text:
Why does a real gas with attractive forces between particles experience an increase in internal energy when expanding isothermally?
I am trying to understand at the molecular level why a real gas with attractive forces between particles experiences an increase in internal energy when expanding isothermally. Here's my thinking:
For a gas, the path-independent differential form of change in internal energy is
\begin{equation}
dU = \frac{\partial U}{\partial V}{ \delta V} + \frac{\partial U}{\partial T}{ \delta T}
\end{equation}
For an isothermal expansion,
\begin{equation}
{ \delta T} = 0
\end{equation}
so \begin{equation}
dU = \frac{\partial U}{\partial V}{ \delta V}
\end{equation}
(For an ideal gas, internal energy is proportional to temperature, so an isothermal expansion doesn't change the internal energy.)
The way I'm looking at the real gas situation is that:

When the real gas expands, say by free expansion into an evacuated chamber, the Δ U = Δ P.E. + Δ K.E.
The increase in volume causes the average distance between particles to increase and therefore increases the potential energy.
The increase in potential energy comes as the expense of a decrease in kinetic energy, because Δ U= 0 = Δ P.E. + Δ K.E., in the same way that throwing a ball upwards into the air causes it to lose K.E.
However, in the case of the gas, this is a very short-lived decrease in K.E., because the decrease in K.E. of the particles lasts no longer until they reach a wall.
Because the walls of the evacuated flask are maintaining an isothermal environment, they transfer heat energy to the particles inside, either by contact conduction, or radiation, thereby restoring K.E. to initial value.
U final = final P.E. (higher than initial conditions) + initial (restored) K.E., and is therefore increased.

(In the case of a volume expansion of a real gas by doing work against a piston, I believe that there would be an even larger transient drop in K.E., immediately counteracted by the introduction of correspondingly more heat because of the isothermal constraint.  So the end-of-pathway conditions are still the same, increased P.E. and restored K.E.)
Is this reasoning correct? The version of Atkins I have asserts that for a real gas
\begin{equation}
\frac{\partial U}{\partial V} > 0
\end{equation}
but doesn't provide a molecular explanation.

Searching for the same thing.

The internal energy change with temperature can be found from one of the Maxwell equations and is
$$ \left (\frac{\partial U}{\partial V}   \right)_T = T\left (\frac{\partial p}{\partial T}   \right)_V -p$$
and  a real gas can be represented  with the empirical van-der-Waals equation $\displaystyle (p+a/V^2)(V-b)+RT$ where a represents the attractive potential and b the molecular volume.
Performing the calculation gives
$$ \left (\frac{\partial U}{\partial V}   \right)_T = \frac{a}{V^2}$$
which shows what you note in your question.
As the gas expands work has to be done to separate the molecules because they have an attractive  potential between them, thus the internal energy has to increase  because the temperature is held constant. If the gas was thermally isolated during expansion it would cool and this is observed in the Joule-Thomson effect.

Won't the gas do work as it expands contributing a factor towards reduction in the internal energy?


text:
Role of methanol in NaBH4 reduction
In reduction of carbonyls (aldehyde and ketones) to alcohols using NaBH4/Methanol, alkoxides are formed with the Boron which usually gets protonated on addition of water/acid.
My question is what is the function of methanol in this reaction as methanol reacts with NaBH4 to form sodiumtetramethoxyborate and hydrogen gas.
Some sources say methanol protonates the alkoxide ion and after doing series of reduction experiments I find that the alkoxide formed only gets protonated to an alcohol on addition of water.
So what does methanol do in this reaction/what is the effect of methanol?

I would say it is a solvent for the reaction.

MeOH does react with NaBH4 but that reaction is slow at ice-bath temperature and the reaction with carbonyls is faster.

whenever we mix NaBH4 and MeOH gas bubbles come out due to formation of H2. So, it does react with MeOH

I believe methanol does protonate the alkoxide. The resulting methoxide coordinates to $\ce{BH3}$ forming $\ce{(MeO)BH3-}.$ The electron donating character of the methoxy group makes the remaining protons more hydride which results in a more reactive hydride source.

Interesting, but a reference maybe?

This is probably the reference in question: Stabilization of NaBH4 in Methanol Using a Catalytic Amount of NaOMe. Reduction of Esters and Lactones at Room Temperature without Solvent-Induced Loss of Hydride, J. Org. Chem. 2018, 83, 3, 1431–1440.

You need methanol to solve NaBH4, other solvents are not very suitable. The active reducing species can be NaBH4 or NaOCHxHx (in which NaBH4 has first reacted with methanol)

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text:
Nitration of (E)-4-Phenyl-3-buten-2-one

What would happen if (E)-4-Phenyl-3-buten-2-one, shown above is nitrated, and is this compound more or less reactive than benzene?
My Attempt: By drawing the resonance structures of the above compound, I noted that the functional group withdrew electron density from the benzene ring. This means that this compound would be less reactive than benzene; my book agrees with me on that.
However, I stated that the nitro group would be added to the ortho or para positions, due to the fact that the sigma complex created is more stable due to the resonating effects of the double bond in the functional group. On the other hand, nitrification at the meta position would result in only three resonance structures in the sigma complex, resulting in a less stable compound. However, my book states that because the functional group is a deactivating group, it is a meta-director, without any further explanation. While I understand this rule is true in the general case, why does this rule still apply in the above compound, rendering my answer wrong?

I don't know why you deleted it, but your book is wrong on this count. Nitration of your compound yields the ortho and para isomers, for the reasons you mentioned. Chem. Ber.  1882, 15, 2861; ibid 1883, 16, 2205; J. Chem. Soc. 1933, 344

@orthocresol Thank you for pointing this out; I thought that I was actually wrong after reconsidering the problem, due to the fact that the additional resonance structure would be very minimally affecting the stability due to its positive charge on both the beta and alpha carbon of the carbonyl group. Please repost the comment as an answer and I will accept it.

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text:
How do colour pigments chemically work?
How do color pigments colour the plastics? Is there any chemical reaction, molecular association, or physical reaction?
What kind of bonding mechanism is there between pigment molecules and plastics?
How could Polypropylen and liquid organic pigments or granular organic pigments join each other? Pigments dont have oil or any solvent.
What would happen in molecular level?

It depends on both the substrate and the pigment/dye. Some are purely physical; some involve chemical reactions. You need to ask a more specific question.

https://en.m.wikipedia.org/wiki/Chromophore?wprov=sfla1

I guess, this is, chomophore,  more about how the molecules absorb the light. I was wondering the bonding mechanism,  if there is chemical, physical or combined bonding mechanism, between the fluid organic pigments and PP material.

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text:
How to prepare a water sample for photometry
The purpose of measuring N P K Ca Mg S is to determine the amount of material consumed by different types of plants in different periods.
The initial solution, which is prepared as a nutritious solution, includes highly accurate and measured doses of substances such as Potassium Nitrate, Magnesium Sulphate, Calcium Nitrate, Potassium Monophosphate, Ammonium Nitrate, and other materials that supply the micro-nutrients required by the plants.
The resulting solution is very clear and i use this solution to irrigate the plant
after 1 or 2 day i re-irrigate but it's done with deionized water
i can measure the runnoff water (drain water) of the pot for the amount of material available (such as NO3,PO4,K,Ca,Mg,SO4) with photometer because water is transparent
The problem occurs when the initial solution of humic acid and fulvic acid is added. In this case, the water released is very dark and it is not possible to use this water in photometry.
I have searched a lot, but the easiest way is to use photometer and chemical reagents to measure the concentration of the material.
my question is :
- how i can remove humic and fulvic acid from water without change in the amount of other materials or low changes (needed material is: N P K S Ca Mg)?
- or how i can measure N P K S Ca Mg in dark water with photometer?
detail:
error measuring up to 15% is acceptable

I simply cannot find a question in your post. Are you asking for different methods to measure those elements? Are you asking something about removing humic and fulvic acids? I suspect you have an interesting question, I just can't find it. Also, this is a confusing statement: "The resulting solution is very clear and after two to three days of water treatment, the water dissolved in the water (drainage water) can easily be measured by photometer." The water dissolved in the water? Anyway, I hope you will try to clarify so you can get an answer...good luck!

hi
sorry,i have described badly
my question is :

how i can remove humic and fulvic acid from water without change in the amount of other materials or low changes (needed material is: N P K S Ca Mg)?
or how i can measure N P K S Ca Mg in dark water with photometer?

detail:
error measuring up to 15% is acceptable


text:
Using Ethylene Glycol as anti-freeze in a biodigester gas storage system
I am busy building a bio-digester to produce biogas for cooking/water heating in a cold climate.
I am using the 3 IBC tank solution.
Tank 1 is the digester itself.
Tank 2 is the gas storage unit.
Tank 3 is the pressure for the gas outlet.
Tanks 2 & 3 will be linked. As tank 2 fills with gas, water overflows into a container with sump pump and is then pumped up to tank 3. When cooking, etc, the return valve is opened, allowing water pressure from tank 3 to push the gas in tank 2 towards the house.
I will be keeping tank 1 warm via a combination of insulation and surrounding the tank with decomposing matter from tree cutting & autumn leaves. I will also be running a hot water line through it from a compost heater.
Tanks 2 & 3 pose a bigger logistical challenge.
I brew beer for a living and use ethylene glycol to keep our fermentors & chillers from freezing over. I was wondering if it could also be used in this sort of set up. Aside from the gas outlet from tank 2 to the house, it is a closed system as far as the water is concerned.
However, I am unsure of how the ethylene glycol would react with the biogas. Is it safe?
The biogas will be produced from a mixture of food waste, cooked malts, and animal manure. It will be run through a water filter to prevent blowback (between the house and tank 2) and a steel wool filter (to remove hydrogen sulfates).
Other than pollution issues, should I be aware of anything else? I am worried about the glycol evaporating into the biogas causing an explosion.

Ethylene glycol is not at all volatile (a lot less than water). The risk of causing any reaction with biogas (which will be mostly methane) is low: the biogas will already be far more likely to go boom than any possible reaction caused by the glycol.

Ethylene glycol is highly toxic... it does not belong in beer. See http://emedicine.medscape.com/article/814701-overview You can buy food-grade propylene glycol, though, e.g. https://www.amazon.com/Bottle-Propylene-Glycol-Kosher-99-9/dp/B01FY4WF54/ref=sr_1_4_s_it

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text:
Are there any significant uses of the compound formed by magnesium and anthracene?
In an unusual reaction, magnesium reacts with anthracene in THF to form a grignard-like compound called, prosaically, magnesium anthracene:

I remember using it briefly in the 1980s. At the time there was speculation about its potential use in both synthesis and as a starting material for hydrogen-storage systems.
Are there any significant current uses of the the unusual complex?

related https://chemistry.stackexchange.com/questions/62254/sodium-naphthalenide https://chemistry.stackexchange.com/questions/62269/structure-of-sodium-naphthalenide not that unusual ;)

It seems like an idea of using magnesium anthracene systems for the $\ce{MgH2}$ production persisted since 1980s [1] till late 2000s, when new more efficient method with better scalability for industrial use was established.
One of the recent reviews in hydrogen-storage applications [2, p. 220] compares the older two-step process of $\ce{MgH2}$ synthesis:

First, metallic magnesium is dissolved in anthracene to form Mg-anthracene solution. Then, the hydrogenation of the Mg-anthracene solution produces $\ce{MgH2}$ with $\ce{CrCl3}$ or $\ce{TiCl4}$ catalyst.

with the most recent direct synthesis from Mg powders and tablets [3]. Their experimental trial involved the following:

Powder and tablet Mg were placed on trays inside a 50 kg hydrogenation furnace. After purging with Ar, the furnace was ﬁlled with 99.9999% hydrogen with a pressure of less than 2 MPa. The temperature of the process was controlled so as to remain near the boundary between the regions $\ce{\{MgH2\}}$ and $\ce{\{Mg + H2\}}$ in the thermal equilibrium diagram, and maintained for 30–45 h. The hydrogen consumption behavior was monitored by a pressure gauge. Afterward, the furnace was gradually cooled down to room temperature. The hydrogenation yield (%) was calculated via the weight difference of Mg samples before/after the process.

$\ce{MgH2}$ obtained with this method showed a hydrogen yield of around 96% with great reproducibility and scalability, and was proved as industry-ready.
But still, magnesium-anthracene · 3THF is used at a lab-scale in fine organic, organometallic, and inorganic syntheses [4, pp. 300-307]:

as a source of soluble zerovalent magnesium;
in preparation of finely-divided metal powders and transition metal complexes;
catalytic synthesis of Grignard compounds.

Bibliography

Bogdanovic, B. Acc. Chem. Res. 1988, 21 (7), 261–267. DOI 10.1021/ar00151a002.
Sasaki, K.; Li, H.-W.; Hayashi, A.; Yamabe, J.; Ogura, T.; Lyth, S. M. Hydrogen Energy Engineering: A Japanese Perspective; Springer Japan: Tokyo, 2016. ISBN 978-4-431-56042-5.
Uesugi, H.; Sugiyama, T.; Nii, H.; Ito, T.; Nakatsugawa, I. Journal of Alloys and Compounds 2011, 509, S650–S653. DOI 10.1016/j.jallcom.2010.11.047.
Fürstner, A. Active metals: preparation, characterization, applications; VCH: Weinheim; New York, 1996. ISBN 978-3-527-61517-9.

I don't think it's zerovalent.

It isn't. Magnesium anthracene has a structure in which the middle ring where the magnesium is attached has lost its planarity, as if it were no longer aromatic. Transferring electrons from the magnesium into the anthracite would do that.


text:
What does negative current reading in galvanic couple mean?
I used a multimeter in a galvanic testing between two metals. I connect one to metal A and the another end to metal B. Then I immersed the 2 metals in a salt solution, securing that they are apart in the solution. It gave me a negative current reading, then when I switched the multimeter probes, it became positive. What does the sign of current indicates?

The negative sign is simply indicative of how you've inserted your multimeter probes (which are labelled for polarity, or should be) into the galvanic battery circuit. In one case, the multimeter "sees" electrons flowing from the positive terminal to the negative one (hence the negative sign); in the other case, the opposite is true.


text:
What is the most complex molecule we can manufacture without the help of proteins? (soft queston)
What is the most complex molecule we can manufacture without the help of proteins?
This isn't motivated by anything but curiosity about the state of chemical engineering. I realize “complexity” is a loaded term, so I'll defer to the judgment of the respondent: number of molecules, complex arrangement of bonds, or whatever other measure might be appropriate.

"Most complex" isn't well-defined.

Complex may not be well-defined, but that doesn't mean it isn't a useful term or that, in most cases, we know it when we see it. Big structures that are simple and repetitive wouldn't, I suggest, count. But even small structures with lots of chiral centres might.

Oh, and a trivial (and possibly ironic) answer might actually be "proteins" since they can occasionally be synthesised without the use of other proteins. Or possibly DNA/RNA which can be built in stages in the lab as a step towards the biological synthesis of non-natural proteins.

When I was taking freshman chem the book mentioned that a bowling ball is basically a cross-linked polymer network, so if a "complex molecule" is just a lot of atoms covalently bound together I am going bowling.

If we limit the answer to synthesized chemical substances with a single, definite chemical structure, and apply a generally accepted measure of complexity, then this is an answerable question. For synthetic organic chemists, vitamin B-12 used to be Mt. Everest.

@matt_black They recently synthesized an entire protein using only D-amino acids and some clever synthetic techniques So if we managed to chemically synthesize mirror-image ribosomes (and associated folding proteins), could we argue that any resultant D-amino systems are purely chemical? Questions...

I guess in the field of inorganic chemistry polyoxometalates (POMs) are comparable with proteins in terms of structural complexity. Macroanionic POMs such as Keplerates $\ce{\{Mo132\}}$

are also capable of further self-assembly in diluted solutions, producing so-called "blackberries" [1] of a size of around $\pu{0.05 .. 0.1 \mu m}$:

Also, see the post Measure the complexity of a molecule and the corresponding paper [2] of Prof. Lee Cronin, one of the most productive polyoxometalate chemist of current era.

Kistler, M. L.; Bhatt, A.; Liu, G.; Casa, D.; Liu, T. J. Am. Chem. Soc. 2007, 129 (20), 6453–6460. DOI 10.1021/ja0685809.
Marshall, S. M.; Murray, A. R. G.; Cronin, L. arXiv:1705.03460 [q-bio.OT] 2017.


text:
Why is the more substituted C-atom attacked in oxymercuration-demercuration?
In hydration of an alkene by oxymercuration-demercuration , for an asymetrical mercunium ion, the more substituted carbon atom is attacked by water
Why is this the case?
Since R groups are electron donating, it should make the more substituted carbon less electrophilic but the reverse is true.

That's all right. Whatever is attacked by an electrophile should be nucleophilic, not electrophilic.

Made a mistake in writing my question. Fixed.

The mercury just electrophilically activates the alkene (kind of like proton). The electrophilic center is the one that is better able to bear a partial positive charge.

The Mercury because of it's + charge withdraws electrons via the inductive effect. The R group stabilizes the electron deficiency on the more substituted carbon atom making it less electrophilic


text:
Magnetic Water Softener
I've been asked to install some kind of technology to a building with 40 rooms to soften the water. Browsing through the product catalog I found this so called "Magnetic Water Softener", however theres no description of how it works, how does it make water soft?

Ostensibly it uses a magnetic field to "pull the ions in the water out" by preventing them to continue flowing after the softening unit. I do not know if this application is proven.

The wiki claimed it is not proven yet but some studies indicated a substancial decrease in water hardness. So it prevents the ions of the minerals from flowing inside it right? Does that mean it needs to be cleaned afterwards to prevent blockage?

Pure crank science, as to me.

It works by pulling money from your pocket. Avoid.

I would be VERY suspicious of the studies that show any positive results, let alone substantial results, if there is no consensus on feasibility. I agree with Ivan Neretin and Toff Minehardt.

Surprisingly these have been sold for over 40 years in industry ( petroleum and chemical). But when you try to get an objective measure the streams are too dirty, too hot, too much Cl , and a dozen other excuses.  Generally they only work when a flow is essentially not corrosive.  Don't believe me ; I expect you should be able to search this in NACE , ( I heard at least one paper presented in the 70's).

One of the most weird things i've read in the product's description is that it never needs maintenance and works for life... Still I don't get it, how should it work in theory?

It doesn't. Save yourself the effort and future troubles and get an ion exchange based softener (typical home softener) or reverse osmosis system.

The magnets might have some use if there are big particles of iron (pumps breaking down), but it would take mega-gauss field to pull tiny particles of colloidal iron oxides out of the flow. People also put magnets in shoes or bracelets to relieve pain... "There's a sucker born every minute, " attributed to P. T. Barnum.


text:
The effect of pressure on equilibrium constant
What will happen to the $K_\text{eq}$ in the following reaction if volume is increased?
$$\ce{2NO(g) + O2(g) <=> 2NO2(g) + energy}$$
From le Chatelier's principle I know that it must be constant unless temperature is changed, however it doesn't make sense because if
$$K_\text{eq}= \frac{[\ce{NO2}]^2}{[\ce{O2}][\ce{NO}]^2},$$
then by increasing the volume we know that equilibrium will shift left and favor reactants. Hence concentration of reactants would increase while concentration of products would decrease. If we plug that back into the equilibrium constant equation we see that if numerator decrease and denominator increase that must yield a smaller number than the initial number. Hence the equilibrium is not constant. Why this occurs?

First step: Don't use Kc... use Kp!
You are focusing on concentration, but as we're working with gases it becomes easier to work with partial pressures.
$K_\text{p}= \frac{\ce{p(NO2)}^2}{\ce{p(O2)}\ce{p(NO)}^2}$
Now, if we remember that partial pressure is pressure $\times$ mole fraction ($p_{a}=P \times \chi_{a}$), and substitute these in we get (and cancelling out the pressures):
$K_\text{p}= \frac{\chi_{(NO_{2})}^2}{\chi_{(O_{2})}\chi_{(NO)}^{2}P}$
Now, if we consider increasing the volume. By doing so (and assuming that temperature is kept constant), then we have decreased the pressure (ideal gas law: PV=nRT).
Now we can see that although the number of moles changes (and hence the mole fractions), our equilibrium constant will remain unchanged thanks to the effect of pressure.
A good guide is here.

It is called STP or standard temperature and pressure(If I remember correctly there are two STP's one for thermo and one for practical lab conditions).  If PV=nRT than ideally pressure is inversely related to volume.  The algebra is simple enough. Remember Keq is always reported at ''standard conditions'', or STP.  Typically 25 degrees celsius and 1 atm(I think SI units would be 1.01 bars but I could be wrong) at a 1 molar concentration.  according to that definition n, T, and P are all fixed and R is a constant. It is only constant under a given set of conditions that you may or may not choose.  As an aside Keq is strictly speaking a thermodynamic parameter so don't forget about entropy as well (there are three or four laws after all)

Volume??? Wow. You may get better answers if you learn about a thermodynamic property called "pressure". After you learn about it, I suggest you use it when asking about the concentration of gas phase reactions. For gas phase reactions usually concentration is measured not in "moles per liter" but as pressure, invoking Dalton's Law of partial pressures. So, your Keq would be in units of (pressure units)-1, say Pa-1.
 Lets say Keq = a²/(bc²) and a+b+c= P. You seem to be saying that the only way you can change P is to change Keq. Well, no.   You have three equations and 3 unknowns, I'm not seeing the problem. (The unknowns are the new a, b, & c...you could call them a', b', c') (Assuming you knew the new P').

Firstly by measuring the gas in terms of pressure that doesn't solve the problem. Secondly your answer does not address the actual question. Thirdly I'm not saying P is the only way to change Keq. I am saying TEMPERATURE is the ONLY way not P. Lastly there is no equations or variables involved in this question. This is a conceptual question not a quantitative question.


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What is the difference between polyatomic ions and compounds?
I don't understand why things like hydrogen phosphate are polyatomic ions but things like potassium nitrate are compounds. Can someone explain to me what the difference is, besides that ions are charged?

There is no other difference.

We can think of polyatomic ions as a subset of chemical compounds. If a compound meets the criteria for a polyatomic ion (see Wikipedia), then you could refer to it as either.

The ions, by themselves, are charged and thus cannot be complete compounds which are electrically neutral.
We sometimes do have names for polyatomic ions as If they are compounds, like cyanidin (which is a cation containing a pyrylium ion function).  But we understand that a "generic" counterion is needed to complete the compound.  For example, "cyanidin" (a polyatomic cation) can't be completely isolated by itself, but it could be incorporated into cyanidin chloride (a compound), which in principle could be isolated.

While you wrote hydrogen phosphate, I believe IUPAC prefers the term to be written without a space as hydrogenphosphate. This is because hydrogenphosphate is not a compound of positively charged hydrogen ions and negatively charged phosphate ions but actually a molecular entity whose sum formula is $\ce{HPO4^2-}$. A rough representation of the structure could be:
$$\ce{(\overset{-}{O}-)3\overset{+}{P}-O-H}$$
This is just a single ion that consists of more than one atom; it is therefore a polyatomic ion. In a crystal structure of e.g. sodium hydrogenphosphate (if you really, really went the extra lengths to determine the position of the hydrogen atoms), you would find sodium ions regularly arranged and $\ce{HPO4^2-}$ entities regularly arranged.
Potassium nitrate on the other hand, is an ionic compound. It consists of two types of ions: positively charged potassium ions and negatively charged polyatomic nitrate ions ($\ce{NO3-}$). Again, if you were to measure the crystal structure (no extra length needed here because all atoms involved have core electrons) you would see distinct potassium ions regularly arranged and $\ce{NO3-}$ entities regularly arranged.
Compounds are those entities that you could, in principle, put in jar and place in a shelf. Potassium nitrate is found in practically all chemistry labs that deal with inorganic chemistry at least partially. Hydrogenphosphate without a counterion cannot be synthesised and thus cannot be sold.
Note that just because the name ends in ide does not automatically mean we are dealing with ions at all. For example hydrogen sulphide is, in fact, a molecule (and IUPAC here definitely suggests a space between hydrogen and sulphide). Again, hydrogen sulphide can be put into gas bottles and technically also put in a shelf (if you lift). And in a simplified Lewis formality, it can be represented as follows:
$$\ce{H-S-H}$$
So, to sum this up:

Compounds are entities that can typically be isolated as they are written without the need of anything else. They can be made up of ions, molecules, molecular ions, zwitterions, metal atoms, etc.
Ions are charged species. Together with counterions of opposite charge, they can form compounds.
Polyatomic ions are charged species that consist of more than one atom. They too need counterions to form compounds.

@Linear Actually that was intended that way, but I’m fine with your change.

Ah, I see. I thought it was as if you were constantly thinking to yourself "don't use the word molecule, don't use the word molecule", and then used the word molecule. :} I guess now I have to say don't mislead the OP in using molecule for an electrically charged particle!! (+1)


text:
Acid-base chemistry - determining stability
I am learning about acid-base reactions and how to identify whether an acid is strong or not by looking at the conjugate base. The more stable a conjugate base the stronger the acid.
However, two of the factors that determine conjugate base stability seems contradictory to me.

I understand the bigger the atom, the more stable the base so $\ce{HI}$ is more stable than $\ce{HF}$.
The smaller the orbital the more stable the base. So $\ce{sp}$ is more stable than $\ce{sp^3}$.

Now why is a bigger atom more stable than a smaller one, but a smaller orbital  more stable than a bigger orbital?

When we say that 'bigger the atom, more stable is the anion', we mean to say that the negative charge would be spread over a larger area. So, the interatomic repulsive forces would be spread over a large area, lowering the energy and stabilising the anion. Going down the group, the jump in size is large enough to compensate the decrease in electronegativity, so we use the 'size of atoms' argument to differenciate between different atoms in the same group.
Conversely, we use the 'size of orbital' argument to determine the basicity of the same atom but in different hybridisations. Hence, the size of the atom would be same and the smaller size of the sp orbital would mean that it would be better attracted by the nucleus making the anion more stable.


text:
What's the smallest Hoberman sphere possible?
What is the smallest Hoberman sphere possible? Could it be possible on an atomic or molecular level?

Image courtesy of Hoberman Associates, Inc
I realize the issue with this will be the scissor joints. I am wondering how small these could be made and arranged. Would it be possible to create a functional Hoberman sphere with carbon nanotubes?
I am also curious if anyone has any opinion or knowledge about if it is possible to make a Hoberman “molecule” or rather a Hoberman sphere made of atoms? If so, what elements would make for the best Hoberman molecule? Could scissor joints be made functional at the molecular level both expanded and closed using something like redox reactions?
Thanks in advance for any answers!

It would improve the question if you add some references what a Hoberman sphere is, and what would distinguish it from any other type of molecular cages. What is a difference between a Hoberman molecule and a Hoberman sphere made of atoms? You have many related questions,  with many unclear details.

Hoberman sphere shares the same topological features being an icosidodecahedron with a family of polyoxometalates called Keplerates $\ce{\{(\color{teal}{Mo|W})_72\color{orange}{M}_30\}}$ ($\ce{\color{orange}{M} = Fe,Cr,V}$, anionic residues and alkali metal cations are removed for clarity) ([1, 2]):

Also, there is a cluster $\ce{[\color{blue}{La}_20\color{green}{Ni}_30(IDA)30(CO3)6(NO3)6(OH)30(H2O)12](CO3)6 * 72H2O}$ (IDA - iminodiacetate $\ce{C4H5NO4}$) consisting of an icosidodecahedron of Ni(II) Ions
encapsulating an inner La(III)-dodecahedron [3]:

The problem is, these structures only resemble the symmetry, but not the expandability of Hoberman sphere. Currently there are many attempts of creating nanomachines with mechanically active molecular assemblies, but I didn't find any working models close to Hoberman structures (yet).
Also, some "spherical" viruses such as Cowpea chlorotic mottle virus (CCMV) with a structure of truncated icosahedron, which is significantly larger than Keplerates ($\pu{28 nm}$ (CCMV) vs. $\pu{3.2 nm}$ (average Keplerate)), can undergo swelling (pH < 6.5) and expansion (pH > 6.5), and show pH and metal ion dependent polymorphism [4, 5]

as a result of deprotonation of carboxyl moieties at the inter-capsomere contacts, leading to electrostatic repulsion that opens out the structure but falls short of complete disassembly through preservation of interwoven
  carboxyl/protein links between capsomeres.

Don't forget that "there's plenty of room at the bottom", so Hoberman sphere can possibly be a real working entity at molecular scale.
Bibliography

Botar, B.; Ellern, A.; Hermann, R.; Kögerler, P. Angewandte Chemie International Edition 2009, 48 (48), 9080–9083. DOI 10.1002/anie.200903541.
Schäffer, C.; Merca, A.; Bögge, H.; Todea, A. M.; Kistler, M. L.; Liu, T.; Thouvenot, R.; Gouzerh, P.; Müller, A. Angewandte Chemie 2009, 121 (1), 155–159. DOI 10.1002/ange.200804496.
Kong, X.-J.; Ren, Y.-P.; Long, L.-S.; Zheng, Z.; Huang, R.-B.; Zheng, L.-S. J. Am. Chem. Soc. 2007, 129 (22), 7016–7017. DOI 10.1021/ja0726198.
Speir, J. A.; Munshi, S.; Wang, G.; Baker, T. S.; Johnson, J. E. Structure 1995, 3 (1), 63–78. DOI 10.1016/S0969-2126(01)00135-6.
Kovacs, F.; Tarnai, T.; Guest, S. D.; Fowler, P. W. Proceedings of the Royal Society A: Mathematical, Physical and Engineering Sciences 2004, 460 (2051), 3191–3202. DOI 10.1098/rspa.2004.1344.


text:
Transition Metals that react with Vanillin and their Complexes
My question is simply whether anyone knows what are the transition metals that the phenol group of the vanillin molecule could react with, and thereby produce discolorations.
This question is particularly relevant to the flavors and fragrances industries, where vanillin is a widely used flavor and fragrance ingredient. It is unfortunately prone to purple discoloration when it reacts with trace amounts of iron (according to chapter 4 of Chemistry and Technology of Flavours and Fragrances).
I am curious to know what is the full list of transition metals that vanillin could react with, and what complexes and their colors could be produced as such.
Thank you.

Are you also interested in complexes with o-vanillin?

Also, I'm not very familiar with the processing of vanillins in the industriy of fragrances and flavors -- are there also other chemicals in the same system that can a source additional ligands (water, nitrate/nitrite, azide, pyridine etc.)? This can also influence complexation and the color.


text:
Why does the proton transfer from the oxygen to the nitrogen atom in imine formation not occur through an intramolecular process?

In my book Organic Chemistry by David Klein, 2nd edition, the book shows that the oxygen atom first takes a proton from an acid, then, the conjugate base of the acid takes a proton from the nitrogen atom, in the second and third steps of the mechanism shown above. However, I believe that an intramolecular reaction where the oxygen atom takes the hydrogen atom would be more kinetically favored, due to the fact that intramolecular processes are much faster than intermolecular processes.
My question is therefore: Why is the reaction pathway proceeding through the acid deprotonation/protonation more favored than the intramolecular reaction?

You'd need 4-membered ring intermediates... Why you even bother when these schemes are more or less arbitrary anyways?


text:
is there any hydrophobic material to be used on paper cups?
Is there any bio degradable hydrophobic material that is not poisonous and easy to spray up on paper cups?

i have made change to the same. please help in finding the material if there any?

The paper cups I have seen already have wax on the inside to prevent water from soaking the paper.

It's a bit late to reply however, paper cup wax can be harmful to your stomach and intestines since it melts and easily mixes with the liquid you'll be drinking. I would still go with cold-warm silicone coating.

I would say it depends on the product being used. As far as I know, silicone is very Hydrophobic and is used to seal off/insulate water. There's also a food safety Grade silicone that can be used. I believe you'll be using that for hot liquids that wont go over 60-70 degrees Celsius? So the answer for me is silicone. Do not confuse with silicon and silica.
Google research showed this:

Silicon– A natural chemical substance (atomic #14), meaning it can’t be divided into smaller particles without splitting atoms. It is the second most abundant element (after oxygen) and when it bonds with oxygen creates minerals called silicates (like quartz, olivine, micas, thomsonite, jadeite, and prehnite)

Silica– A compound made of silicon and another element. It is present in the human body in high amounts and emerging research suggests it can be beneficial for health. I’ve mentioned it before when talking about diatomaceous earth and how I consume it in natural form, but it isn’t the same as silicon or silicone bakeware and isn’t relevant when talking about the safety of silicone.

Silicone– A synthetic polymer created by adding carbon and/or oxygen to silicon. It can exist as a solid, liquid or gel and is often used in medical devices like pacemakers, joint replacements and implants. It is generally considered safe for these uses and is now used to make silicone bakeware.

I need to add that since it's food grade material it shouldn't pose any health issues. It is used for waterproofing and other lubricating uses. It create a film layer on the surface, spray a very thin layer don't spray too much, it's not needed and let that thin layer dry.


text:
What's an "amagat"? How is it different from the Loschmidt Number?
At class, we were (briefly) taught about the Loschmidt Number/Constant ($n_0$), which isn't really different from the Wikipedia definition,

The Loschmidt constant or Loschmidt's number (symbol: $n_0$) is the number of particles (atoms or molecules) of an ideal gas in a given volume (the number density).

Furthermore, it says,

It is usually quoted at standard temperature and pressure, the 2010 CODATA recommended value is  per cubic metre at $0 °C$ and 1 atm.

 (Yes, I know STP is currently defined at 1 bar pressure, and not 1 atm. Since this isn't my main question, I'm bringing it up here: Is the Wikipedia definition of the Loschmidt number inaccurate? Should it be defined at 1 bar, instead?)
Reading on, I saw,

Being a measure of number density, the Loschmidt constant is used to define the amagat, a practical unit of number density for gases and other substances:
1 amagat = $n_0$ = $2.686 7805(24)$ x $\mathrm{10^{25}}$ per cubic meter,
such that the Loschmidt constant is exactly 1 amagat.

To begin with, I've never heard of an "amagat" [<--- I mean the unit, not the law named after the same person]. Yes, this may be the result of my shortcomings as a high-schooler with limited (almost pitiful, really) exposure to physical chemistry...but what bugs me more, is that I still can't wrap my head around it even now.
Including the Wikipedia definition for the amagat before I continue,

An amagat is a practical unit of number density. Although it can be applied to any substance at any conditions, it is defined as the number of ideal gas molecules per unit volume at 1 atm and 0 °C.

We already have the good ol' Loschmidt Number...so why introduce another term/unit that is (seemingly) defined the same way, has the same value and carries the same (SI) units?
Besides, what's with the "...practical unit of number density." bit?

I haven't seen any of these particular units before so I could be mistaken, but I imagine it is similar to how we don't say "I have 3 Avogadro's Numbers of water" we would say "I have 3 moles of water". We very well could describe amounts by explicitly naming multiples of the constant, but we instead defined a new unit so that we could describe things more easily.


text:
How do you explain the formula Fe3O4 using the ionic compound theory?
I don't understand at all how the ionic bonding in $\text{Fe}$$3$$\text{O}$$4$ works. The oxygens all together have $4(–2e)=–8e$ net charge. But we cannot give the three irons equal positive charges to balance this: $8/3$ isn't an integer!
And anyways, what kind of iron is in $\text{Fe}$$3$$\text{O}$$4$? It can only be either Fe(II) or Fe(III), and none of them fits in here.
A diagram demonstrating the chemical structure of $\text{Fe}$$3$$\text{O}$$4$ with supporting explanation would be preferable. I'm only an 8th grader, so advanced chemical concepts won't be comprehensible to me. The lattice structure would be too huge for me to handle, so pointing out a single molecule/piece of that compound will be more helpful.

Iron forms both $\ce{Fe^2+}$ and $\ce{Fe^3+}$ ions. The first combine with oxygen ions to $\ce{FeO}$; the second $\ce{Fe2O3}$.
$\ce{Fe3O4}$ can be thought of as a mixture of both compounds in equal amounts. So $\ce{Fe3O4}$ contains both Fe(II) and Fe(III) ions; twice as much Fe(III) as Fe(II).
In the picture below, showing the crystal structure, green atoms are Fe(II) and brown atoms Fe(III). All Fe(III) atoms are linked with six O atoms, sharing 'half an electron' with each oxygen; all Fe(II) atoms are linked with four O atoms, and all O atoms are linked with four Fe atoms.

(source: Manoj B. Gawande, Paula S. Branco, and Rajender S. Varma, Chem. Soc. Rev. 2013, 42, 3371-3393. DOI: 10.1039/C3CS35480F See also: ResearchGate
)

The lattice structure would be too huge for me to handle, so pointing out a single piece of that compound will be more helpful.

But it's a crystal, not a single molecule. If you insist on one, here you go. But those molecules don't exist in the wild.

There is no single piece. Just think of it this way: some Fe are +2, and some are +3.

@Glorfindel I do understand it's not reality. Diamond is a giant structure, and at the first glance, one will think "All I see is just carbons arranged randomly". But if you point out that each carbon bonds to 4 other, he can understand better. C-C-C-C.

But diamond is C, not C4.


text:
what happens to hydrogen released in addition polymerization?
In formation of polyethylene(free radical method), the free radicals when attached to another ethene molecule release a hydrogen(I think).But addition polymerization does not release and by product, so what happens to that hydrogen atoms?

Consider the reaction stoichiometry.
You begin with ethylene (that is, CH2CH2).
The polyethylene chain is a chain of single-bonded carbons. Hence, it is a  -[...]-CH2-CH2-CH2-CH2-[...]- chain).
As you can se, no hydrogens are removed by the monomer.
So, no hydrogens are released during polymerization.
What is happening in that reaction is an electron transfer, not an atomic transfer: the double bond of ethylene, full of electrons, reacts with the radical and forms a more stable structure with a radical carbon (a radical carbon has three bonds, in this case with C, H and H, and an unpaired electron).
The carbon radical reacts with another ethylene monomer: the "old" radical carbon now forms four bonds (C, H, H, plus the new C from the ethylene monomer), the new ethylene monomer loses its double bond, and forms a carbon radical (again, the radical carbon has three bonds and an unpaired electron).
This is the logic of that tipe of polymerization: no hydrogen is involved.


text:
Solubility of CHCl3 in water
Why isn't $\ce{CHCl3}$ soluble in water?
Why doesn't $\ce{CHCl3}$ participate in hydrogen bonding in water like $\ce{CH3CN}$ as $\ce{Cl}$ can make the $\ce{C-H}$ bond polar like $\ce{CN}$ does?

Chloroform is soluble in water, though only slightly (~1g/100ml in cold water forming a slightly sweet liquid whose mild anaesthetic affects made it a recreational substance in Victorian times before its toxicity was fully recognised).
Moreover, its hydrogen will form (weak) hydrogen bonds which is probably a better explanation for the solubility in water than the polarity of the molecule.


text:
What does it mean if a catalyst support is doped?
I have seen literature refer to catalyst supports being doped with another chemical, such as Nb doped $\ce{TiO2}$. What does it mean if a catalyst support is doped?

If a substance is doped, it means that some small amount of another substance has been introduced into it. More specifically, a substance is doped by a dopant, the latter of which is the minority component. The dopant can be an additive to or a replacement within the bulk material.
In your case, some fraction of $\ce{Ti}$ atoms are replaced by $\ce{Nb}$ atoms (the replacement scenario is the usual one for crystalline materials such as yours). Thus, the physical properties of your substate will be slightly different that those of pure $\ce{TiO2}$.
See this Wikipedia entry for a more complete summary of what I have written here.


text:
What volume does one mole of an ideal gas occupy?
This has been bugging me for a while now...
Obviously, to calculate the volume/space occupied by a mole of (an ideal) gas, you'll have to specify temperature ($T$) and pressure ($P$), find the gas constant ($R$) value with the right units and plug them all in the ideal gas equation $$PV = nRT.$$
The problem? It seems to be some sort of common "wisdom" all over the Internet, that one mole of gas occupies $22.4$ liters of space. But the standard conditions (STP, NTP, or SATP) mentioned lack consistency over multiple sites/books. Common claims: A mole of gas occupies,

$\pu{22.4 L}$ at STP
$\pu{22.4 L}$ at NTP
$\pu{22.4 L}$ at SATP
$\pu{22.4 L}$ at both STP and NTP

Even Chem.SE is rife with the "fact" that a mole of ideal gas occupies $\pu{22.4 L}$, or some extension thereof.
Being so utterly frustrated with this situation, I decided to calculate the volumes occupied by a mole of ideal gas (based on the ideal gas equation) for each of the three standard conditions; namely: Standard Temperature and Pressure (STP), Normal Temperature and Pressure (NTP) and Standard Ambient Temperature and Pressure (SATP).
Knowing that,

STP: $\pu{0 ^\circ C}$ and $\pu{1 bar}$
NTP: $\pu{20 ^\circ C}$ and $\pu{1 atm}$
SATP: $\pu{25 ^\circ C}$ and $\pu{1 bar}$

And using the equation, $$V = \frac {nRT}{P},$$
where $n = \pu{1 mol}$, by default (since we're talking about one mole of gas).
I'll draw appropriate values of the gas constant $R$ from this Wikipedia table:

The volume occupied by a mole of gas should be:

At STP
\begin{align}
T &= \pu{273.0 K},&
P &= \pu{1 bar},&
R &= \pu{8.3144598 \times 10^-2 L bar K^-1 mol^-1}.
\end{align}
Plugging in all the values, I got
$$V = \pu{22.698475 L},$$
which to a reasonable approximation, gives
$$V = \pu{22.7 L}.$$
At NTP
\begin{align}
T &= \pu{293.0 K},&
P &= \pu{1 atm},&
R &= \pu{8.2057338 \times 10^-2 L atm K^-1 mol^-1}.
\end{align}
Plugging in all the values, I got
$$V = \pu{24.04280003 L},$$
which to a reasonable approximation, gives
$$V = \pu{24 L}.$$
At SATP
\begin{align}
T &= \pu{298.0 K},&
P &= \pu{1 bar},&
R &= \pu{8.3144598 \times 10^-2 L bar K^-1 mol^-1}.
\end{align}
Plugging in all the values, I got
$$V = \pu{24.7770902 L},$$
which to a reasonable approximation, gives
$$V = \pu{24.8 L}.$$

Nowhere does the magical "$\pu{22.4 L}$" figure in the three cases I've analyzed appear. Since I've seen the "one mole occupies $\pu{22.4 L}$ at STP/NTP" dictum so many times, I'm wondering if I've missed something.
My question(s):

Did I screw up with my calculations?
(If I didn't screw up) Why is it that the "one mole occupies $\pu{22.4 L}$" idea is so widespread, in spite of not being close (enough) to the values that I obtained?

see https://en.wikipedia.org/wiki/Standard_conditions_for_temperature_and_pressure#Molar_volume_of_a_gas

Just to note it, STP's at $T=273.15\mathrm{K}$ rather than $T=273.0\mathrm{K}$.  I completely understand that folks often truncate the $.15$ for brevity, just looks a little off next to $R$ given with 8 significant figures.

The common saying is a hold over from when STP was defined to be $\pu{273.15 K}$ and $\pu{1 atm}$. However, IUPAC changed the definition in 1982 so that $\pu{1 atm}$ became $\pu{1 bar}$. I think the main issue is a lot of educators didn't get the memo and went right along either teaching STP as $\pu{1 atm}$ or continuing with the line they were taught ("$\pu{1  mol}$ of any gas under STP occupies $\pu{22.4 L}$") without realizing it didn't hold under the new conditions.
Just as a "proof" of this working for the old definition.
\begin{align}
V &=\frac{nRT}{P}\\
  &=\frac{\pu{1  mol} \times \pu{8.2057338 \times 10^-2 L * atm//K * mol}
    \times \pu{273.15 K}}{\pu{1 atm}}\\
  &=\pu{22.41396 L}\\
  &\approx \pu{22.4 L}
\end{align}

And to make matters worse, some examination boards didn't change for quite some time. IB Chemistry only switched to Pa / Bar in the 2016 exams!

A big point of confusion is that it is still taught (at least in the mid-2000's) that STP is defined with respect to $\pu{273 K}$ and $\pu{1 atm}$ of pressure, or $\pu{1.01325 bar}$ of pressure, even though IUPAC changed their definition to be with respect to $\pu{1 bar}$ of pressure. By using the ideal gas law on the old STP definition, you get that the volume is  $\approx 0.0821\cdot 273 \approx 22.4$ liters.


text:
Do the spins of outer orbitals block measurement of spin of inner orbitals?
Do the outermost s1 and s2 electrons block an outside observer from measuring the larger inner group orbitals for elements heavier than Nitrogen (for instance, would Chromium's 4s1 electron or Manganese's 4s2 electrons prevent you from detecting the spin of the 3d5 electrons in either case?)
I edited this into an individual question from the original list of them per moderator instruction.

Have you heard about Aufbau principle?

@PrittBalagopal The Aufbau principle doesn't seem to hold up for the description of Vanadium, Chromium, Maganese and most other elements beyond group 2 (that is, inner orbitals get partially filled then jump to filling the next group of orbitals before going back downward.)

It does hold for vanadium. Ot doesn't hold for chromium because $\mathrm{3d^5}$ is exceptionally stable due to being half filled.

@PrittBalagopal Vanadium is [Ar] 3d3 4s2 - meaning the 4s subshell gets filled before the 3d subshell.

Well yeah? 4s does come before 3d in the Aufbau series. Look here.

@PrittBalagopal Ah, I think I'm following it now.  I was reading it backwards thinking that 3d was getting filled out to 3d3 then 4s was getting filled to 4s2 but judging by Potassium and Calcium this is wrong.  This does still come back to the original question I had though, would the outer suborbitals block the inner suborbitals?  That is, would the 4s2 in Vanadium prevent you from distinguishing between all-up or all-down 3d3 suborbitals in different atoms?  Same question for things like Manganese with 3d5.

With regards to your first question, as you get to higher energy levels, the distance between each one increases and as a consequence the shells get closer together and begin to overlap. This is also seen with 5s and 4d iirc.


text:
For a system containing three reductants, can a combined reduction potential be determined?
I've modelled a system and found that there are three major complexes present at equilibrium, all of which I hope will act as reductants when I add an additional, oxidised species. Each of the three complexes has a different concentration in the solution, and each has an associated reduction potential.
Is there any way to find an overall reduction potential for my reactor, based on some combination of reduction potentials?
I could be missing something fundamental!

There is no such thing as a combined reduction potential. Instead, there are potentials of individual species, and those will tell you which of them is going to get oxidized first.


text:
Anti-Solvent Precipitation Of Lycopene
I have a question about anti-solvent precipitation in organic chemistry:
Is it possible to precipitate lycopene or other carotenes using this method?
I have been extracting carotenes like lycopene by Soxhlet method using acetone as solvent. I was wondering if it's possible to precipitate these compounds using the anti-solvent method.
If it's possible which anti-solvent would be more accurate, in the documents that I have been reading , it was mentioned methanol and other alcohols but I'm not sure if the acetone that I used in the extracting part would cause any interference.

If you have a solution, so long as you don't have too much of your "good", more solubilising solvent,  you should be able to precipitate your terpene.  Maybe not crystallise, but at least bring it out of solution. Lower alcohols and water act as anti-solvents for long chain hydrocarbons like lycopene.
If this is for lab purification scale, another option is to simply evaporate your solvent and dissolve it in your crystallisation solvent of choice. For these longer chain polyenes,  it may require a two solvent system anyway to crystallise the carotene of choice.
Going back to one of your comments, the acetone will not interfere so long as it is the minimum amount. With the Soxhlet method you use, presumably you extract unwanted compounds also, so your experience of how easily these dissolve/precipitate will affect your choice of isolation method and specifically, the anti-solvent choice.


text:
How Does a Precipitate Get a Colour?
If you can use precipitate colours to determine an anion, what makes the precipitate have that particular colour?
Like for example what in sodium sulfate or barium chloride makes a white precipitate when these two combine?

White is the default color. You don't have to do anything to produce it.

Possible duplicate of Why is snow white?

The 'white colour' is due to the fact that the product does not absorb light in the visible part of the spectrum but forms as many fine crystals which in themselves scatter room lights. So the 'whiteness' is due to the scattering of light back towards your eyes after the light is repeatedly reflected of the many facets of the many crystals in the ppt. If you use red light to illuminate then the 'white colour' will become red. If the ppt is coloured then absorption of light occurs at some wavelengths as well as scattering at all wavelengths.
At the surface of a crystal some small fraction of light is reflected, but with millions of randomly oriented crystals numerous reflections occur. The amount of reflection at each surface depends on the difference in refractive index between the crystal and air for the wavelength of light used. If you could add a solvent say dichloromethane which has  a high refractive index and will probably not dissolve the ppt then the crystals can be made to almost disappear as scattering is now far less.  You can do the same thing with frosted glass beads which, when in air, scatter a lot but in water become virtually invisible.

The following answer applies to all questions regarding color.
Color is the result of a substance absorbing certain wavelengths of light and reflecting others.
Different molecules have different shapes (geometry). Different shapes absorb/reflect different wavelengths. In addition to the shapes, the other parameter regarding color is the temperature and pressure.
Closely packed molecules will absorb/reflect light different from loosely packed ones.
2 more things that can "produce colour" rather than reflection of photons hitting it, are:
photons generated from chemical reactions and
photons generated from radiation


text:
Esterification Colour Change
I was in a chemistry lecture about the upcoming laboratory practical on esters that I would have to take. My lecturer had mentioned that we ought not worry about "the pink coloration that will occur" during the esterification process. However, when we asked why a color change would even occur, his reply was very vague. According to him, there is a general consensus that no-one knows why.
We will be asked to produce an ester using ethanoic acid, butan-1-ol, and concentrated sulfuric acid, in the presence of boiling chips.
So is there a known answer to the mysterious pink coloration? I think it's due to some contaminants that might be present in the sulfuric acid, but my lecturer says  the pink coloration is completely random and that everyone in past few years have always used the same solutions.

Please, provide as with exact esterification reaction you are talking about.

What type of boiling chips? If you are using porous carbon boiling chips, then I can verify that pink color as my students observe it occasionally in a similar experiment. My guess is that it is a little bit of sulphonated graphite dissolving into the reaction.

We are adding 15mL of butan-1-ol and 10mL of the ethanoic acid together in a flask. Then we slowly add in 1mL of concentrated sulfuric acid and some boiling chips, which is then refluxed.

I'm not sure about the boiling chips, we might be using an aluminium compound I think?


text:
Why is methyl iodide more dense than ethyl iodide, even though the latter has more carbon atoms?
It is understood that more carbon atoms present in a hydrocarbon, the greater the density of the hydrocarbon.
But then why does $\mathrm{CH_{3}I}$ have a greater density than $\mathrm{C_{2}H_{5}I}$?

Because the iodides are not hydrocarbons, maybe?

@OscarLanzi If so then the effect of iodide should be pronounced on both ethyl iodide and methyl iodide equally and more on ethyl iodide as it is having one more  carbon?

Spacing between iodine atoms in different size molecules.

For the same reason why a bucket of sand will weigh more than a bucket of rocks.
Smaller molecules will result in a more dense, thus more heavy substance.


text:
Proton NMR spectrum of aminocaproic acid
I have completely unpacked the NMR spectrum for the drug aminocaproic acid (structure below) and have found a peak that I myself can not explain. The signals for the drug are simple to understand what corresponds to what but the reason behind one of the peaks having six peaks is unknown to me. I had made the assumption that the furthermost up-field signal would have 5 peaks and be a multiplet, due to it having four nonequivalent neighbors and N+1 rule. If there is a reason behind this can some one please explain.

Could you please specify what ppm region should one look at? The peaks at 2.2 ppm have some kind of marker above, but I'm not sure you are talking about them. Also, check out SDBS entry for aminocaproic acid, at least you can check your assignments.

Coupling to exchangeable protons like OH or NH2 is often not seen ("exchange decoupling") - see https://chemistry.stackexchange.com/q/48232/16683 and this is covered in most organic-NMR texts, usually under the section on spin-spin coupling

You mean why the multiplett below 1.4 ppm has six peaks? Four non-equivalent neighbours make not five peaks, but a dddd signal, which can be up to eight peaks, if nothing overlaps.

Sorry, dddd could have up to sixteen peaks, of course. You could also think about a tt signal. But generally, longer alkyl chains just make multipletts, because the two protons in a methylene group are usually not magnetically equivalent.

The N+1 rule, in its ideal form, assumes that for multiplets arising from $\mathrm{^3J}$ couplings between neighbouring $\ce{CH_2}$ groups,

shifts for the spins within a $\ce{CH_2}$ group are identical
all couplings between spins in two different groups are identical

While condition 1 usually holds for substituted alkanes of the form $\ce{A-(CH_2)_n-B}$, condition 2  does not. This leads to the title "Is the n+1 rule ever obeyed?" in a section of a textbook [1], like a cry of despair. The text elaborates:

At other times, when there is a large difference between $\mathrm{^3J_{AB}}$ and
  $\mathrm{^3J_{BC}}$, distinct peaks, more than five in number, can be seen.
  Deviations of this type are most common in a chain of the type
  $\ce{X-(CH_2)_3-Y}$, where X and Y are widely different in character.

So, the exception is actually closer to the rule, and that is because true magnetic (unlike chemical) equivalence is relatively rare.
[1] Pavia, Lampman and Kriz, Introduction to Spectroscopy, 2nd Ed., Saunders Publishing


text:
Chemically, how stains are removed?what is the principal?
How stains are removed? answering to this question will help me much

I'm afraid this question is too broad. Specifying the type of fabric and source of impurities would help a lot.

I agree - too broad. However, a general answer is that the appropriate solvent is used for the stain. For example, a nonpolar grease stain is typically removed via a combination of a nonpolar solvent and a surfactant, which "traps" the stain molecule

Alternatively, substances such as hydrogen peroxide are applied that chemically react with the color compounds in the stain to make them colorless.


text:
Does gravity affect the rate of a chemical reaction?
For example, on planets with lower gravity, would the rate of reactions be lower because the reactants are slower to mix with one another?

I'm surprised I couldn't find this question on Chemistry.SE. In the meantime Reddit,  RG and Quora have discussions on this topic. Also, related: Can magnetic fields affect a chemical reaction?

Rate of reactions would remain the same. If anything, rate of mixing would slow down, which in some cases would reduce the apparent rate of reaction, and in other cases would not make any difference.

As one can see on the ISS: Not really. What you might look out for though are reactions with compounds of the atmosphere. Take molecular oxygen: If you lower the overall pressure, burning stuff will happen at a much slower rate because at the end all that matters is the pressure of the oxygen. There is no direct connection though - just if you take the same atmosphere. If you however go to very high gravity = very high pressures, you will find completely different modifications (metallic hydrogen for starters) that should have a completely different chemistry

What about gravitational time dilation? Time is running faster at higher gravitational potential, so if you perform a reaction at different gravitational potentials one would run faster. The effect would be very small though.

https://chemistry.stackexchange.com/questions/6595/how-can-enzyme-substrate-reactions-that-adhere-largely-to-quantum-theory-also

It sort of depends what you mean. If by "affect the rate of a chemical reaction", you mean affect the individual steps of some overall reaction as to slow them down, then no, gravity definitely has no effect at all.
This is simply because the gravitational force is so much weaker than the electrostatic force, which will dominate all interactions relevant in chemical reactions, that it does not have sufficient time to make any appreciable difference in the behavior of molecules or atoms.
A kind of standard way of demonstrating this is by imagining a situation in which it is guaranteed that electromagnetism will be stronger than gravity, which is all normal situations in chemistry, and then plugging the numbers into Coulomb's law or Newton's law of gravity.
This, however, is contrived because there are situations where gravity is stronger than electromagnetic interactions. Just never in chemistry. This article does a pretty good job of explaining this, and pointing out that the ratio of forces can really be anything at all.
The key reason gravity will never be relevant to individual chemical processes, and hence never affect the rate, is that the external gravitational field will affect the molecules equally once they are close enough to interact electromagnetically. And once they are interacting electromagnetically, it is guaranteed this force will be much, much stronger than the gravitational interaction between them because both follows the same $1/r^2$ law, so it is just a matter of comparing the proportionality constants which go in front of the law, and inputting some masses and charges characteristic to the situation. The article I linked does provide a context in which the gravitational force dominates the electromagnetic force, so this is possible.
So, no individual rates will not be affected. It's fun to imagine how gravity could affect macroscopic equilibria though, so let's think about that.
Given sufficient time, gravity is very good at organizing things in a particular way. For instance, if you imagine two planets which have atmospheres that at some time are exactly identical, but one planet is much more massive than the other one, I don't think anyone would contend that these atmospheres will look exactly the same after a long time. For instance, the escape velocity of an object is $\sqrt{2rg}$, $r$ being the radius of the planet and $g$ the gravitational constant of acceleration. If $g$ is increased significantly, then you would not lose nearly as much hydrogen from the atmosphere, which could affect the location of certain chemical equilibria in the atmosphere. The atmosphere as a whole would be kept much closer to the planet, which could also be relevant.
All of this is kind of contrived, however. The point is that gravity will only really affect chemistry in how it organizes materials, but cannot affect the individual rates.

Whether gravity can direct affect the rate of homogeneous chemical reactions seems unlikely, but there are a lot of inhomogeneous reactions which will clearly behave differently in the absence of gravity.
Many real world reactions involve phase separation either because there are two miscible liquids involved or because there is a precipitation when some solid products drop out of solution. One of the forces that makes separation happen in such reactions is gravity (less dense liquids float of more dense liquids; dense solids drop to the bottom of the reaction vessel). Such reactions may be very different in microgravity. This is the mirror image of some things chemists already know (some separations don't occur very efficiently even under gravity so we enhance the forces by using a centrifuge to create better separation).
There are practical experiments where the results are different. One example is the production of some specialist optical fibre strands (see this wikipedia article) where the defects in the fibres seem to be much lower when produced in zero gravity. Crystallisation may behave differently in microgravity and this too has been tested with promising results (see this NASA link). Flames are basically chemical reactions but are complicated to understand because the reaction heat also induces gravity driven convection when the warm gases involved interact with the colder surrounding air. Flames behave very differently in microgravity as this NASA link explains. Interestingly, the "cool" flames produced by combustion in zero gravity have radically different chemistry to the ones we are used to seeing on the ground.
So while individual homogeneous chemical reactions might not be influenced by gravity, in practice real reactions and their result are influenced and can be very different in low-gravity conditions. This is new science and we don't yet know whether there will be big economic impacts.

At the gravitational forces we experience on earth, the kinetic energy of molecules and the electrostatic forces between molecules are the dominant factors in determining their average orientations. If you exerted enough gravity on a molecule though perhaps it would behave more like the macroscopic objects we deal with every day in that it would assume positions that minimize its gravitational potential energy. If molecules then had selective orientations in a strong gravitational field it's possible the reaction rates could change since molecules must collide with the right amount of energy in the right orientation to react and you just changed their average orientation.
This is all speculation.

If your "answer" is speculation, then don't post it as answer.


text:
Would a single atom of a metal bond with a plate of that metal?
A question purely for my own curiosity.
I am trying to understand what "holds the atoms in a metal bar" in shape.
A general question about metals, but let's use gold as an example.
Under suitable laboratory conditions

if we isolated a single atom of gold
and dropped that onto a plate of gold

Would the atom fuse with the plate ?
Would the valence electrons of the single free atom "attract that atom into a bond with the crystal lattice" ?
I did some searching and found this:

Metallic bond, force that holds atoms together in a metallic
  substance. Such a solid consists of closely packed atoms. In most
  cases, the outermost electron shell of each of the metal atoms
  overlaps with a large number of neighbouring atoms. As a consequence,
  the valence electrons continually move from one atom to another and
  are not associated with any specific pair of atoms. In short, the
  valence electrons in metals, unlike those in covalently bonded
  substances, are nonlocalized, capable of wandering relatively freely
  throughout the entire crystal. The atoms that the electrons leave
  behind become positive ions, and the interaction between such ions and
  valence electrons gives rise to the cohesive or binding force that
  holds the metallic crystal together.
https://www.britannica.com/science/metallic-bond

I can think of a reason why two gold bars wouldn't fuse together when placed together:

the irregularity of their surfaces would not permit close enough contact ...
(ignoring any surface oxidation)

But I'm curious about the case of a "single free atom".

I'm unsure of how to envision "dropping" a singular atom, since at that point the atom will be in motion throughout the apparatus acting as a gas. So in light of my picture of what the physical scenario is, I would say factors such as velocity at time of contact and surface energies will dictate if the single atom would be absorbed into the crystal lattice.
Related: https://physics.stackexchange.com/questions/87107/why-dont-metals-bond-when-touched-together

I guess you might want to read this: Hevesy, G.; Obrutsheva, A. Nature 1925, 115 (2897), 674–675. DOI: 10.1038/115674a0
and remember the experiment mentioned in all physics textbooks on what happens when lead and gold bars are compressed together.

It is common in the semiconductor industry to evaporate metal to form metal layers.

Of course, the atom can and will stick to the bar, at least if the surface is clean enough.... Is it as in the case of cold, in vacuum soldering that you have mentioned.

Cold welding does happen as shown in this video: https://www.youtube.com/watch?v=mHAhfX7iRjs

Thanks to the folks for the comments and input above. I've started reading the links and references, and they are all pointing to an answer of YES within "suitable qualified environmental conditions". I'll take some time on the weekend to write a coherent summary up as a solution answer and close off the question myself.


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Hybrid thin films containing azobenzene with reversible photoresponsive behavior
I am going to prepare hybrid thin films containing azobenzene with reversible photoresponsive behavior. I have read about different combinations of inorganic layers such as layered double hydroxide (LDH) with azobenzene molecules.
However, I can't understand the photocatalytic role of the inorganic portion in the isomerization of azobenzene. Can insulators be used as the inorganic part instead of semiconductors (if I consider LDH as the doped composite)?


text:
Hydride Donating Ability
We know that some compounds can donate $\ce{H-}$ anion to attain stability. What are the conditions under which it occurs and what should be the extent of stability of the product for this pathway to be viable?

Only some anionic intermediates do that AFAIK.

Many compounds are good hydride donors, if it results in carbonyl bond formation or extended conjugation. Sometimes, a compound can donate H- ion if it results in stable pi bond formation.
(e.g. Tertiary butyl carbanion)

t-butyl carbocation won't release $\ce{H-}$.

@orthocresol.  I meant Carbonion

Ok. I guess you mean something like tBuLi? Never seen that acting as a hydride donor... but I might learn something new here...


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Does sulfur have various allotropes at room temperature?
Recently I had a question in my chemistry test that said:

Which of the following nonmetals does not have multiple allotropes at room temperature?
A) Oxygen
B) Hydrogen
C) Sulfur
D) Phosphorus

The answer was apparently hydrogen, but can't sulfur also be a correct answer? I am judging this just because in my chemistry book it says that rhombic crystalline form exists below $\pu{96^\circ C}$. Monoclinic exists above $\pu{96^\circ C}$. Are there any allotropes of sulfur at room temperature?

yes there are allotropes of sulfur at room temperature. See: https://en.wikipedia.org/wiki/Allotropes_of_sulfur

At room temperature, rhombic sulfur (α-S) is a stable solid comprised of cyclic S8 molecules

"Exists" is an awfully tricky word: it can be understood in a thermodynamic sense or in a kinetic (=everyday) sense, and the meanings are sometimes quite different. Ditto for "stable".
Thermodynamically, there exists only one stable allotrope of any element at any given conditions. All other allotropes are not allowed to exist and must eventually transform to that one. This is true, but the problem is that the transformation may take forever.
Now, your book on sulfur was using this meaning, but your test was not. It was meant to be understood in a kinetic sense, where multiple allotropes may coexist all right. In particular, the high-temperature monoclinic sulfur can be cooled down to room temperature and still remain monoclinic. I can attest to that; I used to hold it with my own hands, as did numerous other chemists. That's what people call metastability.
There is also plastic sulfur and other forms, but that's another story. And as you already know, there is oxygen vs ozone, and white phosphorus vs red, so hydrogen is really the only element here that does not have multiple allotropes.
So it goes.

"Exists" has a very clear meaning, if you specify what, i.e. the element or rather a specific allotrope. What the OP has is just a sloppily written book.


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What is thermal plasma in-flight oxidation technique?
From Vijay, M.; Ramachandran, K.; Ananthapadmanabhan, P. V.; Nalini, B.; Pillai, B. C.; Bondioli, F.; Manivannan, A.; Narendhirakannan, R. T. Current Applied Physics 2013, 13 (3), 510–516. DOI 10.1016/j.cap.2012.09.014:

Since large quantity of energy is consumed to vaporize the solvent in liquid precursors, a novel method that involves thermal plasma in-flight oxidation of $\ce{TiH2}$ has been developed to synthesize nanocrystalline $\ce{TiO2}$."

What does in-flight mean in this context? Obviously it doesn't have anything to do with actual flying, but everywhere I look it says that it is used for something that occurrs or is provided during an aircraft flight.
The article is about photocatalytic inactivation of bacteria by reactive plasma nanocrystalline $\ce{TiO2}$ powder. It suggests a new method for the physical preparation of $\ce{TiO2}$ for a novel reactor.

@andselisk Thanks! Elsevier's DOI linking hub is down for me at the moment, but the citation was enough for me to find it. http://www.sciencedirect.com/science/article/pii/S1567173912003926

Original paper where this method (and, I guess, the term itself) has been introduced was written in Japanese in 1982 [1]. In 1997 there was a great overview published in English by Toyonobu Yoshida, author of the original paper, with a chapter (2.3) devoted to this particular method:

Processes used to manufacture materials by injecting a solid, liquid, or gas into a thermal plasma to use its high enthalpy and reactivity are referred to as in-flight plasma processes (IFP processes).

Due to the nature of the method which utilizes plasma jet with reactants generated via various torch-types, there is actually a flow of chemicals reacting while they "fly", of course, in the plasma area.

Various plasma torches provide various times-of-flight, temperatures and velocities. For example, the direct current jet torch has the following characteristics:

An argon plasma of about $\pu{80 kW}$ that has been generated in air reportedly produces a temperature of about $\pu{11600 K}$ and a velocity of about $\pu{410 m/s}$... The reaction time for reactants injected near the anode at high temperatures is about $\pu{400 \mu s}$ and the gaseous raw materials are
  mainly used for the synthesis of fine powders... The reaction time of the reactants is estimated to be $\pu{1 ms}$.

Bibliography

Yoshida, T.; Akashi, K. Tetsu-to-Hagane 1982, 68 (10), 1498–1502. DOI 10.2355/tetsutohagane1955.68.10_1498.
Ultra-fine particles: exploratory science and technology; Hayashi, C., Uyeda, R., Tasaki, A., Eds.; Materials science and process technology series; Noyes Publications: Westwood, N.J, 1997. ISBN 978-0-8155-1404-6.

Further down in the paper, the following image and accompanying text are present:

Fig. 1. Schematic of plasma reactor (1 – Plasma torch, 2 – Reactor, 3 – Filter and 4 – Exhaust).
As the powder particles enter the plasma jet, they get heated up and
  $\ce{TiH2}$ dissociates into $\ce{Ti}$ particles and hydrogen gas. Both $\ce{Ti}$ particles
  and hydrogen gas react with oxygen to form $\ce{TiO2}$ and water vapor
  respectively; the latter escapes along with the exhaust gas stream. In
  addition to the particle heating by the plasma, the energy released by
  the exothermic reaction between Ti particles and oxygen helps to
  dissociate the formed $\ce{TiO2}$ particles into Ti vapor and oxygen gas. The
  Ti vapor reacts with oxygen at the tail of the plasma jet forming
  nanoclusters of $\ce{TiO2}$ particles.

So what happens is that a stream of $\ce{TiH2}$ particles are sent into the reactor on a stream of gas (in other words, flying into the reactor). The oxidation occurs while the particles are in-flight, not before they hit the plasma or after they're collected, presumably on the filter.
So in-flight here means the same thing it does in other contexts, but the flight is the flight of $\ce{TiH2}$ particles on a stream of gas/plasma.


text:
Major product of this Hoffman elimination?
1-methyl-1-quarternary ammonium cation(with methyl groups) cyclohexane is heated with hydroxyl ion. I want to find the major product.
In the solution, it is given that a hydrogen attached to the methyl group is eliminated.
However,
 1. the methyl group and the cation are not placed in an anti-periplanar configuration, so there would be considerable amount of energy needed (just a guess) to distort the ring;
 2. the hydrogen atom attached to the beta carbon on the ring is equally acidic;
 3. The transition state for OH eliminating the hydrogen attached to the beta carbon is not crowded. I don't think sterics play a role here.
So I feel like the hydrogen attached to the beta carbon should be eliminated.
Why am I wrong?

The anti-periplanar is favored in an E2 elimination, but it is not necessary if an E1cb mechanism is being followed. Further, in an E1cb mechanism deprotonation of the methyl group produces a more stable carbanion. This earlier answer may prove helpful.

If it's a methyl group, there will be a hydrogen anti-periplanar. A figure or specific example would help.

@ron but isn't e1cb a limiting case of the e2 spectrum, with simply less electron density in the pi orbital and more on the beta carbon. Shouldn't it resemble the E2 transition state in which case?

(reference for this claim: Peter Sykes, page 256).

And besides, why would the stability of the carbanion matter? The rate limiting step is the loss of the leaving group

I second @jerepierre's comment - as long as the proton is being taken from a methyl group, there is bound to be one which is antiperiplanar to the leaving group.

Okay, agreed. Post this as an answer so I can accept it


text:
How to remove white coating on the seashells after lemon acid and soap?
What is the way to dissolve oxalates? I have tried to clean the seashells with lemon (or citric, as I was corrected) acid and then I washed them with soap (with pH above 10). There is a white coating on the shells now, which disappears when is wet and appears back when is dry. I suspect I accidentially covered the shells with some oxalate. Please help.

It might just be too late, but I don't see where oxalate might come from if you used citric acid (I guess that what you mean by "lemon acid"). Anyway, I think the white coloring is the remaining fine conchiolin network(pores) remained after the $\ce{CaCO3}$ has been dissolved with an acid. Maybe try to polish the surface in a less visible spot, say, with a Dremel's polishing nozzle, and see whether it'll become clear again when dry. In addition, providing photos before and after the treatment may help.

@andselisk I am very thankful for your comment! I thought oxalate appeared after reaction of citric acid with the soap. I read some info again and understood I was wrong about oxalate. So that white substance is to be the calcium citrate, right? If yes, is there any non-mechanical way to remove it?

I tried to summarize what I think might've happen in an answer, also check out related question: Citric acid chelates calcium carbonate more than ethanoic acid, why does this result in a faster rate of reaction?

Seashells' outer layer (periostracum) mainly consists of conchiolin and polymorphic modifications of $\ce{CaCO3}$ (aragonite, calcite). To sum it up, I see two possible sources of the white coating after treating the seashells with citric acid:

Formation of calcium citrate, which has limited water solubility ($\pu{0.9 g/L}$ at $\pu{20^\circ C}$):
$$\ce{3CaCO3 + 2C6H8O7 -> Ca3(C6H5O7)2 v + 3CO2 ^ + 3H2O}$$
You can try to use acetic acid instead of citric acid in order to produce calcium acetate, which is very soluble in water ($\pu{347 g/L}$ at $\pu{20^\circ C}$). I don't think you can easily dissolve calcium citrate with acetic acid because calcium is complexated by citrate, and this is a very stable formation. The problem here is that the prolonged exposure to acidic medium can further damage the outer layer as more calcite will be washed away.
Conchiolin porous network remained after $\ce{CaCO3}$ has been eliminated with an acid might also appear white when dry and then, when the pores are filled with water, return to it's original appearance. This is rather a speculative theory, but if this is the case, you will need to perform mechanical surface polishing in order to get rid of the porous layer. This can potentially damage the seashells and should be done cautiously.

Thank you for the answer! I will try both and will leave a feedback after.


text:
What does the letter R represent in chemical structures?
The letter $\ce{R}$ appears frequently in chemical structure illustrations. For instance, when I Google keratin, I find Wikipedia's Spanish article about it.
In the image illustrating part of the chemical structure, I see elements like $\ce{H, C, O, N, and S}$ - which I recognize - but I also see $\ce{R}$, which I do not. What does $\ce{R}$ stand for in this context?

Related: What is the correct way to illustrate a bond to something irrelevant in a structural formula?

I'm wondering if we could make this more general so that it would be googleable.  I'm sure it's a common question.

Depending on the context, the letter R can stand for several things. From our friends teaching organic chemistry at UCLA:

R group: An abbreviation for any group in which a carbon or hydrogen atom is attached to the rest of the molecule. Sometimes used more loosely, to include other elements such as halogens, oxygen, or nitrogen. R is an abbreviation for radical, when the term radical applied to a portion of a complete molecule (not necessarily a free radical), such as a methyl group. Should not be confused with R (the gas constant), R (the one-letter abbreviation for the amino acid arginine) or R (a designation of absolute configuration).

In the image from the Wikipedia article you have referenced, the letter R is as defined in the first section of the above quote - that is, it stands for "any group in which a carbon or hydrogen atom is attached to the rest of the molecule."

I thought that $\ce{R}$ originates from German "der Rest" (English "residue"), and the word "radical" was just an adaptation. I wonder if native German speakers could comment on this.

@andselisk That's how German teachers and textbooks use it, yes. Your translation is wrong however,  residue is "Rückstand" in German (e.g. from a distillation). The two words can be synonymous, but not here. "der Rest" better translates as rest of the molecule, R.


text:
Why is the formation of lead (II) iodide precipitate favoured over that of the tetraiodo plumbate (II) complex ion in low concentrations of iodide?
I was wondering why
$$\ce{Pb2+ + 2I- <=> PbI2}$$
occurs in low $[\ce{I-}]$, however when $[\ce{I-}]$ is in excess, $\ce{PbI4^2-}$ is produced with
$$\ce{Pb2+ + 4I- <=> [PbI4]^2-}$$
Why is the formation of complex ion "favoured"? Furthermore, since the formation of insoluble precipitate will reduce entropy, wouldn't the formation of insoluble precipitate be favoured?

@ Ong What do you mean by "Why is there no formation of complex ion "favoured"?"

@Mockingbird Hi there Mockingbird, my auto correct screwed up my sentence a bit. To correct myself, I wanted to know why formation of soluble tetraiodo plumbate(ii) ion complex is more observed than the formation of insoluble lead (ii) iodide (ie why the overall effect of added $\ce{I-}$ is the dissolving of ppt)

When $\ce{Cl^-}$ is in greater concentration, there are just enough ions to favor the 2nd reaction.

@Mockingbird is there any principle behind this?

Well, the rule of thumb is when you need to have grater number reactants for some reaction the greater concentration will work better.

https://chemistry.stackexchange.com/questions/67452/if-copper-is-insoluble-in-excess-naoh-does-it-mean-it-is-soluble-in-naoh-when-n


text:
Why is the methyl group electron-donating via the inductive effect?
I have been reading mechanisms recently, and I find that $\ce{-CH3}$ is a $+I$ effect group. I understand the mechanisms that require this, but can someone explain at the molecular level why is $\ce{-CH3}$ a $+I$ effect group?

The inductive effect is due to hyperconjugation, which is deliberated on in the following post. https://chemistry.stackexchange.com/questions/8412/what-is-hyperconjugation

I know what's HC, but can you elaborate on how HC makes the $+I $ effect?

According to http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch04/ch4-3-2-1.html:
"Hyperconjugation is the stabilising interaction that results from the interaction of the electrons in a σ-bond (usually C-H or C-C) with an adjacent empty or partially filled p-orbital or a π-orbital to give an extended molecular orbital that increases the stability of the system."
How I see "hyperconjugation" is the formation of partial π bonds. Let's say we have a methyl group bonded to a positively-charged carbon atom with a vacant p orbital.
Ideally, it would be best if we could have an atom with a fully-filled p orbital, which is parallel to the p orbital of the this positively-charged carbon atom, to come and interact with this carbon atom. Such an interaction would stabilise the electron-deficient carbon atom. This is essentially the basis of hyperconjugation.
However, in our case, we do not have such an atom with a fully-filled p orbital around. Instead, what we have is a few C-H σ bonds. These C-H bonds are not parallel to the vacant p orbital of the positively-charged carbon atom but they are very close to parallel, around 19.5 degrees away, assuming that the geometry of the methyl group bonded to that positively-charged carbon atom is tetrahedral.
These C-H bonds are thus able to interact with the electron deficient vacant p orbitals of the positively-charged carbon atom, maybe not maximally, but sufficiently. These interactions would be what I would call "partial π bonds" and they stabilise the electron deficiency of the carbon atom. This interaction is also what is known as "hyperconjugation".

*Please note that the illustration is slightly inaccurate because the hydrogen atom would not pop off the methyl group for the formation of a full π bond between the two carbon atoms. This is only illustrated to better show the interaction of the C-H σ bonds with the p orbital of the positively-charged carbon atom.


text:
Raman activity in sulfur
I was wondering why sulfur is Raman active but other elements (e.g. carbon) are not.

Raman activity is a property of compounds, not elements.

Be aware that C is a very bad counterexample. For single atom there cannot be vibrational spectroscopy. But in principle a  pure element that exists in a specific bound forn can have Raman active modes of vibration. C is famous for its allotrooes, diamond graphite fullerenes-nanotubes....

In a review (PDF, CalTech) of the properties of elemental sulfur, the author notes that the most stable form of sulfur at STP is in fact cycloocta-sulfur, $\ce{S8}$, with other polyatomic species also found. Elemental sulfur is commonly formed as a molecular solid, with arrangements of $\ce{S_n}$ rings or chains, and the S-S bonds can be Raman active.
The paper offers depictions of some of those structures, and also tabulates the principal Raman bands for selected allotropes of elemental sulfur.


text:
Optical techniques for kinetics studies
What optical methods can be used to measure the concentration of reacting gases in a reactor during an exothermic and fast reaction? I need to monitor and measure the local concentration of my gases (in the methane production reaction) along a reactor bed in order to be able to clarify the kinetics of this reaction. Which optical techniques are able to make such measurement?

There are a number of techniques but what you would use depends on how you plan to start the reaction. What use to be called Flash-Photolysis and now called Pump-Probe Spectroscopy is the most general and can be used from the femtosecond to multi-second time range. This is a general method as it measured the absorption of all species produced. It works by using a suitable duration and wavelength flash of light, usually from a laser, to initiate the reaction and then at various time delays after this,  the absorption spectrum of any species produced is measured. Absorption can be in the uv, visible or ir ranges. Sometimes Raman scattering or fluorescence can be used instead of absorption. What is used depends on equipment available and the species you need to measure.
If you cannot start the reaction with light then mixing methods can be used, but these are relatively slow, longer than microsecond time scale and more like milliseconds. In the Stopped-Flow method the reactant gases   are mixed as rapidly as possible in the chamber so as to start the reaction then this is monitored with absorption etc. as above. Additionally, there are flow methods whereby the gaseous reactants are mixed at the end of a long flow tube and are monitored spectroscopically at various points as they flow along a tube. The gases are pumped to cause the flow and so distance along the tube is an alias for reaction time.  Each of these methods can be used with gases or liquids.

The reaction is a catalytic reaction and is very fast and exothermic (accompanied with hotspot formation) and we use packed bed reactors for applying the catalyst in it. my purpose was measuring the local concentration of the gas species (methane, carbon mono oxide, carbon dioxide, water, hydrogen) with some optical method e.g. Raman spectroscopy. But I am not quit sure about the details of the system that could actually work.

It is quite routine to measure on the nanosecond timescale, is this fast enough?  Focussed nanosecond lasers are intense and can produce good raman spectra from the focal region so  if you can get your laser to the region you need then this is a possibility. However, sensitivity is low. Have a look therefore at the technique called CARS (coherent anti-stokes raman scattering) and related methods. The book 'Laser Spectroscopy, basic concepts and instrumentation'  2nd ed  by Demtroder  although written some time ago is still good for details on all aspects of lasers and detectors etc.


text:
What is Hydroxy Methylbutyrate-Ca and what is it's relation to butyric acid
First of all, I am nowhere near familiar with chemistry. But I really need help with this question. A certain product contains "Hydroxy Methylbutyrate-Ca" and I need to translate this ingredient in two languages (German and Dutch). After some research I found it very unclear what exactly I am dealing with.
The first thing I found was "beta-Hydroxy beta-methylbutyric acid" $\ce{C5H10O3}$. According to Wikipedia this has a form with calcium:

calcium β-hydroxy β-methylbutyrate monohydrate (HMB-Ca, CaHMB).

Would this calcium form still be considered "Hydroxy Methylbutyrate-Ca"?
The Dutch wiki named the same page "3-hydroxy-3-methylbutaanzuur". "Butaanzuur" would translate(NL > GB) to "Butanoic acid". Is "Butanoic acid" the same thing as "Butyrate" and "Butyric acid"? According to the Dutch wiki 3-hydroxy-3-methylbutaanzuur did have the same composition ($\ce{C5H10O3}$).
Someone told me "Hydroxy Methylbutyrate-Ca" would also be known as "Butyric acid". Someone else told me it turns into Butyric acid once broken down inside the body. What would be correct?
My main goal here is to translate "Hydroxy Methylbutyrate-Ca" but all resources I found have all kinds of names for this substance and I am completely lost. I'm not even sure if some of the dutch equivalents are talking about the same thing as their English counterparts.
My apologies if this is the wrong place to ask for help.

The suffix "-ate" refers to an oxoanionic entity, so "Hydroxy Methylbutyrate-Ca" is definitely not "Butyric acid". Calcium β-hydroxy β-methylbutyrate looks adequate to me. I personally would just directly translate anionic part of Hydroxy Methylbutyrate-Ca into the corresponding language, e.g. Hydroxymethylbutyrat-Ca (German) and Hydroxymethylbutyraat-Ca (Dutch), and call it a day.


text:
Electrons in orbitals with nodes
Node is a region in atomic orbital where there is no possibility of finding electrons. That is no electron can be found there (region of $\psi^2 = 0$)
Take example of say $\mathrm{2s}$ orbital: as electrons cannot be found in node, then how can it spend time in first lobe (spherical) and in second lobe (spherical) separated by nodal region where it cannot go?
And because of course it is electron and its possibility to be found that marks up possible boundary of orbital.

You're talking about quantum tunneling.

To some extent I agree that nanoputian has also asked some what same but there is no answer to above question in the post. I do not really think their being able to cross node like em wave. When em wave passes somewhere it can be found there.

Electron does not cross nodes, nor does it spend part of its time on either side. It just is this way, halfway here and halfway there, every moment since the beginning. Long story short, quantum chemistry is weird.

The electron is on one side of the node or the other only in the sense that it is more likely to be on one side than another. Crossing the node does not require ever being at the node. And being at the node has no physical meaning because you can't know position with perfect certainty by the uncertainty principle.

@PranjalRana Think about a standing wave on, say, a guitar string. There can be a node – a region of zero movement – with movement happening on both sides. That might be a better way to think about it than a ray of light propagating through space.


text:
Maximum hydrogen atom bonds per a single carbon atom
I was solving a chemical formula which includes a carbon, oxygen, and hydrogen compound. And it turned out to be something like this:
$$\ce{C_nH_{4n}O}$$
when $n$ is a natural number. So that, at least ostensibly, can be one of these:
$$\ce{CH4O, C2H8O, C3H12O, C4H16O, ...}$$
But a solution attached to it said

The only compound which has $\ce{C_nH_{4n}O}$ as the empirical formula is $\ce{CH4O}$

so the molecular formula is confirmed as above. So simply my question is this: why is that the only one which makes sense in this situation?

This is the only one which makes sense because the others do not make sense. Try to compose a structural formula for any of them and see what happens.

Well, technically there is ethane hydrate clathrate for $\ce{C2H8O}$, stable   above $\pu{0.1 GPa}$ (Ceppatelli, M.; Bini, R.; Schettino, V. Phys. Chem. Chem. Phys. 2011, 13 (4), 1264–1275.
DOI 10.1039/c0cp01318h).

I would look at it like recursive problem. Starting with the simplest situation:
Carbon has 4 electrons available to form covalent chemical bonds (so 4 simple bonds for example). Hydrogen has one electron - 1 bond.
If you add one more carbon to the situation, you have to bond it somewhere - you need to use the first carbon. Now with this new carbon you added 3 more possible places to add 3 more elements (like hydrogens) but you canceled one hydrogen place on the first carbon. So every time you add another carbon, you can go up only for 2 hydrogen. It is actually called homologous addition or something like that. (Sorry for my english as well)
I ignored the oxygen whole time, cos as far as he is 2 bonded, he never adds anything he either substracts two hydrogen places, or you can instert it between a carbon and a hydrogen.