text:
Why is the electron-nucleus attraction modelled with only electrostatic interactions?
While reading about the structure of an atom, I've encountered (even in some renowned books) the statement that electrons and nuclei are attracted due to electrostatic, or Coulombic, attractions.
However, since the charged particles (electrons and nuclei) are moving, how can the Coulombic force be sufficient to describe their interactions? Surely electrostatics cannot be used in this context. Is it merely an approximation?

Have a look at the formulation of Coulomb's law. Do you see anything there that would result in a change in the magnitude of force between charges had a charge been in (relative) motion with respect to the other? Even if the one of the charges move, it doesn't affect the magnitude of force between the two particles. People normally make a distinction between interactions involving charges at rest, and charges in motion...because the latter is associated with varying electric fields, which would produce a magnetic field as well ;)

It's also relevant to note that the standard treatment of electrons around nuclei is static, and comes from solving the time-independent Schrödinger equations. The standard QM depiction of atoms is stationary (and delocalised).

If I understand the question correctly, OP is somewhat surprised that Coulomb's law is used to describe the interaction between an electron and a nucleus, although it is usually pictured that electrons are moving and Coulomb's law describes interaction between static particles. Should not then the Lorentz law be used instead Coulomb's one?
First note, that electrons do not move around a nucleus in an atom. At least, they do not do so in the classical sense. Yet, a non-zero orbital angular momentum $\vec{L}$ of an electron in an atom gives rise to the orbital magnetic dipole moment $\vec{\mu}_{L}$ and so the magnetic field is indeed generated by the "moving" electron. Plus, an electron has an intrinsic magnetic moment (spin) which also contributes to this magnetic field. The nucleus possibly possess orbital magnetic momentum as well (if has non-zero angular momentum) that couples with its spin which interacts with that of electron. These interactions splits the energy levels of the atom and the resulting hyperfine structure can be measured using radio-frequency spectroscopy.
And in principle, for a complete description of an atom we could use a model in which we treat the interaction between electron and nucleus taking magnetic interactions into account, i.e. similarly to how it is done in classical electrodynamics with the Lorentz law. There is a problem with the Lorentz law though - it is formulated in terms of forces and velocities and this terms are not even present in the vocabulary of quantum mechanics. So, the magnetic field must appear somehow differently in the Hamiltonian, and to make a long story short the full (non-relativistic) Hamiltonian for one electron in an external electromagnetic field is:
$$
\hat{H} = \frac{1}{2m}(\boldsymbol{p} - e\boldsymbol{A})^2 + e\phi \, ,
$$
where $\phi$ is the scalar potential and $\boldsymbol{A}$ is the vector potential. And even in the absence of an external to the whole atom magnetic field, the nucleus itself gives rise to the vector potential $\boldsymbol{A}$.
So, this is certainly true that electrostatic interaction between electron and nucleus is not the whole story. However, electrostatic interaction dominates magnetic ones by far, so that usually the later are just ignored, unless a very high precision is required.

Perhaps better to say that the magnetic interactions are treated as a perturbation atop the zero-order solution defined by the electrostatic potential?

@Wildcat Thanks a lot. Though I'm not in a position to grasp all that you wrote. But surely some points are clearer to me than before. First, that electrostatic forces are not only between "static" charges but even exist between moving charges . Another point is that there are several other factors like magnetic interactions also have a role to play but they are dominated by electrostatic interactions.

It is a misconception that one finds in the physical literature too often. The correct Coulomb energy is not static, but varies with time implicitly as
$$V_\mathrm{Coul} = V_\mathrm{Coul}(\boldsymbol{R}(t)) = \frac{ee}{4\pi\epsilon_0 |\boldsymbol{R}(t)|}$$
with $\boldsymbol{R}$ the interparticle distance. The problem is many people confounds this with the field-theoretic quantity
$$V_\mathrm{field} = \frac{1}{2} \int \mathrm{d}^3 \boldsymbol{x} \int \mathrm{d}^3 \boldsymbol{y} \frac{\rho(\boldsymbol{x})\rho(\boldsymbol{y})}{4\pi\epsilon_0 |\boldsymbol{x}-\boldsymbol{y}|}$$
which is a static quantity, besides divergent (and unphysical). This confusion is not a problem exclusive to quantum theory, one already finds this kind of confusion in classical physics. In Ref. [1] you can find a detailed mathematical proof that the Coulomb potential $\phi(\boldsymbol{R}(t))$ is not reducible to the field-theoretic potentials $\phi(\boldsymbol{r},t)$.
If all of the above was not enough, there are additional misconceptions in quantum theory regarding the time dependence of operators. One finds too often the claim that operators in the Heisenberg picture vary with time, but they do not in the Schrödinger picture. This is not true, and in more advanced textbooks one finds expressions such as
$$i\hbar \frac{d}{dt} A_H(t) = [A_H(t), H_H(t)] + i\hbar \left( \frac{d}{dt} A_S(t)\right)_H$$
where the last term (missing in some textbooks) accounts for the time change of the operator in the Schödinger picture (subindex "S" is the operator in the Schrodinger picture and subindex "H" is in the Heisenberg picture). In what follow I work in the Schrödinger picture and avoid the subindex "S".
The point is that electrons and nuclei, all together attract or repel with non-electrostatic interactions, because electrons are never at rest in atoms.
In the nonrelativistic limit, the interaction between charges is given by the Coulomb potential $V_\mathrm{Coul}$. In the phase space formulation of quantum mechanics one simply uses this expression directly. In the Schrödinger wavefunction formulation one has first to replace position variables with Schrödinguer operators $\boldsymbol{x}\rightarrow \hat{\boldsymbol{x}}$; those operators maintain the implicit time dependence of the classical variables $\boldsymbol{x}=\boldsymbol{x}(t)$, and one obtains a Coulomb operator with a implicit time dependence $\hat{V}_\mathrm{Coul}(\hat{\boldsymbol{R}}(t))$.
This implicit time-dependence is not quoted in the literature, but it is needed to make sense of dynamical variables. For instance if we want to compute the velocity of an electron in the Schrödinger picture, we have first to obtain the velocity operator as
$$\hat{\boldsymbol{v}} = \frac{d\hat{\boldsymbol{x}}}{dt} = \frac{i}{\hbar} [\hat{H},\> \hat{\boldsymbol{x}}]$$
and then use this velocity operator in the ordinary Schrödinger-picture expression
$$\langle\boldsymbol{v}\rangle = \int d^3\boldsymbol{x} \>\Psi^{*}(t)\>\hat{\boldsymbol{v}} \>\Psi(t)$$
REFERENCE
[1] https://journals.aps.org/pre/abstract/10.1103/PhysRevE.53.5373

While there are quite a few answers here already, I would like to approach this without going into complicated theories.
You might have heard that Coulomb's law can't be used in the case where charges move. You were thinking that the Coulomb's law is incorrect for moving charges, but the fact is that Coulomb's law is inadequate to completely describe the forces on moving charges.
Coulomb's law deals with electrostatic attraction. These attractions depend only on the magnitude of charges and the separation between them. However, while charges are moving, different types of interactions come into the picture. Those are called magnetic forces. Like electrostatic forces, their magnitude depend on the magnitude of the charges and the separation between them, but in addition, they also depend on the velocity of the charges, in both magnitude and direction.
Before we start to analyze the magnetic forces, we need to know what a magnetic field is. Magnetic fields, like electric fields, exist and are produced by charges. However these fields are produced by, and can interact with only moving charges. Let's see how the fields are produced a bit later, now I'll show you how a moving charge behaves under a magnetic field.
If we represent the magnetic field by a vector $\vec{B}$, and a charge $q$ moving with a velocity $\vec{v}$, the force the charge experiences is:
$$\vec{F} = q(\vec{v}×\vec{B})$$
The force acts perpendicular to both the field and the velocity.
Moving charges can also produce magnetic fields. The field produced is given by the Biot-Savart law.
$$\vec{B} = \frac{\mu_0}{4\pi}\frac{q(\vec{v}×\vec{r})}{r^3}$$
where $\vec{r}$ is the position vector of the point where you are measuring the magnetic field, with respect to the moving charge as origin.
The combination of both the electric and magnetic fields together are needed in the case of moving charges.

But why aren't the magnetic fields considered in the Bohr model?
Let's see. In the Bohr model, you have a stationary nucleus, and an orbiting electron that follows a circular path. The electron moving in the circular path will be generating a magnetic field, that looks like this:

If you tried to use Biot-Savart law, you should be able to arrive at the conclusion that a spinning electron produces a magnetic field at the nucleus of with a magnitude of,
$$B_{\ce{e-}}=\frac{\mu_0ev}{4\pi a_0^2}$$
or
$$B_{\ce{e-}}=\frac{\mu_0he}{8\pi^2 m_{\ce{e-}}a_0^3}$$
What we have derived above is perfectly valid, and has also been verified. Orbiting electrons do produce magnetic fields, and their magnetic moments have been determined in laboratories. The magnetic moment of a hydrogen atom, is called the Bohr magneton, is a very popular unit to measure magnetic moments in transition metal compounds and coordination compounds.
However, the nucleus is still at rest. A magnetic field cannot interact with a particle at rest, as we have seen earlier. So the magnetic fields do not affect the nucleus, and the simple electrostatic model is enough to predict the properties of the hydrogen atom.
As many others have pointed out, the Bohr model is rather outdated, and that we use quantum mechanics to explain the atom. I'm not very familiar with the quantum mechanical model, so I can't tell you much about that. Nevertheless, do not dwell on the Bohr model too much, it can say some stuff, but not everything.

Hopefully I clarified you completely. Leave a comment if you need more clarification.

Disclaimer, because it wasn't strong enough in the text: The Bohr model is wrong!

Electrons and neutrons are not attracted from strong electrostatic interaction.
Electrons and nuclei are because nuclei contain protons. Coulombic physics are therefore entirely relevant for atoms, because protons bear a 1$e$ charge, while electrons bear a -1$e$ charge.
(Don't confuse neutrons with nuclei!)
Neutrons interact with matter in two main ways: first, within the nucleus, through the strong force, reported to be ~137 times stronger than electromagnetic force. This is what binds neutrons to protons in the first place, in the tiny nucleus. The second is through interactions of magnetic fields. This second way is the only way that neutrons interact with matter outside of the nucleus, and it is what makes them useful for things like neutron-powder diffraction (an alternative to X-ray crystallography).

I want to ask why they are attracted because of electrostatic attractions, even though electrostatics are for charges at rest. I think you got me wrong.

If "they" is electrons and neutrons, the answer is simple: they are not attracted.
If "they" is electrons and nuclei, the answer is that Coulomb's equation is time independent, but it is re-calculated as a function of time, and it appears in the Hamiltonian used to calculate energy in time dependent and time-independent quantum mechanics via the Schrodinger equation.

I'm just a beginner in the field of chemistry and not in a position to understand quantum mechanics . My question is simple. why the term "electrostatic attraction" is used when electrons are not stationary inside atoms but moving.

It's called "-static" because the equation does not depend on time, but rather charge and position. But you can, and we do, use the equation at any point in time. The reason it is applied for electrons is no different than for a moving airplane. Everything is moving. If electrostatic interaction was that exclusive it wouldn't be relevant.

So, it does not matter whether the charge is at rest or in motion. Only that it matters is its position and other charge. Thanks

Yes:    $F = k_e q_1 q_2 / r^2$


text:
Which of these carbocations are more stable?

The line of text that follows this figure in my work-book reads:

Cation (A) is more stable than cation (B), since the former is resonance stabilized but the latter isn't (due to steric effects).

Don't both carbocations show resonance stabilization? More importantly, isn't cation (B) more stable than (A)?

Based on what I've learnt, the following should occur:
1) Cation (A)
The sp2 carbon (in the isopropyl group) acts as a -M group and causes the shift of a pair of π electrons from the benzene ring towards the $\ce{C-C^{+}}$ bond, thereby creating a double bond in its place. So the resulting structure should look like,

The mesomeric effect would also ensure that the positive charge "travels" across the ring, and decreases the electron density at ortho- and para- positions.
The delocalization of the positive charge across the benzene ring should confer some degree of stability to the carbocation.
2) Cation (B)
Once again, a pair of π electrons from the benzene ring would shift towards the $\mathrm{C-C^{+}}$ bond, leaving behind this:

And again, the mesomeric effect would delocalize the positive charge over the benzene ring and create electron deficient centers at the ortho- and para- positions.
However, the presence of two isopropyl groups at ortho- positions (with respect to the orignal charge-carrying isopropyl group) will result in the "supression" of the positive charge via inductive and esomeric effects, hence, conferring additional stability to this molecule.
Hence, cation (B) should be more stable than (A).

Yet, my conclusion is different from that provided by the book. Well, I guess that the presence of three adjacent isopropyl groups (in cation-B) does seem sterically unfavorable, but how would that prevent resonance effects in the molecule (as the book seems to imply)?
So, is my conclusion correct? Have I arrived at it correctly? If the book's right, then why is it right?

Related: Rates of solvolysis of p-dimethylamino benzyl halides; Why can the lone pair not align with the phenyl moiety in 2,6-xylidine?, etc. (more that I'm lazy to find)

What is the book's name?

As we know (to some extent), resonance in a molecule occurs due to sideways overlapping of p-π orbitals.
Now in the molecule B, there is high steric hinderance near the carbocation due to which the plane of the "carbon with carbocation" changes, so that it can feel minimum repulsion from the other groups. This forces the carbocation to remain in the mutually perpendicular plane, which inhibits the resonance.
Thus, this makes A (the less hindered carbocation) more stable than B.

The carbon atom with the positive charge can only participate in resonance with the ring if they lie in the same plane. Due to steric effects/inhibition that particular C atom will not be coplanar with the rest of the ring anymore. Lesser delocalisation of the positive charge makes structure B less stable.

source: Stack Exchange

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

sort: votes


text:
Inductive effect and bond strength
What is the relation between the inductive effect on bond length and strength.
how would one rank the bond strength of
$\ce{CH3-H}$, $\ce{CH3CH2-H}$, and $\ce{(CH3)2CH-H}$
I know that alkyl groups donate electron density by induction, but how does electron density in a bond affect its length or strength?

Just as hyperconjugation stabilizes carbocations and radicals by sharing the electron density of adjacent $\ce{C-H}$ bonds with the unoccupied/partially occupied orbital, hyperconjugation destabilizes carbanions. So more substitution means less acidic and more ionizable. I had a hard time finding a single resource listing bond lengths, but they seem to correlate with bond-dissociation energy. The wikipedia page for methane states that its $\ce{C-H}$ bond length is 108.7 pm and an answer by Jan says ethane’s is 109.4 pm.
$$ \small
\begin{array}{lcc}
\hline
\text{Hydrocarbon} & \mathrm{p}K_\mathrm{a} ^{[1]} & \text{BDE (kcal)}^{[2]}\\
\hline
\ce{CH3-H} & 48 & 104\\
\ce{CH3CH2-H} & - & 98\\
\ce{(CH3)2CH-H} & 51 &  95\\
\ce{(CH3)3C-H} & 53 & 93\\
\hline
\end{array}
$$

 $[1]$  Evan’s pKa Table
$[2]$ Wikipedia: Carbon-Hydrogen Bond

–I effect increase the bond strength and force constant, increase of bond strength is possible only by holding two atoms closer than earlier. hence -I effect reduces bond length and increases bond strength

since the inductive effect leads to ionic character in the bond,the increase in -I effect decreases the bond length.

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

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


text:
What is the mechanism behind the decomposition of enediol phosphate intermediate to methyl glyoxal?
I can't seem to figure out the mechanism for the decomposition of enediol intermediate (in glycolysis) to methyl glyoxal? How does the phosphate group get replaced by a hydrogen?

This reaction can occur only when the $\ce{C-O}$ bond to the phosphate group
  lies in a plane that is nearly perpendicular to that of the enediol so
  as to permit the formation of a double bond in the intermediate enol
  product.

In solution, the enediol intermediate readily breaks down with the elimination of the phosphate at C3 to form the toxic compound methylglyoxal:

The spontaneous decomposition of the enediol intermediate in the triose phosphate isomerase (TIM) reaction to form methylglyoxal through the elimination of a phosphate group.
In order for this elimination to occur, the bond to the phosphate group must lie, as shown above (a), in the plane perpendicular to that of the enediol.
This is because, if the phosphate group were to be eliminated while this bond was in the plane of the enediol as shown in (b), the $\ce{CH2}$ group of the resulting enol product would be twisted 90° out of the plane of the rest of the molecule. Such a conformation is energetically prohibitive because it prevents the formation of the enol’s double bond by eliminating the overlap between its component $\mathrm{p}$ orbitals.
Reference
Voet and Voet Biochemistry


text:
What is the IUPAC nomenclature for the cyclic mono-alkene ions formed after proton or hydride abstraction?
In this blog post, Steven Bachrach discusses the (anti-)aromaticity or lack thereof in the homologous series of anionic and cationic cyclic mono-alkenes formed respectively by deprotonation and hydride abstraction at a vicinal carbon:

$~~~\small\unicode{x201c}$$\small\textit{KassCPcmpds.png}$$\small\unicode{x201d}\textit{ by }$$\small\textit{Steven Bachrach}$$\small\textit{ is licensed under }$$\small\textit{CC BY ND 3.0}$$\small .$
Bachrach uses the construction -enyl anion/cation to name these species, which seems sensible to me.  However, I also vaguely recall seeing generalized use of the -ide and -ium suffixes for singly charged anions and cations. For example, I know that $\ce{NO+}$ is nitrosonium, but I seem to be recalling incorrectly that $\ce{NO-}$ is nitroxide. $($The NIST Webbook page only calls $\ce{NO-}$ the nitric oxide anion, and the Wikipedia article on nitroxide-mediated radical polymerization is discussing radical alkyl N-oxides $(\ce{R2NO^\bullet})$, not $\ce{NO-}$.$)$ On the other hand, $\ce{R3CO-}$ ions are unequivocally alkoxides.
So: What are the systematic IUPAC names for these compounds?

If you search 'cyclohepatrienyl cation' you will find that it's synonymous with tropilium. I tried searching for 'cyclopropenyl cation' and got no synonyms... Maybe worth trying them one by one, and you may stumble in some document describing the whole series.

@user6376297 Huh, good thinking. Tropylium sounds suspiciously like a historically-derived name, though, so I'm guessing any systematic naming won't be related to it. Not sure, though.

Why does the first row of examples follow the pattern of 1a, 2a, 3a ...; yet the second 1c, 2c, 3c, ... rather than 1b, 2b, 3b, ... ?

@Buttonwood #a is for anions, #c is for cations.

Also Is there a preferred way of naming the resonance hybrid in keto-enol tautomerism? might not be unrelated.

Ah, @mykhal, indeed - definitely relevant!

Anions
In general nomenclature, anions may be named by using functional class nomenclature; however, the resulting names are not the preferred IUPAC names. The corresponding subsection of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-72.2.1 Functional class nomenclature
Functional class nomenclature can be used, in general nomenclature, to describe anionic compounds. An anion that can be considered as derived formally by adding an electron to a radical may also be named by adding the class name ‘anion’ as a separate word to the name of the substituent group. The names are formed by using the names of corresponding radicals (not necessarily the name of substituent groups) and the class name ‘anion’ as a separate word. The multiplying prefixes ‘di’, ‘tri’, etc., are added to the class name to denote multiple anions. This type of nomenclature is limited to anions having anionic centers in the same structure. Systematic names (see P-72.2.2) are preferred IUPAC names.

The functional class names for the anions that are given in the question are:

cycloprop-2-en-1-yl anion
cyclobut-2-en-1-yl anion
cyclopent-2-en-1-yl anion
cyclohex-2-en-1-yl anion

The preferred systematic names for anions that can be derived by deprotonation are formed according to P-72.2.2.1:

P-72.2.2.1 Anions derived from parent hydrides and their derivatives
An anion derived formally by the removal of one or more hydrons from any position of a neutral parent hydride is preferably named by using the suffix ‘-ide’, with elision of the final letter ‘e’ of the parent hydride, if any. Numerical prefixes ‘di’, ‘tri’, etc. are used to denote multiplicity; locants identify positions of the negative charges.

Therefore, the preferred IUPAC names for the anions that are given in the question are:

cycloprop-2-en-1-ide
cyclobut-2-en-1-ide
cyclopent-2-en-1-ide
cyclohex-2-en-1-ide

Cations
In general nomenclature, also cations may be named by using functional class nomenclature; but again, the resulting names are not the preferred IUPAC names.

P-73.2.1 Functional class names
Cationic compounds that can be considered as being derived formally by removal of electrons from the corresponding radical may be named by adding the class name ‘cation’ as a separate word after the name of the radical. Polycations are indicated by adding the multiplying prefixes ‘di’, ‘tri’, etc., as appropriate, to the class name. Systematic names formed by using the suffix ‘ylium’ are preferred IUPAC names (see P-73.2.2). When the charge is not localized, the structure is enclosed in square brackets.

The functional class names for the cations that are given in the question are:

cycloprop-2-en-1-yl cation
cyclobut-2-en-1-yl cation
cyclopent-2-en-1-yl cation
cyclohex-2-en-1-yl cation

The preferred systematic names for cations that can be derived by removal of a hydride ion are formed according to P-73.2.2.1.2:

P-73.2.2.1.2 General method
According to the general method, cations formally derived by the removal of one hydride ion, $\ce{H-}$, from any position of a parent hydride are named by adding the suffix ‘-ylium’ to the name of the parent hydride, with elision of the final 'e' in the name of the parent hydride, if present. Di- and polycations formally derived by the removal of two or more hydride ions from the parent hydride are named by using the suffix ‘ylium’ and the multiplying prefixes ‘bis’, ‘tris’, etc. (…)

Therefore, the preferred IUPAC names for the cations that are given in the question are:

cycloprop-2-en-1-ylium
cyclobut-2-en-1-ylium
cyclopent-2-en-1-ylium
cyclohex-2-en-1-ylium

Regarding the anions in the first row of examples, I think rule RC 83.1 (Anionic compounds with anionic centers derived formally by removal of hydrons) apply. Either one uses the "radicofunctional names" (subsection RC 83.1.1):

Anionic compounds that can be considered as being derived formally by addition of electrons to the corresponding radical (see RC-81) may be named by adding the class name "anion" as a separate word after the name of the radical.

which leads to cyclopropenyl anion, cyclobutenyl anion, etc. as in your question.  Following RC-83.1.2. "Anionic centers in parent hydrides", and in particular RC-83.1.2.1. "The operational suffix "-ide"." alternatively states

An anionic center derived formally by the removal of one or more hydrons from any position of a neutral parent hydride is named by replacing the final "e" of the name of the parent hydride, if present, by the operational suffix "-ide", or by adding the suffixes "-ide", "-diide", etc., to the name of the parent hydride.

which would lead to a names analogue to cyclopropenide, cyclobutenide, etc.

For the cations presented in the second row, rule RC 82.2 (Cationic compounds with cationic centers derived formally by the loss of hydride ions) is applicable.  Again, either the name is radicofunctional (subsectio RB 82.2.1) where

Cationic compounds that can be considered as being derived formally by the removal of electrons from the corresponding radical may be named by adding the class name "cation" as a separate word after the name of the radical (see RC-81). Polycations are indicated by adding the numerical prefixes "di-", "tri-", etc., as appropriate, to the class term.

leads to the used naming of cyclopropenyl cation, cyclobutentyl cation, etc..  Or, RC-82.2.2.2. (Cationic centers in parent hydrides other than those described by RC-82.2.2.1.) which accomodates unsaturated compounds stating that

A cationic center derived formally by the removal of one or more hydride ions from any position of a neutral parent hydride or hydro derivative, except as provided by RC-82.2.2.1, is named by replacing the final "e" of the parent hydride name, if present, by the operational suffix "-ylium", or by adding operational suffixes such as "-bis(ylium)", "-tris(ylium)", etc., to the name of the parent hydride.

which would lead to cyclopropenylium, cyclobutenylium, etc.

Please note that the links point to obsolete IUPAC recommendations from 1993. However, that does not necessarily mean that names formed according old recommendations are no longer in accordance with current recommendations.

@Loong Definitely I'll invest time for reading the 2013 edition, including a curious eye of how it evolved over the last 20 years.


text:
Reaction of barium peroxide forming barium carbonate
I stumbled across this reaction:
$$\ce{BaO2 + H2CO3 -> BaCO3 + H2O2}$$
This reaction seems quite similar to the variant with sulfuric acid
$$\ce{BaO2 + H2SO4 -> BaSO4 + H2O2}$$
which is well known and works.
But as carbonic acid is a lot weaker than sulfuric acid, I started to question this reaction. Why does it even proceed in the first place? I thought that the sulfate/carbonate is really insoluble in water which is why no equilibrium forms (or does it?), however barium peroxide is itself pretty insoluble.
So, a) does the first reaction even happen and b) why?

Without going into too much quantitative stuff, barium peroxide is a very strong oxidiser and extremely unstable. It will oxidise just about anything given the chance.

Where did you get the first reaction? Mention your source

I found this here: http://www.chemieonline.de/forum/showthread.php?t=216287 This is a german chemistry forum, and the guy there also just says he found it "in some old book". The method he describes is: "In a cold solution of $\ce{BaO2}$ a lot of $\ce{CO2}$ is bubbled until enough $\ce{BaCO3}$ is precipitated out. This is separated from the $\ce{H2O2}$ and repeated until the desired concentration is reached."

@G.Ünther In the future, please add the original source to your question: it helps you to stay organized,  it helps us to help you, and is overall a good practice:)

Before the precipitation of $\ce{BaCO3}$ starts, the following equilibrium is arranged:
$$\ce{BaO2 + 2H-O-H <=> H2O2 + Ba^{2+} + 2OH-}$$
$\ce{BaO2}$ is but very slightly soluble in $\ce{H2O}$ because the then occurring hydrolysis produces $\ce{OH^-}$, which raises the pH, and hence stops hydrolysis (and dissolving) of further $\ce{BaO2}$. Simultaneously, by the formed $\ce{OH^-}$, the back reaction is fostered. However, these two blockades may be overcome by adding dilute acid, since this latter neutralizes at once the alkaline product $\ce{Ba(OH)2}(aq)$, and so the equilibrium may be continually shifted to the right side.
Hence, it doesn't matter whether the acid is more or less diluted (nor whether it is strong or weak), provided that acid is continually supplied to the extent that it is consumed during the dissolution/hydrolysis process of $\ce{BaO2}$. An aqueous solution of $\ce{Ba(OH)2}$ is an efficient gas detection or gas scrubbing method for $\ce{CO2}$; hence, by bubbling  $\ce{CO2}$-gas through it for a long time, the $\ce{Ba(OH)2}(aq)$ will easily be transformed into $\ce{BaCO3}(s)$. Of course, a gas frit will be used to spread the gas, and the reaction mixture should be constantly stirred.
In the textbook of Sarda/Handa/Arora$^1$ this process with $\ce{CO2}$ is shortly mentioned (without any explanations except that ice-cold water should be used, and that $\ce{BaCO3}$ may be removed by filtration, since the desired product is, of course, $\ce{H2O2}$).

$^1$ Sanda/Harda/Arora, Chemistry, part-2, New Dehli, 2016, p. 885.

"which lowers the pH" You made a blunder.

Shouldn't it be $\ce{Ba^{2+}}$?


text:
Why are optically active compounds abundant in nature?
Enantiomers have most physical properties exactly same, which makes them hard to seperate from one another. Due to large amounts of physical similarities and chemical similarities (when dealing with optical non active compounds) anyone will assume that if I eat any one of enantiomers of some compound I will get same amount of energy and they will perform exact same task in my body.
Given that many optically active compounds are extracted from nature, why do organisms go through pain of separating enantiomers for their consumption? Why do optical isomers are found in nature that are hard to separate in a lab?

They don't separate anything. The symmetry has been broken for a long time. Living organisms pretty much deal exclusively with one enantiomer because that's all they are able to make...

I doubt that eating L-glucose or D-alanine will keep you strong and healthy.

The title is actually a different question than the post, which contains several questions, but I'll answer all of them since they are closely related.
Why are optically active compounds abundant in nature?
Nature needs some kind of complexity to work. To build a living system you need hundreds, thousands or even millions of different molecules, just a few won't do it. If you just draw random molecules you'll end up with a lot of chiral ones. Unless you are sticking to linear chains without functional groups or aromatic (flat) systems you will almost always end up with chiral molecules. It actually would be surprising if there would be no chirality.
Given that many optically active compounds are extracted from nature, why do organisms go through pain of seperating enantiomers for their consumption ?
As Zhe already said in the comment: they don't need to separate those. All the animals and plants basically use only one enantiomer.
Why do optical isomers are found in nature that are hard to seperate in a lab?
Well, you are right. Enantiomers have the same physical properties unless they are in a chiral environment. Organisms on earth are able to produce only one enantiomer because the whole environment is chiral which means both enantiomers have different physical properties.
If we want to separate enantiomers in the lab we are using the same system. We take a chiral phase for chromatography (those are quite expensive) which means both enantiomers behave differently.
We are also able to synthesize enantiomerically pure compounds if we use a chiral environment during synthesis. But this is often expensive and we aren't nearly as good as biological systems doing this.

It's also important to point out that there are many isomerase enzymes that racemize enantiomers. This coupled with the fact that proteins are chiral and react differently with the different enantiomers sets up a natural version of dynamic kinetic resolution. Rather than separate enantiomers as is common in labs, nature forms pools of racemate starting materials which are drawn from selectively by enzymes.

Why earth become a chiral environment ? What is the benefit in doing so ?

@123 First both chiral twins rose from the mud, then one fell off a cliff. Pure chance.

By "chiral environment" I'm assuming you mean an environment with other "chirally imbalanced" chemicals, such as enzymes that only work on D-glucose. Is that correct?

@immibis Yes.  $\qquad$


text:
How to find if the unpaired electron is placed in hybrid orbital or pure orbital in odd electron species?
Odd electron species contain an unpaired electron which can either be placed in hybrid orbital or in pure orbital.
Example: In the case of $\ce{.CF3}$ radical, the unpaired electron is placed in $\mathrm{sp^3}$ hybrid orbital of carbon.
In the case of $\ce{.C(CH3)3}$ radical, the odd electron is placed in pure p orbital of carbon.
How do we determine whether the unpaired electron is placed in hybrid orbiral or pure orbital? What factors govern the same?

When discussing hybrid orbitals, they are justified by the experimentally observed or calculated/simulated molecular geometry. For instance, the $\ce{^.C(CH3)3}$ radical was found to be planar by ab initio calculations (http://www.tandfonline.com/doi/abs/10.1080/00268977600102711). Thus, the molecular shape is best approximated by $\ce{sp^2}$ hybridization and the unpaired electron is most likely to be found in a $\ce{p}$ orbital.
As a simple heuristic under the arguably outdated VSEPR theory, electronegative atoms want to be the furthest away from each other and other electrons as possible due to electric interactions. When considering the $\ce{^.CF3}$ radical, the fluorine atoms are highly electronegative and want to be the furthest they can be from the unpaired electron that holds negative charge. A planar $(\ce{sp^2})$ conformation would make the distance between the unpaired electron and each fluorine atom $90^\circ$. While a pyramidal $(\ce{sp^3})$ conformation would separate the two groups by $109^\circ$, which is much more favorable according to VSEPR theory. The later is what is observed under calculations (http://www.chem.mun.ca/courseinfo/c4420/handouts/handout-03a.pdf) with a $\ce{F-C-F}$ bond angle of $112^\circ$.
So, when considering hybridization problems, you must remember that hybridization fixes geometry. Thus if any plausible argument for a hybridization must be made, there should be experimental or computed evidence of the geometry. One cannot simply look at a molecule and know what the geometry could be with absolute certainty, but we can make a good guess based on simple heuristics, like those of VSEPR, molecular orbital diagrams, previous experience/similar molecules, etc.

Hello, thanks for the explanation, but what about NO , NO2, ClO2 & ClO3?
I understand about NO2 and ClO3 that free radical would reside in hybrid orbital due to electronegativity difference, but what about NO & ClO2? BCOZ, they also have lone pair on themselves along with free radical?.


text:
What exactly is the long chain band?
I have progressed quite enough in infrared spectroscopy, and now I'm looking at the spectrum of alkanes in detail. I found out that straight chain alkanes have this long chain band in the spectrum:

The infrared spectrum of decane. Taken from: Introduction to Spectroscopy, Pavia et. al.
On the other hand, cyclic one's don't have this band:

The infrared spectrum of cyclohexane. Taken from: Introduction to Spectroscopy, Pavia et. al.
I also found this info in my book:

$\ce{CH2}$ - The bending (rocking) motion associated with four or more $\ce{CH2}$ groups in an open chain occurs at about $\pu{720cm^-1}$ (called a *long-chain band).

I am still perplexed. Why can't the $\ce{CH2}$ groups in the cyclic chains undergo rocking vibrations? Why is this specific to the open chained ones?

The 'long chain band' described in your textbook (Pavia) is actually a vibrational mode in which the terminal methyl groups are 'rocking'.
This rocking mode is observed in open chain alkanes,  -(CH2)n-, where the number of methylene units is greater than 4.
Polyethylene is the 'classic' examples, the IR spectra and corresponding vibrational modes are given below:

Why do we see this band?
The reason we see any given IR band is as a consequence of the symmetry of the molecule, with only certain molecular vibrations IR active (we don't, for example, observe the C-C vibrations directly, they aren't active). Your question, more broadly, then becomes 'how to predict the IR active vibrational modes for a given molecule?'.
For a full explanation of the IR spectra in this particular case, using molecular symmetry and group theory, you might try reading J. Chem. Phys 1956, 25, 549.
Why don't we see the band in cyclic alkanes?
In cyclic alkanes, the methyl groups have been constrained to form one continuous ring of methylene groups. As discussed above, the band at ~725 cm-1 is caused by the terminal methyl groups of longer chain alkanes, hence, removing these methyl groups removes their corresponding vibrational modes.


text:
Cannizzaro reaction of furan
I encountered this question in a competitive exam. Can furan undergo Cannizzaro reaction? I though it can only happen with aldehydes.


text:
Thought Experiment: Is it possible to preserve the charge imbalance present during molten lead(II) bromide electrolysis?
Suppose I were to electrolyse some molten lead(II) bromide. According to what I have been taught, the lead ions travel towards the cathode and are reduced at it, while the bromide ions are oxidised at the anode. The movement of the ions creates a charge imbalance (there will be more lead ions near the cathode and vice versa) in the molten fluid.
Suppose I were to let this run for a little while in a vacuum (and siphoning off the $\ce{Br2}$ as soon as it is produced) and non-conducting container. Now suppose I was to quickly put an impermeable non-conducting plate between the cathode and the anode and at the same time remove both the cathode and anode, so separating the molten $\ce{PbBr2}$ into two halves. This will stop the electrolysis but there is already a charge imbalance between the two halves (e.g. the half which originally had the cathode has more $\ce{Pb^2+}$ than $\ce{Br-}$) which has no way to correct itself.
If I cooled down both halves of the mixture now so that they both solidified there would be excess ions on both side which wouldn't be able to recombine. It might even be possible to remove the re-formed $\ce{PbBr2}$ by some method and have just the uncombined ions left on each side. My question is whether this is even possible, and if so what would happen next.

There's no need to "thank in advance". If you do get an answer, upvote it instead of saying thanks. If it completely solved your problem, accept it.

There's not that much of a charge imbalance. The ions move in the melt to quickly restore imbalances.

@PrittBalagopal There's no appreciable excess of ions when the barrier would be put in place. Electrolysis doesn't separate the charges. They are fluid in the melt and adjust to establish a charge balance.

@Zhe I don't know whether I am missing something but in a similar experiment I saw with dissolved KMnO4 on a piece of paper all the purple colour moved towards the anode so an appreciable difference of ions was present in that case, I don't know whether it would be the same in this case or different.

I'm pretty sure this is a misconception. You just can't build up excess charge to any appreciable extent. Electrostatic forces are really strong, and you'll rapidly reach equilibrium...

@Zhe In a molten salt, there's no acid-base equilibrium to neutralize the accumulated polarization, so some imbalance in ionic concentrations has to develop (otherwise there would be no diffusive driving force to bring more of the reacting ions to the electrodes). I think you're right, though: the magnitude of this developed polarization potential is likely to be quite small. I have an answer in progress that will hopefully shed some light onto the magnitude of the phenomenon.

Abdul, your example with the permanganate on the paper is quite different from the $\ce{PbBr2}$ system in your question. The former is more of an electrophoresis experiment than an electrochemical process.

@hBy2Py I think you said this more eloquently than I did: there is a charge gradient, but at steady state, it is going to be small.

@Zhe Huh. Turns out transport in single molten salts is really weird. My answer is becoming steadily more confounded and confusing. This may take a while....

An electret is a dielectric (non-conducting) material that has a semi-permanent electric charge or polarization, the electrostatic equivalent of a permanent magnet. Electrets were made by first melting a polymer or wax containing polar molecules, and then allowing it to re-solidify in a powerful electrostatic field (several kilovolts/cm). The polar molecules in the dielectric align to the direction of the electrostatic field, producing a dipole with a permanent electrostatic bias when the melt solidifies.
Although electrets are only in a metastable state, those fashioned from very low leakage materials can retain excess charge or polarization for many years. The Japanese used microphones polarized by electrets in WWII.
https://en.wikipedia.org/wiki/Electret
So, using the electret as an example, the likely charge imbalance in the solidified lead bromide melt would be very small because the electric field across the molten salt is nowhere near kilovolts per centimeter. It might be measurable, but very small, and probably not very long-lived.


text:
Is it possible to find the ratio of isotopes only given the mean mass number?

Three isotopes of an element have mass numbers $(M)$, $(M+1)$ and $(M+2)$. If the mean mass number is $(M+0.5)$, then the ratio of the amounts of the three isotopes is?

Let amount of $(M)$, $(M+1)$, and $(M+2)$ be $x$, $y$ and $z$ respectively.
$$\frac{(M)x + (M + 1)y + (M + 2)z}{x + y + z} = (M + 0.5)$$
After further simplification, I got $x-y=3z$.
I don't know how to proceed further. Can someone give me a hint, or reassure me that this question can't be solved with the given data?
The answer is given as $x:y:z=4:1:1$.

You have two equations and 3 unknowns, so you can't solve it with just that. Say a, b, c are the fractions (as a decimal) of each isotope...
$$ a(x) + b(x+1) + c(x+2) = (x+\frac{1}{2}) $$
$$a + b + c = 1 $$
The 4:1:1 solution works. Another that works is 3:0:1. Another is 7:4:1. There are infinitely many solutions.

As already stated, the system of equations is underdetermined. But we can get the range of possible solutions. Starting with
\begin{equation}
 xM + y(M+1) + z(M+2) = M + \frac{1}{2}
\end{equation}
and using the normalization constraint
\begin{equation}
 x + y + z = 1
\end{equation}
we get
\begin{equation}
 (x+y+z)M + y + 2z = M + \frac{1}{2} \\
 y + 2z = \frac{1}{2}
\end{equation}
which can be rearranged to
\begin{equation}
 z = \frac{1}{4} - \frac{y}{2} \\
y = \frac{1}{2} - 2z
\end{equation}
We can plug $z$ back into the normalization constraint and solve for $y$
\begin{equation}
 x + y + \frac{1}{4} - \frac{y}{2} = 1 \\
 y = \frac{3}{2} - 2x
\end{equation}
which in turns allows us to express $z$ in terms of $x$
\begin{equation}
 z = \frac{1}{4} -\frac{1}{2}(\frac{3}{2} - 2x) \\
 z = x - \frac{1}{2}
\end{equation}
We further know that $x$, $y$ and $z$ are between 0 and 1. Therefore we can derive:
\begin{equation}
 0 \le y \le 1 \Rightarrow 0 \le \frac{3}{2} -2x \le 1 \\
\frac{3}{4} \ge x \ge \frac{1}{4}
\end{equation}
and similar from $z$ we get
\begin{equation}
 \frac{1}{2} \le x \le \frac{3}{2}
\end{equation}
Combining all constraints we get all possible solutions for $x$:
\begin{equation}
\frac{1}{2}\le x \le \frac{3}{4}
\end{equation}

Maybe strict inequalities are more appropriate, since 1:1:0 and 3:0:1 don't really satisfy the claim that the element has the three isotopes.

It is more general this way. It's just a minor detail, and everybody can decide on its own, depending on how the interpret the question: How do we know it's 3 isotopes? Is it just an assumption or did someone actually do a qualitative analysis of the sample? It's not stated in the question ....
And as long as you only "claim" there are 3 isotopes, somebody can still prove you wrong ;)


text:
Is the middle ring of this organic compound aromatic?
Is the middle ring with 5 atoms of (4Z)-4-[(4-methoxyphenyl)methylidene]-5-methyl-2-phenyl-2,4-dihydro-3H-pyrazol-3-one aromatic?

The carbonyl group could become a negatively charged oxygen, the tertiary amine could then form a double bond with the $\ce C$ atom of the carbonyl group. The secondary amine could contribute one electron to the $\pi$-system. And the $\ce C$ atom of the ethen-group could contribute two electrons if the other $\ce C$ atom becomes positively charged, but I'm not sure if this is likely.

It's hard to know for sure without computing or an experiment. But with the methoxy on the phenyl ring, I am definitely inclined to think that it is more aromatic than not.

@Zhe: Thanks for your comment, I am also inclined to think that this ring is aromatic.

You don't have to force a positive charge on the carbon between the central and left rings.  The methoxy oxygen on the left ring can donate an electron pair to take up the positive charge.  The left ring then becomes quinoid and the central ring becomes aromatic.  Such switching between quinoid and aromatic rings is common in conjugated polycyclic structures.

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


text:
Chemical bonding with f orbitals
For the elements such as rhenium, osmium, iridium how probable is for electrons in the inner 4f orbitals to gain energy and take part in a molecular orbital (such as an $f^2 d^2 s p^3$ hybrid orbital) when these elements are in their 6,7,8,9 oxidation state?

There's no such hybrid and d block doesn't use f-electrons for bonding.

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

You should read papers: here, here and here


text:
Why is ethanol not an oxidizer?
I am attempting the 2016 Canadian Chemistry Contest Part A, and one of the questions concerns the WHMIS symbols and whether a compound is an oxidizer or not.
The correct answer is that ethanol would not require a WHMIS oxidizer symbol. However, I am confused as to why ethanol is not an oxidizer when compounds such as potassium permanganate and nitric acid are when all of them contain oxygen.

As noted in the answers, just because a compound contains an oxygen atom does not mean that it will readily give it up to oxidize another compound. Ethanol is pretty stable as it is and would only oxidize pretty strong reducing agents, so it does not warrant a hazard specification. The best example of a compound that is mostly oxygen (by weight) but rarely considered to be an oxidizer is water.

Ethanol does warrant a hazard specification, just not the one for an oxidizer.  It's (in)flammable.

Airhuff has pointed this out perfectly. In addition, do check out the wikipedia page about redox reactions. It'll help you get a good deal of grip on what oxidation and reduction is.

For something to be an oxidiser. It must be capable of being reduced. The presence of oxygen is no particular indicator of the oxidising ability of a given species.
In potassium permanganate, manganese is in its +7 oxidation state (a very high oxidation state), consequently it is a good oxidant agent as it is able to gain electrons, causing oxidation of another species and itself being reduced.
In ethanol, there isn't the same possibility for reduction as the molecule is already in its lowest oxidation state. Even ethanal  (an oxidation level up) isn't particularly oxidising (despite being able to accept electrons and become formally reduced).

Imho, the lowest oxidation state for a $\ce{C2}$ compound would be ethane and the lowest for carbon is methane. Ethanol is slightly oxidised (but not a lot).

That a compound qualifies as an oxidizer under regulation is not bound to the presence of oxygen in that material, rather to either its innate ability to decompose easily under liberation of oxygen, or to yield compounds that itself liberate oxygen.  Sometimes, oxidizers are simultaneously known as explosives (like peroxides).  Since WHMIS is a Canadian system, details may be found either in English or French.
WHMIS/SIMDUT is to be seen in the context of the GHS system (English, French) by the UN/ONU.  Note that not all oxidizing agents (in the perspective of a chemist), like Fluorine, or Chlorine (ref) need (yet) to carry the corresponding pictogram (ref) of an oxidizer.  (It may be, however, that they are labeled with other pictograms, independently.)


text:
Partial reactions in redox formation of rust
In the iron oxidation that results in rust, the first step is to steal 2 electrons from $\ce{Fe}$. The main thief is said to be the oxygen dissolved in water, witch uses the stolen electrons to form $\ce{OH-}$ according to the reaction $\ce{O2 + 2H2O + 4e-  ->  4OH-}$.
What is really going on in this reaction? I cannot believe that this 3 reactants bump together at the same time and magically  $\ce{4 OH-}$ comes out of the collision. I imagine that there must be some steps between the reactants and $\ce{4OH-}$. Is it that, first, $\ce{O2}$ steals the two electrons from the $\ce{Fe}$ lattice becoming $\ce{2O-}$ and then, later, each $\ce{O-}$ takes an $\ce{H}$ from $\ce{H2O}$ to form $\ce{2OH-}$? That does not seem right to me, because this means that what is happening is:
$$\ce{O2 +  4e-  -> 2O-}$$
$$\ce{2O- + 2H2O  -> 4OH-}$$
But the reaction "$\ce{O2 +  4e-  -> 2O-}$"  implies that $\ce{2O-}$ is more stable than $\ce{O2}$. Is that right?

Welcome to ChemSE.  Please take a moment to familiarize yourself with the tools this web site offers in terms of formatting.  In addition to other sites, mathjax and mhchem (related to the LaTeX package of same name) offer useful assistance for mathematical and chemical expressions; a brief starter is provided here: https://chemistry.meta.stackexchange.com/questions/86/how-can-i-format-math-chemistry-expressions-here/88

The equations you mention are book keeping starting materials, on one side, and products, on the other.  They are not about the mechanism how the interconversion happens.  At least partially related is this https://chemistry.stackexchange.com/questions/76520/why-is-the-notation-for-reactions-not-explicit/76537#76537  And no, while it is already rare two molecules collide with each other such that there is a chemical reaction, Oxygen + Hydrogen + 4 electrons colliding at once to form four $\ce{OH-}$ is even less probable.

Bringing up rust seems to indicate confusion between the concept of corrosion and the concept of oxidation. Practically, iron corrosion is the breakdown of iron as a result of oxidation reactions.
To explain corrosion, understand water can self-ionize producing hydronium, $H_3O^+$, and hydroxide, $OH^-$ as well as ionize small amounts of iron causing it to dissolve. The dissolved iron ions collide with the hydroxide ions to form iron hydroxide.  Iron hydroxide may fall out of the solution as it is not very soluble, or if the concentration is low enough the surplus of hydronium donates a proton to the iron hydroxide converting it to iron oxide.
Oxidation, which is a more abstract concept, involves any reaction which results in the loss of an electron.  Instead of considering corrosion, consider other reactions.  In photosynthesis water is oxidized because it loses an electron.  In combustion, fuel is oxidized because it looses an electron.
Chemical reactions can get extremely complex, so it may be prudent to master the language of chemistry before delving into the complexities of specific reactions.  Many reactions are described with the language of redox (Reduction and Oxidation) chemistry.

Ok, I understand the reduction and oxidation nomenclature and also what happens with iron once it is dissolved in water. What I don't understand is what, exactly, takes the iron electrons form iron so that it ends up positive and get expelled from the lattice to end up in the solution. every explanation says that the iron electrons are taken by oxygen diluted in water. those explanations present the partial equation in my question ( diluted oxygen gas + water + electrons = hydroxide). I still don't understand the mechanism of that partial reaction.

Try rewriting your question omitting the use of the word "I", "me", "my", or anything in the first person.

As Argriculturist has said water can self ionise into $\ce{H+}$ and $\ce{OH-}$, but this takes a long time. So, 4 water molecules will self ionise into 4 hydrogen ions and 4 hydroxide ions
$\ce{4H2O -> 4H+ + 4OH-}$
The 4 hydrogen ions will soon react with the dissolve oxygen to form water molecules.
$\ce{O2 + 4H+ + 4e- -> 2H2O}$
So when we combine the equation, it will form this equation
$\ce{4H2O + O2 + 4H+ + 4e- -> 4H+ + 4OH- + 2H2O}$
And by cancelling the reapeating molecules in both sides, it become like this
$\ce{2H2O + O2 +4e- -> 4OH-}$

This is what I thought, if there're mistakes, please tell

interesting! now, going deeper, for the reaction  $\ce{O2 +4H+ + 4e- -> 2H2O}$ to happen, how the $\ce{4e-}$ got there? in other words, what is the species that pulled the electrons out of the iron lattice? $\ce{O2}$?

Iron is conductive...  Here is a link to how the corrosion process works http://www.rustbullet.com.au/technical/how-it-works/introduction-to-corrosion-and-process/

This site tells it is $\ce{H+}$ that pulls the electrons out of iron ($\ce{2 H+ + 2 e- –> H2}$). But wikipedia (https://en.wikipedia.org/wiki/Rust)  tells that it is oxygen ($\ce{O2 + 4  e− + 2  H2O → 4  OH−}$). Who is right?


text:
Why does hot water eventually turn cold, and why does cold water eventually turn hot?
I get cold water from my tap and put it in my water bottle. Over the day as I drink it, the water looses it coldness and slowly becomes warmer.
On the other hand, if I get hot water (maybe I'm sick or something), over the day it doesn't turn "cold", but it definitely does get "less hotter"
Why?
Is this because of equilibrium?

Yes, what you are describing is a system approaching thermal equilibrium.
The temperature of the water approaches the temperature of it's surroundings as heat is transferred from the warmer object (either the air or the water) to the colder. This is basically true of any two objects in contact with each other; the temperature difference drives the flow of heat from the warmer object to the colder one. At first, when the temperature difference is large, this happens quickly, then more slowly as the difference in temperature gets small.
One complication to this simple idea is that the water is also evaporating, which is a cooling process. Particularly if the surrounding air is dry, the water can evaporate quickly enough to become a fair bit colder than it's surroundings.


text:
Is possible to figure out if a battery is NiCd or NiMH?
Aside from opening it of course. I'm not sure if this is the right place to ask this but I have a rechargeable battery pack for an Xbox One controller and the info on it says that it's a sealed nickel cadmium battery but it also has a recycle symbol with "NI-MH" written underneath. The company's website doesn't seem to have any customer support and I tweeted at them on Twitter but haven't received a response. If it is NiCd, I'm surprised that any batteries of that type are still made. If it were Ni-MH I'm sure they would say it on the packaging. I think I read that NiCd cells don't have a large voltage drop until the end of the charge, so I suppose I could measure that. Does anyone have any suggestions about the battery type or know of a better place I could ask this question?

I still would just open the plastic enclosure and read what the labels on the cells itself say. Measuring the voltage probably won't give you much as both battery types have emf of approx. 1.2-1.3 V per cell. Theoretically Ni-Cd batteries are supposed to keep the charge over moths, if not years, whereas Ni-MH are more prone to the charge leaking, but I doubt an Xbox One player would prefer waiting for so long just to determine the battery type in his controller.

NiCd cells are much more dense then NiMH, so if you have two cells of the same size (e.g. AA), it's easy to determine by "hefting". However, with a sealed battery, this may not be so obvious. You could determine the density by Archimedes' method, but you probably don't want to get the inside of the battery pack wet (unless you can thoroughly dry it), to prevent electrolytic corrosion.
BTW, NiCd cells are much more resistant to damage from over-charge and from complete discharge than NiMH, and are still widely used in portable power tools and other high-drain applications. NiMH provide more total energy by mass, and more environmentally "friendly'.


text:
Air- and light-sensitivity of phenol
I've read that phenol should be stored under inert gas and protected from light.

What reactions does phenol undergo during long-term storage in the presence of air and light?
Is it really necessary to take such precautions when storing or handling phenol?
How does the stability of solid phenol compare to that of phenol dissolved in, say, cyclohexane?

Phenol is oxidized to 1,4-Benzoquinone.

Phenol slowly oxidizes on exposure to air and turns pink in color. At first, quinone (p-benzoquinone) is formed which again reacts with phenol to form a polymerized product phenoquinone which is red in color.

There has been research on the oxidation of phenol like wet-air oxidation and catalytic oxidation but those are out of scope.
Oxidation of phenol using light has been studied using UV/$\ce{H2O2}$/ozone system. From this paper:

The photolysis with 253.7 nm radiation practically does not degrade
phenol. But the combination UV/ozone gave better results than the
using only ozone. In the degradation of phenol using ozone there are
two main reactions: the direct reaction of ozone with phenol and its
products and the reaction of the hydroxyl radical generated. Some
intermediate products as benzoquinone, catechol and hydroquinone were
detected during the degradation of phenol.

So, yes. Necessary precaution are taken such as phenol are kept at brown bottles.

I think solid phenol is less prone to oxidation due to air/light than liquid/azeotropic phenol because of $\ce{\pi}$ stacking, efficient packing of crystals and hydrogen bonding leading to stability.

"Google search" is not appropriate source attribution. BTW you shouldn't reveal urls containing info about your computer, as google urls tend to do.

@Mithoron I removed the Google image URL as per your suggestion. Apart from your suggestion, can you clarify the downvote (assumed you did the downvote)?

Same reason as the new post where you used this incorrect picture of not-an-adduct


text:
On the periodic table: Why are groups of elements organized by 'letter'
Why are the groups of elements on the periodic table organized into areas represented by the letters s,p,d,f,g, and h?
What does this mean?

Letters g and h confuse me a bit. Could you please provide an example of elements that belong to these groups?

As far as I know, g is only hypothetical and h doesn't exist.

The s, p, d, and f stand for "sharp," "principal," "diffuse," and "fundamental," respectively. After f comes g,h,i....and so on...

These areas are called blocks and the letters correspond to the last orbital to be filled by electrons for that block, and they thus exhibit trends or commonalities in chemical properties. Wikipedia gives a pretty good discussion here.

This all is just great, spectroscopic notations and stuff, but from the question it looks like OP encountered a PTE with g and h blocks, which I'd really like to look at.

Possible duplicate of Periodic table and p block elements, and also of What is SPDF configuration?

The letters are related to the electron orbitals, which were originally observed through spectroscopy. The lines shown in the spectroscope were named sharp, principal, diffuse and fine (or fundamental).
With a strong magnetic or electrostatic field, these separate into one, three, five or seven lines, or energy levels. There can be up to two electrons (with opposite spin) with the same energy level, according to the Pauli exclusion principle, so this implies there can be double the numbers above, or 2 s electrons, 6 p electrons, 10 d electrons and 14 f electrons. Thus, the Periodic Table is based on the number of electrons in each elements orbitals.


text:
Is order of reaction defined for nuclear reactions?
I am asking this question after asking a similar one for spontaneous nuclear decay phy.stack. Link
Order of reaction is defined for chemical reaction. But wikipedia says about spontaneous nuclear decay:

Another class of first-order reactions is radioactive decay processes which are all first order. These are, however, nuclear reactions rather than chemical reactions.

Spontaneous nuclear decay is governed by nuclear forces for which it follows an exponential form. There can be deviations from the exponential law at very short time and at very time. But, that's under nuclear physics.
But, has anyone tried to extend the concept of order of reaction to other nuclear reactions eg. transmutation, spallation,fission,fusion etc.
Ben Crowell's comment on my previous phy.stack post:

If you want to make an analogy with chemical reactions, then you should probably be considering nuclear reactions, rather than spontaneous nuclear decay. For example, you could consider the cross-section for neutron or proton transfer reactions as a function of the number of nucleons (mass number) in the target and projectile.

Wikipedia page on nuclear reaction says on rate of nuclear reaction:

If the reaction equation is balanced, that does not mean that the reaction really occurs. The rate at which reactions occur depends on the particle energy, the particle flux and the reaction cross section.

Question:
Has anyone tried to define order of reactions for nuclear reaction? Even though concept of order of reactions might not be that useful for nuclear reactions, has anyone modified it to work with nuclear-chemistry?
P.S. I am not sure if this question is suitable here. If it isn't, then just migrate it phy.stack.

There is nothing to extend or modify; the ordinary definition applies just fine. Most nuclear reactions (except fusion) would be first order.

@IvanNeretin Exactly. It is not the nature of the process that is important when it comes to kinetics description, but rather how many particles participate in a single act of this process. Probably it somewhat bothers people because they are used to render reaction order as concentration's exponent.

@Ivan Neretin Why fusion isn't first order too?

Because it includes two particles reacting, or maybe even more.

What about fission?


text:
Why is cyclooctateraene shaped like a "tub" rather than a "chair"?
I understand that for cyclooctatetraene (COT) to "escape" the horrible prospect of becoming an anti-aromatic molecule, it must adopt a non-planar conformation.

The widely touted conformation of COT is said to be "tub" shaped. My sources (random links Google threw up) also concur on this particular "tub" conformation.
But why isn't it "chair" shaped instead?
Going along the same line of cyclohexane (non-aromatic), wouldn't a "chair" conformation reduce (even if it's a little bit) the steric repulsions due to crowding of hydrogens atoms on one side/face of the molecule?
Or even if "avoiding steric issues" isn't really the big idea here, then I could re-phrase my question as:
What stops COT from assuming a "chair" shaped structure?
The "chair" structure that I have in mind for COT is something like this:

(I don't have access to a decent enough organic chemistry sketcher to have this drawn any better, sorry.)

Well, what you've drawn are either alternating cis and trans double bonds. If you do the tub shape you can have nice cis double bond geometry while if you do the "chair" you need to go towards trans, so in the end very twisted cis bonds.

Interestingly, I can’t find any references to cis-trans-cis-trans-COT. I found some claiming the synthesis of mono-trans-COT derivatives, though. Naively, I assume cis-trans-cis-trans-COT should exist. However, the two are, of course, different compounds (and mono-trans-COTs tend to equilibrate back to all-cis-COT after being synthesised by irradiation).

@Jan If it c-t-c-t COT does exist...then know why people tend to stick with "tubs" rather than "chairs"? [Google images threw up tubs only :/]

The all-cis-form seems to be a stable diastereomer. I actually also found a few crystal structures confirming the synthesis of all-cis from precursors that I would have expected cis-trans-cis-trans to derive from. Well, I don’t know O:)

No, you should assume all COTs are all-cis-COTs unless otherwise specified ;)

@Mithoron no it sn't you cannot build the 3D structure of the "chair" confomer using planar cis double bonds. you would need to "twist" the double bond (which would be extremly high in energy) or go right to the cis,trans,cis,trans configuration. See my post below.

It's all about the 3D structure of double bonds.
If we look at the tub form we see that all dihedral C-C=C-C angles are 0°. the C=C-C angles are 125°, also pretty close to the optimal value. Everything works out fine, there's basically no strain on the whole molecule.

I tried to build the chair form of all cis, but every optimization ended up in either the all cis tube form or the cis-trans-cis-trans chair form. The best structure that looks like the all cis chair form I ended up with is this:

First of all, it's pretty flat, so antiaromaticity would again play a role and the dihedral angles are also ~33° for the double bonds in front and in the back, which is energetically quite bad and unstable.
The cis-trans-cis-trans compound is definitely no monstrosity. I'm working a lot with trans-cyclooctenes and I don't see why this shouldn't be possible if a trans,trans-cycloocta-1,5-diene is possible.

Here we've got 0° for the dihedral angles for the cis bonds and 125° for the trans. Yes, this is far away from the optimal value of 180° but quite similar to what you expect from trans-cyclooctenes (around 135°). I feel like it would be quite reactive and unstable but it could definitely exist. There's just no good way to produce it I guess.
So overall: all angles in the tub form are much closer to the optimal values than in either chair form and the all cis form is quite planar.

Mind a spin-off question? You mentioned: "... it's pretty flat, so antiaromaticity would again play a role ...". I was under the impression that once an annulene with 4n π electrons adopts a non-planar conformation, it totally loses its antiaromatic status. But what you said suggests that antiaromaticity is not so sharply defined with regard to conformation (i.e- antiaromaticity of the annulene depends on the degree of "flatness" of the molecule...rather then simply considering whether the molecule is flat or not). Was this a slip-of-tongue? Or did you really intend that?

Yes that was intended. The Hückel rules, which I guess you are taking the "it has to be flat" rule from, are just very simple basic rules but they tend to be quite inaccurate in some cases. (Anti)Aromaticity depends on the conjugated pi-system, That's why you would normally say it has to be flat. However, if you go out of plane the overlap will be worse but it's still there, you'll still have the aromatic or antiaromatic character, but it's not as strong.

There is no imaginary mode in the 'flat' case?

@Martin-マーチン there are most certainly several imaginary modes in that structure. I wasn't able to optimize to a stable structure that had chair conformation and only cis double bonds. This one occured during optimization and was just taken to show that the bond angles are terrible for such a confomer.

0 degrees? is that a typo?

@NH. no, why do you think so? Keep in mind it's the dihedral angle. 0° basically means planar and cis.

@DSVA : What is an "optimal value" for an angle? IMHO, an optimal structure of a molecule corresponds to a local minimum on the potential energy hypersurface (PES). The PES maps structural arrangements of a set of atoms to the total energy of the molecule. Experimental evidence in favor of one structure may indicate that the structure corresponds to a local minimum whereas others aren't. However, in this particular case the electronic Pi-System is composed of four distinct bonds (unlike in, e.g., benzene) which could be stabilized by $\pi$-electron interaction through space.

@TheFox the optimal structure for the whole molecule, yes that's the minimum on the PES. But that doesn't need to be true for the bondlengths/angles in fragments of this molecule. If we look at a trans double bond in a simple alkyl chain we see that the energy minima is at the planar arrangement. Now if we force the double bond into non planar conformation, for example by making an 8-membered ring, then it is what we call "strained" and higher in energy since it's not the optimal configuration for this part of the molecule.

@DSVA It may well be that the "strainedness" is one portion of the energy difference. However, it is disputable to call  the reason for domination of the tube over the chair. Your argument simply reduces the electronic structure to a background force that is invariant to the actual molecule under consideration. As $\pi$ and especially $\sigma$ electrons play a crucial role for the structure of a compound, I find it doubtful to exclude their role in an argument.

@TheFox I really don't know what to say. I suggest you read some literature about the concept of strain and how it is applied to our understanding of organic molecules, since this doesn't exclude sigma and pi bonds at all. In fact that's the whole point. If you twist this molecule into a chair like confomer you are losing essentially two double bonds due to twist between the carbons.


text:
Proving 1 = 2 by POAC( Principle of Atom Conservation )
I have been told that applying POAC in any equation means conserving the number of moles of an element in both reactant and product .
For example :
$$\ce{H2 + O2 -> H2O}$$
Here when we apply POAC on oxygen we write :
Equation 1 :
No. Of Atoms(or moles) of "O" in reactant =  No. Of Atoms(or moles ) of "O" in product
Equation 2: (obtained from equation 1)
2(moles of $\ce{O2}$) = 1(moles of $\ce{H2O}$)
Where equation 2 gives us the necessary data required in solving questions of stoichiometric calculations .
I understand how equation 2 comes from equation 1 but I don't understand why equation 1 is correct even when the chemical equation is unbalanced ?
As I think equation 1 can be applied only when the chemical equation is balanced .
As here :
No. Of Atoms of "O" in reactant =2
No of Atoms of "O" in product = 1
Thus by equation 1 we can say that  $1=2$ .
And my book says POAC can be applied even when equation is not balanced . So here equation is not balanced and when I apply POAC I get $1=2$ . So what's is wrong with my reasoning?

Atoms cannot be created or destroyed in a chemical reaction, hence we must balance equations for all the types of atoms involved, i.e. there have to be the same number of each type on either side. In your example $\ce{2H2 +  O2 = 2H2O}$

If I am interpreting your question correctly, equation 1 is basically a demand that you balance your equation. So:
$$ H_2 + O_2 = H_2O $$
is not balanced, therefore equation 1 doesn't really apply. I would say that "equation 1" is not even really an equation, more of a demand, or statement that must be fulfilled.

@porphyrin I know my chemical equation is not correct and must be balanced but the point I want to make is while applying POAC we don't need to balance the equation(so says my book ) !! So that is why while writing equation 1,2(applying POAC ) I have not balanced the chemical equation

@iammax If equation 1 is incorrect ( which I too think it is ) then equation 2 is also incorrect but that is what my book , teacher has been applying as POAC  in questions were balancing the chemical equation is difficult (to use mole-mole analysis ) by trial and error

@Freelancer Well, without having seen the exact wording in the book, I would say that it is quite wrong to declare that you don't need to balance the equation to apply POAC. In fact, I don't think you CAN apply POAC to an unbalanced reaction; conserving the number of moles is equivalent to saying the reaction is balanced, as the number of moles on each side is conserved by going to the other side of the reaction. If it's unbalanced, it leads to a contradiction, as you have noticed (the 1 = 2 contradiction). So I think perhaps the book is wrong, or you are misinterpreting what the book says.

@iammax what about this answer here , I think it is saying the same thing as I said about applying POAC in unbalanced reactions  https://socratic.org/questions/what-is-the-principle-of-atom-conservation-explain-with-an-example

@Freelancer In your linked question, their reaction is unbalanced, but they essentially balance it in the calculations. Note the line:
$$ \mathrm{Moles\ of\ } O_2 = \frac{3}{2} x \mathrm{Moles\ }  KClO_3 $$
This step is to account for the fact that the reaction is unbalanced, but in your example in the first post where you "prove" 2 = 1, you don't include a step like this, which is why you are getting a contradiction.

@iammax I think I have understood what you are trying to say, I have written a answer here , based on what I have understood . is it correct and according to the facts ( as I don't want to post anything which is wrong )  ?

As said by many others ,  The equation 1 given above is simply put very wrong  . As it can be applied only when a balanced chemical equation is given to us ,  otherwise we will reach at all sorts of contradictions as in the question above .
However many non- rigorous teachers , textbooks often write equation 1 in POAC (which is always true in a balanced chemical equation ) and then say now equation 2 is obtained from equation 1(which is partially incorrect) . It is true equation 2 comes naturally from equation 1 , but only for a balanced chemical equation.
So then how come we apply equation 2 in POAC in all sorts of question ? And still come up with the right answer ?
The answer is that equation 2 is true in itself always whether the chemical equation given ( in POAC ) is balanced or not . So it is not related to equation  1 for its validity .
How is equation 2 always valid ?
As pointed out by @iammax in the comments actually we (sort of )  essentially balanced the chemical equation given to us in the equation 2 itself .
For example in the $KClO_3$ question linked above , the moles of $O_2$ , moles of $KClO_3$ given to us are actually according to the stoichiometric coefficients , which enable us to use POAC without balancing the reaction (even though we have done that indirectly by using No. of moles of $O_2$ , $KClO_3$).

To conclude we can say after using No. Of moles of reactant , product we have indirectly balanced the chemical equation as the Ratio of their number of moles will be
Equal  to Ratio of their respective stoichiometric coefficients . After which we just use Conservation of Atoms( as the name says it) and multiply the number of moles of the molecule with the number of Atoms present  on which POAC is being applied ( For example we multiply no. Of moles of $O_2$ by 2 when we apply POAC on oxygen atom ).

The reaction you have written $\ce{H2 + O2 -> H2O}$ simply reads hydrogen and oxygen react to form water. It is a way to write incomplete information and it isn't quite correct. Do note that reactions must always be balanced for accuracy.
Principle of atom conservation (POAC) states that in a given chemical reaction (where no nuclear reactions are observed) the number of atoms of all elements remain constant. POAC is really a practical principle based on the fact that (in a chemical reaction where no nuclear reactions are observed) atoms can neither appear out of nowhere nor disappear into thin air. If you were to do an experiment and calculate the number of atoms of hydrogen and oxygen present in the reaction mixture before and after the reaction, it would be equal. POAC may even be applied to complex reactions where there is no need to balance the reaction in order to find the amount of product formed because it may be calculated by equaling number of atoms. Really, it just avoids a few cumbersome steps while solving questions.
You may be misinterpreting POAC. I think a more accurate way to state POAC would be 'observed number of atoms of all elements in a reaction mixture before and after a chemical reaction (where no nuclear reaction is occurring) are equal.' Do note the emphasis 'observed.'


text:
How does the rate determining step determine the overall speed of the reaction if it is the first step in the reaction?
I can imagine it determining the rate of the whole reaction if it is the last step since that means that we will have lots of the reactants needed for the rate determining step, synthesized in earlier steps. Therefore, the rate of this step represents the rate of conversion of these reactants to our final products. But if it is in the beginning , you will have to include the speeds of the other steps in the overall rate which means the overall rate is not dependent on the rate determining step alone. So why is it called a rate determining step if it is in the beginning ?
NOTE : If it is the last step, then we only consider the speeds of the earlier steps until we have excess of the reactants which feed into the rate determining step. As for the rest of the time, we will only have a store of reactants which are waiting for the rate determining step.

If it's the first step that's rate determining, then as soon as that transformation is complete, everything else immediately happens by comparison.

The rate determining step is so slow that we just pretend all the other steps happen instantaneously.  You're right that it still takes time but we ignore it. Generally a RDS is a few orders of magnitude faster than fast steps, so if the first step takes ten hours, you're not going to care if the second step takes a few seconds.

If you aim to arrive as a group while riding a bicyle, the slowliest team member defines the rate of advancement of the ensemble -- regardless if s/he is in front, in the centre, or trailing.  Similar, it does not matter if the lowest rate is observed for the first, or the last reaction, or somewhat in the middle.  Related question (and answers) here : https://chemistry.stackexchange.com/questions/72162/why-does-the-rate-determining-step-determine-the-overall-rate-of-reaction/72172#72172

@Buttonwood I encourage you to add that as an elaborate answer.


text:
Does a zinc metal coating react with coffee in a coffee grinder?
I'm restoring an old coffee grinder and I want to give the grinder mechanism a nice zinc / silver coating.
But I read it is not safe for zinc to be in contact with acidic foods.
When I look at the pH, coffee's is 4.6 – 4.3 which is quite acidic:
https://angelscup.com/blog/taste/coffee-acidity-flavor-ph-acid-reflux-and-low-acid-coffee/
Not sure if the zinc and coffee are likely to react in such a scenario. Should I maybe choose a silver plating?
Before cleaning with citric acid.

After cleaning with citric acid, Al foil and baking soda

[I don't use a coffee grinder] Assuming your beans are fairly dry, then the acidic nature of coffee isn't really a problem. In fact, electroplating metal articles with zinc is otherwise a common method of protecting the metal from corrosion: Galvanization [ https://en.wikipedia.org/wiki/Galvanization ]

and would you go for silver or zinc coating? which is more safer?

Meh, since all you're doing is coating/plating the grinder components, it wouldn't be particularly pricey to use silver (since you won't need much silver anyways), plus, it does a better job at protecting metal from corrosion (assuming you did plate it properly). Zinc should work too, but go for it only if you're pretty sure that you'll store the grinder (component) in a nice, dry place ...and grind thoroughly dried beans with it. Also, nice job cleaning the grinder ;)

My two cents: You should use zinc as a sacrificial coating, so that even if the zinc coat wears off, the iron beneath won't rust. However, use silver if you want an inert coating, that will keep it safe as long as the coating is intact. If a silver coating is broken, corrosion begins immediately.

Maybe i can do zinc and than silver

thats pretty cool

If you plate the pieces with zinc or silver, the main source of zinc/silver in the ground coffee will be from abrasion between the parts and the coffee beans and not from chemical attack.
A few milligrams of zinc or silver (or iron for that matter) won't to do any harm to your health, specially if diluted in many coffee cups along several months or years.
In fact, Zinc was used therapeutically in dosages upwards of hundreds of milligrams per day for months, or more, with no permanent health issues.
Silver, on the other hand, has much lower acceptable exposure limits and while not particularly toxic, I'd not put it in parts of the device that are guaranteed to put metallic silver into the ground beans.

Because the burr ( grinding head ) is steel and maybe the bowl it fits into , why not leave the shaft as bare steel ? As long as the beans are dry there will be very little corrosion; maybe wipe the shaft with a paper towel once a week to  prevent any coffee powder from building up on it. And you could wipe a little vegetable oil on the shaft occasionally.

I use a $ 20 spice grinder for coffee.

I upvoted this because it make perfect sense. Why plate for no functional reason, and wonder about corrosion, metal particles, etc., when steel is simply perfect for this.


text:
Azobenzene heat trapping
Azobenzenes trap heat when exposed to sunlight due to the principle of photo-isomerism. But this occurs in nanoseconds.
I was wondering if we can store this heat for hours and utilize this later on. Is there any way to do that?

One of the two isomerisations may be driven by heat, but more generally speaking, you may switch between cis and trans configuration with light of different wavelengths that may be (and actually already is) used for information storage. (https://en.wikipedia.org/wiki/Azobenzene)  Stilbenes may offer this, too, but azo dyes made the run.

How could you equate light with heat...

This is what photosynthesis does, light energy is converted into carbohydrates that can later on be used as fuel.  In simple systems such as the azobenzenes or stilbenes the photoisomers will thermally convert back to the most stable trans species but do so very slowly so the energy released would hardly be noticeable.


text:
Why is kerosene preferred over other non-polar solvents to store sodium in?
I think we all heard this again and again in our early schooling days:

Sodium is always stored under kerosene so that it doesn't react with air, and corrode itself.

But lately I've been wondering, why kerosene? Why not benzene? Why not cyclohexane? Why not $\ce{CS2}$? What is that makes kerosene preferred?
Is kerosene the cheapest non-polar liquid available? Or is there any particular reason that I'm not aware of?

I always store lithium under mineral oil. I've never tried kerosene though because in case the metal "accidentally" catches fire, being surrounded by kerosene doesn't help O:)

I'm not sure sodium is always stored under kerosene is even a true statement, at least not in the UK/USA.

I expect $\ce{CS2}$ to react with sodium, and quite eagerly. The rest are more or less interchangeable.

@paracetamol I suspect lithium is lighter than kerosene and stays afloat, that's why mineral oil is used for $\ce{Li}$ ($\pu{0.80 g/cm^3}$ (kerosene) vs $\pu{0.53 g/cm^3}$ (lithium)).

@andselisk I mentioned lithium because that's the only alkali metal I've ever dealt with using my ---bare--- own hands. Apparently mineral oil is used to store lithium, sodium and potassium. Also, lithium floats in mineral oil too ;)

@paracetamol Well, a specific mineral oil;)

Regarding the question, I guess any cheap inert water-free/hydrophobic liquid with $\rho < \pu{950 g/cm^3}$ which is also miscible with lubricants or is a lubricant itself (if alkali metal needs to be cut or mechanically processed) will do just fine.

@NotEvans. This statement (sodium is always stored under kerosene) is  pretty much true (mineral oil also mentioned as an alternative in a majority of handbooks) for post-soviet countries.

@andselisk: You say:  “… or is a lubricant itself (if alkali metal needs to be cut or mechanically processed)”.   — But before cutting or processing alkali metals,  one will surely wipe off the kerosene. Why should lubricant be needed for cutting?  ...  Alkali metals are soft like cheese.

@Zeus Good luck cutting a dry $\pu{10 \times 10 \times 20 cm^3}$ brick of sodium with a knife.

@andselisk: Although such a brick is utopic, I would cut this brick with this cheese knife or also by karate chop (with gloves). — Why do you think that luck is necessary for this? What do you mean by “dry”? I have even cut lithium with a knife although it is much harder than sodium, … without “lubricant”.

@Zeus It was a metaphoric expression. You need a guided applied force and a protective layer to prevent metal from rapid oxidation. By "dry" I mean liquid-free. Try your cheese-knife with the brick of the dimensions I mentioned (not the tiny piece from the video) and see how well this goes.

Kerosene is a fraction from the distillation of petroleum, the one with boiling point 180° to 230°, containing hydrocarbons from $\ce{C_11}$ to $\ce{C_12}$. In reaction to what has been said in the commentaries (below the question), I note that mineral oil is a cloudy expression whereas kerosene seems to be the precise name of the more or less broader base material of this cloud.
The best means to store alkali metals, which are all strong reductants, is an alkane because they are inert (as indicated by the name “paraffin”) and cannot be reduced.$^1$ The reason why one does not take the lower boiling fraction (gasoline 40° to 180°, $\ce{C_6}$ to $\ce{C_10}$), cyclohexane (b.p. 81°) or hexane (b.p. 69°) is obvious: it would be too evaporable, which is dangerous since its evaporation when someone forgets to close the bottle sufficiently, would lead to spontaneous ignition of the alkali metal (especially in the case of potassium) by contact with moist air. (Additionally, cyclohexane would anyway be unnecessary expensive.) The reason why the higher boiling fraction (light gas oil, 230 to 305°, $\ce{C_13}$ to $\ce{C_17}$) is not used is that this is much more greasy than kerosene.
It is strange that in the question benzene and carbon disulfide are proposed since they are both more expensive, very toxic and their b.p. is very low. — In the case of carbon disulfide this would be a paradigm of a very dangerous choice for three reasons: 1. concerning the boiling point (46°, extremely fiery), 2. concerning its extreme toxicity and 3. because it is not inert against reductants.

Footnote:
$^1$ In the case of lithium, which is much less reactive than the other alkali metals, the container should ideally be filled level with the top since it is swimming on the kerosene. But anyway the kerosene stays over as a protecting film on the lithium, which obstructs its reaction with moist air, even if it is not submerged.


text:
Ionisation energy
Explain why the 1st ionization energy of boron is less than that of beryllium, whereas the 2nd ionization energy of boron is greater than that of beryllium.

First ionisation enegy
of Beryllium is 900 Kilojoules per mole and that of Boron is 801 Kj/mole, you must be confused by the trend of periodic table in which the size of the atom gets smaller when going from left to right in a period due the force of the Positive nucleus so how can the i energy of Beryllium be more than Boron?
But we are thinking in terms of orbitals now, the game is changed a little bit. The reason is due to their electronic configuration..
Electronic configuration of Beryllium is 1s2 2s2(since the orbitals are filled completely, its going to become stable and when an atom is stable..it is very hard to remove an electron from it and disturb its stability).
Electronic configuration of Boron is 1s2 2s1 2p1(Since the last p orbital is not filled completely then its going to stay unstable and also, the energy of p orbital is higher than s orbitals so its going to be pretty easy to remove an electron from it and make it stable when its configuration will be 1s2 2s1- like Beryllium, stable).
Second ionization energy
After removing the valency electron during first ionisation, you get the following configurations
Beryllium : 1s² 2s¹
Boron: 1s² 2s²
As you can see, Beryllium has less electrons left in its valency shell that Boron has, so the nuclear effect (postitive charge) will have a greater influence on that electron, pulling the entire shell closer, and reducing the ionic radius. When the shell gets closer to the nucleus, it becomes harder to remove an electron from the lower energy level "2s¹" because of the very strong attractive forces between the electrons and the nucleus thus requiring lots of energy. In Boron's case, although there is one electron less than before, and there is now a full sub-shell "2s²", the shielding effect of the inner shells repel the outer two electrons causing the ionic radius to be greater than that of Beryllium, and thus needing less energy to over come to electrostatic attraction between the nucleus and the valency electron.


text:
Is water the only substance that expands in both gaseous and solid states?
I have been told that water is the only substance that expands in both solid and gaseous states!
Is this true? If not, what other substances can do this?

I'm afraid water is not so unique. There are numerous examples of compounds with NTE, there is also Gough–Joule effect. Ga, Ge, Sb, Bi also expand upon crystallization from liquid phase. If you want, I will try so assemble these examples in an answer.

I was wondering if it's the only naturally occurring substance that expands in both gaseous and solid form.  My son says it's because of the polar nature of the molecules.  Is this true, I would love to learn more!

Ge, Ga Bi and Sb all appear to be in the "Poor Metals" group.  Germanium is rather surprising as it was used in Semiconductors and I would think the expansion would be a problem with Junctions?  Perhaps that's why they went with Silicon for Semiconductors.

Actually, Si is also in this "anomaly" list. I guess I also forgot to mention Pu. Silicon substituted Germanium because it's literally dirt-cheap and extremely abundant, and because gifted guys from Fairchild  figured out how to make it pure and how to produce semiconductors on industrial level.

It's a little bit late where I currently am, I'll try my best to provide a more or less complete answer tomorrow, unless somebody else takes care of it.

I assume the liquid state is the "standard." I can't imagine that there is any compound that wouldn't expand in volume when going from a liquid to a gas. So the "curiosity" would seem to be just what liquids expand in volume when solidifying.

@DickPatton I added some info to this question as I promised.


text:
What can I add to a dough ( kind of pizza dough) to keep it fresh after frozen?
I make some "mini pizzas" with a dough recipe i learned in Brasil, I'm interested in freezing the dough after being "half baked" meaning baked for half of the time, so I can bake again after frozen. I was wondering if there is something I can add that will make this dough taste more fresh even after frozen

I'm voting to close this question as off-topic because it would be better asked on Cooking.SE.


text:
Ductless fume hood for acetone?
Administration on my campus has become worried about the lack of a fume hood when using acetone in the 3D Print Club. I'm helping in the process of picking one out.
They do acetone vapor smoothing on their prints, but I know they have a box to do that in. Other than that, it's acetone in a bottle for dissolving support material and plastic residue.

Does there need to be a fume hood with acetone?
If I can get a ductless fume hood, what am I looking for to be sure that it can effectively take care of the acetone problem?

I need to prove to administration that the club is fixing the problem, and I frankly have no idea what to look for in fume hoods.

Honestly, even in an academic chemistry lab, people use acetone to rinse glassware at the sink once its been washed (all done outside of a hood). If I have a lot of washing, can easily get through a couple of litres. The amount it sounds like you'll be going through is insignificant, I see no need for a hood (+acetone is used a lot in DIY applications, again, where no hood possible)

@notevans. If people are rinsing with acetone at the sink, that is probably an environmental safety violation. The wash should go into a waste bottle. Whether or not a fume hood is necessary depends on the concentration of the vapor and the length of exposure. In the US, there are strict laws from OSHA, and you're asking for a hefty fine per violation if you're not in compliance.

@Zhe No no, at the sink, into a waste container, not into the sink literally. In the UK this is very much the norm

@NotEvans. Ah. Sorry. Misunderstood what you meant. Washing at the sink is probably fine. Since waste container and acetone bottle are normally closed, the amount of vapor is relatively limited.

Does there need to be a fume hood with acetone?

Unless you're using very large quantities of it or in a confined space, usually not. I've spent a fair bit of time washing glassware out with acetone at the sink (using a bin to capture the runoff for transfer to a waste container) and I never had issues. Though I don't recommend doing this yourself, I've also had several errant splashes directly on my skin when working with stuff from the hardware store--again, no problems.
Acetone is staggeringly nontoxic: the NIH suggests an 8500 ppm EEL, meaning that in emergency situations, it's considered acceptable to breathe air that is nearly one full percent acetone for up to one hour. Hell, the air in my house contains less water than that sometimes.

If I can get a ductless fume hood, what am I looking for to be sure that it can effectively take care of the acetone problem?

Look at this link. If the acetone is heated in the box, you should be able to meet all the criteria for being able to use a ductless hood:

A limited number of different chemicals should be used.
Acetone only. Check.
No extreme heating should be carried out in the hood, i.e. acid digestion applications.
You only need to boil acetone. That's 56C, not even close to the boiling point of water. Check.
Modest chemical volumes should be used, around 500 mls or less per chemical.
Even if you need more than this (you shouldn't, 500mL of acetone is quite a bit), it'll mostly be sealed in a box, so you shouldn't have nearly that much escaping into the hood. Check.
Moderate chemical exposure times should be maintained.
I don't know exactly how long it takes to smooth a print, but I don't think it can really be more than a few hours. You're probably okay on this one.
Will the chemicals involved in your application be effectively filtered with the available carbon filters?
Yes. Acetone is so common I can all but guarantee that every company will have a filter that can capture it (probably whatever is used for other general organic compounds as well). In Labconoco's case, that's the organic vapor filter.
How often will you have to replace your filters?
I don't know, that'll be your club's responsibility.

Finally, make sure someone doesn't get the idea that the hood will stop all fumes and try to do a perchloric acid experiment using that hood--that will go badly.

Now, all this being said...
What you really need isn't to make this safe, what you need is to convince administration that you're doing things safely. This may have nothing at all to do with whether things are actually safe or not.
At my old school (where I was briefly president of the local ACS chapter), I could have put an experiment into a fume hood with no filter, not plugged the thing in, and gotten an okay from the administrators because they believed that fume hood = safe and no fume hood = unsafe no matter what else was going on.
I say this not to encourage you to try to trick your way out of this, but to point out that, unless your administration is scientific (i.e. actually understands science), then it's fairly likely you're going to have to jump through a few silly hoops in order to satisfy them.
Best of luck!