text:
What is the true formula of pH scale?
I'm learning Acid and Base at school, and I have a confusion in that topic.
The $\mathrm{pH}$ scale is measure of the $\ce{[H+]}$ in a solution. So why is its formula $-\log(\ce{[H+]})$? I don't understand why the formula is not just $\ce{[H+]}$ since the definition of the $\mathrm{pH}$ is that it is the measure of $\ce{[H+]}$ in a solution.
I think that we should write it like the latter since it gives us a easier number to work with. Example: $\mathrm{pH}=2$ instead of $\pu{0.02 mol/L}$.

Please refer to the link: https://www.khanacademy.org/science/chemistry/acids-and-bases-topic Hope it helps!!

@Mérébé you have completely changed your question and now my answer does not seem much relevant. I don't think that one is should reword his questions to such an extent that the answer(s) received no longer look relevant. Also, you are arguing as to why is this scale used for measure of H+ ions which is completely nonsense. Did you ask your teacher why do we use meter (or other units)  as a measure of length and why not something that is, say 0.8 of a meter? So your question is not appropriate.

1) Yes, the pH scale indicates the concentration of $\ce{H+ }$ ion in a solution. But you probably don't know that pH is defined as the negative log(base 10) of $\ce{ H+}$  ion concentration $$\mathrm{pH = -log_{10}[H+]}$$
So, as the $\ce{[H+]}$ increases, its logarithm increases and the pH decreases.
2) No, the concentration and strength of an acid are different. Strength is its property, like $\ce{HCl}$ is strong and $\ce{CH_3COOH}$  is weak. These are related to their extent of dissociation in the solvent. However concentration is basically the amount of acid dissolved in a certain amount of the solvent.
But if the molarity of the acid is same, the higher concentration of $\ce{H^+}$ ions then indicate more dissociation of the acid and hence is a stronger acid.
Don't confuse with the word strength used in titration to denote the concentration of solute in $\pu{g/L}$.
3) This part of your question is not quite clearly stated.
But this might help you: titration of weak acid against a strong or weak base involves the formation of buffer, about which you probably don't know right now.


text:
Help understanding external magnetic field sensitivity of radical pair reaction rate
Two recent papers have been published on magnetoreception in birds. In this case it is the ability of the birds to sense the direction of the Earth's magnetic field potentially for navigation.

Expression patterns of cryptochrome genes in avian retina suggest involvement of Cry4 in light-dependent magnetoreception Atticus Pinzon-Rodriguez, Staffan Bensch, Rachel Muheim J. R. Soc. Interface 2018 15 20180058; DOI: 10.1098/rsif.2018.0058. Published 28 March 2018
Double-Cone Localization and Seasonal Expression Pattern Suggest a Role in Magnetoreception for European Robin Cryptochrome 4 Günther, Anja et al.
Current Biology , Volume 28 , Issue 2 , 211 - 223.e4

The idea has been around a long time; cf A Biomagnetic Sensory Mechanism Based on Magnetic Field Modulated Coherent Electron Spin Motions Zeitschrift fur Physikalishe Chemie Neue Folge, Bd. 111, S. 1-5, (1978)
The two new papers address the role of something called "radical pairs", two molecules, each with an unpaired electron. The large molecules are in the retina and tend to have a preferred orientation with respect to the curved retina, and they also appear to affect the perceived intensity of blue light. This causes a modulation of light sensitivity across the retina dependent on the direction of the magnetic field with respect to the direction of the local normal vector of the retina.
The modulation of the reaction rate by the magnetic field direction has to do with singlet vs triplet states and quantum coherence, but I only say that because I've tried to read Cryptochrome and Magnetic Sensing by the Theoretical and Computational Biophysics Group at U. Illinois, Urbana-Champagne, as well as this Science Alert.
Are there simpler examples of the sensitivity of the reaction rate of "radical pairs" to the orientation of a magnetic field? Perhaps two atoms or two small molecules? Or does this phenomenon generally occur only in large, complex molecules?

below: "This is how a bird might see magnetic fields. (Theoretical and Computational Biophysics/UofI)" From ScienceAlert.

Two atoms or two small molecules have no way to tell up from down.

@IvanNeretin in a gas or liquid that may be true. They could be oriented on a surface possibly, or in a laser trap, so I thought I would remain open to the possibility. The reaction may lead to the production of a photon who's polarization could be detected as well, in which case it might not be necessary to pre-align the reactants, but instead measure something correlated to their alignment following the reaction.

Your question fits under "other chemical applications of magnetism", especially since we don't have a more specific tag to that effect (or so it seems). As for the reaction-mechanism tag, well, let it be.

(This is really a comment but  has become too long)
There are many examples of  triplet states in magnetic fields as studied by electron paramagnetic resonance (ESR or EPR) experiments. See chapter 8 'Introduction to Magnetic Resonance' by Carrington & McLachlan. It is an old book but does contain all you need to know. (A paper I happen to know showing magnetic field effects is Hyde et al. Chemical Physics v151,p239 1991 but there must be many others if you search)
As to the bird experiments, eqn 1 of the paper by Schulten et al. as described in your question describes the situation: scheme copied below.
$$\begin{align}
(D^*&+A)\leftrightharpoons \;^1(^2A^-+\;^2D^+) \stackrel{B}\leftrightharpoons \; ^3(^2A^-+\;^2D^+)\to \;^3T\\ &\downarrow\\ &\;^1X \end{align}$$
What it seems is necessary is (a) The reaction starts with a light induced electron transfer from excited donor $D^*$ to a nearby molecule that is the acceptor $A$. (b) The singlet state $^1(^2A^-+\;^2D^+) $ species produced has energy levels that cross those of one of the  of $m_I=\pm 1$ levels of the triplet states $^3(^2A^-+\;^2D^+) $ three levels at some value of the magnetic field $B$ and so singlet to triplet crossing occurs($^1$). (c) This enables the yield of products $^3T$ to $^1X$ to vary as the value of the magnetic field varies as the animal moves. (d) The protein has to be fixed relative to the bird otherwise this will not work. (This is the 'anisotropy' mentioned)  Being fixed could simply mean being in a cell membrane in one part of the retina. (e) Somehow the ratio of products has to be able to be distinguished.  Clearly this mechanism clearly will not work at night.
So all in all a complicated mechanism, involving subtle processes, and this makes it very interesting.

($^1$) In a molecule as the triplet often differs significantly in energy from the singlet a huge magnetic field is needed to reach the S-T crossing point, but in pairs of molecule the interaction is very small and so the energy difference is v small also. This is advantageous as then small magnetic fields can cause S-T crossing. (Normally the singlet is above the triplet in energy, unlike the in figure, but the crossing is clear nonetheless)
also see Can magnetic fields affect a chemical reaction?

This answer is extremely helpful, you have described some key aspects of the proposed phenomenon in a way that I can begin to understand, which I appreciate very much! I had though that the role of the magnetic field was to disturb the coherence between two states, but you've described it in terms of a level-crossing.

In any event, I wonder if I should just adjust the question to something like "can someone help me understand what might be happening here?" to better match this answer

Both singlet and triplet states $(D^++_A^-)$ are coherent as they are a single spin state not two separate entities $D^+$ and $A^-$. There still has to be a crossing if one has to convert to the other. Also note that the products are no longer coherent with these species.

Okay it's been several decades, I'll go read up on it. Thanks! By coincidence, I've just received this spin-related answer as well.

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text:
What would happen if you heat volatile/flammable material in a vacuum?
Fire is an oxidizing chemical reaction, in the simplest sense it needs oxygen and fuel to burn in order to stay alight. But what reaction would you observe from a flammable substance like gasoline or lighter fluid in a vacuum?

What kind of boxing match would you observe if one of the fighters failed to appear, yet the referee decided to proceed anyway? Not a very spectacular one, I'd say.

In a vacuum volatile compounds will evaporate. They won't react as there is nothing for them to react with.

You should look at a video about fires in vacuum by codys lab

If there's any trace of oxygen, and a spark, the vapors would burn like normal.
If there isn't any oxidising agent, it'll just vaporize and stay that way (maybe the vapors will attain equilibrium with the liquid if the vapor pressure is high enough, basically if there's enough liquid to start with).
That's mostly it... Nothing spectacular :)

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text:
Hybridisation of cobalt in tris(oxalato)cobaltate(III)
Will the hybridisation of of $\ce{Co}$ in $\ce{[Co(C2O4)3]}^{3-}$ be $\mathrm{sp^3d^2}$ or $\mathrm{d^2sp^3}$? The question in which I found it has mentioned the answer to be $\mathrm{d^2sp^3}$ while I was thinking of the other one.
According to me, oxalate is a weak field ligand and it should not cause the pairing of electrons in the half-filled $\mathrm d$ orbitals.

Oxalate is also a chelating ligand, it is not so weak as you might be thinking. It can be thought as very slight strong field ligand. So it is in the competition region of $\Delta _o$ with pairing energy. Unless you have experimental data, you can not tell anything whether it will be inner orbital or Outer orbital complex.

I would avoid using hybridization to describe transition metal compounds. See https://chemistry.stackexchange.com/questions/76726/why-is-it-wrong-to-use-the-concept-of-hybridization-while-studying-complexes

@IanBush Such questions form an important part of undergraduate admission tests here and thus I needed to know what should the answer be. All the textbooks at my level use the concept  of hybridisation.

@PiyushMaheshwari Then all your textbooks are wrong, as it is completely wrong to use hybridisation for transition metal complexes.

@Martin-マーチン Did you notice the words "at my level"? This was meant to convey that I am not really familiar with the deeper aspects of the concept and its limitations. This is because we are not taught about it in so much detail yet. I don't say that my books are correct either. But to know what is wrong with this theory, we also have  to know what is the theory all about.

@PiyushMaheshwari of course I did read that, however, I don't need to try to teach you something that is completely wrong, in order to teach you what is correct. (I'm too short on time to do that right now.) The fact that they still teach this outdated and wrong theory is sad, and if you need it to pass the exam, that does make it worse. Please forgive me, but you should immediately forget this approach on hybridisation as soon as you passed this exam. And further on question everything else in your textbooks.

@SoumikDas The ion's ability to chelate has nothing to do with its place in the spectrochemical series (i.e. how much splitting it can cause). The chelating ability of ligands has got to do with the stability constants of the complexes they form.

https://chemistry.stackexchange.com/questions/94582/hybridisation-of-cobalt-in-trisoxalatocobaltateiii

Valency on Co is +3 As C2O4^2- is a weak field ligand it cant cause pairing of electons but due to chelating effect So, it is d2sp3

https://chemistry.stackexchange.com/questions/94582/hybridisation-of-cobalt-in-trisoxalatocobaltateiii

As explained in The Chemistry of Iron, Cobalt and Nickel: Comprehensive Inorganic Chemistry, at pages 1104 and 1105:

almost all cobalt(III) complexes are low spin ... only with fluoride ions as ligands are high spin complexes ... found

In table 10, the book specifically lists $\ce{[Co(C2O4)3]}^{3-}$ as low spin and cites to J. Chem. Soc. (A) (1966) 798.
Therefore, according to the historical valence bond theory of transition metal complexes, it would be considered $\mathrm{d^2 sp^3}$ for the following reason:
There are (in the context of the theory) two unoccupied $\ce{3d}$ atomic orbitals in $\ce{Co^{3+}}$.
Six pairs of electrons from the oxalate ions are added to the two $\mathrm{3d}$ orbitals and the $\mathrm{4s}$ and three $\mathrm{4p}$ orbitals.

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text:
Confusion about the ICE table

Calculate the pH of a $\pu{2M}$ $\ce{H_3PO_4}$.Calculate the equilibrium concentrations of $\ce{H_3PO_4 , H_2PO_4^{-} , HPO_4^{2-}}$.

First equation:
$\ce{H_3PO_4  <=> H^{+} + H_2PO_4^{-}}\ \ \ K_\mathrm{a1} =7.5\times 10^{-3}\\$
$\begin{array}{|c|c|c|c|} \hline
\ & [\ce{H_3PO_4}] & [\ce{H^{+}}] & [\ce{H_2PO_4^{-}}]\\ \hline
\text{I} & 2 & 0 & 0 \\
\text{C} & -x & +x & +x \\
\text{E} & 2-x & +x & x\\ \hline \end{array}\\$
$$K_\mathrm{a1}=\frac{[\ce{H_2PO_4^{-}}]*[\ce{H^{+}}]}{[\ce{H_3PO_4}]}=\frac{x^2}{2-x}=7.5\times 10^{-3}$$
Solving for $x$ we get $x=\pu{0.118M}$.
$[\ce{H_3PO_4}]=2-x=2-0.118=\pu{1.882M}$
$[\ce{H_2PO_4^{-}}]=x=\pu{0.118M}$
Second equation:
$\ce{H_2PO_4^{-}  <=> H^{+} + HPO_4^{2-}}\ \ \ K_\mathrm{a2} =6.2\times 10^{-8}\\$
The solution manual says:$$K_\mathrm{a2}=\frac{[\ce{HPO_4^{2-}}]*[\ce{H^{+}}]}{[\ce{H_2PO_4^{-}}]}=\frac{0.118\times[\ce{HPO_4^{2-}}]}{0.118}$$
$[\ce{HPO_4^{2-}}]=K_\mathrm{a2}$
However I'm not sure why didn't we use an ICE table for the second equation as follows:
$\begin{array}{|c|c|c|c|} \hline
\ & [\ce{H_2PO_4^{-}}] & [\ce{H^{+}}] & [\ce{HPO_4^{2-}}]\\ \hline
\text{I}& 0.118 & 0.118 & 0 \\
\text{C}& -x & +x & +x \\
\text{E}& 0.118-x & 0.118+x & x\\ \hline \end{array}\\$
Is it because $x$ is negligible with respect to 0.118? Or should the concentrations precalculated above be in the equilibrium row of the ICE table?

Yes, x << $\mathrm{0.118 M}$ that it is not gonna hurt even if it is ignored. But for a rigid perfectionist,
$$K_\mathrm{a2}=\frac{[\ce{HPO_4^{2-}}][\ce{H^{+}}]}{[\ce{H_2PO_4^{-}}]}=\frac{x\times (0.118+x)}{0.118-x}$$
Hence we get a quadratic equation,
$$x^2 + 0.118000062 - 7.316\times 10^{-9}=0$$
Solving for $x$, we get $x = 6.199993484584665\times 10^{-8}$ which is approximately equal to the value of $K_\mathrm{a2}$.


text:
How to compare the reactivity of sec-butyl iodide and n-butyl bromide for SN2?
Steric crowding is greater for the secondary halide, and this is important as the neucleophile will attack the back lobe of carbon atom. But the relative ability to leave is more for iodide than bromide.
So how can I compare the $\mathrm{S_N2}$ rates for sec-butyl iodide (2-iodobutane) and butyl bromide (1-bromobutane)?

It doesn’t matter. Steric hindrance is the more-to-worry about factor. See it this way; A group can leave only if the backside attack can happen. If that doesn’t even happen, no point in comparing the relative capability to leave.


text:
Non-Scanning Mass Spectrometry
Does anyone do mass spectrometry with a magnetic sector to separate ions by mass and an array of detectors in a line to measure the ions? If not, why don't they?
Something like this this setup for measuring the spectrum of light?

Yes, this is done in Medium Energy Ion Scattering, for example.

Probably not any more for mass spectrometry. Time of flight (TOF) methods are far better (both in mass resolution and signal to noise) as it is very easy to measure nanosecond time intervals. Also, only a single detector is needed and no expensive magnet.

Triple quadrupole mass spectrometers sort of work this way when they are in "MRM" mode.  But they only detect a single fragment ion from a single parent ion; they don't really monitor multiple ions with several detectors separated in space.  (They can switch very rapidly from looking at one parent/fragment pair to looking at another one, though.)

It's also worth noting Ron Heeren's work on imaging mass spectrometry.  He has developed instrumentation with spatially resolved detectors for e.g. qTOFs.  Those instruments have the feature of spatially separated (pixelated) detectors for ions, but rather than use that spatial separation to assess m/z, he and his group use the spatial separation to measure the spatial location of the origin of the ion, i.e. to do MS imaging. https://en.wikipedia.org/wiki/Ron_Heeren

Yes, this is called distance-of-flight mass spectrometry. It offers some advantages for large (e.g. biological) molecules and because of its high speed for identifying multiple targets, in chemical and biological warfare.

Indeed mass spectrometers that disperse the ions in space according to there m/z do exist. The main question is how the detection is done: the detector needs to have a rather large surface compared to the detector in a standard mass spectrometer. In the early times of mass spectrometry, these were called "mass spectrographs", using a photographic plate as the detector. Standard detectors (multi-channel plates, conversion dynodes, electron multipliers) either are not well adapted to a large surface collection or are expensive to produce with large surfaces. Thus currently, only some specific instruments such as isotope ration mass spectrometers still rely on using several collectors and a magnetic sector, placed at an adequate position to measure simultaneously the various isotopes of a given element. This brings much better accuracies on the abundances.
Dispersion is space is not the only way to achieve simultaneous detection of several m/z values (thus achieving a non scanning mass spectrometer): dispersion in time or in frequency can also be a method to measure simultaneously the m/z of ions without scanning the m/z range. Time of flight mass spectrometers disperse the ions in time (the smallest m/z arriving before larger ones) and FT-MS instruments (Orbitrap or FT-ICR, when using, respectively, an electrostatic field or a magnetic field) use the dispersion in frequencies to separate ions and measure simultaneously the frequencies of the ions present through their image current. The limitation of these mass analysers is that although they do not technically scan the entire mass range (all m/z that are present are detected), they do work in a sequential manner: ions are only admitted in the dispersion region at precise times. If an ion storage region is added ahead of the mass analyser, even with a continuous ion source it is possible to store the ions until the next analysis event and lose very little ions.
Thus as other methods do exist which paralellize ion detection over the full m/z range, the mass spectrographs have mostly gone out of the market.


text:
Reaction between a carbene and n-pentane
Does the following reaction occur when n-pentane reacts with methylene or does it form some other simpler products due to the reactivity of the carbene?
$\ce{CH3-(CH2)3-CH3 + :CH2-> CH3-(CH2)4-CH3}$
EDIT:
Do higher alkanes (pentane, hexane, octane....) show a similar reactivity with methylene or do the products depend upon the alkane?

I would expect a wild mixture of different products. Besides, carbene would rather attack just about anything else instead of a saturated alkane.

But in lower alkanes it inserts itself in between a (C-H) bond

Well, yeah, if it has nothing else to attack, then maybe.

Carbenes:
The reactions of carbenes with alkenes are well documented (e.g., cyclopropane formation by intramolecular cycloadditions to alkenes). Yet, the reactions of carbenes with alkanes have different story.  Carbenes can undergo insertion into a $\ce{C-H}$ bond. The the mechanism involves hydrogen transfer to carbene from alkane through cyclic intermedeate of $\ce{C*-H--CH2--C*}$ to substitute $\ce{-H}$ by $\ce{-CH3}$ (read). That mechanism has close resemblance to the mechanism suggested for the intramolecular cycloaddition of carbenes to alkene (ref 1). However, the reactivity of carbenes is depend on whether the excited state of the carbenes at singlet or triplet state 1 (courtesy from Ref.1):

Singlet carbenes can insert in a concerted manner, with the orbitals overlapping constructively provided the carbene approaches side-on, whereas triplet carbenes insert via a two-step radical pathway. However, very few triplet carbene insertions have been observed. Nonetheless, insertion between two carbons are unlikely, according to two separate groups from Yale and Princeton Universities (ref 2). Accordingly, when bicyclo[1.1.0]butane (also 1,3-dimethyl version) was treated with carbene, it did not insert between bridged carbon atoms, which is the most unstable bond, to give expected bridged methylene group (e.g., ref 3). Princeton group has very interesting on  different approach, but similar result: see ref 4.

References:

Pierrede Frémont, Nicolas Marion, and Steven P. Nolan, "Carbenes: Synthesis, properties, and organometallic chemistry," Coord. Chem. Rev. 2009, 253(7-8), 862–892; https://doi.org/10.1016/j.ccr.2008.05.018
K. B. Wiberg, G. M. Lampman, R. P. Ciula, D. S. Connor, P. Schertler, and J. Lavanish, "Bicyclo[1.1.0]butane," Tetrahedron 1965, 21(10), 2749-2769; https://doi.org/10.1016/S0040-4020(01)98361-9
W. von E. Doering and J. F. Coburn, Jr., "1,3-Dimethylbicyclo[1.1.0]butane," Tetrahedron Lett 1965, 6(15), 991-995; https://doi.org/10.1016/S0040-4039(01)99513-9
Gong-Huey Shiue, Ulf Misslitz, Xin-teng Ding, Maitland Jones, Jr., and Armin de Meijere, "Reaction of carbenes with bicyclo[2.1.0]pentane," Tetrahedron Lett 1985, 26(44), 5399-5402; https://doi.org/10.1016/S0040-4039(00)98218-2

n-Hexane is one of the 3 products formed. Carbenes are high-energy, reactive intermediates.  Due to their extreme reactivity they usually react indiscriminately. In the case of methylene (the carbene in your diagram), it will insert into each of the different $\ce{C-H}$ bonds that are present in n-pentane.  Because of its reactivity, the 3 products will be formed in a statistical ratio that is roughly equivalent to the number of different hydrogens present in the starting n-pentane, 6:4:2.

If you'd like to learn more about carbenes, for example

singlet and triplet carbenes
their addition to carbon-carbon double bonds

then take a look at these earlier answers - ref_1, ref_2


text:
Electrical conductance of NaCl solution
I'd like to ask whether EC of a salt solution of iodized table salt differs from that of uniodized and to what extent?

Too little to be noticed.

And what is the point of this question?

Is salt an electrolyte?
Common table salt (NaCl) is an electrolyte, and when this is dissolved in water to form salt water, it becomes sodium ions (Na+) and chloride ions (Cl-), each of which is a corpuscle that conducts electricity.
Let's go back to conductivity. Conductivity is an index of how easy it is for electricity to flow. In water, it is the ions that pass electricity from one to the next. This means that the more Na+ and Cl- contained in water the more electricity is carried, and the higher the conductivity.
Here's a chart:

(Don't mind the sugar water side)
Source linked here.

If something is not clear enough, feel free to ask.


text:
Why does aqeous fluorescein shine when exposed to UV light?
I know that (aq) fluorescein "absorbs" UV light and releases light in the visible spectrum , but how does it do that. It is a really interesting and useful organic chemical and I just wanted to know what goes on in it so that it lights up.

The process is called fluorescence. Some substances, such as calcium sulfide, $\ce{CaS}$, absorb light and release it slowly in a process called phosphorescence, glowing for some time after the source of light is removed.

Your answer is usefull!


text:
Why is fluoride good for your teeth?
The main component of tooth enamel is calcium hydroxyapatite. I know fluoride ions can replace hydroxide ions in the enamel matrix, slowing the rate of enamel demineralization. Why does hydrofluoric acid dissolved in water have no adverse health effects on other areas of the mouth? I suspect that its because the fluoride is in low enough concentrations to help your teeth and not harm the gums.

related https://chemistry.stackexchange.com/questions/2395/why-is-tin-ii-fluoride-more-effective-in-turning-apatite-into-fluorapatite

Why does hydrofluoric acid dissolved in water have no adverse health effects on other areas of the mouth?  || It definitely does. Fluoride in significant concentrations can cause sedimentation of calcium, which may lead to very bad things (tm) happening to heart. However 1) fluoride is NOT used as a fluoric acid and 2) it is used in very low concentrations.


text:
Is teflon a poison if ingested?
Nowadays, a lot of frying pans are made with a non-stick coating. That's usually teflon, or chemically speaking polytetrafluorethylene - $\ce{(C2F4)}_n$. Can you be poisoned if some of that teflon gets in your meal by scraping the pan or in any other way? And if so, what happens (chemically speaking) in your body after ingestion?

Teflon is very inert and small pieces in the body will not be absorbed, so no. The only way to suffer from it is to overheat the pan enough to thermolyse the teflon back to the monomer, which is harmful (though less so to people than birds).

@matt_black Thanks, your comment was useful. Would you mind forming it into na answer?

I hope I am not breaking any rules of this forum by directly quoting, but here you go:
"The cold polymer is harmless on ingestion, in contact
with the skin, or implanted in living tissues. When
heated above 250° C. it begins slowly to give rise to
small amounts of decomposition products, some of
which are toxic.
The accidental inhalation of fumes from the overheated
polymer by workmen produces an influenza-like
illness which has been described elsewhere. The
illness follows a latent interval of a few hours and
resolves within a day or two with no subsequent illeffects.
This may happen, for example, when the
polymer is heated to comparatively high temperatures
(350-400 C.) in an oven, an extruder, or some other
equipment used to fabricate it, but it is easily prevented
by the intelligent use of exhaust ventilation."
In: Harris, Kenwin. “Toxicity of Polytetrafluoroethylene.” British Medical Journal 1.5285 (1962): 1146.
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC1958415/pdf/brmedj02864-0082a.pdf

Well we can conclude not to overheat it or we can poison ourselves!


text:
In what way could benzoin give Tollen's test?
In Q29 of Joint Entrance Exam (JEE) 2016 India, the official answer key mentions that benzoin gives Tollen's test. However, I saw this post which says that it doesn't: Why do α-hydroxy ketones give Tollens' test?
Benzoin:

I'm very confused now. If benzoin does give the test, how? And what is the mechanism/intermediate involved?
I thought of all possible reactions benzoin may undergo in basic medium:

I did its aldol condensation but the product was too crowded and seemed quite unlikely. It didn't have any aldehyde group.
Another possibility seemed oxidation to benzil, but I'm not sure. I don't think $\ce{Ag+}$ is strong enough.
I was thinking about cleavage of central $\ce{C-C}$ single bond — as the hydroxyl $\ce{O}$ donated its lone pair to the carbon it is attached to, but am not sure on this too.

I couldn't find any related reaction of benzoin or any rearrangement on the internet.

Of your three "thoughts", you are correct that an aldol reaction is not an option. Not only is the product "crowded", but the reaction is reversible. [BTW, an aldol condensation occurs when water is eliminated from the initial aldol product. In the case of a benzoin aldol product, elimination of water is impossible.]
Your second premise is a good one. Benzoin (1) in the presence of alkaline Tollens' reagent, can be enolized (blue arrows) and the hydroxyl group deprotonated (red arrows). The enediolate 2 is a prime candidate for oxidation by Ag+. This species is the dianion of the enediol 6 tautomer of benzoin. The enediolate 2 is formed in the acyloin condensation of ethyl benzoate (7).
A one-electron oxidation of the enediolate 2 produces resonance stabilized radical anion 3. Protonation of 3 may occur at this point but a second one-electron oxidation produces the diradical 4, which is benzil (5). Under more vigorous alkaline conditions, Liebig's benzylic acid rearrangement of benzil occurs to afford benzylic acid (8), which itself is known to oxidize to benzophenone (9) under a variety of oxidative decarboxylation conditions. Neither of these reactions, i.e., the formation of 8 and 9, is expected to occur under the mild conditions of the Tollens' oxidation. Accordingly, your third idea is unlikely under the reaction conditions.

Can you please specify the source of your answer? In particular of the fact that benzil does form??

And.. Not all alpha hydroxy ketones will do this(i.e. get oxidised ).. Am  i right on this??

I cannot attest that ALL acyloins react with Tollens to give alpha-diketones. Here is an example: Tetrahedron Letters, (1969), 33, 2823.  Acyloins are readily oxidized by Bi2O3, Cr+6, V oxidants, etc.

here a double bond in conjugation with 2 phenyl rings is being formed, and also a benzyl radical ... i cannot open the Tetrahedron Letters link as i don't have access .. can you send me the paper on email (vishesh.jain331990@gmail.com)  ..? @user55119

have you found any other compounds or acyloins that give a positive test ..? @KavitaJuneja

https://youtu.be/xlaKvKfNidA
This clearly shows the thing


text:
Ester vs ether links: Why are archaeal cell membranes more resistant to high temperatures and acidity than bacterial and eukaryotic membranes?
The phospholipids composing the cell membranes of eukaryotes and bacteria contain ester links between the glicerol and the fatty acids. Archaeal membranes, instead, are composed of phospholipids containing ether links between a glicerol and an isoprene chain:

My microbiology teacher explained that these differences, mainly the ester vs ether links, were determinant to maintain the archeal membranes stable in environments with high temperatures and acidity, but she didn't go into much detail on why. I would like to know the specific chemical reasons why.
(Disclaimer: Considering I'm interested only in the organic chemistry involving this situation, and not in specific archeal phisiology or biology in general, I found it adequate to post this question here in the "Chemistry" cathegory.)

hmm... Ether should undergo SN2 reaction with an acid to produce alcohol, right? (confuzzled)

Put simply esters are easily cleaved by nucleophiles, whether enzymes or simpler species, and ethers are not.

This is an interesting question. It might be because temperature rise favours elimination (separation between the glicerol and fatty acid, which would increase enthropy), and a deprotonated fatty acid is a much better leaving group than an alkoxide (deprotonated alcohol), because the negative charge is less localized. Therefore the high temperatures archaea are subject to, sometimes over 100 C, could release the fatty acids and destroy the membrane. I`m not sure about this, though. Also, a carbonyl group tends to be more reactive, especially as an electrophile.

@JoaoBrinco I think you can try to put this into an answer. On a side note, would the fact that archaebacteria membranes have connected monolayers also be a factor to heat resistance?


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Zinc in a single displacement reaction with hydrochloric acid
I understand that on the metal activity series, zinc can displace hydrogen from acids, but not from cold water. But is there a structural reason for this. For example, why can zinc displace hydrogen from hydrochloric acid, but not from water?
Thanks

Note that $\ce{Zn + H2O -> Zn(OH)2 + H2}$ when hot water is used. However, the $\ce{Zn(OH)2}$ layer formed doesn't allow the reaction to proceed. See here: https://socratic.org/questions/do-magnesium-aluminium-and-zinc-can-react-with-water

The zinc replaces the hydrogen in that is reduces (gives electrons to) $\ce{H^+}$, forming $\ce{H2(g)}$ and $\ce{Zn^2+(aq)}$ according to this reaction: $$\ce{Zn(s) + 2H^+(aq) -> Zn^2+ + H2(g)}$$ This reaction relies on the prescence of $\ce{H^+}$.
The hydrogen in acids exists as $\ce{H^+}$. Acids are also ionic compounds that involve hydrogen as the donor of electrons. For example, hydrogen gas and chlorine gas react to make hydrogen chloride: $$\ce{H2 -> 2H^+ + 2e^-}$$ $$\ce{Cl2 + 2e^- -> 2Cl^-}$$ $$\ce{2H^+ + Cl^- -> 2HCl}$$  The reverse of this happens when the hydrogen chloride is dissolved into water.  this forms hydrochloric acid.The compound breaks apart into hydrogen and chlorine ions:$$\ce{H2O + HCl -> H+ + Cl-}$$ Hydrochloric acid completely dissociates, meaning that every $\ce{HCl}$ molecule yields one $\ce{H^+}$ ion.
Essentialy, acids increase the amount of $\ce{H^+}$ in solution. Normally, in pure water, concentrations of $\ce{H^+}$ are very low. The hydrogen in $\ce{H2O}$ is covalently bonded to the oxygen, so it is difficult for the hydrogen to react. In order to liberate a hydrogen ion from $\ce{H2O}$, this must occur: $$\ce{H2O -> H^+ + OH^-}$$
This reaction happens infrequently in water, so concentrations of $\ce{H^+}$ are very low. Therefore, zinc barely replaces hydrogen in water.


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Why do you not need sun protection whe the UV index is 3?
In Australia, they recommend sun protection when the UV index is above 3.
Why is this? Surely whenever it's above 0 there are still UV rays hitting your skin?
Would 10 minutes in the sun with a UV index of 6 be the same as 20 minutes in the sun with a UV index of 3? If so, surely 10 minutes in a UV 6 index would be bad, and therefore 20 minutes in UV index 3 would also be bad?

The scale is linear, but I would bet it doesn't linearly correlate to sun burn due to other factors that aren't included like your body's natural rate of healing. I would imagine below a certain threshold, the UV index is low enough where the rate of damage is less than the rate of repair or at least the difference between the two is very small.

Something like you do not need gloves to handle a body at 40 degree Celsius :)

All depends -- an albino would use it all the time, darker-skinned people might not need as much.


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Triphenylphosphine vs. trimethylphosphine in reduction reactions
Triphenylphosphine is used typically as a reducing agent for organic peroxides as it is a good nucleophile. So, if I wanted to computationally model a reaction like: $$\ce{PR3 + ROOH -> OPR3 + ROH}$$
Can something like a trimethylphosphine be used instead of triphenylphosphine, since it is an even better nucleophile, and the strength of the P=O is what's driving the reaction/reduction? Or am I missing some information?

Yes it can. It also has the advantage that the trimethyl phosphine oxide produced is water soluble and thus easier to remove from a reaction mixture than triphenylphosphine oxide which is notoriously difficult to get rid of. I have used it for a Staudinger reduction of an alkyl azide when triphenyl phosphine was unreactive.
However, trimethyl phosphine is more expensive, pyrophoric, has an unpleasant odour and is toxic requiring full personal protective equipment. Safety data here: https://www.alfa.com/en/content/msds/english/30143.pdf


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Does heating of potassium ferrocyanide with carbonate and carbon yield potassium cyanide?
Does the following reaction occur? If so, at what temperature? Would $\pu{640 °C}$ be enough?
$$\ce{2 K4[Fe(CN)6] + 2 K2CO3 + C -> 12 KCN + 2 Fe + 3 CO2}$$

Disclaimer: This answer is just for education purpose only. DON'T TRY TO MAKE POTASSIUM CYANIDE ON YOUR OWN if you don't have profound laboratory experience. Potassium cyanide is extremely toxic and would even cost your life.

Yes, your reaction of producing potassium cyanide has been discussed in a forum:

...Dry fusion of pure $\ce{K4Fe(CN)6}$ is a bit tricky, if you will
  try to do this on air contact you will end with with cyanate and
  cyanogen beeing evolved. To prevent (this) oxidation process,
  $\ce{K2CO3}$ and carbon works the best - you don't need protecting
  atmosphere you just must cover reaction vessel with lid to reduce free
  air flow, reaction products CO and CO2 will form protecting buffer of
  gas under the cover and prevent oxidation of product to cyanate,
  remember you don't need completely airtight vessel (reaction gases
  must evolve freely) just cover crucible with ceramic lid and heat it
  up in electric furnace. I made $\ce{K4Fe(CN)6 + K2CO3 + C}$ process
  for many times now and produced a lots of cyanide using this method,
  temperature program i've used ~30 min to 600C, then 1 hour at this
  temp, rise to 650 in 30 minutes, and one more hour at 650. If your
  crucible is high and you sit it long enough cyanide separates from Fe
  and carbon particles and solidifies as shiny snow white layer at the
  top layer, however if this has't happened and resulting mass is
  black and feromagnetic this is fine too, you can separate cyanide by
  dissolving and filtering.

Other methods are also possible like simply decomposing potassium hexcyanoferrate or with any combination of potassium hexacyanoferrate/ potassium carbonate/carbon reaction.
$$\ce{K4Fe(CN)6 + K2CO3 -> 5KCN + KCNO + CO2 ^ + Fe}$$
$$\ce{K2CO3 + C + NH3 -> 3KCN + H2O}$$

A great idea at the end of reaction(1 hour for one mole of KCN should be long enough) is to quickly put the crucible in an 8 ounce metal can modified with gas inlet at bottom and exhaust port near top.The can must have a lid that seals tightly. Now flush the can with nitrogen gas for about 1 minute to flush out oxygen!This will stop oxygen reacting with KCN to make KOCN cyanate.Then reduce nitrogen flow to several bubbles per minute.The crucible is processed when cool.

I often think that such disclaimers at publicly posted and easily abuseable info do not really release authors from their moral responsibility. One should remember Google does not care about search reasons.

Definitely a +1. One minor point is the reaction may not occur with the stoichiometry given. Heating carbon with CO2 could promote a side reaction forming CO. You would add more carbon to the mixture to compensate.

Best is the following reaction since there is no fire because there is no fuel to burn!
$$\ce{K4[Fe(CN)6] + K2CO3 + C -> 5KCN + KOCN + FeC + O2}\tag{R1}$$
The black solid is $\ce{FeC},$ which is slightly magnetic. Since $\ce{K4[Fe(CN)6]}$ melts at $\pu{300 °C}$ and decomposes at $\pu{400 °C},$ best is to heat at $\ce{390 °C}$ for 1 hour, then ramp up the temperature to $\pu{700 °C}$ to force the transition of cyanate to cyanide:
$$\ce{KOCN + C -> CO + KCN}\tag{R2}$$
Heat at $\pu{700 °C}$ about 20 minutes and then shut off the oven. With thermal gloves, take stainless steel crucibles out of oven to cool quickly. When about $\pu{120 °C},$ add $\pu{100 ml}$ water and let sit for 20 minutes.
Then use screwdriver to crack the solids off the crucible bottom and walls. Let the liquid settle and decant most of it. Filter the black iron carbide. pH of the liquid is about 12 and the test with iron(II) sulfate and iron(III) chloride forms beautiful Prussian blue!


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How does Hofmann elimination work in this reaction?
In my Chemistry class, we were given this question to solve, and I'm not sure I fully understood the solution.

I understand that since the substrate is bulky, Saytzeff elimination can't happen, so it has to be Hofmann. What I don't understand is how it happens here. Where does the base attack the substrate? Essentially, I'd like to know the mechanism of the reaction.
Also, our teacher made us write that had all three substituents on the nitrogen atom been methyls, Saytzeff elimination would have been followed as there wouldn't have been any other choice. However, even if all three substituents were methyls, wouldn't the substrate still be bulky? Why would it follow Saytzeff elimination?

Wow, this is an amazing question. The expected reactivity is strongly dependent on the exact structure.
For starters, trimethylammonium is about the size of a tert-butyl group. So, expect an A value around 4.9, which is very large. So, put that group equatorial on the ring.

Once you do this, you see that there are only 2 hydrogen atoms that are anti-periplanar for an E2 elimination. One is on the axial methyl group (red). The other is on the N-ethyl group (blue). Only these two can easily be deprotonated for elimination. Of these two hydrogen atoms, the one on the axial methyl is very sterically crowded. Specifically, any base has to get around the ring structure and the encroaching flag pole hydrogen atoms. On the other hand, the hydrogen atom on the ethyl group sees no such steric encumberance.
Of the three elimination products, I expect to see:

Almost exclusively ethylene
Maybe a small amount of the methylidenecyclohexane
No methylcyclohexene

EDIT:
@Martin-マーチン was kind enough to compute the structure below. Notice the structural features we've already discussed: the ring conformation and the position of the anti-periplanar hydrogen atoms.


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How can I safely store highly outgassing self-pressurizing chemicals?
I have a standard copper etching solution of 1 part $\pu{10M}$ $\ce{HCL}$ and 2 parts 3% $\ce{H2O2}$ (as described in numerous places online).
I previously had this stored in a tightly-sealed glass bottle, as described in the instructions I followed and general acid storage instructions.
However, my solution seems to be outgassing hydrogen or chlorine gas quite significantly, and when I opened the sealed container a few days later it violently liberated much of its contents into my face. (I was wearing a partial face shield and immediately went under the shower, so no ill effects were observed).
I know that for some chemicals that self-pressurize, such as cryogenic nitrogen, the storage method is essentially constant venting - is that what I'll have to resort to?

It's hydrogen peroxide related. It slowly decomposes, especially if iron cations or some other catalysts are present. Cheap hydrochloric acid contains some.  Any hydrogen peroxide solution should be stored in a fridge.  Any highly outgassing solution should not be stored at all.

@permeakra This was over the course of only two days, and it built up a significant amount of pressure...would peroxide decomposition be that fast? My working theory was that it was either some remnants of copper left in solution reacting to form chlorine gas, or hydrogen outgassing as some others using this solution have described...

It can be really fast, up to and including explosions in borderline cases. 2days is a lot of time for something like this. If you want a safe solution for etching copper, consider iron chloride based solutions. They look dirty, but they are (reasonably) safe, efficient and (to some extent) reusable. Do care to find absolutely airtight container for solid iron chloride, though.

@permeakra Yeah, I've tried ferric chloride, I've just found it pretty slow.

In some cases, electrolytic etching also works.

Bottles for 30% or stronger $\ce{H2O2}$ usually have vented caps to prevent sudden out-gassing, splashing and boil-over that you experienced, as well as to prevent bursting the container. The vent may be as simple as thinned spot, which may be scored or slitted, or it may have a polythelene, PTFE or other sponge or "paper" separator that helps retain liquids while allowing gas to escape.

In either case, I would not suggest storing a mix of $\ce{H2O2}$ and $\ce{HCl}$ for more than a few hours for a few reasons.

As gases evolve, the solution becomes weaker and less effective an etchant.
Stored on a shelf, the gases will corrode the shelving surprisingly quickly. You would not want shelving with caustic or toxic chemicals to collapse.
If gases given off are other than $\ce{O2}$, you probably don't want to breathe them for long.

Ferric chloride is a slow etchant at room temperature, but gentle warming speeds etching greatly (PDF). At 54 °C, $\ce{FeCl3}$ removes ~35 um of copper per minute, so standard PC boards ("0.5 oz" - "1 oz") should be etched in ~1 - 2 minutes.
The biggest nuisance I find using $\ce{FeCl3}$ is that it tends to stain containers, tables and hands, but Iron Out and similar bisulfite or hydrosulfite stain removers can be used on cleanup.

stain removers can be used on cleanup || Citric acid doesn't work?

Also, links are broken

@permeakra, thanks, link fixed (hard to copy PDF URL from Google). As far as citric acid, it might work on fresh stains on an impermeable surface, but once the reduced iron gets into things, it's hard to remove.


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Why Lysine is abbreviated as 'K'?
It does not has a 'K' in its name. It does not have a potassium in its chemical composition.
Would be nice to know why Glutamine is Q too, some of these abbreviation seems simply random.

From here
Back in 1960's Dr.Dayhoff was shortening the names of amino acids so that ancient computers could save computing time.

By the time Dr. Dayhoff got to lysine, there were not too many letters left, so she used the letter K, explaining that K is at least near L in the alphabet.

As for Glutamine.

For the remaining 5 amino acids, Dr. Dayhoff was reaching somewhat to find an easy-to-remember connection between the single letter and the amino acid.  She assigned aspartic acid, asparagine, glutamic acid and glutamine the letters D, N, E and Q, respectively, noting that D and N are nearer the beginning of the alphabet than E and Q, and that Asp is smaller than Glu, while Asn is smaller than Gln.

Q=glutamine, N=Asparagine, and E=glutamic acid had all been used by others earlier, http://cccc.uochb.cas.cz/26/2/0531/ (article submitted 1959).  K=lysine and D=aspartic acid may have been new though.

Šorm et al. were the first to publish a specific single letter abbreviation system, in On proteins. LXIII. Lower structures in the chains of proteins Collect. Czech. Chem. Commun. 1961, 26, 531-578.
In this publication, an earlier, previously unpublished, system of Hans Neurath was also published.

If practical symbols are to be proposed, they must conform to the following requirements: (l)
  23 different one-letter symbols of amino acids occurring in protein structures have to be chosen, (2)
  only simple symbols currently accessible in printing-works with the possibility to be written by
  common typewriters have to be used, (3) the symbols must offer the possibility of expressing
  whether the bond of the sequence is determined or, alternatively, so far undetermined, within the
  space of one single letter, and (4) the usage of symbols must be of international applicability.
The usage of figures is excluded by requirement (1), and the usage of Greek or other uncommon
  letters and symbols by requirement (2). The most appropriate manner seems to be the use of the common
  letters of the Latin alphabet, which conforms to requirement (4) and permits to distinguish
  according to (3) between determined and undetermined sequences by choosing either capital or small
  letters. To express a determined sequence of the peptide chain we propose to use capital letters
  according to Table VIII, and, further, to put small letters at places where brackets and commas have
  hitherto been used. Our symbols are derived from the English names of amino acids. Since, however,
  identical letters commonly recur at the beginning of the names they could not be consistently
  utilized in our symbols. Nor could a thorough system be compiled involving two-letter symbols
  of the first two letters of the name (e. g. glutamic acid, glutamine, glycine); moreover this would be
  contradictory to the intended system of one-letter symbols. Therefore, we have compromised and
  the first letters are proposed as symbols only in some instances (asparatic acid, glutamic acid, cystine
  (half-), histidine, methionine, proline, serine, tryptophan, valine), in many other cases characteristic
  symbols derived either from the spelling (ala- l, arg- r, lys- i, leu- u, ileu- w, phe- f) or from the
  graphical representation (glu- g, glu(NH$_2$)- q) were chosen.

Šorm was the first to use "Q" for glutamine.  (Neurath had used "E' for both glutamine and glutamic acid.)
Neurath used "O" for lysine and Šorm used "I".
No one unified system was used until selection by IUPAC.


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Could creating ashpalt roads actually be net CO2 reducing?
I remember reading a popsci book, which at some point mentioned, that the problem with reducing $\ce{CO_2}$ emissions was to actually trap carbon in long-term permanent structures.
It then half-jokingly concluded that one of the counterintuitive green activities was to make books and... build more roads; presumably because the asphalt roads are long-lived and asphalt contains a lot of carbon.
Would the idea have some practical consideration from standpoint of actual carbon content? Never mind for the moment other aspects, such as that creating asphalt gives off greenhouse gas emissions itself, etc, but how much carbon would approximately there be in, let's say, a ton of typical asphalt? And could there be any comparison with other human or natural activities that bind carbon in large volumes into long-term objects?

Pretty much. But then again, what is the alternative to building roads? We could just as well sit there and do nothing. The carbon in these compounds is already trapped anyway.

My desk isn't a mess covered with books and paper.  It's actually an advanced carbon storage unit.


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Domestic cleaning products and types of solvents
I have a floor cleaner which washes and vacuums the floortiles.   The instructions advise not to use solvent cleaners with it.  My understanding is that even water is a solvent.  The manufacturers/agents will just try to sell me their product.  I want to use cheap domestic solutions (eg. Sugar Soap, Pinoclean, Bar Keepers Friend).    Can anyone advise please? At present I use just hot water, sodium bicarbonate ($\ce{NaHCO3}$) and vinegar of a small amount of washing up liquid.   Thanks.

The manufacturer likely means that solvents shouldn't be use.  Solvents likely stands for non water based detergents and stuff such a varnish removers...

There are two ways of cleaners. Ones that will "grab on to" the dirt and bacteria to wash it away e.g. soap and shampoo. Then there are ones that will dissolve the dirt and kill the bacteria such as bleaches. They won't do anything dangerous together, they'll just not work.


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Which is a better measure of polarising power of a cation?
Recently, I have come across literature stating that polarising power of a cation is proportional to the inverse of the surface area $(4\pi r^{2})$ while some associate it with the charge density of the cation, and thus state that it is proportional to the inverse of the volume $(\frac 43 \pi r^{3})$. I have been wondering which is a better measure of the polarising power. My knowledge of physics is limited so I would like to ask if anyone could kindly enlighten me on the matter.

I feel if you cite the literature (and relevant pages) as well, it would also  be of help to future readers of those particular textbooks :-)

Define "better".

@IvanNeretin "Better" would mean being able to give a more accurate approximation of the ion's polarising power.

Define "polarising power".

"Polarising power" would refer to the ability of a cation to distort the electron cloud of a neighbouring anion.


text:
Hydrogen bond strength of trichloromethane and chloromethane
Chloromethane has a larger dipole moment than trichloromethane. Does that mean that the hydrogen bond (C-H...O type) between chloromethane and acetone is stronger than that of trichloromethane and acetone?

It has nothing to do with dipole moment of molecule, but of the C-H bond. In chloroform it's more polarised, so bond has about 2 kJ per mole. With chloromethane h-bond would be much weaker.

@Mithoron Why chloroform is more polarised than chloromethane? According to this, chloromethane has a larger dipole moment, does it mean it is more polarised?


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How to visualise the total electron density from a Gaussian 16 calculation?
I am trying to generate a 3D visualisation of the total electron density of a molecule. Using Gaussian 16, I have optimised this structure, with DFT, and formatted the resulting .chk file into a .fchk using the formchk utility. I have used the cubegen utility:
cubegen 0 fdensity=scf molecule.fchk molecule.cube -5 h <0.20_3d_cube.xyz

to generate a .cube file with 0.20 angstrom spacing about a cube of the molecule.
However, the resulting .cube file fails to visualise any total electron density in visualisation software. Gaussview 6 fails to create a surface, prompting an error:

Surface build failed! Processing mesh, No data to plot. IsoValue may be out of range

VMD allows the visualisation of an isosurface, but visually nothing appears.

Can you try with npts -1 instead of -5? According to the gaussian manual it has to be set to -1 in case you are using the cubefile2 option

The syntax appears to be in order. I have never used this approach before, i.e. specifying the box manually. Have you tried something simpler, like cubegen 0 fdensity=scf molecule.fchk molecule.cube -4 h? Did that work? You should be able to visualise the scf density from within GV only reading in the output, checkpoint, or formatted checkpoint file.

You can also visualize the electron density directly in Avogadro using just the fchk file. See this video for a short (3 min) tutorial. youtube tutorial video

When GaussView gives you the error

Surface build failed! Processing mesh, No data to plot. IsoValue may be out of range

it means that the data in your .cube file doesn't have values of density at that isovalue. So when it tries to plot a surface, it cannot plot anything.
The IsoValue (default is 0.00040, I believe) means that you construct a surface at every place in the density where the density equals the IsoValue ('iso' = 'same'). So for the default, it opens up your .cube file and looks for the cube coordinates that have density at (or near) 0.00040, and plots a surface that connects these points (this is your 'mesh').
Most likely, your density values are too small large to plot an isosurface at this value. Try making the default density isovalue in GaussView to be something smaller larger and go from there. (Practically, I'd keep shifting an order of magnitude until a surface builds, e.g. try a value of 0.004 next).
Even better, you should look into the .cube file itself to see what your volumetric data looks like. The first few lines define the grid spacing and dimensions of the cube, but then you will see a bunch of floating point values. Here's an example .cube I took from this site (also a good reference):
CPMD CUBE FILE.
 OUTER LOOP: X, MIDDLE LOOP: Y, INNER LOOP: Z
    3    0.000000    0.000000    0.000000
   40    0.283459    0.000000    0.000000
   40    0.000000    0.283459    0.000000
   40    0.000000    0.000000    0.283459
    8    0.000000    5.570575    5.669178    5.593517
    1    0.000000    5.562867    5.669178    7.428055
    1    0.000000    7.340606    5.669178    5.111259
 -0.25568E-04  0.59213E-05  0.81068E-05  0.10868E-04  0.11313E-04  0.35999E-05
      :             :             :           :            :            :
      :             :             :           :            :            :
      :             :             :           :            :            :
        In this case there will be 40 x 40 x 40 floating point values
      :             :             :           :            :            :
      :             :             :           :            :            :
      :             :             :           :            :            :

Look at the value of your floating point numbers at the end. As a starting point, pick a number for your isovalue that is about an order of magnitude less than your largest values, and adjust from there.
If your values are all at or near zero (so, practically speaking, on the order of 1E-8 or lower) then you need to revisit how you are generating the .cube file, since your cube is not capturing any density---then it's a cubegen issue not an isosurface issue.


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Which compound is responsible for the borax bead test?
When borax is heated, the reaction that takes place is :
$$\ce{Na_2B_4O_7.10H_2O \ce{->} 2NaBO_2 +  B_2O_3 + 10H_2O}$$
And the compounds that are responsible for the colors are $\ce{M_x(BO_2)_y}$ salts, like $\ce{Co(BO_2)_2}$. My question is, does this $\ce{BO_2}$ come from $\ce{NaBO_2}$ or from $\ce{B_2O_3}$ or do both of them react?

A little chemical detective work gives a plausible answer.
Say the metal is cobalt and it's oxidation state is +2.  Then, with the cobalt calcined to the oxide, you have two possible acid-base reactions:
$\ce{CoO + 2 NaBO2 -> Co(BO2)2 + Na2O}$
$\ce{CoO + B2O3 -> Co(BO2)2}$
Both reactions look good, but upon further review the first one gives something we recognize as a powerful base, more likely to be a reactant than a product.  Moreover,  a quick check here reveals that boric oxide has a low enough melting point to liquefy in the flame, favoring the second reaction kinetically.
So the second reaction above is really the only plausible one.

Does the NaBO2 have no use then? Is It just present due to the decomposition of borax into NaBO2 and B2O3 and because NaBO2 cannot be separated from B2O3 in the glass bead?

In the case of $\ce Cu(BO2)2 $ when this compound is heated in a reducing flame, the $\ce NaBO2$ present in the bead reacts with it forming $\ce CuBO2$ and $\ce Na2B4O7$ OR it may form metallic Copper and $\ce Na2B4O7$


text:
What noble metal is most resistant to oxidation by diatomic oxygen in air at room and elevated temperatures?
To clarify, by noble metals in this case I mean all platinum metals + gold, no copper, silver or rhenium. These noble metals are described as having low reactivity, but which one is least reactive or most "noble" of them all?
I am chemistry noob and I have feeling this might greatly depend on what type of element/molecule is trying to react with given noble metal so to make this simple I will ask only about the element most resistant to *oxidation by diatomic oxygen in air at room and elevated temperatures.
If we had all these noble metals side by side in identical shape, let's say cube with highly polished sides, and we heated them in air atmosphere to 100, 500, and 1000 degrees Celsius, which one would oxidise least?
Background and current understanding:
Here is my current knowledge and thoughts on why some noble metals might be more noble than other.
This is just window into my head so you can see what I think so you are able to correct what is wrong, approve what is right and expand on what is missing.
My current understanding of what makes noble metal noble, is that it's somewhere in middle of periodic table, their electron shells are half full. Elements on the left side are strongly reactive because their shells are only sparsely filled, so they give away electrons easily, they have low ionisation  energy.
Elements on the right side, excluding noble gasses, are strongly reactive because they have almost full electron shell and they have tendency to tear out and "steal" electrons from less electronegative elements. That is why noble metals are in middle, they don't want to give away or get inside electrons very much.
That would be my first trait of noble metal, the more in middle it is, the more noble it is. Since periodic table is 18 numbers wide, 18 is even number so there is no integer exactly in middle, the middle is between 9 and 10.
That means elements in group 9 and 10 should be more noble than those in 8 and 11 group because they are closer to middle, so rhodium, palladium, iridium, and platinum should be more noble than gold, ruthenium, and osmium.
Another thing that I believe is that the heavier elements don't hold their electrons as strongly as lighter elements because the inner electron shells shield the valence electron shell from attractive force of nucleus. If this is true, then noble metals in period 5 should be more noble than those in period 6, that is ruthenium, rhodium, and palladium more noble than osmium, iridium, platinum, and gold.
Third factor in my opinion is first ionisation energy, the higher it is, the more it should be resistant to reacting with oxygen. From highest to lowest first ionisation energy: Gold = 9.2255 eV, iridium = 8.967 eV, platinum = 8.9587 eV, osmium = 8.4382 eV, palladium = 8.3369 eV, rhodium = 7.4589 eV, ruthenium = 7.3605 eV. If this is true, gold should be most noble, and ruthenium least noble.
I read one study about oxidation resistance at high temperatures, the study said rhodium was most resistant, even more than iridium or platinum. I also heard Platinum doesn't oxidise in air at any temperature bellow melting point, that seems contradictory to the paper I read.Furthermore, spark plugs are mode from iridium and platinum but not rhodium.

The spark plug application may depend on availability/cost as well as chemistry.

@OscarLanzi - quite true,  and add resistance to erosion by the spark as well. Way more variables than just 'reactivity' going on for any engineering application.

Let's see.  Pt (78) has 1st ionization energy = 8.96 eV, Pd (46) = 8.34 eV.  Uh ... How do heavier atoms hold their electrons most loosely within their groups?  Relativistic effects can blow up the expected periodic trends.

The answer to your general question (as usual) is "depends". And the setup in your last sentence is very specific, and arbitrary on the border to irrelevant. What is it you really want to know? What are you trying to do?

I want to know,what noble metal is most oxidation resistant in air at 100,500 and 1000 degrees celsius.I gave three temperatures becose I think some noble metals might loose their "nobility" faster than others as temperature rises.If that is not the case,then it can simply be reduced to,what noble metal is most oxidation resistant in air.

Oscar Lanzi I am aware of that contradiction,in fact I put it there on purpose  in hope someome will explaim to me why dont the heavier metals have lower ionisation energy.These noble metals seem very weird to me,for example,Platinum have only one electron in highest shell,and the second highest have 17,that means one is missing,but it doesnt act like alkali metal like Lithium that too have just one electron in valence shell,thats really strange.

There should be always a space after punctuation mark!

@wavscientist I've made some adjustments to your post to make it clearer that you've clarified your question to focus on oxidation by diatomic oxygen in the air, and voted to reopen.

Ruthenium, Rhodium, Palladium, Osmium, Iridium, Platinum, Gold

... I will narrow this question down to oxidation by diatomic oxygen in air at room and elevated temperatures.

According to this criterion, gold is the most oxidation resistant of the above metals.
First of all, gold oxide cannot be obtained by reaction of gold and oxygen ($\ce{O2}$) at standard pressure. Gold oxide ($\ce{Au2O3}$) decomposes to gold and oxygen above $\pu{160°C}$.

Ruthenium reacts with oxygen (air) above $\pu{700°C}$ to give $\ce{RuO4}$. Ruthenium powder burns in air.
Rhodium reacts with oxygen (air) above $\pu{700°C}$ to give $\ce{Rh2O3}$.
Palladium powder burns in air.
Osmium powder reacts with oxygen (air) at room temperature to give $\ce{OsO4}$.
Iridium reacts with oxygen at high temperatures to give $\ce{IrO2}$ or $\ce{IrO3}$. Iridium powder burns in air.
Platinum powder burns in air.

So up to 160 celsius,Gold is most oxidation resistant?

Correct. It does not directly react with oxygen.

Does the Gold start to oxidise at 160 C or is it merely the temperature point at which the Gold oxide starts to separate into Gold and oxygen? What is even this separation thing,it reminds me of self-healing materials,except instead of healing a crack,it "heals" oxidation.Also,is the fact that Gold is most oxidation resistant  result of Gold  having the highest first ionisation energy of the noble metals?

Gold does not directly react with oxygen. Neither above nor below 160°C.

Whereas Osmium is the biggest pain in the butt. If you leave the pure stuf alone in a room it will happily start making $OsO_4$ and turning the walls green ...

Using your comment clarifying that the conditions include high temps of 100, 500, and 1000 degrees celsius, this is my answer:
At high temperatures, the noble metal that is most resistant to oxidation by diatomic oxygen in air is ruthenium. It can resist oxidation at temperatures up to 1000°C in air.
Ruthenium is a noble metal that has a very high resistance to oxidation, due to its stability and chemical properties. It has a high melting point of around 3900°C and low coefficient of thermal expansion which makes it to withstand high temperatures without showing degradation. Additionally, ruthenium has a relatively high level of resistance to chemical reactions and corrosion. These properties make ruthenium particularly useful in high-temperature applications, such as aerospace, electronic and chemical industries.
Other noble metals such as platinum, palladium, rhodium can withstand lower temperatures compared to ruthenium, for instance, platinum and palladium resist oxidation up to about 900°C, while rhodium can resist up to 800°C.
It's important to note that the oxidation resistance of a metal at high temperature is also affected by the specific conditions of the environment, such as the presence of other gases and the oxygen concentration.

The Ru melting point seems to be wrong,


text:
What are the best cleaning agents for carbon residues on glassware?
Introduction
Recenly I wanted to isolate some ascorbic acid (Vitamin C or chemically - $\ce{C6H8O6}$). I've put the filtered solution (I had to filter it because of the binder) in a flask and heated it. It didn't go as planned because then I didn't know that ascorbic acid decomposes at high temperatures, but I still decided to keep the product of the reaction. I left it on a hotplate at a high temperature and it all went back to carbon because it was there for too long (It was dehydrated).

The cleaning agent that I know of - $\ce{NaOH}$
I used a saturated solution of $\ce{NaOH}$ to clean it but there were still some flakes stuck on the wall that I couldn't reach.
It worked well for me - cleaned ~90% of the residue.

The question
Are there any other cleaning agent that'll work better than aqeous Sodium Hydroxide?

Are there any other cleaning agent that'll work better than aqeous sodium hydroxide?

Yes, but I would not recommend them to anybody running experiments at home!
Back in the days in the lab we used arm-deep tanks filled with a saturated solution of potassium hydroxide in isopropanol to soak dirty flasks. You need long (!) chemical-resistant gloves and once in a while, the whole mess has to be discarded and replaced. Down the sink is not an option ;-)
And there is piranha solution, a mixture of 30% hydrogen peroxide and concentrated sulfuric acid. While is is effective, it is infamous for going off with a bang, see this report.
To sum it up: Stay with the sodium hydroxide solution. Don't risk your health and don't screw up your own environment to rescue a spoiled flask.


text:
Placing of lone pairs in axial/equatorial position
If the lone pairs of elections are put on axial or equatorial position such as to minimize the intera-electronic repulsion, and for the same reason in $\ce{SF4}$, the election pair is kept on the equatorial position(at 90 degrees to only two domains), why then in $\ce{BrF5}$ the lone pair is kept on the axial position, when it is at 90 degrees with 4 domains?

If you see the structure of $BrF_5$ which is square pyramidal, and with the lone pair it is octahedral, you will find that there is no other option. All the bonds are equiangular from each other. So, it is equivalent to put the lone pair in axial or equatorial position.

There is no axial or equatorial position in an octahedral structure. All places are equivalent and you can put the lone pair on position.

Not necessarily equivalent - not if there's Jahn-Teller effect.


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Producing sulphur dioxide for using in a reaction
I am doing an investigatory project on studying the coloured reactions of d-block and f-block elements. I have included two reactions that require sulphur dioxide as a part of my study.

$$\ce{5SO2 + 2KMnSO4(purple) + 2H2O -> 2H2SO4 + 2MnSO4(pale pink) + K2SO4}$$
$$\ce{SO2 + K2Cr2O7(orange) + 3H2SO4 -> K2SO4 + Cr2(SO4)3(green) + 3H2O}$$

Now, my problem is I need an effective method to produce sulphur dioxide to be able to use for these in a school laboratory environment. I thought of using ferrous sulphate which on heating gives $\ce{SO2}$,
$$\ce{2FeSO4 ->[{$\Delta$}] Fe2O3 + SO2 + SO3}$$
But this reaction also produces $\ce{SO3}$. Will that screw with any of the above reactions? Also, are all the above reactions feasible in a school laboratory? Do suggest if there are better methods and how I can make it to react with the other reactants.

Note: Sulfur dioxide causes dyspnea. Prolonged inhalation of it may causes respiratory problems. So, reaction should be done under laboratory supervision.

The most common laboratory method of producing sulfur dioxide is the reaction of copper and sulfuric acid, both available in laboratories.
$$\ce{Cu(s) + 2H2SO4(l) → CuSO4(aq) + 2H2O(l) + SO2(g)}$$
Process: To a round bottom flask, 100 g of copper (small pieces, turnings, etc.) are placed. From a dropping funnel 100 ml of concentrated sulfuric acid is slowly added dropwise. The reaction flask is gently heated and obtained sulfur dioxide is dried by passing through a wash bottle with concentrated sulfuric acid. (Process taken from here. Read more to know how the reaction can be controlled and the precaution needed)
Alternatively, sodium sulfite can also be used.
$$\ce{Na2SO3 + H2SO4 -> SO2 + Na2SO4 + H2O}$$
Process: Sodium sulfite is placed in the bottom of the 300 ml Erlenmeyer flask of the apparatus. The dropping-funnel is filled with dilute sulfuric acid, made by pouring one volume of concentrated sulfuric acid into an equal volume of water and cooling the mixture. The glass elbow is connected with a gas washing-bottle containing concentrated sulfuric acid, and the dry gas is collected by displacement in two or three empty cylinders. On allowing the dilute sulfuric acid to drop slowly upon the sodium acid sulphite, a very regular evolution of sulfur dioxide is obtained.

For full setup, check here.

Thank you, are normal lab conditions enough for the reactions 1,2 that I have mentioned in my question? And can you mention the downsides of using ferrous sulphate?

Probably yes. They are normal high school reactions and has been used for demonstration purposes. But take care of sulfur dioxide. It can be harmful if inhaled.

Okay then, I will either stay away and ask the teacher to do it or will undertake the necessary precautions. Thank you.

@MollyCooL ferrous sulfate decomposition also produce sulfur trioxide which is also a gas and can be difficult to separate from sulfur dioxide. Moreover, sulfur trioxide is way more nastier than sulfur dioxide.

Yes okay, Just wanted to know if SO3 screws the reactions in someway other than being harmful.

@MollyCooL SO3 will not screw the reaction in any way. It is a part of products(gas) evolved.  To separate SO3 and SO2 can be a hefty task.

Wet chemistry can be so old fashioned! The gist of the reaction is the reduction of the Mn or Cr by tetravalent sulfur. How about bypassing the gas formation and introducing the tetravalent sulfur as bisulfite ion (from NaHSO3 solution)?
5SO2+2KMnSO4(purple)+2H2O --> 2H2SO4+2MnSO4(pale pink)+K2SO4
This scheme (plus a bit of added H2SO4) then becomes
5NaHSO3 + 2KMnO4 + H2SO4 --> 2Na2SO4 + NaHSO4 + 2MnSO4 +2H2O
The same color change will occur.
If the second reaction is properly balanced, it becomes
3SO2+K2Cr2O7(orange)+H2SO4 --> K2SO4+Cr2(SO4)3(green)+H2O
By replacing the SO2 with NaHSO3, this scheme then becomes
3NaHSO3 + K2Cr2O7 + 3H2SO4 --> K2SO4 + Na2SO4 + NaHSO4 + Cr2(SO4)3 + H2O.
The overall reaction is the same, but simpler. You have the advantage of using more available and much safer chemicals. And the NaHSO3 solution will still give you the fragrance of SO2, but without total dyspnea.

Yes, I went ahead using sodium bisulphite but I got a mere tinge of green rather than a complete green solution!

The pH in both cases is different. Cr III oxide is green; its hydrate is violet or blue-gray green, according to CRC Handbook. Cr III sulfate is violet, red or blue violet. These color differences might be small shifts in an absorption band, depending on pH or impurities or who knows what! You could try measuring pH and recording color of solution. Interesting!


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What makes bond break apart from heat?

What makes bonds break apart from heat?

I am wondering if anyone could explain the reasons why new bonds are formed in the absence of energy. I have found that they form in order to lower the overall energy, presumably because electrons move closer to the nuclei via bonds.
But in a reaction like $\ce{2 K (s) + 2 H2O (l) -> 2 K+ (aq) + 2 OH- (aq) + H2 (g)}$, I wonder what makes the bonds break? I just don't understand why $\ce{2 K}$ would suddenly split to form new bonds unless I'm missing some sort of energy being put into the system. Maybe potassium is able to attack $\ce{H2O}$ since it is a polar molecule?
But, I truly can't find a reaction mechanism that explains these phenomena.

You're asking completely different thing in head then in body of your post.

You have a two part question so I'll answer both of them:

Question - What makes bonds break apart from heat?

"If you start to heat the system, two things can happen:

The atoms gain enough energy to move into higher, excited states in the potential well. Eventually the additional energy may be high enough to make the total energy positive. The nuclei can escape the potential well - the bond is broken.
The electrons that participated in forming the bond, i.e. those on orbitals overlapping both nuclei, get excited into different, higher energy orbitals. This makes the potential well shallower, and the nuclei can escape it easier - the bond becomes weaker and easier to break. Eventually the potential well can even become so shallow, that no bound state can exist." - Source

Question - Why do alkali metals ($\ce{Li, Na, K ...}$) break bonds in water ($\ce{H_2O}$)?

"All alkali metals melt as a part of the reaction with water. Water molecules ionise the bare metallic surface of the liquid metal, leaving a positively charged metal surface and negatively charged water ions. The attraction between the charged metal and water ions will rapidly increase the surface area, causing an exponential increase of ionisation. When the repulsive forces within the liquid metal surface exceeds the forces of the surface tension, it vigorously explodes."- Source

I hope that my answers were useful to you.


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Why does s-p orbital separation increase with increasing atomic number?
More specifically, why does s-p energy separation increase:

Across the period
Down a group

I do know that penetration of electrons increase with lower angular momentum and lower principal quantum number. An increase in penetration means that it has a higher shielding ability.
If we have two elements in period 2 and period 3 respectively, than the $\ce{3s}$ orbital would be less effective at shielding than the $\ce{2s}$ orbital, and thus the $\ce{3p}$ orbital would be higher in energy than the $\ce{2p }$ orbital would be relative to their s-orbitals? Is this the right justification?


text:
Why does the half time decrease in a 0th order reaction?
A --> B
rate = k
Half time = [A]0/2k
From the half time equation, since there are no variables, i don't see how the half time of a 0th order reaction can change? It's dependent on the initial concentration of A, which is constant, so how can the half time be decreasing as shown in the graph below.

The only explanation I can imagine is that A0 is not the initial concentration at the start of the reaction, but the concentration at each the start of each half time.

A half life time is a property of a first order reaction. You are misapplying the concept.

For the first half life(say,$T^1_{\frac{1}{2}}$ )and the second half life(say, $T^2_{\frac{1}{2}}$) the initial concentrations of the reactant to be used for calculating are different. For, the $0$th order reaction $\ce{A -> B}$, the integrated rate equation will be, $$\ce{[A]_t = [A]_0 - kt}$$ where $k$ is the rate constant.So, at the first half life($t$ = $T^1_{\frac{1}{2}}$), $\ce{[A]_t = \frac{[A]_0}{2}}$, So, we have $T^1_{\frac{1}{2}}$= $\frac{[A]_0}{2k}$ But for the second half life ($t$ =  $T^2_{\frac{1}{2}}$),  $\ce{[A]_t^, = \frac{[A]_0}{4}}$ and $\ce{[A]_0^' = \frac{[A]_0}{2}}$, Putting these values in the equation, we have $T^2_{\frac{1}{2}}$ = $\frac{[A]_0}{4k}$ Thus we can see that for any $nth$ half life in a $0$ th order reaction, the $[A]_0$ which we are habituated to use in $1st$ half life is not the initial concentration(i.e. at $t=0$) but for the nth half life, $[A]_0$ is the concentration of the reactant after completion of $n-1$ half lives i.e $\frac{[A]_0}{2^{n-1}}$, and thus using this fact a general formula for nth half life (i.e. $T^n_{\frac{1}{2}}$) in ) $0$ th order reaction will be  $\frac{[A]_0}{2^nk}$.


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Stretching frequency of nitro and nitrito
I got the values of the stretching frequency of nitro and nitrito ligands from this webpage.
I understand why stretching frequencies are close in nitro and well separated in nitrito. How to explain why nitrito has slightly higher stretching frequency?

Please consider adding a comment when you down vote

First of all, in the webpage, I think there is some shuffling between Nitro and Nitrito first in the headline. But, according to the contents, Nitro has closer stretching frequency, but Nitrito has distinguished stretching frequency.Let's take that one as the correct data.

The stretching frequency of nitro ($\ce{<-NO_2}$ )corresponds to the stretching of the bonds between $\ce{N}$ and $\ce{O}$, which are in between double and single bond, due to resonance, and the bond order between these $\ce{N}$ and $\ce{O}$ becomes almost equivalent in both the two  bonds.
Thus the stretching frequency of $\ce{-NO_2 }$ corresponds to partial double bond character between N and O and almost equivalence of two bonds.  On, the other hand, the two streching frequency of nitrito ($\ce{<-O^ -N=O)}$ corresponds to one $\ce{N -O}$ pure single bond and  one $\ce{N=O}$ pure double bond as there is no delocalisation in this case. So, due to that pure double bond stretching frequency of $\ce{N=O}$, one streching frequency of nitrito is higher than Nitro, which is consisting of only partial double bonds.


text:
Colored Flames Experiment
I am going to be doing a gender reveal for my sister's pregnancy. To do this, we wanted to light a huge fire pit and throw in a chemical to change the fire into the color (red/pink for girl, blue/violet for boy).
From my research, I found that strontium chloride is the best option for red/pink and copper I chloride is the best for blue. I found that Iron Filings make great gold sparkles, as well as magnesium shavings make for a great silver spark.
It is my goal to have my sister and brother in law throw a satchel full of chemical-enriched wax pucks into the fire pit fire and have the fire first glow many different colors, a rainbow of colors until the other colors die down and the pink or blue for the gender remain.
To do this, I was going to create many wax discs with many different colored chemicals and methanol (to create a more vivid color) by first melting wax, adding methanol and the chemical, mixing and setting to dry. The pink or blue one would be the LARGEST wax puck/disc and would, hence, color the longest after the rest of the colors died down.
I was going to add all the discs and the iron filings into a satchel with the largest disc (the gender color in it's own satchel so it remains last), and then when they throw it in, it will create this spectacle.
My question is, does this sound like a viable option and experiment?
Is using methanol in the wax an unnecessary step?
Is it ok to mix methanol with wax? Will it still harden?
How can I insure that the blues, yellows, oranges, greens will die down and the gender color (lets say RED in this case/girl), remains?
How long will it take the color to shine through? How can I make sure this is done within the course of a minute or so?
How much of a chemical should I use to ensure it is seen in a large fire pit (about 3-4 feet in height)?
Any further advice? Tips? Tricks?

Did you end up pulling this off?? We'd love to but some of your chemical disks for our gender reveal!

Colorful flame
To create a colorful flame you could just buy some packages of Colorful flame which is not toxic.
To buy it on amazon click here.

Gender part
For the gender part you could treat some sawdust with aqeous Copper(I)-chloride ($\ce{CuCl}$). Let that sawdust dry and throw it in the fire (after some time passes so they can be amused by the "Colorful fire"). It'll create a blue colour which would be great if its a boy.
However, if its a girl you could do the same thing, but instead of the Copper(I)-chloride you would use Strontium(II)-chloride ($\ce{SrCl_2}$) or Strontium(II)-nitrate ($\ce{Sr(NO_3)_2}$).
I don't know how long it'll  last but it depends on how much you put in the fire. (More stuff lasts longer)

Precautions

Be careful! You could be burned or something could catch on fire.

Definitely don't breathe in the fumes from the fire.

Don't use methanol because its a poison if ingested.

And the general advice: Wear safety gear!

Sources linked here and here.

How is there aqueous copper(I) chloride at the needed concentration? That solid is only sparingly soluble. Copper (II) maybe?

Well I was going for the blue color, but Copper(II)-chloride will burn in a blue-greenish color.


text:
Gluing bakelite together
I have an older stainless steel saucepan with what I've always called a Bakelite handle.  The pan fell, the handle broke.  If possible, I'd like to glue it.  The break is near where the handle attaches to the pan.  So, the repair would ideally bond Bakelite-to-metal stud as well as Bakelite to itself.  (pic below.  NB: the color is off: it's actually 'light black', there is no blue.)
If it matters, the pan doesn't ever go in the oven, but is used to pour boiling water.
If anyone knows what glue to use, I'd appreciate hearing.  If I should just forget about it, knowing that would help too.

Arts and Crafts would probably be a better place to ask this question. Not much chemistry involved here.

It would be better to fabricate a new handle than to rely on a glued joint -- consider what would happen if the handle came loose while carrying a pan filled with boiling water!

@DrMoishePippik Yeah.  On further thought, this was not a good plan.  I'm done with it.


text:
Why does NOCl dissociate as +NO and Cl-?
$\ce{N}$ is more electronegative than $\ce{Cl}$ (albeit by a very small difference) but it should ideally dissociate as $\ce{Cl+}$ and $\ce{^-NO}$ (similar to the dissociation of $\ce{Cl}$ in $\ce{HOCl}$). Moreover, due to the stronger bonding between $\ce{N}$ and $\ce{O, N}$ should also be pulling electrons even more tightly to itself due to the very electronegative $\ce{ O}$  by its side.
I was doing a problem on the electrophilic addition of $\ce{NOCl}$ on alkene and ended by getting the wrong product (according to the book) because I took $\ce{Cl+}$ and $\ce{^-NO}$ in the mechanism at the very beginning. Could someone explain?
I've attached pictures of what I have done and what the book has done too. Please correct me if I am wrong.
What I did:

What the book did:

Note: Electronegativity of Cl >$\approx$ Electronegativity of N.

@MollyCooL Could you further explain why? Doesn't N form hydrogen bonds but not Cl? Shouldn't that be considered as a push for N to be more electronegative.

The regiochemistry of Markovnikov addition of ClN=O to propene gives the answer. BTW the chloronitroso  cmpd. your book gives as a answer likely tautomerizes to the oxime of 2-chloropropanal.

Of course N supports H-bonding but Cl doesn’t. That in no way means N is more electronegative. @NehaMalcom

Regarding your drawing: a positive charge "(+)" never attacks a double bond or lone pair. Always the other way around.

by molecular orbital theory the bond order of NO+) is 3 which makes it very stable and therefore it can be easily formed from NOCl

Firstly, the correct comparison to make to determine which is more likely of $\ce{NO+}+\ce{Cl-}$ and $\ce{NO-}+\ce{Cl+}$ is the ionization potential vs. the electron affinity of the the two species. The electronegativities are correlated with these other metrics, but it is important to emphasize that those are really the important things.
Relatedly, nitrosonium $\ce{NO+}$ is stabilized by resonance and so this is a considerably stabilized cation (i.e. IP not too high). On the other hand, the IP of $\ce{Cl}$ is huge while the EA is small; it would much rather get an electron than lose one. The hyponitrite anion $\ce{NO-}$ actually usually dimerizes for reasons outside the bounds of the question. So $\ce{NO+}+\ce{Cl-}$ is way lower in energy.
In general, a red-flag should go up when you see halogens with positive charges. They do happen, but very rarely, especially at the level of chemistry in an intro organic course.

Actually there can be halogens with positive charges.  But, only if these are pi-bonded to something else so as to keep their full octet, like a contributing structure to a $\ce{CR2Cl+}$ ion.

You don't even need to get that exotic. There are $\sigma$-bonded examples. ICl (iodine monochloride) is a bench stable source of $\ce{I+}$ and is a superb reagent for turning Aryl-TMS to Aryl-I. I stand by that halogens with positive charges should set off red-flags, especially at intro organic (which is the level of most Chem SE users). You will be served best by treating such invocations with caution.

Not a positive formal charge, though, which is what I thought you meant.  Nucleophilic attack on ICl could be Sn2.  In any event "short of an octet" is a bigger problem for halogens, even fluorine, than a positive charge per se.

The original question involved invoking fully dissociated $\ce{NO-}$ and $\ce{Cl+}$ so that it what I'm talking about with respect to charges. ICl is principally an electrophilic reagent (although I'm told it can add to double bonds, I've only ever used it for electrophilic aromatic substitution).

NOCl cannot be considered as NaOCl for the calculation of oxidation staes owing to the following :
(i)Na is highly electropostive and it exhibits +1 state always
(ii) O is neither peroxide or superoxide in NOCl ; it is normal oxide and it is in -2 state
(iii)O exhibits positive oxidation state only with F
(iv)N is next to O in electronegativity and it exhibits -3 to +5 states
(v)NOCl adds to unsymmterical alkenes obeying Markovnikov's rule i.e., NO+ attacks double bonded carbon containing more hydrogens to create a secondary carbocation which is attacked by chloride ion to give the major product
(vi)So the calculation is x - 2 = +1 and x =+3
In nitrosyl chloride, the oxidation state of N is +3


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how to avoid bubble formation in hexane
I have faced this problems when I was working in lab with hexane for the first time.
I was trying to guide dispersion of nano-particles in hexane solution to a fluid cell (vacuum chamber) with a syringe and a flexible tubing (about 2mm or 3mm diameter). All the materials that I use are hexane resistant.
The fluid cell is kept on very low pressure, so the goal is not to have any bubble formed in it.
When I inject hexane fluid, midway before it reaches the chamber, the vapor bubble are formed ( I think because of low boiling point of hexane ) , which is bad I don't want any bubbles go in the chamber.
However, I figured that if the flow is not horizontal or parallel to the ground (i.e when i inject vertically up ward to chamber) I can avoid the bubble flow.
I also thought to do the injection process fast, however the whole flow has to be very slow for the experiment.
 I would like to know what exactly causing it and what are your suggestions to avoid the bubble formation in a horizontal flow.
I was thinking maybe adding some other solution with higher boiling point would be a case to avoid it. But I don't know exactly what solution. And also it is important to keep in mind that hexane solution contains magnetic nanoparticles and I don't want any effect (like aggregation and so on) of added solution into the dispersion nanoparticles.
Thank you,

If you continuously evacuate the chamber, the partial pressure of hexane will be lower than at equilibrium. So hexane will evaporate and probably form bubbles. This is how nature works.

Probably didn't explain well enough and vacuum chamber is a misleading term, i mean fluid cell. The fluid cell is in constant pressure prior to the injection of the fluid. Two independent process. The bubble formation inside the flexible tubing is formed even without attaching it to the fluid cell (or chamber).

As you state, hexane has a low boiling point. If the bubbles are hexane vapor, as opposed to an air leak, then cool the fluid and apparatus more. You might pre-chill the hexane in a dry ice bath. The sublimation temperature of $\ce{CO2}$ at 1 atmosphere is about 195 K, and the melting point of a mix of hexane's various isomers is about 179 K, so there should be no problem with crystals blocking the flow.
Be careful with the cold -- frostbite is not painful at first, but is as damaging as a burn!


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Paramagnetic relaxation in 13C nmr
I'm trying to make my 13C{1H} (hydrogen decoupled) nmr more quantitative. From the literature I've noticed that researchers either

increase the delay time in the experiment (which I can't do, as it takes too long to collect the spectra)
run the decoupling pulse only during data acquisition (to reduce the nuclear Overhauser effect), and/or
add a paramagnetic relaxation agent to the sample.

I want to try a paramagnetic relaxation agent (I'm working with $\ce{CDCl3}$ as a solvent).  The most common one seems to be $\ce{Cr(acac)_3}$.  Are there better relaxation agents out there (i.e. more effective at lower concentration)?  Also, the literature I've read shows all sorts of concentrations of the relaxation agent... would a low (0.01 M or less) concentration work, in your experience? Do these agents lead to too much spread in the peaks?
Also, will using a relaxation agent ruin the nmr tube for further use?

Paramagnetic relaxation in $\mathrm {~^{13}C}$ NMR:
Quantitative $\mathrm {~^{13}C}$ NMR requires full relaxation of all carbons at each scan.  Adding paramagnetic relaxation reagents could speed up the relaxation.  The most common reagent you can find online is chromium acetylacetonate, $\ce{Cr(acac)3}$. However, the amount you need to add to your solution is very critical. It is tricky that if your added amount to your solution is too little, $\mathrm T_1$ of the $\mathrm {~^{13}C}$ will not be short enough. I you added too much of $\ce{Cr(acac)3}$, the signal will be too much broadened due to too much shortening of $\mathrm T_2$, thus you will lose resolution. Typically used recommended amount by experts is Between $\mathrm{0.1wt\%}$ and $\mathrm{0.5wt\%}$ (https://blogs.umass.edu/weiguoh/?p=71). With that amount, the solution will have a light purple color [1] (courtesy of https://blogs.umass.edu/weiguoh/?p=71).

Above picture is a picture of a standard sample with $\mathrm{0.5wt\%}$ $\ce{Cr(acac)3}$, which is an excellent reference for you. It'd provide you the easiest way to avoid the mess of weighing $\ce{Cr(acac)3}$. What you have to do is to look at the color of your solution.  Following the color of the picture, you can add $\ce{Cr(acac)3}$ a little by little so that your sample has a similar or slightly lighter color than this.
To read about othe reagents download this thesis (available online): https://www.diva-portal.org/smash/get/diva2:793434/FULLTEXT01.pdf


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Is the mass of a mole identical for different elements?
For example, if I have 2 moles of iron and 2 moles of copper, do I have an equal mass of each?

Have you tried answering this yourself? Please tell us where you got stuck by [edit]ing.

Well you can calculate this on your own! (I'll just point you in the right direction)
To calculate the mass of the two you'll have to multiply their molar mases by the amount of mols (in this case 2 mols). At the end compare the two mases!
(The molar mass is equal to the relative atomic mass, but only for monoatomic elements)


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Does tetrachlorodibenzo-p-dioxin (2,3,7,8) have a nonzero dipole moment?
I've looked at and drawn the structure, and I originally felt that it would be nonpolar - as I believe all the dipoles would cancel out, but I'm not 100% sure. I might be too dependent on the fact that it looks symmetric.
Also, is it true that the overall molecule shape would be planar? And is the hybridization of the oxygen atoms sp3 like I think they are?
Edit: Thanks to a commenter, I have changed the question to ask about the dipole moment in hopes of being more clear. I'd also like to add that this is a homework question that I am stuck on (for general chemistry l) and the teacher hasn't been the most thorough which is why I am here trying to teach myself.

Because of the molecule symmetry its dipolar moment is null and you're right about oxygen hybridization. The two planes perpendicular to the molecule allows you to conclude that there is no net dipolar moment in there...

Polarity is a property of solvent, equalling polarity with dipole moment is tragic misapplication of concept.

I mean, I didn't come up with the question and that's one of the ways which is taught to figure out molecular polarity, hence coming here to ask for help...

I think loosely using polar/nonpolar to describe a molecule qualitatively is ok and people do it on a daily basis, but it can be ambiguous/debatable, see for example the discussion here. In this case I think it is perfectly clear from context what you mean, but a more precise wording (that leaves no room for nitpicking) would be something like "Does xyz molecule have a nonzero dipole moment?"

Here is another example of problems it causes https://chemistry.stackexchange.com/questions/65188/polarity-index-vs-dipole-moment

The best answer for this seems to be in Structure, Energy, Vibrational Frequencies, and Potential Energy Curve of
2,3,7,8-Tetrachlorinated Dibenzo-p-dioxin: Ab Initio MO Studies J. Phys. Chem., 1996, 100 (12), pp 4810–4814.
This research finds that though planar is the lowest energy conformation, the energy for bending with respect to the O-O axis from 180 degrees to 170 degrees is only 0.008 kcal/mol.  The molecule flaps like a butterfly.
The article says:

At ambient temperatures, nonplanar conformations
  will be significantly populated for this potential. This prediction
  accounts for the experimental results that the electric dipole
  moment of the DD molecules is reported to be 0.64 D. [reference 24]

Where "DD" means the unsubstituted dioxin and where reference 24 is Higashi, K.; Uyeo, S. J. Chem. Soc. Jpn. 1941, 62, 396.


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Why should hydrolysis break down fat into fatty acid and glycerol?
Fats and lipids are composed of carbon, hydrogen and oxygen. Both esters form different structures at different room temperatures.
Now fats and lipids are generally insoluble in polar solvents(water) and soluble in organic solvents(hexane, ether).
If this happens to be true, why should the actions of hydrolysis(chemical breakdown of a substance with the use of water) breakdown fat into fatty acid and glycerol?
Just like the enzyme lipase, uses hydrolysis to breakdown fats. Why should a polar solvent dissolve fat?

"Insoluble" doesn't mean "has zero solubility", it means "it's solubility is low or very low".

I suggest a subtle rethink here. Fats are not hydrolysed to fatty acids (long chain carboxylic acids) and glycerol ; triglycerides are. Fats tend to be solids; heavier homologues of the corresponding hydrocarbon oils.

Lipase enzyme breaks down fats into fatty acids and glycerol, and water is essential for enzyme action/activation.


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How does the molarity of a solution change on boiling off the solvent?
The dilution equation $$M_1V_2=M_2V_2$$ is an equation that can be used to calculate how adding water can impact the volume and molarity of a solution. So I was wondering if this equation would apply if you were to remove the amount of solvent through means such as boiling.
For instance, suppose I have a $\pu{100 mL}$ solution of $\ce{HCl}$ that has water as its solvent and has a molarity of $\pu{3.0 M}$. Would the molarity become $\pu{6.0 M}$ if were to boil out $\pu{50 mL}$ of water of the solvent, or would it have a different $M$ value?

That's a bad example. By boiling this solution, you will remove not just water, but also HCl.

It is probably better to refer to a better form of your equation, i.e. $$c = \frac{n}{V} \implies n = c\cdot V.$$ Where $c$ is the concentration (or if in $\pu{mol/L}$ molarity; better not to use $M$ for this as it commonly refers to the molecular mass in $\pu{g/mol}$), $n$ is the amount of substance, $V$ is the volume. Keeping $n$ constant and varying in $c$ and $V$ will therefore give you $$c_1\cdot V_1 = c_2\cdot V_2.$$ The way in which you change the volume is (in principle) irrelevant. (It should be noted, that the solvent and solute should endure this kind of change.)