text:
Are periods the horizontal numbers on the periodic table and groups the vertical numbers?
I am writing this post so that I can ensure for my upcoming SUC3U0 (Chemistry, Grade 11) course test of nomenclature is correct. It's a Review of my Grade 10 General Science Course.
My teacher stated something odd:
She said on the periodic table, the vertical numbers are called Groups and horizontal are periods. Now from my previous understanding, its supposed to be the other way around. I then showed her a periodic table from ChemLabs and Wikipedia disapproving her point and affirming my point of the groups being horizontal and periods being vertical but she says I am wrong and the sites are wrong.
Below is a picture from Wikipedia that I annotated. It shows in blue what my teacher says (Periods at top, Groups Vertically); and we can see what is stated by the creator of the diagram, which again contradicts this point.
Picture below:

This is also posted at https://www.reddit.com/r/chemistry/comments/16j0yi3/my_teacher_states_that_periods_are_the_horizantal/ for reference.

First understand the concept of a period and then it make more sense why a period is a period and why a group is a group. In mathematics, any function which repeats itself after a certain period (say, time period), it is called a periodic function. I am sure you are familiar with sine and cosine functions. They repeat themselves after every 2$\pi$. A clock repeats itself twice in 24 hours. In the same way, the way elements are arranged in rows repeat their properties. Now also note that most Western languages start reading from left to right. So start with atomic numbers, 1,2,3, and so on from left to right. Hydrogen, to helium (first two gases), then Li, Be, B, C, N, O, F, Ne, stop here and note that the next element is Na, which is very similar to Li in chemical properties. So the chemical properties are repeating themselves, after a certain number of elements. This is why the set {Li, Be, B, C, N, O, F, Ne} forms a so called "period". Repeat the same exercise, till you reach K.
Now, Li, Na, K, Rb have very similar properties, both chemical and physical. This is why it is natural to call them a group. Look at the periodic table, this set is vertical.
In short, horizontal rows = periods;
vertical columns = groups.

My teacher states that periods are the horizontal numbers on the perioidc table and groups are the vertical

Groups and periods refer to sets of elements, and these sets have numeric labels (and sometimes names).
Groups
Here are two examples of groups. Group 17 (the halogens, boxed in magenta) comprises six elements, while group 8 (sometimes called the iron group, boxed in purple) comprises four elements. In the periodic table, elements in a given group form a column (top to bottom, or vertical).

Periods
Here are two examples of periods. The second period (boxed in magenta) comprises 8 elements, and the sixth period (boxed in purple) comprises 32 elements. In the periodic table, elements in a given period form a row (left to right, or horizontal). Period 6 appears in two rows in the short-form periodic table shown below, but is in a single row in the long-form periodic table.

As I have written in comments in the very beginning,
there are 18 vertical groups and 7 horizontal periods.
You may have noticed that  e.g. on Excel spreadsheets,

horizontal rows of cells are numbered in a vertical column on the left,
vertical columns of cells are numbered in a horizontal raw the top.

It is similar for the Periodic table of elements. That could be the primary source of misunderstanding on either or both sides, if one talks about objects (groups, periods) and the other about object numbering.
The best would be probably to focus, what a group and a period means generally, not being related to the Periodic table nor chemistry at all. Then you can imply what are groups and what are periods in the Periodic table of elements and you would not have to rely on what you are told.
That is better than to memorize what is what, without understanding what all are talking about.
Imagine this fruit arrangement on a table as a 6x6 set:

Positions
1
2
3
4
5
6

1-smallest
apple
pear
orange
banana
grapes
mango

2
apple
pear
orange
banana
grapes
mango

3
apple
pear
orange
banana
grapes
mango

4
apple
pear
orange
banana
grapes
mango

5
apple
pear
orange
banana
grapes
mango

6-biggest
apple
pear
orange
banana
grapes
mango

Groups are collections of members that are similar, or have some common properties.

In our fruit collection, 6 apples make the apple group, 6 pears make a pear group, etc. Each group is sorted by size, the smallest first.
In the element case, H, Li, Na, K, Rb, Cs, Fr make the group of alkali metals, having similar chemical properties. H is an outlier, but that is not important now.
Similarly Be, Mg, Ca, Sr, Ba, Ra have similar properties, making the alkali earth metal group.
We can check other groups, column per column to the far right, where He, Ne, Ar, Kr, Xe, Rn have similar properties too, making the group of noble gases.

If members of a collection are arranged in a pattern that is regularly repeated, a period is one such a repetition.

The line apple to mango consists of different kinds of fruits, but the fruit pattern repeats every 6 fruits. The set apple-to-mango makes a 6 member period, being followed by the repeated pattern of another period apple-to-mango, that is very similar to the prior one.
At the Periodic table, each row of the table is a period, starting with alkali metal, reacting with water, progressively going through elements loosing metallic properties and ending with a noble gas, that is usually chemically inert.
You can see periods get longer every second period, as more groups are being added. But initial and ending patterns of the periods remain.

You can see then yourself on the Periodic table that there is 7 periods arranged in rows and 18 groups, arranged in columns.


text:
Can all equations be balanced using ion-electron method?
Consider the reaction of a mixture of $\ce{C2H4}$ and $\ce{CH4}$ with $\ce{O2}$. Using the hit-and-trial method, this is the balanced chemical reaction:
$\ce{C2H4 + CH4 + 5 O2 -> 3 CO2 + 4 H2O}$
The above reaction can be easily balanced using hit-and-trial or the conventional arbitrary coefficient method. However, I attempted to balance it using the ion-electron method, but it seems like it cannot be balanced using that method.
My attempt:
Ox-half: $\ce{2 C2H4 + 2 CH4 -> 6 CO2 + 40 e}$
Red-half: $\ce{10 O2 + 40 e -> 5 CO2 + 10 H2O}$
However, when we add these, we end up with this, which is obviously incorrect:
$\ce{2 C2H4 + 2 CH4 + 10 O2 + 40 e -> 11 CO2 + 10 H2O + 40 e}$
Why does this happen?

For future reference: for the body of questions, answers, and comments, chemistry.se offers to use mhchem as a comfortable method to add chemical equations.

Though the reactions are now easier to read, please revise the reactions.  In a balanced reaction equation, the number of atoms on the left to the arrow equates the number of atoms on the right of arrow; the same applies to charges (see especially the half reaction labeled as oxidation).

Well, technically this equation would be a sum of two equations.

Your equations are wrong for many reasons.
First, $\ce{CH4}$ and $\ce{C2H4}$ do not react together, even if you would like them to react. Both react with $\ce{O2}$ and both produce $\ce{CO2}$ and $\ce{H2O}$.
Second, $\ce{CH4}$ and $\ce{C2H4}$ react independently with $\ce{O2}$. These reactions occur independently. So it is nonsense to introduce both $\ce{CH4}$ and $\ce{C2H4}$ together in the same equation. The amounts of $\ce{CH4}$ and of $\ce{C2H4}$ reacting with $\ce{O2}$ may be different and are not related to one another. Both oxidation equations are : $$\ce{CH4 + 2 O2 -> CO2 + 2H2O} \tag{1}$$ $$\ce{C2H4 + 3 O2 -> 2 CO2 + 2 H2O \tag{2}}$$
Pedagocially speaking, these equations can be decomposed into half-equations. For the oxidation of $\ce{CH4}$ in $(1)$, one gets : $$\ce{CH4 + 2 H2O -> CO2 + 8 H+ + 8 e-} \tag{3}$$ $$\ce{O2 + 4 H+ + 4 e- -> 2 H2O} \tag{4}$$ Doubling the second equation $(4)$ and adding it to $(3)$ produces $(1)$ $$\ce{CH4 + 2 O2 -> CO2 + 2H2O} \tag{1}$$  For the oxidation of $\ce{C2H4}$ as in $(2)$, the half-equations are $$\ce{C2H4 + 4 H2O -> 2 CO2 + 12 H+ + 12 e-} \tag{5}$$ $$\ce{O2 + 4 H+ + 4 e- -> 2 H2O} \tag{6}$$ so that multiplying the second equation $(6)$ by $3$ and adding it to the previous one $(5)$ gives (2) :
$$\ce{C2H4 + 3 O2 -> 2 CO2 + 2H2O} \tag{2}$$

I apologize for any inconvenience, and I appreciate your patience. I will be deleting this question as I believe it may not contribute to the understanding of concepts for others. Thank you for the clarification!

One last question, why didn't you consider CO2 in product in reduction halves where oxygen gets reduced in both reactions?

Since you're trying to be exact, I'll point out that you're giving the "overall" reactions. Oxidation of the hydrocarbons doesn't occur in one step. It involves numerous intermediate species.

@MaxW Thank you for your observation. You are right. The given equations should be decomposed into a lot of successive equations. For example, half-equations (3) and (5) are the result of a lot of intermediate steps. Half-equations like (3) and (5) have only pedagogical values.

source: Stack Exchange

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

sort: votes


text:
Source(s) of extreme Raman scattering of FeOOH under intense laser excitation?
RRUFF's Raman spectrum of FeO(OH) as Goethite:

Bears no apparent similarity to the extreme Raman scattering of this sample of purported FeOOH under 785nm laser excitation with 100x objective that appears at 50% of 300mW but not at 25%:

What are plausible sources of such extreme shifts in FeO(OH)?

As the 8000 cm$^{-1}$ shift is so big wrt. any vibration,  are you measuring 2nd order spectra.

And third order but only in the Stokes-spectral lines -- not in the anti-Spokes shifts.

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


text:
How can I calculate the electrostatic energy between two fragments?
I have a 2-fragment molecular complex with partial charge transfer from fragment 1 to fragment 2. The binding energy is ~0.3 eV and I want to evaluate what fraction comes from electrostatic attraction. I have the wavefunction file and I was wondering if there is a way I can evaluate the electrostatic attraction between the two fragments on either Gaussian or Multiwfn.
I can think of a simple approach of performing some type of atomic charge assignment (eg using NAO) and calculating the Coulomb attraction by treating each atom as a point charge, but I was hoping to be able to account for distributed electron density in the calculation.

EDA NOCV comes to mind, but I'm too long out of the game to judge correctly. But look it up and judge for yourself.


text:
Can this equation be balanced using ion-electron method?
Reaction:
$\ce{C2H4 + O2 → CO2 + H2O}$
I want to balance the above reaction using the ion-electron method.
My attempt:
Ox-half: $\ce{C2^(^-^2^)H4 + 4H2O → 2C^(^+^4^)O2 + 12H^+ +12e^-}$
Red-half: $\ce{2O2^(^0^) + 4H^+ + 8e^-→ CO2^(^-^2^) + 2H2O^(^-^2^)}$
After making the electrons gained and lost equal and adding them, and finally summing up, the carbon atoms and the oxygen atoms get unbalanced. How is it supposed to be balanced?

Your red-half itself is not balanced. All in all, this method is hardly ever used in organic chemistry, though still technically applicable.

@IvanNeretin How would we balance it then?

The way kids do. How did you balance reactions before you ever heard of the ion-electron method?

Yeah, but I wish to use this approach.

It feels pretty much like chopping trees with a shovel, but if you insist...
$$\ce{C^{(-2)}2H4 -> 2C^{(+4)} + 4H+ + 12e- }\
\ce{O2 + 4e- -> 2O^{(-2)}}$$

Does this answer your question? Can all equations be balanced using ion-electron method?

Asking again about the same thing isn't a good idea...

@AyushNaman - I'm definitely old school, but what you calling the ion-electron method is known to me as an oxidation-reduction reaction.

The reduction half equation starts from $\ce{O2}$ and goes to $\ce{H2O}$, with the help of $\ce{H+}$. So it is something like $$\ce{O2 + .. H+ + .. e- -> 2 H2O}$$As $\ce{2 H2O}$ contains $4$ $ \ce{H}$ atoms, these atoms must be provided at left by $4$ $\ce{H+}$ ions, and, as a consequence by $4$ electrons. This makes the final reduction half-equation $$\ce{O2 + 4 H+ + 4 e- -> 2 H2O}$$ There is no carbon atom here.


text:
How to estimate the hazardousness of the smoke of a lipo fire?
As generally known lithium polymer batteries (lipos) may catch fire, for example as consequence of deep discharge, overcharge, (internal) short circuit etc.
As a hobby rc car driver, I am thinking about the dangers of a possible lipo fire. To prevent the fire one could use a lipo box like batsafe. However this doesn't filter toxic fumes.
Now there are different opinions found on the net about those fumes, one (as in the faq of bat-safe) says that the fumes dissipate over time. So to say, after a possible fire inside of the box, you just have to open your windows for a few hours or days and everything is gone.
Other people say that parts of this toxic mixture could make a complete house uninhabitable, because material will distribute all over your house and emit toxic fumes, which will make you ill over time (see here)

But what really causes damage is the extreme smoke development.
Besides highly toxic and corrodes your lungs with every breath
irreversible.
No use for a batsave or ammunition box, the smoke will find its way,
these things are only good to seal off fire from the rest of the
apartment. The only exception would be a connection to the outside.
This extremely dense smoke will push you through all the
cracks,through all the boxes, through all doors (even closed) in the
house......
.....The result is an uninhabitable apartment that must be completely
gutted.
At least that was always, without exception, the statement of the
damaged ones as the forums were still full of reports about lipo
fires.
Translated with www.DeepL.com/Translator (free version)

However I don't see how to estimate which opinion is true. So I try to post this as a kind of fermi question to chemistry stackexchange and hope for some insights.
You may assume that four 3s lipos like this burn inside of a lipo box (which prevents the fire) in a basement of a house in a room with 100 m^3 air and the possibility to open larger windows or a door, if needed after the accident.

There isn't stuff like lead of Hg in such batteries. Electrolyte won't burn. That leaves solvent and polymer - not entirely pleasant, but at least some of them aren't all that  bad to burn.

In general, I wouldn't worry too much. Not any more than some random piece of plastic catching fire.


text:
Does lime juice "alkanize" in the body?
My brothers and I have a dispute. They claim lime juice alkanizes in the body. I contest as an acid it remains acidic in the body. My guess is they are confusing "lime water" with "limewater", the latter which IS made from a base and used in tortilla making.
Whom is correct here? I thought this place best to ask.

Your guess is most likely correct.

https://chemistry.stackexchange.com/questions/105503/sugar-solutions-have-a-neutral-ph-in-themselves-but-it-makes-your-body-acidic https://chemistry.stackexchange.com/questions/122718/is-it-wrong-to-allege-that-lemons-and-limes-are-alkaline

This is a pseudoscientific diet idea. Best ignore it.

Your brothers are misinformed and should stop getting their nutritional knowledge from Tik Tok

Btw, is "alkanize" even a word? There is "alkalize".

There could be two sources of this misinformation.

Confusion of the words lime (a citrus family fruit) and lime-water (a highly alkaline corrosive solution of calcium hydroxide).

The name lime for the citrus fruit originally comes from Arabic and it does have many cognates in South Asian languages. The lime of calcium hydroxide is connected to German and Latin! Orthogonal meanings and origins.

Second source of confusion could be citrate salts or buffers. Citrate salts of sodium or potassium are indeed mildly alkaline.

In short, drinking lemon juice will not increase alkalinity, it may cause acidity instead and may cause pain in stomach ulcers if there are any. Citrate salts are traditionally used in antacid formulations where citric acid and sodium bicarbonate powder is mixed with water as an antacid. Nature has a very good system of controlling internal pH.


text:
Bromate ion point group?
Would the bromate ion have C3v point group due to the trigonal pyramidal geometry? I see a C3 rotational axis and a vertical plane of symmetry. Am I missing anything in terms of the double bonds or are those the only symmetry operations in this molecule?

As its not planar it should be should be C3v, the double /single are just a representation of bonding.


text:
Term symbols in transition metals and lanthanides
Hello I was researching a little about term symbols of lanthanides and transition metals and have got two questions.
It is known that lanthanides don’t follow LS-coupling, but instead JJ-coupling. Does that mean that the term symbols for the rare earth (III) cations in the famous Diecke diagram are formed by JJ-coupling? I am asking because the ground state term symbols for lanthanides are found just like the transition metal ground states:

My second question is the following: In the term schemes for transition and lanthanide metals, all filled shells and subshells are discarded, so we are only looking at d-d-transitions and f-f-transitions of their respective term symbols if I understood correctly.
But below the f- and d-orbitals there are other orbitals (for example 2s) which could also do electronic transitions (at much higher energies of course). How are they described?

For readers, see also Wikipedia articles Angular momentum coupling - Spin-orbit coupling and Spin-orbit interaction // Possible source of the diagram on researchgate.net

You can have a look at the NIST webpages, especially on the compendium on atomic spectroscopy and the atomic level database.

As you write, the filled orbitals do not contribute as spin ang mom is zero, those lower energy ones with unpaired electrons as part of an excited state will add in in just the same way as any other.

So not only can we describe the outermost electrons with the term symbols, but any other electron in an lanthanide, for example electrons in the 4s shell? Just like main group elements


text:
Can Acyl halides react with PCl5, if not, why?
I was searching for reaction of $\ce{PCl5}$ with carboxylic acids and found this mechanism:

Here a corresponding acyl chloride is formed from a carboxylic acid.
Now I doubt if this is the final product as this acyl chloride's carbonyl group should also react further with $\ce{PCl5}$ just like it happens in Aldehydes and ketones.

So I searched for reaction of acyl halides with $\ce{PCl5}$ and I can't find any results showing the mechanism or even discussing what can be the product.
I believe that further reaction should happen on the carbonyl group of the acyl halide forming gem-dichloride replacing the double bonded Oxygen.
This direct replacement can't occur on the carbonyl group of the carboxylic acid as it's -OH group also have involvement (working as acid and releases H+ which changes the dynamics of the reaction) in the reaction while halide group of acid halides don't interfere that much and so acyl halides should react like aldehydes or ketones with $\ce{PCl5}$ as discussed in this answer by following mechanism:

So With all things looking perfectly fine then why don't this reaction occurs? What are those factors preventing this reaction?

Actually, certain $\ce{MX_n}$ molecules can be strong halide abstractors, such that they can pull off the halide in $\ce{RCOX}$ to form an acylium cation. I don't think $\ce{PCl_5}$ is strong enough, but things like $\ce{SbF_5}$ and apparently $\ce{SbCl_5}$ are.

The carbonyl group of an acyl chloride will be less nucleophilic than a ketone because of the electron-withdrawing effect of the Cl making your first step of the second scheme less likely to occur

@Waylander if the carbonyl group can be enough nucleophilic for reaction with PCl5 in Carboxylic acids (EN Oxygen = 3.44) then it should also be in acid chloride (EN Chlorine = 3.16)

@NicolauSakerNeto this looks like a fact rather than a comment for this question.
In my understanding*

The first step of the reaction with carboxylic acids shows the answer.  Nucleophilic attack on the phosphorus is enabled by electrons from the second oxygen donating in to the central carbon.
Those electrons come from the free methyl group in your second mechanism via a keto/enol tautomerization, but they aren't present in most acyl chlorides. When they are present the reaction proceeds; eg. the reaction of niacin with phosphorus pentachloride.

One possibility is that the reaction mechanism springs a leak when you have a halogen attached to the carbonyl group.
When the chloride ion from the $\ce{O–\overset{-}{P}Cl5}$ function attacks the carbonyl carbon, either $\ce{POCl4-}$ or $\ce{POCl3}$ must act as a leaving group (depending on when the second chloride ion initiates its attack). If there is a better leaving group, the nucleophilic substitution might go the wrong way preventing the intended breakage of the carbon-oxygen bond.
An aldehyde or ketone presents no such problem because a hydride or alkide(?) ion is not going to get displaced in a nucleophilic reaction. But a halide ion could be displaced from an acyl halide instead of breaking the carbon-oxygen bond.

What I conclude from your answer is that $\ce{-Cl}$ is better leaving group than $\ce{-O-\overset{+}{P}Cl3}$, but then shouldn't $\ce{Cl-}$ not be able to push out $\ce{POCl3}$ in the second scheme? If it was relatively a good leaving group then it should be removed as soon as it attacks.


text:
Clear polystyrene bottle holder repaired with butyl acetate glue. Is it safe to use in the fridge?
I've successfully repaired my fridge's bottle holder, which is made of clear polystyrene, using butyl acetate glue.
While the glue smelled at the time I was gluing, the repaired bottle holder no longer smells.  The couple drops of glue which ran now appear to be made of solid polystyrene.  They and are not liquid or sticky.
Are there any health and safety reasons which should stop me from using the repaired bottle holder in the fridge?  (There is often uncovered food in the fridge, but no unpackaged food in direct contact with the repaired bottle holder.)

Acetate? It was acrylate, wasn't it? Likely this: https://en.wikipedia.org/wiki/Butyl_cyanoacrylate

@Mithoron The glue's paper insert reads: "Warning Humbrol Polycement contains butyl acetate" and doesn't mention any other chemicals.  https://www.amazon.co.uk/dp/B000JCE690  https://en.wikipedia.org/wiki/Butyl_acetate  (If you're the down-voter, would you mind removing it?)

Butyl acetate is just a solvent, and if the thing doesn't have its smell, its volatility shouldn't be a problem now.

You should edit it with details and better wording, like "glue containing butyl acetate". Still, questions like that are kinda problematic. Borderline "personal medical" thing. How do we know if someone isn't for example allergic, or glue got spoiled, or whatever...

Butyl acetate is found in small quantity in fruits such as apples, and is considered safe, in most regions, for use in nail polish.
However, this site specifically excludes medical recommendations. (Personally, I would not recommend eating the plastic bottle holder, even if you enjoy the solvent odor ;-)


text:
Understanding Reaction Rates and equilibria
As a high school student grappling with the interplay of reaction rates, Le Chatelier's principle, and equilibria, I have a question about a system that hasn't yet achieved equilibrium.
Specifically, if the reaction rate to the right exceeds that to the left, should I expect more products or more reactants to be formed as the system approaches equilibrium? On one hand, the higher reaction rate to the right suggests more products are being produced, but on the other hand, doesn't the reaction rate to the right need to decrease for equilibrium to be reached?

Note that as soon as reactions starts, two processes start to occur. Conversion of reactants into products and conversion of products back into reactants.

It's not that reverse reaction starts occuring only after equilibrium point. It is already occuring with the forward reaction. Past the equilibrium point, the reverse reaction rate is more than forward reaction rate contrary to before equilibrium where forward rate is more than reverse rate.

You cannot always expect more product in the right side of equilibrium. If the rate if independent of concentration of reactants and products, such as in zero order reaction, conc. of product need not be more than reactants. It's just their respective rates.

In the equilibrium reaction is A⇌B, "if the reaction rate to the right exceeds that to the left" then B is produced from A at a greater rate than A is produced than B at that moment, so yes, "the higher reaction rate to the right suggests more products are being produced." It is also correct that the reaction rate to the right (relative to the rate to the left) needs to decrease for equilibrium to be reached because after equilibrium is reached, the two rates will be the same.

At $t=0$ the forward rate $A\to B$ is at a maximum, as $\ce{[B] = 0}$ (assuming only A at the start) and after a second or so some B is present and rate $A\to B$ has reduced a bit as there is less $A$ and the rate of $B\to A$ has increased but is still small. Some A is, of course reformed, but that is always accounted for in the forwards rate. As time passes the rate of $A\to B $ falls further and $B\to A$ continues to rise until they are equal and equilibrium has ben reached. No further change in rates occurs, unless of course you change something, such as temperature etc. The position of equilibrium (amount of A vs B) depends on the ratio of rate constants.

Specifically, if the reaction rate to the right exceeds that to the left, should I expect more products or more reactants to be formed as the system approaches equilibrium?
Let's assume that at the beginning there are no products. Thus at the beginning of the reaction there can't be any reverse reaction. Also at the beginning the forward reaction rate is at its peak, since the reactants are at their maximum concentration.
On one hand, the higher reaction rate to the right suggests more products are being produced, but on the other hand, doesn't the reaction rate to the right need to decrease for equilibrium to be reached?
Now as the forward reactions starts two things happen. First the forward reaction rate, not the rate constant but the overall reaction rate, starts to slow because the reactants are being used up. The second thing that happens is the reverse reaction starts. Since equilibrium hasn't yet been reached, the forward reaction rate is faster than the reverse reaction rate.
But as the forward reaction drives the system to equilibrium the forward reaction rate continues to slow because the reactant concentrations are decreasing and the reverse reaction rate continues to speed up because the product concentrations are increasing. Equilibrium is reached when the two reaction rates are finally equal.
Note that if there are no products initially then the reverse reaction rate can never be faster than the forward reaction rate. Thus the system is always going towards equilibrium.


text:
Is ionisation enthalpy physical property or chemical property?
Ionisation enthalpy means energy change when an electron is removed.
According to Wikipedia

A chemical property is any of a material's properties that becomes evident during, or after, a chemical reaction; that is, any quality that can be established only by changing a substance's chemical identity.

Chemical changes are defined as changes that involve breaking or formation of bonds.
Ionisation is not involving any breaking or formation of bond.
But when we are removing electron from any atom we are changing its chemical properties. Properties of Sodium cation are not as same as Sodium metal.
So my question is whether ionisation enthalpy physical or chemical property?
Also please tell me about whether electronegativity is physical or chemical property?

What if definitions of both property domains overlap? Similarly, there is no sharp boundary between chemistry and physics.

Like, the ionization enthalpy has aspects of both physical and chemical property.

The question is void. All chemical properties are physical properties, too.

As Ivan says by other words, chemistry is the part of physics that physicists do not want to get their hands dirty with.

Ionization enthalpy is defined as the energy required to separate an outermost electron from the atom to completely ionize the atom.
Perhaps you are familiar with NCERT textbook class 11 Part-2  from which this is a table

Here the heading says 'atomic' and physical properties of alkali metals. No need to mind 'atomic' as it is used to refer to metallic and ionic radii. That means the ionization enthalpy is referred to as a physical property here. So ionization enthalpy is indeed a physical property.
The reason being that a chemical property is based on how the substance behaves chemically in response to some other chemical. Sodium is a chemical and it can be conferred certain chemical properties toward other chemicals such as oxygen, nitrogen, water, etc. But when it becomes sodium ion, it is not a chemical in the way sodium is. $\ce {Na}^+$ doesn't have a really independent existence. It will be either as aqueous solution or molten state or as counter-ion in salts. $\ce {Na}^+$ has more to do with the standard potential which again is a physical property as per the above table.

Yeah but if we want to measure ionisation enthalpy we would need to chemically react our sodium and if we follow what Wikipedia says then it should be chemical property.


text:
If charge is quantised, how can It delocalize?
As I was trying to understand resonance, I got to know that it was imaginary phenomenon and it is just a tool to explain mechanisms and structures that are unexplained by Lewis dot structures or structures where bonds are localized.
Hence its proper description is now done using molecular orbital theory which is beyond scope of studies for new students. Now after that I got to know that delocalization of electrons releases energy or "increase stability" of system, there are two explanations on Stack Exchange:

One of the explanation on Stack Exchange uses "Particle in box" concept to explain the phenomenon, that is: larger the length, less is the energy.

The other explanation involves saying "delocalization causes charge cloud over large volume decreasing charge density, hence decreasing energy of system".

My question is: How can charge distribute over a volume if it is quantized? Does it come in discrete packets in form of electrons?
If this has any relation to probability of charge occurrence, the first explanation cannot be used, hence is there any rough, simple and intuitive explanation rather than using quantum mechanics and "Particle in box" etc.?

It is about probability of charge occurrence, like everywhere in context of orbitals and chemical bonds in quantum chemistry.

@poutnik,If it is about probability of charge occurence,hence second explaination of charge density cannot be used,so is the "particle in box" type only valid explaination here?can't it be rougly simplified ?

Review your post, namely the length of sentences,  usage of punctation and spaces. Near whole your post is one giant single sentence. As the result, it is difficult to read.

@Poutnik followed the suggestion,is it readable now?

Well, there is this concept of using dots/full stops at the end of sentences and using spaces after commas, instead of a new paragraph. Note also the general guidance that the average length of sentences in scientific texts should be 20-25 words, and the length of a particular sentence should better not exceed 35 words.

Try to apply the question in the title on a hydrogen atom. Then of HF molecule with the polar bond and partial charges of both atoms.

Hence partial charge is just a notation to describe movement of electron from its original neutral point?is it correct?@Poutnik

Electrons are always delocalized. The way and complexity how depends on the complexity of the system where they are delocalized. // You should really learn to write proper English sentences, if you want to communicate in English.  Long sentences, full of commas without spaces, where one forgets the beginning before reaching the end, are very hard to read.

ohh, is it something related to this explaination : in hydrogen atom, though it being neutral, at atomic scales, it is consisting of partial positive and partial negative charge which on macroscopic scale is neutral, it is not neutral like a neutron, but the net sum of charges is zero, there is still a finite distance between nucleus and electron, hence it is not intrinsically neutral as neutron

when bond forms between hydrogen and fluorine, the initial distance between electrons of hydrogen and its electron increases due to electronegativity which I suppose we call partial charge on hydrogen//sorry for poor editing skills, will try to keep that in mind

Let us continue this discussion in chat.

Would be good if you explain more clearly why you feel that delocalization and quantization are mutually exclusive or competing concepts. Quantization arises due to electron wavefunctions behaving as standing waves (in the time-independent case). Delocalization refers to the shape or distribution represented by that wavefunction.

Quantum theory states everything is unlocalized until it is observed. Consider the double-slit experiment, which applies to photons and electrons, to atoms and even molecules... Through which slit does the "particle" go? Both. What is a particle when there is no observer? A wave?
There is a probability of observing an electron at a location. That does not mean that the electron (or its charge) is broken into pieces, only that when it is observed, in its entirety, there is a chance that it is found here, or there.
When a photon of light goes through two slits, it does not change frequency and have half the energy go through each opening; the interference pattern is created by the probability of detecting the entire photon, at one instance, in a location.
We are distributed, all one with the universe.

I somehow didn't get the meaning of the last sentence but I would like to know. Could you please tell a bit elaborate about it...

@Proscionexium, All particles are waves. See https://www.scientificamerican.com/article/giant-molecules-exist-in-two-places-at-once-in-unprecedented-quantum-experiment/ and, just possibly, https://www.youtube.com/watch?v=Slwkped573s


text:
What is the effect of temperature on titration between Iron(II) sulfate and potassium permanganate
I'm conducting an experiment in school, the titration of Iron(II) sulfate dissolved in H2SO4 against potassium permanganate.  In this, I am varying the temperature that the analyte is heated to and finding the titration value.  I'm finding a general upward trend as increase temperature, but I can't find the theoretical values for this or the way to calculate them.
Any assistance on how to calculate the theoretical titer required at different temperatures would be greatly appreciated.

Provide more details about the experiment, like used concentrations, temperatures and titration volumes. // One of reasons, that can be difficult to quantize theoretically, is thermal decomposition of the permanganate, leading to higher spent volumes.

Temperature has no effect on the stoichiometry of chemical reactions. Only the kinetics of the reaction is modified by the temperature. And Poutnik is right in stating that permanganate is not very stable at high temperature : it gets thermally decomposed when hot.

Users Poutnik and Maurice make excellent points. However I'll raise a different point. The Nernst equation is temperature dependent. So it is possible that changing the temperature could effect if a reaction occurs. Also for a reaction like copper being dissolved with nitric acid,  the nitric acid can undergo two different reductions. So changing the temperature would effect the "overall stoichiometry" as the ratio between the two reactions changes.


text:
Identifying Key Chemical Series, Pharmacophores, and Scaffolds in a Large SDF Dataset
fellow computational chemists!
I have encountered a challenge that I'd like to discuss and seek advice on. Suppose I have a substantial dataset in SDF format containing thousands of molecules. My goal is to identify the major chemical series, pharmacophores, or scaffolds within this dataset. The primary motivation behind this task is to reduce a large dataset into a more manageable one that retains representative structures. This curated dataset will then serve as a reference for subsequent hit selection processes in my research.
The overarching objective is to ensure that my hit selection process captures sufficient chemical diversity from the larger dataset for subsequent in vitro testing. This is crucial for the success of my research, as it helps guarantee a broad exploration of chemical space and increases the likelihood of discovering novel compounds with desirable biological activity.
I would greatly appreciate any insights, suggestions, or experiences you might have to share regarding this process. Have you encountered similar challenges in your computational chemistry work? Are there specific tools or techniques you've found particularly useful in identifying key chemical features in large datasets? Your input will be invaluable in helping me streamline my research and improve the diversity and representativeness of my hit selection process.
Thank you in advance for your contributions to this discussion!

Probably a better fit at [mattermodeling.se].


text:
How is temperature inside a bomb calorimeter constant?
I've looked around and have been unable to find an answer.
When determining molar reaction enthalpy, you first perform the reaction in the steel bomb then measure the temperature change of the water to calculate the molar energy released from the reaction. Then you add $\mathrm{\Delta(PV)}$ to $\mathrm{\Delta U}$ to calculate molar reaction enthalpy. $\mathrm{\Delta(PV)}$ is then said to equal $\mathrm{\Delta n\,RT}$, where temperature is constant.
But, if for example the reaction raises the temperature in the steel bomb from 25 °C to 40 °C and eventually the temperature equilibrates so that the temperature of the water is 28 °C and the temperature inside the steel bomb is 28 °C, doesn't the $\mathrm{\Delta U}$ value underestimate the energy change of the reactants since the temperature inside the bomb is raised as well and shouldn't $\mathrm{\Delta(PV) = \Delta(n)R\Delta(T)}$?

Yes, any increase in the reaction product mixture temperature results in a small inaccuracy in the calculated heat of reaction.  But, it is really pretty small, and you can correct it by calculating the amount of heat that needs to be removed for the products to be at the same temperature as the reactant mixture.


text:
Can a heated gas leave a container through downwards direction?
If I have a flask full of heated gas, with an opening at the BOTTOM of the flask and connected pipe facing downwards, can the heated gas can leave the flask through that opening?
And for an alternate case: a flask full of heated gas, with an opening at the BOTTOM of the flask and connected pipe facing downwards as well as a TOP opening, does the heated gas always leave the flask only through the top? Or can it leave through both?

It depends on if you consider only convection or if you count with diffusion too.

I suppose it depends on how hot the gas in your flask is, no? This might be an extreme example of an answer to your first question: https://www.nasa.gov/sites/default/files/images/624229main_1969-07-16-5_full.jpg

Molecules are always moving. Even at a (theoretical) temperature of absolute zero, there is still motion due to zero-point energy. One can watch liquid helium move up and out of a container, at 2 K!
On the average, molecules move at approximately the speed of sound (well, within a factor of two, or so). At room temperature and pressure, the molecules are scooting about randomly at roughly 500 m/s, bouncing off each other and the surface of any object. Eventually, they spread throughout a container, and out any opening, top or bottom, by the process of diffusion. Wait long enough, and they'll exit anywhere, and outside molecules can enter through that opening, until the inside and outside are in equilibrium.
However, the molecules do not leave, all at once, at 500 m/s, because they get in each other's way. If the hole is relatively small, as molecules try to leave, they are blocked by those already in the hole. If the gas inside a container is less dense (i.e., more buoyant) than that outside it, and the hole is on bottom, then external air pressure slows that diffusion, so, in a narrow-necked flask of hydrogen or helium, or just hot air, turned with the opening at bottom, it would take longer for the gas to diffuse out (and for the more dense gas outside to enter) than if the flask were upright.
The same would be true if one left dry ice in an upright flask -- it would take longer to diffuse out, to be replaced by air.


text:
Gibbs Free Energy and dependence on temperature and pressure
So, I have read that change in Gibbs free energy is defined under constant pressure and temperature, i.e
$$\Delta G= \Delta H -T \Delta S $$
but for an ideal gas change in entropy of system can be written as
$$\Delta S= nC_pln(T_2/T_1) + nRln(p_1/p_2)$$
this formula implies that change in entropy at constant temperature and pressure is 0.
Also for an ideal gas change in enthalpy is only a function of temperature so for an ideal gas at constant temperature and pressure change in enthalpy would be 0.
But as I said above $\Delta G$ is defined at constant temperature and pressure, so wouldn't change in gibbs free enregy for an ideal gas always be 0?
PS I have read some related answers but they have not helped me solve this query.
Some Read answers -
https://physics.stackexchange.com/questions/716364/how-can-the-gibbs-free-energy-equation-be-at-constant-temperature-and-pressure
Why at constant pressure and temperature Gibbs energy change of a process can be negative?

change in Gibbs free energy is defined under constant pressure   G is defined regardless of conditions as G = H - TS, but only at T=const (p=const is not required) it can be written as ΔG=ΔH-Δ(TS)=ΔH-TΔS.

But finally for us to write $\Delta G= \Delta H -T \Delta S$ doesn't your statement imply that the conditions should be constant pressure and temperature

No, it does not. G as well as H are especially useful and usually used at isobaric conditions (like A and U at isochoric ones), but the above formula does not have isobaric requirement, being directly derived from G definitions and isothermic requirement.

Ok, so the equation is valid for isothermal conditions, I guess many places it is written wrong then, couple books I have, have written it as both constant temperature and pressure.

Well, it is valid for that too. If something is said to be valid for 2 simultaneous conditions, it does not generally imply it cannot be valid for just one of the conditions.

But wouldn't under those conditions $\Delta G$ be always 0?

Let us continue this discussion in chat.

There is happening no change in a (not reacting) ideal gas system at isothermic and isobaric conditions.
Therefore, unless no chemical reaction is happening, $ΔG =  0$
Similarly, for reversible processes at isothermic and isobaric conditions:
\begin{align}
ΔS&=Q/T\\
ΔH&=Q_p\\
ΔG&=ΔH-TΔS=Q_p-Q_p=0
\end{align}
If there is ongoing spontaneous process, like chemical reaction, $ΔG \lt 0$. We then say the process is exergonic. Note that a process need not to be exothermic ($ΔH \lt 0$) to be exergonic, if balanced by the system entropy increase. See e.g. dissolving of $\ce{KNO3}$ or $\ce{KClO3}$, which is endothermic but spontaneous.


text:
UV-Vis Absorbance against wavelength graph for finding max wavelength
I have plotted the absorbance against wavelength graph to find the optimum wavelength for maximum absorbance of p-nitroaniline. However, the curve (dark blue) of the most concentrated solution (sample 7 - 80.0 x 10^-5 M) looks a bit weird. There are 2 decrease in absorbance at around 350 nm and 370 nm before the maximum absorbance at around 385 nm. p-nitroaniline was diluted with DMSO and it was mixed with enzyme solution in the cuvette to duplicate conditions in the enzyme assay. p-nitroaniline is the product of an enzymatic reaction and there is a second part of the experiment regarding the enzyme reaction kinetics. All the absorbance were measured using UV-Vis spectrometer.
I am guessing if it has anything to do with my blank?

A blank would make the same patterns for all samples, not proportional ones as on the plot. The peak may consist of multiple merge peaks. If you measured spectrum in gaseous phase, you would probably see a peak forest.

If you can, check the manual for the spectrometer to see if there is an automatic switch-over in light sources at 350 nm, i.e., deuterium lamp for wavelengths below 350 nm and tungsten lamp for longer wavelengths.

@EdV, A good spectrometer should not cause a dip from light switch until and unless the mirrors are messed up, and why there are two dips?

Joe, What is the wavelength step size and what is your cuvette material? Also, it seems your dips are concentration dependent, do you see any particulates or aggregation?

@AChem I have seen the switch-over issue before, in a teaching lab instrument. Sadly, teaching instruments can be subjected to hard use or abuse and not maintained properly. No clue about the other dip.

Don't forget that absorbance $\text{abs} = \log \frac{I_0}{I}$ for incident radiation intensity $I_0$ vs radiation intensity $I$ passing the sample. Thus, with $\text{abs} = 1.0$, there is only 10% remaining, at 1.5 about 3%, and with 1.8 around 1.5%. As a rule of thumb, to stay in the linear range of detectors of (especially) the more affordable UV-Vis spectrometers (teaching lab), it is better to limit the concentration of the sample to the effect to stay below $\text{abs} \approx 1.3$. With multiple solutions of different concentrations, you still can plot $\log{}\epsilon = f(\lambda{})$.

If the purpose of the exercise was to localize $\lambda$ of maximal $\epsilon$ (e.g., of the sample of $\pu{20E-5 mmol}$, or $\pu{40E-5 mmol}$), than record the spectra multiple times, average the recordings (for better signal/noise ratio) and tap into derivative spectroscopy (was mentioned here w/ link to Agilent); especially the second derivative for its sharp positive-zero-negative transition can be more indicative, than the first derivative.


text:
Reactivity in my standard solution of Cu, P, Al, Ca, Fe, K, Mg, Mn, Na and S in 7,8% HNO3 and H2O2
I have a problem with my standard for ICP-OES. It contains the following elements Cu, P, Al, Ca, Fe, K, Mg, Mn, Na and S in a solution of 7,8% HNO3 and H2O2.
I cannot make it work for reading P and most worryingly the high concentration standard bottle expands and makes a high hiss when opened after a while indicating a reaction that makes a gas.
Can anyone suggest what elements cross reacts? We have measured all elements in other combinations without problem.

To get any help, make readers feeling they know your scenario as well as you do.

What is precisely the composition of your solution ?  Apparently the solution is acidic and contains the cations $\ce{Cu^{2+}, Al^{3+}, Ca^{2+}, Fe^{3+}, K^+, Mg^{2+}, Mn^{2+}, Na^+}$ and the anions $\ce{NO3^-, PO4^{3-}}$. What is the concentration of $\ce{H2O2}$ in the solution ? Don't you think the "hiss" is not simply due to $\ce{O2}$ produced by the catalytic decomposition of $\ce{H2O2}$ ? It is not a cross reaction. It is well known that the ions of $\ce{Fe, Cu, Mn}$ act as catalyzers in the spontaneous $\ce{H2O2}$ decomposition.

Are you following a standard protocol that has been shown to work and that you have tried before? What is the exact concentration? Are other elements giving the expected readout?

@Buttonwood, P can easily have ppm or below ICP detection limits.

The knowledge of classical chemistry remains important, irrespective of the sophistication of analytical techniques or the cost of the instrumentation. It is important to understand the incompatibility between iron and phosphorus, which form highly insoluble phosphates. Similarly, Mg, Cu, Al will all interact with dissolved phosphate with their insoluble phosphates. Consequently, the utility of a dissolved phosphate in the presence of these metallic ions to save analysis time goes to waste. This explains the poor limit of detection for phosphorus, which likely is resting at the bottom of the standard stock solution. A second concern is the catalytic decomposition of  $\ce{H2O2}$. A repetition of the entire analysis is needed because the whole standard solution is messed up.
Until and unless these metallic ions were strongly chelated (perhaps not) by the standard solution manufacturer, the results of these ICP-OES analyses are not reliable at all.


text:
Alcohols and bromine
As far as I know, bromine can act as an oxidising agent. But when propan-1-ol is mixed with bromine water, there is no reaction.
Why bromine cannot oxidise alcohols? Is the reaction available if a catalyst is added?

One question is if it can, another question is how quickly.

Who says bromine does not oxidise alcohols? Bromine is known to directly oxidise primary alcohols. According to this JACS paper ethanol is oxidised to a mixture of acetaldehyde, acetic acid and ethyl acetate depending on amount and concentration of bromine used


text:
Standard electrode potentials
I know that many substances have their own standard electrode potential and they can be arranged in he electrochemical series. For example, the standard electrode potential of iron is -0.440V.
Since organic compounds can also undergo redox reactions,like the oxidation of alcohols, I would like to know whether organic compounds have their own standard electrode potentials as well.

Well, yeah, but this aspect isn't delved all that much, I think. https://en.wikipedia.org/wiki/Table_of_standard_reduction_potentials_for_half-reactions_important_in_biochemistry

https://en.wikipedia.org/wiki/Standard_electrode_potential_(data_page)

When you search for values check to what standard the values apply, standard hydrogen electron (SHE) is usually used in tables in chemistry texts , but using the calomel electrode or pH7,  as used in biochemistry, are not uncommon.

Be aware these potentials are thermodynamic quantities while reactions of organic molecules have often slow kinetics, so there is no chance to measure such potential.

It just requires patience, potentials are measured at zero current draw. Consider the quinhydrone electrode to measure pH. Many quinones have well defined potentials. I have wondered how an EKG works.

But quinones are rather exception, based on direct electron transfer. OTOH, some organic redox systems are quite obscure. Note that the concept of potentials is formally applicable to gaseous states too.  See e.g. researchgate.net - Redox potentials of CH4 oxidation reactions

Organic compounds equally have their standard electrode potentials, and their determination in solutions (e.g., by cyclic voltammetry, see for example the primer by Elgrishi et al.) is one method to eventually record the gaps between HOMO and LUMO of dyes used in organic electronics (organic solar cells, OLEDs, etc).  Beside characterization of a material, it equally represents an entry in organic synthetic electochemistry to substitute otherwise unpractical reaction conditions.  The Baran group, for example, developed an e-Birch protocol which (contrasting to the original conditions of a Birch reduction) is easier to manage (no elemental metals, no need to condense molecular ammonia) and to scale because the electric potentials applied can be tuned to the actual needs.  An elder essay by Baran hence is titled «Synthetic Organic Electrochemistry: Calling All Engineers».
(1) Elgrishi, N.; Rountree, K. J.; McCarthy, B. D.; Rountree, E. S.; Eisenhart, T. T.; Dempsey, J. L. A Practical Beginner’s Guide to Cyclic Voltammetry. J. Chem. Educ. 2018, 95, 197–206. doi 10.1021/acs.jchemed.7b00361 (open access).
(2) Peters, B. K.; Rodriguez, K. X.; Reisberg, S. H.; Beil, S. B.; Hickey, D. P.; Kawamata, Y.; Collins, M.; Starr, J.; Chen, L.; Udyavara, S.; Klunder, K.; Gorey, T. J.; Anderson, S. L.; Neurock, M.; Minteer, S. D.; Baran, P. S. Scalable and Safe Synthetic Organic Electroreduction Inspired by Li-Ion Battery Chemistry. Science 2019, 363, 838–845. doi 10.1126/science.aav5606 (entry PubMed).
(3) Hayashi, K.; Griffin, J.; Harper, K. C.; Kawamata, Y.; Baran, P. S. Chemoselective (Hetero)Arene Electroreduction Enabled by Rapid Alternating Polarity. J. Am. Chem. Soc. 2022, 144, 5762–5768. doi 10.1021/jacs.2c02102 (entry PubMed.
(4) Yan, M.; Kawamata, Y.; Baran, P. S. Synthetic Organic Electrochemistry: Calling All Engineers. Angew. Chem. Int. Ed. 2018, 57, 4149–4155. doi 10.1002/anie.201707584 (entry PubMed).


text:
Pressure in Boyle's law
According to Boyle's law, PV = k as long the temperature and amount of gas remain constant.
In this equation, Is P the pressure inside the container or outside?
I'm confused because I just saw a video on the internet that showed a bag of chips becoming inflated when it was taken to higher altitude. Now in this case the pressure inside the bag didn't change yet the volume changed (let's also assume temperature also remained constant as we moved up)
I know that the bag inflates because the atmospheric pressure decreased as we moved up so the air inside the the bag started exerting a greater force on the walls of the bag but again doesn't this mean pressure increased? It also means that the pressure was directly proportional to volume now??

See, Boyle law is applied on an equilibrium system. Suppose you increase the external pressure reversible and infinitisimally, the volume will decrease. But to reestablish equilibrium, internal pressure will increase. So you have seen here that $ V \propto \frac{1}{P}$. So it is indeed the internal pressure that is considered.

In your example, the volume of the bag increased, and the pressure inside the bag decreased.  So what's the problem?

Why should gas volume depend on external pressure?
It depends on the internal pressure, which can be:

equal to the external pressure, as in the case of a container with fully adaptable walls.
different from the external pressure, as in the case of a container with stressed walls.
independent of the external pressure, as in the case of pressurized cylinders.

If the chips bag is being taken to higher altitude, internal pressure becomes higher than external pressure, pushing the bag walls against the lower pressure. As the result, the bag expands and increased volume decreases the internal pressure until both pressures are equal again.
At some point, the bag can become fully bloated. Further lifting of the bag causes bag material stress. This leads to the internal pressure becoming bigger than the external pressure, as $p_{\text{ext}} + p_{\text{stress}} = p_{\text{int}}$.

A stop in a grocery store in Leadville, CO (3000+ meters) reveals highly stressed bags of chips. Care is required handling them.

The gas laws apply to the P, V, n and T of the contained gas. [Engineers also have a different value of R for different gases]. The external pressure can have an effect on the container and then has an effect on the variables. The only real effect is if work is done against an external pressure; then the external pressure determines the work done. A common example is a balloon expanding when it is being inflated. The pressure is one atmosphere until the rubber is stretched. Then the external P has little effect on the pressure unless the strength of the rubber is exceeded. A rigid container will not respond to a change in external pressure.
The atmosphere is a nonrigid container. Air is held in place by the pressure of the surrounding air and by the downward component of gravity. There are two components to the pressure the random motion of molecules and the downward component of gravity. Air heated by conduction from the ground increases its molecular motion and pressure. It will expand and do work against the atmosphere as it expands. This results in an apparent lower density and, since the gravity component of the pressure is less, a faux lower pressure, resulting in wind. If the air is above warm water, there is an added factor. The evaporating water molecules increase the n in PV=nRT increasing the pressure. Less massive water molecules move faster and the expanding air again has a lower gravity component and lower density and winds result. The added complication is that the adiabatically cooled water vapor condenses releasing the heat of vaporization increasing the wind.
The gas laws do apply to these dynamic situations. The math must be some complicated set of differential equations.

Pressure of air inside the bag of chips > The atmospheric pressure at high altitude.
This imbalance cause the bag to expand, but (this is Boyle's law) as the bag expands the pressure of air inside the bag falls ... until it becomes equal to the atmospheric pressure.
It seems a bit counterintuitive because a balloon expands with "increased pressure", but Boyle's law applies to a fixed mass of gas and blowing a ballon is basically adding more air (increasing the mass of gas).


text:
Understanding Hybridisation in Cyclic Compounds
I had a doubt while solving this problem.
At first glance of these compounds, I felt that all of them have sp3 hybridisation (for nitrogen), since all of them had 3 bonded pairs and 1 lone pair around the nitrogen atom.
But then I saw in a website that both 2 and 3 have sp2 hybridisation. Can you explain why this occurs?

Only (3) - it's conjugated, 2 is not.

The design of the question offers you four choices where choice 4 states nitrogen in both case 2 and 3 (formally) were $\text{sp}^2$ hybridized.
Case 3 however differs from case 1 and 2 by that the lone electron pair of nitrogen no longer is isolated (as in open chain/alkyl amines), but is in conjugation with both $\ce{C=C}$ double bonds.  Because the skeleton is cyclic, of about planar geometry, and the total of participating $\pi$ electrons equals to 6, pyrrole is an electron rich Hückel aromatic system ($4n + 2$, here with $n = 1$).  Thus, 3 it is the only example among the structures displayed where nitrogen is $\text{sp}^2$ hybridized.*  Because the orientation of the lone orbital of nitrogen affects the chemistry (e.g., $\text{p}K_a$), I recommend an additional reading about e.g., imidazole:

(image credit to chem.libretexts.org)
* Actually, atoms do not hybridize.  Hybridization is one of the mathematical concepts in chemistry, here: how to mix atomic orbitals to yield molecular orbitals of certain shape.

Only option three would have $\ce{sp^2}$ hybridization for nitrogen as it’s Nitrogen has π-bond in conjugation. Therefore the lone pair of nitrogen is delocalized thus in the newly formed resonating structure Nitrogen does not have 1 lone pair and 3 $\sigma$-bonds but 3 $\sigma$-bonds and one π-bond… Giving $\ce{sp^2}$ hybridization.


text:
Lithium aluminium hydride
Lithium aluminium hydride is an important reducing agent in organic chemistry.
Since I cannot find the half-equations telling how this reducing agent performs its action, I am wondering, is lithium a dummy role in redox reactions involving the reducing agent? Or does it contribute to the stability of the compound?
So, are compounds like sodium aluminium hydride available to perform the same thing?

There is also sodium analogue, but cation makes some difference. It coordinates to hydrides in the compounds and also to lone pairs in org. compounds.

Sodium Aluminium hydride has different solubility characteristics to LiAlH4 which somewhat reduces its utility but it does similar reactions  https://en.wikipedia.org/wiki/Sodium_aluminium_hydride

Contrasting to reductions with elemental sodium (or other metals), the reduction by $\ce{LiAlH4}$ is not an electrochemical one by mere electron transfer only.  Instead, the complex anion ($\ce{[AlH4]^-}$) acts e.g., on the $\ce{C=O}$ double bond of a carbonyl group as a $\sigma$-bond nucleophile to yield an alcoholate.  The subsequent (light acidic) aqueous workup protonates to eventually yield an alcohol:

Mechanistically similar is the reduction with $\ce{NaBH4}$ (and related compounds, e.g. $\ce{Na[BH3CN]}$, sodium cyanoborohydride) where again the reduction is not by sodium, but by the hydride.


text:
Does a slippery liquid leak through pipe fittings?
We use a 6% potassium hydroxide (KOH) solution for electrolysis.  This solution is known to be "slippery".  Does this mean that it will leak/seep through NPT pipe threads more than plain water?

@JonCuster, it's not slippery becuase it removes oils, but because it saponifies the fats and oils, changing skin to soap. That is not at all good for tissues. Sorry if this comment sounds caustic.

@DrMoishePippik - sigh, yes, you are correct. But because it is messing with those oils it is also ripping them out of the near surface layers...

Potassium hydroxide ($\ce{KOH}$) makes skin feel slippery because it saponifies oils in your skin, changing them to soap. It is called caustic potash because of the damage it does to tissue.
However, though $\ce{KOH}$ is caustic to skin, it does not affect most metallic pipe or, at 6%, PTFE ("Teflon") tape dope, nor is it much more likely to leak out (it may lower surface tension somewhat, so leakage would be faster if there is a hole). It could cause leaks if there are any glass, aluminum, brass or other materials which it attacks. See this corrosion chart from Dixon.
BTW, PTFE does dissolve in potassium, and other alkali metals, in liquid ammonia, $\ce{NH3}$. See O. Lanzi, S. Godfrey answers.


text:
Would the partial pressure of O2 in the system affect the rate of the decomposition of H2O2?
In my Bio 1 lab we studied the way catalase catalyzes the decomposition of H2O2 and at the end it asked me what conditions would affect the rate of the reaction. It was the type of question that obviously wants you to answer in a specific way, in this case, that a high pH or high enough temperatures would denaturalize catalase since we discussed that in class. I turned it in with that but now I have the question of whether LeChatelier's principle applies here. The decomposition is not a reversible function, so it doesn't really reach an equilibrium, but I couldn't help but think that a high enough atmospheric O2 pressure would "push down" on the H2O2 and not allow it to release any more oxygen.

The simple answer is that the equilibrium of this reaction is shifted too far towards the products to reverse with any ordinary amount of pressure.
For a very rough back-of-the envelope estimate, the decomposition of hydrogen peroxide has $\ce{\Delta G} = -120\ \rm{kJ/mol}$ at standard temperature and pressure. Oxygen is not an ideal gas, but let's just use that as an approximation for the moment. Let's also pretend that the $\ce{H2O2}$ and $\ce{H2O}$, being liquids, are completely unaffected by the pressure change.
Increasing the pressure of an ideal gas by a factor of 10 changes its Gibbs free energy by $nRT \ln(10) = 5.2\ \rm{kJ/mol}$. So you would have to increase the pressure by a factor of more than $10^{23}$ before the reaction became endergonic. That's somewhere in the stellar core or white dwarf  pressure range. At those pressures, some unwanted side reactions might happen... specifically, unwanted nuclear side reactions.
Clearly we've left the realm of validity of our assumptions from the prior paragraph. Still, it's safe to conclude increasing the pressure is not a practical method of reversing the decomposition of hydrogen peroxide, even to a small extent.


text:
Where can I find large glass covered stir bars?
I am looking to use a large stir bar, on the order of 100mm, that is not coated with any Teflon or PTFE. This is because my application is measuring PTFE in water and a stir bar can effect the results. I cannot seem to find one over 38mm. I would be willing to use any other material than PTFE/Teflon.

What about searching for longer but narrow rod/cylinder magnets and sealing them in a glass tube, done by glassware workers or using a lab gas burner?

The trouble is that magnets made of steel are not magnetic any more at higher temperatures than Curie point ($773$°C).

Why should that be a problem? The magnet would not be heated. Alternative is e.g. LDPE coating. // The question is rather if there is a big enough and strong enough stirrer and/or what is the expected length of the magnet itself. If the rotating magnet of the stirrer is too short, the stirrer may not be able to power the stirring.

You don't describe the setup in enough detail (e.g. container is a flat beaker, or round bottom flask, temperature/pH/composition of the medium). At larger scale, and especially for more viscous media, overhead stirrers are more efficient than stirring rods (without particular endorsement, examples of such a «drilling machines», and mounts, some can be manufactured of borosilicate glass, too). A 10cm rugby egg can shatter your flask quickly.

@Liz Be aware that what is being asked for often differs a lot from what is really needed. But what is really needed cannot be determined without detailed knowledge of the problem scenario.

Fisher Scientific have glass-covered stirrer magnetic bars up to 60mm  https://www.fishersci.co.uk/shop/products/glass-encased-stir-bars-3/p-8007716

As @Poutnik states, it's very simple to make such a stirrer.

Obtain one or more magnets (such as this) to make the needed length.
Obtain glass tubing of appropriate inside diameter. Soft glass is easier to melt, but prone to shatter for large pieces due to thermal stress. Borosilicate glass can be melted with a propane torch, with patience.
Cut the glass tubing about 15 or 20 mm longer than the magnet.
Melt one end of the tube shut.
Insert magnet, possibly wrapping in plastic wrap or paper to cushion it and to keep from sliding readily. Push it into the tube so that it is at least 10 mm from the open end.
Melt opposite end shut. The heat should not affect the magnet, just 10 mm away inside the tube, because glass is a poor conductor of heat.

Such a stirrer would be more fragile than a commercial product, but is far less expensive, and can be re-sealed into another piece of tubing, if broken.
You might look at some classic book on lab glass work, such as Bolas, Frary or Shenstone. Glass blowing (flame working) is useful in chemistry, and can create better equipment at lower price than can be bough commercially.

En route to check if there is a scanned copy of the references (there is e.g. an entry on archive.org), I notice your link to the book by Frary (this one?) and Shenstone (this one?) equally use the ink to the book by Bolas on gutenberg.org.  Unclear if intentional, or accidental, feel free to add the links to archive.org.

@Buttonwood, now corrected; all available gratis on Gutenberg. Thanks for catching those errors!


text:
Recovery of antimony trichloride
I have bought 250 g of old antimony trichloride bottle (very cheap), it was never opened because it is sealed from manufacturer. But it was probably not stored well, or seal is not good, because there is a liquid above crystals. I have read at Wikipedia that with air moisture it can react and produce antimony oxychloride and hydrochloric acid (hydrolysis):
$$\ce{SbCl3 + H2O → SbOCl + 2 HCl}$$
Is there any method for recovery of $\ce{SbCl3}$?

"Antimony trichloride is best purified from old or partially hydrolyzed reagent by subliming it under vacuum and condense it on a cold finger. " from https://www.sciencemadness.org/smwiki/index.php/Antimony(III)_chloride

Thank you, this will separate SbCl3 from other stuff in mixture, but is it possible to convert also SbOCl or other products of hydrolysis back to SbCl3 somehow? Or is it worth to extract elementary antimony at least?

99.8% pure Sb metal is available for ~US$75/kg, https://www.amazon.com/Antimony-Pound-Metal-MS-MetalShipper/dp/B01F5SBAEG/ref=sr_1_2 . Though it might be an educational experiment to extract it, economically, not feasible. Consider in the 250 g, about half is Sb, and consider cost of reagents.

@Tom  Antimony trichloride can be prepared from Antimony oxide and hydrochloric acid (prep here https://www.prepchem.com/synthesis-of-antimony-trichloride/). I think this should work for the oxychloride as it must be an intermediate in the reaction.


text:
Reason behind paramagnetic substances having color
I've read in many books that paramagnetic substances are coloured while diamagnetic substances are colourless (yes there are some exceptions too). But nowhere i could find a reason for this phenomenon. I have an explanation so please tell me whether it is valid.
Paramagnetic substances have an unpaired electron. So the energy required to excite an unpaired electron is relatively less. Due to this the energy of visible light is sufficient for excitation and hence we are able to see the color when electron de-excites.
Please let me know if this is correct.
Thank you.

The generalisation that paramagnetic compounds are coloured compared to diamagnetic ones is not a good generalisation. Consider, for example,  that manganese sulphate (paramagnetic) is white to pale pink but potassium permanganate (diamagnetic) is an intense purple colour.

True but in that case the colourisation is due for the color of respective anions, correct? In a situation where there is a colourless anion, is my explanation valid?

There is IMHO rather very weak correlation than causation. Paramagnetic molecular entities have unpaired electrons, that MAY BE more inclined to absorb visible light due smaller difference of excited energy levels. It is far from being a rule.

@matt_black Additional example: the nice blue crystals of copper sulfate pentahydrate, $\ce{CuSO4 * 5 H2O}$  vs the white/gray powder of copper sulfate without any crystal water, $\ce{CuSO4}$.

@GnanadeepSai I chose my example to illustrate that the colour of the ion containing the metal doesn't depend strongly on the magnetic state. The specific colour of a metal cation or anion varies a lot and doesn't correlate much with unpaired electrons. Also true for many non-metal compounds. Nitric oxide (NO) is not coloured despite being paramagnetic, as is oxygen (though very faintly coloured as a liquid).

@matt_black I get it now, thank you.

As pointed out in the comments, this is not a good generalization.
You correctly stated that paramagnetic complexes have (at least) one unpaired electron, while diamagnetic ones have a closed shell. Your idea of colour is incorrect, what you describe is fluorescence. Generally speaking, we do not observe color because a compound emits light of a certain energy. We rather observe what is left from the light that shines on a compound. The absorbed part is the part we cannot see anymore.
Whether a complex is coloured or not depends on the energy difference of the ground state and possible excited states, and whether a transition between these is spin-allowed or not. If there is a spin-allowed transition that has an energy in the range of visible light, we observe a colour.
Consequently, there is no inherent connection between colour and magnetism.
I can, however, imagine how this coincidental correlation might arise: In complexes with an even number of d electrons, there is always the possibility to have a paramagnetic or a diamagnetic ground state configuration, e.g. the $d^8$ example in the diagram. When the orbital energy difference $\Delta\varepsilon$ is sufficiently small, the paramagnetic configuration will be the ground state. If it is large, the diamagnetic one is the ground state. The energy difference $\Delta\varepsilon$ also correlates with the energy of an electronic transition: the larger the difference, the shorter the wavelength we observe in a spectrum.
I assume that the energy difference at which the diamagnetic spin state becomes the ground state lies in the region where electronic transition have such high energies that their absorption does not happen in the visible region anymore and thus the complex appears colourless. This would justify what you found by connecting the electronic transition energy with the magnetism via an orbital energy difference.
I want to emphasize that an orbital energy difference is not the same as a state energy difference and serves purely as conceptual aid. Neither orbitals nor their energy differences are observables.

Okay so there is no relation with magnetism but we can conclude from the energy difference between orbitals?

Yes, you can get an estimate of the colour of a complex from the orbital energy difference, if I correctly understood your comment.

I can't believe I'm the only other person to give your post a +1. In any event welcome to the community, you seem good at this, can you possibly attempt my question on colors? https://chemistry.stackexchange.com/questions/172258/colored-inorganic-liquids-why-is-scl2-red-and-s2cl2-orange


text:
Can zinc carbon batteries be made in the way gen did in dr. Stone?
So in episode 23 of the anime dr. Stone https://www.youtube.com/watch?v=faQ2EVj7SrQ a character named gen was making zinc carbon batteries by using manganese and I’m pretty sure it wasn’t manganese dioxide, just elemental manganese. He put the manganese directly on a zinc sheet and wrapped it around it leaving the top open, and then inserted a carbon rod and covered the rest of the manganese with a ring of something and added what i think is some adhesive on that ring to keep the manganese from falling out, at least that is what I understood.
Is this what he actually does or did I get something wrong?
Will it work?
Can zinc carbon batteries work using manganese instead of manganese dioxide?
Usually there has to be some sort of insulation between the zinc and manganese right? He didn’t, he put the manganese directly on the zinc, would it work? Regardless of whether it was manganese or manganese dioxide, would it work without any insulation?
And there was no electrolyte, would it work without it? And if not, does the electrolyte have to be ammonium chloride or can i use any thing else like sodium chloride?

Why have you thrown out of the window everything you know while watching it? Think what is in the zinc-carbon battery, why it is there and what happens if it is not there.

@Poutnik I know that in normal zinc carbon batteries  zinc is the anode and gets oxidised while manganese dioxide gets reduced and carbon plays as a cathode but doesn’t get involved in the reaction, not sure if it would work with elemental manganese. And not really sure about the electrolytes either, that’s why I asked

Contact of 2 metals can provide several mV of contact potential at the best.

@Poutnik so you mean that without insulting the manganese and the zinc not much sufficient voltage would be generated ?

You could make a thermopile with Zn and Mn in contact at ends, but if in direct contact along edge, this is called a short circuit, with virtually no voltage across it. Some cartoons went to effort for accuracy, but not that, if you're reporting correctly.

@DrMoishePippik so you mean that without insulating the zinc from the manganese the battery wouldn’t work, and that the show was wrong, right?

From your description, correct. An electrolytic cell requires a spacer. E.G.: Zn-paper-Mn. For a battery, Zn-paper-Mn-Zn-paper-Mn-Zn-paper-Mn... the Zn and Mn do contact, but not immersed in electrolyte.

The filling inside the zinc foil is black, it's definitely manganese oxide and not manganese metal. And since the manganese oxide holds its shape instead of crumbling to dust, it's not hard to assume it's already moistened with electrolyte. With a bit of googling, you can find someone who made a functioning cell based on this design, though they do use a paper spacer.

For a battery to work, there has to be a chemical reaction.
If the battery is made of zinc, manganese, and carbon, these do not react with each other. So not only is this not a working battery, it's not even a working hot pack.
If the "manganese" is actually manganese dioxide (that's what it looks like in the video), it would be able to oxidize zinc, and a reaction could occur. However, the reaction can only occur at the $\ce{Zn}$/$\ce{MnO2}$ interface, producing boring old heat. Electrons can't be pushed down a wire to do useful work, because that would create a charge imbalance between the anode and cathode. Without an electrolyte, there is no way for ions to flow to fix that imbalance.

Without any medium for ions to flow such as an electrolyte that seems to be missing, electrons won't move.
On the other hand, if elemental manganese was there, there is absolutely no chance of any reaction to occur as zinc and manganese don't directly react, which is obviously needed.
If Manganese dioxide ($\ce{MnO2}$) is used, then only heat will be liberated which is not what a battery is made for.
There is some effort here to make it seem like an actual battery but unfortunately, it's not and thus these can't be made.

Chem+Math Expression formatting reference:
MathJax Basics / Chem+Math expressions/formulas/equations / Upright vs italic / Math SE Mathjax tutorial
// MathJax is preferred not to be used in CH SE Q titles.


text:
How does a pyrrole react with its C2?
First of all, thank you for reading my question. There is a mechanism I read where the pyrrole reacts with its C2 but I can't draw the arrows so I don't know if I am correct to think that :

the ester: is a deactivating group because it has a Mesomeric electron-withdrawing effect (M-) so it deactivates the C4
the methoxy: is an activating group because of its electron-donating effect (M+) so it activates the C4
--> and M+>M- so it will react on C4
But the pyrrole likes to react on its C2 but it is deactivated by the ester (because in para)
Please explain me how does it work I am very lost...

All of your mesomeric effects may serve to slow the rate of the reaction but reaction occurs as shown.

Actually I think I understood why I was lost, because I was thinking of the reaction like it is SEar but here it's in SNar, right ?

It sounds to me that you are memorizing too much and not going with the facts. You're given the answer. Now "push mechanistic arrows" accordingly.


text:
Why does alpha-carbon substituents increase the rate of E2 reactions?
The reason for more reactivity of more substituted alkyl halides for E2 Reaction is said to be due to the more stability of its more substituted alkene character having transition state but here (https://home.iitk.ac.in/~madhavr/CHM102/Lec13.pdf) at page 6 it is mentioned that:

The reaction rate increases with the increase in number of alkyl substituents on the carbon with the leaving group ($\alpha$-Carbon)

However, it further says that the reason is the relatively more stable transition state (which has partial double bond) due to the more alkyl substituents (probably Hyperconjugation effect). But my question is that Will not the stability of the same transition state increase if the $\beta$-carbon is also more substituted? Why it specifically mentions for the $\alpha$-carbon to be more substituted?

You are on to something. Check out the product ratios here: https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkyl_Halides/Properties_of_Alkyl_Halides/Introduction_to_Alkyl_Halides/Elimination_by_the_E2_mechanism. The halide with 2 beta methyls and only one proton gives the same product ratios as the halide with one despite one less proton albeit not equivalent. The original kinetic data should be consulted. Substitution at the beta carbon changes steric and entropic effects. [An error in the first example in the link, cyanide is a weaker base than carbonate but a stronger base than bicarbonate.]

From your provided page, I can only understand the same thing that the transition state has double bond character and that's why "alkene stability differences will be reflected in the transition states of elimination reactions, and therefore in the activation energy of the rate-determining steps". It doesn't answers the question about the effect of beta carbon substitution. But confuses more on the point that if there are e- withdrawing groups at beta then it should help C-H breaking but donating grps stabilize TS. And "Substitution at[...]entropic effects" should only matter for bulky bases.

Also the page is hard to follow due to poor readability and styling:')

Electron withdrawing on the beta carbon increase E2 rates On the alpha carbon increase rates by about half of the beta [My PhD thesis unfortunately not published].  Read between the lines a bit. With one beta the Hs are not equivalent one gives cis the other trans with different rates adding a second group removes one of the rates  but the product ratios ae the same suggesting  the rate for one H has increased. research similar situations


text:
Why does a hydroxyl group decrease the acidity of an Aromatic ring?

We are currently learning about resonance and induction, but I am confused why the addition of the hydroxyl group increases the pKa. As oxygen is more electronegative, should it not increase the acidity via induction? If it is to do with resonance, why doesn't the acidity also decrease when the 4th carbon is bonded to a chlorine; should this not be able to donate electrons via resonance as well?

Consider the effect of increasing electron density of the ring, what boosts both electrophilic substitutions and basicity of the conjugate base.

Assuming you’re referring to the acidity of the protons on the aromatic ring:
The OH is an electron donating group by resonance. By deprotanating your forming an anion on the ring. As the ring is increased in electron density, the anion is less stable than if there was not an electron donating group.
By having a less stable conjugate base, the aromatic ring is less willing to loose a proton/less acidic


text:
Cyclopentadiene planar
Cyclopentadiene has a sp3 carbon yet it is said to be planar. I am not able to understand and haven't been able to find a credible source. Some sources say it to be non-planar

Can you give a reference that says cyclopentadiene is planar? The Cyclopentadienyl anion is planar, see  https://en.m.wikipedia.org/wiki/Cyclopentadienyl_anion, but not to my knowledge the neutral molecule. The difference is important

https://www.sciencedirect.com/science/article/abs/pii/0166128083800219 here read the abstract. There are several like this

Quoting from the given reference "Cyclopentadiene has a planar ring".  That is not the same as "Cyclopentadiene is planar" - it means the 5 carbon atoms are all in a plane, not the whole molecule. Also note that is an old paper which uses quite primitive methodology by the standards of today.

But isn't this correct that Each element within the ring must have a p-orbital that is perpendicular to the ring, hence the molecule is planar.  This means that the ring cannot contain a neutral sp3 carbon.

"Each element within the ring must have a p-orbital that is perpendicular to the ring, hence the molecule is planar." I don't follow this logic at all. In fact I don't even know how to respond. What stops the 2 hydrogens on the unique carbon being out of a plane defined by the 5 carbon atoms (assuming planarity)? Also I say again the theory you referred to is quite primitive by today's standards, and even wikipedia notes that the barrier to a non-planar geometry is small (see the Diels-Alder section) , thus that paper may well be incorrect.

Yeah, agree with you. This is not logical yet it is said so that it is easy to check the planarity of a six-membered ring. This also helps - https://socratic.org/questions/how-would-you-know-if-a-cyclic-organic-molecule-is-planar

Last thing I will say - https://pubchem.ncbi.nlm.nih.gov/compound/Cyclopentadiene might help, click on animate in section "1.2 3D Conformer". The carbon ring is/may well be planar. The molecule as a whole is not due to the hydrogens on the unique carbon. The reference you gave only claims the former, not the latter.

But here won't the carbon be strained as it is sp3 and and the bond angle is too large

Analysis just based on ideal local geometry

[OP in comments] But here won't the carbon be strained as it is sp3 and and the bond angle is too large

In a regular pentagon, the angle at each corner is 108 degrees (tracing the pentagon, you have to turn 72 degrees at each corner: 360 / 5 = 72). In a regular tetrahedron, the angle is about 109 degrees. So just considering the bond angles, cyclopentane could be flat, while there is a good reason for cyclohexane to buckle and assume a chair conformation, again based on bond angles alone.
In pentadiene, you have four carbon atoms that should have a planar geometry with bond angles of 120 degrees. The fifth carbon atom, just considering the ideal geometry, then also has to be in the plane. The angles don't fit a perfect 360 rotation along the ring, so there will be some modest ring strain if the carbons are in a plane, or some non-ideal geometry around the double bonds if the carbons are not in a plane.
Analysis taking into account neighboring carbons
It turns out that there is no free rotation around single bonds. Ethane is staggered more often than it is eclipsed. This is the reason that cyclopentane is puckered (envelope or twist conformation). The paper mentioned in the comments to the question states that cyclopentane is in an envelope conformation, cyclopentene can become completely flat without a large energy penalty, and cyclopentadiene is flat (in terms of the carbon atoms, as nicely discussed and illustrated in the other answers).

There are two confusions here
The first source of confusion is the caused by the fact that the most common occurance of the five membered cyclopentadienyl ring isn't cyclopentadiene but the cyclopentadienyl anion commonly found in organometallic sandwich compounds like ferrocene. Her, the cyclopentadienyl unit is flat (and aromatic like benzene) as is the free ion. So many leap to assume that the this is what is being referred to if someone claims the free, neutral, cyclopentadiene is "flat". but this is a distraction and not really relevant to whether the neutral molecule is flat.
The second source of confusion is the assumption that sp3 carbons must render the molecule non-planar. But there is only 1 sp3 carbon in the 5-membered ring, the other 4 carbons are sp2 hybridised and are conjugated so driving the whole ring to be planar. This is easy to see if you build a model as non-planarity of the ring would demand that two of the sp2 carbons had angles to the sp3 that would not be "flat".
Pictures might help clarify this. A simple skeletal drawing gives some insight:
It isn't the ring that is non-planar, it is the hydrogens| attached to the sp3 carbon. The planarity of the 4 sp2 carbons "holds" the sp3 carbon in a plane as there is nothing forcing the sp3 carbon out of the plane defined by the other 4 carbons.
This might be clearer in a 3D picture. Here are two views of the 3D version, the first as a space filling view:
The second, a ball and stick 3D model viewed from the side:
If you built this model with a standard physical modelling kit, it would also be obvious that the conformations of the ring where the sp3 carbon was out of the plane of the other 4 carbons was more strained.
Neutral cyclopentadiene is flat: it is only the hydrogens attached to the sp3 carbon that don't sit in the plane of the ring.


text:
Anti-bumping granules
Anti-bumping granules are usually small stones. I presume that the principle of these granules ensuring smooth boiling, is that it blocks the upward motion of solution molecules.
Usually the stones have an irregular shape. I am wondering, if the anti-bumping granules are in regilar shapes, like a sphere, a cube, will there be some influences to the anti-bumping action of the granules?

The idea is to provide nucleation sites, so reasonable surface area and surface roughness are desirable. Small bubbles have higher internal pressures than large bubbles, other things equal, so Henry’s law works against the initial formation of bubbles. But tiny cavities in the surface of the “boiling stones” allow the bubbles to get initiated. You can see this in an old beer glass with beer in it: small bubbles stream up from such tiny cavities or rough spots in the glass.

Cross posted: https://physics.stackexchange.com/q/781455/313612.

Nucleation (for boiling, effervescence,  or crystallization) can also be induced by sound, e.g., the "clicker" in sodium acetate hand-warmers... or for a fun geyser, put a cold, freshly opened bottle of soda in an ultrasonic bath or ultrasonic humidifier reservoir. Also, scratching a glass container with a glass rod produce both sound and imperfections to nucleate crystallization, https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/03:_Crystallization/3.05:_Procedural_Generalities/3.5E:_Initiating_Crystallization .


text:
Why do the dipoles in dichloroethane point in opposite directions?

I understand why the dipoles would align in a polar solvent to maximise dipole-dipole interactions but wouldn't the replulsion due to like charges destabilise the molecule? Is it a MO effect instead? An undergraduate level response would be appreciated.

Suppose you just have H-Cl molecule, can you draw the dipole moment directions?

Yes, they're aligned, in the same direction.

No, what is the direction for H-Cl's dipole moment? There is only one dipole moment so the question of alignment does not arise. Just think about H-Cl. What is the direction of dipole moment arrow? It is a vector, right?

Yes it starts at H and ends at Cl

When dipoles are free to move, all else being equal, they will tend to align "tail-to-tip". That is the case, for example, when the dipoles belong to different molecules in a liquid. But the two polarized bonds within a single molecule of dichloroethane are not free to move to arbitrary positions. The two carbons are stuck together by a bond, and the only degree of freedom is rotation around that central carbon-carbon bond.
This rotation doesn't change the distance between the two positive carbon atoms. It also doesn't change the distance between either carbon atom and either chlorine atom. So the only relevant effect of this rotation is to change the distance between the two negatively charged chlorine atoms.
Therefore, the most stable conformation is the one where the negatively charged chlorines are furthest apart, since that minimizes the repulsive force.


text:
Why is sodium zincate two sodium ions and two oxygen ions but only one zinc ion?
$2$NaOH + Zn ---> Na2ZnO2+H2
I have always been fascinated by this reaction because it never made sense to me why sodium zincate has such a strange composition. Why does it have two sodium and oxygen ions but only one zinc ion? What is really happening here?

Who says it has oxygen and zinc ions? BTW, it is more commonly $\ce{2 Na+ + [Zn(OH)4]^2-}$

Your answer is false.

Care to respect the community rules? BTW, you confuse answers and comments.

The reaction of metallic zinc $\ce{Zn}$ with $\ce{NaOH}$ may happen in the two successive reactions :

Oxidation of $\ce{Zn}$ from $0$ to the usual +$2$ values in a redox reaction due to the extremely rare $\ce{H+}$ ions present in basic solutions
$$\ce{Zn + 2 H^+ -> Zn^{2+} + H2}$$
Reaction of zinc ions $\ce{Zn^{2+}}$ with $\ce{OH^-}$ according to one of the two following equivalent equations :
$$\ce{Zn^{2+}  +  4 OH- ->  ZnO2^{2-} + 2H2O} $$
$$\ce{Zn^{2+}  +  4 OH-  ->  [Zn(OH)4]^{2-}}$$
as the final zincate ion may be witten $\ce{ZnO2^{2-}}$ or $\ce{[Zn(OH)4]^{2-}}$ according to the chosen handbook. Both formula are found in the literature.

Shouldn´t the reduction of zinc in basic media be written as $\ce{Zn + 2H2O -> Zn^2+ 2OH^- }$?

@PAEP. It is of course possible to replace my first equation by the following one, which is an oxidation of zinc (and not a reduction) similar to yours : $\ce{Zn + 2 H2O -> Zn^{2+} + 2OH- + H2}$

How does this answer my question?

Your question was : Why does it have two sodium and oxygen ions but only one zinc ion? The reason is that one $\ce{Zn}$ atom must react with two $\ce{O}$ atoms. If one $\ce{Zn}$ atom reacts with only one oxygen atom, it does not produce an ion. It forms an oxide, which is $\ce{ZnO}$, but this oxide is not an ion and does not contain any sodium atom.


text:
Purify ethanol from benzene
In reference to my previous question: Separation of methyl acrylate from ethanol
I have successfully get rid of acrid odor by treating ethanol with sodium hydroxide and filtering out the precipitate, but after distillation, there is "thinner" odor, possible benzene, toluene or similar aromatic hydrocarbon. Also, the remainder after distillation has orange color (before distillation it was pale yellow), which is possibly due to tetracene.
The question is the same, how can I get rid of these aromatics, probably benzene, easy way? Distillation is here not very good method because of similar boiling points (80.1 °C benzene, 78 °C ethanol).
I also do not know how can benzene get into ethanol, maybe it is by-product of previous reaction with sodium hydroxide with something in UV resin?

Now I am not quite sure it is benzene as contaminant (or maybe it is not the only one). For some time at the beginning of distillation, I observe something is boiling at 74 °C, +- 1 °C. and after that it raised to 78 °C for ethanol. What could it be, if it has "thinner" odor?

The melting points of benzene and of ethanol are quite far apart: ~279 K and ~159 K, respectively. Also, solubility tends to decrease at low temperature.
On a small scale, greatly chilling the mix to separate it would not be economically feasible, but on a large scale, an efficient countercurrent heat exchanger would require little energy to chill, separate and then return the effluents to room temperature.
In cold climates, applejack and slivovitz is often made by "freeze distillation".
BTW, many warm-blooded animals have a rete mirabile, a system of blood vessels acting as a countercurrent heat exchanger to get oxygen to limbs, while retaining heat in the body core.

Thank you for answer, these are all physical methods, but is there any chemical method to separate them? I mean to add reagent(s) to mixture which will react with benzene but not with ethanol and then it can be easier to separate via filtration, sedimentation etc.?

Anyway, I will try this "freeze distillation". I will simply put the  mixture in freezer and try to separate solid parts via cold filtration, which can be done in freezer as well. Is that correct?

I am still curious about this "freeze distillation", I think it would not be so easy, because otherwise it would be enough for absolute ethanol to freeze e.g. 50% ethanol and separate away the ice from water. I think that mixture will have different behavior and different freezing point. Same could be for mixture of ethanol with benzene.

Benzene ethanol have a low boiling azeotrope about 10degrees lower you are possibly seeing this get a better column. Apple jack is made by keeping a barrel of hard cider outdoors The barrel freezes from the outside in by Spring there is a liquid core with eutectic ethanol and apple flavorings in the center ready to be tapped. This worked in the northern climes and may be a lost art with global warming. Freezing probably will not work check the ethanol-benzene phase diagram to see why. [or if it will work]

@jimchmst, not shuah its a lost aht... down east, them revenuahs don't see a still  ;-)


text:
Substitution or elimination when a chloroalkene reacts with NaOH in ethanol?
NaOH + EtOH will eliminate the Cl atom forming a double bond. At least, that's what I think. 3 could also be a viable answer since the OH can also attack the said double bond (this is probably not correct) and cause an OH to be added forming the compound in 3.

This is an example of the subtle difference between different bases that are strong in water but may not be equally strong in other, less polar solvents. Alcoholic potassium hydroxide certainly does eliminate a proton from the substrate leading to product (4), but sodium hydroxide in ethanol is just a little weaker and uses nucleophilic substitution to form (1) instead.

Supporting data? Perhaps 4 less ring strain and conjugated double bond. This is a better to see the products and determine the mechanism than quess the mechanism and guess the products.


text:
What is the rate of hydration in the following compounds
I came across a question :
What is the order of rate of hydration in the following compounds:

My attempt to this question was checking the stability of carbocation. As III has a possibility of 3* carbocation with dancing resonance (hence most stable) it would have the highest rate.
How do I decide between (I) and (II) as both have 2-degree carbocation possibility with dancing resonance
Carbocation formed at (I) would have 4 alpha Hydrogen whereas carbocation formed at (II) would have 3 alpha hydrogen hence according to this logic I am getting the order II<I<III.
But we can also argue that (II) will have 2 positions at which carbocation can form hence higher rate that (I).
Which factor would dominate and why?
(PS: answer given in my book is I<II<III)

This could be a tricky one. I'm not sure if sterics wouldn't hinder delocalisation on cyclopropyl. Huh, why do q. about this are asked so much - for me it's pretty exotic.

The fact that II can form two secondary carbocations would be the most significant factor as it provides a second mechanistic pathway which would drastically speed up the reaction.
Also notable is that:

In II the double bond is more electron rich by hyper conjugation and hence will more easily take a proton so the carbocation will form faster (and carbocation formation is the rate determining step)

In II by forming the cation of the right carbon of the double bond, the cation benefits from having four neighbouring C-H sigma bonds to stabilise it by hyperconjugation.