text:
Decomposition of phosphine solution in water when exposed to sunlight
Quoting from my chemistry textbook:

The solution of phosphine in water decomposes in presence of light giving red phosphorus and hydrogen gas.

I could not find any relevant details on the Internet. I wish to know the reactions behind it, and some similar examples, if any.

What's to explain? Many things decompose under sunlight or heating, and PH3 molecule is not particularly stable. AsH3 is yet more fragile. Ever heard of Marsh test? That's an example for you.

@IvanNeretin "Many things decompose under sunlight or heating," But $\ce{NH3}$ does not, and neither does water, and probably many more chemical species that we see everyday, do they? So, I think OP's confusion is fair. Please let me know if I misinterpreted your comment.


text:
What is the "blue spirit" (or "red spirit") in mercury-free thermometers?
I surmise it's some kind of alcohol, but is there a standard chemical used for these?
One of them just exploded in my students' lab! It was the kind with blue fluid in it.

https://en.wikipedia.org/wiki/Alcohol_thermometer

exploded, as in "ignited/fireball" or as in "overheated and glass broken due to pressure"?

@dlatikay the latter. Loud pop. No combustion that I know of

As per the Wikipedia page on Spirit Thermometers:

The liquid used can be pure ethanol, toluene, kerosene or isoamyl acetate, depending on manufacturer and working temperature range. Since these are transparent, the liquid is made more visible by the addition of a red or blue dye.

1,4-bis(butylamino)anthraquinone - commonly known as "Oil Blue 35" - may possibly be used as the blue dye.

Another organic solvent used may be pentane. Perhaps the common ones contain toluene dyed with a red aniline dye .
Other spirit filled thermometers variants include:

Blue fill contains isoamyl benzoate;
Red fill may contain kerosene or pentane with aniline dye

Aniline red

Credits
Acknowledgements

Spirit-Filled Glass Thermometers: Document

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
Why does a neutral atom of beryllium not have any electrons in a p orbital?
Here's what I understand about quantum number and orbitals, please correct me if anything is wrong:
Electrons enter into these different types of orbitals because they have a higher/lower amount of energy, and we know that the number of electrons varies from element to element along with the energy levels of those electrons in a pretty linear way, conveniently allowing us to organize them into blocks on the periodic table where their highest energy level electron is in one of the many types of orbitals (spdf). I know the energy shell occupied by an electron (the principle quantum number) corresponds with the angular momentum quantum number which determines the magnetic quantum number, which determines the type of orbital an electron is in. I know that all of that is because the electrons in higher energy shells/levels have more energy are thus more likely to be in an orbital such as d or f as opposed to an s or p orbital, for example.
Note that I don't know why the orbitals exist in the exact shapes they do (although I imagine it has something to do with the electrons speeding up or slowing down at certain points making it more likely for us to discover it in a certain spot), or what the threshold is for the amount of energy required to make an electron move into the next type of orbital (d to f, for example).
But, I have one question: why does a neutral beryllium atom with 4 electrons in 2 energy shells/levels not have any electrons in a p orbital?
My reasoning is as follows:
If $n = 2$ where $n$ is the energy shell occupied by beryllium's fourth electron, then $l = 0,1$ and $m_l = \text{either }0 \text{ or }-1,0,1$. If $m_l = -1,0,1$. Then, why isn't this fourth electron in a p orbital?

https://chemistry.stackexchange.com/questions/152/why-is-the-2s-orbital-lower-in-energy-than-the-2p-orbital-when-the-electrons-in

You are correct in all your analysis in the first two paragraphs (except that the "energy levels of those electrons in a pretty linear way, " follow the $(n+l)$-rule so $\ce{4s<3d}$ in energy, but it doesn't matter). But I don't understand your final question. Two electrons go into $\ce{1s}$ subshell. The remaining two electrons go into $\ce{2s}$ subshell. Is there any more electron left for the $p$ orbital?

In the ground state of the atom the orbitals fill up with electrons going into the lowest energy orbitals. Up to two in s orbitals and up to  6 in p orbitals.  The 4 electrons in Be go into the 1s and 2p orbitals first.  The 2p orbitals are at slightly higher energy and so do not have any electrons. If you were to excite the Be atom then it is possible to have an electron in a 2p orbital.

https://chemistry.stackexchange.com/questions/47633/are-empty-orbitals-degenerate

I seem to have forgotten that each orbital has 2 electrons in it, not one. Bit of a facepalm here... sorry guys! Thanks for the answers.

This is funny because from the Aufbau principle, this is all correct, but quantum chemistry predicts some admixture of p orbitals into the ground state. I miss general chemistry...

Beryllium has 4 electrons in neutral atom.
Lowest energy levels are filled first. So firstly 1s2 (ist shell now filled) That leaves 2 left over.
So they go into the 2nd shell and the 2s orbital.
Therefore electron configuaration is 1s2 2s2. That's it, no electrons left over to put into 2p.

We need to first consider n, the quantum number for the shell, and then l (~the classical angular momentum of the electron), which tells us the shape of the electron density. The first shell (n=1) contains two s electrons with paired spins. That shell is closed, and you begin over again with n=2: the possibilities for n=2 are l=0 (s electron shape, spherical) and l=1 (p electron shape, lobed). Then, since a lobed shape can point in different directions, when l=1, ml can equal +1, 0 or -1 , defining px, py, pz electron shapes, each pointing in one of three orthogonal directions.
So the filling process puts the first two (lowest energy) electrons in the 1s level (when full, it is 1s2); then the next level is 2s (filling to 2s2), then 2p. Since we have run out of electrons, the 2p levels do not get filled, so we don't have to consider 2px, 2py or 2pz designations.
Now, for boron, there will be a p electron. Whether you call it px, py or pz will depend on how you define the directions in a molecule. In an individual free boron atom, 2px, 2py and 2pz all have the same energy and are degenerate.
This is all theoretical. If you ever heated up beryllium so that there were free atoms, they would be so energetic that quantum levels up to 10 or 100 would be occupied, and there would be beryllium ions floating around too!

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
How does recrystallization work to purify if compounds have equal solubilities, but one is present in a greater quantity?
Every article or text I've read on crystallization states that it works by taking advantage primarily of differences in solubilities, but I've also been told that it can purify compounds with equivalent solubilities if one is present in a greater quantity.
For instance, if there is a solution of 90% Compound A, and 10% Compound B, both with equal solubilities, a crystallization would supposedly purify it. But how? I've been told that, if say, 20% of the product was left behind in the solution, then it would consist of equal ratio of Compound A and Compound B, causing the formed crystals to be almost pure Compound A (saying a 1:1 ratio of A:B gets left behind). I was told that this ratio that gets left behind in the solution depends on the relative solubilities of the compounds, and not the quantity that which they were present.
Is this accurate? How does this work exactly? I've read dozens of articles on crystallization and none of them seem to explain this sort of instance well, yet it seems very widely used in preparative chemistry.

I don't see what is there to explain. I seems you don't understand what solubility is or something else?

@Mithoron Intuitively if both have equal solubility why would they not both precipitate out equally? Why would Compound A precipitate out selectively? Wouldn't intuition dictate that the precipitate contain 90% of A and 10% of B, if the solubilities equal? I'm told this isn't the case, and that goes against intuition... Do you see what I mean?

Well then of course you don't understand solubility - only surplus over solubility value crystallises, not all of compound.

@Mithron Clearly I don't. Usually solubilities are affected by other compounds dissolved in solutions, at least in instances of salting out and such. If B is soluble at 1g/L in a solution, and A is the same, and 1g of each is added to said solution, would they both dissolve? Intuition about dissolved compounds affecting solubility would suggest that only 1g max of the 2 compounds (total) could be dissolved.Or would 2g total?Or is this an incorrect understanding of solubility? I'm basing this off my understanding of how salting out works, decreasing solubility of one compound by adding another.

https://chemistry.stackexchange.com/questions/72700/can-other-substances-be-dissolved-in-a-saturated-solution

@Mithron yeah... the top answer in that link is exactly the point I was making/asking about. That is my understanding of solubility, that adding compounds to the solution will effect the solubility of compounds added in the future, but all the answers I am getting here suggest that isn't the case with crystallization. Though I suppose the second answer in that link contradicts the first somewhat... see why I am confused?

Check out third answer then... sigh First approximation is that different solutes have no influence on each other at all - it's what you should get to know in very beginning, and that's why I felt you may have no idea about solubility at all. Such things like salting out, common ion, and others come later.

I understand now. So basically there is no hard rule about whether dissolved solids in a solution will affect future solids to be dissolved? Sometimes they do, and sometimes they don't? I was operating under the assumption of how salting out works, that dissolved solids generally always affect the solubility of other compounds being dissolved. I guess my knowledge was incomplete about this. But can you see why I was confused in the first place?

sigh If those compounds were, say, benzoic acid and deuterated benzoic acid, then they would crystallize together just as you imagined, and obviously trying recryst. would be nonsense. That would be more like coprecipitation. Recryst. compounds need to be significantly different and similarity in solubility rather random. Salting out is by no means ordinary, and there's also salting in.

I think the statement "I've been told that, if say, 20% of the product was left behind in the solution, then it would consist of equal ratio of Compound A and Compound B, causing the formed crystals to be almost pure Compound A (saying a 1:1 ratio of A:B gets left behind)." is very confusing.

Such recrystallization procedures are often used in organic chemistry where A and B would both be dissolved in hot solvent, but a large amount of A would precipitate as the solution cools. The rub here is that as A precipitates it will also adsorb some of the B. So to purify A further, often a second, or third, crystallization may be necessary.
So imagine that 20 g of each of A and B are soluble in some quantity of the cold liquid. You start out with 100 g total of solute, so 90 grams of A and 10 grams of B dissolve in the hot solution. So all of the B would be soluble in the given quantity of cold solution, but only 20 grams of A. So 70 grams of A should precipitate.
Notice that in order for this method to work all of A and B must dissolve in the hot solution. The idea is that crystals of compound A prefer to have A in the crystals rather than B. But if A has been coating grains of B, then if the whole solid particle is not dissolved, a lot of B will still be in the solid matter which didn't dissolve.

So, to put this in mechanical terms, would you say that because A is present in a much larger quantity, it's formed crystals will have greater surface area, and thus crystalize faster than the less present compounds? In a sense, would it be the surface area that facilitates the fractionation of the ratios, so to speak? I'm basically just trying to understand the variable which causes the ratios to shift from solution to crystal.

No that is the wrong notion. Let's say that 70 g of B are soluble in the hot solvent, but only 20 grams of B in the cold solvent. So I get the solvent hot, dissolve 10 grams of B. How much B will crystallize out when the solution cools?

50g. My confusion seems to relate to how compounds affect the solubility of each other. In your example in your original response, why would 30g total stay dissolved in solution? (20g A 10g B)... if the solution can only hold 20g of either of those compounds, how can 30 stay dissolved? Rather, if 20g of B is already dissolved in the solution, and A has an equal solubility, how would any additional A be able to become dissolved, without precipitating B? My understanding is that solubilities depend on the solvent having nothing else dissolved in it -  Based on my understanding of "salting out"

HUH? I only dissolved 10 g of B in the hot solvent. How can I be getting 50 grams of ppt in the cold???

In the original comment the notion is that 20 g of A and 20 g of B are soluble in the cold solvent. Since there are only 10 g of B, all of the B is soluble in the cold solvent.

So in your original example you're saying that the fact that there is already 20g of A dissolved in the solvent would not prevent 10g of B from dissolving? I think I'm understanding my confusion. I was under the impression that solubilities are always affected by other compounds being dissolved in solution as with "salting out", but other comments have suggested this isn't the case.

RE: I was under the impression that solubilities are always affected by other compounds being dissolved in solution - That is true to some extent. How much depends on the compounds. But in general that would be a secondary effect, not the primary one.


text:
Concentration of minor products in the nitration of nitrobenzene
I was baffled by the concentration of minor products in the nitration of nitrobenzene.
This is the picture from Solomon Frhyle's book, page $695$:

Since nitro is electron withdrawing and meta directing, the high concentration of meta isomer is justified. However, I had predicted the order to be $m>p>o$ because both nitro groups are electron withdrawing and if they are close together they will cause destabilisation by increasing charge density at adjacent places. On the other hand in the para isomer, the nitro groups are far away so development of partial positivity at opposite ends shouldn't be a big problem.
Could someone explain why the para isomer is less in concentration and how my reasoning is incorrect?

I wouldn't be too concerned about small differences in the yield of the minor products. If these are isolated yields, then they have been scaled to 100%. How easy is it to isolate ortho vs. para isomer? There is a 2:1 statistical factor in favor of the ortho isomer.

These are only values given for example, possibly even from single experiment; you seem to be trusting simple rationalisation way too much.

Maybe some intramolecular attraction that we are unaware of? Was anything mentioned in the book about that? @Mithoron

@user55119 Could you please explain what you meant by "If these are isolated yields, then they have been scaled to 100%" and "There is a 2:1 statistical factor in favor of the ortho isomer."?

@Gaurang there are two ortho positions and only one para.


text:
Why is the effect of ring activators more dominant than ring deactivators?
In 3-nitroaniline, if it were subjected to an electrophilic substitution reaction, the electrophile will most likely attack at (b) or (d) as mentioned in my textbook.

To me, it seems that the only role of the nitro group is that the electrophile will not attack at (a) due to steric hindrance. So my question is, why is the -M effect of the nitro group ($\ce{-NO2}$) discarded and only the +M effect of $\ce{-NH2}$ plays a role? Nitro group is meta directing, so why wouldn't the product form at (c) as well? It is as if the nitro group doesn't even exist. In better words, why is the effect of the ring activator, $\ce{-NH2}$, more dominant than the effect of the ring deactivator, $\ce{-NO2}$?

You answered the question yourself. The process is electrophilic aromatic substitution. Clearly, the resonance contribution of the amino group carries the day over the electron-withdrawing nature of the nitro group.

Yes but why? @user55119

Whatever is going to react with the aromatic ring is an electrophile (electron-loving: NO2+, Br+, R(CO)+, etc.). Electron donation via resonance from the amino group increases electron density at the o,p-positions. Resonance of the nitro group deactivates the electron density of the ring. Therefore, the amino group will dominate. In a competitive  electrophilic aromatic substitution reaction, the electrophile reacts faster with aniline than with nitrobenzene. The relative rates of reaction would be aniline>m-nitroaniline>nitrobenzene.

$\ce{-NH2}$ activates the ring due to its +M effect, increasing electron density in the ring at (b) and (d). On the other hand, $\ce{-NO2^{-}}$ causes de-activation because of its -M and -I effect and the electron density would be higher at (c) relatively.
Now since in activated rings, the rate of electrophilic aromatic substitution is higher than those in deactivated rings, the effect of $\ce{-NH2}$ as ortho-para director is seen more during product formation and not of $\ce{NO2^{-}}$ majorly. The electrophile will attack at (c), but since its rate will be quite slow, it would not form the major product.

What about the  -M effect of $\ce{-NO2}$?

It would just not take place because $\ce{-NO2^{2-}}$ is highly unstable.

$\ce{-NO2}$ is a very strong ring deactivator, stronger deactivator than $\ce{-C=O}$ even, right? @Tanuj

Yup ! The strongest I would say.

@raajsuriya You see , $\ce{-NO2^{-}}$ is strongly deactivating , so it decreases rate of electrophilic aromatic substitution.

Nitro group is meta directing, right? So why won't the product form at (c) as well? @Tanuj

Let us continue this discussion in chat.

Just edited your answer, is it good?


text:
Correct monomer for polycondensation?
I'd like to formulate the synthesis reaction of this polymer:

I assume this is a normal polycondensation reaction, and the monomer would then be:

Is this correct?

Try bis-phenol A and phosgene or its equivalent.

Why you didn't check out https://en.wikipedia.org/wiki/Polycarbonate ?

Yeah, this is totally a polycarbonate, thanks... I guess it was late. However, I would still be interested in knowing whether or not the monomer I wrote down could perform a nucleophilic attack at the carboxyl C atom of a second molecule, form a tetrahedral intermediate and condense off a water molecule. Probably not (otherwise why go for phosgene), but why?

The monomer you drew is not stable; carbonic acids decarboxylate under standard laboratory temperatures and pressures to the alcohol. It's also very difficult to get phenol derivatives to react with CO2 at the oxygen atom, rather than the ortho carbon atom.


text:
Electrodes involving sparingly soluble salts (Electrodes of the second kind)

Consider the sparingly soluble salt, $\ce{AgI}$ ($K_\text{sp} = 8.52\times10^{-17}$) and a silver electrode (Standard Reduction Potential = $\pu{0.799V}$). $\pu{0.02L}$ of $\pu{0.001M}~\ce{KI}$ solution is added to a silver nitrate solution with a silver electrode immersed. After connecting to SHE, the reading is $\ce{0.068V}$. Hence, what is the concentration of silver nitrate added at 298K?

Clearly, we should start off by writing the relevant Nernst equation, as follows
$$E = \pu{0.799V} - \frac{8.314 \times 298}{1 \times 96485}\ln\left(\frac{1}{[\ce{Ag+}]}\right) $$
From here we can substitute $E = \pu{0.068V}$ and obtain a result of $[\ce{Ag+}] = 4.331 \times 10^{-4}$
However, I am not sure how to proceed from here. An unknown amount of $\ce{AgI}$ precipitated and so we cannot calculate the initial amount of silver nitrate added. Although we know the amount of $\ce{KI}$ added, most of it presumably precipated due to low [$\ce{I-}$] from $K_\text{sp}$, suggesting excess silver nitrate. However, the still low concentration of silver nitrate presents problems when we use approximations.

I am not sure how you assumed the temperature to be $\pu{298 K}$. It isn't given in the problem statement...

Thanks for the formatting! I believe the 298K was mentioned earlier in the parts of the problem


text:
What is the product of magnesium reacting with carbon?

$$\ce{Mg + C -> X}$$
  $$\ce{X + H2O -> Y}$$
  Determine unknowns X and Y.

Now the problem is I think the answer should be $\ce{X=MgC2}$ and $ \ce{Y=C2H2 + Mg(OH)2}$. But the given answer was $ \ce{X}= \ce{Mg2C3}$ and $\ce{Y}=\ce{C3H4} + \ce{Mg(OH)2}$
I tried to google the answer but where some website say $\ce{C3H4}$ is correct some say $\ce{C2H2}$ is.  Funny thing is in the exam just after that one we got the same question and this time the $\ce{C2H2}$ was the answer (and I wrote $\ce{C3H4}$)

Magnesium forms more than one carbide.  The question is ill posed.

@OP: Were there any other conditions in the question? Or was the question literally word-for-word your equations?

Yes, both type of carbide is possible. At a certain condition, one type of carbide predominates over the other.

@JavaScriptCoder The equation arrow did have a delta sign over it indicating that it's heated but I think it won't help?

This asnswer is intended to extend my comment. Most of the information is taken from this site.

Both types of magnesium carbide($\ce{MgC2, Mg2C3}$) is formed when acetylene gas is passed over magnesium. At different temperatures, one type of carbide predominates over the other. At 570°C-610° C, $\ce{MgC2}$ converts to $\ce{Mg2C3}$:
$$\ce{2MgC2 -> Mg2C3 + C }$$
Above 610° C, $\ce{Mg2C3}$ decomposes:
$$\ce{Mg2C3 -> 2Mg + C}$$
Both types of magnesium carbide can be hydrolysed:
$$\ce{MgC2 + 2H2O -> Mg(OH)2 + C2H2}$$
$$\ce{Mg2C3 + 4H2O -> 2Mg(OH)2 + C3H4}$$

For more information on the structure of magnesium carbide, see this old paper from 1943. Also, how magnesium carbide behaves at high pressure is discussed in this new paper(2015). Another type of magnesium carbide, $\ce{Mg2C}$ exist at high pressure and it along with $\ce{MgC2}$ behaves as semiconductor while $\ce{Mg2C3}$ behave as insulator-metal transition. (OP is requested to see both papers if interested)

Just summarizing; So X would by MgC2 and then if enough heat is provided it would convert to Mg2C3. Then Y will depend on the temperature at which the reaction is carried out, right?

Yes, that's the point.


text:
Determine the geometry of each nitrogen atom in formamidinium ion

The geometry at A seems to be straightforward: $\ce{N}$ is surrounded by $\ce{C}$ and two $\ce{H}$'s. By VSEPR, this is trigonal planar.
For the geometry at B: $\ce{N}$ is surrounded by $\ce{C}$, two $\ce{H}$'s, as well as a lone pair. Once again by VSEPR, I determined this was trigonal pyramidal.
However, the answer was given as A, which means that my process for the geometry at B was incorrect. What mistake did I make and how should I fix it?

https://chemistry.stackexchange.com/questions/4390/how-to-rationalise-the-resonance-structures-and-hybridisation-of-the-nitrogen-in https://chemistry.stackexchange.com/questions/4390/how-to-rationalise-the-resonance-structures-and-hybridisation-of-the-nitrogen-in

The hybridisation of any atom is determined by the total count of its sigma bonds and localised lone pairs. This is because, while localised lone pairs occupy hybrid orbitals (as in ammonia), delocalised lone pairs occupy pure $\ce{p}$ orbitals instead. Hence, the latter is not counted in determining hybridisation.
The $\ce{N}$ atom labelled "B" has its lone pair in conjugation with a pi bond. This is perfect for delocalisation/resonance to take place. Hence, that lone pair is not counted in determining hybridisation. Thus, it is also a trigonal planar geometry.

Interestingly, notice that the formamidinium ion has two equivalent canonical resonating structures. The lone pair of the atom B on delocalisation, results in a lone pair on A and a formal positive charge on B, which is just the reverse of the structure shown.

Thank you for the prompt and clear answer! To clarify: Since it is a resonance structure in which the lone pair can either be on A or B, the lone pair is described as "delocalized," and can therefore be excluded from the VSEPR determination. Is that correct?

@JohnLou It isn't necessary that the lone pair must be on A or B only. As long as the lone pair is delocalised, it is excluded from VSEPR determination.

This is a kind of a trick question (but it's tricking you to teach you an important point, so it's kind of justifiable).
The question says that the structure it shows is "a Lewis structure of the formamidinium ion," which is technically correct.  But note that it doesn't describe it as the Lewis structure of the formamidinium ion, since the structure shown is really just one of the possible resonance structures of the molecule.
Indeed, it's easy to see that the arrangement of the atoms in the molecule is symmetric, and that the structure shown can be converted to its mirror image, with the nitrogens A and B swapped, just by shuffling electrons around.  This symmetry strongly suggests that both the Lewis structure shown and its mirror image should be equally major contributors to the resonance, and that the actual minimum energy configuration of the molecule will be a symmetric hybrid of the two structures (and possibly other minor contributors), with both nitrogens having the exact same geometry and bonding.*
This symmetry argument rules out the asymmetric answers B and C, leaving only two plausible choices: either both nitrogens are trigonal planar, or they're both trigonal pyramidal.  Since the Lewis structures contributing to the resonance feature a C=N double bond, their resonance hybrid must presumably involve some degree of π bonding between the central carbon and the nitrogens.  This suggests that both nitrogens (and the central carbon) should be sp² hybridized, and thus planar.

*) In principle, the molecular geometry corresponding to the symmetric resonance hybrid structure could also be a local energy maximum, e.g. due to steric repulsion, with the true minimum energy geometry being asymmetric and (arbitrarily) favoring one of the mirror image Lewis structures over the other.  Off the top of my head, I can't think of any specific example of a molecule where that happens, but I'm sure they must exist.  For simple small molecules like this, though, it's usually a pretty safe bet that if a symmetric resonance hybrid can exist, it's probably the minimum energy configuration.


text:
Work equation in thermodynamics
As known, work done on the system is: $$\mathrm{d}W = -P_{\text{ex}}\mathrm{d}V $$
However, I have some ambiguity about this equation:

Why is there $P_{\text{ex}}$? Pressure of a gas contained in a cylinder is by definition the force that gas exerts on a piston divided by a area of a piston. But, by the Newton's third law piston exerts the same force on a gas, and so, why is the infinitesimal work not $\mathrm{d}W = -P\mathrm{d}V$?
Some can argue, that we use whatever pressure the thing doing work is pushing against. But I don't accept it. It would be correct if I regarded a gas and a piston as a one system. However, system of interest is only a gas, on which a piston exerts the force with the same magnitude as a gas exerts on a piston, which is $P_{\text{gas}}$.

My calculations:
$$ m\frac{\mathrm dv}{\mathrm d t} = F_{\text{ext}} - F_{\text{gas}}=S(P_{\text{ext}}-P_{\text{gas}}) = S\bigg(P_{\text{ext}} - \frac{nRT}{V_\mathrm{i}-Sx} \bigg)$$
$$ mv\frac{\mathrm d v }{\mathrm dx} = S\bigg(P_{\text{ext}} - \frac{nRT}{V_i - Sx}\bigg)$$
After intergration:
$$ \frac{mv^2}{2} = \int_0^x{SP_{\text{ext}}\mathrm dx} - \int_0^x{\frac{SnRT \mathrm  dx}{V_i - Sx} } = P_{\text{ext}}\Delta V - nRT\ln\bigg(\frac{V_\mathrm{i}}{V_\mathrm{f}}\bigg) $$
Rearrange, we get:
$$ P_{\text{ext}}\Delta V = W = \frac{mv^2}{2} + nRT\ln\bigg(\frac{V_\mathrm{i}}{V_\mathrm{f}}\bigg)$$
There, on the right side, first term is just an energy gain of a piston. The second term is familiar, it's work on the system in reversible process.
So work done on the system is the same as it would be if process was reversible. Why then infinitesimal work is $\mathrm d W = P_{\text{ext}}\mathrm d V$ and not just $\mathrm d W = P_{\text{gas}}\mathrm d V$ ?

because that is how it is defined. This is the convention taken.

This is a good question that deserves a thorough reply. If someone wishes to answer, I recommend reading doi: 10.1021/ed3008704, and 10.1021/ed043p233.

Many a times, we are dealing with reversible processes. In such cases, we are dealing with infinitesimal changes - changes so small that we can assume the system to be at equilibrium with its surroundings at all points along the path it is following. This would mean mechanical equilibrium - $P_{ext}=P_{gas}$.

In a reversible process, the gas pressure is spatially uniform within the cylinder, and is described globally by the ideal gas law.  However, in an irreversible process, the force per unit area at the piston face is not equal the force per unit area at other locations within the cylinder.  Furthermore, the ideal gas law does not describe the behavior of the gas because viscous stresses contribute to the force per unit area for a rapid irreversible deformation.  So, even though Newton's 3rd law is satisfied at the piston face, unless we specify the force per unit area externally (e.g., manually), we will get the wrong answer if we try to calculate the pressure at the piston face using the ideal gas law.
In applying the equation $W=\int{P_{ext}dV}$ to calculate the work, $P_{ext}$ is supposed to be the force per unit area exerted by the surroundings on your system, at the interface between your system and the surroundings.  So, if the gas is your system, $P_{ext}$ is the force per unit area exerted by the inner face of the piston on your gas (and by your gas on the inner face of the piston).  In your example, if the cylinder is vertical and we do a force balance on the piston, we get:  $$m\frac{dv}{dt}=P_{ext}A-mg$$assuming there is vacuum on the outer face of the piston.  If we multiply this equation by the piston velocity v = dx/dt and integrate, we obtain:$$m\frac{v^2}{2}=\int{P_{ext}dV}-mg\Delta x$$Therefore, the work done by the gas on its surroundings (up until an arbitrary time) is given by:  $$W=\int{P_{ext}dV=mg\Delta x}+m\frac{v^2}{2}$$When the final thermodynamic equilibrium state of the system has been realized, the piston will no longer be moving (including any oscillations of the piston, which will have eventually been damped out by viscous stresses) and the work will then be determined by:$$W=\int{P_{ext}dV=\frac{mg}{A}\Delta V}$$
If, instead of vacuum, there is some constant force external to the piston (say, $P_{atm}$ equal to the atmospheric pressure), the previous results change instead to:$$W=\int{P_{ext}dV=mg\Delta x}+m\frac{v^2}{2}+P_{atm}\Delta V$$when the piston is still moving and $$W=\int{P_{ext}dV=(P_{atm}+\frac{mg}{A})\Delta V}$$at final equilibrium when the piston has been damped to rest.

There is no difference and the equation (first law) will yield the same answer to the problems you may come across - particularly based on the first law of thermodynamics. The difference arises in deciding the point of reference for the work done. We have dW = +PdV or -PdV (note that the two Ps are different): you just have to decide where do you look from.
Am I the system and work is done on me, or am I outside and doing work on the system?
Once you choose your point of reference, that is, the convention you desire to follow - you should stick to it and do all your calculations according to it. As in most of these cases, the reason is entirely historical and has no conceptual difference whatsoever.
People who prefer using dW= +PdV, where P is the pressure of the gas, are usually physicists, and engineers – who want to know "what the system can do for us" in practical applications. This is the section of the community that prefers to be in the work-done-by-the-system camp.
On the other hand, work-done-on-the-system seems to foster the view of an experimenter or theoretician operating on a system from outside. Hence, this is usually the view accepted by chemists, i.e. dW = -PdV, where P is the pressure of the surroundings and dV refers to the infinitesimal change in the volume of the system. (Clearly, if dV>0, that is, volume increases, the work done is negative. This is because the surroundings 'oppose' the system's expansion)
Edit:
I found this really important to mention -
Pressure due to gas and pressure due to surroundings are not always equal.
They are equal, at all times during the process, if and only if the process is quasi static, that is, 'reversible'.

If I understood the question correctly, it's less about the sign than it is about the proper term for pressure. The usual equation $\delta w = -P_\pu{ex}\pu{d}V$ has some additional requirements that usually go unstated in textbooks.


text:
Reduction of nitro group to amine

In the above reaction, I do know that $\ce{SnCl2/HCl}$ reduces $\ce{-NO2}$ group to $\ce{-NH2}$ but which one of the two $\ce{-NO2}$ will be reduced first by $\ce{SnCl2/HCl}$? Why so?
And how does $\ce{NH4HS}$ react and again which one of the two $\ce{-NO2}$ will be reduced first by $\ce{NH4HS}$ and why so?
NOTE: 1eq represents 1 equivalent.

Partial duplicate of https://chemistry.stackexchange.com/questions/34066/selective-nitro-reduction-of-poly-nitro-compounds/53698#53698

Hyperconjugation accounts for the somewhat greater negativity of the 4-nitro-group compared with the 2-nitro-group in 2:4-dinitrotoluene; in consequence, selective reduction of the former by alkaline sulphides produces mainly 2-nitro-4-aminotoluene, although some 4-nitro-2-aminotoluene is also formed. In acid medium the 2-nitro-group becomes relatively negative in comparison with the 4-nitro-group, owing to the inductive (+I) effect having the chief role, with resultant preferential salt formation to give the very positive $\ce{-NO2H}$ group at 2-position and preferential reduction of it by acid stannous chloride.

Quoted from Coloration Technology 62 (1946) 114
As for how ammonium sulphide reacts see this answer:  Selective nitro reduction of poly nitro compounds


text:
Does computational chemistry get the molecular orbitals of dioxygen wrong?
The molecular orbitals of $\ce{O2}$ are typically shown as follows, with every orbital filled by a spin alpha (or up) and spin beta (or down) electron (except the homo levels which are singly occupied).

Calculations using for instance Hartree-Fock and density functional theory however produce alpha and beta orbitals with different energies - no degeneracy of alpha and beta orbitals. For instance, the B3LYP/6-311G** HOMO/LUMO levels (taken from the computational chemistry database) are as follows: alpha HOMO (-0.31186), alpha LUMO (0.19755), beta HOMO (-0.46536), beta LUMO (-0.11702). Clearly the HOMO levels are not degenerate, at odds with what we expect from the above molecular orbital diagram. Similar mismatch between the alpha and beta orbitals exists for orbitals below the HOMO, as well as using other levels of theory.
Is computational chemistry getting the orbitals of $\ce{O2}$ wrong? Am I missing something? Conceptually I can't grasp that alpha and beta orbitals have different energies since at least for the case of filled orbitals when the electrons pair up I expect them to have the same energy.

The short answer is, yes, computational chemistry gets them wrong because we don't have a way to calculate orbital energies exactly. In the end, the root cause is the general problem with quantum many-fermion systems - we can't solve the Schrödinger equation exactly.
Let's have a look to different computational chemistry methods to understand why that's the case. In general, more sophisticated methods will reduce the energy gap between alpha and beta electrons, but as we will see, none will guarantee that it doesn't exist.
Hartree-Fock methods
Hartree-Fock methods are inherently approximate: when we solve a Hartree-Fock calculation, we are solving not a system of N electrons, but N coupled systems of electrons that only interact indirectly. Of the energy terms in the Hamiltonian, HF accounts for the kinetic energy of each electron, its interaction with all the nuclei, the coulombic repulsion between electrons, and the exchange energy with other electrons with the same spin (because they are fermions), but we ignore electronic correlation. In HF, electrons only interact with each other indirectly - by contributing to a background external potential in the resolution of each electron's particular Hamiltonian. This makes it necessary to solve the equations iteratively using a mean field approach - what we usually call "self-consistent field" or SCF. Correlation requires explicitly considering more than one electron at the same time, so it's not compatible with the core HF model.
But, even if it's approximate, shouldn't we still get the same, not exact, values for alpha and beta electrons? - sure we do, in systems where the error introduced by the HF assumptions is the same for alpha and beta electrons.
For instance, if you run an unrestricted HF calculation on singlet $\ce{O_2}$, where you have the same number of alpha and beta electrons, you'll get exactly the same values for the alpha and beta orbitals (and these will be exactly the same as you'd get for a restricted HF calculation of the same system).
However, for triplet $\ce{O_2}$, alpha and beta electrons are exposed to slightly different interactions - since correlation depends on the spin of the interacting electrons, ignoring it affects alpha and beta electrons differently.
Post-Hartree-Fock methods
Post-Hartree-Fock methods attempt to correct the HF limitations by first resolving a Hartree-Fock system and then adding a correction for electronic correlation. The problem, however is that the corrections aren't exact.
Møller-Plesset perturbation methods take the HF wavefunction and then add a numerical correction to the energy in a perturbative fashion - you apply the HF method to double-excited (MP2), triple-excited (MP3), and so on versions of the same ground state HF wavefunction and then introduce a correction to the energy of your orbitals (not to the wavefunctions themselves) based on that. It has several problems - you are only accounting for dynamic correlation (how electrons influence the movement of other electrons within your HF wavefunction), and, because excited estates are not variationally limited, there's no guarantee that you will not overestimate your correction and will end up with an energy below the exact energy. But even if those problems could be completely corrected, since part of the correlation is not accounted for, it will still give slightly different energies for alpha and beta orbitals in a system with $\mathrm{S > 0}$.
Configuration interaction (CI) methods apply the HF method not only to the ground state but also to a number of excited states, and then find the correct  solution to the system variationally, as a linear combination of those configurations. This method is in principle exact - and it's variational, so we are guaranteed not to underestimate the energy - the problem is that it is a series expansion, in which we add a term for each possible configuration state function (CSF). Still, even a full CI taking into account all the possible CSFs won't account for all of the correlation - CI is only exact in the limit of a complete basis set. In practice, expanding the series is computationally very expensive, and it is truncated relatively early - full CI is only practical for the simplest of systems.
When it's applicable, however, difference in energy between alpha and beta electrons should be negligible.
Coupled cluster
Coupled cluster (CC) has a theoretical foundation that is distinct from HF and post-HF methods; essentially, it takes a wavefunction (usually a Slater determinant) and performs a number of excitations on it, producing a coupled cluster description of the system that, like full CI, is in principle exact.  However, like CI, CC is a series expansion - you have terms for single, double, triple, etc. excitations, and we would require all possible excitations to be taken into account to fully consider electron correlation. As with CI, we usually truncate the series quite early - the usual "gold standard" of computational chemistry, against which all other methods have usually been benchmarked in the last couple of decades, is CCSD(T) - meaning that we consider the first and second term, and the third term only partially. This is usually considered to be as exact as it gets in computational chemistry, but again, it is prohibitive for all but the simplest of systems.
Like with full CI, CC differences between alpha and beta electrons will be negligible.
Density functional theory
I left DFT for the end because it's an entirely different beast, theoretically speaking, but in the end the problem comes down to the same issue: you don't account for correlation correctly, and that flaw impacts alpha and beta electrons differently for $\mathrm{S > 0}$.
Density functional theory, in principle, shouldn't have the same problems Hartree-Fock-related methods have, since we only with a single, global, three-dimensional density function $\mathrm{\rho(r)}$; and, as per the second Hohenberg-Kohn theorem, the ground state electron density is unique and variationally limited, so in principle we could keep applying the functional to different three-dimensional functions until we find the absolute minimum - and that's our exact ground state density.
There, are, however, two big problems.
First, in practical terms, we need a way to translate the electron density to wavefunctions so we can understand what's happening at the electron level. For instance, the very values for orbital energies that you mention for a DFT calculation are not inherently in the density function that DFT deals with; there's no way to distinguish the first from the second occupied orbitals in the density function. What we do is introduce a formalism to be able to translate the density function to orbital terms - that's the Kohn-Sham formalism, which boils down to treating the global particle density function, which distributes all N particles in space, as the sum of N single-particle functions, each corresponding to one electron, and which interact with each other only indirectly through a background external potential, which accounts for the coulomb, exchange and correlation interactions, so we update densities and potentials iteratively in another mean field, self-consistent approach. You'll notice that this is pretty much the same thing we did with the wavefunction in Hartree-Fock.
Two considerations, however. First, the Kohn-Sham formalism allows us to map the total density to orbital occupations and from there calculate a number of properties that we couldn't derive from the total density function, but it guarantees only that the sum of the single-electron densities is equal to the total density function for the ground state - not that the depiction of a particular orbital is accurate. So, values for specific molecular orbitals (such as orbital energy) should be taken with a pinch of salt. In fact, in practice, orbital properties from DFT are most accurate around the HOMO-LUMO gap, while values for core orbitals or virtual orbitals can diverge quite wildly from reference values. This is so because orbitals are constructed in a way that best fits density, and not one that best fits the wavefunction. You can find a lot of discussions about whether Kohn-Sham orbitals are physically meaningful and how they relate to the actual orbitals of the system.
As you can imagine, if you have important spin effects in your system, spin deviation in these properties are almost guaranteed.
But there's also an advantage over Hartree-Fock: in HF, we had to drop correlation because we couldn't incorporate it - it required treating more than one electron wavefunction explicitly. But, in DFT, wavefunctions are an afterthought - what we are minimising is the total density function, so there's nothing that prevents us from introducing correlation. We could even ditch the Kohn-Sham formalism and calculate correlation directly in terms of the total density function. So, aside from the problem of breaking down the total density function into a number of one-electron densities, we could in principle at least get an exact total alpha electron density and a total beta electron density, couldn't we?
Unfortunately, no, because we don't know how the functional for electrons looks like. And, if we move from wavefunctions to density, we also lose the ability to calculate exchange interactions exactly. The only system for which we can calculate these exactly is the free electron gas; so all the functionals we regularly use are proposals for the exchange or correlation terms, or both, that approximate the exchange-correlation interactions but that don't represent them exactly.
And as with all the other cases, if we are not capturing all the exchange and correlation effects, that will impact alpha and beta electrons differently in a system with $\mathrm{S > 0}$.

This could be due to my misunderstanding, but how do we know the true picture has degenerate alpha and beta orbitals? Can we really say anything about orbitals of real compounds, since orbitals are just a construct we use to make forming an multielectron wavefunction easier?

Yes, you are right - however, the decomposition of the total electronic wavefunction into an alpha wavefunction and a beta wavefunction, which would lead to different orbitals for alpha and beta electrons, is no less of a construct - it's a different decomposition of the same 3N-dimensional wavefunction into N single-electron wavefunctions.

then I guess my question is how we know that the correct limit of  using alpha and beta orbitals is that the pairs will be degenerate? If that's the case, why not just use ROHF which ensures that from the beginning? I guess I don't see why we should trust the simple MO theory picture over the computational picture.

@Tyberius Even if we can't solve the schrödinger equation analytically, we know that energetically degenerate eigenstates can exist, which can be mathematically deduced by arguments of symmetry. Then, if our MO picture is symmetrical, does it not follow that their eigenvalues in a hypothetically analytical Hamiltonian would be degenerate?

For example: a doubly degenerate $E_{1u}=\Pi u$ irrep from the $D{\infty h}$ point group should be doubly degenerate in the Hamiltonian, right?

When Blaise comments that our MO picture is symmetrical I think that hits the key point.  The actual system has broken spin symmetry - 7 electrons in one spin vs 9 in the other.  Hence we shouldn't expect the the MOs to retain this symmetry.  Of course the Hamiltonian is agnostic to spin up vs down electrons so two degenerate spin-symmetry-broken solutions exist by inverting the spin of one another.

A triplet species, as your MO scheme shows, has two unpaired alpha spin electrons, which is why you don't have to look at the beta orbitals at all but you have to look, if the $\alpha$-HOMO and $\alpha$-HOMO-1 are equivalent or not.
After optimizing the geometry of triplet oxygen using the method you states, B3LYP 6-311G**, we end up with a bond length of $\pu{1.2058 Å}$ and are given the following energies:
 **********************************************************************

            Population analysis using the SCF density.

 **********************************************************************

 Orbital symmetries:
 Alpha Orbitals:
       Occupied  (SGG) (SGU) (SGG) (SGU) (PIU) (PIU) (SGG) (PIG)
                 (PIG)
       Virtual   (SGU) (PIU) (PIU) (SGU) (SGG) (SGG) (PIG) (PIG)
                 (SGU) (PIU) (PIU) (DLTG) (DLTG) (DLTU) (DLTU)
                 (SGG) (PIG) (PIG) (SGU) (PIU) (PIU) (SGG) (PIG)
                 (PIG) (SGU) (SGG) (SGU)
 Beta  Orbitals:
       Occupied  (SGG) (SGU) (SGG) (SGU) (SGG) (PIU) (PIU)
       Virtual   (PIG) (PIG) (SGU) (PIU) (PIU) (SGU) (SGG) (SGG)
                 (PIG) (PIG) (SGU) (PIU) (PIU) (DLTG) (DLTG) (SGG)
                 (DLTU) (DLTU) (PIG) (PIG) (SGU) (PIU) (PIU) (SGG)
                 (PIG) (PIG) (SGU) (SGG) (SGU)
 The electronic state is 3-SGG.
 Alpha  occ. eigenvalues --  -19.28009 -19.27992  -1.31462  -0.83929  -0.56862
 Alpha  occ. eigenvalues --   -0.56862  -0.55311  -0.31192  -0.31192   <------
 Alpha virt. eigenvalues --    0.19733   0.66242   0.66242   0.69937   0.71138
 Alpha virt. eigenvalues --    0.77370   0.78062   0.78062   1.23343   2.23120
 Alpha virt. eigenvalues --    2.23120   2.53472   2.53472   2.80188   2.80188
 Alpha virt. eigenvalues --    2.83167   3.40715   3.40715   3.75295   4.50288
 Alpha virt. eigenvalues --    4.50288   4.64528   4.80485   4.80485   5.97229
 Alpha virt. eigenvalues --   49.48209  49.67839
  Beta  occ. eigenvalues --  -19.24954 -19.24923  -1.25730  -0.75086  -0.51327
  Beta  occ. eigenvalues --   -0.46530  -0.46530
  Beta virt. eigenvalues --   -0.11708  -0.11708   0.24099   0.70896   0.70896
  Beta virt. eigenvalues --    0.71935   0.72389   0.79954   0.83108   0.83108
  Beta virt. eigenvalues --    1.27043   2.27711   2.27711   2.62147   2.62147
  Beta virt. eigenvalues --    2.87800   2.89802   2.89802   3.46348   3.46348
  Beta virt. eigenvalues --    3.78751   4.58054   4.58054   4.66543   4.88322
  Beta virt. eigenvalues --    4.88322   6.00448  49.51237  49.70844

You can clearly see, that the $\alpha$-HOMO and $\alpha$-HOMO-1 are degenerate with both an energy of $\pu{-0.31192 Eh}$. If you look at both MOs you can easily see why, because they are qualitatively the same combination of 2p AOs but with different orientation.

As symmetry was turned on, I guess, this needs to be the case, but the result stays the same, if it is turned off.
So, does computational chemistry get the molecular orbitals of dioxygen wrong? No, the MOs are right.

But actually you have a different question: Why are the energies of alpha and beta MOs in unrestricted open shell calculations not identical?
There are two methods to treat open shell systems. One is restricted open (RO*) and the other is unrestricted (U*). To not expand it too much, the main difference is, that as restricted closed shell methods, the restricted open shell methods force the alpha and beta orbitals to share the same MOs. Then, of course, they also would have the same energies.
But the energies that you show are from an unsrestricted open shell calculation. If you use this method, alpha and beta MOs are not forced to be in the same spatial MOs. And this freedom leads to different energies alpha and beta MOs.

Wow, can I ask you which software you used for the 3D representations?

Sure, that’s IQMol.


text:
Calculation of molarity using dilution equation
Using the formula $C_1V_1 = C_2V_2$ I am supposed to calculate the molarity of dilutions.
The original solution is $\pu{0.100 M}$ at $\pu{10 mL}$. I diluted it by mixing a $\pu{7.50 mL}$ stock solution with $\pu{2.50 mL}$ water, and made another one by mixing a $\pu{5 mL}$ stock solution with $\pu{5 mL}$ water.
Intuitively, I should get a higher molarity for my first dilution since it is higher in concentration, but when I calculate using
$$0.1(10)=C_2(7.5)$$
for the first dilution and
$$0.1(10)=C_2(5)$$
for the second, the second dilution's molarity is higher than the first. I know I am doing something wrong but I just can't figure out what.

It is a quite common mistake. Look that in first and second volume the number of molecules per unit of volume is the same, as the solution is homogeneous..; so the first equation C1 = C2, and how it is written is wrong.

You're mixing up the terms and formulae. Instead of rote learning the formulae, remember that the moles of the solute will remain the same in a solution - before and after dilution.
With that in mind, let's inspect your $\pu{7.5ml}$ stock solution. Before dilution, it had $\pu{7.5ml}\times\pu{0.1M}$ moles of the solute. After dilution, it has an unknown molarity. Let that molarity be $x ~\pu{M}$. Then, the new moles are $\pu{10ml}\times x~\pu{M}$. Equating and solving, you get $x=0.075$.
If you do that again for the second solution (I give it you as a homework), you'll get $x=0.05$. This easily verifies your own intuition that you should get "a higher molarity for the first dilution since it is higher in concentration".

It should be $7.5\cdot0.1=(7.5+2.5)\cdot C$ then find the value of C which is the concentration of 1st solution....for the 2nd one similarly, $5\cdot 0.1=10\cdot C$ then similarly calculate concentration....

Nice answer! Could you elaborate on your answer (because currently OP might not understand your answer).


text:
Can the pyrophoric tendencies of a pyrophoric substance be suppressed in a pressurized environment?
It's common knowledge that the pressure of an environment influences many properties of a substance, like the melting point, boiling point, or triple point. Pyrophoric substances ignite spontaneously at temperatures below $\pu{55^\circ C}$, like caesium for example.
I wish to ask if the tendency for pyrophoric substances to ignite is influenced at all by pressure? Obviously, these substances would not spontaneously ignite in a vacuum, but would their ignition temperature change and/or be dependent on the pressure of their environment? Is there a pressure apart from a vacuum that could cause the substance to lost its pyrophoric properties?

How do you imagine changing triple point? Are you only asking about elements and phase changes or pressure induced reactions?

Depending on the pyrophoric substance, there can be differeing effects from changing air pressure. Strange as it may seem, sometimes combustion is slowed at very high pressure. For example, ethane, which reacts explosively with oxygen at STP, oxidizes slowly in a series of steps at ~100 atmospheres (~107 Pascals).
That said, it depends on the substances involved, pressure and presence of catalysts or other substances. Aluminum in a ball mill produces pyrophoric dust, but adding graphite during milling suppresses ignition.


text:
Volume of NaOH needed to precipitate copper(II) but not calcium
I have $\ce{NaOH}: C = \pu{0.1M}, V_{\ce{NaOH}}$ that I want to progressively add in a solution (in water) of:
$$\begin{align}
\ce{Cu^{2+}}&: \pu{0.01M}; & K_\mathrm{sp} &= 10^{-18.6} \\
\ce{Ca^{2+}}&: \pu{0.01M}, & K_\mathrm{sp} &= 10^{-5.26}
\end{align}$$
with initial volume: $V_\mathrm{init} = \pu{0.1L}$. I want to add $\ce{NaOH}$ as long as $\ce{Cu(OH)2}$ precipitates, without causing any precipitation of $\ce{Ca(OH)2}$.
Knowing  that the equilibrium will be displaced to the left and I will have precipitation, I have done:
$$
\begin{array}{ccc}
\ce{Cu(OH)2 &->&Cu^2+ &+ &2OH^-}\\
&&0.001&&0.1V_{\ce{NaOH}}\\
&&0.001-x&&0.1V_{\ce{NaOH}} - 2x\\
\end{array}
$$
Also, since I want to achieve $99.999\%$ precipitation of $\ce{Cu(OH2)}$ I have got: $$x = 99.999\cdot10^{-5}$$
Using all that in the expression of $K_\mathrm{sp}$ I get :
$$10^{-18.6} = \left (\frac{0.001 - 99.999 \cdot 10^{-5}}{0.1+V_{\ce{NaOH}}} \right)\cdot\left(\frac{0.1\cdot V_{\ce{NaOH}} - 2\cdot99.999\cdot10^{-5}}{0.1 + V_{\ce{NaOH}}}\right)^2$$
However for this expression I get a value of $V_{\ce{NaOH}}$ that is extremely small. Is there anything wrong with my reasoning and if yes, how can I determine the volume for which i get the precipitation I want?

You need to precipitate $\pu{9.9999\times 10^-4 mol}$ of $\ce{Cu^{2+}}$, hence you need to add $2(\pu{9.9999\times 10^-4 mol}) = \pu{1.99998\times 10^-3 mol}$ of $\ce{NaOH}$, this is:
$\displaystyle{V_\ce{NaOH} = \frac{n_\ce{NaOH}}{c_\ce{NaOH}} = \frac{\pu{1.99998\times 10^-3 mol}}{\pu{0.1 mol L-1}} = \pu{19.9998 mL} \approx \pu{20.0 mL}}$
There are $\pu{1\times 10^-8 mol}$ of $\ce{Cu^{2+}}$ left in solution, i.e. $\ce{[Cu^{2+}]} = \pu{8.33 \times 10^{-8} mol L^-1}$. Then, $\displaystyle{\ce{[OH-]} = \sqrt{\frac
{K_\text{sp}}{\ce{[Cu^{2+}]}}} =  \pu{1.74 \times 10^{-6} mol L-1}}$.
The final concentration of $\ce{Ca^{2+}}$ is $\pu{0.8333 mol L-1}$, so $Q_\text{ps} = \ce{[Ca^{2+}][OH-]^2} = \pu{2.52 \times 10^{-12}} < K_\text{sp}$, i.e. $\ce{Ca(OH)2}$ won't precipitate at this point.

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


text:
What do "degrees of freedom" refer to?
Can you explain simply what are the degrees of freedom in chemistry? I don't fathom it very intuitively from Wikipedia.
I have almost no background in physics. I understand that it refers to all the variables that can describe a system, e.g. when talking about the motion of the molecule, the degrees of freedom refer to the ways in which the molecule can move. I understand that an atom can only have translational motion, whereas a molecule can rotate, vibrate and translate. But how this gives the molecule more "freedom" is beyond me.

Related: Why do methane molecules have 15 degrees of freedom?; Does a diatomic gas have one or two vibrational degrees of freedom?

It’s the minimum number of “variables” you need to fully describe the state of a particle or object under consideration. Things can move along 3 axis like x,y,z but sometimes they can rotate or vibrate as well, giving them more degrees of freedom. Read about generalised coordinates to get a better idea.

To find the number of degrees of freedom of a system, ask yourself the question, "in how many ways can I describe the motion of this system?"
Let us begin with monoatomic gaseous molecules. Visualise them as minute spheres.
Image source: http://vle.du.ac.in/mod/book/print.php?id=13264&chapterid=29093
Do you think there are any other ways to describe its motion? No, there aren't, but you might argue that it can also rotate about its own axis, so why do we exclude that? If that's the case, refer to this answer (tl;dr: The moment of inertia of an atom about its own axis is negligible due to the extremely small size and can be safely ignored.)
Now, let's switch to diatomic gases:

Source: https://physics.stackexchange.com/a/210955/168273
The diatomic molecule too is free to translate in any three coordinates of space. Also, it can rotate about the two axes perpendicular to the axis joining the two constituent atoms.
Again if you are wondering about the rotation about the axis joining the two atoms, apply the moment of inertia argument here too.
Alternative definition: The number of terms appearing in the energy equation of the molecule is called its degree of freedom.
$$E_\text{monoatomic}= \frac{1}{2}mv_x^2+\frac{1}{2}mv_y^2
+ \frac{1}{2}mv_z^2$$
$$E_\text{diatomic}= \frac{1}{2}mv_x^2+\frac{1}{2}mv_y^2
+ \frac{1}{2}mv_z^2 + \frac{1}{2}I_x\omega_x^2+\frac{1}{2}I_y\omega_y^2$$
where $I$ denotes the moment of inertia about an axis and $\omega$ the angular velocity about the same axis.
Do you see how this agrees with our qualitative analysis? There are $5$ terms in the energy equation of the diatomic molecule and $3$ in the case of the monoatomic one.
But there's a problem, at higher temperatures, under the provision of sufficient amount of energy the molecules can vibrate too, leading to an increment in the number of terms in our energy equation.
From Wikipedia:

The number of vibrational degrees of freedom, or vibrational modes, of a molecule is determined by examining the number of unique ways the atoms within the molecule may move relative to one another, such as in bond stretches or bends. This can be determined mathematically using the $3N-6=n_\text{DOF}$ rule, where $N$  is the number of atoms in the molecule and $n_\text{DOF}$ is the number of vibrational degrees of freedom. Note that for linear molecules, the rule instead becomes $3N-5=n_\text{DOF}$, meaning a linear polyatomic made up of $N$ atoms will have one less vibrational degree of freedom than a non-linear polyatomic with $N$ atoms.

Why are degrees of freedom important in thermodynamics?
According to the equipartition theorem an ideal gas distributes its energy equally in all degrees of freedom.  The average energy of a molecule in a gas associated with each degree of freedom is $kT/2$ where $k$ is the Boltzmann constant and T the temperature.
Task: Use the equipartition theorem to derive the concise internal energy equations for a diatomic and monoatomic gas molecule at low temperatures (i.e. ignore the vibrational degrees of freedom.)
Besides, knowing the degrees of freedom of the gas under investigation, we can find out its approximate molar specific heat capacities $C_p$ and $C_V$ which we often use in our thermodynamics calculations.

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


text:
Find the half equivalence point

How to find the half equivalence point knowing the pH, molarity, titrant added at equivalence point?

This a fairly straightforward and simple question, however I have found many different answers to this question. I originally thought that the half equivalence point was obtained by taking half the pH at the equivalence point. However, I have encountered some sources saying that it is obtained by halving the volume of the titrant added at equivalence point.

Yeah it's not half the pH at equivalence point your other sources are correct

Half equivalence point is exactly what it sounds like. It is the point where the volume added is half of what it will be at the equivalence point.
There are 3 cases

Strong Acid vs Strong Base: Here one can simply apply law of equivalence and find amount of $\ce{H+}$ in the solution. Similar method for Strong base vs Strong Acid.
Weak Acid vs Strong Base: Here one can apply Henderson-Hasselbalch equation as a general result one gets pH = pKa.
Weak Base vs Strong Acid: This case is similar to WA vs SB and one can easily use Henderson-Hasselbalch for bases again one gets pOH = pKb


text:
Transesterification of vegetable oils to produce biofuels - problems in the lab
I'm an undergraduate doing a small research project on biofuels.
I have 12 hours of lab time to investigate something about biofuels. I've chosen to explore how the fatty acid content of different vegetable oils affects the quality of the biofuel produced.
To investigate this I've chosen to do several different things in the lab. To determine the fatty acid content of each vegetable oil I chose I decided to compare their respective 'acid numbers'.
The acid number (or neutralization number or acidity) is defined as the mass of potassium hydroxide (KOH) in milligrams that is required to neutralize one gram of chemical substance.
To do this I decided to do a simple acid-base titration. I made up a standard solution of KOH to a concentration of approximately 0.1M. I did some qualitative tests to determine which alcohol was best to dissolve my vegetable oil - I chose propan-1-ol to be my solvent and I used 125ml along with 20ml of a recorded mass of vegetable oil and phenolphthalein indicator in the conical flask. Because acid number requires a mass of oil I did not pipette the oil into the conical flask, I simply recorded the mass on a top-pan balance. This does result in a larger error for my acid number value but it avoids a difficult time pipetting oil and then cleaning up afterwards - especially with my limited time in the lab. I then titrated against my 0.1M standard solution of KOH and reached the endpoint incredibly quickly - my titre value was less than 1.00 cm3. My
first question here is should I dilute my standard solution to a much lower concentration, as such a small titre would lead to a large apparatus error? There is already a large amount of apparatus error in this experiment because I weighed out the mass of oil rather than pipetted a known volume and then used density etc.
For my investigation I also have to synthesise a few biofuels. In my first lab session I tried out an acid catalysed transesterification reaction. I would like to stick to transesterification reactions as it's a common process in industry and in labs and can easily done in my undergraduate teaching lab. The method I used is as follows1:

Combine 8 cm3 of an alcohol with a few drops of concentrated sulfuric acid (approximately 0.15 cm3) in a 100 cm3 B19 round-bottomed flask and stir well.
Add 20 cm3 of a vegetable oil.
Reflux for approximately 1 hour
Allow to cool and transfer mixture to a seperating funnel. Allow the layers to seperate and discard the lower layer (containing glycerin). Wash the top layer with 2-3 15 cm3 portions of 4M NaCl solution.
Check the final washing with pH paper to ensure the pH is neutral.
Transfer the biodiesel to a clean, dry conical flask and dry the liquid with anhydrous magnesium sulfate.
Filter by gravity to remove the magnesium sulfate.

In italics is what went wrong essentially and why I think it did. When washing with 4M NaCl I could not get the pH to stabilise, I used almost 50 cm3 and yet still it was at a pH of 1-2. This makes me think I added too much concentrated sulfuric acid. I remember adding in several drops, where in reality one small drop is all thats required for 0.15 cm3. As such when seperating out my biofuel, I lost a lot of it so my yield will be poor. I may be able to still rectify it so I can do some sort of analysis on my biofuel so my first three hours weren't a complete waste, but I am unsure how. Should I wash with more NaCl (potentially reducing my yield to almost zero) or should I try to use a stronger base like NaOH or KOH? I'm unsure as to whether these will interfere with the product that I have made. This same issue is referred to in the following paper (page 5):  downloads.hindawi.com/journals/jchem/2010/689051.pdf
I am also unsure as to why neutralisation is needed in the transesterification process?
On the other hand if that dosen't work I could switch to a base catalysed method for biodiesel production which is faster and can be done with an exact mass of KOH rather than adding conc. acid dropwise. On the other hand some papers seem to suggest that this method would be worse for vegetable oils with a low fatty acid content. However, I am just comparing fatty acid content to the quality of biodiesel produced with one method - I can discuss this aspect in my final report.
[1] - This procedure was adapted from Bladt, D. Murray, S. Gitch, B. Trout, H. Liberko, C.J. Chem Ed, 2010, 88, 201-203.

First, I believe that using mass was the appropriate way to go. Volume measurements are subject to error due to temperature changes (most glassware is calibrated at $\pu{25 ^\circ C}$, I think). Furthermore, density is also dependent on temperature (since volume changes with temperature), but mass doesn't. Also, are you measuring the density yourself, or using it from tables? For vegetable oils, I would expect the density to vary depending on the batches, so using a theoretical value might not be the most accurate choice. In the end, you need the mass, so measuring the mass would give you the least amount of error.
For your $\ce{KOH}$ solution, you can dilute it to have a larger volume used (or you can weigh more oil, of course). I wouldn't dilute it more than 10-fold for this case. Another source of error is your low accuracy for the concentration of $\ce{KOH}$, $\pu{0.1 mol L-1}$. If you're trying to minimize error from here too, I would suggest standardizing your $\ce{KOH}$ solution with potassium hydrogen phthtalate, a common primary standard for bases, and calculate the concentration with a tenth of a mililmolar acuracy (e.g. $\pu{0.1023 mol L-1}$). Furthermore, use this solution freshly prepared, as hydroxides react with $\ce{CO2}$ to form carbonates.

Thank you. I don't think I have enough time in the lab to standardise with potassium hydrogen phthalate but I will discuss this in my project report.

Sure. I had also answered that neutralizing the acid in thé transesterification shouldn't be a problem (I'd use a milder base to avoid hydrolysis of your newly formed esters, such as potassium carbonate), but I don't see how the low yield might be a result of this issue.


text:
What are the magnetic properties of [Ni(S2C2Ph2)2]^z's 3 oxidation states?

I was doing some research on Wikipedia and I came across the article on non-innocent ligands. It talked about the chemical $$[\ce{Ni(S_2C_2Ph_2)_2}]^z$$ as an example. It mentioned that the transition metal complex was diamagnetic when it was neutral(z=0). I decided to figure out how many unpaired electrons the central nickel atom would have at various formal oxidation states. I found the following:
$$
\begin{array}{ccc}
\ce{Ni(II)}&(z=-2)&=2~\text{unpaired}\\
\ce{Ni(III)}&(z=-1)&=3~\text{unpaired}\\
\ce{Ni(IV)}&(z=0)&=4~\text{unpaired}\\
\end{array}$$
I don't understand why the complex is diamagnetic when its charge is 0. The radical ligands would be coupled with each other anti-ferromagneticly but there would still be 4 unpaired electrons in the central nickel atom. Wouldn't they make the complex paramagnetic? Could someone please explain to me why it is diamagneic and tell me whether the other 2 oxidation states are para or dia-magnetic.

$\ce{Ni^{2+}}$ is a $d^8$ metal. In a square planar orbital splitting, all $8$ electrons are paired, hence it is diamagnetic. Even when the complex is neutral, the electron loss is at the ligands (by definition, non-innocent), and they are antiferromagnetically coupled, which means that their spins "cancel" each other. Since there are two, the net moment is zero.
All complexes are $\ce{Ni(II)}$, so assuming all complexes are square planar, all $d$ electrons are paired up and the magnetic moment from these is 0. Both the $z=-2$ (no radicals) and $z=0$ (antiferromagnetic coupling) area diamagnetic. The $z=-1$ complex is paramagnetic. There is a single unpaired electron in one of the ligands that can't couple to the other one, where all of them are paired up.


text:
Why do oxygen and fluorine form a covalent bond?
Why do oxygen and fluorine form a covalent bond? I cant find an in-depth enough explanation online.

Because when you construct molecular orbitals, you find that most of the electron density is captured by the anion due to the electronegativity difference between the atoms forming the bond.

Well really the question is not why but why not?
Oxygen and fluorine readily form covalent bonds with other non-metals such as in $\ce{H2O}$, $\ce{CO2}$, $\ce{HF}$, etc. There is no reason stopping oxygen and fluorine from forming a covalent bond between each other.
It is perhaps better to ask the more general question "why atoms form covalent bonds with each other in the first place".
Simply put, atoms with lone electrons (called radicals) are very reactive and unstable. As well, atoms want to try to get 8 electrons in their outer shell so that they can obtain noble gas configuration and become even more stable. So when two atoms come together, one way to satisfy the above 2 goals is to share electrons with other.
In the case of $\ce{O}$ and $\ce{F}$, $\ce{O}$ has a 2 lone electrons and has 6 valence electrons while $\ce{F}$ has 1 lone electron and has 7 valence electrons. So what happens is 1 $\ce{F}$ atom comes and shares its lone electron with a lone electron of $\ce{O}$ to form a pair of electrons; called a covalent bond. Now the $\ce{F}$ atom has 8 electrons so it is happy. Then another $\ce{F}$ atom comes along and repeats the same process so that $\ce{O}$ now also has 8 electrons and is happy. So we are left with the molecule $\ce{OF2}$.

thank you for comment, but would you be able to explain it in more depth

Explanation of what?? There nothing in your post about why you ask it and even if you ask about them making bond with each other or with other elements. @Nano I don't get why you answer without clarification on account of OP.


text:
Does the pre-exponential factor (Arrhenius constant) depend on temperature?
I was taught that the the pre-exponential factor $A$ in the Arrhenius equation \eqref{arrhenius} does not increase with temperature.
$$k = A\cdot\mathrm \exp\left(-\frac{E_\mathrm a}{RT}\right)\tag1\label{arrhenius}$$
But, a component of the pre-exponential factor is the number of collisions, which presumably increases with temperature. Should the pre-exponential factor therefore not increase with temperature as well?

Sort of related: Units in modified Arrhenius equation? How exactly is activation energy defined?

Your intuition is correct. The factor $A$ changes with temperature.
This article details how the value of $\ce{A}$ for an elementary, bimolecular reaction between $\ce{P}$ and $\ce{Q}$ can be derived to be:
$$A_{\ce{PQ}}=N_\ce{P}N_\ce{Q}d^2_{\ce{PQ}}\sqrt{\frac{8k_\mathrm{B}T}{\mu}}$$
The RHS is clearly a function of temperature. Without going into the details,[a] it suffices to remember that $A$ is a function of temperature because it is related to molecular collisions, which themselves are a function of temperature.
However, it is worth noting this paragraph on Wikipedia:

...under a wide range of practical conditions, the weak temperature dependence of the pre-exponential factor is negligible compared to the temperature dependence of the $\mathrm{e}^{(-E_\mathrm{a}/RT)}$ factor[b] (my emphasis); except in the case of "barrierless" diffusion-limited reactions, in which case the pre-exponential factor is dominant and is directly observable.

Given this, it may be within sufficient experimental errors to make the assumption that $A$ does not vary with temperature. However, it is just that, an assumption. In reality, $A$ does vary with temperature.

[a]: This isn't the exactly correct expression though. As the article itself notes, "Often times however, when the term is determined experimentally, $A$ is the preferred variable and when the constant is determined mathematically, $Z$ is the variable more often used. The derivation for $Z$, while mostly accurate, ignores the steric effect of molecules."
[b]: Of course, here Wikipedia is using the actually computed values of $A$ (and not vague estimated formulae like the one above; as actually $\sqrt{T}$ grows faster than $\mathrm{e}^{-1/T}$)

That Libretexts article is horrible (I find the website to be very hit or miss, it used to be better). Symbols come and go randomly :( I will try to find a better source some time soon. Most phys chem books cover this properly.

The Arrhenius equation for the rate constant is an empirical relationship and used only because it fits many sets of data. I'm assuming that you mean a temperature dependence in the prefactor $A$ as $T$ is already in the exponential.
Collision theory in the gas phase predicts an equation of the form
$$k = a T^{1/2}\exp\left(\frac{-E_\mathrm a}{RT}\right),$$
where $a$ and $E_\mathrm{a}$ are fitted to the data. (If data is measured over a narrow range of temperatures (as is often the case) the $T^{1/2}$ has a small effect and $aT^{1/2}$ may appear to be constant with temperature and hence the 'normal' Arrhenius equation.) The $T$ in the exponential causes most of the change with temperature.
A more sophisticated derivation can be done using partition functions, which are defined in statistical mechanics, in which case the equation has the more general form
$$k = aT^{\beta}\exp\left(\frac{-E_\mathrm a}{RT}\right),$$
where $a$ and $\beta$ depend on the particular reaction type (e.g. atom molecule, or molecule-molecule), $\beta$ can have positive and negative values usually in the range $-3/2$ to $2$. These values depend on knowing some properties of the transition state, which is the geometry of the species at the top of the barrier separating reactants to products. However, experimentally, these geometries are generally unknown. Thus there are three parameters to fit the data, $a$, $\beta$ and $E_\mathrm{a}$.
Just to mention in passing that, in solution, some reactions are 'diffusion controlled' reactions, whose rate constants are controlled by the viscosity of the solvent, not the activation energy which in these reactions is very small. A detailed theory due to Kramers describes the dragging ('friction') effect of a solvent on the reacting molecules. In the gas phase so called 'unimolecular reactions' are modelled by what is called RRKM theory (Rice-Ramspburger-kassel-Marcus) based also on statistical methods that describe how much energy the reactants have when they collide.


text:
Which one of the following is optically active?

To be optically active the presence of chiral centres are very important but if there exists a plane centre or axis of symmetry then the molecules are optically inactive.  How to determine the optical activity when more than one ring are present?  Is it sufficient to check those two mentioned condition?

Rings are irrelevant. The rules for many rings are the same as with one ring or without rings at all. Besides, symmetry axes are also irrelevant.

How to determine the optical activity when more than one ring are present? Is it sufficient to check those two mentioned condition?

As Ivan Neretin accurately noted above:

the rules for determining optical activity of a molecule are the same for all molecules, whether it has one ring, several rings, or no rings at all.
axes of symmetry are irrelevant in determining the optical activity of a molecule. We only have to consider center and plane of symmetry.

That said, the given molecules have several chiral centers, which is a great sign that the molecule may be optically active.
If you will try to find the plane of symmetry, you won't find any. While the rings may appear planar on the paper, they (even the fused ones) actually exist as conformations in 3D.
Moreover, the center of symmetry is a point that requires identical groups equidistant from itself; a quick glance over both the molecules shows there are no such centers.


text:
Why does adding water to a saturated solution increase the number of ions present in the solution?
In my book, it is stated that when some water is added to a test tube having a saturated $\ce{Ag2SO4}$ solution, with its solid at equilibrium, the number of $\ce{Ag}$ ions in the solution increases.
I don't understand it. If $\ce{Ag2SO4}$ solution is saturated, it means that no more solute can dissolve, and if we add water, the amount of dissolved solute should not change.
Also, if the amount of dissolved $\ce{Ag}$ increases, would the $K_\mathrm{sp}$ value also increase?

The concentration is fixed. Not the amount of dissolved ions present in a solution.

$\pu{K_{sp}}$ is a constant at a given temperature. It doesn’t increase with increase in $\ce{[Ag+]}$ ions.

Related question here

The number of $\ce{Ag}$ ions increases, but the number of $\ce{Ag}$ ions per volume of water does not increase.

Consider a saturated solution of $\ce{Ag2SO4}$. The following equilibrium is attained.
$$ \ce{Ag2SO4 <=> 2Ag+  + SO4^2-}$$
Adding water accounts to increasing the volume of the solution. Hence more amount of solute can be dissolved since solubility of a salt depends on the amount of salt dissolved per unit volume. Therefore, there is an increase in the no. of ions produced in the solution. But as Avnish Kabaj mentioned, $[\ce{Ag+}]$ remains the same.
Also as MollyCooL said, $\pu{K_{sp}}$ doesn't change as it is a constant at a particular temperature.

We have to add a fixed amount of solute to a fixed amount of water to get saturated solution so if we add more water to solution and don't change the amount of solute so the amount of dissolved ion should not change

@MarvaJami - It depends. Is there more of the solid $\ce{Ag2SO4}$ or not? So if 1 liter dissolves 0.1 grams of something, then 2 liters would dissolve 0.2 grams (total).

Will if the solution is saturated it means we have 0.1 gram dissolved ions in 1 liter so if I add 1 liter water and don't add more agno3 the number of ions would remain the same am I right?

Yes it wont change then, but what if there is a solubility equilibrium established where you have both the sakt and the ions, then on adding water, the equilibrium will be disturbed and more amount of salt can be dissolved


text:
Electrolysis of concentrated sulphuric acid
In my textbook it is given that for electrolysis of dilute sulfuric acid  at anode following reactions can occur:
At moderate concentrations
$\ce{2H2O ->  O2 + H+ +4 e-}$
And for high concentrations
$\ce{2SO4- ->  S2O8^2- +2 e-}$
SRP value for first reaction is less than second and hence the first reaction should take place. But why does that not hold for high concentrations?

I was taught that at high concentrations, there are too many sulphate ions surrounding the anode. Hence, they get oxidised even if their SRP values disagree. But, I am not sure if this is correct.

Why anyone would think that "standard" value has importance in highly non-standard environment?

Do you mean it becomes kinetically preffered? and we have something like overpotential?

There's not much water around in conc Sulfuric acid

Related: https://chemistry.stackexchange.com/q/75108/5026

This is because in concentrated sulfuric acid solution equilibrium condition will favour more water molecules than oxidation of water into protons. So there will be an over potential required (to go against the equilibrium) , that is extra potential beyond the theoretical reduction potential derived from thermodynamics to complete the reaction. So the total potential required will be theoretical Plus overpotential. In case of oxidation of sulphate reduction potential will be much less that for water ,thus oxidation of sulphate happens.


text:
Why does the reaction between potassium permanganate and oxalic acid result in a yellow/brown solution?
The reaction between potassium permanganate and oxalic acid is a redox reaction and produces $\ce{Mn^2+}$ ions, the final solution however goes yellow/brown.
If $\ce{Mn^2+}$ ions are colourless, and its complex ions are pink, where does the yellow/brown colour come from?

The reaction between $\ce{MnO4-}$ and $\ce{C2O4^2-}$ is performed under acidic conditions.
In acidic conditions only, $\ce{MnO4-}$ gets reduced to colourless $\ce{Mn ^{2+}}$ as, $$\ce{MnO4- + 8H+  +5e- -> Mn^2+ + 4H2O}$$
But if you don't maintain proper acidic condition of the medium, it continuously absorbs $\ce{H+}$. Eventually, due to lack of protons, it gets reduced to brown coloured $\ce{MnO2}$ under neutral conditions, where it needs less no of protons to get reduced than acidic conditions, as $$\ce{MnO4- + 4H+ + 3e--> MnO2 + 2H2O}$$ That's why you see the brown colour which is due to $\ce{MnO_2}$.


text:
The Ionization of Water
When we put $\ce{HCl}$ in a container of water and let's say that the concentration we put is $10^6$ molar. Let me understand this calculation:
$[\ce{OH-}]=x$
$[\ce{H+}]=10^6+x$
$x(10^6+x)=10^{-14}$
The $[\ce{OH}]$ is $x$, but what about the $[\ce{H+}]=10^6+x$ why not $10^6-x$? Because the quantity of the $[\ce{H+}]$ does not increases after we add the $\ce{HCl}$ but decreases. So why $+x$ and not $-x$?

10^6 is absurd, almost to the point of being offensive.

what is absurd about it? I have a question with 2.19×10^−7  how is that better?

Being physically possible is a big difference

"the quantity of the [H+] does not increases" you probably meant the quantity of H+ released by water...

Sorry but why a concentration of 10^-6 is physically impossible while 10^-7 is?

Do you mean $10^6$ or $10^-6$?

sorry now I saw the error.

Consider the reaction,
$$\ce{HCl -> H+ + Cl-}$$
Assuming the initial $\ce{[HCl]}$ is $\pu{C M}$, after reaction takes place, we have $\ce{[H+]}$ as $\pu{C M}$.
Also its an aqueous solution. Hence we have to consider the following equilibrium:
$$\ce{ H2O <=> H+ + OH- }$$
Let initial $\ce{[H2O]}$ be $\ce{a M}$. At equilibrium, we have $\ce{[H+]} = \ce{x M}$. Since we have more $\ce{H+}$ from the ionization of water, therefore $\ce{[H+]_{net}}$ = $\ce{C +x M}$ and not $\ce{C-x M}$
Note: If $C \geq \mathrm{10^{-6}~M}$, $\ce{[H+]}$ coming from $\ce{H2O}$, ie. $x$, can be neglected w.r.t $C$ unless and until you’re a rigid perfectionist.

How does this answer the question: " what about the $[\ce{H+}]=10^6+x$ why not $10^6−x$?"

I have shown its plus and not minus. Haven’t I?

You have. But you haven't explained why $10^6-x$ is invalid.

We have $\ce{[H+]}$ to be increasing and hence the ‘plus sign’. I don’t get why ‘minus sign’ comes into picture. Wait, is the OP relating anything to common ion effect?

"I  don’t get why ‘minus sign’ comes into picture" I know. I also agree with you completely. But, the OP doesn't. And this answer needs to explain the OP so... :)

Your whole question is so muddled that it is difficult to know where to begin.

When we put HCl in a container of water and let's say that the concentration we put is $10^6$ molar.

$10^6$ molar HCl is totally absurd. The most concentrated HCl solutions will be around 12 molar. HCl is a gas at room temperature and pressure. To solidify HCl you must cool the gas to −114.22 °C. HCl has a molecular mass of 36.46 g/mol, thus neglecting any water, $10^6$ molar HCl would have $10^6$ moles of HCl in a liter volume, or a mass of 36460 kilograms per liter. Gold is only 19.30 kilograms per liter.
The other thing is that HCl ionizes in water according to the reaction
$\ce{HCl + H2O -> H3O^+ + Cl^-}$
Since pure water is only about 56 molar, there isn't nearly enough water to ionize $10^6$ moles of HCl.
Let's talk about something more reasonable, 1.0 molar HCl.
We can assume that all the HCl ionizes then the $\ce{[H^+]= 1.0 molar}$.
Since
$\ce{[H^+][OH^-] = } 10^{-14}$
$\ce{[OH^-] = } \dfrac{10^{-14}}{\ce{[H^+]}} =  \dfrac{10^{-14}}{1.0} = 1.0\times10^{-14}$
Now there is an assumption here about significant figures. So when we note that the HCl is 1.0 molar, we mean +/- 0.05 moles, or two significant figures. Very, very few chemistry experiments have results good to 4 significant figures.
In reality the starting pure water had $1.0\times10^{-7}$ molar $\ce{[H^+]}$ and $1.0\times10^{-7}$ molar $\ce{[OH^-]}$. Since the final solution had $1.0\times10^{-14}$ molar $\ce{[OH^-]}$, then the added HCl obviously reacted with about $1.0\times10^{-7}$ moles of $\ce{[OH^-]}$. But this amount is so small compared to the 1.0 moles of $\ce{[H^+]}$ that it is insignificant.


text:
What causes the anti-siccative effect of soot (amorphous carbon) as a pigment in oil paints?
Oil paint containing carbon black (soot, charcoal) pigment is known to polymerize ("dry") a lot slower than if it has other pigments. I understand this is not just due to the absence of pigments which are themselves siccatives (drying agents), but is also a direct "slowing" effect of the carbon black in the paint. The pigment supplier Kremer Pigmente confirms that it is a "slow drier", and the book "Artists' pigments, c. 1600-1835" by Rosamond D. Harley indicates:

They [authors on the matter] invariably state that a drier should be added to it to accelerate its exceptionally slow drying time.

What is the nature of this effect? Why does carbon black slow the drying of oil paints?

According to the Wikipedia page on oil-drying agents (emphasis added):

An oil drying agent, also known as siccative, is a coordination compound that accelerates (catalyzes) the hardening of drying oils, often as they are used in oil-based paints. This so-called "drying" (actually a chemical reaction that produces an organic plastic) occurs through free-radical chemical crosslinking of the oils. The catalysts promote this free-radical autoxidation of the oils with air.

The exposed surfaces of carbon black are highly functionalized, in many cases with oxygen-containing moieties.  The Birla Carbon website indicates that this surface chemistry "is a function of the manufacturing process and the heat history of a carbon black and generally refers to the oxygen-containing groups present on a carbon black’s surface." Per panel 3 of the figure at Asahi Carbon's page, "Carbon black properties," these functional groups can include at minimum carboxyl, hydroxyl, and carbonyl (aldehyde/ketone) functionality.  The section of the Wikipedia page on carbon black describing its surface chemistry indicates that quinonic, lactonic, and phenolic groups may also be present.
Most of the above-mentioned functional groups can participate in radical reactions, including reactions with the active species involved in paint hardening.  These side reactions of the carbon black siphon off the active radical species in the paint into pathways that do not contribute to the hardening/drying process.  The Lamp Black page on the Natural Pigments website provides support for this (emphasis added):

However, [lamp black] is not generally recommended for use in oils, because carbon is an anti-oxidant that greatly slows the drying time of oil paint.


text:
How to test for halides without ammoniacal silver nitrate solution?
If I have an halide e.g magnesium iodide, how can I test what halide the compound contains?
I understand there is a method using silver nitrate and ammonia but what other methods are there?

In case of iodides I'd always prefer oxidizing them with $H_2O_2$ + acid to form iodine and then add hexane or heptane. Iodine will dye the non-polar layer purple.

@Justanotherchemist Thanks for your reply. I was looking for a more general method to identify should the compound be unknown.

Most methods are not that specific. Silver is quite useless it will precipitate with almost anything. For the fluoride and chloride I also like converting them to the corresponding $CrO2X2$. It's probably more uncommon but if simple halides are mixed with dichromates (toxic!) and conc. $H2SO4$ and heated it will form characteristic fumes and liquids that can be further tested for chromium using the peroxo-complex. If nothing else in there reacts with the dichromate I could probably distinguish fluoride from chloride. But it's quite dangerous.

ChemGuide gives another method:

If you add concentrated sulphuric acid to a solid sample of one of the halides you get these results:
  $$
\begin{array}{c|c}
\text{Ion present}&\text{Observation}\\\hline
\ce{F-}&\text{steamy acidic fumes}\\
\ce{Cl-}&\text{steamy acidic fumes}\\
\ce{Br-}&\text{steamy acidic fumes contaminated with brown bromine vapour}\\
\ce{I-}&\text{some steamy fumes, and lots of purple iodine vapour}\\
\end{array}
$$

Note that those "fumes" are of the corresponding hydrogen halides. Also, in the case of the iodide ion, "various red colours in the tube" are also observed.
This reaction occurs because concentrated sulphuric acid acts as an oxidizing agent. It isn't able to oxidize fluoride or chloride ions, but definitely oxidizes $\ce{Br-}$ and $\ce{I-}$ to $\ce{Br2}$ and $\ce{I2}$ respectively (details here)
Do note that, to distinguish between the fumes of $\ce{HF}$ and $\ce{HCl}$ above, you'll need to dissolve them in water and perform the silver nitrate test!


text:
Why does iodine solution turn transparent on adding base?
Yesterday in biochemistry lab when I added sodium hydroxide to iodine solution. The iodine turned transparent. Can someone please explain why is this happening?

On adding acid the NaOH neutralized and the iodine turned black again.

OK, how is this off topic :-( ?

@OscarLanzi Probable violation of homework policy, OP's probably going to make lab report, but doesn't know things which should be known even before start of experiment.

@Mithoron The question is based on genuine curiosity.
I tried googling the topic but could not find a relevant result. So I posted the question here to understand the underlying chemistry.

Iodine disproportionates, forming iodide and iodate ions, in the presence of base.  But it can be restored by neutralizing the base.  Neutralization will drop the hydroxide concentration low enough so that the disproportionation does reverse.
Let's punch in some numbers using the equilibrium constants given here.  We have
$$
\begin{array}{ccc}
\ce{I2 + 2OH- &<=>& I- + IO- + \ce{H_2O}}&K=30\\
\ce{3IO- &<=>& 2I- + IO3-}&K=10^{20}
\end{array}
$$
To superpose the reactions we take three units of the first one plus one unit of the second, canceling out the unstable hypoiodite ions.  So, rendering that three units of the first reaction mean three factors of that value of $K$, we end with:
$$\ce{3I2 + 6OH- <=> 5I- + IO3- + 3H2O}, K=2.7×10^{24}$$
If we consider a hypothetical case where all the iodine species are in their standard states, including iodine as the element, then the equilibrium hydroxide ion concentration is  $\pu{8.5×10^{-5} M}$.  Taking the base 10 logarithm and adding $\mathrm{p}K_\mathrm{w}=14$ then gives a $\mathrm{pH}$ just below $10$, so a neutral solution gets your elemental iodine back.


text:
Why does water electrolysis use direct current?
Every source I've found regarding water electrolysis says that it's done using direct current (rather than alternating current), but I haven't found a single source that mentions why.
Obviously, using AC will result in the hydrogen and oxygen being mixed together, so if the goal is to produce separate gases then DC is required. But is that the only reason, or is there some chemical process that works better with DC?


text:
How to compare SN2 rate of cyclopropane with bridged cyclic compound?
How to compare the SN2 rate of iodocyclopropane and 1-iodobicyclo[2.2.1]heptane?

Backside attack is must, but as the cyclopropane ring is planar it's quite impossible for back side attack. Then the bridge compound is non planar, we know in bridge compound both SN1 and SN2 substitution are bad, so I just confused in these two.

I don't think SN2 will take place at all.

Concerning the rates of SN2 reaction, the rate of reaction for cyclopropyl iodide will be relatively higher than 1-iodonorbornane (name of given bridged compound).
The cyclopropyl iodide (same as 1-iodocyclopropane) has one hydrogen atom substituted by the iodine atom. If you look into the proper stereochemistry of the compound, this $\ce{I}$ atom can be above the plane of the cyclopropane ring or below the plane. Angle between the above-planar atom and below-planar atom is approximately 115 degrees. So, the attacking nucleophile can attack from above the plane if $\ce{I}$ is below the plane, or vice-versa. This attack would be best if leaving group $\ce{I^-}$ and the nucleophile are in almost 180 degrees position. Here the repulsion due to lesser bond angle and with the hydrogen will decrease the rate, but SN2 is still possible to a low extent in the cyclopropane ring.
Now to the bridged compound. Firstly, there is very very little chance for backside attack. Secondly, for occurrence of SN2, there should be a Walden inversion of configuration. But in the briged bicyclo compound, the ring is in a very compact structure, and if it tries to undergo any inversion of configuration, a huge strain will be created in the bicyclo-compound, which is thermodynamically, very unstable. So, the rate of SN2 is almost nil for 1-iodo norborane.