephelometer. I'll point out though in some sort of indeterminate solution that $\ce{H2S}$ would precipitate many heavy metals.

I have addressed your comment by an edit in the thread. If light is NOT an option, it follows that a nephelometer is not likely a viable option either, and certainly show NOT be a rationale for dismissing electrical resistance paths.


text:
Why is the carbon monoxide triple bond stronger than the nitrogen-nitrogen triple bond?
The $\ce{CO}$ triple bond has a bond enthalpy of 1072 kJ/mol while the $\ce{N2}$ triple bond has a bond enthalpy of 945 kJ/mol, at least according to my source. The molecules are isoelectronic, so I couldn't think of any way to explain it using MO theory. I've read many explanations that did not make much sense, including

Of course because they are different elements and nitrogen nitrogen bond is balanced and oxygen is more electronegative than nitrogen so it will have a greater bond strength

So my question is why is the $\ce{CO}$ triple bond stronger than the $\ce{N2}$ triple bond?

I don't exactly see what answer would you want. The two molecules are different, so one has to have stronger bond. This rationalisation you mention indeed doesn't make much sense. Unless you want to get into serious computational chemistry, there probably isn't much to talk.

Somewhat related https://chemistry.stackexchange.com/questions/66919/why-do-post-hartree-fock-methods-fail-to-predict-the-direction-of-the-dipole-mom https://chemistry.stackexchange.com/questions/30797/how-can-the-dipole-moment-of-carbon-monoxide-be-rationalised-by-molecular-orbita

To add to Mithoron's point, one molecule is heteronuclear, and one is homonuclear. So at a high level, you're comparing apples and oranges.

I agree in part with Mithoron and MaxW's comments: these are different molecules, not even sharing a mutual atom (e.g. C-O vs C-C), so direct comparison is restricted.
However, Pauling's concept of electronegativity does explain why, in general, the heteronuclear A-B bond is stronger than the average of the homonuclear A-A and B-B bonds. the difference in the electronegativity of the two atoms in the bond lend a somewhat ionic character to the bond, strengthening it.

Perhaps useful to remind readers that bond strengths specifically refer to homolytic cleavage of bonds.


text:
How can I remove the bitartrate anion from an organic bitartrate salt?
I have a compound, dimethylaminoethanol (DMAE) bitartrate, which I originally purchased as a supplement and possible smart drug. I bought an absurd amount of it. It didn’t do anything for me as a supplement. I understand that DMAE, not the bitartrate salt, is commonly used in cosmetics, & I’d like to play with it for that purposes, maybe mixing it with glycerin, etc. Is there a simple way to strip off the bitartrate anion and have it precipitate or sublimate or something, leaving the DMAE behind? Or otherwise separating the DMAE out from the salt?
I have a strong preference for any other reactants and reaction products being non-toxic

I see on Wikipedia: "Dimethylethanolamine (DMAE or DMEA) is an organic compound with the formula (CH3)2NCH2CH2OH. It is bifunctional, containing both a tertiary amine and primary alcohol functional groups. It is a colorless viscous liquid. It is used in skin care products."
Now the fact that it is used in skin care products does not make it safe: the SDS  (https://www.fishersci.com/store/msds?partNumber=AC116180010&productDescription=N%2CN-DIMETHYLETHANOLAMINE+1LT&vendorId=VN00032119&countryCode=US&language=en) shows these pictograms:

Uh oh! Oh, well.
Well, let's forget that for a minute. DMAE is a liquid, boiling at 134C. Bitartrate anion is very insoluble if attached to potassium (potassium bitartrate is cream of tartar). If you added a stoichiometric amount of potassium bicarbonate (KHCO3, quite safe) to an aqueous solution of DMAE, you would precipitate almost all of the bitartrate, leaving the DMAE in water. Filter off the ppt and distill off the water and then distill and collect the DMAE. It will probably be a little wet and you will hate the overwhelming fishy smell.
Oh, and use a respirator (from the SDS: "Follow the OSHA respirator regulations found in 29 CFR 1910.134 or European Standard EN 149. Use a NIOSH/MSHA or European Standard EN 149 approved respirator if exposure limits are exceeded or if irritation or other symptoms are experienced.")
Good luck. Actually, I think you will have better luck if you do something else.

You have persuaded me that this is a bad idea. Thank you. Though I admit that I am a little puzzled by something acutely toxic if swallowed being readily available as a supplement, and often used in cosmetics as a severe skin irritant.

Preumably used at very low concentrations in cosmetics as it causes burns

This might work:
Make a solution of NaOH in absolute ethanol, e.g. at a 1 mol/L concentration. Suspend the DMAE bitartrate in ethanol, and add 2 equivalents of the NaOH solution slowly to the suspension, making sure it doesn't heat up too much. After everything is added, stir or agitate it for a while. Then filter off the solids and wash them with some ethanol. Add the washings to the ethanol solution. From this solution, let the ethanol evaporate, and only DMAE should remain. The solids are sodium tartrate.

SAFETY - I'd be very wary of using NaOH one step before collecting the product which is to be used in some cosmetic formulation. It is highly doubtful that the OP has the right equipment/techniques to test the collected product for a NaOH impurity.

I assume people attempting any chemistry know about safety. Would you put safety statements in the electronics forum when people ask questions about voltages higher than 40 V and would you vote down answers that are "unsafe"? IMO personal responsibility and knowledge of basic safety is assumed on an exchange like this.


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How can we tell if a salt will melt or decompose?
Take $\rm NaCl$ for instance, a salt that will melt when it reaches its melting point, and compare it with $\rm NH_4NO_3$, a salt that doesn't melt, but instead decomposes to $\rm N_2O$ and $\rm H_2O$ on heating to a high enough temperature.
How can we determine if a salt will simply melt, or if it will decompose on heating?

Chemistry is in large extent empirical science. Even if there are various models able to compute various things, for many cases the easiest is just to remember or look up things. An extreme example is melting point of mercury. The about proper value had to be computed by relativistic quantum chemistry and the needed computation power was available just recently. For stability of salts, there may help thermodynamic tables for compounds, but melting points are empirical knowledge.

Some info can be semi-derived  semi-empirical, based on already learnt behaviour patterns of chemicals, it may be  called "chemical sense". But it comes with experience.

Long story short, we tell that by knowing chemistry.

By heating it up and seeing the results? And if it's explosive ,we'll do it in a fortified chamber.

As noted, chemistry is an experimental science.  That being so, experimental data allow for a few rules.  Given that salts are often formed from reactions of acids and bases, we should not be surprised to see acid-base chemistry enter into these considerations.

Salts with three or more elements are more likely to decompose than those that are only binary.  Ternary salts often contain within their ions acidic and basic components from which they could form, such as $\ce{CaO + CO2}$ in calcium carbonate or $\ce{MgO + SO3}$ (or, $\ce{MgO + (SO2 + (1/2) O2})$ in magnesium sulfate.  In a binary salt such as sodium chloride the original acidic and basic functions have generally been expelled to other compounds, such as water if $\ce{NaCl}$ is formed from $\ce{NaOH + HCl}$.

Within the above criteria, salts are more likely to decompose if the acidic or basic component (such as ammonia among bases) or both are volatile, as entropy becomes more favorable upon heating for the evolution of gases.  Silicates (acid = $\ce{SiO2}$) on or in Earth exist under conditions where carbonates (acid = $\ce{CO2}$) would decompose.

Salts with stronger acid and base components are more difficult to decompose, because more energy is required to reverse the acid-base combination.  Magnesium sulfate has a higher decomposition temperature than zinc sulfate because magnesium oxide is a stronger base than zinc oxide.  Similarly, magnesium sulfate requires a much higher temperature to decompose than magnesium carbonate; the latter decomposition requires evolving weakly acidic $\ce{CO2}$ whereas the former would have to evolve strongly acidic $\ce{SO3}$ or couple the acid-base separation with the thermal decomposition of $\ce{SO3}$ to $\ce{SO2 + (1/2) O2}$.  Sometimes redox reactions can facilitate decomposition if they convert stronger acids and bases to weaker ones, such as ammonium nitrate being decomposed to water and nitrous oxide rather than ammonia and nitric acid.


text:
Why does the oxidizing strength of the chlorates decrease with increasing oxidation number?
Here is a table of the standard reduction potentials of the chlorates from wikipedia:

Intuitively, one would expect that the more positive the oxidation state, the more oxidizing that compound would be. However, as we see, the standard reduction potentials become more negative as the oxidation state of the $\rm Cl$ atom becomes more positive. So, why does the oxidizing strength of the chlorates decrease with increasing oxidation number?

Even from this table you can see it's not exactly true and your idea with correlation with ox. state false. That's pretty much it.

The chlorate system is very complex. I doubt that the open-circuit potentials can be actually measured in solution, but rather are calculated from heats of formation. If the term "oxidizing strength" is used to compare systems, it should be rigorously defined; in this case, it seems that "oxidizing strength" is defined as "open-circuit voltage". But this could be a poor definition - even woefully poor.
In particular, the comparison of the open-circuit voltages does not include the potential current derivable from the reaction of a mole of each oxychloride. The output energy of the electrolytic cell (if you could just test each reaction, one at a time) would be the output voltage times the number of electrons transferred. By that standard, perchlorate will consume 7 times as many hydrogens as hypochlorite, and thus give 7 times as much current, with only a slightly lower voltage.
A comparison of the heats of reaction (just for the acidic equations) proves to be much more dramatic:
2 moles hypochlorous acid (total heat of formation = -56 kcal (CRC Handbook) produces   2 moles of H2O (total heat of formation = -136 kcal)
2 moles chlorous acid (-28 kcal) --> 4 moles H2O (-272 kcal)
2 moles chloric acid (-46 kcal) --> 6 moles H2O (-362 kcal)
and 2 moles perchloric acid (-62 kcal) --> 8 moles H2O (-544 kcal).
The heats of formation of the chlorine acids are similar, but the heat of the reactions is much higher for the more highly oxidized acids.
A mental picture using water: imagine 4 dams each 200 feet high and 200 feet wide. The lake behind the first is filled with water 163 feet deep and is 2 miles long. The second lake is 164 feet deep and 6 miles long. The third is only 147 feet deep, but 10 miles long. The fourth is 142 feet deep but 14 miles long. A pressure gauge at the dam would indicate the height (similar to the open circuit potentials above), but if the dam is used to provide power, the fourth dam/lake will outshine the other three.
Assuming that all the electrons don't react simultaneously, the open circuit potentials are a measure of the activation energy of the first step of the overall redox reaction: in order to get zero current, you just have to stop the first step. This activation energy will be sensitive to many factors, and we occasionally oversimplify, ignoring the succeeding steps by using non-quantitative words like "strength". The total oxidation strength will be a combination of all these steps, and the open circuit potential may not reflect this combination.

More than 50 years ago, Pauling has also observed this regular tendency observed among the different chlorinated acids. But he suggested that it may be due to a tendency to get the highest possible number of Cl-O bonds. The oxidative power could depend on the number of free doublets around the chlorine atom. Perchlorate has no free doublet : it is not a good oxidant. It does not need any more O atom to reach the maximum number of bonds Cl-O. Hypochlorous ion has 3 available doublets on the Chlorine atom. It is the strongest oxidant. It needs a lot of O atoms to get the maximum number of bonds Cl-O.
Funny enough, I am not convinced about this interpretation. It even seems somewhat illogical to me. I would be pleased to read any comment about this strange tendency.
See : L. Pauling, College Chemistry, W.A. Freeman & Co, San Francisco, 1947.

Technically, the statement "Hypochlorous ion has 3 available doublets. It is the strongest oxidant" is not quite correct. It is HClO2, see the prior related thread at https://chemistry.stackexchange.com/questions/117725/why-is-hypochlorite-a-stronger-oxidizing-agent-than-other-oxychlorides and comments by Mathew Mahindaratine.


text:
What is a secondary shell in an enzyme?
I've searched the internet for a definition of what the different shells in enzymes/proteins are and haven't found a good answer. This Nature article, which is about an enzyme that catalyzes Kemp Elimination, has the sentence:

The Val12Met and Phe77Ile mutations, also in the second shell, cause changes to internal cavities that cause a slight rotation in the backbone at Trp50 that also favors the alternative configuration.

What is the second shell, and how many different shells are there?

The authors mention three residues that they consider "second shell", 12, 77, and 102. They are shown in the figure below in green:

Residues Trp50 and Glu101 are directly involved in ligand binding, shown as space-filling in yellow. As you can see, they are part of the central beta barrel. The residues in green are also part of the beta barrel, but the side chains are not involved in ligand binding because they point to the outside. For that reason, the effect of mutating these residues on ligand binding must be indirect by repositioning the residues that point inside.

What is the second shell, and how many different shells are there?

This is not a well-defined concept. The term is more common for hydration of ions (first shell of water molecules makes direct contact, second shell contact first shell and so on), or in the context of coordination spheres (https://en.wikipedia.org/wiki/Coordination_sphere). When researcher write papers, they come up with descriptions that fit their structures. A famous example is to describe polymerase active site secondary structure as hands, with a palm, thumb and fingers, see e.g. https://marlin-prod.literatumonline.com/cms/attachment/c9a1e727-5b4b-45ce-81c8-8a24c1aea2ef/gr1.jpg. Some of these descriptions become common (such as the beta-barrel mentioned above) and some do not (my research advisor once described an enzyme as hat-shaped - it did not stick, maybe because hats come in many different shapes and forms, and barrels are more defined).


text:
Does the hybridization of sp3d2 type exist?
Is the hybridization of $\ce{XeOF4}$ $\mathrm{sp^2}$ rather than $\mathrm{sp^3d^2}$, because the latter type of hybridization does not exist?
Every answer in the Quora thread "How can we find the hybridization of $\ce{XeOF4}$?" concluded that the hybridization of $\ce{XeOF4}$ is $\mathrm{sp^3d^2}$. However, the very last answer posted by Steven Fawl, Professor of Chemistry at Napa Valley College (1985-present) reads: "Where to begin? Let me start here - every answer thus far is wrong! This molecule is not $\mathrm{sp^3d^2}$ hybridized. It is actually $\mathrm{sp^2}$ hybridized and I am going to prove it." The issue that I have with his post is that it only cites a couple of Wikipedia articles as his sources, and it contradicts every chemistry book that I have read by claiming that $\mathrm{sp^3d}$ and $\mathrm{sp^3d^2}$ hybridizations do not exist. The justification he gives for the latter is the following quote from this Wikipedia article:

In 1990, Magnusson published a seminal work definitively excluding the role of $\mathrm{d}$-orbital hybridization in bonding in hypervalent compounds of second-row (period 3) elements, ending a point of contention and confusion. Part of the confusion originates from the fact that $\mathrm{d}$-functions are essential in the basis sets used to describe these compounds (or else unreasonably high energies and distorted geometries result). Also, the contribution of the $\mathrm{d}$-function to the molecular wavefunction is large. These facts were incorrectly interpreted to mean that $\mathrm{d}$-orbitals must be involved in bonding (Ref.1 and Ref.2).
For transition metal centers, the $\mathrm{d}$ and $\mathrm{p}$ orbitals are the primary valence orbitals, which are only weakly supplemented by the $\mathrm{p}$ orbitals (Ref.3). The question of whether the $\mathrm{p}$ orbitals actually participate in bonding has not been definitively resolved, but all studies indicate they play a minor role.

Indeed, the abstract of Ref.2 states that:

No evidence is found to support the traditional notions of $\mathrm{sp^md^n}$ hybridization.

But this article is not publicly available and it may not reflect the current scientific consensus. Making a long story short: Are the authors of Ref.2 and Steven Fawl correct? If so, is the bonding of every hypervalent molecule analogous to $\ce{XeOF4}$ (removing 1, 2 or 3 electrons from the valence shell of the central atom and $\mathrm{sp^n}$ hybridize it)?

References:

Eric Magnusson, "Hypercoordinate molecules of second-row elements: $\mathrm{d}$ functions or $\mathrm{d}$ orbitals?," J. Am. Chem. Soc. 1990, 112(22), 7940–7951 (https://doi.org/10.1021/ja00178a014).
David L. Cooper, Terry P. Cunningham, Joseph Gerratt, Peter B. Karadakov, Mario Raimondi, "Chemical Bonding to Hypercoordinate Second-Row Atoms: $\mathrm{d}$ Orbital Participation versus Democracy," J. Am. Chem. Soc. 1994, 116(10), 4414–4426 (https://doi.org/10.1021/ja00089a033).
Gernot Frenking, Sason Shaik (Eds), eds. "Chapter 7: Chemical bonding in Transition Metal Compounds," In The Chemical Bond: Chemical Bonding Across the Periodic Table; Wiley-VCH: Weinheim, Germany, 2014, pp. 175-218 (ISBN 978-3-527-33315-8).

It does exist - in old textbooks and heads of those who read them. Particularly Quora users, apparently :> Not sure what better proof you want then two papers and relatively new book.

There's like a few dozen answers here succinctly debunking any significant mixing in compounds like the one you mention here, so I guess you can find as much evidence as you wish.

related https://chemistry.stackexchange.com/questions/54737/how-are-the-hybrid-orbitals-of-sulfur-hexafluoride-shaped

https://chemistry.stackexchange.com/questions/66698/theoretically-which-d-orbital-participates-in-sp3d-and-sp3d2?r=SearchResults

https://chemistry.stackexchange.com/questions/75009/number-of-f-br-f-angle-of-90-degree-in-brf5?r=SearchResults

First, hybridization is a model used in Bond Valence Theory to explain the bond nature between some atoms in some molecules. Actually, the current binding theory is the Molecular Orbital Theory, in which we no longer use hybridisation. But, we study hybridisation because it is a simple and useful tool to explain a lot of compounds.
But, what the professor said in your question, it is not that the sp3d2 hybridization does not exist and cannot be used to explain any molecule. What he is saying is: In hypervalent compounds, there is evidence that the d functions do not participate in the bond.
After reading your question, I have searched a lot about the binding in hypervalent compounds. And what I found was that the nature of bond in hypervalent compounds like XeOF4 has been the subject of a lot of controversy for a long time, because in the represetative atoms, the energy gap between the nsp and nd orbitals is too large, so the d orbitals cannot hold extra electrons. And, nowadays we have evidence through ab initio calculations that the d-functions do not participate in bonding in hypervalent molecules.
However, this is a recent evidence, so, if you have to awnser in a test about the hypervalent molecules, use the d-orbitals to explain them!
I am leaving some links with some articles and books that I read about this subject:
Chemistry of Hypervalent Compounds
The three-center-four-electron (3c-4e) bond nature revisited. An atoms-in-molecules theory (AIM) and ELF study
New bonding concept for Hypervalent molecules, including electron poor and electron odd compounds


text:
Are there any molecular modeling programs that can build materials made up of three or more elements into a cylindrical shape?
I would like to build a material similar to what is in the image below.
I have tried programs such as Material Studio, Quantum ATK, and VNL, but most of them can only do materials with 2 elements.

Any references attached to the image (thus also materials and methods)?

@BuckThorn: I don't think so. He got this from the Materials Modeling Stack Exchange facebook group: https://www.facebook.com/groups/700426147050999/

Various custom software exists to create nanotubes.
The following appears suitable:
http://www.jcrystal.com/products/wincnt/
A Wolfram Demonstrations application may also be of some use:
https://demonstrations.wolfram.com/NanotubeBuilder/
Furthermore, you may have further considerations like how to generate molecular mechanics parameters. For example, you may be interested in how to generate the nanotubes to be compatible with AMBER, or GROMACS. The following may be of use if that is the case:
https://github.com/bio-phys/cnt-gaff
http://www.gromacs.org/Documentation/How-tos/Carbon_Nanotube
I should also mention that VMD and Avogadro, which are more easy to use, can create carbon nanotubes. VMD can create boron-nitride nanotubes
http://www.ks.uiuc.edu/Research/vmd/plugins/nanotube/

Thanks @cody Aldaz. However, the programs you mentioned are for carbon nanotubes. They can't be used to create nanotubes of materials made up of three or more elements such as barite.

Thanks everyone. I later found out that you can use Quantum ATK. It has a tube wrapper feature under the coordinate tools which enables one to build nanotube of any material.


text:
A question about geometry of benzenetetrol
In the 3d conformation of 1,2,4,5 - benzenetetrol, how could i determine the direction to which hydrogen atoms of hydroxyl groups would point? and if there were tertiary butyl groups in positions 3 & 6 would it be more stable for hydrogen atoms to point in the direction away from t-butyl groups as a matter of steric hindrance ?
!
Source

Consider whether there would be hydrogen bonding between the OH groups

@Waylander well this may partially answer the first half of the question, but if i needed this molecule in the stable form which you mean then it would be better to do the reaction in low temperature in order to preserve hydrogen bonds

@Waylander but it is not a one molecule it is present with other molecules so which forces will win intramolecular hydrogen bonding or intermolecular hydrogen bonding

Intra pretty much always wins over inter

@Waylander  so i suppose that that the conformation would be like 'swastika' symbol, but what would happen if there were t-butyl in 3 and 6 position?

The melting point of 1,2,4,5-benzenetetrol is 210C (https://www.chemicalbook.com/ProductChemicalPropertiesCB41307665_EN.htm), so there is very probably significant intermolecular attraction between molecules, probably in three dimensions, which would involve the hydrogens pointing up and/or down, and not in-plane.
For a crystal structure, https://pubchem.ncbi.nlm.nih.gov/compound/1_2_4_5-Benzenetetrol suggests that 3 hydroxyls are planar while one is as out of plane as you would expect for cyclohexanol. Believe it or not. (I don't believe it.)
The (predicted) NMR spectrum shows two peaks, one very sharp for the aromatic protons, the other fairly sharp for the hydroxyls, but no experimental data seems to be available. The sharpness of the predicted spectrum suggests that the prediction is that free rotation is likely in solution.

do you know how to force the hydrogen bonds to stick to a specific direction?

I'll take a wild guess (because I don't "know"). If you can find a substrate where you get strictly monolayer adsorption, it is likely that the adhesive forces would be thru the electrons of the aromatic system plus oxygens, so the hydrogens would be pointing away from the adsorbent (which could be graphite or graphene). Now if the adsorbent had a polar surface, like clay, the hydrogens might be preferentially attracted. But i am not aware of any method that could prove the orientation.

what about steric hindrance (i mean by putting t-butyl in 3,6 position ) wouldn't that push the hydrogen of hydroxyl groups away from t-butyl and also prevent the formation of strong intermolecular hydrogen bonds ?


text:
Can all NMR spectrometers perform special techniques such as COSY, NOESY, HETCOR,..?
Do you need special spectrometers for:

probing nuclei other than hydrogen or carbon?
performing 2D spectroscopy such as COSY, NOESY, HETCOR, HSQC,..

Or are these techniques usually available on standard NMR spectrometers?

Can you give a bit of a background for this question? It would seem strange that you don´t know the answers (or have an expert at hand to ask) if you know these techniques and are thinking about using them.

What do you consider a standard NMR spectrometer? Could you be more specific (it depends on the probe in the instrument, e.g. triple resonance probe)?

@Karl I agree with you that the question as such rises indeed the (meta) question about the background of the OP.  On the other hand, there is a wide spectrum (no pun intended) of groups.  Some still drop the NMR tube in a card-board box and basically await the spectrum to be physically printed; others expect each member pass an introduction and then to record 1D/2D, Hoye's no-D and kinetic NMR on their own.  Speculation: maybe the OP transients to such a later group.

@Buttonwood Of course, it´d just be good to know about that, to give an answer that actually fits. I hate preparing a talk for an unknown auditorium. ;)

@Karl I aggree.

@Karl: however, keep in mind that here on sx people with other background may end up reading/being refered to this question as well.

@cbeleitesunhappywithSX I´d much rather serve one kind of reader properly. My thing below seems to be liked, but it´s a quick stab in the dark, not a profound answer.

@Karl: that's fine, and my comment is really only a comment. As a vibrational spectroscopist who's never really had anything to do with NMR (outside the 1d 1H and maybe 13 C student practicum level, that is), your answer reads understandable to me, so IMHO it is written for a sufficiently broad audience. I'll delete both comments soon in order to declutter.

It depends on a lot of things, software-wise

some of these techniques are simply a software option
you can implement pulse sequences yourself, if that´s possible with your control software
you need to have the right software to transform and evaluate the data
... or write it yourself (not exactly rocket science if you know your way with Matlab/Octave etc.)

and also regarding your hardware

you can have several probes for one spectrometer

they can be single channel, or dual ("HX" usually), triple ... channel
the "X" channel can be tuneable over a wide(r) range, or you are limited to e.g. carbon

of course you need a transceiver with enough channels, which can be used at the wanted frequencies
and a suitable preamp for every channel
some techniques need more special hardware, like CPMAS, pulsed gradients, cryo etc.
low field spectrometers on the market (afaik) have no user exchangable parts, so you are stuck with the hardware you bought

But to answer your immediate question: I guess with basically any living (as in "not undead") spectrometer, all those you named above should be possible, provided you can tune the X channel to the het nucleus you are after. They are the basic standards.
If you want to know a lot more, 200 and More NMR Experiments by Berger and Braun is probably your book.

There are multiple aspects to consider here.

In terms of the magnet, the ones suitable to record a $300$ or $500\, \pu{MHz}$ $\ce{^1H}$ NMR spectrum may be well used to record 1D or 2D homo- and heteronuclear spectra as well.
Other nuclei than $\ce{^1H}$ or $\ce{^{13}C}$ routinely are recorded in 1D with these standard spectrometers as well; for example $\ce{^{15}N}$, $\ce{^{19}F}$, $\ce{^{31}P}$ often relevant in samples equally relevant for biochemistry (heterocyclic chemistry, proteins); or inorganic chemistry (e.g., $\ce{^{117}Sn}$, $\ce{^{119}Sn}$, $\ce{^{195}Pt}$).
Recording spectra with nuclei other than H and C however often is performed with dedicated probeheads. As an example, you are used that DEPT-135 / DEPT-Q / APT spectra are recorded to seemingly yield singlet signals only by proton broadband decoupling.  A standard two-channel probehead however is not designed to equally decouple $\ce{^{19}F}$ beside $\ce{^{1}H}$, for example, and consequently your 1D $\ce{^{13}C}$ spectrum still show the couplings with $\ce{^{19}F}$:

(source)
An other example is recording NMR spectra for three types nuclei at once by triple resonance with three scales, as seen in biochemistry to elucidate protein structures. (An entry to this vaste topic might be Wuethrich's Nobel lecture 2002, here.)

(source)
Displaying only the mount of the heads still visible on the bottom of the NMR magnet, this would not be possible with a standard probehead:

(source)
but would recorded comfortably with a triple-resonance head like this

(source)
Recording NMR spectra requires adjustment of pulse sequences (e.g., delay times and phases). Typical parameters are literature known (e.g. APT for $\ce{^{13}C}$ in the Journal of Magnetic Resonance), so the software of the spectrometer may internally look up and adjust thes for the more frequently used ones experiments.  Occasionally, you find blogs about these techniques, too (e.g., University of Ottawa, PNA)

Can you check the first sentence in your last point, I can´t make head or tail of it. ;)

@Karl Indeed, the phrase required an edit.


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Combining enthalpies of formation and combustion in the context of Hess's law
I know that you can use enthalpy of combustion to work out formation, and Formation to work out enthalpy change. But can you use enthalpy of formation and combustion data is the same Hess's cycle to work out enthalpy change?
For example, we are trying to work out the enthalpy change for the reaction
$$\ce{A -> B}.$$
The alternative route is
$$\ce{A -> X -> B}.$$
We have been given combustion enthalpy data for $\ce{A -> X}$ and formation enthalpy data for $\ce{X -> B}.$
Could you work this out? Or does it always have to be the same type of enthalpy data in the alternative route?

Hess's Law is path independent. Try Addition!

Formation is from the elements. Combustion produces CO2 and H2O. So if you have the formation enthalpy of CO2 and H2O handy, you are all set.


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Why is the ionic product of water $K_w$ the same in acid, water, and alkali?
Seen on the web:

The concentration of [H+] and [OH−] both vary based on the the composition (acid/alkaline) of the solution, but the remarkable thing is that their product does not. When [H+] goes up, [OH−] goes down, in the same proportion.

The word "remarkable" was used to describe the ionic product of water (Kw). Ideally, it should not seem more remarkable than anything else, there should be simple explanations that make sense. Why is Kw constant?

You seem to be implying that something that has an explanation that makes sense can't be remarkable?

Also you should give where you found the quote. "On the web" isn't helpful nor a valid way to give credit.

"Bohr was inconsistent, unclear, willfully obscure and right. Einstein was consistent, clear, down-to-earth and wrong. "  John Bell on the Bohr-Einstein debates.  I'm afraid in real physics "simple explanations that make sense" don't always work, even for genius.

@AaronStevens I'm implying that explanations and descriptions are different (but related) things. Keppler observed, for example, his third law, empirically. Surface tension on water, another empirical observation. That it is a result of the gel-phase of water, or "ionized ice", still not very broad consensus, socially.

Another way to put that, people tend to conflate cause and effect. They'll observe an effect, and, a bit lazily, ascribe it as the cause. Lazy thinking. Common in the history of ideas, "generatio spontanea".

The equilibrium reaction for the auto-dissociation of water is:
$$2\text{ H}_2 \text{O}(l) \leftarrow \rightarrow \text{H}_3\text{O}^+(aq) + \text{OH}^-(aq)$$
The associated equilibrium constant $K_w$ is:
$$K_w=[\text{H}_3\text{O}^+]\times [\text{OH}^-] \approx 10^{-14}$$
(Strictly speaking the expression is:
$$\frac{[\text{H}_3\text{O}^+]\times [\text{OH}^-]}{[\text{H}_2\text{O}]}$$
But because $[\text{H}_2\text{O}]=\text{constant}$ for dilute solutions we can use the first expression.)
The reaction rates for the reactions are, where $k$ are the rate constants:
Forward reaction:
$$\text{H}_2\text{O}(l) \rightarrow \text{H}_3\text{O}^+(aq) + \text{OH}^-(aq)$$
$$v_f=k_f[\text{H}_2\text{O}]$$
Reverse reaction:
$$\text{H}_2\text{O}(l) \leftarrow \text{H}_3\text{O}^+(aq) + \text{OH}^-(aq)$$
$$v_r=k_r[\text{H}_3\text{O}^+][\text{OH}^-]$$
We can now show easily that:
$$\boxed{K_w=\frac{k_f}{k_r}}$$
Because in our case the rate constants $k$ are only dependent on temperature (see Arrhenius) that means that $K_w$ is a temperature (only) dependent constant.

One can also say that any equilibrium constant $K$ is related to the change of free enthalpy ${\Delta G°}$ of the particular reaction :
$${\Delta G° =  RT lnK}$$  And this ${\Delta G°}$ does not depend on any other compound present in the solution.


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SPE recovery greater than 100%?
I am evaluating a paper Gan et. al. 2013 (https://doi.org/10.1016/j.chroma.2012.11.081) Table S4 (see image below) and I am focusing on evaluating the percent recovery for using different solid-phase extraction (SPE) cartridges. Investigators were measuring different artificial sweetener concentrations in water samples. This table suggests that their sucralose (SUC) % recovery was 106% when they used 1 mM TRIS buffer and eluted with methanol and they used a Waters Oasis WAX SPE cartridge. I am assuming n = 3 means they did a sucralose measurement 3 times. My ultimate question is, how can something have recovery greater than 100% after SPE? Does the precision and error inflate this recovery value to be over 100?

106% is obviously an experimental error of some sort. Exactly why heaven only knows.

Exactly as MaxW said. It is important to remember that true values, and experimental estimates of them, are different things. Especially with only three replicate determinations.

MaxW and Ed V, thank you for your quick feedback. Am I stupid to assume that despite having a 106%, it would mean that they still had an acceptable or adequate % recovery greater than 70%?

I think it is fine. If it had been 106% with 30 replicates, I would be suspicious: maybe the recovery protocol was not quite correct or whatever. But 30 replicate determinations would be highly atypical: too costly in terms of time, money, materials, etc.

@EdV Thank you very much again for your feedback and taking the time to answer my questions.


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Gaussview GMMX grid seach: job completed before going through all the structures
I am recently using the GMMX add-on in gaussview to conduct conformer search. I was using the grid search method and the progrom suggested the number of structures is 729 (See picture below). I submitted the job and it completed normally. However when I checked the log file, the "Nmin" is 496 (See picture below). I think "Nmin" should indicate the number of times of minimizations. So what I am confused about is why it is not 729?
If it's some built-in algorithms to skipping some structures, I got even smaller Nmin when I was using smaller bond resolution.
Thanks!

I am not familiar with GMMX, but two things come to my mind: 1. Does increasing the energy window (eg. To 100 kcal/mol) include other structures? 2. Could some structures be excluded because of symmetry (and hence redundancy?)

This question may fit on the new Materials Modeling SE


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Spectrophotometry: shifting absorbance curves vertically so isosbestic points match
We're measuring reaction kinetic parameters of a hydration reaction (Co complex) by measuring the absorption spectrum a different time intervals, and calculating the ratio of product to reactant by comparing the absorbance measured to the absorbance spectrums of the product and reactant in pure form. Water baths at 6 different temperatures were tested and for certain graphs, the results are coherent, with curves meeting around the isosbestic point (where the absorbance of both product and reactant are the same, first graph). However for certain graphs there is a significant vertical shift on certain curves which renders the results incoherent (second graph)
I had the idea to shift the stray curves down to match the isosbestic points of the product and reactant, as this should technically be constant whatever the reaction progression (at a given concentration). I realize this will render the use of the Beer-Lambert law pointless as we are completely fabricating our results, however what is important is the ratio of product and reactant, relative to other curves, which will be conserved, and these will actually be comparable once the isosbestic points match.
I just wanted some outside input
-Is this unscientific? Will my TA wonder what the hell i've done and take points off, or is this process justifiable
-Should i just process and present my data without trying to correct the error.
-What could this baseline error be caused by? since the errored curves seem to be shifted up, i'm suspecting condensation on the cuvette walls increased the absorbance, as the water baths went up to 50 degree. Although I dont remember noticing condensation on the cuvettes.
Thank you for any input.

For clarification: Is the instrument a dispersive one (reference and sample cell measured simultaneously)?  Just a couple of weeks ago, there was a similar question about overcompensating a baseline, and fluctations in the light source's intensity over time -- which a dispersive instrument may easier compensate than a single beam instrument -- was one possible cause. (https://chemistry.stackexchange.com/questions/128248/why-is-absorbance-in-uv-vis-spectrometer-negative-even-after-calibration/128251#128251)

thank you for the link, the instrument was in fact dispersive. The reference solution was not changed throughout the measurements,

Could you provide a bit more information? For example, how many water baths were used to get the 8 curves in the bottom figure? In the upper figure (at room temperature?), only the yellow curve is offset a small amount. But the bottom figure has rather large jumps: 0 min curve high, next three low, rest high again. Were these all collected by you or were lab partners involved?

@EdV An evidence against working within the same lab group could be that the the ratio of Abs for t = 0 around  460 and 620 nm are both about 1.3, but the absolute ABS values recorded for this time do not equate to each other.  Otherwise, the first plot _(except the yellow trace) would be fine.

It's very hard to be precise, but I would not expect condensation at the  higher temperatures, more likely here is that small bubbles form on inside of the cell wall that come and go with time. Unless you can repeat all you can do is to use the isobestic points from the top graph to correct the lower one, always remembering in your report you must describe why you had to do this.

@ibsrewed Was the reference cell filled with the blank solvent?  You ensured both sample and reference at the same temperature, correct? Was the direction of the cells always the same -- sometimes using the cell with the pathway mark facing the lamp does not yield the same readout as using the cell with the pathway mark facing the detector (plastic cells more than the ones of quartz, but likely you used always the same sample cell, didn't you?).

@Buttonwood Yeah, good point! I would like to see the lab write up given to the students, and literature reference(s), but, lacking that, it could just be operator error(s) or the bubbles mentioned by porphyin!

@ibsrewed I speculate the conditions for the second plot are such that the reaction run about six times as fast to completion, than for the first plot.  With the old rule of thumb that per 10...15 K temperature increase the reaction rate would double, are the time readings converted to minutes, but based on seconds? Recalling my TAing, the most accurate students for recording UV kinetic data had several dry runs to learn who would consistently need the same time between taking the sample, putting the cell into the UV / closing lid and hitting the button within $\pm 5,\pu{s}$ per cycle.

@ibsrewed In this perspective, perhaps something delayed the anticipated line of events; e.g. once there were droplets on the wall of the cell, but there wasn't a tissue to dry the cell, hence delaying putting the sample into the slot, thumbling to get the file name into the software, etc. pp. You need to have practise rounds (almost drill-like) especially if you collect the data manually; without chatty disturbance from / with other colleagues or the TA until all data for one of these plots is recorded. And plot these plots early (e.g., graph paper) to see the problem early, in the lab.

thank you for your input. each graph shows the evolution of absorbance spectrum at the given time intervals at a specific temperature. sorry I didn't add the actual captions. The methodology for each measurement was relatively straight forward, extracting a given volume by pipette, placing into cuvette and usually within a minute the spectrometer measurements started, each lasting about 1.5 minutes. Multiple reactions were going on at the same time, a few times we had a "queue" of 2 cuvettes however this should not change the fact that the curves should go through isosbestic point.

this would just affect the accuracy of the kinetic parameters, since the reaction kept proceeding in the cuvette, although it was relatively slow at room temperature. The first plot was at 35degC, the second at 50degC. We cant say for sure whether the temp of the reference solvent cuvette and sample cuvette were the same however given the size of the cuvette I would have assumed it would cool relatively quickly.

anyway we'll proceed to calibrating the curves around the isosbestic point, i've only to decide which one, the right or the left, or do an average shift around the two. We'll play around and see which gives the nicest results. also the cuvettes used were just as translucent in both directions (no corrugated sides like ive seen some are) and we were careful not to leave fingerprints so they were always held on the top.


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Why doesn't diamond conduct electricity but silicon does?
Diamond's structure is a large network covalent crystal, where there are strong covalent bonds between neighboring tetrahedral carbon atoms. The explanation I was given was:

The electrons are held tightly in the covalent bonds of the structure, so they are unable to move around. Thus, diamond cannot conduct electricity.

Why is it then, that in silicon (also a network covalent solid), which has an identical crystal structure to diamond, is known to conduct electricity relatively well and even dubbed a semiconductor? Aren't the electrons held tightly in the covalent bonds between neighboring silicon atoms?

Actually, diamond is a high band gap semiconductor. For most purposes, it gets treated as an electrical insulator. Look up “Electronic band gap” in wiki and scroll down to filling the bands. Intrinsic Si has a thermally-generated population of electrons in its conduction band and holes in its valence band. With diamond, the band gap is over 5 eV, so there are almost no electrons and holes in the relevant bands. Doped diamonds, like the famous Hope blue diamond, are extrinsic high band gap semiconductors.

I just searched for “semiconductor” here and there are 158 questions or answers. Lots to learn reading through these!

Yes, a high band gap semiconductor and when doped with Boron, a diamond electrode for wastewater treatment, see https://infoscience.epfl.ch/record/79837 .

40 years ago we considered diamond an insulator. Now it is a wide band gap semiconductor. Either way, it doesn’t conduct that well compared to silicon, much less a metal. But it doesn’t have to in order to be useful.

@JonCuster As per your comment to the answer here! Nice to know that AlN can be usefully doped. Thanks!

@EdV, Nice comment, and it might be moved to an Answer, perhaps with a reference link.

@DrMoishePippik Thanks, but I thought this one was really more for the physics stack exchange, or what Murray Gell-Mann snarkily called “squalid state physics”!

I'm not a solid-state physicist. Yet, this is how I understand this problem, so bare with me.
According to this article:

Every solid has its own characteristic energy-band structure. This variation in band structure is responsible for the wide range of electrical characteristics observed in various materials. In semiconductors and insulators, electrons are confined to a number of bands of energy, and forbidden from other regions. The term "band gap" refers to the energy difference between the top of the valence band and the bottom of the conduction band. Electrons are able to jump from one band to another. However, in order for an electron to jump from a valence band to a conduction band, it requires a specific minimum amount of energy for the transition. The required energy differs with different materials. Electrons can gain enough energy to jump to the conduction band by absorbing either a phonon (heat) or a photon (light).

This forbidden gap plays a major role in determining the electrical conductivity of material. The materials can be classified into three types based on the forbidden gap. they are:

Insulators: The materials, which does not allow the flow of electric current through them are called insulators. The band gap between the valence band and conduction band is very large in insulators, which is approximately equal to $\pu{15 eV}$. Normally, in insulators the valence band is fully occupied with electrons due to sharing of outer most orbit electrons with the neighboring atoms whereas its conduction band is empty, i.e., no electrons are present in the conduction band.
Semiconductors: The material, which has shown electrical conductivity between that of a conductor and an insulator is called a semiconductor. The band gap of semiconductors is is very small, and it falls between that of valence band and conduction band. The value of this forbidden gap is about $\pu{1-3 eV}$. As a rule of thumb, if the material has $\Delta E_\mathrm{gap} \le \pu{3.0 eV}$, it is considered as a semiconductor (Semiconductors- Band Gaps, Colors, Conductivity and Doping).
Conductors: In a conductor, the valence and conduction bands are overlapped each other. Therefore, there is no forbidden gap in a conductor (the value is generally negative). Thus, the valence band electrons to move into conduction band needs only a little amount of applied external energy.

Experimental values of $\Delta E_\mathrm{gap}$ for diamond and silicon are $\pu{5.48 eV}$ and $\pu{1.17 eV}$, respectively (Ref.1). Keep in mind that band gap get narrowed when going down on a group in periodic table. For example, $\Delta E_\mathrm{gap}$ of germanium is $\pu{0.74 eV}$ (Ref.1).
Therefore, silicon and germanium behave like semiconductors, while diamonds fall into insulator category in general (yet, please refer to  Ed V's comment above for better definition).
References:

Fabien Tran, Peter Blaha, "Accurate Band Gaps of Semiconductors and Insulators with a Semilocal Exchange-Correlation Potential," Phys. Rev. Lett. 2009, 102, 226401, 4 pages (DOI: 10.1103/PhysRevLett.102.226401).

+1 Excellent answer: at the right level of detail, without getting bogged down in the little band diagrams, Fermi levels, and on and on.

Thanks for the great answer, but I am curious as to why the band gap decreases going down a group?

Below germanium, you have tin and lead. Lead is a metal (no band gap), but tin has two allotropes: white tin is ordinary metallic tin, but gray tin has the diamond, Si, and Ge crystal structure and is not a metal. See the wiki article on ‘tin’. Metals have no band gap, but gray tin does.

@Ed V: Great comment to answer OP's question. I appreciate it. :-)

Diamond is made of Carbon. The valence Shell of the carbon atoms in a dielectric Crystal like diamond are filled. Conductivity can only oceur for electrons in a electronic shell that is partially filled.So Diamond crystal with no defects can not Conduct electricit. On the other hand Silicon reacted as a semiconductor.So Silicon have the ability to conduct electricit. But diamond has a much higher band gap than Silicon.So diamond  doesn’t conduct electricity but Silicon does.


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Could metal hydride storage tanks for hydrogen store helium as well, given their repective atomic radii?
With atomic radii very similar, could helium dissolve into metallic foam about as well as hydrogen, thus making metal hydrite hydrogen tanks suitable for helium as well?

Absolutely not!

He can be conveniently stored in pressurised steel bottles or, slighly less convenient, in liquid form. There are no mobile applications that need significant amounts of helium.

Hydrogen does not just insert itself physically into the metal.  Although only a few metals form stoichiometric hydrides with predominantly ionic character, most of them (and all those involved in hydrogen/hydride storage) form some type of bond, generally involving saturation of the hydrogen 1s orbital with ionic or polar covalent bonding.  Helium simply does not have such capability for the same reasons it does not bond with much else, and it is generally not significantly soluble in metals.
An introduction to various methods to store hydrogen is given by Wikipedia.  The
US Department of Energy reviews some proposed materials here.


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How to deal with the unit "Coulomb per equivalent"?
In a question that I came across they defined $1\,\pu{F}$ as 96500 coulomb per equivalent. Now as far as my knowledge is concerned $1\,\pu{F} = 96500\,\pu{C/mol}$. The question involved asks to calculate the current that needs to be passed (or may be a function of that) wherein I mistook the $n$ factor and the answer was just half of what it was supposed to be. However that taking $1\,\pu{F}$ as "coulomb per equivalent" would lead me to a antoher unit (equivalent per mole) in order to get the desired unit in the question.
In this context I would like to:
a)  Get an intuitive explanation into the number of equivalents (just like moles is Avogadro number of substances).
b)  Know what the unit "equivalents/mole" means (is it the same thing as $n$ factor?) How does the $n$ factor treat number of moles and equivalents so differently?

This is one of the best and unanswered question that I have come across.I intend to give some reputation for a better answer to the question.How to do that?Kudos @Rajorshi Koyal.

I am unable to completely understand what you said here: "However that taking 1F as "Coulomb per equivalent" would lead me to a unit equivalent per mole in order to get the desired unit in the question." Can you please elaborate on this?

@AumkaarPranav I edited the question actually I meant C/mol instead of C/equivalent. That is all.You might ignore that line also.

There is confusion in the 1st sentence. 1F = 1V / 1 C is 1 Farad, the unit of electrostatic capacitance. F without a quantifier is Faraday's constant equal to the charge equivalent to a molar equivalent. I.e. to 1 mol, If 1 electron is exchanged, or to 1/n moles, if n electrons are exchanged. F = Na . e, the multiple of the Avogadro constant and the elementary charge.

In redox reaction, one equivalent of a substance is simply the number of moles of it that accepts or donates one mole of electrons. Notice how equivalent is not defined as an independent quantity like mole. One equivalent of electrons is simply one mole of electrons.

@Poutnik References accessible to mine (e.g., https://en.wikipedia.org/wiki/Coulomb and https://en.wikipedia.org/wiki/Farad) state an inverse definition regarding the farad.  Instead of 1F = 1V / 1C (ratio potential over charge, your current / initial comment), they state 1F = 1C / 1V (ratio charge over potential); thus, possibly an accidental error.

@user586228 This question is basically asking about how to think of equivalent and it's relation with mole. I believe it has been asked (and answered) on this site earlier too. Anyway, you can put a bounty on a question after 2 days.

@Buttonwood  Sure, thanks. You are obviously right.

Does this answer your question? What is an equivalent?

For an electrochemical reaction, you count the atoms / ions by mol, and use the coulomb as a counting unit of charge. For a more intuitive explaination of the $n$ factor in the Faraday equation, try this analogy:
The summer olympics include swimming in a pool with lanes $50\,\pu{m}$ long.  Among the typical competitions are runs about $50$, and $100\,\pu{m}$.  Now, for the sake of the analogy, think the swimmers were representatives of ions.  Each of the swimmers could represent

a $\ce{Cl-}$ to yield $\ce{Cl}$, a change by one unit of charge.  So they swimm only one lane in one competition, $n = 1$; the single discharge happens when the swimmers clap the wall with their hand.
a $\ce{Cu^{2+}}$ wanting to yield $\ce{Cu}$. As they start with two charges now, two discharges in this single race are required ($n = 2$). Their first discharge is when they contact and turn at the opposite wall ($50\,\pu{m}$), their second and final discharge when they clap the other wall with their hand ($100\,\pu{m}$).


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Reaction of grignard in presence of Halogen
What happens when a Ketone  say $$RCOCH3$$  on reaction with grignard reagant in presence of X2 (halogen) what happens?? Can someone please explain

In the presence of X2? The Grignard will react with the elemental halogen first.

Grignard reacts with the elemental halogen. Ketone will be left un-reacted with. If you have insufficient halogen in reactant side then the leftover  Grignard will react with the ketone to for the tertiary alcohol to quench all the reactants.
However I am enlisting the products Chlorination of Grignard reagents, RMgX, (where X = Br, I) gives considerable quantities of RI and RBr together with RCl. Bromination of RMgI results in RI and RBr. Bromination of RMgCl leads to RBr, and iodination of RMgX (where X =Cl, Br), to RI.


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Are there ionic solids that conduct electricity?
We are taught in school that ionic substances don't conduct electricity, and when they do, it is either because they are in a molten state or because they are in solution. I understand these concepts.
Are there any exceptions? By exceptions, I am looking for ionic compounds that may not conduct electricity well, but they conduct electricity better in comparison to other ionic compounds. Additionally, what makes such a compound a better conductor of electricity?

Mentioning that you are looking for ionic substances that conduct electricty in 'their solid state' would make the question more clear. The standalone statement that ionic substances don't conduct electricity is false as (like you mentioned) their molten state (which still is the same substance) conducts electricity.

Does not fit perfectly into your question but you can have molecules and macromolecules whose ions - often present in pending groups - have mobility, or at least leads to electrons / hole with mobility. Example of the first case is PEDOT, examples of the second kind are the Os of OLEDs.

related https://chemistry.stackexchange.com/questions/50552/what-makes-ionic-conduction-fast-or-slow https://chemistry.stackexchange.com/questions/26863/what-is-the-meaning-of-superionic

Yes there are  - take a look at the second paragraph in the answer to https://chemistry.stackexchange.com/questions/67481/does-aluminium-oxide-ionize-after-melting-point/67493#67493

Yes, there are.
Ionic conductivity greatly varies between compounds. For some of them is much higher then for others. They are called fast ion conductors, or superionic conductors.
Well known example is silver iodide. While it may be called a "salt", it's probably more covalent then ionic, that is, before it's heated to 146 °C, when its structure changes and Ag ions are "set loose" to conduct electricity. In result conductance rises a few thousand times.
If rubidium iodide is added to $\ce{AgI}$ we can get rubidium silver iodide which is even better, and is sometimes classified as one of "advanced" superionic conductors.
Many substances can be superionic in right circumstances, even water ice can, if temperature and pressure are high enough.

Yes.
Another example is VO2. Over 67 °C its crystal structure changes to rutile with a band gap of ~0.1 eV. Dopants of other substances can reduce the temperature at which it becomes a conductor.

It's arguable if it is actually ionic. While oxides of transition metals are typically treated as such, it's a drastic simplification. VO2 conducts via electrons not ions.


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Is hydroxychloroquine an anti-psychotic drug?
Is hydroxychloroquine an anti-psychotic drug or psychotic drug?

Hydroxychloroquine certainly crosses the blood/brain barrier and there are reports of it having psychiatric side effects example here. It is not prescribed for psychiatric conditions details here


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Why does increasing the concentration of the salt bridge in a galvanic/voltaic cell increase voltage?
I recently set up an experiment with 1M half cells made up of zinc and copper nitrate. They were connected with a sodium nitrate salt bridge of varying concentrations. As the concentration doubled 5 times from 0.125M to 2M the voltage also increased. I couldn't find much explaining why this is however there is another question on this site which is quite similar and I imagine the answer may explain my problem as well. Link is attached underneath. Could increasing the concentration allow for a higher flow of electrons through the salt bridge due to lowering internal resistance and therefore increasing voltage?
(Why does increasing number of salt bridges increase voltage of electrochemical battery?)
Any help would be greatly appreciated. Thanks

How do you measure the voltage ? If it is with an analogical apparatus, the cell will produce some current, and the voltage will decrease from the theoretical value. And then if you improve the connection between the two compartments, the internal resistance decreases and this will produce an increase in the voltage. If you use a digital voltmeter, this effect sill not be observed, as practically no currant is consumed.

Electrons do not flow through a salt bridge, only ions do so. Increasing the ion concentrations in the salt bridge lowers its resistance, which matters if substantial current flows. As @Maurice noted, a digital voltmeter results in very low (negligible in your case) current draw because the input resistance of the digital voltmeter is typically 10 M ohms.

Based largely on a prior answer to the question: 'Why is it important to use a salt bridge in a voltaic cell? Can a wire be used?', the explanation centers on maintaining the charge balance (electrical neutrality) of the battery cell.
The electrons are efficiently transported via the electrodes connected by wires as sourced from the half-cell:
$\ce{Zn -> Zn(II) + 2 e-}$
However, to balance the charge in this half-cell, NO3- must enter from the salt bridge (and conversely, K+ must enter the other half-cell to balance the charge in both that cell and the salt bridge).
Not surprisingly, the less efficient part of this cell is the salt bridge itself creating resistance. Raising the concentration in the salt bridge lowers relative water content (a poor conductor and source of resistance) and increases the potential rate of diffusion of the respective NO3- and K+ ions, resulting in a corresponding increase of the cell efficiency (as witnessed by an increase in voltage). Some supporting comments per Wikipedia:

If no salt bridge were present, the solution in one half cell would accumulate negative charge and the solution in the other half cell would accumulate positive charge as the reaction proceeded, quickly preventing further reaction, and hence production of electricity.[1]

Also:

The conductivity of a glass tube bridge depends mostly on the concentration of the electrolyte solution. At concentrations below saturation, an increase in concentration increases conductivity. Beyond-saturation electrolyte content and narrow tube diameter may both lower conductivity.

And with respect to filter paper bridges, more details:

Conductivity of this kind of salt bridge depends on a number of factors: the concentration of the electrolyte solution, the texture of the filter paper and the absorbing ability of the filter paper. Generally, smoother texture and higher absorbency equates to higher conductivity.

Another interesting cited desirable property of the salt bridge is that the mass ratio of the bridge electrolyte relative to half-cell electrolyte should be small.

@Karsten Having a high ion concentration in the salt bridge and having a gel-based salt bridge (to minimize convection mixing) slows down the process of redox-active ions making their way through the salt bridge.