text:
How many covalent bonds are there in carbon suboxide?

Calculate the number of covalent bonds in $\ce{C3O2}$.

Structure:

I am confused. I can count covalent bonds as 8 by counting single-single bonds, but also I can say that there are 4 "double covalent bonds". Which is correct?

Eight covalent bonds is correct

@Maurice Any Sources  from where I can confirm?

Four double bonds is equivalent to eight covalent bonds.

@Maurice But “Double Covalent Bond Is A Type Of Covalent Bond” so probably could be counted as 4 too .

@ Rasputin. A double bond is not a type of covalent bond. It is two bonds !

@Maurice The physical reality is that there are 4 bonds. The character of those bonds is partly theoretical. The term "covalent bonds" does not specifically or uniquely apply to just single bonds, at least in most terminology that is widely used.

The first step is to ensure all speak by the same language. It must clarified and widely agreed what is and what is not a bond. Then in the second step we can start counting.  It is quite possible the result would be there are 4 bonds(1) but 8 bonds(2), where (1,2) means 2 different bond definitions.

Widely used terminology might not match what an examination wants
It is possible that your question wants the answer 8 in this context, but whether this answer would be widely accepted by anyone familiar with bonding theory is doubtful.
In the early days of theories about bonding Lewis described bonds as consisting of two shared electrons. You are probably familiar with "Lewis Structures" which are simple pictures of bonding where electrons are represented with two dots between atoms. This is also how the theory of bonding is initially introduced to students. In this view the number of bonds in the molecule is, indeed, 8. But this–vastly over simplistic theory–is not good for explaining the bonds in many real compounds. Hence the dispute about the correct terminology.
A good theory of bonding needs to take quantum mechanics (and the shape of electron orbitals) into account and those were not understood in the early 20th century when Lewis developed his explanation of bonding.
But we don't need any quantum obfuscation to see why Lewis theory doesn't work well and why it is better not to pretend that bonds consist of pairs of electrons. Take benzene as an example.  You could claim that the bonds in benzene contain 18 electrons so make up 9 bonds. But this means alternating double and single bonds. But that can't be right as all the bonds in benzene are the same length. It is better to say than benzene consists of 6 bonds each of which have somewhere between the character of a single and double bond.
This inability of the Lewis model of bonding to explain real world structures is very common. Nitrate ions are one example where all the bonds have characteristics of more than a "single" bond but less than a "double" bond. Many boranes have bonds between 3 atoms but only two electrons to share among them.
So equating bonds to shared pairs of electrons is not helpful. Nor is using terminology that defines a bond that way. So real chemists should be reluctant to count bonds that way. Whether this is the answer your question is looking for is less clear as you might still be at the stage where they have taught you Lewis structures and nothing else.

I actually am taught Lewis Structure , VSEPR, And Molecular Orbital Theory ( using the LCAO method)..(A bit simplistic approach valid for less than equal to 20 electron species)...I know about resonance and the case you mentioned and I agree that the examination does require “8” to be the answer....I wouldn’t count bonds in this way at all but the examination wants what it wants...I just wished to clear up the confusion!

@ Matt black. Congratulations for your clear and clever development. I am not joking ! It is nice and bright.

That is a really good answer! Kudos. I would like to bounty it, but that is not yet possible; I'd appreciate a reminder. To the matter at hand: carbon suboxide is probably an equally terrible example to count bonds as for the case with benzene. It has a very shallow bending potential, which indicates a much more complex electronic structure than what is possible to explain with Lewis-like structures. And as far as I remember there is/was some debate from different points of view. Certainly interesting, certainly too much for the undergrad level.


text:
Reduction of epoxide
Does $\ce{LiAlH4}$ and $\ce{NaBH4}$ reduce an isolated epoxide? I have two books where both are saying differently. One says it can and other says cannot. On searching the web there was no definite answer. Can they reduce epoxide to diol or something else ?
I am aware of the reduction of the other groups in the picture:

Sources:  NCERT: yes; JEE Advanced 2016 paper solution by Arihant: no; multiple coaching stuff: both yes/no

Is that a 5 bond carbon on the epoxide on the left?

Please correct the molecular diagram of the starting material.

this is NOT an isolated epoxide

The correct question (taken from the 2016 JEE Paper is):

Lithium Aluminium Hydride reduces epoxides1, hence (A) would not give the required product. LAH also reduces ethers to primary alcohols, which is an easier way of eliminating the option (A). $\ce{BH3}$ in THF can reduce the acid, as explained in this question, hence the requisite product would not be obtained.
$\ce{NaBH4}$ cannot reduce esters, acids or epoxides in the absence of catalysts such as $\ce{PdCl2}$ or $\ce{AlCl3}$ and diglyme, hence (C) would be a correct option here.
Can you tell what Raney Ni would do here?
Raney Ni Reduces carboxylic acid and esters at elevated temperature (eg: $\pu{350 K}$); and no reaction with epoxide. So, it can be used for selective reduction of aldehyde at room temperature ($\pu{298 K}$).
Use this resource to help yourself out and as a reference for such reactions in the future.
References:

Murphy, David K., et al. “Reduction of Epoxides. III. Lithium Aluminum Hydride and Mixed Hydride Reduction of Some Secondary-Tertiary Epoxides.” Journal of the American Chemical Society, vol. 91, no. 10, May 1969, pp. 2649–53. doi:10.1021/ja01038a042.

Oh okay! I wasnt aware of the catalyst thing. Would NaBH4 reduce an isolated epoxide without catalyst PdCl2 ?

No, it would not

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text:
Isomerization reaction involving carboxylic acid derivatives

I have tried searching for what type of organic reaction this is in my textbooks and ended up with nothing. Thus, I could not fully understand and refer to the mechanisms that the solution has explained.
So, please tell me the types of organic reactions (i.e. name the reactions involved for me) that are involved in this question. If possible you can describe how the mechanism of this reaction. Thank You!

Change the question title to a more specific one and also think about it yourself, what do you think should be happening?

@RobinSingh Sorry, I suck at organic chem. But though I understand the mechanism, it feels completely random to me due to the shifts which are happening and I can't reason why it is so. At least knowing the name of the reactions involved, I can refer to the Internet to learn the mechanisms and how it works myself. If you know the reactions involved, then please do share it with me.

@rash it's not a single reaction, rather a set of reactions, all of which are intramolecular (first one is acid-catalysed nucleophilic addition to carbonyl group, then the intramolecular acid-base, then the rearrangement). Try analysing each step and see where you are getting confused.

This is somewhat like NGP (neighboring group participation ), but with carbonyl.

tl;dr A slightly anthropomorphic approach with some basis in general organic chemistry:

Step 1: The protonation of the carbonyl (An acid-base reaction)

Here, we have three options – we can protonate the nitrogen (deactivated by the resonance with oxygen), the alcohol group ($\mathrm{sp^3}$) or the carbonyl group ($\mathrm{sp^2}$). As you can see this is basically an acid-base reaction and therefore the strongest base (having highest electron density) would react the fastest and this means since the carbonyl has the highest electron density here the carbonyl is protonated.
Step 2: The intermolecular attack ($\ce{O}$ is very sneaky)

Well, we began with $\ce{HCl}$. Time for the $\ce{Cl-}$ to make an appearance? Not really. As you can see, we have an alcohol group in the compound which has electrons to donate. The carbon on the (protonated) carbonyl group is highly electropositive (It was positive enough when there was no positive charge on the oxygen and now it just got more electropositive). Now, the oxygen with the lone pair makes an appearance. It attacks the carbonyl carbon and at the same time the π electrons move to the previously protonated oxygen making it neutral.
Step 3: An intermolecular proton transfer (Another acid-base reaction?)

Now, as you may have learnt, oxygen is not the most electropositive element. If possible, it doesn't stay positive. Nitrogen comes to the rescue. Another acid base reaction takes place where the newly active amine group deprotonates the oxygen forming $\ce{>NH2+}$.
Step 4: The final rearrangement (We're done here(?), you can figure it out)

We still need the final isomer and we also need to remove the positive charges on the compound before we end. Therefore, the ring breaks and the electrons move to the nitrogen removing the positive charge on it. This also means a carbocation is formed in the carbon where the alcoholic group is attached which stabilises it using resonance to form a $\ce{C=OH+}$ bond. (try drawing this step).
But we aren't done.
Step 5: The true final step (A small deprotonation)

Nobody needs any positive oxygens, do we? Therefore the oxygen is deprotonated. (Note: this is in equilibrium since medium is acidic). All $\ce{H+}$ used has been released which also makes this a catalytic reaction.
There might be one final question in the back of your mind (for the curious).
Why not go back to the starting compound?
This is because as @user55119 very well pointed out in a comment to the other answer. My additions in italic.

In the absence of acid, heating amino ester (a) would give the starting material, the hydroxyamide because esters are more labile (easily broken down or displaced) than amides. The reason this reaction can occur is that the amino group is protonated in acid which shifts the equilibrium.

Welcome to the realms of organic chemistry.

I upvoted your answer, very well explained. Just wondering, in step 2, what is stopping water from attacking and just hydrolysing the amide? After all they use 'dil. HCl', which I take to be aqueous HCl. Is the spatial proximity of the OH group sufficient to make this intramolecular reaction highly favoured compared to an intermolecular one, even though H2O is about 55 M in the mixture? Arguably, the relative stereochemistry of the two groups on the cyclopentane ring might play a role, too. So perhaps there's more to this than meets the eye...

@user6376297 https://chemistry.stackexchange.com/questions/136985/mechanism-of-three-membered-lactone-hydrolysis-in-base/137019#137019 this might help..

@user6376297: Even if there is no $\ce{HCl}$, this intramolecular acyl transformation is possible if end product is more stable. For example, acetylation of 3-O-hydroxy group of ascorbic acid is eventually lead to 3-O-acyl ascorbic acid if leave in water or organic medium for a while.

I have commented previously that the success of this reaction depends on the protonation of the free amino group of answer (a).

In the absence of acid, heating amino ester (a) would give the starting material, the hydroxyamide because esters are more labile than amides. The reason this reaction can occur is that the amino group is protonated in acid which shifts the equilibrium.

The specific p-nitrophenylbenzamide of this question could not be found in a Chemical Abstracts search but a study from 70 years ago was uncovered that substantiates my previous comment. Fodor and Kiss[1-3] treated hydroxybenzamide 1 with 2.0 equivalents of absolute ethanolic HCl at room temperature producing a mere 9% yield of amine hydrochloride 2 along with substantial recovery of benzamide 1. With an excess of HCl, equal amounts of 1 and 2 were isolated. Titration of salt 2 with aqueous NaOH to a phenophthalein end point formed the free amine of 2 which, within minutes, underwent O $\Rightarrow$ N migration. The structure 3 is the presumed intermediate.The authors also explored the trans-isomer of 1, which gave less than 1% of the trans-salt of 2 with 2.0 equivalents of HCl. The O $\Rightarrow$ N migration was also slower than in the cis-series.

G. Fodor and J. Kiss, Nature, 1949, 163, 287.
G. Fodor and J. Kiss, Nature, 1949, 164, 917.
G. Fodor and J. Kiss, J. Am. Chem. Soc., 1950, 71, 3495.

You need to understand something in organic chemistry. Not every question is going to be part of some named reaction that you just have to apply. For example in this question, the basic concepts of organic chemistry is used nothing more.
First step, you look at the molecule and think, hmmm what can $\ce{H+}$ do? There are two possible things:

The nitrogen could take it forming $\ce{NH2+}$ but see that this does not lead anywhere that is you cannot proceed (also the nitrogen is right next to the carbonyl so the carbocation will also not be stable). Note that protonating the OH group also leads to nothing. (a rearrangement  is possible but as none of the options have it we can eliminate it)

The lone pair on the oxygen (of carbonyl) takes up the $\ce{H+}$, forming a postive charge on it and this is a no-no as oxygen wants electron. So the double bond shifts onto the oxygen to form the OH group but this forms a positive carbocation and then … we see that an intramolecular attack can take place from the OH on the other side (O has lone pair of elecrons!) forming a five membered ring!

Then as mentioned in your diagram, an intramolecular proton exchange so as to quench oxygen's positive charge (a standard acid base reaction). Now from this part, see that we can regain the lost carbonyl group by opening up the formed ring in a rearrangement (the bond energy of carbonyl group is high). That's it!

What an appreciable answer this is! Not only did this give the solution to this single question but it also kind of gave me a headstart on how to proceed further in organic chemistry. Thank You!

For formatting, See here and here. For a more detailed MathJax guide, look here, minor other details

In the absence of acid, heating amino ester (a) would give the starting material, the hydroxyamide because esters are more labile than amides. The reason this reaction can occur is that the amino group is protonated in acid which shifts the equilibrium.

If this question did not have any options, what would you have said about the rearrangement?

@RobinSingh A hemiacetal is formed.Hemiacetals are unstable in acidic medium and hence the rearrangement must happen.

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text:
% coefficient of variation (%CV) for log-linear and log-log regression (calibrations)
I have a data set comprising the peak areas of an analyte measured in spiked calibration samples at 'known' and different concentrations levels. For each concentration level I have 5 replicates. My goal is to generate a calibration curve from this data and then, for each concentration level, determine the precision of the 'estimated concentration' expressed as percent coefficient of variation ($\%CV$).
If a linear relationship existed between the measured peak areas and 'known' concentrations, then for each concentration level I would simply calculate $\%CV$ as the standard deviation of the estimated concentration at a given 'known' concentration level and divide by the mean of the same estimated concentrations, before multiplying by 100. For my data set, however, I observe an inadequate linear (i.e. straight line) fit between measured peak areas and 'known' concentrations. Furthermore, there is heteroscedacity of the residuals when fitting a linear model.
To address the above, I've performed a $\log_{10}$ transformation of BOTH peak area and 'known' concentration. An adequate linear fit is observed. I would now like to calculate the precision (coefficient of variation, $\%CV$) of the estimated peak area based on this model.
According to the article cited below, the %CV for log-transformed data would be calculated as:
$$\%CV (\text{estimated concentration}) = 100\%  \cdot \sqrt{10^{\ln(10)\theta^2_{\log}} -1}$$
Where (if I understood correctly): $\theta^2_{\log}$ is the variance of the $\log$-transformed data.
So, I would specifically like to know: is the formula proposed by Canchola, et al. appropriate in the case where BOTH the response (i.e. peak area) and predictor (i.e. 'known' concentration) variable have been transformed?
In my mind, seeing as I would consider the variable of the estimated concentration on the $\log_{10}$-transformed scale, the formula outlined by Canchola, et al. should be fine.
Finally: if I had only $\log_{10}$-transformed the peak areas and then estimated the concentration (i.e. log-linear relationship), would I need to use the Canchola, et al. equation?
Referenced article: Jesse A. Canchola,  Shaowu Tang, Pari Hemyari, Ellen Paxinos, Ed Marins, "Correct use of percent coefficient of variation ($\%CV$) formula for $\log$-transformed data," MOJ Proteomics & Bioinformatics 2017, 6(4), 316-317 (DOI: 10.15406/mojpb.2017.06.00200).

While the question may be ok for this site (and I personally consider it interesting) it strikes me as better suited for crossvalidated SE.

I agree with Buck that this appears to be more about statistical analysis and less about chemistry. Another alternative, if you want an answer specifically from the perspective of quantitative experimental science (rather than from pure mathematicians), is https://mattermodeling.stackexchange.com/. Statistics questions are fine there, but the answer would still come from a chemist or physicist.

Thank you both for your recommendations. I'll take this on over to crossvalidated then.

Following the advice of @Buck Thorn and @user1271772, I asked this same question over at the Cross Validated Stack Exchange.
There, I received a very good answer and am providing the link for the interested reader: % coefficient of variation (%CV) for log-linear and log-log regression (calibrations)
To summarise: yes, the formula is appropriate where the predictor (concentration) and response (peak area/other metric) have both been log-transformed.


text:
Why is vapor pressure constant at boiling point?
At non-boiling point, I understand that when you add heat, temperature increases and also the amount of gas increases, so by using $PV=nRT$ at constant volume, the increase in $n$ and $T$ would increase $P$.
Thus, vapour pressure increases with temperature.
However, at boiling point, if you add heat, temperature does not increase and the energy is used to generate more gas molecules.
So if you are at a equilibrium where there is $\pu{1 mol}$ of $\ce{A(l)}$ and $\pu{1 mol}$ of $\ce{A(g)}$, after you add heat and shift equilibrium to $\pu{0.5 mol}$ of $\ce{A(l)}$ and $\pu{1.5 mol}$ of $\ce{A(g)}$ wouldn’t the vapour pressure increase even though temperature is the same by the ideal gas law– $PV=nRT$, where increase in $n$ increases $P$?
So why is vapour pressure is constant at boiling point?

Of course it would increase, if you keep the volume constant (as would T, by the way). Who said it wouldn't? The pressure is constant in a constant-pressure setting, but that's kinda trivial.

At the boiling point under constant pressure, the temperature is constant. Seems you are trying to make sense of this by looking at (likely sloppily written) definitions in a book. Keep the experiment in your mind!

By the way, "vapour pressure" is only defined for $T < T_{Bp}$. At $T >= T_{Bp}$, vapour pressure makes no sense.

https://www.chem.purdue.edu/gchelp/liquids/vpress.html
@Ivan Neretin The graph here seems to indicate a fixed vapour pressure for each temperature, even at boiling point? Thanks

@Karl From the graph I found in the link under my comment, it seems at though vapour pressure exists beyond boiling point as well? Isn’t vapour pressure jus the pressure exerted by the gas molecules of that particular substance? Thanks

Yes, there is a fixed vapour pressure for each temperature.  If you heat the closed vessel from your example, some of the liquid will evaporate, T will increase, and so will P.

@Ivan Neretin, but at boiling point isn’t the temperature fixed until all of the liquid is boiled off? In a sense, the heat is not used to increase kinetic energy of the molecules hence T remains constant. The heat is used to convert liquid molecules to vapour phase, so why in the graph the vapour pressure indicates that it is fixed at boiling point even when like you said, vapour pressure should increase when you continue add heat at boiling point? Thanks

It is fixed if you keep the pressure fixed. Otherwise it isn't.

Imagine you heat water. When reaching 100°C, the water boils at ordinary pressure. If you carry out this operation in a closed syringe, the volume of the gas will steadily increase, the number of moles of gas will also increase. You may continue heating, the volume of gas will increase, but the pressure remains at 1 atm. and the temperature remains at 100°C. At the end, when all liquid water has been vaporized, heating the vapor even more will increase its temperature and its volume. But the presence of two phases prevents the temperature from increasing by heating.

@Barry No. It is the equilibrium partial pressure of a gaseous compound above a liquid containing the same compound. Without the liquid, vapour pressure is undefined.


text:
How to differentiate between iron and sodium flames?
According to my textbook, when a flame test of an iron salt is performed, it produces an orange, mostly yellow flame. Sodium salts also produce a yellow flame.
As the colours of these two flames are too similar, how do I differentiate an iron flame from a sodium flame?

You can watch for spectral lines of iron and sodium. Note that solid  particles containing iron will likely produce rather continuous emission spectrum.

Sodium flame is yellow, but all the light is due to two lines in the yellow region. If you look a sodium flame through a purple glass (cobalt glass), the yellow color is absorbed, and you do not see the sodium flame any more. The flame looks dark and colorless through a cobalt glass.
On the contrary iron flame looks yellow, but it is made of a huge number of lines belonging to all regions of the spectrum. The sum of these colors look yellow, but it's a visual effect. The flame contains all colors. If you look at such an iron flame through a cobalt glass, the flame is visible. It looks purple or violet, but it is visible.

A diffraction grating or a hand-held spectroscope (you can make one with a broken CD) can be an easier way to distinguish them based on the same principles.

Very good question. I believe you are talking about ordinary Bunsen burner flame which is considered a low temperature flame by atomic standards. Good temperatures for atomic emissions are on the order of the temperature of the surface of the Sun (around 10,000 K).
So you are right, it is very "difficult" to visually distinguish the apparent flame color of the two elements. Also keep in mind, that there is no 100% pure iron salt. Sodium is everywhere!! So even iron flame is contaminated with sodium flame. The yellow color of sodium atoms comes from atomic emission at 589 nm.
The yellow color of iron flame comes from its oxide FeO which emits exactly at 579 nm. Our eyes are extremely sensitive to yellow light range, because is a very small range in the visible spectrum. There are no huge number of lines in the visible spectrum in such low temperature flame of Bunsen burner.

Reference: Analytica Chimica Acta, 115 (1980) 121
A real emission spectrum of iron has literally thousands of lines but you need to create temperature close to the Sun's surface with a plasma or an electrical arc.

Surface of Sun: not quite 10,000 K, closer to 6000.


text:
Balancing a reaction where dichromate ion reacts with ethanol to give Cr(III) and carbon dioxide
I have only started to learn balancing redox reactions, and a recent question has me confused.
The question is :

Balance in both acidic medium and basic medium:
$$\ce{Cr2O7^2- + C2H5OH -> Cr^3+ + CO2}$$

I usually look for symmetries between the reactant and product sides of a reaction to pair reactants with the corresponding products to form the half reactions, as per the first steps of balancing a redox reaction. I'm having a difficult time doing that here; what am I supposed to pair Cr with? I would personally pair it with the $\ce{Cr2O7^2-}$ molecule but that's just by process of elimination because I would naturally pair $\ce{C2H5OH}$ with $\ce{CO2}$ due to the stronger resemblance they have with each other.
Another matter of concern is the answer in the answer key; the answer in acidic solution is, apparently:
$$\ce{2Cr2O7^2- + 16H+ + C2H5OH -> 4Cr^3+ + 2CO2 + 11H2O}$$
In a basic solution:
$$\ce{2Cr2O7^2- + 5H2O + C2H5OH -> 4Cr^3+ + 2CO2 + 16OH-}$$
It seems like in the answer for the reaction in acidic medium, the left side of the reaction has a charge of +14 and the right side has a charge of +12.
However,  in the basic medium the answer has the left side with a total charge of -2 and the right side -4. How can these answers be correct if the charge isn't balanced?
How do you balance a redox reaction lacking a molecular reactant on the product side? I am referring to the atom or molecule that changes oxidation state when I say molecular reactant.

I see you put in a lot of effort into making those reactions look like actual reactions. There is a simpler way to do that using $\ce{}$ which is used for chemical compounds.. If you wish you can look at the actual code by hitting [edit]

For formatting, See here and here. For a more detailed MathJax guide, look here, minor other details

Thank you, I saw that I could use HTML and just got blindsided and went with that

Also, I think that there was a typo which I may have accidentally corrected that may have removed part of your question. It was $\ce{Cr^3+}$ and not $\ce{Cn^3+}$

You made a couple bad edits IMO. For one, the changed title is not the question I'm asking, I'm asking the original title question that I had because I assume that this is not the only reaction with the same problem. There were also other unnecessary pedantic edits that changed the meaning of my question in an adverse way; it was supposed to be Cn3+ unless this the reaction at the bottom of the page on this link under the section titled "Practice Problems" was erroneous and contained the typo

You basically slaughtered my question and I give up now. That's not the question I had at all, that's not even the reaction I have a question on, you edited the reaction to a different reaction. Unless the it's a typo in the original reaction, found here https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Redox_Chemistry/Balancing_Redox_reactions/Balancing_Redox_Reactions%3A_Examples

If I were to answer why I did so it would be equivalent to giving an answer for the question. Also, for the fact that I changed $\ce{Cn^3+}$ to $\ce{Cr^3+}$ is because there is no such reaction, you did get that right.. You don't need to get this defensive..  If you were to know how to balance this reaction, wouldn't that solve your questions as well?

Let us continue this discussion in chat.

I'm only getting defensive because I'm trying to defend my question that I want an answer to, please don't take as an insult; I'm not on offense, I'm on defense here. BUT if you are right, if that was a typo on the original reaction, because it was Cn3+ on the website then that would solve my problem and I would probably know how to balance it. But my other question about unbalanced charges still remains, just throwing that out there

In basic medium you should be dealing with chromate and not dichromate. In half reactions balance Cr and C first. In acid if you need oxygen, use H2O minus 2H+. In basic medium for each oxygen 2OH- minus H2O and for H in base, H2O minus OH-. Then balance each rxn. in electrons. If unequal, "cross multiply". Simplify reaction. Done!

Let's get the simple question out of the way first:

How do you balance a redox reaction lacking a molecular reactant on the product side? I am referring to the atom or molecule that changes oxidation state when I say molecular reactant.

You can not balance it then. Also, you wouldn't be given such a reaction and is always a typo when they've said that it is a redox reaction.

Now, onto the bit more complicated question.

Another matter of concern is the answer in the answer key; the answer in acidic solution is, apparently:
$$\require{cancel}\ce{2Cr2O7^2- + 16H+ + C2H5OH -> 4Cr^3+ + 2CO2 + 11H2O}$$
In a basic solution:
$$\ce{2Cr2O7^2- + 5H2O + C2H5OH -> 4Cr^3+ + 2CO2 + 16OH-}$$
[sic] It seems like in the answer for the reaction in acidic medium, the left side of the reaction has a charge of +14 and the right side has a charge of +12.
However,  in the basic medium the answer has the left side with a total charge of -2 and the right side -4. How can these answers be correct if the charge isn't balanced?

A simple explanation would be that the excess charges that you feel are there are actually not. Try breaking the given reaction in the answer into each atom with their own oxidation state. I'll do the one for acidic medium, and the other will be up to the reader.
$$\ce{2Cr2O7^2- + 16H+ + C2H5OH -> 4Cr^3+ + 2CO2 + 11H2O}$$
Here, we start from the reactant side, we assume the following for simplicity:

$\ce{C-H}$ bond leads to positive charge on $\ce{H}$ and a negative charge on $\ce{C}$
$\ce{Cr-O}$ bond leads to positive charge on $\ce{Cr}$ and a negative charge on $\ce{O}$
$\ce{C-O}$ bond leads to positive charge on $\ce{C}$ and a negative charge on $\ce{O}$

\begin{array}{lc} \hline
&\text{Reactants} &\text{Products} \\ \hline
\ce{Cr} & \ce{Cr^6+} & \ce{Cr^3+} \\
\ce{O} & \ce{O^2-} & \ce{O^2-} \\
\ce{H} & \ce{H+} & \ce{H+} \\
\ce{C} & \ce{C-,C^3-}  & \ce{C^4+} \\ \hline
\end{array}
Now, we split the actual equation into this form:
$$\ce{4Cr^6+ + \cancel{22\mathrm H^+} + C- + C^3- +\cancel{15\mathrm O^{2-}} -> 4Cr^3+ + 2C^4+ + \cancel{15\mathrm O^{2-}} + \cancel{22\mathrm H^+}}$$
Which simplifies the reaction to:
$$\ce{4Cr^6+ +C- +C^3- -> 4Cr^3+ +2C^4+}$$
Adding the charges on both sides we get:
\begin{align}
\text{Charge on reactant}=(4\times 6 + (-1) + (-3)) = 20 \\
\text{Charge on products}=(4\times 3 + 2\times4)= 20
\end{align}
Therefore the charge on both sides are equal ...
Note: The reaction in basic medium uses $\ce{CrO4^2-}$ and not $\ce{Cr2O7^2-}$
$$\ce{4CrO4^2- + 7H2O + C2H5OH -> 4Cr^3+ + 2CO2 + 20OH-}$$

@Safdar: Probably better to use chromate in basic medium. https://chemistry.stackexchange.com/questions/134075/why-is-chromate-stable-in-basic-medium-and-dichromate-stable-in-acidic-medium

Basic environment would lead to either $\ce{CO3^2-}$ either $\ce{HCO3-}$, but not $\ce{CO2}$.

Dichromate (orange) is present in acidic medium while chromate (yellow) prevails in basic medium. Here are stepwise solutions in both media. Acidic Medium:
Recognizing that chromium is the oxidant, balance the chromium atoms.
$$\ce{Cr2O7^2- -> 2Cr^3+}$$
Because the reaction is conducted in aqueous acid, only water and protons are available to balance the half reaction. Balance the oxygens using $\ce{H2O}$.
$$\ce{Cr2O7^2- -> 2Cr^3+ + 7H2O}$$
Now balance the protons.
$$\ce{14H+ + Cr2O7^2- ->[{+6e}] 2Cr^3+ + 7H2O}\label{1}\tag{1}$$
The carbon in ethanol is being oxidized.
$$\ce{C2H5OH -> 2CO2}$$
Balance the oxygen with water added to the left side.
$$\ce{3H2O +C2H5OH -> 2CO2}$$
Add protons to the right side.
$$\ce{3H2O +C2H5OH ->[{-12e}] 2CO2 + 12H+}\label{2}\tag{2}$$
Multiply equation $\eqref{1}$ by 2 to balance electrons in both half reactions.
$$\ce{28H^+ + 2Cr2O7^2- ->[{+12e}] 4Cr^3+ + 14H2O}\label{3}\tag{3}$$
Add equations $\eqref{2}$ and $\eqref{3}$.
\begin{align}
\ce{28H^+ + 2Cr2&O7^2- + 3H2O +C2H5OH \\ &->[{+12e}] 4Cr^3+ + 14H2O + 2CO2 + 12H+}\label{4}\tag{4}
\end{align}
Simplify duplicates on either side of equation $\eqref{4}$.
$$\ce{16H^+ + 2Cr2O7^2- +C2H5OH -> 4Cr^3+ + 11H2O + 2CO2}\label{5}\tag{5}$$
Equation $\eqref{5}$ is the balanced oxidation in acidic medium.
Basic Medium:
In basic medium the chromate anion is dominate. Only water and hydroxide are available to balance oxygen and hydrogen.
$$\ce{CrO4^2- -> Cr^3+}$$
Every oxygen atom requires two hydroxides where they are needed and one water on the other side of the equation ($\ce{2OH - H2O = O}$). Hydrogens are balanced.
$$\ce{4H2O +CrO4^2- ->[{+3e}]Cr^3+ + 8OH^-}\label{6}\tag{6}$$
Now deal with ethanol.
$$\ce{C2H5OH->2CO2}$$
Balance oxygen.
$$\ce{6OH^- + C2H5OH->2CO2 + 3H2O}\tag{7}$$
Balance hydrogens ($\ce{H2O - OH^- = H}$). Water on the right. Hydroxide on the left.
$$\ce{12OH^- + C2H5OH->[{-12e}]2CO2 + 9H2O}\label{8}\tag{8}$$
Multiply equation $\eqref{6}$ by 4 to balance electrons.
$$\ce{16H2O +4CrO4^2- ->[{+12e}]4Cr^3+ + 32OH^-}\label{9}\tag{9}$$
Add equations $\eqref{8}$ and $\eqref{9}$.
\begin{align}
\ce{16H2O +4Cr&O4^2- + 12OH^- + C2H5OH \\ &->4Cr^3+ + 32OH^- + 2CO2 + 9H2O} \label{10}\tag{10}
\end{align}
Simplify equation $\eqref{10}$.
$$\ce{4CrO4^2- + C2H5OH + 7H2O -> 4Cr^3+ + 2CO2 + 20OH^-}\label{11}\tag{11}$$
Equation $\eqref{11}$ is balanced in atoms and electrons.
Postscript: I suspect that this reaction, in acid medium at least, is not a known reaction. If it were, how could a Kuhn-Roth C-methyl determination possibly be accomplished? The process depends upon the quantification of acetic acid.

You can use $\tag{}$ for tagging equations and a combination of $\label{<label-name>}$ and $\eqref{<label-name>}$ to reference equations. I suspect you know this, but I mentioned it just in case.

@Safdar: After 3 years on ChemSE, this was my first attempt at inline equations. I can learn from your edit.

@Safdar Close to 6 years here and I don't think I even heard that ^ can be done with MathJax, you really dig this stuff ;D


text:
Plotting the data using Beer-Lambert law
For the spectroscopy lab we had a stock dye ($\pu{1.00E-4 M}$ Erioglaucine Blue) that we had to dilute. We had to dilute $\pu{1 mL},$ $\pu{3 mL},$ $\pu{5 mL},$ $\pu{8 mL},$ and $\pu{10 mL}$ in $\pu{100 mL}$ of water and measure their absorbance $A.$ I obtained the following data:
$$
\begin{array}{ccc}
\hline
V/\pu{mL} & A & λ/\pu{nm} \\
\hline
1 & 0.127 & 629.0 \\
3 & 0.342 & 626.2 \\
5 & 0.550 & 627.6 \\
8 & 0.850 & 628.3 \\
10 & 1.064 & 626.9 \\
\hline
\end{array}$$
How do I get the concentration in $\pu{mol L^-1}$ in order to make the graph? If I could get one example from this data set I should be able to calculate the rest for myself. I just need a head start.

I corrected a few typos, terminology ("concentration in Moles" is meaningless) and tried to make the data somewhat more readable, but I'm still not sure what graph you are talking about. Do you mean a calibration curve? Also, what about getting the molar absorptivity for your dye?

Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

@ Jasmine. You must absolutely measure your absorbances at the same wavelength. If you change the wavelength for each sample, you cannot use your measurements for any graph or calculations whatsoever. The only hope is to admit that the absorbances at $626.2$ nm and $629.0$ nm are not significantly different. Is it the case ?

Jasmine, wert @Maurice's comment: did you record the apsorption spectrum at one of the concentrations? What difference does it show in the wavelength range your measurements cover? Also, what is the spectral resolution of your instrument? Do you really have 4 significant digits in your data? Why did the wavelengths change between measurements?

You need to be aware of the dilution formula in order to plot concentration on the x-axis and absorbance on the y-axis:
$C_iV_i = C_fV_f$
So your initial concentration is your stock concentration and your initial volume is listed in the table. Your final volume is $\pu{100 mL}$. Can you determine $C_f$? With that you would be able to calculate all 5 different concentrations.
Keep in mind, that you need to fix the wavelength of measurement. Since the spectrum is very broad, $\pu{1 nm}$ will not make a huge difference. You may not have perfect linearity.

@Jasmine: I just want to make a point for M. Farooq's answer (kind of icing on the cake).  You want the concentration in $\pu{mol L−1}$. Since the concentration of your original stock solution is already given in  $\pu{mol L−1}$, you do not have to convert units of $V_i$ and $V_f$ during the calculations. Just make sure both in same units so they will be cancelled out from either side of the equation.


text:
Mechanism of reaction of alcohol with NaBr + H2SO4
My teacher told that the mechanism for reaction of alcohol with $\ce{NaBr + H2SO4}$ follows SN1 when alcohol is 2°/3° and SN2 when it is primary or methanol. He told that the reactivity of alcohol also follows the order 3° > 2° > 1° which is that of SN1.
But I couldn't understand why is it the case. The nucleophile is $\ce{Br-},$ which is anionic nucleophile—so it should always follow SN2.

@Aniruddha Deb , it's not just NaBr , it's NaBr + H2SO4 , which would protonate the -OH group into a good leaving group.

Primary alcohols will form primary carbocation which is not very stable , also the nucleophile here is a strong anionic nucleophile which would favour SN2 .

@AniruddhaDeb Read carefully the post you linked. OP's teacher is right.

OH is a poor leaving group. After protonation, it gets converted to one of the best leaving groups which leave the substrate immediately. This results in a carbocation

Can people please stop proposing primary and methyl cations in their mechanisms?


text:
Oxidation of alcohols by Ceric Ammonium Nitrate
I have realized that reaction with ceric ammonium nitrate essentially results in coordination of alcohol with $\ce{Ce(IV)}$ and my teacher referred that we should be able to proceed it to form ketone by dissociating $\ce{C-H}$ of the alcohol.I have the following doubts:
A) Precise mechanism as I am literally lost as to why the $\ce{C-H}$ being as strong bond is dissociated. what is the driving force here?
B) What is the final oxidation state of cerium as it seems that $\ce{Ce(III)}$ is produced but we initially had $\ce{Ce(IV)}$. Is there an electron transfer mechanism here?

It is a redox reaction.

In this reaction, the ion $\ce{Ce^{4+}}$ is a powerful oxidant. It needs one electron to be transformed into the more stable ion $\ce{Ce^{3+}}$. This electron may come from the central function $\ce{-CHOH-}$ of any secondary alcohol $\ce{R-CHOH-R'}$. The bond between $\ce{O}$ and $\ce{H}$ is able to loose $1$ electron and $1$ $\ce{H+}$ ion. Same thing happens with the bond $\ce{C-H}$ belonging to the same group $\ce{-CHOH-}$. As a consequence, two electrons are emitted with two $\ce{H+}$ ions, and the remaining organic molecule becomes a ketone $\ce{R-CO-R'}$. The two electrons react with two ions $\ce{Ce^{4+}}$. The two half-equations are : $$\ce{Ce^{4+} + e^- -> Ce^{3+}}$$ $$\ce{R-CHOH-R'->  R-CO-R' + 2 H+ + 2 e-}$$ so that the final equation is : $$\ce{2 Ce^{4+} + R-CHOH-R' -> 2 Ce^{3+} + R-CO-R' + 2 H^+}$$

ok i get the mechanism of this, the driver is the powerful oxidising nature of ce4+ right? however this mech makes me ask one more qn are you implying that 2 moles of CAN is required? my highschool book say it require 1 mol per alcohol. can you refer a book or a site which says otherwise?

@ Napstablook. Your high school book is wrong. I am sorry to tell it this way. It is impossible to oxidize one mole alcohol with one mole Cerium(IV).


text:
Why is the temperature graph sigmoidal in simple distillation of components with very different bps?
I'm writing an organic chemistry lab report on distillation and I thought I understood this, but the more I look at my lab textbook, I realize I don't.  My textbook 1 explains that the temperature of a mixture will rise continuously while it boils and distills if it is composed of two components with similar boiling points, but take on a sigmoidal shape when they are

When a liquid mixture is distilled, often the temperature does not
remain constant but increases throughout the distillation. The reason
for this is that the composition of the vapor that is distilling
varies continuously during the distillation (see Figure 14.3B).

I don't understand this sentence: "The reason for this is that the composition of the vapor that is distilling varies continuously during the distillation." I understand why it is true, but not why it is an explanation for the shape of these graphs. Let's say I'm distilling volatile component B from non-volatile component A. If the boiling point of the mixture is the point at which the vapor pressure of the mixture equals the atmospheric pressure, what does the composition of the vapor have to do with anything?  The relevant pressure is the atmospheric pressure, right?  I would think that as component B is boiled off, the vapor pressure of the mixture would decrease, because the mixture has less B, by Raoult's law, Ptotal = PA + PB = PA°NA + PB°NB, and as its vapor pressure goes down, its boiling point goes up.  I don't see why this isn't happening continuously.
Is the relevant pressure not the atmospheric pressure, but some additional pressure exerted by the vapor in the flask?  (It's been a long time since I took general chemistry.)

Pavia, D. L., Lampman, G. M., Kriz, G. S., & Engell, R. G. (2011). A small-scale approach to: Organic laboratory techniques. Belmont, CA: Thomson Brooks/Cole

Because being Pa and Pb very different you have an abrupt change when Na tends to zero. Mathematically that is continuous of course.

In a mixture of little chloroform with much methanol (the second the component of higher boiling point), the boiling temperature actually may decrease till you reach the azeotropic point (e.g., https://en.wikipedia.org/wiki/Azeotrope), despite the composition of your liquid is more and more rich in methanol.  Similarity of boiling points of the two components isn't the cause for this slope here actually contradicting the claims illustrated by the illustration in the centre.  Plot c) equally depends on the rate of distillation vs. the volume still to distill in the lower round bottom flask.

(Continuation) And if you perform a distillation of two liquids with well separated boiling points, especially with a column, you may glad to observe a drop of the head temperature (at the top of the column) prior to the arrival of the next fraction.

You must understand... in an experiment, the sigmoidal shape will never have infinite slope... This would be a discontinuity... the sigmoidal shape (rightmost) and the "curve" (middle image) are essentially representing identical phenomena... there is a transition from one pure component to another. The only difference is how abrupt the change is... and this is determined by the endpoints/boiling points for each component. The more abrupt the change from one to another pure component, the steeper the slope of the change. I hope this helps. I can speak to this further, if necessary.

I think the down vote is a bit unfair. This basically echoes my comment. It should not have mentioned very actual experiments as for the comment by Buttonwood, for instance. And simply a very long condenser can make the plot discontinuous without physical meaning. We shall stay realistic.   The plots are about the physics not about the apparatus.

I noticed the book is about experimental techniques :)


text:
Easiest/Fastest way to obtain a good quantity of fine Iron powder?
I am wanting to make a reasonable quantity (10-100g) of iron powder that I can use to visualise magnetic fields (by putting the powder in a thin layer between two transparent materials).
The finer the powder and more consistent the size the better. Ideally I would like a particle size 1 micron or smaller.
I've used a very fine grit sandpaper of about 2000 grit and this had a great result but took an eternity to produce.
I also made iron powder via electrolysis with an iron anode/cathode, in an aqueous solution of iron sulphate. I used an old repurposed PC PSU to supply 3V and however many amps the PSU could output for the electrolysis.
This produced some iron flakes which seemed a bit too large for what I wanted but it made a lot of iron particles quickly and with little effort.
Main question:
Do you know of any good processes I could use to make fine iron particles using electrolysis or another chemical process?
Related side questions:
What's a good liquid medium for suspending iron particles?
Is iron the best material to use for a viewing magnetic fields? How does it compare to other materials like Cobalt,Nickel,magnetite,Tungsten (Paramagnetic), NdFeB dust, etc?
A magnetic field viewer with silvery white particles (as opposed to black particles) could look nice. What might be a good way to turn the magnetic powder more silvery instead of black?
Is there a way to produce a good quantity of very fine, needle-like rods of iron? This could also work well for visualising magnetic fields as I believe the 'needles' would want to align to the magnetic fields.
I appreciate any help you can offer me!

One can buy iron powder of various diameters from standard chemical suppliers. One supplier has 250g of 10 micron powder for about $65.

A less laborious path to very fine iron dust (in fact, likely  pyrophoric iron) perhaps follows from first purchasing or preparing iron(II) oxalate (a path is detailed in a reference provided below). Next, thermally decompose the salt and finally, stabilize the nano-iron product with say mineral oil.
The details on the thermal decomposition in an atmosphere of hydrogen is discussed in this paper: Pyrophoric Nanoparticles and Nanoporous Foils for Defense Applications. See also this discussion which includes a suggested preparation for Iron(II) oxalate together with a video here.
Lastly, follow a similar process as is outlined in this US Patent 3,520,676 which claims, to quote:

Pyrophoric powder of metals (such as iron, cobalt, nickel and alloys thereof) is stabilized to prevent its spontaneous ignition. Such stabilization is achieved by Wetting the metal particles with a high-boiling organic liquid (such as an ester of carboxylic acid, mineral oil, silicone oil or fatty acid) and holding the wetted particles in the presence of oxygen until a thin oxide layer is formed on the surfaces thereof.

in order to stabilize the very fine iron product for your intended use.
A particular advantage to this approach is the low cost, and actually, not a great need for a large amount of the nano-particles for your demonstration purposes.

Fast and easy , maybe not cheap. Look up Non Destructive Testing such as Magneflux Corp. They supply iron oxide and iron powders to perform magnetic flaw inspection . Also  fluorescent liquid. The expense may be caused by minimum quantity. If you are in an oil/gas producing area there are likely pipe ( OCTG ) inspection companies that will sell/give you a small amount ( such as Tuboscope ). And, more sophisticated auto garages and machine shops may have mag particle inspection equipment and supplies they would be willing to sell.  You are sort of reinventing the wheel.


text:
Why is there carbon lining on the steel vessel in Hall Heroults process?
I read This answer, and completely understand that why graphite anode is used , but I have read that the vessel acts as the steel cathode but with lining of carbon. What is the need of this carbon lining ?
And this in turn raises the question , what actually is the cathode , steel or carbon ?
Thanks !!

An obvious answer is to why the presence of carbon, per a source Chapter 4: The Role of Carbon in Metallurgy on the historical extraction of iron from its ores, for example, to quote:

Carbon plays a crucial role in the chemical preparation of metals, particularly iron. ... This chapter discusses carbon as an intermediary product used in very large quantities to reduce and insulate metals and semi‐conductors from their natural oxides.

In a few words, carbon is a reducing agent, an insulator and a good oxygen scavenger.
Also, given the electrolysis nature of the Aluminum process, it is a cheap and very noble cathode. The employment of other noble metal cathodes would be more expensive and likely add impurities.


text:
Why is change in internal energy not equal to zero during a phase transition?
Many books state that when there is no change in temperature $\Delta U$ is equal to zero. Why is this not applicable for phase transitions? I have had this question for a while now and have been unable to find an answer to it

"when there is no change in temperature ΔU is equal to zero." This is true of ideal gases, not general at all!

Such books forget to mention, or you may have not noticed such notes, that $\mathrm{d}T=0 \implies \mathrm{d}U=0$ does not apply to phase transitions because there is heat exchange at constant temperature, related to different energy of different phases at the same temperature.
E.g. by water boiling, water gains about 5.5 times more energy than by heating it from its freezing to boiling poit.


text:
How do i experimentally determine the vapour pressure of a solution?
So i wanna dissolve different salts in water and measure the vapour pressure of the resulting solution. I am doing this to compare the freezing point depression of different cations, used with the same anion and in the same concen for the experiment, as well a smaller vapour pressure implies greater freezing point depression. How would I go about this? I know the apparatus available to us is a gas pressure sensor but not much about how to utilize it.
Note: If possible, I dont want to use Raoult's Law.

How are you going to find out the freezing point depression by measuring the vapour pressure? And please give your own thoughts about how this could be done. There is a big problem if you have nothing but a pressure sensor. What is it?

It seems to me he wants to relate ebulioscopy and kryoscopy via respective constants. But it seems to me like  "scrabbing himself behind his right ear by the left hand" .

@Poutnik Ja, and like a physchem lab course where they haven't given out experiment descriptions. Novel idea, that.

@Karl I remember doing the experiment described in my answer, determining water p=f(T) curve in the first year of my undergraduate studies within the lab about laboratory techniques. ( 1984 )

@Poutnik The most interesting part here was always the quantitative(ly correct) error analysis. ;-)

@Karl  Yes, e.g. proper linearization for the linear regression.

@Karl I wont be directly finding out, but dissolving the salt decreases vapour pressure right? So im gonna compare the vapour pressure for different solutions. The greater the dip, the lower the freezing point.

@Poutnik I have no idea how to find vapor pressure tbh... how would you do it? im only in high school so i know none of those constants xD. What do you mean by "scrabbing himself behind his right eye by the left hand". The general idea is to form 5 solutions with diff salts of equal concen and measure their vapor pressure. the lower the vapour pressure, the greater the freezing point depression.

Why not to directly measure decreased freezing point ?

@Raze XIX The ebulioscopic/cryoscopic constant ( Wikipedia ) relates molality of solution and change of voiling/freezing point. Byt the "scrabbing...." I mean aiming for a complex setup to measure pressures instead of easy setup to measure temperature. Also, instead of changed vapour pressure, it is easier to measure the changed boiling point.

@Poutnik because i need to use really small concentrations and the freezing point depression constant is really low, like 1.89. hence the freezing point depression or elevation is only... like 0.2, so there is too much error. what do u suggest i do? thanks

Create a vacuum enduring  apparatus, equipped with a heated flask with the solution, with constrained capillary air inlet immersed into the liquid ( provides boiling centers to avoit overheating ) and pressure sensor, attached to a vacuum source.
Combination of applied heating and vacuum, there will be established "laboratory grade equilibrium", where the measure pressure is the saturated vapor tension at the measured temperature.


text:
Why do PES graphs list 4s after 3d?
Coming straight off of some electron configuration worksheets, our teacher started to teach us how to read photoelectron spectroscopy (PES) graphs and I'm having a bit of trouble with the two.
During the previous lecture on electron configuration, we were told that electrons occupied the most stable energy level available. That came in this order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, etc.

But then, during the photoelectron spectroscopy lecture, we were told that the 3d electrons have higher ionization energy than the 4s electrons and, in fact, that this pattern continues, just counting up the orbitals: 1s, 2s, 2p, 3s, 3p, 3d, 4s, 4p, 4d, 4f, etc.

Why do electrons fill orbitals in the former pattern, but leave orbitals in the latter pattern?
Why would the 4s orbital fill before the 3d if it is less tightly held by the nucleus, and thus less stable?
I assume I have been taught some simplified versions and my teacher has my interests at heart, but I'm a bit confused.

For a given atomic kernel, the orbitals energies are not fixed like a set of shelves. Their energy depends on filling themselves and also on filling the other orbitals by electrons, because of kernel shielding and mutual repulsion.
The mutual energy is $4\mathrm{s} \lt 3\mathrm{d}$ when empty. But as kernel charge gets increased and $\mathrm{4s}$ and then $\mathrm{3d}$ are being filled, their energy order gets switched.
The typical case is the 3d energy drop $\ce{K -> Ca -> Sc}$, where the order is:
$$\mathrm{K: 4s \lt 4p \lt 3d}$$
$$\mathrm{Ca: 4s \lt 3d \lt 4p}$$
$$\mathrm{Sc: 3d \lt 4s \lt 4p}$$
well described on  charts in SE Chemistry Q what-is-d-orbital-collapse, which are related to the effective kernel charge acting upon electrons on various orbitals.


text:
Wittig reaction in presence of carboxylic acids
The following reaction came in a recent test:

According to this reaction, the acidic hydrogen would protonate the ylide, and the standard wittig reaction would not take place. This source agrees with the same results.
However, according to El-Batta et al.1, the Wittig reaction in water can take place even in the presence of acidic groups, such as carboxylic acids and phenols:

It is also interesting that the Wittig reaction in water works well with substrates having acidic unprotected functional groups present; hence the presence of a carboxylic acid group (vide infra) or phenolic groups, such as p-hydroxybenzaldehyde, gives the corresponding cinnamate product 16 in 92% after heating the reaction for 1 h at 90 °C (entry 16). In sharp contrast, the same reaction in refluxing CH2Cl2(4 h) has been reported to provide only 8% of 16

Which one of these references is correct? Does the Wittig reaction really proceed in the presence of acidic groups? I'm looking for a good reference, along with an explanation on what occurs when a carboxylic acid group is present on the substrate.
References:

El-Batta, Amer, et al. “Wittig Reactions in Water Media Employing Stabilized Ylides with Aldehydes. Synthesis of α,β-Unsaturated Esters from Mixing Aldehydes, α-Bromoesters, and $\ce{Ph3P}$ in Aqueous $\ce{NaHCO3}$.” The Journal of Organic Chemistry, vol. 72, no. 14, July 2007, pp. 5244–59. doi:10.1021/jo070665k.

Here exactly 1 equivalent of the ylide is given so I guess that matters.

By the principle of least motion, one would expect the deprotonation to occur faster than the Wittig reaction, as the former requires less nuclear rearrangement, especially considering the fact that you have a single equivalent of the ylide, so it would be anticipated theoretically that the deprotonation occurs preferentially in this case

You are comparing an unstabilized ylid, Ph3P=CH2, with a stabilized ylid, Ph3P=CHCO2Et. The latter can be stored as a solid; the former can't. The former is protonated by RCO2H. The latter, if protonated, is easily deprotonated by a weak base.

This works because the acidic starting material is going to be deprotonated as there is excess bicarbonate present. Because of the adjacent carboxylate group, the pKa of the phosphonium species is going to be around 6-7  pKa Table here so within the range where bicarbonate can deprotonate it. The reaction is between the anion of the starting material and the anion of the phosponium species in aqueous phase. This will only work for readily deprotonated phosphonium species and aqueous soluble aldehydes in the presence of excess base.

This and @user55119 's comment clarified it for me. Thanks!


text:
Purification of Red Phosphorus
The Boring Stuff
I'm well acquainted with various fields of science & engineering, but my foray into the world of chemistry is relatively recent.
I'm in a location (probably most of us are to some degree) where many chemical reagents are highly scrutinized and difficult to legally obtain from legitimate supply houses in any appreciable quantity. To do so is considered highly suspicious, requires a business account, legal documents, and so on. In short, it's a headache and it's very uncomfortable for amateur chemists/experimentalists.
And possibly rightfully so. While googling chemical (i.e. red phosphorus) purification methods, it quickly becomes apparent why. Most of the information available is in the form of poorly written (often dubious) discussions between amateur illicit drugs manufacturers.
But I digress. Despite all this, I'm determined to continue my exploration and extraction of phosphorus in it's various allotropes.

The Interesting Stuff
The only readibly available consumer class product (that im aware of) which contains red phosphorus is the humble safety-match striker pad. Here they're only something like $2 for twenty, so it seemed like a good place to start.
The sample can be collected in a bowl with the use of a solvent and/or sufficient agitation. Pretty easy. Although naturally it contains some undesired impurities, particularly including fine particles of glass frit, an unknown adhesive, the odd stray paper/cardboard fibres, and probably traces of a few various other things.
Washing with solvents and decanting seems to work to a degree, however on drying it cakes and adheres to its container almost like a paint. It's not the pulverulent powder described in the literature. In suspension with various solvents, fine particles can be seen while agitating, yet a somewhat viscous milky yellow substance rapidly adheres to the sediment upon settling.
Some sources recommend alkaline and acidic solutions (e.g. HCl, H2SO4, H2O2, NaOH, etc.) for removing various impurities. While other sources claim acids and bases like these will react with the desired red phosphorus, creating undesired molecules such as phosphorus pentoxide and ortho-phosphoric acid.
Furthermore if the desired effect is to reduce the organic compounds to a fine black elemental carbon powder, the challenge of separation still remains.
However, in my experience, when mixing with H2SO4, it seems no phosphorus is left in the beaker at the end.
It will apparently crystalize when heated under pressure, so that seemed like a possible solution, but supposedly the black allotrope posseses different properties, is highly unreactive, and can't be converted back to the red allotrope, so might not be viable.
Does anyone have a solution or suggestion for how to purify the red phosphorus? Thanks in advance.

References

Liquid-Liquid Techniques in Phosphorus Chemistry by
H Harnisch

Process for Stabilizing Red Phosphorus Against Oxidation by G Nordblom & M Silverstein

Purification of Red Phosphorus by P Miller

So in short, you want to extract the red P on match striking pads, using easy-to-obtain chemicals and apparatus.

If it comes down to extracting phosphorus from matchsticks, you can watch this wonderful video by NileRed: https://www.youtube.com/watch?v=5ZrfNAHDjWU&t=1s

@AniruddhaDeb I want to purify it; separate it from the organic components, particularly the adhesive. it seems to have a real affinity for the red phosphorous resulting in a cloudy colloid or suspension. it's proving quite difficult to separate the two.

@NilayGhosh thank you. i've seen it before. and several others. my results were similar, but i can't seem to separate it from this adhesive goop.


text:
At what pressure will hydrogen start to liquefy at room temperature?
I want to increase a fixed-size object's internal gas pressure by generating hydrogen in it, but I could not find the proper phase diagram for it. So I am wondering how high pressures I can get.

As far I know, on such extreme pressures, there is no such fixed boiling point any more. But possibly someone knows it better.

Here's your phase diagram: https://www.engineeringtoolbox.com/hydrogen-d_1419.html

The term "permanent gas" has been invented exactly to describe your situation.

Apparently, the provided answers are enough, yet just wondering, given the danger of hydrogen, why do you want to inject hydrogen? Can you consider an alternative, like nitrogen?

$\ce{H2}$ cannot be liquified at room temperature, whatever the pressure. Generally speaking, all gases can only be liquified when the temperature is under its critical value.

The critical temperature of hydrogen is 33.20 K, and so it becomes a supercritical fluid if the pressure is great enough.

Yes. But a supercritical fluid is not a liquid. It has no surface, where a boat can float.

Yes you're right

The critical temperature of Hydrogen is $\pu{32.938 K, resp.  -240.21 ^{\circ}C}$. Above this temperature, it cannot be liquified.
So to answer your question, you can get as high pressure as you can produce and the container can withstand, as there is no condensation reducing the pressure.
WARNING: An accidental explosive container rupture can easily cause severe injuries or even death.
The industrial liquifying process involves:

Cooling down with liquid nitrogen. The purpose is to reach temperatures below about $\pu{200 K}$ where Joule--Thomson-(Kelvin)-coefficient becomes positive, which enables cooling by Joule-Thompson effect via throttled gas expansion.
Liquifying hydrogen by further cooling down by the Linde process ( Hampson-Linde cycle. For gases other than hydrogen and helium, this is possible even at room temperature ( neon is at the edge ).

Does it turn into a solid eventually?

Any gas does. Hydrogen would need hundreds of GPa,

@melanieshebel "It is not common to use the ’s with non-living things."

@Poutnik It's fairly common, but the alternative for non-living things would be to use "of" or "of the" to indicate possession, like "...critical temperature of hydrogen..."

@MelanieShebel I was just going to use it. But grammer sites Ihave visited advice not to use the Saxon genitive to things.

As others have said, hydrogen cannot be liquified above its critical temperature, which my source (Wolfram Alpha chemical database*) says is $\pu{32.97 K} = \pu{-240.18 ^\circ C}$
However, with sufficient pressure, the molecules can be squeezed together until they have a liquid-like density**, and are thus no longer considered a gas, but rather a supercritical fluid.   The pressure required to reach this point is called the critical pressure which, for hydrogen, is $\pu{1.239 MPa} = \pu{179.7 psi}$.
Hence, while you can't change hydrogen into a liquid at room temperature, you can change it into a (supercritical) fluid.
Incidentally, there are two established storage technologies for pure hydrogen in vehicles.  One involves cooling the hydrogen below its critical temperature and liquifying it.  The other stores it at ambient temperature, at pressures of $5\,000$ to $10\,000\ \mathrm{psi}$.  In the ambient temperature cases they are not storing compressed hydrogen gas, but rather compressed supercritical hydrogen.
*Their sources are:  https://www.wolframalpha.com/input/sources.jsp?sources=ChemicalData&sources=ElementData
**Achieving a liquid-like density is not the formal definition of the transition from gas to supercritical fluid, but it does, I think, paint a helpful physical picture.

@Baris Note for the OP: liquid like density is still quite low, compared to the liquid at less extreme conditions. E.g. liquid water ( and water vapor as well ) near water critical temperature and pressure have both density about 200 kg/m3. For hydrogen, it would be much lower.

Hmm, formally, you are probably right, but I have never heard calling hydrogen with p>1.239 MPa as hypercritical fluid (HCF).( for 4He 0.227 MPa ). In my undestanding, speaking about HCF has IMHO sence for temperatures not too far from the critical one. Hydrogen at such a pressure and room temperature is much closer to ideal gas than to hypercritical fluid.

@Poutnik  Regarding your second comment:  I think that's precisely why it's interesting to note this, since no one does typically recognize that these ambient-temperature containers are holding hydrogen in the supercritical fluid state.

What I had in mind is 1000->200kg/m3 density drop for water, versus analogical density drop for liquid hydrogen. Hydrogen critical density is 30.3 kg/m3 engineeringtoolbox

I still insist if the state is not near enough to the critical point, speaking about supercritical state is very formal, not being related to specific supercritical properties found near the critical point, caused by very significant ideal gas deviation.

@Poutnik   Supercritical fluid extraction with CO2 is an important industrial process, and (depending on the application) can be done at several hundred bars, well above the 74 bar critical pressure.  When chemical engineers call this process "supercritical fluid extraction" they are not being "formal".  Rather, they are describing it for what it its.  Not sure what you are trying to argue here.  When I think of a formal approach to describing the critical point, I think of, for instance, the divergence of the compressibility and the appearance of critical opalescence.

Note the CO2 critical temperature and its deviation from ideal gas.

Yes, SCF is often done near its critical temperature, but your argument was that it needs to be close to the critical point to talk about supercritical properties, and if the critical pressure is 74 bar, and you're  at hundreds of bar, you are well away from the critical point, regardless of temperature.  Again, your arguments aren't making sense to me.  Not sure what you're trying to argue here.  What is your background on this subject?

Let us continue this discussion in chat.

Hydrogen can be stored at relatively high density (in some cases actually higher density than pure liquid hydrogen) at ambient pressure, by reversible formation of electropositive metal hydrides. One of the simplest such proposed storage compounds is [magnesium hydride]. Various Synthesis methods are reported by Wikipedia:

In 1951 preparation from the elements was first reported involving direct hydrogenation of Mg metal at high pressure and temperature (200 atmospheres, 500 °C) with MgI2 catalyst:[1]
$\ce{Mg + H2 -> MgH2}$
Lower temperature production from Mg and H2 using nano crystalline Mg produced in ball mills has been investigated.[2] Other preparations include:
• the hydrogenation of magnesium anthracene under mild conditions:[3]
$\ce{Mg(anthracene) + H2 -> MgH2}$
• the reaction of diethylmagnesium with lithium aluminium hydride[4]
• product of complexed $\ce{MgH2}$ e.g. $\ce{MgH2.THF}$ by the reaction of phenylsilane and dibutyl magnesium in ether or hydrocarbon solvents in the presence of THF or TMEDA as ligand.[5]

The first method with magnesium anthracene offers espcially good reversibility, as the magnesium anthracene can be formed under mild conditions from the metal and the anthracene recovered after reaction with hydrogen to form the hydride.
Cited References
1.
Egon Wiberg, Heinz Goeltzer, Richard Bauer (1951). "Synthese von Magnesiumhydrid aus den Elementen (Synthesis of Magnesium Hydride from the Elements)" (PDF). Zeitschrift für Naturforschung B. 6b: 394.
2.
Zaluska, A; Zaluski, L; Ström–Olsen, J.O (1999). "Nanocrystalline magnesium for hydrogen storage". Journal of Alloys and Compounds. 288 (1–2): 217–225. https://doi.org/10.1016/S0925-8388(99)00073-0.
3.
Bogdanović, Borislav; Liao, Shih-Tsien; Schwickardi, Manfred; Sikorsky, Peter; Spliethoff, Bernd (1980). "Catalytic Synthesis of Magnesium Hydride under Mild Conditions". Angewandte Chemie International Edition in English. 19 (10): 818. https://doi.org/10.1002/anie.198008181.
4.
Barbaras, Glenn D; Dillard, Clyde; Finholt, A. E; Wartik, Thomas; Wilzbach, K. E; Schlesinger, H. I (1951). "The Preparation of the Hydrides of Zinc, Cadmium, Beryllium, Magnesium and Lithium by the Use of Lithium Aluminum Hydride1". Journal of the American Chemical Society. 73 (10): 4585. doi:10.1021/ja01154a025.
5.
Michalczyk, Michael J (1992). "Synthesis of magnesium hydride by the reaction of phenylsilane and dibutylmagnesium". Organometallics. 11 (6): 2307–2309. https://doi.org/10.1021/om00042a055.


text:
Monosodium Diphosphate (NaH3P2O7) Preparation
I looked on wikipedia for an answer but I couldn't even find a page for this salt, a subsequent google search didn't further enlighten me.
Does any one know how monosodium diphosphate can be prepared?

The famous Holleman's book says that the pyrophosphoric acid $\ce{H4P2O7}$  makes only two salts with alcaline elements, and it is, with sodium : $\ce{Na2H2P2O7}$ and $\ce{Na4P2O7}$. It is due to the fact that the two first values of $p$K$_a$ are nearly equal. So, apparently, the monosodium salt $\ce{NaH3P2O7}$ does not exist.

Thank you Maurice. So it only coexists in solution?

$\ce{NaH3P2O7}$ is an intermediate acid-salt formed during the neutralization of pyrophosphoric acid ($\ce{H4P2O7}$) which becomes unstable as pH rises and leads to formation of trisodium diphosphate, $\ce{Na3HP2O7}$:
$$\ce{H4P2O7 + NaOH ->[pH < 3] NaH3P2O7 + H2O}$$
$$\ce{NaH3P2O7 + NaOH ->[pH = 3-9] Na2H2P2O7 + H2O}$$
$$\ce{Na2H2P2O7 + NaOH ->[pH > 9] Na3HP2O7 + H2O}$$
This acid-salt can also be produced by the reaction of trisodium phosphate and pyrophosphoric acid:
$$\ce{Na3PO4 + H4P2O7 -> NaH3P2O7 + Na2HPO4}$$
Notes and references:

Study of crystallization processes in some rare earth vanadate, tungstate and phosphate systems under hydrothermal conditions, K Byrappa and B Nirmala, Indian J. Phyx. 73A (5), 621-632 (1999) (PDF)
Its CAS number is 13847-74-0
The first dissociation of pyrophosphoric acid gives the anion, $\ce{H3P2O7^-}$. The dissociation constant is pretty large compared to 2nd, 3rd, and 4th dissociation of the acid.

$$\ce{H4P2O7 + H2O <=> H3O+ + H3P2O7¯ ~~~~ K_{a1} \sim 10^{–1}}$$


text:
Removing Iron Oxide from a Floppy Disk?
Don't ask me why, but I did an experiment where I tried to remove the magnetic iron oxide layer from a floppy disk by soaking it in HCl for a week. I put pieces of floppy disk in 0 - 1 M HCl, then measured their absorbance at 700 nm. HCl concentrations at 400 mM or more were able to remove about half of the absorbance, but not more than that.
We can even visibly see the change in color intensity.

This suggests that something is preventing the acid from dissolving the remaining iron oxide, and that simply increasing the concentration beyond 1 M won't help. However, I can still scratch the surface and remove some of the color, suggesting that it isn't buried throughout the plastic film, but still accessible on the surface. But I don't want to spend forever scratching the iron oxide off floppy disks, so I would still like a simple chemical method to remove the metal. Does anyone have any suggestions for removing it?

The magnetic iron oxide is $\ce{Fe3O4}$ and it may be considered as made of $1$ Iron(II) oxide $\ce{FeO}$ plus $1$ Iron(III) oxide $\ce{Fe2O3}$. The first oxide $\ce{FeO}$ is easily soluble into diluted HCl solutions. The second oxide $\ce{Fe2O3}$ is much more difficult to dissolve into diluted HCl solutions. The dissolution is better in concentrated HCl. Why don't you try with more concentrated HCl, like HCl $10$ M ? Have you ever measured the absorption of the solution ? It would give you more useful measurements, because the absorption would be only due to the dissolved Iron ions.

@Maurice I have not tried measuring the absorbance of the acid solutions. I couldn't see a visible difference between the 0 M HCl and 1 M HCl, so it didn't occur to me to try it. I'm a little reluctant to increase the HCl to really high levels because I suppose that the ester bonds in the polymer (Polyethylene terephthalate) might be hydrolyzed. But there's no harm in trying.

it is rather surprising that you see a loss of matter on the disk, and no color in the solution. It may be due to the fact that the only dissolved compound is the ferrous atom, which gives nearly colorless solutions. I would suggest you to add some Hydrogen peroxide to the HCl solutions, just to see if the solution takes a brown color. It should ! And I doubt that the polyethylene terephthalate will be so quickly hydrolyzed. Usually this plastic substance needs high temperatures to be hydrolyzed.

An alternative strategy might be to dissolve the polymer and leave the inorganic parts as solids. How about putting the disk in acetone?

@matt_black I need to keep the polymer intact. This experiment is about removing the iron oxide so I can try to label the polymer with other compounds. If I can't remove the iron oxide from real floppy disks, I'll try to cut out disks from blank PET and make my own.

Increase the activity coefficient with CaCl2/HCl to act on Fe3O4 together with also nano-Fe  per the reaction (14) here https://pubs.rsc.org/en/content/articlepdf/2017/ra/c7ra04312k to quote: Fe + 2 Fe(3+) = 3 Fe(2+). No conc HCl or H2O2 required, added O2 may also assist.

I have to say thank you to Maurice for his suggestion to use more concentrated HCl. I took the same pieces of floppy disk that I used before and increased the acid concentration by a factor of 10 and let them sit for another week.
At first I was disappointed, because there was no visible change in the plastic or the solutions. But when I took the pieces out of the acid and cleaned them off with an alcohol soaked kimwipe, the remaining Iron Oxide came right off. It's not perfectly clean, most likely because I had taped the pieces into my notebook after the first treatment, and I didn't clean the tape residue off before the second treatment. This is also probably why the remaining absorbance at 700 nm is so variable. But I think this method works as long as I don't get the plastic covered in tape goo. The acid treated plastic also might feel a little less stiff than the 0 M control, but this is subjective.

I wonder what adhesive is used to adhere the iron oxide to the plastic film. If you knew that, you might be able to dissolve it and recover the iron oxide intact (not dissolved). The concentrated acid apparently did a better job of dissolving the adhesive than the dilute acid.


text:
IUPAC Nomenclature of a dioic acid
I have to name the following organic cpd according to IUPAC.
My attempt:
I am unable to select a parent carbon chain as there are 2 carbon chains possible and I don’t know how do we proceed in such a type of compound.
According to the key the answer should be 3,3’ -[oxy bis(ethyleneoxy)]dipropanoic acid
Now, this seems certainly wrong to me as there is no double bond in the carbon chain so ‘ene’ must not be there, or am I missing out on something.
PubChem gives the name of the cpd as 3,3'-((Oxybis(ethane-2,1-diyl))bis(oxy))dipropanoic acid
I would be glad if somebody could explain the reason behind this nomenclature.
Thank You!

Note that the retained name “ethylene” for $\ce{-CH2-CH2-{}}$ was still included in the 1998 Guide to IUPAC Recommendations. According to the current (2013) IUPAC Recommendations, the preferred prefix is “ethane-1,2-diyl”.

okay i get it now why ethylene is used. But i still don’t understand the nomenclature of the ether part. Could you please explain that along with the choice of parent carbon chain?

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), a structure with two or more identical units that are linked by di- or polyvalent groups can be named in two ways:

by ordinary substitutive nomenclature, in which one of the parent structures is chosen as the senior parent structure and the remainder of the structure is expressed by substituent prefixes
by multiplicative nomenclature, in which two or more parent structures are connected by symmetrical or unsymmetrical single or concatenated substituent groups

Both methods yield correct names, which may be used in general nomenclature. However, the IUPAC recommendations define a preferred IUPAC name (PIN), which is the name preferred among two or more names generated from two or more IUPAC recommendations.

P-45.1 MULTIPLICATION OF IDENTICAL SENIOR PARENT STRUCTURES
P-45.1.1 Multiplicative nomenclature is senior to substitutive nomenclature for generating preferred IUPAC names to express multiple occurrences of identical senior parent structures, other than alkanes, in the name of the parent structure (see P-51.3.1). In most cases multiplicative names are shorter than regular substitutive names. A preferred IUPAC name is generated by multiplicative nomenclature when the following criteria for its use are met (see P-51.2.3).
(1) the linking bonds (single or multiple) between the central substituent group of the multiplicative group and all subsequent structural units are identical; and
(2) the multiplicative groups, other than the central multiplicative group, are symmetrically substituted; and
(3) the locants of all substituent groups on the identical parent structures, including suffix groups, are identical.
When these conditions are not met, substitutive nomenclature generates preferred IUPAC names.

The given structure fulfils these requirements; therefore, the preferred IUPAC name is generated by multiplicative nomenclature as follows.

P-15.3.1.3 Multiplicative name formation
Multiplicative names are formed in accordance with the number of occurrences of identical structural units as defined in P-15.3.1.1 and the relationship of the linking multiplicative substituent group to the identical structural units.
When a compound contains identical structural units as defined in P-15.3.1.1 linked by a symmetrical simple, compound, complex, or concatenated multiplicative group (a di- or polyvalent substituent group), it is named by stating successively:
(a) the locants for the positions of substitution of the linking multiplicative substituent atom or group to the identical parent structural unit (the locant 1 is omitted when alone in the name of a mononuclear parent hydride);

Here: 3,3'

(b) the name of the linking multiplicative substituent atom or group;

Here: oxybis(ethane-2,1-diyloxy) for $\ce{-O-CH2-CH2-O-CH2-CH2-O-{}}$

(c) the numerical prefix ‘di’, ‘tri’, etc.; and/or ‘bis-‘, ‘tris-‘, etc., with no elision of the final vowel before the name of the identical parent structural unit;

Here: di

(d) the name of one of the identical structural units including the principal characteristic group and substituents, if any, enclosed in appropriate enclosing marks (see P-16.5).

Here: propanoic acid

The numbering of the identical parent structural unit is retained and, when there is a choice, the locants of the point of substitution by the linking multiplicative substituent groups on the identical parent structure are as low as possible.

Here: 3,3'-(…)dipropanoic acid
Therefore, the preferred IUPAC name is 3,3'-[oxybis(ethane-2,1-diyloxy)]dipropanoic acid.

By way of comparison, the Blue Book contains the similar example 2,2'-[oxybis(ethane-2,1-diyloxy)]diacetic acid.


text:
Is there more efficient reaction for generating hydrogen rather than HCl+Al
I want to generate hydrogen to acquire known pressure inside of an object.
I have calculated required mole number for hydrogen using the ideal gas equation by giving wanted pressure, volume and temperature. I was thinking about using the reaction 2Al + 6HCl = 3H2 + 2AlCl3. But amount of produced hydrogen moles are not efficient in my case.
Is there a reaction which is more efficient and easy to do like with Al + HCl?

If the "object" was iron/steel , hydrogen could be charged into it easily by electrolysis.

@blacksmith37 Sadly, The energy is a limitation, it should be done with least energy.

Aluminum has a disadvantage. It is covered by a thin, transparent and waterproof layer of aluminum oxide $\ce{Al2O3}$, also called alumina. When dipped into an acidic solution, a piece of aluminum does not react immediately, because this alumina layer has first to be destroyed by the acid, and it may take one minute or two $$\ce{Al2O3 + 6 HCl -> 2 AlCl3 + 3 H2O}$$Once this $\ce{Al2O3}$ layer is destroyed, the reaction Al + acid is efficient and produces plenty of Hydrogen.
If you do not want to wait, you may replace aluminium by another metal. The most efficient metal is Magnesium, which reacts strongly and quickly with dilute solutions of HCl : $$\ce{Mg  +  2 HCl ->  MgCl2 + H2}$$ You may also use Zinc, which reacts also strongly (specially if there is one drop of Cobalt chloride in the HCl solution) $$\ce{Zn  +  2 HCl -> ZnCl2 + H2}$$
If, for some reason, you would prefer aluminum, you could replace HCl by sodium hydroxide NaOH. The reaction of NaOH on Aluminium is much more energetic than HCl on Aluminum, because the alumina layer is quickly destroyed:$$\ce{Al2O3  +  2 NaOH + 3 H2O ->  2 Na[Al{OH)_4]}}$$ and then follows the violent reaction:$$\ce{2 Al + 2 NaOH + 6 H2O -> 2 Na[Al(OH)4] + 3 H2}$$


text:
Calculating the cell potential of a half cell from two half cell equations
I understand the premise of the question and how to solve it but I don't quite understand why I'm a factor of 2 out:

Construct a potential diagram (at $\mathrm{pH} = 0$) for the reduction of aqueous $\ce{HNO2}$ to $\ce{NO}$ and then to $\ce{N2O}$ given that $E^0$ for the $\ce{HNO2/NO}$ and $\ce{NO/N2O}$ couples are $\pu{+0.98 V}$ and $\pu{+1.59 V}$ respectively. Calculate $E^0$ for the following half-reaction:
$$\ce{2HNO2 + 4H+ +4e- <=> N2O + 3H2O}$$

So by constructing the two half-equations:
$$\ce{HNO2 + H+ +e- <=> NO + H2O~~~~~~~~~~~~(E^\circ= +0.98V)}$$
$$\ce{2NO + 2H+ +2e- <=>N2O +H2O~~~~~~~(E^\circ= +1.59V)}$$
And so by adding 2 times the first equation to the second, you obtain the desired overall equation. And so you would add the individual reduction potentials to obtain the overall potential:
$$\pu{+0.98 V + 1.59 V} = \pu{+2.57 V}$$
The answer stated however is $\pu{+1.29 V}$, ie, half of my answer.
Why do we need to divide by $2$ at the end here? The standard reduction potential doesn't depend of stoichiometry, and when doing similar questions combining two half cells into a full equation (with no electrons in the equation), simply subtraction or adding the equations is sufficient. Why the difference here?

The title seems rather confused to me.

Check how are related the potential of a half reaction and it's Gibbs energy charge. Then you are 1 step from recognizing the rules for deriving one halfreaction  potential from other ones.

Conceptually related? https://chemistry.stackexchange.com/questions/139495/cell-voltage-with-different-n-factors/139507#139507

@Maurice: might be an issue in calculations. I get $\pu{0.1285 V}$ though, which IIRC rounds to $\pu{0.128 V}$..

https://chemistry.stackexchange.com/questions/9454/deriving-a-reduction-potential-from-two-other-reduction-potentials

@Mithoron yes thank you. My key downfall is that potentials are only additive if they make a complete redox reaction. If the final equation is a half cell, you have to convert to Gibbs energies.

@Josh Mitchell Rather, they are additive, if there is the same number of exchanged electrons. The relation of potential, charge and Gibbs energy simply follows the formula of electrostatics: dE = U . dq(charge) . So the alternative way to Gibbs energies are half-reaction potentials with weights of the number of exchanged electrons.

Josh, you and others have missed the key point. It is not a simple question! They are asking you to draw a potential diagram not Ecell, basically a Latimer diagram.
So take the approach of using a Latimer diagram. The book answer is correct. Nitrous acid when goes to nitrous oxide, the potential must be +1.29 with respect to SHE. Check the web for solving Latimer diagrams.
You would work on the portion starting from nitrous acid to nitrous oxide.

By definition : $\ce{\Delta $G = - zEF$}$, where $z$ is the number of electrons in the half-reaction, and $F$ is the Faraday (about $96'500$ Cb)
In the first half-reaction , $z = 1$, and $E = +~ 0.98$ V. So $\Delta G_1 = - ~0.98~ F$ (in Joules)
In the second half-reaction, $z = 2$, and $E = +~1.59$ V. So $\Delta G_2$ = - $2·1.59 F$ = - $3.18 F$ (in Joules)
For the final process, which is the sum of twice the $1$st and $2$nd equation, the free energies must be added. $\Delta G_f$ = $\ce{2 \Delta $G$_1} + {\Delta G_2} = 2(-~ 0.98~ F) + (- 3.18~ F) = -~ 5.14~ F$ (in Joules)
But the total number of electrons is $2+2·1 = 4$.
So the corresponding potential is : $\ce{E_f} = -\frac{\Delta G_t}{4~F} = \frac{5.14~ F}{4~ F} = + ~1.29$ V

I thought of this method and got this answer but it's different from the textbook's so I assumed an error. Just a textbook typo then?

hang on because surely if you just put these species in a Latimer diagram and worked out the potential for HNO2 to N2O you get (0.98+1.59)/2 = 1.29V, not 1.39.

Yeah I think you've made a mistake because the Gibbs energy for the first reaction should be 2*-0.98F as you've multiplied that reaction by 2 to get the overall equation balanced. Following the rest through you get 1.29V.

Thank you Josh Mitchell. I forgot the factor 2. I have edited and corrected my initial text. Thank you !

"…and $F$ is the Faraday (about $96'500$ Cb)" I'm sorry, what?

@ Andselisk. Yes ! F is the Faraday, and is equal to 96500 Cb. The product of Coulomb by Volt gives Joule.

@Maurice No, $F$ is the Faraday constant (about $\pu{96500 C mol-1}).$ Wrong constant name, notation (prime?) and unit symbol. Also, none of the units you've spelled out should be capitalized.

@ Andselisk. OK. Thank you. I must admit that I have difficulty in using capitalized or not capitalized letters. Sorry !


text:
Finding reactants in a azo-coupling reaction whose product is given

Find the reactants of the following azo-coupling reaction:

My solution would look like this:

My reason for the above answer
The amino Group is more activating than the nitro group and therefore the diazonium salt had to be on the left ring.
Is this correct? If not, what is the correct explanation?
Edit: To get to the azo-coupling product shown above I should use a high temperature, or? So, the reaction has to be thermodynamically controlled because the azo-group is localized on the  carbon.

You have 2 amino groups in your starting material. Do you think there will be any problem of selectivity in the initial diazotisation?

If the diazonium salt would have been made on the naphthol molecule, this molecule would have had two amino groups. And the two amino groups would have produced two diazonium salts. This could not be used to obtain the wanted final diazo compound.

Well, no it's not good at all; in particular your rationale is completely backwards.

In the future, please use a 2D chemical sketcher like ChemDoodle or ChemDraw online to draw structures: they're easier to use and produce better results

The comments summarized the reasons well: the diazotization reaction for the initial structure would produce a mixture of $\alpha$-azo and $\beta$-azo compounds, which would give a mixture of products. Also, diazonium salts couple with activated aromatic systems. The diazonium salt shown would not couple with nitrobenzene, which is highly deactivating and does not favour electrophilic aromatic substitution.
Also, having an electron withdrawing group on the diazonium salt favours the reaction. Thus, the correct solution would have the azo group on the smaller, less activated benzene ring rather than the larger, more activated naphthalene ring.

But isn't $\alpha$-position on napthalene more "prone" to EAS than $\beta$?

@RobinSingh yes it is, and major product would be coupling opposite to phenol group. However, this looks more like a textbook problem of determining which ring would have the azo group initially, so I suggested the above. If you have to practically synthesize the first molecule, then there are a lot more reactions involved.

It would probably be a good idea to summarize the comments at that point, and (or at least) link to them. There could be new comments later, or older ones deleted, it's better to have it all in one place.


text:
Artificial Muscle | HASEL's Reaction
I was watching a TED-talk on artificial muscle, HASEL, where the inventor demos that an insulated oil in the presence of electric potential field gets displaced due to Maxwell stress inside oil.
Question: What are the chemical properties specifically of the oil that allows such contraction to occur? Can oil be replaced with other fluid of different chemical properties to amplify the contraction effect?

https://science.sciencemag.org/content/359/6371/61

@Mithoron +1 thanks for adding credible reference.

The oil is simply an inert, nonconducting, unpolar liquid. No chance to improve on that.

Btw. there is nothing "self-healing" in this contraption, and calling electrostatic attraction "Maxwell stress" is just an embarrassing try to make an nice little undergrad engineering project sound "scientific".

@Karl yes thats what I was wondering as well, how could it be self-healing. Looks exactly like what you described. Anyways you mentioned no chance to improve on that..but can actuation by means of contracting bags of oil be made significantly stronger like electromagnetic actuators? Induction/electromagnetism discharges too much heat. So this could be viable alternative if actuation is strong enough with reasonable form factor. Your thoughts on this?

The loss on this one here goes up as soon as you want to become faster, because of the viscosity of the oil, which produces heat. Also the loss goes up when you come close to the maximal deformation.

@Karl I makes so much sense now. And since oil is a poor conductor of heat, the performance might be suboptimal compared to metallic motors that dissipate heat faster. I didn't understand why the loss will go up when it come close to maximal deformation, shouldn't loss be linear?

A thinner (flatter) channel generally produces higher shear fields at the same volume flow rate, i.e. more friction. The linked paper unfortunately doesn't show the relevant graphs. That's Science magazine, mediocrity in paper form.

@Karl, best comment ever :) thanks for these useful comments.

@Karl What if a bigger channel is used? will the large volume produce lower shear fields? Arent there any ways to minimize friction and improve efficiency by clever design?