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Redox and anode/cathode in lead acid battery

I have watch a video about lead acid battery and I am very confused about the way the half reactions are labelled. I have learned that oxydation is the loss of electron (so it's the half reaction with the electrons on the right side) and reduction is the gain of electron (so electrons should be on the left side of the half reaction). This question labels the half reactions the way I would have labelled them, but in the video, the reaction are labelled the other way around. Why?

I am also confused about the the way the electrode are labelled. I learned that the anode is the electrode where the oxydation occurs (which means the electrode where the electrons leave or where the conventional current enters). This is consistent with this image from wikipedia. But this is not what the video shows, why?

Anode is the element where oxidation occurs. So Pb is the anode and it is oxidized. Of course PbO2 is the cathode, and it is reduced. As a consequence, the 2nd drawing is wrong. It shows  Pb as the cathode and PbO2 as the anode. PbO2 cannot be oxidized. it cannot be an anode, whatever the nature of the currant (the movement of the electrons is contrary to the sense of the electric currant)

Thanks for this @Maurice. And what about the first printscreen.

The first print screen is correct.

@Maurice, thanks for your help. I have tried again to understand but I am sorry, I don't see it. You wrote that Pb is oxidized and that the first printscreen is right but it indicates that the half reaction with Pb is a reduction (and that PbO2 is oxidized). Why? Thanks again!

I confess you are right. The first half-equation goes from PbO2 to PbSO4 consuming 2 electrons. So it is a reduction, as Pb goes fro  +4 to +2, using 2 electrons. The half-equation is correctly written, but it is a reduction, and not an oxidation. So the name (or title) of this equation is wrong. The second half-equation goes from Pb to PbSO4, releasing 2 electrons; Pb goes from 0 to +2. So it is an oxidation, and not a reduction. The chemical equation is correctly written, but its name (or title) is wrong. So the names (or titles) of the first drawing are wrong.

@Maurice, thanks so much for having taken the time to explain it. That makes sense now! Please copy your comments in an answer so I can select it.

Let's speak about the first drawing.
The first half-equation goes from $\ce{PbO2}$ to $\ce{PbSO4}$ consuming $2$ electrons. So it is a reduction, as Pb goes from $+4$ to $+2$, using $2$ electrons. The half-equation is correctly written, but it is a reduction, and not an oxidation. The second half-equation goes from $\ce{Pb}$ to $\ce{PbSO4}$, releasing $2$ electrons; $\ce{Pb}$ goes from $0$ to $+2$. So it is an oxidation, and not a reduction. The chemical equation is correctly written, but its name (or title) is wrong. So the names of the first drawing are wrong.
Let's go to the second picture.
Anode is the element where oxidation occurs. So Pb is the anode and it is oxidized. Of course $\ce{PbO2}$ is the cathode, and it is reduced. As a consequence, the 2nd drawing is wrong. It shows $\ce{P}$ as the cathode and $\ce{PbO2}$ as the anode. $\ce{PbO2}$ cannot be oxidized. It cannot be an anode, whatever the nature of the currant : the movement of the electrons is contrary to the sense of the electric currant.


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Analyzing alternating metals in a polymer matrix with 1 micrometer depth and only 10% metal
This is a theoretical question that I am struggling to get started on for a seminar discussion. If you were attempting to design an instrument with all the basic components (source, analyzer, detector, sample holder) what type of each component would you use and why? The sample is alternating layers of heavy metal in a polymer matrix with 1 micrometer depth and 10% metal
My instinct is that if this is a metal we should use the photoelectric effect and potentially look at scattered electrons. If we are looking at electrons some time of photomultipler could amplify the signal but how would the different electrons from different metals be analyzed? What type of source would be most appropriate?

Could you described the alternating layers in more detail? Thickness? Composition?

This is a hypothetical question so no, the information on thickness is in the question and it is alternating metal layers

There have been x-ray diffraction crystals for a wavelength dispersive instrument made by alternating carbon and metal layers which were good for low energy x-rays. But such layering is only useful in a narrow range.

could you elaborated on what you mean by wavelength dispersive instrument?

Photoelectrons are pretty low energy so won’t get out from very deep in the sample. Why not RBS?

My notion is that the layers repeat and are consistent, though the metal layer and the polymer layer might not be the same thickness. Thus you are looking for Bragg diffraction. You have two choices for energy source, x-rays and an electron beam (a moving charge has a wavelength).

What would be the benefit of x-rays over an electron beam? Presumable the electron beam uses the PE effect but can this penetrate layers? X-rays I assume will diffract differently off of different metals right?

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How does an MRI work?
This is my interpretation of an MRI:
There's an external magnetic field and most of the hydrogen nuclei will align themselves with that field (which is the lower energy state) but some are not aligned (which is the higher energy state). A chemist can then direct a wide range of radio waves with different wavelengths. The specific wavelength that causes the hydrogen nuclei to go from their lower energy aligned state to their higher energy unaligned state can be determined by the wavelength of the energy emitted by the photon when the hydrogen nuclei return to its ground state. And that wavelength determines the environment. Is this accurate?

That's more like a very old form of NMR. Modern NMR is a tiny bit more complicated than that. MRI, on the other hand, is substantially more complicated, because you need to figure out exactly where that molecule you detected was; it's not enough to know its chemical environment, you also need to pinpoint its location. Why are you asking this, and what's your background? Is there something specific you're looking for? It's unlikely that we can give you a general treatise on MRI here. I'd suggest looking for a book.

@orthocresol I'm a grade 12 chemistry student so I assume the explanation I'm looking for is quite basic. Just looking for a brief, simple explanation that will help me gain some insight. Thanks

Sam, there is no emission of a photon by a proton. It is a major misconception you will in webpages on NMR.

@M.Farooq When a photon strikes a hydrogen nucleus and causes it to rise to a higher energy level. Will the nucleus not emit a photon when it returns to its ground state?

Actually, I had asked this from a very big name in NMR spectroscopy. He said "no". It is a myth created by MRI people. I have discussed this question in SE Chemistry.

@M.Farooq I would argue for the existence of photons, they are by definition the "force carriers" of the electromagnetic field. It adds unnecessary confusion to argue about this very useful idea. There is an assumption here that because MRI is an amazing advanced tool it cannot be explained in terms of simple concepts. I think in this particular case that is incorrect.

Protons do not flip their spin 1/2 without accepting spin from or passing it to a spin 1 photon. Both spin energy levels are about equally distributed, but with slight energy difference levels, belonging to photons of radio frequency. Lower energy spin absorbs such a photon,  higher level one emits such a photon. Chemical neighbourhood affects specifically the local value of magnetic induction and therefore the proportional spin energy level difference. It is similar to electron levels in UV/VIS, but in different energy, frequency and wavelength region.

@BuckThorn, I agree it is a very complex phenomenon. The person who told me that photon "emission" should not be included in the NMR detection was one of the leading editors of Concepts in Magnetic Resonance. I respect his view point. I had this misconception as well. He said that photon myth was spread MRI people. There are many articles on this topic.

@BuckThorn, I think NMR receiver coil is like an antenna. I don't think antenna engineers like to invoke photons. I had a similar dilemma in microwave emission spectroscopy. What is the horn antenna detecting?

@MFarooq It just feels like barking up the wrong tree and creating unnecessary confusion to argue about photons, because single quantum transitions are useful to understanding the QM basics underlying NMR. Granted the picture of a single discrete H nucleus undergoing a transition quickly loses importance as you dwell into details of NMR.

Also, those nuclei don't know you placed a loop around them (to a first approx), and photons are associated with the electromagnetic field one way or another, even if they might lose their meaning as spatially or temporally localized entities.

As pointed out by Poutnik, the population difference between two energy levels is very small (the energy difference is very small compared to kT).

The simplest way to detect where a spin in a sample is is to place this in a magnetic field that changes in direct proportion to depth. Then the frequency at which spins respond to the r.f. radiation can be interpreted as a depth. Clearly it is going to be far more complex to get 3D images but in essence is the same; the detected signal has to be encoded with a position.

That's a reasonable description of chemical shift based resolution in NMR, and also describes an important aspect of how one-dimensional imaging is performed. There is an additional important concept, phase. MRI relies on phase-shifting signals from different voxels (volumes of the sample) using pulsed magnetic-field gradients. These gradients are applied after the nuclei have been excited. They are also applied during detection of the signal. The gradients alter the magnitude of the field in which individual nuclei find themselves, and therefore the energy (or resonant frequency) of the photons emitted. Different energies can be mapped to different locations in the sample.
The following image shows an example MRI pulse sequence (source: Wikipedia). Time is shown along the abscissa. There is one RF channel and three gradient channels. The RF channel is used to excite the signal (first pulse labelled $\pi/2$) or trigger spin echoes ($\pi$), while the gradient channels encode position by shifting the phase prior to acquisition (channels Gy and Gz) or the frequency during (channel Gx) signal acquisition.

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Questions About an Appropriate Textbook on Electrochemistry for Self-Study
Apologies if this is a somewhat frequent request (I promise I did scour various sources before posting, but none quite applied to me), but I was hoping for some insight from the community regarding an electrochemistry textbook that I could self-study.
To give some background, I am an electrical engineering student with a minor in physics (and I have in the past taken courses in organic and physical chemistry, although that was some time ago so I forget much, if not all of that) who is interested in learning more about batteries. I am ultimately interested in their operation as storage for large-scale renewables, but I would love a deep understanding of the electrochemistry involved beforehand. I've never been happy with the hand-wavy explanations that I've seen in the past, so I wanted something fairly rigorous.
My thought was to read McQuarrie's Physical Chemistry: A Molecular Approach first (this would be a good accompaniment to my actual courses in quantum and statistical mechanics at any rate, presumably with more of a high-level focus on the chemistry which would be an interesting angle). If this is a bad idea, please let me know!
Where I am more lost is in the next step, in picking up a book that looks virtually only at electrochemistry. I've come across three books which are listed below, but I am not sure about which would be appropriate or if anyone has some insight as to why I should pick one over another.

Electrochemical Systems by Newman
The venerable Electrochemical Methods: Fundamentals and Applications by Bard and Faulkner (I'll probably get this at some point no matter what, but I'm interested in what would be a best touch with electrochemistry after McQuarrie).
Physical Electrochemistry: Fundamentals, Techniques, and Applications by Eliaz and Gileadi.

While Bard and Faulkner are respectable electrochemists, Electrochemical Methods: Fundamentals and Applications their book is not meant for the faint hearted. It is not meant for self-study either. Have you checked Bockris and Reddy's Modern Electrochemistry Vol 1 and Vol 2. It has more of a talking style.


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Does mercury react with polycarbonate?
Looking around at various material compatibility charts, they seem quite divided on the compatibility of metallic mercury and polycarbonate. Some sources (here, here, or here) indicate that polycarbonate is resistant to any sort of degradation by mercury. However, other sources (here, here, or here) suggest the exact opposite, that polycarbonate and mercury are severely incompatible. Why the disagreement?
Few of these charts explain their methodology or sources, and I've yet to find anything describing the claimed reaction in any detail. Mercury isn't particularly rare in chemistry, many people recommend storing mercury in plastic containers for safety, and polycarbonate media bottles are available from several major labware manufacturers. Therefore, it seems unlikely that such a reaction wouldn't be documented  on the internet (apart from a few table entries).

All Hg is the same, are all polycarbonates the same?

That's a great point. It looks like there are a variety of possible additives, formulations made with different starting monomers than the usual bisphenol A, and variations in molecular weight. Unfortunately it seems that most manufacturers don't like to disclose these exact details, and most of the compatibility charts don't list the particular brand or product line of polycarbonate tested.

So is that one possible answer to our question of "Why the disagreement?"


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Why is the aluminium boiling point higher than silicon? And why is silicon's melting point higher than aluminium?
Why is the Aluminium boiling point higher than silicon? And why is silicon's melting point higher than aluminium?
$\ce{Si}$ has 4 bonds which are covalent in nature. $\ce{Al}$ is metal thus possesses ionic bond.
Where have I not understood the concept?

The concept itself does not seem to be awfully useful in predicting these differences.

You can't directly compare the melting of aluminum with the melting of silicon.  Silicon converts from a nonmrtallic structure to a metallic one when it melts, aluminum is metallic all the time.

For one aluminium does not for ionic bonds with itself; those are metallic bonds.

BTW, though the BP of Al is 2743 K, it's vapor pressure is fairly high at lower temps... see https://en.wikipedia.org/wiki/Aluminium . Thus, Al metalization is easily done in vacuo perhaps 1,000 K cooler. So BP is not quite so sharp, in this case. Now gallium is ab bit more extreme...

I'm afraid that the entire premise of the question is wrong.   Both the melting point and the boiling point of Si are higher than than those of Al.  Al:  mp = 660 C, bp = 2519 C.   Si: mp = 1414 C, bp = 2900 C.

Aluminium is a metal thus formed of metallic bonds. The positive Al cations are regularly arranged in a lattice and surrounded by a sea of free/delocalised negatively charged electrons (therefore metals can conduct electricity). The negative charge of electrons and positive charge of cations have strong attraction forces which require a lot of heat energy to break. However, silicon being a non-metal forms simple covalent bonds with other silicon particles. Covalent substances have weak intermolecular forces between their molecules so they require less energy. However, covalent atoms are hard to break from each other because the pairs of shared electrons are attracted to both positively charged nuclei.

As u said covalent atoms are harder to break due to mutual sharing of electrons, the silicon would be expected to have a  higher boiling point however it doesn't. Aluminium has a high boiling point, but a lower melting point than silicon. I guess this all can be imputed to their structures perhaps.

Where this answer slightly misses: silicon has its diamond-like covalent structure only as a solid. Once melted it becomes metallic like aluminum, so a different comparison applies.


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Why are some liquids hotter than room temperature gases if liquid and solid molecules move more slowly than gases?
We learned in school that state change is a matter of molecules moving faster and faster. And we learned "heat em up, speed em up, spread em out". Why then can some liquids have higher temperatures than some gases?

Note that "quick questions" without explicitly expressed solving effort are not very welcome. It may lead to hesitation to answer or even closing such a question.  If you ask on Chemistry SE site, the site policy expects you to elaborate the question e.g. by searching in textbook+online resources, writing what you already know, what was found, understood or tried for the answer to be found.  Be aware the network of Se sites aims for a  collection of  high quality answers AND questions.

You are also comparing different stuff. Besides the intrinsic motion of molecules, consider different interactions among them. It is the latter point, if not overcame, that dictates if you have a liquid or a gas.

Interactions between molecules determine whether something is a gas, solid or liquid not just temperature
You have missed an important component of why some things are gases and others are liquids. While, in general, the hotter things are the more the molecules are spread out and the faster they will move, the specific reason why some things are liquids and other are gases depends on the interactions among those molecules.
Water molecules, for example, are fairly 'sticky' to other molecules so have to have a lot of energy to make them a gas. Argon atoms are not at all 'sticky' and it takes far less energy to make them gaseous. So argon is a gas at room temperature and water is a liquid. Other molecules similar to water (like hydrogen sulfide) have weaker attractive forces ('stickiness') so are gaseous at the same temperature as liquid water.
Another factor is the mass of the molecules. Heavy molecules with a given kinetic energy (or temperature) move more slowly than small molecules. So may also need more energy to make them a gas. Methane is a gas at room temperature but pentane is a liquid.
So the average energy (or temperature) isn't everything. You also need to know how sticky a molecules it to know whether it will be a gas or liquid at a given temperature. Hence some things are liquids, some are solids and others ar gases despite having the same temperature.

Of course heating a gas increases the speed of the molecules. But the speed and the existence of the gas molecules do not depend exclusively on the temperature. The mass and the interatomic forces have a greater importance.
First, light molecules are easier to put in the gas state than heavy molecules. That is why :

the lightest molecule $\ce{H2}$ is a gas already at very very low temperatures $(-252°C = 21 K)$

the relatively light molecule $\ce{CH4}$ is more difficult to vaporize $(-161°C = 112 K)$,

the heavier molecule $\ce{n-C4H10}$ is still more ($-1°C = 272 K$)

the extremely heavy molecules containing more than $20$ Carbon atoms are much more difficult to vaporize.

But this effect of the mass is increased if there are attractions between the molecules. For example water molecule has a molar mass ($18 g/mol$) not very different from methane $\ce{CH4}$ ($16 g/mol$}. But its boiling point is much higher : $100°C = 373K$). On the absolute temperature scale, it is more than three times the boiling point of $\ce{CH4}$ ($112 K$). This is due to special interaction forces called hydrogen bonds, that maintains water molecules in liquid (or even solid) state at temperatures between about $-100°C$ and $+ 100°C$
Furthermore, ionic compounds like $\ce{NaCl}$ are extremely difficult to vaporise $(1413°C)$, because of the huge attractions between $\ce{Na+}$ and $\ce{Cl-}$ ions.

How often is water liquid at atmospheric pressure and -100 C?

@Matt black. OK. I should have written "in condensed phase" instead of liquid. I'll edit this change.


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When calculating the distance between molecules in a solution, is it reasonable to approximate a sphere around each molecule?
I am attempting to calculate the average distance between molecules in a particular concentration of a solution. My first instinct was to treat the solution as a system of spheres surrounding the solute molecules, the size of which would be dictated by the concentration. Obviously such a calculation of the distance between molecules is primarily dimensional analysis, but is it a realistic treatment of the system, or is there a more appropriate way to arrive at the distance? A spherical treatment, in my mind, has the drawback of not accurately reflecting the distribution of solvent in the solution: a matrix of identically-sized spheres has vacant space, which a solution clearly would not have.

Which distance do you want to compute? The rdf? The distance of closest approach? An average distance? Why would the solvent matter if small enough? As for packing, you can always expect some empty space, look up "sphere close packing".

I am attempting to calculate an average distance between molecules.

Wouldn't that depend on the dimensions of your container?

Unless you're thinking about rather high concentrations, that's the way everybody does it. And large molecules which are not in good approximation spherical and still soluble are a bit of an oddity.

Be more specific: What kind of molecules, and what vol% concentration?

They are fairly small molecules: benzaldehydes; and the molar concentration is intentionally quite dilute: roughly 0.1 M.

And, sorry for the homework, what is the largest error that you could make by treating the molecules as spherical, and what is the average distance between pointlike molecules at that concentration?

No worries. I think I didn’t say it clearly enough but the spherical approximation was for the solvent surrounding the molecule. The average distance, using the spherical approximation, is roughly 3 nanometers.

Ah, OK. You are aware that neighbouring molecules in a condensed phase are in direct contact? I don't understand where you want to go with this.

I don't see why solvent spheres shall enter in a rough calculation. Treat the molecules as spheres homogenously distributed in a volume.

My guess is that what you are looking for is an estimate of the distance to the nearest neighbor assuming molecules are evenly distributed, right?

@thornewilder You say you are evaluating the distance between solvent molecules, don't you mean solute molecules. If you are looking at, say, benzaldehyde at low concentration in, say, ethanol then what you care about is the average distance between benzaldehyde molecules not ethanol molecules (so the solute not the solvent). IS that right?

Above all, does it exist another model for solving this problem ? Spheres have an advantage. They can be used to make calculations. What else ?

The nearest neighbour probability of being at distance $r$ is $ w(r)=4\pi r^2n\exp(-4\pi r^3n/3)$ where there are $n$ solute molecules / unit volume which applies when the size of the molecule is small compared to their separation.

@BuckThorn That is correct. I am attempting to calculate the distance between two solute molecules, and I am assuming that they are evenly distributed in solution.

The distribution of molecules in solution is not uniform but has a distribution. The calculation of nearest neighbour distances from one molecule to another has been worked out a long time ago see,  S. Chandrasekhar, Rev. Mod. Phys. 1943, v15, p1.
Let $w(r)$ be the probability that the nearest neighbour occurs between distance  $r$ and $r+dr$. This must be the probability than no molecules exist closer to the victim molecule that one occurs at a distance $r$ and that this molecule exists in the shell $r$ to $r+dr$. Thus,
$$ w(r)=\left(  1-\int_0^r w(r)dr \right)4\pi r^2 n$$
where $n$ is the average number of molecules / unit volume. (1 mol/m3 $\equiv 10^3N_A/10^{27} = 0.6023 $ molecules / nm$^3$).
This looks hard to solve because $w(r)$ is also inside the integration, so it is necessary to isolate the integral first (by dividing both sides by $4\pi r^2 n$) and then differentiating both sides to find
$$\displaystyle \frac{d}{dr}\left(\frac{w(r)}{4\pi r^2 n}\right)= \frac{d}{dr}\left(1-\int w(r)\right) =-4\pi r^2 n \frac{w(r)}{4\pi r^2 n}$$
and the last step we effectively multiply by 1. The result is
$$ w(r)=4\pi r^2n\exp(-4\pi r^3n/3)$$
The average distance between molecules is
$$\langle r \rangle =\int_0^\infty rw(r)dr =\Gamma (4/3)/(4\pi n/3)^{1/3} =0.55396n^{-1/3}$$
where $n$ is in number of molecules/volume then $n^{-1/3}$ is in units of length.

An easy way to estimate the intermolecular distance is to break up the container into a regular cell array, each cell containing one solute molecule. First invert the concentration to arrive at the volume per molecule. For instance, if you have a 1M concentration, then the molar volume is 1 L/mol. If you divide the molar volume by Avogadro's number you arrive at the volume per molecule. Next each cell, assumed cubic, has sides of length
$$d=\left(\frac{10^{24}}{cN_A}\right)^{1/3}=1.18c^{-1/3}$$
with the distance in nm when the concentration is in mol/L. This is also the distance between the centers of neighboring cells, therefore the estimated average distance between nearest molecules.
The shape of the molecule is not really very relevant unless the concentration is high. Note if solvent molecules are smaller or equal in size to the solute then it's ok to proceed this way. If either (1) the solvent molecules are larger and exclude some of the volume in the container so it becomes unavailable to the solute, or (2) the solute-solute or solvent-solvent interact very strongly, then you need to take the solvent into consideration. The approach applies best to dilute solutions. The approach will of course also fail if the concentration of solute is so high that it would no longer be regarded as the solute. This happens when the intermolecular distance becomes comparable to the size of the molecule.

Post-mortem: Porphyrin's result is exact provided the particle distribution is uniform and excluded volume is ignored. My estimate is comparatively pretty bad (off by a factor of ~2). It should not have been surprising. After all the particles on average can only get closer than my estimate. If you have a regular grid, moving one molecule from its position on the grid will bring it closer to another one, making it its nearest-neighbor and reducing the nearest-neighbor distance estimate. Therefore my estimate is in effect an upper bound.
Inspired by the better answer, another rough geometric estimate might be made by using the radius of a sphere having the volume per molecule:
$$d=\left(\frac{3\cdot 10^{24}}{4\pi cN_A}\right)^{1/3}=0.735c^{-1/3}$$

Where does $\ce{10^{24}}$ come from ? What is its unit ? What is C ? In what unit ?

@Maurice c is the molar concentration, the factor $10^{24}$ converts from L to $nm^3$


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Etching PCB - dealing with anhydrous ferric chloride
I'm looking into PCB etching,which requires ferric chloride (There are other etchants, but they're pretty hard to come by where I live and too expensive if I can find them).
The issue is that where I live, only anhydrous ferric chloride is sold (Which is stupid. They don't even label it properly as anhydrous - got exposed to HCl fumes for 2-3 hours, thankfully the doctor found nothing abnormal with me and I got sent off with some anti allergy meds).
I'd like to know what equipment I'll need to deal with this stuff safely. I always use nitrile gloves with these, but in particular what kind of respiratory protection will I need? I plan on just having 4 PC fans blow these fumes outside the window. If I must, I will invest in respiratory protection (It's very expensive where I live, the NIOSH approved ones at least). I'm assuming a 3M 6002 will be enough.
Please keep in mind I will be using this stuff at home. If you have any suggestions regarding alternatives, do mention them. If I can find it, I'd rather use that than this.
If this isn't the proper place to ask these questions, please direct me to a place more suitable for these questions.

$\ce{FeCl3}$ ( and similarly $\ce{AlCl3}$ ) has very different properties, comparing its anhydrous and hydrate form. They cannot be easily transformed one to the other. Heated hydrate decompose to the oxychloride or oxide eventually. Anhydrate put in water undergoes violant hydrolysis.
I suggest obtaining the hydrate, or preparing it yourself, e.g. reacting of iron, $\ce{HCl}$ and $\ce{H2O2}$ ( this mixture of the 2 reagens was an alternative Cu etching solution, when $\ce{FeCl3}$ hydrate was not easily available in the communist Czechoslovakia, but is more aggressive then $\ce{FeCl3(aq)}$ ).
Another option is to let react the anhydrate with limited water outdoors and that adding concentrated HCl. For the dissolving itself it would be the best to put in directly to the acid:water 1:1, but it would develop dangerous fumes.
Avoid mixing concentrated stuff, it would turn into a chlorine chemical weapon. Start with 10% peroxide and 1:1 acid, calculate the ratio. Tune it later at will. Keep the etching solution rather mild than strong, to simulate etching performance of FeCl3. My experience with etching $\ce{Cu}$ circuit boards was that too strong solution was often etching metal under the protection layer. Be aware of proper after treatment, to prevent future corrosion.
Usual gloves, coat and eventually shield. Well vented place is needed, at least for first experiments, until you know the behaviour ( I mean rather passively - like outdoor - and just temporary for initial experiments. when it is settled, it is not much needed. ).

Do you have any idea what kind of PPE will be required for HCl and H2O2? I've seen it used. Unfortunately I don't have access to HCl, but I do have access to tile cleaning acid, if that works.

You say well ventilated, how well ventilated? Again, I'm using this stuff at home, the best I can do is use a bunch of PC fans and blow it out a window. I can't afford a fume hood of any sorts as a student.

Another thing I'd like to clarify - while I was using anhydrous FeCl3 (The brown-green powder is anhydrous, I'm assuming), the first couple of PCBs did etch just fine. Why is that so if anhydrous FeCl3 is unsuitable for etching?

Did you perform stoichiometric estimation  ?

No, I just followed a guide I found online - add the powder to water till dark brown and then etch PCBs. That was about it.

Typical thickness of Cu is reportedly 1 ounce per square feet of PCB. 1 oz is then about 0.5 mole of Cu, what requires  1 mole of FeCl3, what is about 160 g FeCl3(aq).

Yup, all copper clad boards will have 1 or 2oz per square feet. It's the standard in most cases. Why does that matter?

So you need 160-320 g FeCl3 per sq ft of etched out copper. Of course , less, as you need to keep some. So a half of it as a guess. OTHO, you cannot count with 100% performance, as it get slow in the end. So the initial numbers may be applicable.

Let us continue this discussion in chat.

That seems about right. But my question was regarding the efficacy of anhydrous ferric chloride. Is my assumption that the brown green powder is anhydrous wrong?

What is the reason you look for anhydrous stuff if the recipe calls for water?

@Alchimista I didn't try to look for it, it's all that's available here in India. The recipes I find just say "FeCl3 powder", and the powder in those recipes tend to have a dark brown-green color, so I'm assuming it's anhydrous.

@Poutnik It's supposedly available as a yellow powder according to what I've read.

Yes, you are right - it is yellow

@ShaheAnsar ok you will be fine with the more yellow one.. It is just that you find the more or less anhydrous form.

@Alchimista I really wish I could find the solution or the hydrate, but it's pretty rare to come across where I live.

One final question - regarding the solution I've already made (Regular tap water with anh. FeCl3 added to it), are there any decomposition issues I need to be aware of? I've heard this stuff can give chlorine gas off and I have my previous mixed batch in an airtight plastic container (4-7 days old). At this point I've chosen to switch to sodium persulphate, and am just looking to dispose this stuff off.


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What is b in the Gompertz model?
I am wondering how to use the Gompertz model in modelling drug diffusion:
$$x(t) = x_\mathrm{max}\exp(-a\exp(b\log t)).$$
$x_\mathrm{max}$ is to be taken as $100,$ $a$ is the undissolved proportion at time $t = 1.$ But what is $b?$ I've seen it variously described as half of the max asymptote, the $m$ of the linear portion of the graph, or the $m$ divided by $\mathrm e.$

Your model looks different from others I've seen. I would guess that a and b are related to the max rate of change of the function during the growth phase.

I can see at Gompertz function they use the formula below. See the link for Graphs of Gompertz curves, showing the effect of varying one of a,b,c while keeping the others constant

$$f(t)=a\mathrm{e}^{-b\mathrm{e}^{-ct}}$$
where

''a'' is an asymptote, since $ \lim_{t \to \infty} a\mathrm{e}^{-b\mathrm{e}^{-ct }}=a\mathrm{e}^0=a$
''b'' sets the displacement along the ''x''-axis (translates the graph to the left or right). When ''b'' = log(2), f(0) = a/2, also called the halfway point.
''c'' sets the growth rate (''y'' scaling)
''e'' is Euler's Number (''e'' = 2.71828...)

The halfway point is found by solving $f(t) = a/2$ for $t$.
$$t_{hwp} = -\frac{\ln(\ln(2)/b)}{c}$$

Regardless of if the equation $$x(t) = x_\mathrm{max}\exp(-a\exp(b\log t)).$$ in the question is the true one, analyzing the equation as is, being aware ift has different shape than the above.:
It can be rewritten as $$x(t)=x_\mathrm{max} \cdot  \exp{\left(-a \cdot t^b\right)}$$
If $\ce{b=1}$, the equation reduces to the classical equation of the kinetic of the first order and the half-time $t_{1/2}=\frac{\ln 2}{a}$, where an asymptote at any point in time crosses x axis after $t_{1/2}$.
$$x(t)=x_\mathrm{max} \cdot  \exp{\left(-a \cdot t \right)}$$
The parameter $\mathrm{b}$ affects the time evolution of the diffusion kinetic model..
$b = 1$ keeps the relative decrease with time constant.
$b \gt 1$ accelerates the relative decrease with time.
$b \lt 1$ decelerates the relative decrease with time.
Illustrative table about the $b$ value influence.
$\begin{array}{|c|c|c|c|}  \hline
 & b & b & b \\  \hline \hline
t & 0.8 & 1 &1.2 \\
0 & 100 & 100 & 100 \\
1 & 50.0 & 50.0 & 50.0 \\
2 & 29.9 & 25.0 & 20.3 \\
3 & 18.8 & 12.5 & 7.5 \\  \hline
\end{array}$
$t$ was padded by small value to avoid log 0. Results were rounded.
$a$ was set to $\ln{2}$ to have $t_{1/2}=1$ for $b=1$.

The  Gomperts rate eqn for a species population $n$ can be written as $\displaystyle \frac{dn}{dt}=-kn\ln\left(\frac{n}{a}\right)$ with initial values at time zero, $n=n_0$ and $a>0,\; k>0$. The solution is
$$  n=a e^{\ln(n_0/a)e^{-kt} }$$
At long times $t\to\infty$ the inner exponential tends to zero and so $a$ represents the limiting or maximum population $n\to a$. The rate constant $k$  limits how quickly the final population is reached.
The equation can also be written as $\displaystyle n=a(n_0/a)^{e^{-kt}}$ or $\displaystyle \ln(n/a)=\ln(n_0/a)e^{-kt}$ which may be easier to understand.


text:
The formation of ice crystals on/in frozen food that has been defrosted and then frozen again
I made a mistake. I put my ice cream in the refrigerator instead of the freezer and left it there for a night. I realized my folly the next day.
I put it in the freezer. When I took it out, I noticed it had a bunch of ice crystals inside it and on its surface. This intrigued me (not that it is a new discovery of mine or anything). I was left with the question: Are these ice crystals the product of added water, or is it just the phase change of water already present in the ice cream.
Okay, so I have to hypotheses:
Hypothesis 1
When the ice cream was defrosted, it became more liquid and as such, the bonds of the chemicals were weakened. This allowed the water that was mixed with the milk to break loose a little. When the ice cream was put back inside the freezer, it cooled down at too quick of a pace, making it so that some of the water didn't have time to fall back in line before it underwent a phase change.
Hypothesis 2
While the ice cream was hotter and more liquid, its permittivity increased. Atmospheric water, of which there is probably much of inside a refrigerator, entered the semi-melted ice cream and over-saturated it with water. When it was put back into the freezer, this over-saturation led to the crystallization of excess water.

So, which one of these hypotheses are most likely, if not surely, correct or most correct? Or is none of them correct? Or is the phenomenon caused by a combination of the two hypotheses?

Industrial ice cream makers make it frozen much faster than household freezers. They freeze faster and ice cream is disturbed to take out heat and to keep the emulsion well dispersed. As result, frozen water based emulsion droplets are very small.
When melted and standing, emulsion slowly separates either and oily phases.
She put in household freezer, it is getting frozen very slowly. That helps further emulsion separation and ice have enough time to form big crystals.
Note that it happens for frozen vegetables as well, if melted and refrozen again. Without any added water.

Okay, so the first hypothesis was the most correct? Thanks for the answer btw!

Not really,there is no too quick freezing. Too slow freezing together with prior emulsion separation.

Yeah that's true, forgot that part of the hypothesis.


text:
Whats the driving force behind simple diffusion? Is it just the count of molecules?
If I were to have two separate containers with solutions of different concentrations with a small opening, most of the molecules would flow down the concentration gradient. Does this occur just because more molecules can pass through the opening or is there a force pushing them out?
My second question is if the molecules could pass through the opening under normal conditions even against the gradient?

Does this answer your question? Diffusion Phenomena : Why does diffusion occur?

All particles are constantly moving randomly due to their kinetic energy, and are thus constantly colliding into each other and the walls of their container (if any).
Particles in gases move rapidly and randomly and are spaced far apart, particles in liquids slide slowly over each other, and particles in solid vibrate around a fixed position.
Hence, diffusion can be defined as the overall movement of particles, from an area of higher concentration to an area of lower concentration, down a concentration gradient, due to their random motion.
Illustration of particles in a gas:

Source: wikipedia.org/wiki/Kinetic_theory_of_gases#/media/File:Translational_motion.gif
This explanation relies on Kinetic theory and Brownian motion.

So it’s just a random event when a particle goes down it’s concentration gradient?

@igorbernat For 1 particle it is random, for the huge set it is statistics.

@IgorBernát Be aware molecules are by no means aware there is anything like concentration gradient.

Yeah, this is the thing I’m  confused about. How do molecules ”know” they are doing something that’s entropically positive?

@IgorBernát they don't. They eventually move from an area of high concentration to low concentration due to their random movement. This is why diffusion is such a slow process.


text:
Type of diagram and ordering of bonds used to depict a glucose molecule

I find these type of diagrams

easier to consult when building molecules using a 'molecular building kit'. Is there a name for this type of diagram?

For the glucose molecule (in the above diagram), I noticed the order of H, and OH bonds are not in the same order (sometimes H to the left, and sometimes to the right). Does that represent anything? Will it change the way one would build this molecule using a kit?

The representation of the Fischer projection is not the standard presentation. Rotate your screen by 180 degrees. What is on the bottom should be on the top. This Fischer projection is of L-glucose not of THE glucose, D-glucose. To create D-glucose, switch the positions of the OH's and H's on the central four carbons. Better Fischer projections of D-glucose may be found on the internet.

So the diagram your referring to is called a Fischer projection and represents the non-cyclic form of glucose (usually when glucose is in the presence of water, it reacts to form it’s cyclic structure).
When looking at the Fischer projection, take note of the four carbons in the middle of the linear structure. These are chiral carbon atoms. The way the other atoms are attached to the carbon atom matters. Switching the position of an H or OH group changes the geometrical orientation of that chiral center, making the molecule different.
To make this a little more clear, think about how you have a left and right hand. Although they appear the same, the way your fingers are attached matter, i.e. you can’t put your right hand into a lefty glove because of the way your fingers are attached to your hand.


text:
In what case would water attack the least substituted carbon in an oxymercuration–demercuration reaction?
I understand that oxymercuration–demercuration is textbook Markovnikov. However, in a case like this, where the $\ce{OH}$ is formed on the least substituted carbon (secondary) rather than the most substituted (tertiary because of the methyl group), what could be the explanation?
I thought that $\ce{H2O}$ always attacks the most substituted carbon which is why $\ce{OH}$ forms there.

This is not a simple alkene. The presence of the ester makes a big difference

So would this be an example on one of the exceptions for oxymercuration-demercuration where Markovnikov doesn't apply? Because of the presence of an ester?

Think about which carbocation is more stable in this case: the one on the more substituted carbon or the one on the less substituted carbon. Hint: the ester is an Electron Withdrawing group.

Oh ok! I think it'd be because the tertiary carbon is closer to the ether, and the ether has the o=c which is an electron withdrawing group. So it will cause a withdrawing effect, rendering a bond there to be unfavourable? It would be more favourable in this case for the OH to be formed on the secondary carbon...


text:
Effect of water vaporation on saturated solution
At a high temperature, water is evaporated from the saturated solution by heating. What is the solution losing exactly by the fact that the water left the saturated solution? Does it affect its concentration...solubility?
I have a two thoughts. First, as water evaporated, the amount of dissolved substance 'inside the water' in the solution a.k.a concentration is decreasing. Thus, if the concentration decreased, based on the substance the solution may turn into unsaturated solution.
However, I have another thoughts that water only carries the 'overall' volume of the solution. Meaning, decreasing the amount of water it will decrease the volume of the solution while the amount of substance inside the solution will remain constant, thus, the concentration will increase . These thoughts leaving me a doubt in my understanding.

@Poutnik thats why I need someone to confirm this. How?

@Poutnik so that means the water just affecting the total volume of the solution? e.g. if 10 liter of water evaporated from 50 liter saturated solution (after the temperature increased) then the initial volume would be 60 liter?

The answer has been updated.

Salts do not precipitate from unsaturated solutions.
When saturated salt solution is being evaporated, water content decreases. The solution keeps being saturated, as while water is evaporating, the excessive amount of salt is  precipitating. There is less salt in the solution, because there is less of saturated solution.
10 L of saturated solution does not contain 10 L of water. By other words, evaporating 10 L of the solution does not mean evaporating 10 L of water. If you get data about solubility and density you can easily calculate how much of salt and water is contained in 1 L of the saturated solution.

That's actually my main question and I tried to connect the logic from the question I posted here. Let's say again, 10 liter of water evaporated from 50 liter saturated solution (after the temperature increased to T) and the solubility of saturated solution is 20 at that T temperature. What do we know from this? I couldn't relate the evaporated water and the solubility of saturated soln. Does this have a connection to precipitation formed? I apology if I kept bombing you with questions.

Technical note:  If you need to clarify an answer, than comment the answer. If you rather clarify and further elaborate your question, modify the question instead. That gives you more space and options to formulate and format the text.

By solubility 20 you mean 20g/100mL water, right? As that is the standard way. Then you need the density of such solution. There is many data pages or online calculators to do so.Then from this density and density of the solid salts you can calculate, how much of water evaporated and how much of salt precipitated if the total volume decreased by a given values. The key here is law of mass conservation. Law  of volume conservation does not exist.

When water is evaporated from a solution, the water molecules leave the solution so there are more molecules if the substance left in the solution, so therefore overall the amount of water decreases from the solution and hence there are more substance molecules dissolved than water molecules in comparison to the solution prior to hearing and as the amount of substance increases relatively,the concentration of the substance increases in the solution


text:
comparision between basic nature of CH3O- and -OH
when comparing the basic nature of $CH_3O^-$ and that of $OH^-$. I found two arguments. one favoring the more basic nature of $CH_3O^-$ and the other favoring the basic nature of $OH^-$. so which of them is true. and why isn't the other condition correct.
the conditions are as follows.

since $CH_3OH$ is a stronger acid when compared to $H_2O$, its corresponding conjugate base should be weaker. that is $CH_3O^-$ is a weaker base when compared to $OH^-$
since there exists a methyl group on the $O^-$ the electron density on $O^-$ increases hence it is a better base when compared to $OH^-$ since H doesn't have any +I effect /

Point 1) seems out of nothing

What makes you think CH3OH is a stronger acid than H2O ?

Their pKa values


text:
Insight into covalent bonds from electrostatic perspective
I try to understand, from the electromagnetic side only, how the molecule could be more stable than each atom, and concluded that it could not. Where is my mistake?
Here is an illustration :
By using a ruler in order to have an "example" of numerical values, we conclude, by taking the numerical values of distance in the electric field formula (I removed the common factor with $\epsilon_0$, $\pi$, etc.), that the repulsion force is bigger than the attraction force.
So where is the mistake in my reasoning ?
(I try to stay at a very basic approach of fundamental law of electromagnetism, so not using LCAO with quantum mechanics)

Quantum mechanics was necessary even to explain the existence of an H atom alone, it is not surprising that you cannot describe H-H. Besides that, if you allow for charge distribution one might perhaps find a field distribution that leads to a stable configuration. Note that this goes already - even in loose terms - in the direction of somehow distributed partial negative charges. It would be a rough way towards orbitals. I am very critical about the question in the box as for the stability of the molecule is due to none of them.

@Alchimista : Thank you. What do you mean by "charge distribution" : I put the electron here as "punctual" particle : do you mean replace it by a field=a non punctual particle ?

yes. Kind of "an orbital". I feel you got my point.

You should look into the quantum effect sometimes called the 'exchange' interaction  as this adds extra terms missing in any classical description.

Such questions troubled physicists in the early 20th century. Their inability to add more than one electron to the planetary Bohr-model was one of the reasons that quantum mechanics was developed. Nevertheless, continuing to ask this type of questions may help us sharpen our thinking and develop our intuition, as far as it goes
(you might also take a look at the simpler Gedanken-experiment about Helium explored here).
In the case at hand, we will not be able to find an analytical solution because this is a four-body problem. However, if you did want to portray the molecule with point charges in a plane then it might be more fruitful to draw an ellipse with —for argument's sake— two electrons in opposite positions ‘chasing each other’. The time-varying force balance will depend on the specific configuration as illustrated with two sketches below.

The intuitively most favourable position would be figure A. In this case, the sum of the attractive forces from the electrons on each proton is higher than the repulsive force between the protons: i.e. there is a net attraction. The least favourable case would seem to be figure B where the electrons pull the protons away from each other in addition to their natural repulsion. One might imagine that the actual force balance is some sort of convoluted average between these extremes — but calculating any such point charge model with multiple electrons has so far proven to be a dead end.
Surprisingly, the parameters of the $\ce{H2}$ molecule can be calculated using classical electrodynamics provided we approach the electron not as a point charge but as a membrane of distributed charge, as proposed by Mills. The shape of the molecule in his theory turns out to be the 3D version of the ellipse pictured above, with the (large) electrons physically encompassing the (small) nuclei.

The internuclear distance is calculated at $\sqrt2$ times the Bohr radius and the binding energy, i.e. the difference between 2 separate $\ce{H}$ atoms and an $\ce{H2}$ molecule, works out to 4.5 eV — in line with the experimental value. It is this more advantageous energy level that makes $\ce{H2}$ a stable molecule.

Randell L. Mills, "Classical Quantum Mechanics" Physics Essays, Volume 16: Pages 433-498, 2003 [equation 156 gives the binding energy of $\ce{H2}$]

I assume you mean from the classical electrodynamics side only, not from quantum electrodynamics side. By the former, even H atom cannot exist, as the electron would fall along a spiral curve on the nucleus, continually emitting radiation being radially accelerated.
For a hydrogen molecule, both electrons move around both protons, they are not dedicated to the one they came with.
Consider a similar case of charged macro objects, where applying the pre-relativistic classical electrostatics is justified. Imagine 4 respectively charged balls in gravityless vacuum.
The scenario in the bottom picture is one of the less preferred positions, having rather higher energy due short distance of negative charges. But there is plenty of different positions where the whole system would have lower energy than if they were 2 and 2 separately.
Note that even classical system have no general analytic solution for the motion equations for more than 2 objects.

Thank you. For the system of atom itself, nucleus and its electron, I was not putting the forces on the pictures, sine as you say, this is quantum mechanism that dictates that. So I was only isolating the forces that communicate between the two distinct atoms.

But inter-atom interactions have equally quantum character as nucleus electron interactions. Additionally, aside of classical coulombic interaction, there is Pauli spin interaction


text:
Proton NMR splitting in 2-substituted furan
I have recently run proton NMR on 1-(furan-2-yl)ethan-1-ol I synthesised in the lab.

My 400 MHz spectrum consists of the following peaks:

7.36 ppm (dd, J = 1.9, 0.9 Hz, 1H)
6.31 ppm (dd, J = 3.3, 1.9 Hz, 1H)
6.22 ppm (dt, J = 3.2, 0.9 Hz, 1H)
4.87 ppm (q, J = 6.6 Hz, 1H)
1.53 ppm (d, J = 6.6 Hz, 4H).

For the peak at 6.22 ppm, I am wondering why a doublet of triplets is observed rather than a doublet of doublets. I understand the long-range splitting that results in the dd, but am just a little surprised by the dt.
I attach a picture of the doublet of triplets.

Maybe you have a ddd with two similar Js. The additional doublet might be due to coupling to the proton at 4. 87 ppm. Does that one show an additional small splitting?

Can you run a 2D NMR experiment? ¹H-¹H COSY would clear this right up.

The compound is commercial, available from Merck. Unfortunately they don't show NMR spectra for free anymore. You can try a simulation here. See also ChemSpider.

I agree with @BuckThorn. I suspect the coupling in the quartet is just not fully resolved. You may be able to play around with window functions (or apodisation as it's referred to in Mestrenova) to try and tease out this small coupling. Suggestion: try selecting gaussian and exponential, set the exponential value to -2 Hz and the gaussian value to  0.2 GF, or something similar. See if the quartet is still a quartet.

@orthocresol you are correct! Thanks for the advice on the window functions. Once adjusted, the quartet of doublets was clear! What exactly do the different window functions do?

Great! I'll convert my comment to an answer with more detail.

I agree with Buck Thorn's comment. I suspect the coupling in the quartet is just not fully resolved. I'm not 100% sure why, but one possible reason is because there is further unresolved coupling with the OH peak: instead of being resolved as a splitting in the spectrum this just broadens the individual lines. Or it could just be that that proton relaxes faster.
You may be able to play around with window functions (or apodisation as it's referred to in Mestrenova) to try and tease out this small coupling. My suggestion would be: try enabling/selecting both "Gaussian" and "Exponential", set the exponential value to -2 Hz and the gaussian value to 0.2 GF, or something similar. See if the quartet is still a quartet. [From the comments, it seems that this has worked and that the quartet is actually a quartet of doublets under the hood.]
The window functions are basically mathematical functions that you multiply the FID by prior to Fourier transformation. Peaks in the FID generally have a damped sinusoidal form. The faster a signal decays to zero, the larger the corresponding linewidth will be, i.e. the broader the peak will be.
The two "big" window functions for 1D processing are exponential and Gaussian window functions. I won't go into too much detail because these are very well-described elsewhere. See e.g. Chapter 5 of Keeler's Understanding NMR Spectroscopy (2nd ed.), or Glenn Facey's blog posts on Exponential Line Broadening and Resolution Enhancement (which also include some helpful pictures). Googling for either "window function nmr" or "apodisation nmr"  will also throw up a huge number of documents.
The exponential one basically multiplies the FID by another decaying exponential. This is largest at the start and decreases as time passes, so effectively causes peaks to look as if they're decaying even faster. This leads to broader lines, but because the decaying exponential is largest at small $t$ where the signal is large, it "emphasises" the parts of the FID that have more signal. Effectively, you lose spectral resolution but gain some signal-to-noise.
A Gaussian window function (or a combination of Gaussian + exponential, which is more common) starts small, reaches a maximum in the middle, and tails off again towards the end. In a way it has an opposite effect to the exponential: by trying to "compensate" for the natural signal decay, it makes the peaks look as if they are decaying more slowly, i.e. you get narrower lineshapes. On the other hand, you lose signal-to-noise because you de-emphasise the start of the FID which has the greatest signal.


text:
Halogens chemical reactions with each other
https://en.wikipedia.org/wiki/Halogen
Can Halogens react with each other?
Examples :
Chlorine + Flourine
Flourine + Iodine
Iodine + Bromine
Chlorine + Bromine
and so on.
Permutations and Combinations.
Chlorine + Flourine =
Chlorine + Bromine =
Chlorine + Iodine =
Flourine + Bromine =
Flourine + Iodine =
Bromine + Iodine =
What will be the chemical reactions & applications?

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

There is plenty of information about all these compounds if you look at "Interhalogen compounds" through Google. All these compounds are known :
AX- type : ClF, BrF, BrCl, ICl, IBr
AX3-type: ClF3, BrF3, (ICl3)2,
AX5-type: ClF5, BrF5, IF5,
AX7-type: IF7.
Their synthesis, structures and properties are nicely described, much better than what I could do here.
And please note that the symbol F describes the element fluorine, and not flourine. Its compounds are fluorides, and not flourides.


text:
Will the entropy change be equal if a state change is brought about reversibly and irreversibly?
Since entropy is a state function so it doesn't depends on the path followed. So will the entropy change be equal if a state change be first brought about reversibly and then irreversibly.

The Q/A model of StackExchange network does mean the question title is the question and the question body is the answer. The title is to be a summary question or word expression and the body is expected to contain the full question elaboration. But it is not really a question if you already knew an answer, so why it was posted ?

In my text book it was mentioned that in irreversible process the entropy change of system+the entropy change of surroundings taken together is greater than zero and in reversible process it is equal to zero, so i was getting a bit confused. That's why I asked if the entropy change would be same Or not. Can you explain the above point to me that why in irreversible process it happens so?

If the initial and final states of systems are the same for both reversible and irreversible process, the entropy change for the systems will be the same.
But the entropy change for the surrounding will not be the same, as the final states will be different. The surrounding for the irreversible process will end with higher entropy.
Therefore the total entropy change  is zero for reversible processes, but positive for irreversible ones.
The irreversibility condition applies not only to system-surrounding interactions, but for system internal processes as well.

Yes, that is how we determine the entropy change of a system for an irreversible process .  We determine the entropy change for a reversible path between the same two end states.

Yes, they will be the same. As you know, entropy is a state function and it only depends on the initial and final states. Whether the process is reversible or not does not affect the entropy change (i.e. the entropy change does not care about the path).


text:
Deposition of saturated solution and affect of evaporated water
20g of water is evaporated from 200g of saturated solution of KNO3(aq) at 60°C by heating and the solution was cooled to 20°C. How many grams of KNO3 were deposited? The solubility of KNO3 in 100g of water is 32 at 20°C and 109 at 60°C.
I have a doubt on "20g water is evaporated from 200g KNO3" statement. I'm assuming that as 20g evaporated that means the initial saturated solution has mass of 220g. This is at 60°C. I also assume the solubility in g/L.
Now, at 60°C the max amount of solution that can dissolve is 109. Thus, if we get 220g this leaves $220-109=111g$ of solvent that is not dissolved.
Then, if the saturated solution is cooled down to 20°C, the solubility will decrease to 32g/L while the amount of precipitation we have remains the same. This leaves $111-32=79 grams$ of KNO3 precipitated. Therefore, the final amount of KNO3 deposited is 79g.
Did I treat the amount of water evaporated in a correct way? I'm not sure whether the water is affecting the total concentration or 'overall' volume of the saturated solution.

You would serve a good service to yourself and readers, if you evaluate everything via symbolic algebraic expressions and equations. This way you would easier learn the principles and as bonus, both you and readers would easier spot eventual mistakes done.

if 20g is evaporated from 200g, how could be the initial mass 220g ?? I is like saying I have $200, I give you $20 and I initially had $220. Strange, huh?

I'm new here and I was looking for that, I appreciate your info.
I'm thinking before the solution get evaporated so initial mass is 220g and I feel this is really wrong because I never cover the topic relating to evaporated water in solubility and I couldn't find the source on internet explaining how it can relate to it's solubility.

Solubility for given T is the constant. No evaporation related operations affect it.

So what is the 20g evaporated water tells us?

You cannot be serious. It says 20 g water was evaporated and the mass of the condensed part of the system decreased from 200g to 180g.

There's no reason for me to joke, I just want to understand more about this. I do not have any physical or digital source regarding this topic and I only utilize internet. However, the topic covered in this area only covers by the definition of concentration, solubility, and type of saturated soln. I couldn't find the answer of my question making me unable to answer my practice question. How can i proceed with 180g? Is my working for the rest correct?

It seems to me you are struggling with the basic concepts from which you are supposed to design the answer. As the first step, forget evaporation and focus on concepts of solubility constants and terms related to solute content: %content w/w, w(weight=mass)/v(volume), v/v, molarity, density. Search Wikipedia. When you get it, all the evaporation task will be a piece of cake to you.

I have bombard my self with the basics especially the conversion and how mass, moles, M, V, P, concentration, temperature relate to each other. I asked here after I'm stuck for days letting my brain to be in diffuse mode and learn something else and I came to the point where I cant spend more time to solve this 1 question. Furthermore, I considered the volume and density in this question however, the density is not given. I treat the question as I'm in the test condition and I can't look up on internet about the density and I must use the values that solely given from the question.

You have all info needed there. The key point is to relate solubility in x/100mL(about 100g) versus content y g in 100 g of solution. y=x*100/(100 +x). As the task, in opposite to your prior question, does not consider volumes, but masses, you need not to consider densities.

The purpose of this site is to help to understand how to solve tasks, not the solving itself.

Thank you for sticking with me, I figured it out.

Unfortunately, your calculations are a little bit wrong because of the poor choice of the total initial mass. It is $200 g$, and not $220~ g$. So here are the right calculations.
As $100 g$ water dissolves $109g$ $\ce{KNO3}$ at $60$°C, producing a total mass of $209g$, you may deduce that $200g$ of your initial solution contains $(200/209)·109g = 104.3 g$ $\ce{KNO3}$ and $(200/209)·100g = 95.7g$ water.
Now you loose $20 g$ water by evaporation. The amount of $\ce{KNO3}$ is still $104.3 g$. Remaining water is $95.7g - 20g = 75.7g$ water. At the final temperature $20$°C, $100g$ water can only dissolve $32g$ $\ce{KNO3}$. As a consequence, $75.7$ g water can dissolve $(75.7/100)·32g = 24.2 g$ $\ce{KNO3}$. The rest is deposited in the flask. It is $104.3 g - 24.2 g = 80.1 g$.
Please note that the mass has a unit, the gram. It is incorrect to say "the solubility of $\ce{KNO3}$ in $100g$ of water is $32$ at $20°$C and $109$ at $60°$C". You should have said that "the solubility of $\ce{KNO3}$ in $100$g of water is $32 g$ at $20$°C and $109$ g at $60$°C

Thank you for confirming this. I literally just solved the question after Poutnik gave me the hint and I was hesitating to confirm my answer.

For the note, that's original from the question. I was wondering why it did not state any unit as well.

Please, both OP and responders, always use as the first step a general algebraic solution, for 3 reasons.: 1/better grasp of principles for OP. 2/ better orientation and spotting eventual mistakes. 3/solution reusability and permanent value.  Use eventually MathJax for mathematical/chemical expression/formula formatting. E.g. $$\frac{[\ce{SO4^2-}]}{[\ce{HSO4-}]}$$ looks like $$\frac{[\ce{SO4^2-}]}{[\ce{HSO4-}]}$$


text:
How can mass spectrometry be used to determine concentration?
Which forms of mass spec will determine accurate concentrations of metal ions in a sample? Preferable not using GC/MS but can MALDI ToF or quadrupole do this somehow. Also, not relative concentration but actual concentration.

As indicated in the Wikipedia page, Inductively Coupled Plasma Mass Spectrometry (ICP-MS) is the technique of choice to quantify metal ions in a sample. In summary, this technique is based on the atomization and ionization of the sample metal atoms in a plasma torch, and the ions are then separated based on their mass, usually with a quadrupole mass filter, and then directly quantified on the detector. External calibration with standard solutions of known metal concentration or isotopic dilution then affords access to absolute concentrations.


text:
why does the gap between the energy levels of a hydrogen atom keep decreasing as we move further and further away from nucleus?
why is the gap between the first two orbits greater than the gap between second and third orbit?
shouldn't the gap keep decreasing as the radius keeps increasing? Can someone explain this ?

Whatever the radius, the allowed energy levels are proportional to $1/n^2$.
The first gap is proportional to $1/1^2 - 1/2^2 = 0.75$.
The second gap is proportional to $1/2^2 - 1/3^2 = 1/4 - 1/9 = 5/36 = 0.139$. This is smaller than $0.75$.
The third gap is proportional to $1/9 - 1/36 = 1/12 = 0.083$. It is still smaller.

So mathematically speaking, if we look at the equation for the energy levels of the hydrogen atom below,
$E=- \frac{ \mu e^4}{8 \epsilon^2_0 h^2 n^2}$
where n is the principal quantum number, $\mu$ is the reduced mass, and the rest are constants, we see that as the principle quantum number increases to infinity, the energy eigenvalues approach zero and the spacing decreases (1).
This arises from the electrostatic forces of attraction between the nucleus and electron. By consulting Coulomb’s law,
$F = \frac{k q_1 q_2}{r^2} $
we see that the magnitude of the electrostatic force is proportional to the square of the distance.
Therefore, as the energy level is closer to the nucleus, the spacing between each quantized level, gets greater because of the $r^2$ term.

The effect is a consequence of the Coulomb law for the interaction of the proton and electron; potential energy $\displaystyle V(r)=-\frac{e^2}{4\pi\epsilon_o}\frac{1}{ r}$ producing (quantised) energy $\displaystyle E(n)=-hcR\frac{1}{n^2}$, where $n=1,2\cdots$. (That the energy does not depend on angular momentum quantum number $l$ is an accidental effect of the Coulomb potential). If another potential, say linear, quadratic etc.  were to be used a different dependence on energy would be obtained.
The lowest energy is $-hcR$, but why is it not zero? The answer is that the uncertainty principle restricts this because as the energy gets lower the electron is confined into a smaller space, i.e. is closer to the proton, this means that the kinetic energy increases and so does the uncertainty in the momentum. So the energy is a compromise between the kinetic and potential energy and this applies to all levels. The existence of atoms would then seem to be a consequence of the uncertainty principle.


text:
Meaning of 'constant T and p'
In chemical thermodynamics, the condition "constant temperature and pressure" appears almost everywhere. However, I felt a small ambiguity in the phrase so I wanted to clear my doubts. When we say "constant $T$ and $p$" in chemical thermodynamics, which of these two statements do we actually mean?

$T$ and $p$ are the same at the beginning and end of the reaction
$T$ and $p$ are the same throughout the entire reaction

If we care about path-dependent variables, then there is a major difference between the two meanings.

It means that the temperature and pressure may change during the transformation. But th initial and final values of $T$ and $p$ are the same.

All time if process-wise, initially and finally if state-wise.

@Poutnik if that's the case then it's definitely an ambiguous phrase

@juanperez I suppose it is true, but only if it is cut out of its context and if one is not familiar with TD states and processes. For the starters, they can think constant all the time. When they learn state variables do not depend on the path, they will learn that just states matters.

The key point is you will have to think whether the process is reversible or irreversible. You will have to be careful with the context.

As a chemical engineer, my perspective on this differs from those of the other responders.  I regard a process occurring at constant temperature and pressure as one in which

The system is held in contact with a constant temperature over the portion of its interface with the surrounding through which heat transfer is occurring, not necessarily the initial temperature of the system, throughout the process

The external pressure (or more precisely, the external compressive stress) at the portion of the system interface with its surroundings at which displacement work is being done is held constant (not necessarily at the initial pressure of the system) throughout the process.

Why is it enough to know only about the nature of interface for chemical engineers?

@Burian It may not be enough to solve a specific problem, but it is what we define as a constant-temperature process or a constant-pressure process.


text:
Fundamental vibrational modes for QM9 molecules
QM9 dataset contains about $134000$ small organic molecules with following properties computed using DFT approaches:
A = "rotational_constant_A"
B = "rotational_constant_B"
C = "rotational_constant_C"
mu = "dipole_moment"
alpha = "isotropic_polarizability"
homo = "homo"
lumo = "lumo"
gap = "gap"
r2 = "electronic_spatial_extent"
zpve = "zpve"
U0 = "energy_U0"
U = "energy_U"
H = "enthalpy_H"
G = "free_energy"
Cv = "heat_capacity"

In many papers, fundamental vibrational modes $\omega_1$ is also mentioned along with these properties. I was wondering:

Can we calculate $\omega_1$ given the properties above?
Are rotational constants used for computing $\omega_1$?
Do we need DFT calculations to compute $\omega_1$ and zero point vibrational energy (ZPVE)?

Only for a diatomic molecule and only assuming the vibration is harmonic. In this case, $\mathrm{ZPVE}=\omega_1/2$. If its anharmonic or has rovibrational coupling, even the expression for a diatomic doesn't not allow you to determine $\omega_1$ from just the ZPVE. With polyatomic molecules, the ZPVE has contributions from all the vibrations, so you can't just extract a single vibrational frequency.
Rotational constants are not used in computing $\omega_1$ (unless you are considering forms of rovibrational coupling).
You can determine $\omega_1$ from other electronic structure methods and even classical molecular dynamics, but assuming you don't want to compare against experimental data, you would need to repeat these DFT calculations to get the corresponding $\omega_1$. For this dataset to include ZPVE, the underlying calculations would have had to determine all the vibrational frequencies of the molecule, but they seem not to have included this data in the QM9 set.


text:
Unsure whether this classification of Volatile Organic Compound and/or Particulate Matter is correct
I am asking on behalf of a friend so please bear with me!

Instructions:

For each chemical, classify whether it is a VOC (Volatile Organic
Compound) and/or PM (Particulate Matter 2.5 or Particulate Matter 10).

A chemical can sometimes be both VOC and PM, they are not mutually
exclusive.

If it is not VOC or PM, then label it as Other.

Use the following definition for VOC: "Volatile organic compounds
(VOC) means any compound of carbon, excluding carbon monoxide, carbon
dioxide, carbonic acid, metallic carbides or carbonates, and ammonium
carbonate, which participates in atmospheric photochemical reactions."

Use the following definition for PM2.5 and PM10: "PM10 means
particulate matter with an aerodynamic diameter less than or equal to
a nominal 10 micrometers" "PM2.5 means particulate matter with an
aerodynamic diameter less than or equal to a nominal 2.5 micrometers"

The main trouble is that while I can classify the VOC's, I am not at all sure about the PM's and others.
I have already marked some chemicals with "1" meaning that it is a VOC, but I am unsure of whether it is also a PM, if it is not a PM then just put a "0" in that cell. For example, for "2-ACETYLAMINOFLUORENE", if this is not a PM, then it should be VOC = 1, PM = 0, Other = 0.
Some other chemicals are completely empty such as "CHLORAMBEN" and "HYDRAZINE". For these, I simply have no idea how to classify them.
Could anyone check what I have done so far, and provide some hints/guidelines on what to do with the rest, especially on how to classify a PM please?
Thank you so much!

2-Acetylaminofluorene is a solid, mp 192-196 C. It satisfies the definition of being a VOC by virtue of being a compound of carbon, not excluded by the definition. It could be fine enough to be classified as PM 2.5 or 10 if it is powdered; more information would be necessary about the source of the compound. It is also carcinogenic, but that does not appear to be involved in the questionnaire.
Chloramben can be looked up and is also a solid with a high melting point: a VOC and possibly a PM.
Hydrazine is only dangerous, but not a VOC, and a liquid with a bp of 114 C. Not a PM. I don't think a spray of droplets would qualify as a PM; they would just evaporate.
Diisocyanates: VOCs for sure. Hazardous for sure. If you allow a pot of diisocyanates to evaporate, that's all: no PM. However, I can imagine a source of sprayed diisocyanates (in a paint?) that would result in polymerized particulate matter: then it could be a PM. Thus the classification would depend on the source of the material - the environment and the application.
Polycyclic aromatic compounds: most often encountered in smoke from various sources: VOC for sure, probably also PM because they could be larger agglomerates than single molecules.
This could be a fun exercise for a chemist to look up and play with mentally, to get to know and understand a large number of chemicals. For a non-chemist, this could be very difficult.


text:
Is it easier to mix a solution that is in the liquid or in the vapor phase?
In the article on dilution, Wikipedia has the following statement:

The same direct relationship applies to gases and vapors diluted in air for example. Although, thorough mixing of gases and vapors may not be as easily accomplished.

I was wondering if there is any supporting evidence for the second statement in the quote (or for the opposite statement). I would think that gases mix quickly, given how quickly gas expands into a vacuum (I am thinking of the classic picture of half a container filled with gas, and a divider that is removed). On the other hand, I don't know how effective mechanical mixers are for the gas phase. For the liquid phase, it is common to mix large volumes with a magnetic stir bar whose volume is comparatively small.
Is it easier to get a homogeneous mixture for a liquid phase than a gas phase? I am excluding immiscible liquids from the question because they will never become a homogeneous solution. To focus the question a bit, I am interested in the rate of mixing; to make it comparable, the mole ratios should be equal in the comparison of the liquid phase case and the gas phase case. Mixing could either be by diffusion, or sped up by a mixer appropriate for the system.

It might refer to problems other than achieving homogeneity. See for instance https://en.wikipedia.org/wiki/Gas_blending

I think that's comparing pears and apples, and to the statement on wp, I say "balderdash". ;-)

The overriding factors are solubility and viscosity. Since, AFAIK, all gases are miscible, and have much lower viscosity than liquids (except perhaps He II), then it would be easier to get a uniform gas mixture with the same amount of work. That said, they might be talking about getting a uniform mixture in a specific situation, e.g. IC engine cylinder bore or burner nozzle.


text:
Carbohydrate Chemistry for Candy Making
I am trying to understand the different behaviors of sugar syrups when it comes to candy making. Specifically, maple syrup vs molasses.
In making maple candy, pure maple syrup is heated to soft ball stage, which is 235F or about 113C. It is then cooled without stirring to 175F (79.5C), then beaten until it becomes opaque and thick, indicating sugar crystals have begun forming. When it cools to room temp, it has a somewhat fudgy consistency and is fairly brittle. If you do the same process with molasses, you get a sort of proto-taffy. In fact, you can pull it, and it will incorporate air, but doesn't quite reach taffy consistency (taffy typically also includes a fat and an emulsifier, and is also cooked just above soft ball stage to 244F so this is probably why).
I've been able to find some resources on the differences in chemical composition. This paper: https://doi.org/10.1016/0963-9969(96)00000-2 tested 80 maple syrup samples and determined the average syrup was 68% +/- 4% sucrose, with less than 1% each fructose and glucose. In contrast, the Encyclopedia of Food Sciences and Nutrition (Second Edition) available at https://www.sciencedirect.com/topics/neuroscience/molasses lists molasses as 30-40% sucrose, 4-9% glucose, and 5-12% fructose. Since glucose syrup is added to other candies (including taffy, ice cream base, and fondant) to prevent crystallization, I assume the higher level of glucose is what's preventing the molasses from forming crystals and making a candy similar to maple candy?
Maple syrup also has a significantly higher water percentage, 32% on average compared to molasses at 17-25%, so the cooking process to bring maple syrup to temp is longer than it is to bring molasses to temp. Additionally, maple syrup is mildly acidic, with malic acid being the largest contributor. Is it true that with a mild acid the cooking temperature is not high enough to hydrolyze the sucrose to glucose and fructose?
What ultimately drives the crystallization process? If, for example, you added a significant measure of table sugar to molasses to bring the total sucrose percentage up to 75%, and then cooked it as with maple syrup, could you reasonably expect it to crystallize into a solid candy? Would it have to be a higher percentage because of the glucose present? Or is the higher native glucose level an insurmountable issue in terms of the chemistry?
ETA: Is there a food-safe reaction that can be done in a home kitchen to synthesize the glucose and fructose in molasses to sucrose?
(a final note, I have tested this with 3/4 cup of molasses to 2/3 cup sugar, and again got something closer to a very soft and sticky taffy. It is possible that this was simply not enough sugar, and that it needs to be portioned by mass not volume.)
Any insight is appreciated, thank you!

The chemistry of naturally occurring mixtures in food is very complex.Some substances could slow or prevent crystallization, others could speed it.What you're doing by experimentation may be the best way to find optimal techniques. As for synthesizing sucrose from fructose and glucose, this is not feasible in the kitchen.,

Sure, but ingredients get pricey and some basic ideas beyond blind experimentation might be useful to direct the experimental methods. Are there guiding principles that govern sugar crystallization in solutions? I'm assuming crystallization of glucose does happen, mostly because it is possible to buy anhydrous glucose crystals, but it doesn't seem to be the primary mechanism in confectionery, and possibly isn't attainable at home.
And thanks, that was my guess on sucrose synthesis.


text:
Does helium freeze at absolute zero?
Does helium freeze at absolute zero? In theory, it should, since all random motions of particles stops at absolute zero, right?

It actually doesn't. I'm not talking about ordinary pressures, I'm talking about absolute zero at any pressure.

Not at atmospheric pressure.  You need at least 25 atmospheres pressure to solidify helium-4 and at least 29 atmospheres (0.3 K) for helium-3.  See Wikipedia.
At low pressure you fail to solidify the helium because of weak interatomic interactions combined with quantum mechanical zero point energy.  Higher pressure favors forming a denser structure with an ordered lattice arrangement, thus a crystalline solid.


text:
What does the number 4-88 mean in "Polyvinylalcohol 4-88"?
Polyvinylalcohol is sold in different forms, one of those is 4-88.
What does that number mean?

The value $\bf{4-88}$ can be described as follows:

$\bf{4}$: This represents the viscosity (in $\pu{mPa s}$) of a $4\%$ solution in water at $\pu{20 ^\circ C}$. This value is somewhat proportional to the molecular weight of the polymer chain.
$\bf{88}$: This represents the percentage of moles of acetate ester that were hydrolyzed, since polyvinyl alcohol is manufactured by hydrolysis of polyvinyl acetate. This means that it is an indicator of its purity.

Source: you can check the catalogue at Sigma-Aldrich (Mowiol is a brand name for PVA) and see how those numbers correlate to those parameters.


text:
Reaction of phosphorous acid (H3PO3) and potassium hydroxide (KOH)
For the reaction of phosphorous acid ($\ce{H3PO3}$) and potassium
hydroxide ($\ce{KOH}$), write (a) the balanced molecular equation
and (b) the net ionic equation. Note that phosphorous acid is
a diprotic acid.
$$\ce{H3PO3(aq) + 2KOH (aq) -> 2 H2O(l) + K2HPO3 (aq)} \tag{a}$$
$$\ce{H3PO3(aq) + 2 OH-(aq) -> 2 H2O(l) + HPO3^2-} \tag{b}$$

The answers are above. Can somebody explain why it is $\ce{K2HPO3}$ in (a)? I am familiar with monoprotic acid, and this is my first encounter of diprotic acid. Thanks.

Note that Chemistry SE site policy is using just plain raid for titles, for indexing and searching reasons.

Instead of writing H_3PO_3 in MathJax, write e.g.  \ce{H3PO3 + 2 KOH -> 2 H2O + K2HPO3}. \ce{} invokes mhchem   MathJax extension. It additionally uses expected upright font. See $$\ce{H3PO3 + 2 KOH -> 2 H2O + K2HPO3}$$

@Poliom. Diprotic acid react always with 2 OH- ions. You should know the first and most important diprotic acid : the sulfuric acid $\ce{H2SO4}$. It reacts according to : $$\ce{H2SO4 + 2 KOH -> K2SO4 + 2 H2O}$$

I’ll point out that calling it phosphorous acid is wrong; it’s phosphonic acid.

You have made the assumption that all the hydrogens are attached to oxygens. This is not a good assumption; one of the hydrogens is bonded directly to phosphorus and is not removed by OH-.
A better description of phosphorous acid is HPO(OH)2 with one oxygen doubly bonded to phosphorus  wikipedia

Phosphorous acid, or phosphonic acid, is actually far from unique.  You might want to look at what seems to be carbonous acid, $\ce{H2CO2}$.  But it's actually formic acid and its formula is generally rendered as $\ce{HCOOH}$ or even $\ce{H-C(O)-OH}$ -- one hydrogen atom attached to oxygen and the other attached to carbon.  Similarly to the phosphorus compound in the question, even strong bases generally react only with the former hydrogen atom, at least in water solution, so for instance potassium formate is $\ce{KCHO2}$ or $\ce{HCOOK}$.  When you get into organic chemistry you will encounter a lot of compounds like formic acid where some hydrogen atoms can dissociate as protons and others, not.


text:
The unit of relative atomic mass
The mass of Carbon-12 atom is 12.0000000 a.m.u. For relative isotopic mass, e.g. Mg-24 atom is 24 compare with the c-12 as standard. However, why is it unitless? Shouldn't it be 24 a.m.u.?

Notice the term relative, meaning ratio of 2 quantities of the same unit. Note that the mass of the atom in units of 1/12 of 12C mass is not the same as it's relative mass, compared to this unity mass.

What is the unit of relative volume ?

Related: Units of mass on the atomic scale

Also note that the mass of a carbon-12 atom was not 12.0000000 amu. It was about 1.0003 amu (using the old definition of the atomic mass unit in physics) or 1.00006 amu (using the old definition of the atomic mass unit in chemistry).

I understand your query. So a atomic mass is equal to molar mass that is mass in grams per mole of the substance. it has the units grams per mole. This is because definition of mole says that it is the number of atoms of carbon in a sample of 12g of carbon-12.

@Stack3002 Equal numerically only and only for a particular atomic mass unit. Atomic mass is the mass of an atom, in whatever units ( kg, amu,Da),   (Atomic) molar mass is the mass of a mole of atoms.

My comment addresses only his query. He is referring to relative atomic mass when talking about mass in a.m.u.

Relative atomic mass is a dimensionless quantity.
It is relative mass of a specie with respect to the mass of carbon-12.
So the units of masses get cancelled and hence we have a 'unitless' quantity.
Just like refractive index doesn't have a unit.
Physical quantity that are a result of ratio don't have units as they get cancelled.

I am not sure they can be call quantities anymore.


text:
How is indigo carmine used in Lowenthal-Permanganate titration as a redox indicator?
In a Lowenthal-Permanganate titration, cider tannins are quantified by titrating the cider with indigo carmine indicator and sulfuric acid against potassium permanganate. Tannins are antioxidants, so that means they give up electrons to reduce the potassium permanganate. The endpoint of this titration is when the blue analyte turns a yellow colour. According to Wikipedia, indigo carmine turns yellow when reduced.
What I don't quite understand is why the indigo carmine didn't reduce when it was in the presence of the antioxidative tannins before the potassium permanganate was added and why it turned yellow only after potassium permanganate was added.

Could there be an error of switched colours ? Titration blue to yellow makes little sense, analysts always try light to heavy color transitions. Check independent sources.

Thanks for the comment @Poutnik, I actually carried out this titration myself and the colours were correct. I based my procedure on procedures used in academic journals such as this one: https://onlinelibrary.wiley.com/doi/pdf/10.1111/1750-3841.14713.

Hm, I was just assuming that as an option.

Can it be gallates are not strong enough to reduce indicator, which is oxidized farther by permanganate ? Similarly as high pH discolour phenolphthalein ?

It's very possible I'm just not understanding this fully, but if the indigo carmine was oxidized by permanganate, wouldn't it be blue?

I mean $$\ce{IC(red, yellow) <=>[oxidation][reduction]IC(ox, blue)->[Strong oxidation][KMnO4]IC(oxox,yellow)}$$ As permanganate is heavy weight for most organic dyes, even  redox indicators with low redox potential see https://en.wikipedia.org/wiki/Redox_indicator

Hmm, that's a good suggestion. Thanks for the help! I'm in high school, perhaps this is why it's confusing for me :(

After the endpoint of this titration you have an excess of permanganate, so your indicator is going to be oxidized, not reduced. It is confusing because most everything you find about indigo carmine as a redox indicator is about its reversible reduction to its leuco form, which is indeed yellow. However, like most dyes, indigo derivates can also be irreversibly degraded ("bleached") by strong oxidizers like acidic permanganate, and that is what's going to happen here.
The tannins do not reduce the indigo carmine before or during the titration, as that reaction would be either unfavorable or very slow, and are not supposed to. They do react rapidly with the much stronger oxidizer permanganate as it is added. Once all tannins are consumed, the excess permanganate will attack the indigo carmine, oxidizing it to, presumably, isatin sulphonic acid which in dilute solution is also yellowish.

I did some brief looking but given that the carmine is also pH indicator that remains blue at below 11.4 pH it would appear as though your sulfuric acid is preventing the reduction from occurring in your indicator. Under acidic conditions KMnO4 is a strong oxidizer where carmine is a weak oxidizer under these conditions. However I am not sure how the carmine gets reduced to yellow upon equilibrium. Perhaps the pH changes. It appears as though permanganates acid conjugate is strong so its a weak base. Just throwing out ideas there though.
http://chem.lapeer.org/Chem1Docs/IndigoCarmine.html
https://www.sciencedirect.com/topics/earth-and-planetary-sciences/permanganate


text:
Why don't proteins form branched polymers?
My reference book (Princeton Review for SAT Chemistry Subject Test) mentions that:

Proteins and carbohydrates are both polymers; however, only
carbohydrates commonly form branched polymers. Glycogen and cellulose
are both carbohydrate polymers made up of glucose monomers; glycogen
is a highly branched polymer while cellulose is primarily
straight-chained.

But I don't get why proteins cannot form branched polymers. Proteins have both an amino group and a carboxylic acid group, and I understand that the two groups join together with the elimination of a water molecule during polymerization.
Branched polymers are formed when a hydrogen atom (or any other substituent) is replaced by another monomer unit, and I don't see why this can't take place in the case of proteins. Could someone please explain?

Hmm how you imagine branching there? BTW there's lots of posttranslational modification in there, so your premise is a bit flawed.

If we have an amino acid (a monomer group) with say 2 amino groups and a carboxylic acid, then there can be branches right...

Note that branching is not unheard of, for instance glutathione is an abundant short branched peptide formed by condensation of the carboxylic group on a glutamic acid side chain to the main chain amino group of a cysteine aminoacid. This requires a dedicated enzyme both for synthesis (glutamate cysteine ligase) and degradation of the gamma peptide bond. Such gamma peptide bonds could serve as the basis for branching but afaik occur only in special cases.
An important example of branching also occurs during ubiquitylation, in which ubiquitin is attached to a protein's C-terminus via lysine side chains to target the protein for degradation. A variety of polyubiquitin chain arrangements (including multiple branching) have been identified.
I suppose one can only speculate from its absence (or non-ubiquity) that branching either (1) does not generally confer a significant evolutionary advantage or that (2) there are evolutionary barriers that have impeded development of advantageous biochemical machinery for generation and regulation of branching.

Proteins don't commonly form branched polymers because they are formed, in all known organisms, by template-directed synthesis.  This templating machinery, the ribosome and the accompanying system of tRNA "charging" of amino acid monomers, does not provide a mechanism for incorporating branches. And, the template, the messenger RNA, is itself unbranched.
The other answers are correct that sometimes branching is introduced after template directed synthesis (i.e. post-translationally).  The $\epsilon$ amino group of lysine, the thiol group of cysteine, and the OH groups of serine, threonine, and tyrosine are all places that this branching could occur.  But it has to be introduced after template-directed synthesis.
In contract, polysaccharides are not formed by template-directed synthesis.  The biosynthetic enzymes that make polysaccharides in the first place are the same ones that introduce branches, they are perfectly happy to do so.

Proteins are built from monomers with only two linking groups
The monomers that make up proteins join in very specific ways. The general structure is:

where R can be a variety of possible groups. To form a protein the monomers join when the amino group reacts with the carboxylic acid group. this gives a product with a specific orientation and which still contains one free amino group and one free carboxylic acid. Another monomer can react at either end but, inevitably, the core chain of the backbone is linear (at least in the sense that no branches are possible).
Glucose, on the other hand, has several groups that can react in condensation reactions:

so, depending on which groups react, there is the possibility for branches to form in the core chain of the polymers that result.
It is worth noting that a limited from of other links can form with proteins, though these would not normally be called "branches" as they don't affect the core chain making up the protein. These can occur when the R group in the amino acid has some sulfur groups which can form S-S cross links by reacting with other amino acids elsewhere in the protein chain.
But the basic idea explaining why proteins don't form branches in their core backbone is that, the monomers have only two sites where polymer links can form. Branched chain polymers need each monomers to have more than two sites for forming polymer links.

Theoretically, there could be amidic bridges via basic and acidic aminoacids like lysine and glutamic acid. But it would be rather by means of organic chemistry than biochemistry, as AFAIK enzymes are not focused this way. Ribosomal synthetic factories use a linear way, but I cannot exclude such option at post synthesis step.

What about proteins that have more than one amino group, like L-tryptophan or arginine? Wont this provide more than two sites for forming polymer links? I am a high school student, so pardon me if im missing something obvious.

@WolframtheTungsten Yes, there are some amino acids where the R-group has a carboxylic acid or an amino group. But these are nowhere near the backbone. In principle, if the reactions were random, branching might be possible. But, in living systems, proteins are not assembled randomly but by precision biological machinery which controls which monomer is added next and this machinery is designed around the backbone of an amino acid and will completely ignore the side groups.