text:
Chemical formula of Lactic acid
Formula of Lactic acid is given as C3H6O3, can't this be divided by the common factor3, and written as CH2O?

This was supposed to be a molecular formula. You don't divide molecules.

FWIW, CH2O is the correct "empirical formula" for lactic acid, as you would calculate from elemental analysis.

Basically, you can simplify or "divide" a molecular formula only if it really makes sense.

Basically, a given molecule has a unique (molecular) formula. However, some molecules have a tendency to associate to form bigger molecules. Only in that case can you simplify or "divide" a molecular formula. For example, elemental sulfur (S) is usually found in nature as S8 and then you can consider its true formula is S. Also P4S10 is a dimer of P2O5; you have P4S10 in your flask but you can consider P2O5 in your calculations.

You have initially tagged your question "structural formula", however $\ce{C3H6O3}$ is somewhat ambigous in this respect, and besides lactic acid can represent its isomer 3-hydroxypropionic acid, methoxyacetic acid, and almost completely different compounds like dihydroxyacetone, dimethyl carbonate, glyceraldehyde or trioxanes, not speaking of the second one $\ce{CH2O}$ (formaldehyde or ethylene oxide). Well, these are not structural formulas at all.


text:
Acidity of amides
Structure (b) and (c) has two NH groups . So will they lose hydrogen ions from both the N atoms or just from one of the N atoms? Which is the strongest acid? Please explain.

You need to make some effort to answer this before we will offer advice

Obviously Pthalimide

How? Could u please explain. I mean fine, it has got 2 strongly EWGs attached to it. So it will be highly acidic. But will not b) and c) lose 2 H ions in the same time?

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


text:
How to determine net charge of a polycation using Wade's rules?
In order to determine the net charge of polycations, which adopt a Wade-like structure (octahedral, trigonal bipyramidal, ...), Wade's rules can be used.
For example, the charge $x$ of the octahedral cluster $\ce{[Bi6]^x+}$ should be determined:

It is known that a Wade-cluster overall "needs" $2n+2$ electrons (to fill the $n+1$ energetically favoured, bonding MOs).
Bi atoms contain three valence p-electrons (the two s-electrons do not take part in the cluster-bonding). So in a $\ce{Bi_n}$ cluster, there are $3n$ electrons already available for bonding.
But only $2n+2$ electrons are needed: We need to remove $2n+2-3n=-n+2=-(n-2)$ electrons to obtain the favoured $2n+2$ electron count in the cluster.
So for example the $\ce{[Bi6]^x+}$ should have $-(6-2)=-4$ electrons "removed". We would arrive at a $\ce{[Bi6]^4+}$ net charge... ...but in reality, there is "only" a $\ce{[Bi6]^2+}$ - with two electrons more than determined according to the Wade's rules.

EDIT:  In contrast to that, the rules aboves apply to $\ce{[Bi5]^3+}$ (Thanks to @orthocresol).
Is there a misconception in my above thoughts? I can not find an illogical step - Maybe Wade's rules can not be applied to the polyatomic clusters (my textbooks suggests this nevertheless)? Or I should have take the s-electrons into account?

As far as I can tell, your calculations are correct. A closo structure would require $2n+2$ cluster electrons (or $4n+2$ total electrons), which is satisfied by the formula $[\ce{Bi}_n]^{(n-2)+}$. As such, for $n  = 6$, a closo octahedral structure would correspond to the ion $\ce{[Bi_6]^4+}$ as you have outlined.
The contradiction seems to be that the ion $\ce{[Bi_6]^2+}$ is not closo, but rather nido, according to e.g. Hampel & Ruck:[1]

The $\ce{[Bi_6]^2+}$ cation, a nido cluster with 16 skeletal electrons, has the shape of a distorted octahedron with an opened edge.

These require $2n$ cluster electrons for complete population of the bonding molecular orbitals, consistent with the formula $\ce{[Bi_6]^2+}$. I do not currently have access to Holleman-Wiberg (nor would I understand the German version!), so I cannot check what they have written, but it is possible that it is a simple typo.
Interestingly, some of the higher bismuth polycations also display a tendency to have "extra" electrons. For example, Greenwood and Earnshaw[2] describe the following cations:
$$\begin{array}{cccc}\hline
\text{Cation} & \text{Formal oxidation state} & \text{Cluster structure} & \text{Point group symmetry} \\ \hline
\ce{Bi+} & 1.00 & - & - \\
\ce{Bi3+} & 0.33 & \text{Triangle} & D_\mathrm{3h} \\
\ce{Bi5^3+} & 0.60 & \text{Trigonal bipyramid} & D_\mathrm{3h} \\
\ce{Bi8^2+} & 0.25 & \text{Square antiprism} & D_\mathrm{4h} \\
\ce{Bi9^5+} & 0.56 & \text{Tricapped trigonal prism} & C_\mathrm{3h} (\sim\!D_\mathrm{3h}) \\ \hline
\end{array}$$
The rationalisation which they provide for this is

[The application of Wade's rules] would account for the stoichiometries $\ce{Bi3+}$ and $\ce{Bi5^3+}$ but would also lead one to expect $\ce{Bi8^6+}$ and $\ce{Bi9^7+}$ for the larger clusters. However, these charges are very large and it seems likely that the lowest-lying nonbonding orbital would also be occupied in $\ce{(Bi+)_n^2-}$. For $\ce{(Bi+)8^2-}$ this is an $\mathrm e_1$ orbital which can accommodate 4 electrons, thereby reducing the charge from $\ce{Bi8^6+}$ to $\ce{Bi8^2+}$ as observed. In $\ce{(Bi+)9^2-}$ the lowest nonbonding orbital is $\mathrm a_2''$ which can accommodate 2 electrons, thus reducing the charge from $\ce{Bi9^7+}$ to $\ce{Bi9^5+}$ as observed.

References:

Hampel, S.; Ruck, M. Synthesen, Eigenschaften und Kristallstrukturen der Cluster-Salze Bi6[PtBi6Cl12] und Bi2/3[PtBi6Cl12]. Z. Anorg. Allg. Chem. 2006, 632 (7), 1150–1156. DOI: 10.1002/zaac.200500379.
Greenwood, N. N.; Earnshaw, A. Chemistry of the Elements, 2nd ed.; Butterworth–Heinemann: Oxford, U.K., 1997, pp 590–591.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Help understanding the pi-properties of the NCS- ligand
Course knowledge
When considering the NCS- ion, I take all atoms to be sp hybridized. This means that there are 8 electrons that are accommodated in the $\pi$-system.
There are two orthogonal 3-orbital 4-electron pi systems: similar to the allyl anion, each made of an antibonding, non-bonding and bonding set. The bonding and non-bonding orbitals are occupied in this system. As the non-bonding orbitals are both full of electrons, NCS- therefore can act as a $\pi$ donor.
The above is what I have been taught, however I am confused with this rationalisation when comparing the positions of NCS- and SCN- in the spectrochemical series.
Spectrochemical series
SCN- is a weak field ligand, which comfortably suggests that it is a $\pi$-donor. However, NCS- is relatively high field, and is therefore a $\pi$-acceptor - which disagrees with what I've been taught. As sulfur is less electronegative than sulfur, it is higher in energy and therefore contributes more to the anti-bonding orbital.
Questions
Wouldn't this mean that SCN- would be the better $\pi$-acceptor rather than NCS-? Or is it the case that since sulfur is larger than nitrogen, the better orbital overlap with the transition metal favours donation?
Additionally, how can we say that the pi-donation from the non-bonding orbitals wins out over the pi-acceptance from the anti-bonding orbitals? Is it because the anti-bonding orbitals are involved in the ligand bonding whereas the non-bonding orbitals are not?
Also, why is NCS- a pi-donor (according to what I have been taught) even though it is relatively high field?

Although the thiocyanate ligand acting through the nitrogen atom is higher field than water (a $\pi$ donor), it is still weaker than $\sigma$ donors, such as $\ce {NH3}$. Thus, it should not be classified as a $\pi$ acceptor.


text:
Directional emission from molecules
This might be a silly question (probably is).
When a molecule absorbs it has a transition dipole moment, a measure of how strongly a certain transition will interact with light will depend on how closely a the light is to resonance with this dipole moment. The same in reverse applies for emission.
Typically in a bulk material, these dipoles are randomly spread, so luminescence is in all directions. Some molecules can be aligned, like in LCDs.
Is there a way I could align some luminescent materials such that they might emit preferentially in a certain direction? Has anyone done any work on this, what would this be called?

To get directional emission from a molecular substance, just make it into a LASER. Actually, the predecessor to the LASER, the first MASER, was based on microwave emission from ammonia. A $\ce{CO2}$ LASER is quite efficient and fairly easy to make, and the atmosphere of Mars is a natural, solar-pumped laser. So if you want directional emission from molecules, it's not even necessary to align them -- just provide suitable end-reflectors, or even a linear medium with enough gain.
That said, your question is about aligned materials... and there is a strong correlation between alignment and emission. An interesting example is found in topological insulators, which conduct only in certain directions and locations. It is possible to make a LASER from topological insulators. Some of these are as simple as $\ce{Bi2Se3}$, and others are more complex such as bis(methylene) adamantyl carbocation.

A laser emits radiation in the same direction (and phase) where the incoming light stimulating it is going. Mirror pairs are just a convenient  way to favour the normal of the mirror planes as a direction.

@Karl, yes: " a linear medium with enough gain". Fiber LASERS, Mars CO2 and others do their "magic" without mirrors.

My point was that those don't favour a special direction. Fiber lasers are used as amplifier for light signals that already come in the right (axial) direction afaik, not the original source.

In addition to the energy requirement for absorption of a photon there is also an orientational requirement that depends on $\cos^2(\theta)$ where  $\theta$ is the angle between a molecule's transition dipole and that of the polarisation direction of the photon. (btw. cos squared has the shape of a 'p' orbital).
This requirement means that with linearly polarised light, photo-selection of randomly oriented molecules in solution occurs, i.e only those molecule with their dipole more or less oriented parallel with the polarisation are excited. If the exciting light is present in a short laser pulse, (picosecond, femtosecond) this provides a way of measuring how rapidly molecules undergo rotational diffusion. This occurs in few tens to hundreds of picoseconds for most organic molecules in mobile solvents such as cyclohexane.
Suppose now that the molecule are in a very viscous solvent so that fluorescence occurs before any significant rotational motion then the emission will be directional. This will have a cosine squared distribution but the direction will be determined by the emission dipole's orientation in the molecule and this can be at a different angle to that of absorption, for example, emission may be from a different excited state to absorption.   Similarly molecules oriented/aligned in a liquid crystal or in a normal crystal will show directed emission after excitation with polarised light. The extent of the latter will of course depend on how the molecules are arranged in the crystal.  Surface oriented molecules may also, in principle,  show directed emission, but most often their motion on the surface and surface imperfections will limit this.
If the block of excited molecules just produced has polarised light of the right wavelength passed through it then stimulated emission can occur and this is now directional. This simple arrangement is effectively a laser amplifier. If this arrangement is placed between suitably focussing mirrors then an laser oscillator is produced. The geometry of the laser cavity now governs the properties ( diameter, profile, divergence) of the laser beam produced. Directionality is now ensured, for example, lasers are routinely reflected off a retrorefector placed on the moon so as to accurately measure the earth-moon distance, something that has been done since the 1970's.


text:
Molarity of a substance in a solution when more fraction is given
Mole fraction of ethanol in a solution with water is $0.040$.
What will be the molarity of ethanol in water? Density of water can be assumed to be $1\ \mathrm{g/mL}$.
This is how I tackled the problem:
$0.040 = n(\mathrm E)/(n(\mathrm E) + n(\mathrm W))$
where $n(\mathrm E)$ is number of moles of ethanol and $n(\mathrm W)$ is number of moles of water. This is the first equation.
I assumed that there is $1\ \mathrm L$ of solution in total and that the molarity is to be found at STP.
That gives:
$(n(\mathrm E) + n(\mathrm W)) \times 22.7 = 1$
This is the second equation.
Two equations, two unknowns. Solving yields, $n(\mathrm E) = 0.00176$ and because the total volume of the solution is $1\ \mathrm L$, molarity is $0.00176\ \mathrm M$.
But that answer is wrong! Probably because I am assuming molarity is to be found at STP.
Here is an answer I found online:
Assume there is $1\ \mathrm L$ water. This implies $1000\ \mathrm g$ or $55.56\ \mathrm{mol}$ water.
Using $\text{eq(1)}$ from my answer, $n(\mathrm E) = 2.32$.
And then the answer concluded by saying that as there is $1\ \mathrm L$ water, molarity is the same, $2.32\ \mathrm M$.
How can this be? Molarity is moles over total volume not just the volume of the solvent.
Am I missing something or the online answer is wrong? Is my answer correct? If not, what's the correct solution?

Curious, where did the 22.7 come from?

@MaxW The volume of 1 mole of an ideal gas at 273 K and 1 bar.

@user64082 You might also want to check your units. Especially your values fore the amount of substance $n$ sometimes are correctly expressed in mol, but also wrongly in M = mol/l or without any unit. Besides, the quantity ‘amount of substance’ shall not be called ‘number of moles’, just as the quantity ‘mass’ shall not be called ‘number of kilograms’.

@Loong Thanks for correcting me. I have edited my question so that n(x) is strictly a number without units.

@user64082 - the solution is a liquid. The molar volume of a gas has nothing to do with this problem.

@MaxW Oops! Missed big-time on that one. Extremely sorry.

For a mass of 1000 grams of solution we can find the number of moles of water and ethanol. Let $x$ be the total number of moles of ethanol and water in 1000 grams of solution. Then
$$(0.040x)46.07 + (0.960x)18.015 = 1000$$
$$x = 52.254$$
So there are $0.040 \times 52.254 = 2.09$ moles of ethanol in the 1000 grams of solution, and $0.96 \times 52.254 = 50.16$ moles of water.
However there is no way to compute molarity since the density of the ethanol-water solution itself is unknown.

The density of the solution is not 1 g/mL.

Ahhhh... misread the problem statement.

No problem. Happens to everybody :)

@user64082 -  I edited my answer. The problem is unsolvable as given. The only way to solve it would be to look up the density of ethanol-water mixtures. Are you sure that the problem didn't ask for molality rather than molarity?

Double-checked and the question did ask for molarity (M). It is question 1.29 from Unit 1 of class 11 NCERT Chemistry.

Again, the problem is unsolvable as written, unless you look up the density of the particular ethanol-water mixture.


text:
How does DIBAL-H react with acid anhydrides?
I know DIBAL-H reduces carboxylic acids and some derivatives, but couldn’t find out about its reaction with acid anhydrides. Do these react? And what are the by-products (if any)?

Welcome to Chemistry.SE! Questions are the central focus of this site, but we ask that questions demonstrate an attempt to solve/understand the question to maintain quality.  Please take a minute to look over the help center and tour page to better understand our guidelines and question policies.

I haven't been able to find any references for this but my instinct is that Dibal-H will react with anhydrides, probably to give aldehydes

@Waylander ... yes, that’s what I thought, too, but I couldn’t find a reliable source to verify


text:
Rearrangement with sodium iodide (ring opening)

I'm not able to figure out the mechanism for this reaction. I don't see how sodium iodide in acetone could lead to rearrangement, and surprisingly ring opening!
What is this reaction called, if it has a special name? How does it occur?
P.S. In general, as far as I know, sodium iodide in acetone promotes excellent SN2 substitution, so all bromine atoms should be replaced by iodine. After this, because 1,2-diodo compounds (vicinal di-iodides) are unstable, elimination takes place (removal of iodine) and formation of alkene occurs (that's what I've seen happen, usually).

Do you have a reference to the book / paper you got the reaction from? Looks pretty interesting

I'm sorry, I don't have a reference - but it appeared in one of my senior's lecture notes, so it must be right

@NotEvans. For once the reaction actually has some precedence that looks similar (the starting material is not exactly the same), but the reference is in German: Angew. Chem. 1977, 89 (12), 909–910. My first instinct is a series of electrocyclic reactions, but I'm not sure. I think the SM drawn in the question is too unsaturated.

A paper I found discusses a similar reaction: an "exotic" dehalogenation of vicinal dihalides.
Actually, it should be noted that the solvent used here is methanol, which might imply a different reaction path.
https://pubs.acs.org/doi/abs/10.1021/ja01607a037?journalCode=jacsat

And I agree with @orthocresol about the starting molecule being too unsaturated

If we take the starting material as the correct one, and perform the reaction, I think only four Bromine atoms can be eliminated out of eight, and the remaining four Bromine atoms will be very difficult to eliminate. So, I think the final product should be logically (1E,3Z,5E,7E,9Z,11E)-2,5,8,11-tetrabromo-6a,12a-dihydrooctalene .

I think the final product will also contain some Bromine atoms attached to it. and the no. of double bonds will be one less than that of the compound given in the picture. I may be wrong but with all of my known possibility of Organic reactions, I can actually reach to the final product somewhat different from that given in the picture  and I don't think the final product given in the picture can ever be got from the given starting material. I will be happy if someone else can find flaws in the logic  given below.

$\ce{NaI}$ in acetone is the reagent used for Finkelstein reaction which is nothing but a standard substitution reaction with $\ce{I-}$ ions. But in presence of vicinal dibromides, or any special structure where there is a chance of elimination of hallide ions even further, the iodide ions can actually make the molecule undergo preferably an $E_2$ elimination (in most of the cases).

In this starting compound also, there are substitutable $\ce{Br}$ atoms present. So, the iodide ions first performs a substitution reaction, and then undergo an elimination via bond migration. This same procedure occurs for two times. and finally we are left with a product which has $\ce{Br}$ atoms attached to $sp^2$ carbon atoms, and therefore they are highly reluctant to undergo any substitution or elimination reaction. So, that's why, logically the end product should be, $\text{ (1E,3Z,5E,7E,9Z,11E)-2,5,8,11-tetrabromo-6a,12a-dihydrooctalene}$. Here is the structure of the possible end product.

$\textbf{Mechanism:-}$

As far as I see, there can not be any more further reaction to the final compound. According to me, the given compound in the picture should not be the final product if we take the starting material to be the correct one. Instead, the end product in the mechanism should be the one produced with higher yield at the end.

At least two more bromine halogene atoms can be eliminate to form a double bond in the annealed part. It is the fate of two last halogene atoms that is surprising. It seems that NaI is working as a reducing agent for those two.

@Soumik Das can i ask 1 thing, if there is 1,2 dichloro-ethane, and if it reacts with NaI in acetone + heat is given, then is elimination the major product? this question is much similar to this one but i still have this doubt..

@maverick there's always a competition between substitutions and elimination in these type of reactions. If you provide heat i.e. higher temperature, I think the chance of elimination gets increased, although you still might get a slight amount of substitutited products as well along with mostly eliminated products.


text:
cluster of milk
I have a problem with milk when we are making yogurt in factory. When the milk pass through the machine it clogs the filter of the machine because the cluster that formed on the filter. My question is why that formed and what is a solution to the problem.

Possible duplicate of Why does milk flake?

I don't think you will find many food engineering specialists here.


text:
Do all flames require the same minimum concentration of oxygen to sustain themsleves?
Assume you have burned some substance A in an airtight container until the fire goes out due to insufficient oxygen. You then place some burning substance B in the container through an airlock. Are there substances A and B such that the substance B would continue to burn for a significant amount of time using the oxygen left behind after substance A burned out?
In other words, do all materials require the same concentration of oxygen to sustain a flame?

Not all flames require an oxygen atmosphere.

No, different substances will cease combustion at different concentrations of oxygen. The concentration needed can be described by the limiting oxygen index. This is the reason that the use of pure compressed oxygen can be so hazardous; not only will materials burn that would not have burned before, but now materials that do burn require a lower temperature/energy input to ignite.
You can understand this conceptually when you consider where the heat goes. On combustion in air some of the heat must be used to heat the $\pu{79\!\%}$ of the air that isn't oxygen and that heat is wasted in the inert gas and the heat is diluted. That means only ~$\pu{20\!\%}$ of the heat goes into activation energy to propogate the combustion. When the oxygen is more concentrated, less heat goes into inert gases and more goes into oxygen to propagate the combustion and vise-versa.

Since A.K. has already answered the question you're really asking, I feel it's reasonable to interpret your question literally and answer, "No, because it's possible to have flames with no oxygen at all."
The flames in this Periodic Videos video  are from iron burning in fluorine. The experiment is done in the open air (wot no fume cupboard?) so the later glowing and sparking is due to the hot iron reacting with atmospheric oxygen. Later, the video also shows charcoal, cotton wool, sulphur and even iodine burning in fluorine. Again, some of those (especially the cotton wool) get hot enough that they continue to burn in atmospheric oxygen after the fluorine has been turned off, but the initial flames are from the reaction with fluorine.

Indeed, there are reactive metals that can burn in much less reactive gases, such as carbon dioxide, nitrogen and water vapour. Hence if material A is a hydrocarbon, and consumes (nearly) all the oxygen in Steve B's vessel, and material B is, say, lithium, then material B can burn in the products of combustion!


text:
Is it possible to apply reaction progress kinetic analysis on non catalytic reactions?
I was wondering if anyone familiar with reaction progress kinetic analysis knows if it is possible to apply the method on non catalytic reactions, and if so provide a few references with relevant examples.

Apparently the method works normally with non catalytic reactions according to the example presented in a Mettler Toledo webinar (Titled: Reaction Progress Kinetic Analysis, https://www.mt.com/us/en/home/library/on-demand-webinars/automated-reactors/kinetic.html) where a simple Diels-Alder reaction is analyzed by RPKA.


text:
Why do the Metal Ions in an Electrochemical Cell Not Pass Through a Salt Bridge?
I understand that the purpose of a salt bridge is to maintain the neutrality of the solutions in an electrochemical cell, but why do the ions    being oxidized and reduced not pass through the salt bridge.

For example, in this electrochemical cell, why do the silver ions not pass through the salt bridge and become reduced in the copper sulfate solution. Wouldn't this cause less resistance in the circuit, or does the salt bridge prevent the metal ions from passing through?

https://chemistry.stackexchange.com/questions/19540/why-the-salts-in-a-salt-bridge

https://chemistry.stackexchange.com/questions/19540/why-the-salts-in-a-salt-bridge

Actually the purpose is rather what you say - prevent mixing, otherwise you could just put electrodes next to each other.

They do pass.

The salt bridge is kind of first-in, first-out. So the ions that are already in the bridge come out first. The oxidized/reduced ions follow later.


text:
Give the order of nucleophilic addition reaction in the following -

I am confused that whether we should take only inductive or hyperconjugation effect also into consideration while finding the order. As the answer in the book is only according to inductive effect . Will hyperconjugation effect come into play or not ?
If we take hyperconjugation effect then order must be 2>4>3>1.

The question is: how will these molecules react to the presence of a nucleophile (which would add to the more electrophilic center, i.e. the carbonyl carbon atom)?
The more the carbon atom of the carbonyl group has a partial positive charge, the more it is susceptible to be attacked by a nucleophile, so the problem is to compare the relative positive charge on the carbonyl carbon atom between those molecules.
Remember that an alkyl group has an electron-donating effect AND that a longer alkyl group has more electron-donating effect.
With all that, you should be able to answer your problem.

Yes I understand, but i am asking that won't hyperconjugation have any role . As more the hyperconjugation structures lesser will be the positive charge on carbon hence less reactive.

Structure 4 have more hyperconjugation than 2, while 2nd have more inductive effect. And hyperconjugation> inductive


text:
Electrolysis of water using 240V AC
I remember back in my high school, when we started electrolysis, I became quite interested in it. I decided to do a home experiment on electrolysis of water. I grabbed my graphite pencils, a glass of water made of steel, two wires and some water. I wrapped the wire on my graphite nibs(on one end) and immersed the other end in water. I connected the wires to 240V AC supply(that was the supply which powers our home.)
Observations:
Initially, bubbles formed on both of the electrodes(I think they must be a hydrogen/oxygen mixture since the cathode and anode switches in between due to ac)
The glass of water became very hot with the passage of time(which was due to resistance heating i suppose)
The interesting part came when I added a few grams of table salt to water. Within seconds, the graphite rods turned red and there were fumes of gas(most probably it was steam cause the fumes were white.) There might have been sparks in the plug as well, but I don't remember for sure. I immediately pulled the plug out before it got any worse.
I didnt try anything of that sort again as i knew i could have done worse than that. But it doesnt cause any harm in knowing what actually happened when I added salt.
Why did the graphite turned glowing red?
What was the composition of those fumes?
What more could have gone worse?

There are easier ways to kill yourself.

@IvanNeretin That I know. What i don't know is: the chemistry behind it.

What you are trying to do is called a redox reaction (from REDuction-OXydation) in which one molecule will be oxidized (i.e. give one electron) and another will be reduced (i.e. receive one electron). The propensity of two different molecules to exchange electrons is called "redox potential" and measured in Volts.
In the case of water, its dissociation into hydrogen and oxygen has a potential of 1.23V(1), so a 1.5V battery would be enough (however a rechargeable battery usually has a potential of 1.2V which would not be enough).
In your case, you apply a 240V potential, which will not only promote many other reactions but also heat the water up because of the Joule effect and make it boil.
(1) I purposely ignored the sign not to confuse you.

I am comfortable working with reduction potentials and electrochemistry, but the question was: why did the graphite rod became glowing red within seconds of adding NaCl.
It could have glowed red during electrolysis of water but it only did upon adding NaCl. Why?

Adding NaCl will make water MUCH more conductive so the graphite rod will warm up so much (at the surface) it will glow. Actually, very pure water does not conduct electricity at all. Only its impurities do.

What more could have gone worse?

Not much I think, though adding NaCl and with a 240V potential, you will also make chlorine gas, which is extremely toxic. I really suggest that you don't work with 240V current... or 110V because it can be deadly

I haven't put my life on risk in such a way since..

I am not worried about you. I am worried about science!


text:
Which of the following is most likely to undergo hydride shift?
For the following question the answer is given as option (a) . However both,option (a) and (c) has alpha H atoms and both will form 3° carbocations after rearrangement. So what makes (a) more favorable than (c)?

Ring expansion will be favored in (c) over a hydride shift.

https://chemistry.stackexchange.com/questions/28939/ring-contraction-in-a-carbocation-due-to-ring-strain-and-back-bonding

Besides Avinsh's comment, there is another factor.  When you have a carbocation center next to a cyclopropane ring or cyclobutane ring, the "bent bonds" of the ring partially overlap the "vacant" orbital on the cation center.  To the extent that this occurs, the formally primary ion (c) is something like a more stable tertiary one.  Ion (a) does not have this factor because the cation center is already inside the ring instead of next to it.

Thank you..but could explain more precisely what do u exactly mean by" bent bonds"?

https://en.m.wikipedia.org/wiki/Bent_bond.

The cyclobutyl cation itself (a) is probably non-classical, too, although I'm not sure how much the methyl group affects it


text:
Doubts on IUPAC name 1,1'-methylenebis(4-isocyanatobenzene)

Wikipedia states  IUPAC name of the above compound as 1,1'-methylenebis(4-isocyanatobenzene) .
I have a confusion. Shouldn't isocyanate group be given preference over the methylene group? I assume the name as 4,4'-methylenebis(1-isocyanatobenzene) .

Similar is the case with DDT.  Wikipedia states its IUPAC  name as 1,1'-(2,2,2-Trichloroethane-1,1-diyl)bis(4-chlorobenzene) .
Why is the numbering of carbon with three chlorine atoms done $1$ ? Moreover,  can it be  $1,1'$ in the first place in place of $4,4'$ ? I guess the $\text{diyl}$ term indicates that it is connected to two different carbons(of benzene)  .
To sum up,  I want to know which chain should I select as parent chain and why .

Related: What is the rationale of the IUPAC name for DDT?

@Loong is methylene diisocyanato diphenyl also related?

Methylene (or xx-diyl) is not a substituent. Having two outgoing bonds, they must be part of the core structure, which takes precedence. Simply speaking.

It's of course not a duplicate, but most of the quoted text in Loong's answer (linked above) is applicable to your case.

@Loong..  Would you please simplify your answer?  Some terms and language in the quoted section of your answer seems to be confusing to me (that's not vague,  but the language is too unfathomable to me as I'm a non-english ) ... Would you please provide an answer as to why it's 1,1' etc...in a simple language?

The isocyanato group is named by prefix (like halogeno-), it's not a principal characteristic group named by suffix. It can be said that these groups have lower priority than the “joint” position.
See the different numberings (examples from the IUPAC nomenclature of organic chemistry, 2013):

P-15.3.2.4.1
(…)

1,1′-oxybis(4-bromobenzene) (PIN)

4,4’-oxybis(2-bromobenzoic acid) (PIN)

Therefore it is named 1,1′-methylenebis(4-isocyanatobenzene), not 4,4′-methylenebis(1-isocyanatobenzene).
(You will probably argue that here, the oxy might have higher priority than bromo group or methylene connector, but “connector priority” is not taken into account.)
See also more rigirous answer to Q: What is the rationale of the IUPAC name for DDT?

One rule of nomenclature is that you should get the simplest name and, in both of your examples, the simplest is to acknowledge the symmetry of each molecule. So the "parent hydride" is methane in your first example and ethane in the second one because there is no simple way to describe an "alkyl group with 2 substituted phenyls attached".


text:
How to evaluate a non-ionized compound activity coefficient?
I know that the activity coefficient of compound such as $\ce{NaHCO3}$ is greater than unity averaging 1.13 for seawater ionic strengths, but how can we evaluate this value?

Two separate conceptual problems here.
First, Debye-Hückel is a model for the approximation of electrolyte activities, but activities are conceptually distinct from concentrations for all other thermodynamic properties of a system, too, so activity coefficients can be derived from the apparent difference between any other thermodynamic property that relies on activities and its expected value due to concentration. For non-electrolytic solutions, activity coefficients are usually estimated from "contribution methods" - methods that apply contributions based on the empirical thermodynamic properties of large ranges of compounds, mathematically treated in a way that separate "contributions" can be statistically derived from the data, so that any new compound's activity coefficient can be estimated by the relevant contributions depending on its structure, environment, etc. These methods typically look at a wide variety of compounds and properties as a way to avoid singular effects and provide a good broad-range estimation of activities. You can read the Wikipedia articles on MOSCED and UNIFAC as a brief introduction to how these methods operate.
$\ce{NaHCO3}$, however, is ionised at seawater ionic strengths - it's a strong electrolyte that decomposes into $\ce{Na+}$ and $\ce{HCO3-}$, with separate activity coefficients. It is common, however, to provide activity coefficients of 1:1 salts as a mean activity coefficient, $\gamma_\pm$, which is the geometric average of the cationic and anionic activity coefficients:
$$ \gamma_\pm = \sqrt{\gamma_+\gamma_-}$$
or, for a general binary salt $\ce{A_xB_y}$,
$$ \gamma_\pm = \sqrt[x+y]{\gamma_{A}^{x} \gamma_{B}^{y}} $$
I suspect the reported activity coefficient for $\ce{NaHCO3}$ is therefore this mean coefficient:
$$ \gamma_\ce{NaHCO3} = \sqrt{\gamma_\ce{Na+} \gamma_\ce{HCO3-}} $$

Thank you very much for your good answer,
However, I'm asking about the activity of a non-ionized, I meant, like if we have a precipitant of  CaCO3 for instance, what's its activity?, please refer to: http://ocean.stanford.edu/courses/bomc/chem/lecture_02.pdf    (the table in Page 10)

Looking at your source, it's not clear to me what these values refer to or how they are derived. It seems that they are experimental values (given that they are presented in contrast, and approximated by, the Davies model discussed just above), but, to begin with, these activity coefficients aren't presented with correct dimensionality, and $\ce{Na}$-containing salts are strong electrolytes of which only a negligible amount will remain as bounded ionic pairs. I'm not sure where these values come from but since all ionic pairs have the same value makes me suspect they aren't actually (...)

(...) empirical but derived from electrostatic considerations (the only model the text seems to consider), which tend to assign identical values to compounds with the same charge, as explicitly discussed by the text with regards to Davies.

For instance, Pitzer's 1980 paper, which seems to be an older source for a value that is oft-referenced in more recent studies, gives a value of around 0.6 for ionic strength ~0.7 and 25ºC, something that is in line with the free ion activity coefficients of $\ce{Na+}$ and $\ce{HCO3-}$ but quite different from the quoted value of 1.13. The experimental measurements Pitzer quotes are also in the 0.5-0.6 range although they do not measure exactly the same system.

Thank you, But I know that it's possible to find activity coefficient greater than 1, refer to the following question: https://chemistry.stackexchange.com/questions/8605/activity-coefficient-greater-than-1

It is definitely possible. So? What in what I have written implies it isn't possible?

I meant, that according to Debye-Hückle equation , it's not possible to find an average activity greater than 1, according to the equation you've previously mentioned, however, I need to find a simple equation to find out such like this activity coefficient.
I think Pitzer is a quite hard equation!

Debye-Hückel is valid only for very dilute solutions, with very low ionic strength (not to mention that for a neutral species Debye-Hückel predicts $\gamma = 1$). Activity coefficients larger than one typically appear only at high ionic strengths - characteristically quite larger than even seawater's. It is simply not possible to use a treatment of activity coefficients that is 1) simplistic, 2) accurate, and 3) applicable to concentrated electrolytic solutions. You can try this for a step-by-step discussion, though.

Thank you for the link you have shared, I'm already making a comparison between the activity coeficient equations (Helgeson, D&H, Ex.D&H, Davies, Güntelberg), and as you said, D&H has a limitation of 0.01 M ionic strength,
I understood, that you suspect the values of activity coef> 1, as they are the same, 1.13.... I would say that if we use i.e. Davies equation for monovalent ions, it will lead to a same value as well as bi or trivalent.
The values (1.13) in the reference I've shared, is it possible to be attained using a certain equation?
Thank you Very much for your answers!


text:
π-backbonding in tungsten complexes
I need to find in which of the following complexes the M-CO bond is stonger:
[WO3(CO)3] and [W(CH3)3(CO)3].
According to my knowledge this is chiefly determined by the π-backbonding and how good donors or acceptors the present ligands are. CO is a great σ-donor as well as a decent π-acceptor while CH3- functions as a donor. Now apart from the final answer, I'm not sure about the oxygens. Is there a dioxygen ligand alongside with an oxo ligand? If that's the case I think the second compound would present a stronger M-CO bond since only the CO ligands can receive charge from the metallic centre. Is this approach correct?


text:
Which of the following molecules are enantiomers?
My book says that only molecule 7 is enantiomer of molecule 1. Why is this? Why can't molecule 1 and 2 be enantiomers?

Please use the built-in image uploader - we even have a special server for the images! 1 and 2 are certainly not enantiomers. Only one chiral centre out of three differs in configuration. Mirror images should have opposite configurations for all chiral centres, as you cannot reflect one chiral centre without reflecting the other two.

How about 1 and 7? Isnt there also just one different chiral centre (carbon attached to chlorine)?

Ooh... my earlier explanation isn't the best, I apologise. The issue is that those apparently chiral centres on the cyclohexane ring aren't truly chiral. Maybe the best way to see it is if you rotate the cyclohexane ring by 180° along a vertical axis. You'll find that the CH2OH at the top, which was originally pointing below the plane, is now coming out of the plane - an apparent "inversion" of configuration, but you haven't actually reflected it. Therefore the absolute configuration at these centres doesn't matter (but the relative configuration, cis/trans, does).

one note: 5 and 7 are identical

Using orthocresol's suggestion, try 1 and 5 are identical. Either one is the mirror image of 7. Structures 2, 3, 4 and 6 have no enantiomers present.

1 and 2 enantiomers? Maybe you confuse enantiomers with diastereomers and/or epimers.


text:
Rate of a reaction
We know that the rate of a reaction $\ce{aA + bB -> cC + dD}$, the rate of the forward reaction is given by $r_\mathrm f = k_\mathrm f[\ce A]^p[\ce B]^q$ where $ a\neq p$ and $ b\neq q$ according to Chemical Kinetics.
However when studying Chemical Equilibrium, $\ce{aA + bB <=> cC + dD}$ when we write the forward and backward reaction rates and equate them, we write them as
$r_\mathrm f = k_\mathrm f[\ce A]^a[\ce B]^b$ where the exponents are equal to the stoichiometric coefficients.
How is this the case?

In your example for equilibrium there is a key assumption which you were either not taught, or glossed over. It is the assumption that both forward and backward reactions are elementary, which makes the use of the stoichiometric coefficients valid. For a generic multi-step reaction the exponents are not necessarily equal to the stoichiometric coefficients.

We are actually taught equilibrium a year before kinetics. So will a non elementary multi step reaction never achieve equilibrium OR will we define the new $K_c as K_c = \frac{[C]^w[D]^x}{[A]^y[B]^z}$ where $w,x,y,z \neq$ stoichiometric coeffecients?

(1) Of course it will reach equilibrium. Everything does! (2) the equilibrium constant $K$ is still defined using stoichiometric coefficients. It is just that you can no longer derive it in that fashion, by saying that $K$ is a ratio of rate constants. There are other, more rigorous, ways of deriving it. This definition of the equilibrium constant using stoichiometric coefficients is necessary to fulfill all the other thermodynamic relationships, such as $\Delta G^\circ = -RT \ln K$, see e.g. my comments here.

Sir, your comments lead to a lot many pages. Could you give me a link to the direct proof of the above expression. I don't have the Peter Rock's book.

You should look in a proper physical chemistry textbook, it can be somewhat involved. https://chemistry.stackexchange.com/a/42532/16683 is the standard proof but if you are new to the topic then you will find it to be lacking in detail. If you don't know what activities or chemical potentials are, or how they relate to $\Delta G$, then I suggest coming back to it at a later stage in your education. Or, of course, reading a physical chemistry textbook, which should explain all these concepts.

Yup I think that you are correct. I don't understand a thing. Thanks for the answer.


text:
How is Volume an Extensive Property?
Based on the following definitions, can anyone explain why volume is an extensive property?
Extensive Property: Properties that depend on the amount of matter in a sample.
Volume: the amount of space an object occupies.
It has clearly been shown that volume depends on the AMOUNT OF SPACE and not the AMOUNT OF MATTER (that is mass!). So why do textbooks describe volume as being an extensive property?
Note: I do not believe it is an intensive property either, such as color. I am just wondering why it is labeled an extensive property?

Welcome to Chemistry.SE!  Please take a minute to check out the help center and tour page to better understand our guidelines and question policies.

Think of it this way. Extensive properties are like price: you take twice as much, you pay twice as much. This is clearly the case with volume.

The definition is very short, imprecise and incomplete actually imo. An extensive property is directly proportional to he amount of matter, at constant conditions. Twice the mass, twice the volume.

"It has clearly been shown that volume..." this has not been shown, it's is the definition of volume. If two definitions clash, then one of them must be illogical. Or in this case, misunderstood. ;-)

I'm voting to close this question as off-topic because it's about elementary logic applied to physics.

For solids and liquids volume occupied is strictly proportional to mass
Liquids and solids occupy volume. This volume is proportional to the amount of substance (or mass). The volume is determined by the mass and the constant of proportionality is the density which is specific to the substance. Density, mass and volume are all directly related for a given substance (and could be thought of as being different ways to measure the same thing). Therefore all are extensive properties.
This isn't true for gases as they fill whatever container they are in and the volume is not a measure of how much stuff of the substance there is.

If we maintain a constant external pressure, however, then extensivity holds for gases also.

@a-cyclohexane-molecule But, of course, the volume of a gas at constant pressure depends on the number of moles of the particles but not the mass of the particles.

Only for ideal gases. In real gases, volume depends on more factors, but it is still true that volume is proportional to mass for a given gas under certain $P$, $T$ conditions.


text:
How do you handle manganese nitrate hexahydrate?
So I wanna do an experiment using manganese nitrate and I bought manganese (II) nitrate hexahydrate $\ce{Mn(NO3)2.6H2O}$. But when I received it, it was kinda watery already (most part is still crystallized but there was also liquid in the bottle). After inspecting the bottle, it says that I need to refrigerate at 5 degree Celsius. But after I refrigerate it becomes icy hard, I cannot take anything out. So I am now wondering how to handle this compound. I am afraid if I use the liquid part it will have various concentration, and it is barely impossible to use the solid part because it's so hard. Thank you.

As long as you can you DON't use nitrates of multicharged cations, period. Most of them are extremely hygroscopic and cannot be weighted easily. If you really need to use nitrate solution, the best way is to prepare a solution and obtain the exact concentration using titration. Than said solution with known concentration can be stored in a closed bottle/flask and used as needed.

Manufacturers of chemicals can make life interesting for first-time users. What a boon the CRC Handbook is! Without even listing $\ce{Mn(NO3)2.6H2O}$, the CRC Handbook can solve your problem. It gives the melting point of $\ce{Mn(NO3)2.4H2O}$ as $\pu{25.8^\circ C}$. Undoubtedly, the melting point of the hexahydrate is even lower, but maybe not by much. In hot water, the tetrahydrate is "infinitely" soluble.
Heat your jar of hexahydrate to 30C in warm water until all the crystals are melted and withdraw what you need with a pipette. You might plan ahead and fill several small bottles with the hexahydrate for later use. I doubt that the material needs to be stored cold.

Manganese nitrate hydrate can be converted into anhydrous manganese nitrate by using a dehydrating agent like phosphorus pentoxide or dinitrogen pentoxide. First gently heat the hexahydrate around 100-110°C in vacuum to form the dihydrate and then heating around 80-90°C with the dehydrating agent to get the anhydrous product.
\begin{align}
\ce{Mn(NO3)2•6H2O &->[110°C,vacuo] &&Mn(NO3)2•2H2O + 4H2O}\\
\ce{Mn(NO3)2•2H2O + N2O5 &->[80-100°C,vacuo] &&Mn(NO3)2 + 2HNO3 + H2O}
\end{align}
The mechanism of this reaction is found in this e-book:

Anhydrous nitrate is mostly prepared by dehydration of solid hydrate
  in vacuum dessicator at room temperature over phosphorus pentoxide or
  dinitrogen pentoxide.[...], a solvate is formed $\ce{Mn(NO3)2.N2O4}$
  and by heating at 90°C, the dinitogen tetroxide can be removed to give
  the unsolvated compound. Thermal decomposition of the hydrated
  manganese(II) nitrate in vacuo also yields the anhydrous salts[...]

But you should be careful if you want to direct heat the hexahydrate. Overheating may cause the hexahydrate to fully decompose into manganese dioxide.


text:
Aromaticity of heterocyclic compound
This is a question of JAM 2017. According to the answer key the answer is 4. But I found only the last 3 as aromatic . So which one is the 4th aromatic compound? Please explain

Are you sure about the boron compound?

No Sir..I was wrong... It's non aromatic.. I got the correct answer . 1,2,4 and 5. @waylander

Actually not all of the last three are aromatic.  Look carefully at the third compound; the heteroatom is boron, not nitrogen, and so does not contribute the electron pair that would be needed to make six pi electrons in an aromatic ring.  That compound  is instead antiaromatic.
So JAM 2017 is apparently rendering both of the first two compounds as aromatic.  This, however, is dubious.  The first compound in the list is a pyrone, which not everyone would call aromatic.  To make an aromatic contributing structure in pyrone we would have to render the carbonyl group as its zwitterionic structure ($\ce{C^+-O^-}$) and then distribute the positive charge around the ring which has an electronegative oxygen atom.  We'd have to weigh the favorable effect of aromaticity against the unfavorable effect of putting a positive charge into an electronegative ring.
As for the second compound, we would need to "polarize" two carbonyl groups and concentrate two positive charges into the small ring, and that is not happening even with aromaticity.  There is a saving grace, however: if the $\ce{N-H}$ protons are tautomerized onto the oxygen atoms, converting the "keto" structure to its "enol" form, then the tautomerized molecule is 2,4-dihydroxy-1,3-diazine, which would be unambiguously aromatic.
What it boils down to is that aromaticity is often not the simple yes/no question often suggested in introductory chemistry texts.  Given this ambiguity, one could potentially defend an answer of 2, 3, or 4 (but not the last three for the two larger numbers) to the JAM 2017 question.

Actually in keto form of uracil it's maybe second most important mesomeric structure is the one with + on nitrogens and is aromatic.

Curious about the source.  Wikipedia does not refer to a carbonyl polarized structure (but does indicate that both tautomerization and stability of the given structure without Huckel aromaticity occur).

Meh, it's standard "amidic" stabilisation, but it does lead to aromatic mesomeric structures. also even full 2+ charge on aromatic cycles is possible. Ron made an answer about it, mayybe wold findd it...

https://chemistry.stackexchange.com/questions/28882/determining-aromaticity-of-compounds


text:
Predicted number of signals in proton NMR spectroscopy
Given the molecule,.
I can predict that there will be 4 signals present in the NMR spec. Problem is, the quiz says 5 is the correct answer. My question is, are alkene hydrogen not chemical-shift equivalent?

No, they are not. They have different relationships with the rest of the molecule.

Does that mean it shows up as two separate peaks?

Two 2J doublets, each further split up by allylic 4J couplings. They would have different reactivity in chemical reactions, so they obviously cannot be equivalent! CS and reactivity both are directly related to the local electron density distribution.

Yes, the two alkene hydrogens are different. As to whether they are readily resolved may depend on the magnitude of the field of the NMR.

@user55119 Ja. Typical setup for an AB spin system.

Since there isn't free rotation around the C=C bond in the alkene, the two hydrogens, Ha and Hb are experiencing different electronic environments. Whenever two hydrogen atoms are experiencing non-identical electronic environments, they are going to possess differing chemical shifts in proton NMR. If you heat this compound, and it doesn't decompose, I would expect the two signals from Ha and Hb to coalesce into one as the hindered rotation about the alkene C=C bond becomes unhindered.
https://chem.libretexts.org/Textbook_Maps/Organic_Chemistry/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_(Soderberg)/Chapter_05%3A_Structure_Determination_II/5.2%3A_Chemical_equivalence

Is the lack of free rotation due to double bonds having both Siqma and Pi bonds, essentially "locking" the hydrogens in place?


text:
What happens to chromium (as Cr2O7-) in the plasma of the ICP?
I have recently been analyzing some biological samples for chromium using a standard Cr(VI) solution for external calibration.
Cr standards are typically supplied in a 2% nitric acid matrix. That concentration corresponds to a pH of 0.44. Under that pH, most species of Cr would have oxidized into the form of $\ce{Cr2O7-}$ or $\ce{CrO4-}$; yet in literature, all measurements carried out on ICP-MS for Cr have been in the +1 oxidation state.
Could this be from the rapid reduction from the free electrons in the argon plasma?


text:
What are standard controllers for automatic pH control?
My background in control theory comes from electrical engineering (robots, motors, etc.) and I want to build a control system for automatic pH control.
Can I directly apply my knowledge from classical control to chemical control? E.g. can I use a PID controller in a similar way as I am use to, or are there special pitfalls/best practices in chemistry that I should be aware of?

Short system description: My plan is to have two pumps that can add either base or acid to a solution under constant agitation (shaker plate), using a standard glass electrode sensor for measurements.

Welcome to Chemistry.SE!  Please take a minute to check out the help center and tour page to better understand our guidelines and question policies.

Certainly. Thermometer and heater, or pHmeter and a pump each for acid and base, the difference is just that you can fire in two directions in the latter case, which should make the control circuit easier to balance.

I do have 2 pumps, one for base and one for acid

Automatic pH control is not that easy as is shown by the fact that automatic titrators, the equipment that you want to build, are relatively expensive.
Besides titration, automatic titrators usually have also a pH stat function, which is for maintaining a certain fixed pH.
In first place, note that to  have a good control, typically you must know which type of system you are working with.
Normally, to maintain a constant pH only one reagent is added, either acid or base.
This is because chemical systems typically become more acidic or more alkaline when they proceed, but do not oscillate.
When you add both reagents, they usually start to compete: a bit of acid is added, it passes the set point, base is added, it passes the set point in the other direction, etc.
This happens especially when the response of the system is slow and the mixing is not good (shaker plate is not a good idea, use a stirrer instead).
The way to solve this is by adjusting very well the concentrations of the reagents, the delay times, addition rates, etc., and this is different in every system.
You also should know in which part of the pH curve you are working to set the concentration of the reagents and the rate of addition.
The pH curve has three very different zones, a vertical part and two horizontal parts.

Depending on where you are, the conditions must be different.
Note also that the inflection point becomes less obvious when weak acids or bases are involved.
Finally, you might need to integrate also a temperature compensation since, depending on the $\Delta\epsilon$ you have set, the pH change with temperature will be to high.
In short, a PID might work in some specific system, but for a general purpose system you will probably need some more sophisticated control like a PLC.


text:
NaOH being added to Formic Acid (Please catch my error!)
I am adding 0.55 mL of 0.1M NaOH to 10 mL of 0.5M Formic Acid.
Here are the values I calculated to use.
Moles H+ : 0.5 * (10mL / 1000) = 0.005 mol
Moles OH- : 0.1 * (.55mL / 1000) = 0.000055 mol
volume : (10mL / 1000) + (0.55mL / 1000) = 0.01055 L
Moles Total : moles H - moles OH = 0.005 - 0.000055 = 0.004945 mol
M Total : moles Total / volume = 0.004945 mol / 0.01055 L = .46872 M
Moles Salt : 0.000055 mol
M Salt : moles salt / volume = 0.000055 mol / 0.01055 L = 0.00521 M
Calculations to find the pH
pH = -log(1.80E-4) = 3.745
3.745 - ( -log(M Salt / M Total) ) = 3.745 - 1.95407 = 1.791
However, this value is too low as the lowest pH of 0.5M Formic Acid is ~2.03.
If the volume of NaOH were to say, 0.5 mL, then the M Salt / M Total value would be less than 1/90, and the pH could then be found by
-log( sqrt( 1.80E-4 * M Total ) ) = 2.036
...which makes more sense, but doesn't apply to the initial problem.
Any idea of where I'm going wrong?

It looks like you misused the acid dissociation constant, which is apparently the $K_a=1.80×10^{-4}$ figure you have.
Properly, you should do the following:
1)  Let $x$ be the hydrogen ion concentration $[\ce{H^+}]$ from the formic acid that actually disdociated.
2)  Since you ended up with $0.469 \text{M}$ total acid the un-dissociated acid $[\ce{HCOOH}]=0.469-x$.
3)  The formate ion comes from both the salt ($0.00521 \text{M}$) and the dissociated acid ($x$), so $[\ce{HCOO^-}]=0.00521+x$.
4)  Plug these into the equilibrium relation
$\frac{[\ce{H^+}][\ce{HCOO^-}]}{[\ce{HCOOH}]}=K_a$
and solve for $x$.  The equation reduces down to a quadratic equation so you should have an easy formula for $x=[\ce{H^+}]$.
5)  Get the pH from the result of Step 4.
Good luck!

Thank you! This was just what I was looking for.
This method can always be used to find the pH, correct?

It's for the pH of an acidic solution, actually.  If you have a basic solution (like patrially neutralizing ammonia with hydrochloric acid), use the above method, with base instead of acid, to find the hydroxide concentration first, then use the water dissociation product (typically rendered as $[\ce{H^+{aq}}]×\ce{OH^-{aq}}]=10^{-14}$) to get the hydrogen ion concentration.


text:
How can I interpret the CH4+ structure?
I am trying to teach myself about Electron Ionization, and am very confused about the formation of the CH4+ Ion.

Do the dotted lines between the C and the H, and between the two Hydrogens indicate a resonance structure of sort? I have taken organic chemistry, but cannot wrap my head around this.

https://chemistry.stackexchange.com/questions/68849/how-does-a-5th-hydrogen-bind-to-carbon-atom-in-ch5 https://chemistry.stackexchange.com/a/32326/9961

Dotted lines indicate three center bonds, which can be viewed as a sort of resonance.

The dotted lines in this structure could be thought of as partial bonds. In other words, none of those three bonds has a bond order of 1 with two electrons attributable to that bond.
Another way to look at this structure is that the dashed lines represent multicenter bonding. A common example of multicenter bonding is diborane $\ce{B2H6}$. The structure of diborane features 3-center-2-electron bonds (often abbreviated 3c2e bonds - in this notation, a typical single sigma bond is a 2c2e bond) with two hydrogen atoms shared by both boron atoms:

These bonds in $\ce{CH4+}$ could be considered a 3-centered-3-electron bond as originally each C-H bond would have had two electrons. The radical cation is one electron short, so there are only three electrons shared between those three atoms.

Afaik there's no 3c3e - there's normal 3c2e in CH4 cation and unpaired electron. https://www.ncbi.nlm.nih.gov/pmc/articles/PMC23401/


text:
What can possibly turn litmus paper white if there is even no chlorine present?
I did a hydrocarbon test on cyclohexane with bromine solution. The mixture was put inside a dark place. To be exact, a cupboard filled with laboratory equipments. I also put a piece of wet blue litmus paper on top of the tube.
It should've stays blue because there was no reaction occured, right? But the litmus paper turned white. What can possibly cause it?

Perhaps, bromine also has bleaching capabilities?

Chlorine turns litmus paper white by oxidizing it.  There are, of course, other oxidizing agents (like bromine) that can do the same thing; chlorine is just the best known because we associate it with commercial bleaches.


text:
Expression for fugacity coefficient derived from a pressure-explicite EOS
In the book, Thermodynamics for Process Simulation, the authors propose to derive an expression for the fugacity coefficient from a pressure-explicit equation of state, as an example among many others.
Assuming, $P = P(T, v)$ or $z =\frac{Pv}{RT} = z(T, v)$, the calculations lead to,
$$\left(\frac{\partial \ln \phi}{\partial v}\right)_T = \left(\frac{\partial z}{\partial v}\right)_T-\left(\frac{\partial \ln z}{\partial v}\right)_T + \left(\frac{1}{v}-\frac{P}{RT}\right) \tag1$$
where $\phi$ is the fugacity coefficient.
What I don't understand is, when they integrate, they get,
$$\ln \phi = z - 1 - \ln z + \frac{1}{RT}\int_{\infty}^v \left(\frac{RT}{v}-P\right)\mathrm{d}v \tag2$$
Can someone explain me how they find $z-1-\ln z$ by integrating the first two terms of $(1)$ between $\infty$ and $v$?

When integrating the first two terms, it leads to $$\left[z - \ln z \right]_{\infty}^v$$ which I understand, but then I don't get the final result.
I suppose it's linked to the values of $z$ when the volume is very large and when the volume is "$v$" but it does't make sense for me.
I would have assume to get $$\left[z - \ln z \right]_{\infty}^v = v - \ln v - \left[z(v^{\infty}) - \ln z(v^{\infty}) \right]$$ and nothing else.
But I am missing something probably obvious.
Thank you in advance for your help!

When $v\rightarrow \infty$ you get the ideal gas eos, so $z\rightarrow 1$. This gives you the result, i.e. $1$ at the lower limit (subtracted) and $z-\ln z$ at the upper one.

Mindful of the guideline "don't give answers in comments", I've converted my comment to an answer, trivial though it is.
When $v\rightarrow \infty$ you get the ideal gas eos, so $z\rightarrow 1$. This gives you the result, i.e. $1$ at the lower limit (subtracted) and $z-\ln z$ at the upper one.
When you write down the integrated form, $z-\ln z$, you need to remember that it means that you are evaluating the function $z(v)$ at the two limits $v=\infty$ and $v=v$, and using those values in that expression; you should not be trying to set $z=v$ at the limits, which seems to be (partly) what you have written at the end of your question.
Edit following OP comment.
More details. At constant $T$, $z=z(v)$ is a function of molar volume $v$,
determined by the equation of state, $P(v)$ at the given value of $T$. This equation is unknown, but we can be sure that,
in the ideal gas limit $v\rightarrow\infty$, $z(v)\rightarrow 1$.
Integrating both sides of the equation from the ideal gas limit to
the desired volume $v$, and using $v'$ as the integration variable, gives
eqn (2) of the question in its full form:
$$
\left| \ln \phi(v')\right|_{v'=\infty}^{v'=v}
= \left|  z(v') - \ln z(v') \right|_{v'=\infty}^{v'=v}
+ \frac{1}{RT}\int_{\infty}^v \left(\frac{RT}{v'}-P(v')\right)\mathrm{d}v'
$$
I think it's clearer to distinguish between the integration variable $v'$
and the "upper" limit of integration $v$,
but many people would be happy just to use $v$ instead of $v'$.
Anyway,
on the left,
we know that, in the ideal gas limit,
the fugacity coefficient $\phi(\infty)=1$,
so $\ln\phi(\infty)=0$ and we are just left with $\ln\phi(v)$.
On the right,
similarly,
we substitute in the upper and lower limits for $v'$.
We know $z(\infty)=1$, so
the function being evaluated is $z(\infty)-\ln z(\infty)=1$ at the lower limit,
and $z(v)-\ln z(v)$ at the upper limit.
So the final answer is
$$
\ln \phi(v)
= z(v) - \ln z(v) - 1
+ \frac{1}{RT}\int_{\infty}^v \left(\frac{RT}{v'}-P(v')\right)\mathrm{d}v'
$$
An important point is that the integration variable
is the molar volume $v'$ (or $v$ if you prefer), not $z$.
In evaluating the result, the integration limits $v$ and $\infty$
are substituted for $v'$, the argument of the function being evaluated.
It is incorrect to set $z=v$, or $z=\infty$.
(This should be even more clear in this case,
since $z$ is a dimensionless quantity, whereas $v$ is not).

Why shouldn't I try to set $z = v$ ? I am just reasoning as I always did using integration. And I understand now what you pointed out that I didn't get but then if $z \rightarrow 1$ why not to write $[z - \ln z]_{\infty}^{v} = [v - \ln v] - [1 - \ln 1]  = v - \ln v - 1 + 0$ ? Because then it doesn't make sense to subtract two infinities. I just don't understand why it is then correct to assume the relation with the $z$ instead of $v$ correct.

You are not integrating the variable $z$ between two limits $\infty$ and $v$; you are integrating a function of $v$ between these limits. And actually, you already did the difficult part, namely the integration, so you just need to evaluate the resulting function at those limits. In this case $$\left | z(x)-\ln z(x) \right |_{x=\infty}^{x=v}$$ where I'm deliberately using a dummy variable $x$ for clarity. If this is still not clear, I can edit my answer to go into more detail.

I've added more detail to my answer anyway. Let me know if things are still not clear.

Thank you for your answer. I'm still a little confused but I will work on it and I will probably get it.


text:
Can we oxidize aliphatic alkane to produce carboxylic acid?
Toluene can be oxidized to form benzoic acid. do aliphatic alkanes give similar kind of reaction?

Roughly speaking, aliphatic alkanes don't have any reactions other than combustion.

Not in any useful manner

It's tricky, because the bond enthalpies of sp3 C-H bonds are high (so they are hard to break) and because alkanes typically have a large number of CH bonds so selectively oxidizing any one of them can be challenging. Generally speaking the more electron-rich CH bonds (i.e. tertiary positions) are oxidized more easily that less-substituted, less electron rich ones.
TFDO will do it, see here: https://en.wikipedia.org/wiki/Oxidation_with_dioxiranes and there are a bunch of other methods (just try googling "aliphatic CH oxidation").
It's quite a hot area of research, since if you can control it then in theory you can turn feedstocks from crude oil into more useful functionalized molecules.


text:
Why is carbon monoxide not a greenhouse gas?
Carbon monoxide is considered to be a toxic gas and, given that it is usually found along with $\ce{CO2}$, why it is not considered a greenhouse gas?

A burger is considered to be a fatty food and it usually comes with a coke, then why is the burger not a soft drink?  Correlation between CO2 and CO, and the fact that CO is toxic, have no bearing at all on whether it is a greenhouse gas. I think the answer is because CO has too short a lifetime to contribute directly (see https://en.wikipedia.org/wiki/Greenhouse_gas#Non-greenhouse_gases) but I'll leave it to an atmospheric chemist to explain properly rather than just citing wikipedia...

@PCK - The absorption/emission spectra of the two are different as well...

$\ce{CO}$ is not considered a primary (or significant) greenhouse gas due to the weak (but non-zero) absorption of energy in the infrared. However, it does increase global warming by reacting with certain chemical species in the atmosphere which in turn lead to an increase in concentration of primary greenhouse gasses, most notably methane and ozone (see final paragraphs cited below). Thus, it is not completely accurate to dismiss it as a contributing factor.
From the National Oceanic and Atmospheric Administration's page on $\ce{CO}$ as a greenhouse gas:

Carbon monoxide (CO) is not considered a direct greenhouse gas, mostly because it does not absorb terrestrial thermal IR energy strongly enough. However, CO is able to modulate the production of methane and tropospheric ozone.

Carbon monoxide does contribute to global warming indirectly, as noted by the American Chemical Society's page entitled "Which Gases Are Greenhouse Gasses?":

Several other gases can have indirect effects on atmospheric warming. This occurs when chemical reactions in the atmosphere produce or destroy greenhouse gases, including tropospheric ozone. Indirect effects also occur when a gas influences atmospheric lifetimes of other gases or affects atmospheric processes like cloud formation that alter Earth’s radiative energy balance by increasing Earth’s albedo. Gases that can cause these indirect effects include carbon monoxide ($\ce{CO}$), oxides of nitrogen ($\ce{NO_{\rm x}}$) and non-methane volatile organic compounds (NMVOC).

The above is elaborated in the text Global Climate Change Linkages: Acid Rain, Air Quality, and Stratospheric Ozone (James C. White, Editor. Elsevier (1989), p. 106):

As has been noted, ozone also initiates reactions that generate the highly chemically active oxidants, including hydroxyl. The lifetime of radiatively active methane in the atmosphere is determined by hydroxyl abundance. Carbon monoxide and methane are both sinks for hydroxyl, $\ce{CO}$ being the more reactive of the two. Greater concentrations of carbon monoxide in the atmosphere lead to lesser concentrations of $\ce{HO\cdot}$, which in turn lead to higher concentrations of methane.
Carbon monoxide and ozone are linked in the non-urban atmosphere that is low in hydrocarbons through the interaction with $\ce{HO\cdot}$, as described above, and $\ce{NO2}$. Hydroxyl is destroyed by $\ce{CO}$, increasing the ambient concentration of $\ce{NO2}$. The resulting increased proportion of $\ce{NO2}$ to $\ce{NO}$ leads to a higher steady state concentration of ozone. Carbon monoxide, a non-radiatively active species, enhances the greenhouse warming by raising the levels of methane and ozone.

In summary: $\ce{CO}$ is not a direct greenhouse gas due to low absorption in the infrared but is an indirect greenhouse gas because it increases the concentration of direct greenhouse gases (methane and ozone most prominently) by reacting with atmospheric $\ce{HO\cdot}$, which otherwise would decrease the concentration of those gasses. The toxicity part is not germane to the greenhouse gas issue.


text:
Why are Colloids Heterogeneous?
I understand that the definition of a heterogeneous mixture is a mixture in which 2 or more phases are present, i.e. you can see the mixture is not uniform. I know a homogeneous mixture is a mixture in which only one visible phase is present, but why are suspensions and colloids considered heterogeneous, for you only see one phase as the definition states? I also know that homogeneous mixtures are usually referred to as solutions, but if only solutions are homogeneous then why do we define homogeneous as anything with 1 phase; isn't this misleading? Is there a more accurate definition of a homogeneous and heterogeneous mixtures?

You're taking "see" a bit too literately. For example milk has suspended fat globules which you can't "see" with your naked eye.

A solution of table salt (or any other low Mw solute) in water is surely homogeneous, because all inhomogeneieties have a similar (on a log scale) size and diffusion coefficient as the solvent molecules. For a polymer solution, you have to stretch this definition quite a bit already. For a colloidal solution even more.

I know a homogeneous mixture is a mixture in which only one visible phase is present, but why are suspensions and colloids considered heterogeneous, for you only see one phase as the definition states?

The question is about what is going on at the molecular level. In a homogeneous mixture, the components are mixed all the way to the particular level. One way to see this with your eyes is that these mixtures are clear, not turbid, and have no phase separation. Grape juice is a solution, while milk is not (it is turbid) and neither is a mixture of oil and water (it often has two colors, and if you mix it vigorously, it turns turbid).

I also know that homogeneous mixtures are usually referred to as solutions, but if only solutions are homogeneous then why do we define homogeneous as anything with 1 phase; isn't this misleading?

Solutions are not the only homogeneous mixtures, they also exist in the solid state, like certain alloys. In the gas phase (if the sample is small enough and you wait a bit), all mixtures are homogenous. So a solution is a subset of possible homogenous mixtures.

Is there a more accurate definition of a homogeneous and heterogeneous mixtures?

If you can take smaller and smaller samples (larger than a couple thousand particles, though) and you always get the same composition, it is homogenous, otherwise heterogeneous.


text:
How many equal Xe–O bond length are present in (XeO_6)-3?
I tried to solve this problem by drawing a structure but the structure did not give me any answer and I was not able to draw the proper structure please help me out.

what level of theory are you looking at? VSEPR? MO? I don't want to delve into anything MO (those stuffs that abuses hydrogen atom solution. meh. not a believer.).  With VSEPR, if you show what you have done so far, I can take a look.

@Argyll are you serious about MO theory? You don’t have to believe in it, but that would put you at odds with the majority of chemical literature published in the last few decades. If you’re joking, it’s hard to tell.

On topic, I have some suspicion that this odd-electron ion does not actually exist. $\ce{XeO6^4-}$ does, though.

This compound exists.

Protonation of perxenate ion yields $\ce{(HXeO6)^3-}$. See: https://en.m.wikipedia.org/wiki/Perxenate

!. Ions aren't compounds 2. Such ion/radical is most probably only transient intermediate in redox reactions. 3. If it's some task then you probably misunderstood or there was misprint.

Probably a misprint because I got it from a book

@orthocresol: majority of chemical literature????? you are overblowing what makes up MO theory here.

As for $\ce{[XeO6]^{3-}}$, it's not unlikely that the question asks for an answer based on MO even if the actual ion hasn't been found to exist.

@Argyll He actually doesn't, at least not much. MO theory is the most advanced model and most 'correct' model of bonding that doesn't require advanced math to use. As such, it is widely used, even if often it is used to provide 'downgraded' description of model of higher level of theory. Thus, absolutely every advanced general chemistry and advanced organic chemistry textbooks uses MO theory at some point.

@permeakra: And that is not the majority of chemical literature in past decades. If such a claim was true, I expect to run into MO theory frequently in whatever random selection of chemistry literature. But obviously not. My interest actually isn't in chemistry now. I used to skim through papers broadly when I explored what topics I may like. In fact, I didn't run into MO at all.

If such a claim was true, I expect to run into MO theory frequently in whatever random selection of chemistry literature. || you actually would run into it a lot more often than into VSERP, which is of mostly historic interest.

Most likely, the structure is a distorted octahedron. This means, that the electron structure of the compound can be derived from the orbital diagram of $\ce{XeF6}$ octahedron (see here What is the molecular structure of xenon hexafluoride?)
Electron count in $\ce{XeF6}$ and $\ce{[XeO6]^{3-}}$ differs by 3 electrons, so we have to remove three electrons from the diagram found in the reference. Thus, the HOMO are two orbitals of $e_\mathrm{g}$ symmetry populated by 3 electrons. When two orbitals of the same energy have different electron population, a Jahn-Teller distortion occurs. The orbitals of similiar symmetry with similar occupation can be found, for example, in $\ce{Cu^{2+}}$ compounds. They typically show a geometry of  distorted octahedrons with 2 opposite ligands significantly further from the central atom, than the 4 other ligands.
So, with quick and dirty treatment suggests that there are 2 different bond lengths in the ion $\ce{[XeO6]^{3-}}$.
VSERP is unlikely to be of use in this case.


text:
Can vicinal dihalide produce conjugated alkene in strong base?
In Klein's Organic Chemistry (3rd edition), p. 410, we learn that vicinal dihalide can produce alkyne in strong base. And on Reaxys I see many reactions of this sort. For example:

However, why isn't the conjugated alkene, which should be more stable, the major product under such conditions?
Somebody once gave me an explanation, to the effect of "the hydrogen geminal to the halide is more acidic, hence more easily eliminated." However, Smith's Organic Synthesis (4th edition), p. 121 says the following:

[In the following reaction] $\mathrm{H_a}$ is more acidic than $\mathrm{H_b}$ since it is attached to the less substituted carbon. The base should react with
  the more acidic hydrogen atom, which would lead to the less substituted alkene, 3-ethylhex-1-ene, but that is not the major product.

This suggests that we cannot base our logic on the acidity of the hydrogen atoms.

Comprehensive Organic Transformations (2nd edition) p. 467 gives examples of converting 2,3-dibromobutane to buta-1,3-diene.

On the other hand, Reaxys has many examples of producing alkynes from vicinal dihalides, even when potassium t-butoxide is used.
According to Carey's Advanced Organic Chemistry (5th edition, volume A), p. 554, the direction of E2 elimination is determined by whether the mechanism is more similar to E1 or to E1cb. If it is similar to E1, then the thermodynamically more stable product dominates; if it is similar to E1cb, then the direction is determined by the more easily removed hydrogen atom.
On page 556, Carey attributes the Hoffman rule of t-butoxide to its stronger basicity, not to its bulky size, and states that stronger bases drive the mechanism toward the E1cb side.
Therefore I think the problem might be understood this way: If a strong base such as $\mathrm{NaNH_2}$ is used, then the elimination direction is determined by which hydrogen atom is the easiest to remove. Since bromine acidifies its geminal hydrogen, the alkyne product dominates. If a weak base is used, the diene product might predominate.

I can't find a pka value for a simple alkyl bromide, but bromoform has a pKa of  about 14, while methane is ~50, so clearly having bromine atoms on an alkyl carbon does acidify it. So it looks like the premise of the argument in Smith is somewhat flawed.

Nonetheless, I'm guessing the text goes on to explain that E2 reactions have product-like transition states, and as a result the Saytzev (more-substituted) product is generally formed. So in this case, both the pKa argument and the alternative Saytzev selectivity argument give the same result - H(a) will eliminate preferentially over H(b).


text:
Is it possible for a bicyclo compound to have three bridges of equal length?
My teacher told me that the value of $x$ is always greater than or equal to $y$ and that y is always greater than $z$ in a bicyclo compound, where $(x,y,z)$ are the lengths of the bridges.

But in this compound $x = y = z$.
Can someone clarify?

That's when the "or equal" part steps in.

Do you mean that z is equal to or less than y?

Yes, it should read like that.

For the ignorant among us what are the definitions of X, Y and Z?

The compound is bicyclo (4,4,4) tetradecane I think. So the x,y and z refer to the numbers 4,4, and 4 which are the number of carbons in each part of the rings excepting the two "anchor carbons." So for a different bicyclo compound you'd write (4,3,2), going high to low, not any other combination such as (3,2,4).

Why not? The simplest bicyclic bridged compound, bicyclo[1.1.1]pentane, is a well-known example of such a molecule for its use in synthesizing the even more strained propellane. I believe your teacher was trying to prescribe the rules of bicyclic nomenclature, i.e.

In between the prefix and the suffix, a pair of brackets with numerals denotes the number of carbon atoms between each of the bridgehead atoms. These numbers are arranged in descending order and are separated by periods.


text:
Production of oxirane mechanism
Regarding oxirane production developed by Wurtz, is there an accepted mechanism?

I carried out a research but couldn't find anything. My approach was this
Is that correct?

The only additional clue, I've found, is that heat must be applied for the reaction to take place.

Deprotonate using hydroxide before doing an intramolecular SN2

I see. So there is no way something like what I suggested could take place, right?

You should not make a primary unstabilized carbocation unless you have a very good reason to do so. The short answer here is, no, I don't believe this is a likely mechanism for the reaction.

https://chemistry.stackexchange.com/questions/81896/stereochemistry-synthesis-of-an-epoxide

If you were thinking of running the reaction using Wurtz conditions with elemental sodium then the sodium will react with the OH preferentially.

authors: Αντώνιος Κελεσίδης


text:
Stable oxidation state for Thallium and indium
The first Ionisation Enthalpy for Thallium is ~ 590 KJ/mol whereas that for Indium is ~ 558 kJ/mol. But why is +1 oxidation state more stable for Tlthan In?
What are the other criteria that explain the stable +1 oxidation state of Thallium?

What happens to $\ce{In+^1}$? How is it unstable?

In+1 readily oxidises to In+3 .


text:
How to purify sodium chloride table salt that contains sodium ferrocyanide
I wish to obtain pure sodium chloride crystals from ordinary table salt.  But all proprietary table salts seem to contain the anti-caking agent sodium ferrocyanide which I should like to remove.

If you're looking for high purity salt, don't buy table salt. Buy it from a chemical supplier, for example: https://www.sigmaaldrich.com/catalog/product/sial/s7653?lang=en&region=US

Dissolve and recrystallise.

And it's supposedly soluble in acteone. Or so wikipedia says. Just wash it out.

Pass a salt solution (uniodized table salt) through a column packed with an anion exchange resin in the chloride form (e.g. Dowex 1x8). The ferrocyanide ions will be replaced by chloride via mass action. This mild process will not release cyanide, as the ferrocyanide ion is very stable.
At some point during such use, the resin will start to become saturated with ferrocyanide and would have to be regenerated back to the chloride form. This would require chemically pure salt, if course.
I agree with Zhe that it would probably be easier and cheaper to just buy pure reagent grade sodium chloride. But the principle is worth illustrating in case you have a special reason to purify it yourself.

Don't.  Ferrocyanide is a good way of tying down cyanide ions so they don't do their usual bad things to you.  Trying to purify the salt could release the cyanide ions.

How should that happen, except by adding strong acids?


text:
Why do lattice structures form into powders?
For example, when you react magnesium with oxygen over heat, it forms a lattice of ions. Why does a strong lattice of ions like MgO turn out to be a powder, rather than a solid block?
(Sorry if the answer is obvious; I'm only just being introduced to this)

Each particle of the powder is a big solid block indeed.

And you can make a solid block - it is all a question of how it is processed.

I'm voting to close this question as off-topic because it has nothing to do with chemical properties.

This is simply because the combustion of magnesium doesn't occur orderly enough to create one big solid block of magnesium oxide.
When magnesium reacts with air, the way of formation of the oxide lies at the mercy of the chaotic motion of oxygen due to it's entropy in the air. Often, the molecules react simultaneously at various locations in the air around the part of the ribbon undergoing combustion, creating several micro-crystals, which fall down due to gravity on reaching a certain weight. These micro-crystals appear as powder in the macroscopic scale.


text:
Heat of combustion among substituted cyclohexanes

Arrange in ascending order of heat of combustion:

I know heat of combustion is inversely proportional to stability. Also, to solve I did try making chair conformations of the substituted cyclohexane.
In same type of, say, 1,3,5-conformations one can do comparisons by seeing equatorial axial positions, as bulky group tend to be in equatorial positions. But how to compare and arrange when there is 1,2,3 and 1,2,4 and 1,3,5 substitutions?

Ashish: Welcome to SE! You are quite correct that heats of combustion and formation are inversely related. Since all of your compounds are trimethylcyclohexanes (C9H18), their sum will be equal to the heat of combustion of 9 moles each of graphite and hydrogen. To approach the solution of this problem, we can start with methylcyclohexane and evaluate the effect of adding ultimately two more methyl groups. In the somewhat busy diagram below, all numbers are in kcals/mol. Known heats of formation (NIST) are located inside the ring. Calculated heats of formation (ΔHfo) are in red and values over the arrows are the incremental changes in heats of formation (ΔΔHfo). The more stable conformation of all compounds is assumed. Equatorial (e) and axial (a) assignments are made.
There is no methyl/methyl interaction in the formation of 1,4-isomers A (cis) and B (trans) or 1,3-isomers E (cis) and F (trans). The addition of an equatorial methyl (B and F) is worth -7.0 and -7.1 kcal/mol, respectively, while addition of an axial methyl (A and E) has a value of -5.2 kcal/mol. Note that the difference between an axial and equatorial methyl is worth 1.9 kcal mol. The axial methyl necessarily makes the ΔHfo value less negative than addition of an equatorial methyl. However, there are gauche butane interactions in 1,2-isomers. Because C is less stable than D, the addition G --> C (cis) is -4.1 kcal/mol and G --> D equals -6.0 kcal/mol.
Using a value of -7.0 kcal/mol for the addition of an equatorial methyl, the conversion G --> R gives a value of ΔHfo = -51.0 kcal/mol [NIST = -51.48 kcal/mol]. Note that 1,3,5-isomer U's value is what is expected given the difference between a non-interacting equatorial vs. axial methyl group.
The reader should be able to corroborate the values assigned to P, Q, S and T. The stability (heats of formation) of P through T are R>S>T>Q>P. Therefore, the heats of combustion are in the order R<S<T<Q<P.

The order of stability is
R>T>S>Q>P
Then the heat of combustion will be
P has the highest heat of combustion and  R the less.
The difference in  kj/mol are:
If R is 0 then T has 5 more kj/mol, R 14 kj/mol, Q 20 kj/mol and P 34 kj/mol.
This are due to the steric strains.

Do you mean "S 14 kj/mol"....


text:
PCl3 vs SCl2 boiling point
Why does sulfur dichloride have a lower boiling point than phosphorus trichloride? Is it to do with the higher number of electrons causing stronger id-id interactions in PCl3, or do we need to compare pd-pd interactions also?

I think you're looking down the wrong path to explain this. Boiling point can often be explained by looking at the interactions between molecules such as hydrogen bonding, dipole-dipole interactions, and london forces, as well as overall molecular mass. I would explain this phenomenon by looking at two data points for each molecule. The dipole of $\ce{SCl2}$ is 0.4 Debye and the dipole of $\ce{PCl3}$ is approximately 0.780 Debye.$\ce{^{1,2}}$ Based on  this alone, I would expect $\ce{PCl3}$ to have a higher boiling point. If you also look at the mass of each molecule, 102.97 g/mol for $\ce{SCl2}$ and 137.33 g/mol $\ce{PCl3}$, it becomes pretty clear which substance one should predict to have the higher boiling point. It's $\ce{PCl3}$.
Ref 1. http://openmopac.net/PM7_accuracy/data_molecules/sulfur%20dichloride_jmol.html
Ref 2. https://cccbdb.nist.gov/diplistx.asp


text:
How do I obtain Chrysoidine from Benzene?

So, basically I can convert Benzene to Phenol using Dow's synthesis process and then use $\ce{NH3}$, $\ce{ZnCl2}$, $\ce{HNO2}$, $\ce{NaNO2}$ under 0-5 °C controlled temperature and convert the phenol to Diazonium Chloride Salt. Then with Coupling Reaction, I can convert the Diazonium Chloride to Aniline Yellow. But, Chrysoidine has another $\ce{-NH2}$ group attached to the Ortho position. So, an explanation would be greatly appreciated as to how to obtain Chrysoidine from there on.

why go via the phenol? Easier surely to nitrate benzene then reduce to aniline.

@Waylander Yes, I know that pathway to Aniline as well. However, I wanted to know whether Aniline Yellow, upon reacting with Nitric Acid and then being reduced by Sn and HCl, yields Chrysoidine. Or will the -NO2 group just get attached to the Meta position?

I think aniline yellow is likely to nitrate ortho to the aniline giving you the wrong structure. Also the Sn/HCl reduction may well reduce the azo double bond.
Here is a route from benzene.
From benzene a simple nitration gives nitrobenzene. Divide your nitrobenzene, reduce one portion to aniline by hydrogenation over Pd or Pt catalyst. Nitrate the second portion to 1,3-dinitrobenzene procedure here. Reduce the 1,3-dinitrobenzene by hydrogenation over Pd or Pt catalyst to give m-phenylenediamine. Diazotise your aniline to the diazonium and react this with the phenylenediamine. This route was used industrially to prepare Chrysoidine in the 19th century reference here.

My gratitude, sir. This helped a lot. :D

Please accept the answer if you are happy with it.


text:
What's the effect of temperature on buffering capacity?
I did some research online but I still couldn't get any conclusion. Some said that the buffering capacity decreases when temperature increases because of the increase in ionisation of the weak acid molecules and therefore, some of the conjugate base ion will be used to neutralised the $\ce{H+}$ ion from the ionization of the weak acid molecules. Hence, only a small amount of strong acid is needed to complete remove the conjugate base ion and to break the buffering capacity. But, the problem is that the acid molecules ionise into both $\ce{H+}$ and the conjugate base so doesn't it mean that the number of conjugate base ion will remain the same since the $\ce{H+}$ ion is just reacting with the same conjugate base ion which they were initially bonded together ?
The other assumption is that temperature has no effect on buffering capacity since buffering capacity apparently only depends on two factors: concentration of the buffer solution and the ratio of pH to pKa. So even if the pKa changes due to the change in temperature, the pH also changes accordingly whereby pH increases when pkA increases and vice versa. So the ratio will remain the same no matter what.
So could anyone please give me a solution?

Mathematically
For a buffer solution containing a weak acid and its salt with a strong base, buffer capacity is given by $$\beta = 2.303 ( [\ce{H+}] + [\ce{OH-}] + \frac{C_{buff}.K_{\mathrm{a}}.[\ce{H+}]}{([H^+]+K_{\mathrm{a}})^2})$$
It should be noted that the third term in the expression, $$\frac{C_{buff}.K_{\mathrm{a}}.[\ce{H+}]}{([\ce{H+}]+K_{\mathrm{a}})^2} = \frac{[\ce{HA}][{\ce{A^-}}]}{[\ce{HA}]+[\ce{A-}]}$$  which is  independent of temperature ( if you are working at same concentrations at any temperature ) and the only temperature dependent terms are $\ce{[H+]}$ and $\ce{[OH-]}$ and
$${[\ce{H+}] = \frac{K_{\mathrm{a}}.C_{acid}}{C_{salt}}}$$ and $${[\ce{OH-}]= \frac{K_w.C_{salt}}{K_a.C_{acid}}}$$
For the sake of simplicity let us consider the maximum buffer capacity at a given temperature, which occurs when $${C_{salt} = C_{acid}}$$
Now, $${\beta= 2.303( K_a + \frac{K_w}{K_a} + \frac{C_{buff}}{4})}$$
The acid dissociation constant $K_{\mathrm{a}}$ varies with temperature approximately as ${e^{\frac{-\Delta H_{ion,acid}}{RT}}}$ and $K_{\mathrm{w}}$ as ${e^{\frac{-\Delta H_{self-ion,water}}{RT}}}$.
Differentiating $${\beta= 2.303( K_a + \frac{K_w}{K_a} + \frac{C_{buff}}{4})}$$ with temperature, and substituting values of ${\Delta H_{ionization,acid}}$ and ${\Delta H_{self-ionization,water}}$, for most of the acids, we get a graph like this:
It increases up to a certain temperature, depending on the acid and then it starts to decrease.
Experimental Observations
For Histidine and related amino acids the variation is shown below:

For further details read, Effect of Temperature on the Buffering Capacities of Histidine-Related Compounds and Fish Skeletal Muscle[1].
Reference:
[1]: Abe, H.; Okuma, E. Effect of Temperature on the Buffering Capacities of Histidine-Related Compounds and Fish Skeletal Muscle. Nippon Suisan Gakkai Shi 1991, 57 (11), 2101–2107.

You question is too broad to answer since there is no "universal answer." In chemistry there is a "tends too" answer and then a list of exceptions.
By and large the root of the question centers around the change in the equilibrium constant as a function of temperature. The equilibrium constants do depend on temperature, but how much "tends" to depend on the particular equilibrium under consideration. However the Arrhenius equation can offer some insight.

The other assumption is that temperature has no effect on buffering capacity since buffering capacity apparently only depends on two factors: concentration of the buffer solution and the ratio of pH to pKa. So even if the pKa changes due to the change in temperature, the pH also changes accordingly whereby pH increases when pkA increases and vice versa. So the ratio will remain the same no matter what.

This is the better argument. For example, if you start with equal concentrations of conjugate acid and base, and change the temperatures, that ratio will not change much, and the buffer capacity would be similar. The pH will be different as the pKa will change, but you already mentioned that.
There are small secondary effects: The volume of the solution might increase slightly, slightly decreasing the buffer concentration and its buffer capacity. Also, the change in pKa will shift the equilibrium a bit, resulting in the change of pH and in a slight change in the ratio of conjugate acid and base.


text:
Doesn't the isothermal irreversible expansion/compression of an ideal gas law violate the ideal gas law?
Let's say a cylinder has some ideal gas. We use a piston to suddenly expand the gas in isothermal condition from $1\ \mathrm{atm}$, $20\ \mathrm L$ to $0.5\ \mathrm{atm}$, $50\ \mathrm L$.
Since the gas is ideal, $pV = nRT$ implying that $nRT = 20\ \mathrm{L\ atm}$ initially. Later $pV$ changes to $25\ \mathrm{L\ atm}$. However, $nRT$ has to continue to be $20\ \mathrm{L\ atm}$ as the number of moles and the temperature remain constant.
Doesn't this violate the ideal gas equation, as $pV$ and $nRT$ are no longer equal. Does this mean that the irreversible expansion or compression of an ideal gas can not be isothermal unless the number of moles change (through some chemical reaction)?

If it is isothermal, it will be 0.5atm in 40L

You can’t control all of p, V, and T at the same time.

Well, there is some misunderstanding or mistake, but I'm afraid OP didn't make it clear enough to tell what it is. There may be some options: 1. misprint of volume 2. isothermal/adiabatic confusion 3. heavily non-ideal gas 4. ?

How would this ensure that the pressure was 0.5atm? If the piston controls the volume it doesn't also control the pressure.

@orthocresol That's what I thought as well. I found a similar question in my chemistry textbook so I was wondering if such a process requires a chemical change (resulting in the change of $n$) or not.

The ideal gas law describes the behavior of an ideal gas exclusively for thermodynamic equilibrium states.  During an irreversible expansion or compression, the gas passes through non-thermodynamic-equilibrium states, except at the very beginning and at the very end.  In these irreversible non-equilibrium states, the ideal gas law does not correctly describe the gas behavior.  For rapid expansions and compressions, the force that the gas exerts on the piston depends not only on the gas volume, but also on the rate of change of gas volume.  So yes, in irreversible/expansion of an ideal gas, the ideal gas law is violated.
That said, if we can control the force between the gas and the piston by some external means (say by using an automatic control system in conjunction with a flush mounted pressure transducer in the inner piston face), we can calculate the amount of work that the gas does on the piston (without trying to resort to the ideal gas law) even for an irreversible expansion/compression.  This then would enable us to use the first law of thermodynamics to calculate the final state of the gas (at least for an adiabatic process).

Hmm, I interpreted the question to not be about the process but rather the initial/final states, but I'm probably wrong.

That explains what happens during the process. However, at the end of the process, $pV$ and $nRT$ are not equal either.

If the system is again at thermodynamic equilibrium in the final state, they certainly are equal.  What gave you the misconception that they wouldn't be?

@ChesterMiller So does that imply that the gas is not at thermodynamic equilibrium at the final state I've mentioned (where $pV ≠ nRT$) and the process will continue for more time till it attains the equilibrium?

The final state is thermodynamic equilibrium, and, in that state, pV=nRT

Doesn't this violate the ideal gas equation, as pV and nRT are no longer equal?

Yes, it does.

Does this mean that the irreversible expansion or compression of an ideal gas can not be isothermal unless the number of moles change (through some chemical reaction)?

Yes, that's right. I think it is quite misleading to speak of the process "...irreversible expansion or compression..." rather than the initial and final states, which is what you're appealing to when you use the ideal gas law.
As you've argued, regardless of process, there is no way an ideal gas with $(p, V, n, T) = (1\,\pu{atm}, 20\,\pu{L}, n_1, T_1)$ can get to $(p, V, n, T) = (0.5\,\pu{atm}, 50\,\pu{L}, n_2, T_2)$ if $n_1=n_2$ and $T_1=T_2$.

Didn't you mean 50 L in the second state (in line with what the OP speculated over)?  I think that with 40 L, the final state can be attained.