Why does Fe(CO)₄ adopt a tetrahedral, as opposed to square planar, geometry?

The hypothetical $\ce{Fe(CO)4}$ complex is isoelectronic to the observed $\ce{[Ni(CN)4]^{2-}}$ complex and therefore would also be square planar.[1,2] Both are 16 electron complexes, i.e. one of the valence orbitals of the central atom remains unoccupied. Your assessment for $\ce{Fe^{\pm0}}$ as $\mathrm{d^8}$ is absolutely correct, also that carbonyl is a good ligand causing a strong field and hence a strong splitting. In a tetrahedral complex there would be triply degenerated HOMO, which would only be occupied by four electrons. This arrangement cannot be more stable than a planar complex. However, the iron tetracarbonyl complex as a singular entity has not been observed (to my current knowledge). The observed structures for the compound are various dimers and trimers.[3]

There is also a neutral iron carbonyl complex, but it binds five instead of only four ligands. Therefore $\ce{Fe(CO)5}$ is a trigonal bipyramidal 18 electron complex.[4]

The observed $\ce{[Fe(CO)4]^{2-}}$ complex[5] is isoelectronic to $\ce{Ni(CO)4}$ which is a tetrahedral 18 electron complex.[1,4] Without going into more detail, the iron in this complex is of course $\ce{Fe^{-II}}$ and therefore $\mathrm{d^{10}}$. A square planar arrangement would no longer be more stable and a complex with a higher symmetry and more degeneracy is preferable.

References

J. Demuynck, and A. Veillard, Theor. chim. acta,
1973, 28 (3), 241-265.
ChemSpider.com
Raymond K. Sheline, J. Am. Chem. Soc., 1951, 73 (4), 1615–1618
Dario Braga, Fabrizia Grepioni, and A. Guy Orpen, Organometallics, 1993, 12 (4), 1481–1483.
Henry B. Chin, and Robert Bau, J. Am. Chem. Soc., 1976, 98 (9), 2434–2439.

Note
Carbonyl is nowhere close to a bulky ligand. Apart from halides and hydrides it might as well be the smallest possible ligand.


Why is the water solubility of CO worse than that of CO2?

The fact that $\ce{CO2}$ does not have a dipole moment does not really mean it's nonpolar.  What $\ce{CO2}$ has is a quadrupole, two opposing dipoles that occupy different portions of space (opposite sides of the carbon atom).
Ordinarily we expect an electrostatic interaction with a quadrupole to be relatively weak, but in $\ce{CO2}$ the component dipoles are strong and at close range the water molecules can "home in" on one end of the molecule and interact selectively with the corresponding component dipole.  Thereby the electrostatic interaction between water and the $\ce{CO2}$ quadrupole is stronger than the interaction between water and the weak $\ce{CO}$ dipole.


What is intermolecular distance in water?

Not all data are provided in explicit form. They are often not frequently needed and can be often deduced from other data with using of high school knowledge.
You can calculate the volume occupied by 1 water molecule ( in average ) from the water molar mass $\pu{M}$, water density $\pu{\rho}$ and the Avogadro constant $N_\mathrm{A}$ as
$$V=\dfrac{M}{\rho \cdot N_\mathrm{A}} $$
Then you can calculate the side length $a$ of a cube of this volume:
$$a={\left( \dfrac{M}{\rho \cdot N_\mathrm{A}}  \right)}^{1/3}  $$
As a rough number, is is equal to
$a  \approx { \left( \dfrac{\pu{0.018 kgmol^-1}}{\pu{1000 kgm^-3} \cdot \pu{6.022e23 mol^-1}}  \right)}^{1/3}  \approx \pu{0.31 nm}$
what can be taken as the typical "center to center" distance of water molecules within the order-like rough precision.
The more precise average distance would have to calculate with the size and shape of the water molecule ( $\ce{O-H}$ bond length about 0.096 nm, the bond angle 104.5 degrees) and to make the integral average over the directions. But it would be extremely complicated, considering also difficulties with molecule borders and hydrogen bonds. In some sense can be said the distance is zero or even "negative", as the molecules form bonds ands overlaps.


I have searched and searched, oh how I have searched.

Do you know what I always tell my mom when she asks me to find something in the Internet she was not able to find herself? I ask her: "Are you sure that the thing you are looking for even exists?"

  I am looking for a 3 dimensional visualization of a whole (moderately
  complex, hydrogen is just a ball) atom that includes 3 dimensional
  orbital geometry.

Hydrogen atom is not "just a ball".
There is no orbitals. In short, except for one-electron systems (such as the hydrogen atom) orbital description is just an approximate model of the reality.
Usually when one asks to visualize a physical object, he/she wants to visualize its physical "boundaries" with respect to some medium. The notion of such "boundary" simply looses its meaning in the microscopic world: atoms do not have "boundaries" like macroscopic objects do.

  Please. The universe is weird and I want to see.

Sorry, but you can not do this: you can not visualize some things in the Universe, you can not even imagine them. Our visualization & imagination abilities (unlike an atom) has some boundaries, because they are based ultimately on our senses which in turn has their boundaries.

As Paul Dirac said on that matter:

  […] the main object of physical science is not the provision of
  pictures, but is the formulation of laws governing phenomena and the
  application of these laws to the discovery of new phenomena. If a
  picture exists, so much the better; but whether a picture exists or
  not is a matter of only secondary importance. In the case of atomic
  phenomena no picture can be expected to exist in the usual sense of
  the word 'picture', by which is meant a model functioning essentially
  on classical lines. One may, however, extend the meaning of the word
  'picture' to include any way of looking at the fundamental laws which
  makes their self-consistency obvious. With this extension, one may
  gradually acquire a picture of atomic phenomena by becoming familiar
  with the laws of the quantum theory.


What is the simplest way to create a precipitation reaction at home?

Here's one easy way :

Make a solution of salt ($\ce{NaCl}$) in water, and another solution of Silver Nitrate ($\ce{AgNO_3}$), which you can buy easily) in water.

Upon mixing the two, $\ce{AgCl}$ (Silver chloride, a white compound) will instantly precipitate.


This paper (J. Sep. Sci. 2011, 34, 1489–1502) lists 1,4-Dinitrobenzene as having a very small dipole moment (0.01 Debye). No error is given on the dipole moments, but they do list compounds with exactly zero dipole moment (naphthalene for example).


What reaction occurs when soaking a piece of meat in apple cider vinegar?

This "cooking" phenomenon happens because of the basic building blocks of a meat (protein) and how they can be broken down (which is called "denaturation"): heat (traditional cooking), pH changes (your vinegar question) or enzymes (bromelin or papain, which are typical ingredients of meat tenderizer)

When exposed to these factors, the proteins that make the bulk of the meat "break", and the result you can observe is very similar, which is the "cooking". Since it varies with the exposure level to the denaturating agent, this is why you only see the "cooking" happening on the surface of marinaded meats.

pH changes is also an alternative "cooking" method used in some dishes, e.g. ceviche, which is raw fish finely chopped that is soaked in lemon juice to "cook"

Hope it helped :)


How to determine the pH of a mixture of two weak acids?

Below a more general approach.

Suppose that we have two weak acids $\ce{HA}$ and $\ce{HB}$.
The initial concentrations are $C^0_\ce{HA}$ and $C^0_\ce{HB}$, and their constants are $K_{\mathrm{a},\ce{(HA)}}$ and $K_{\mathrm{a},\ce{(HB)}}$.
Suppose yet that volumes, $V_\ce{HA}$ and $V_\ce{HB}$, are additives.
So we have:

Reactions

\begin{align}
  \ce{HA + H2O &<=> H3O+ + A-} &
  K_{\mathrm{a},(\ce{HA})}
    &= \frac{\ce{[H3O+][A-]}}{\ce{[HA]}}
  \tag{1}\label{eq:KAcidHA}\\
  \ce{HB + H2O &<=> H3O+ + B-} &
  K_{\mathrm{a},(\ce{HB})}
    &=\frac{\ce{[H3O+][B-]}}{\ce{[HB]}}
  \tag{2}\label{eq:KAcidHB}\\
  \ce{2 H2O &<=> H3O+ + OH-} &
  K_\mathrm{w} &= \ce{[H3O+][OH-]}\tag{3}\label{eq:KWater}
\end{align}

Mass balance

\begin{align}
  C_\ce{HA} &= \frac{C^0_\ce{HA} V_\ce{HA}}{V_\ce{HA} + V_\ce{HB}}
  &&=\ce{[HA] + [A-]}\tag{4}\label{eq:MassBalanceHA}\\
  C_\ce{HB} &= \frac{C^0_\ce{HB} V_\ce{HB}}{V_\ce{HA} + V_\ce{HB}}
  &&=\ce{[HB] + [B-]}\tag{5}\label{eq:MassBalanceHB}
\end{align}

Charge balance

$$\ce{[H3O+] = [OH-] + [A-] + [B-]}\tag{6}\label{eq:ChargeBalance}$$

Replacing \eqref{eq:KAcidHA}–\eqref{eq:MassBalanceHB} equations on \eqref{eq:ChargeBalance}, we have:
$$
\ce{[H3O+]} =
  \frac{K_\mathrm{w}}{\ce{[H3O+]}}
  +\frac{C_\ce{HA} K_{\mathrm{a},(\ce{HA})}}{\ce{[H3O+]}
  + K_{\mathrm{a},(\ce{HA})}}
  + \frac{C_\ce{HB} K_{\mathrm{a},(\ce{HB})}}{\ce{[H3O+]}
  + K_{\mathrm{a},(\ce{HB})}}
\tag{7}\label{eq:GeneralEquation}$$

or as polynomial

\begin{align}
\begin{split}
 &\ce{[H3O+]}^4\\
+&\ce{[H3O+]}^3 (K_{\mathrm{a},(\ce{HA})}
  + K_{\mathrm{a},(\ce{HB})})\\
+&\ce{[H3O+]}^2 \left[
  K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})}
  - ( C_\ce{HA} K_{\mathrm{a},(\ce{HA})}
    + C_\ce{HB}K_{\mathrm{a},(\ce{HB})} )
  - K_\mathrm{w} \right]\\
-&\ce{[H3O+]} \left[
  (C_\ce{HA} + C_\ce{HB}) K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})}
  + K_\ce{w} (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})}) \right]\\
-& K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})} K_\mathrm{w}\\
=&\ 0
\end{split}\tag{8}\label{eq:GeneralPol}
\end{align}

This single equation will exactly solve any equilibrium problem involving the mixture of any two monoprotic acids, in any concentration (as long as they're not much higher than about $\pu{1 mol L-1}$) and any volume. Depending of $K_\mathrm{a}$, $K_{\mathrm{a},(\ce{HB})}$, $C^0_\ce{HA}$, and $C^0_\ce{HB}$ values, we can yet obtain a simpler version[1].

The \eqref{eq:GeneralPol} equation can simplified considering[2] that $K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})} K_\mathrm{w} \ll \ce{[H3O+]}^4$:

\begin{align}
\begin{split}
 &\ce{[H3O+]}^3\\
+&\ce{[H3O+]}^2 (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})\\
+&\ce{[H3O+]}\left[
  K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})}
  - (C_\ce{HA} K_{\mathrm{a},(\ce{HA})}
    + C_\ce{HB} K_{\mathrm{a},(\ce{HB})})
  - K_\mathrm{w} \right]\\
-&\left[
  (C_\ce{HA} + C_\ce{HB}) K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})}
  + K_\mathrm{w} (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})
  \right]\\
=&\ 0
\end{split}\tag{9}\label{eq:GeneralPolSimp1}
\end{align}

The \eqref{eq:GeneralPolSimp1} equation can simplified considering[2] that $K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})} \ll \ce{[H3O+]}^3$:

\begin{align}
\ce{[H3O+]}^2 + \ce{[H3O+]} (K_{\mathrm{a},(\ce{HA})}
  + K_{\mathrm{a},(\ce{HB})})
  - (C_\ce{HA} K_{\mathrm{a},(\ce{HA})}
    + C_\ce{HB} K_{\mathrm{a},(\ce{HB})}+K_\mathrm{w})
= 0\tag{10}\label{eq:GeneralPolSimp2}
\end{align}

The \eqref{eq:GeneralPolSimp2} equation can be solved as usual.

$$\ce{[H3O+]} =
\frac{
  - (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})
  + \sqrt{
    (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})^2
    + 4 (C_\ce{HA} K_{\mathrm{a},(\ce{HA})}
      + C_\ce{HB} K_{\mathrm{a},(\ce{HB})}+K_\mathrm{w})}
}{2}$$

Or using the initial concentrations

$$\ce{[H3O+]} =
\frac{
  - (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})
  + \sqrt{
    (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})^2
    + 4\left(%\displaystyle
      \frac{C^0_\ce{HA} V_\ce{HA} K_{\ce{a},(\ce{HA})}
        }{V_\ce{HA} + V_\ce{HB}}
      + \frac{C^0_\ce{HB} V_\ce{HB} K_{\ce{a},(\ce{HB})}
        }{V_\ce{HA} + V_\ce{HB}}
    +K_\mathrm{w}\right)}
}{2}$$

Replacing $C^0_\ce{HA}=\pu{0.01 mol L-1}$, $C^0_\ce{HB}=\pu{0.01 mol L-1}$, $V_\ce{HA}=\pu{0.050 L}$ and $V_\ce{HB}=\pu{0.050 L}$, and using $\text{p}K_\ce{a}=3.75$ for formic acid and $\text{p}K_\ce{a}=4.756$ for acetic acid[3], we have
$$\ce{pH}=3.05.$$

Addendum

@MaxW made some warnings about the assumptions which simplificate ($\ref{eq:GeneralPol}$)(see comment) and ($\ref{eq:GeneralPolSimp1}$)(see comment) equations.
To completeness and with the aim of show the limitations of present approach, consider the following graphics.
The ($\ref{eq:GeneralPol}$), ($\ref{eq:GeneralPolSimp1}$), and ($\ref{eq:GeneralPolSimp2}$) equations were solved using $\text{p}K_{\mathrm{a},(\ce{HA})}=\text{p}K_{\mathrm{a},(\ce{HB})}$, $C^0_\ce{HA}=C^0_\ce{HB}$, and $V_\ce{HA}=V_\ce{HB}=\pu{50.0 mL}$.

As we can see, the ($\ref{eq:GeneralPolSimp1}$) equation can be used on any $\text{p}K_\ce{a}$ values, inasmuch as the concentration be major of $\pu{1E-6 mol L-1}$.
The ($\ref{eq:GeneralPolSimp2}$) equation becomes problematic when dilute solutions or strong acids solutions are used.

Thus, when applied to book-examples, which $\text{p}K_\ce{a}$ values from 3 to 6, and the concentrations from $1.0$ to $\pu{1.0E-4 mol L-1}$, the assumptions are valid and the equations can be used, providing good approximations of $\ce{pH}$ values.

Acknowledgement

Thanks to @MaxW for draw attention to limitations of assumptions.

Reference

Edited based on first @MaxW's comment.
Edited based on second @MaxW's comment.
Haynes, W. M.;  Lide, D. R.; Bruno, T. J., CRC Handbook of Chemistry and Physics. 97 ed.; CRC Press: 2017; pp. 5–88.


Chromic acid oxidation of aldehydes and primary alcohols

There are many $\ce{CrO3}$ oxidation reagents (Jones, Sarett, Collins, PCC, etc.), however, most are selective and stop the oxidation of a primary alcohol at the aldehyde stage.  On the other hand, the Jones reagent is relatively unselective and will oxidize a primary alcohol to the corresponding carboxylic acid - if enough reagent is present. It takes two-thirds of an equivalent of Jones reagent to oxidize one equivalent of the alcohol to the aldehyde and an additional two-thirds equivalent to oxidize the aldehyde to the acid.  Also the Jones oxidation is run under acidic conditions (typically dilute sulfuric acid).

I would have though that A, B (the gem-diol will equilibrate with the aldehyde under acidic conditions) and D were correct answers.  It seems like the computer answer is either in error or it is a "trick" question (seems unlikely for computer-based learning).  The "trick" would be that since an excess of reagent was not specified only 1 equivalent is assumed.  Therefore there is not enough reagent to carry out two sequential oxidations; as a consequence only the aldehyde or gem-diol could be completely oxidized with one equivalent of reagent. But then A or B is not an acceptable answer either.


The typical binding energy per nucleon $(E_\mathrm B/A)$ of most nuclides is about $5{-}8\ \mathrm{MeV}$. Such values are higher than the released energy of a typical radioactive decay. Therefore, emission of neutrons doesn’t occur during most usual radioactive decay processes.

In order to make a neutron source, you have to find a suitable target (not $^{208}\mathrm{Pb}$) and a corresponding nuclear reaction that releases a neutron and that is also energetically favourable, for example
$$^9\mathrm{Be}\ (\alpha,\mathrm n)\ ^{12}\mathrm C$$
$$^9\mathrm{Be}\ (\gamma,\mathrm n)\ 2\ ^{4}\mathrm{He}$$
Note that $^{12}\mathrm C$ and $^{4}\mathrm{He}$ are very stable nuclides with a relatively high binding energy per nucleon so that these reactions can release a large amount of energy, but still you need to irradiate the beryllium targets with high-energy radiation (high-energy $\gamma$ radiation of more than $E=1.665\ \mathrm{MeV}$ e.g. of $^{124}\mathrm{Sb}$ or $\alpha$ radiation) to release neutrons.

Nevertheless, neutron emission can occur during radioactive decay if the neutron excess is large and the beta decay energy is high enough, i.e. at the edge of the valley of beta stability. At very high values, spontaneous emission of neutrons becomes possible.

More important, however, are so-called neutron precursors, e.g. $^{87}\mathrm{Br}$ or $^{137}\mathrm{I}$, which have a very high decay energy so that the product nuclides ($^{87}\mathrm{Kr}$ and $^{137}\mathrm{Xe}$, respectively) can get enough energy to make neutron emission possible in the next step.
$${}^{87}\mathrm{Br}\xrightarrow[{E_\mathrm{max}=6.83\ \mathrm{MeV}}]{\beta^-}{}^{87}\mathrm{Kr}^*\xrightarrow[\quad]{}{}^{86}\mathrm{Kr}+\mathrm n$$
$${}^{137}\mathrm{I}\xrightarrow[{E_\mathrm{max}=5.89\ \mathrm{MeV}}]{\beta^-}{}^{137}\mathrm{Xe}^*\xrightarrow[\quad]{}{}^{136}\mathrm{Xe}+\mathrm n$$
Such reactions are responsible for the so-called delayed neutrons, which are very important for most nuclear reactors – without this effect, controlling a nuclear chain reaction would be difficult.

Of course, the most important neutron sources are nuclear reactors, which generate neutrons (about 2–4 neutrons per fission) in a controlled nuclear chain reaction.


Tin in silver nitrate, any soln color change?

The solution will remain colourless until silver particles will participated from the solution.

Silver nitrate solution is unstable, its decomposition is accelerated by light:

$\ce{2 AgNO3(l) -> 2 Ag(s) + O2(g) + 2 NO2(g)}$

That will change the color of the solution to brown. After silver participation, the solution will be colorless again.

A tin strip added to the nitrate solution will dissolve into tin(II) nitrate and will consequently react with silver nitrate accordingly:

$\ce{Sn(NO3)2 + 2AgNO3 + H2O -> Sn(OH)4 + 4HNO3 + 2Ag}$

Analogous silver reactions are described by Tollens' reagent, mirror preparation and gelatin silver process.


How much potassium permanganate is needed to completely oxidize ferrous oxalate?

See here.

The "green salt" is $\ce{K3Fe(C2O4)3.3 H2O}$.
\begin{align}
\ce{MnO4- + 8H+ +5e- &-> Mn^2+ + 4 H2O}\\
\ce{C2O4^2- &-> 2 CO2 + 2e-}\\
\ce{2 MnO4^- + 16 H+ + 5 C2O4^2- &-> 2 Mn^2+ + 8 H2O + 10 CO2 (g)}\\
\end{align}

One expects all the oxalate is oxidized to $\ce{CO2}$ and $\ce{Fe(II)}$ is oxidized to $\ce{Fe(III)}$.


Ortho-effect in substituted aromatic acids and bases

I'd like to throw a tentative explanation for the ortho effect into the ring:

In the molecules in question, an interaction between the protons of the methyl group and the lone pair of the amine nitrogen and the negative charge on the carboxylate, respectively,  can be assumed.

In the first case, the electron density on the N atom is (slightly) reduced and thus the basicity of o-toluidine.

In the latter case, a similar interaction provides additional stabilisation of the carboxylate. As a result, o-toluic acid is more acidic than the isomers.


Steric hindrance does indeed interfere with the stability of carbon tetraiodide, as does poor covalent bonding overlap between the very differently sized carbon and iodine atoms.  Even so, carbon tetraiodide is definitely established and in use as an iodinating agent.  Storage below normal room temperature is recommended.

Carbon tetraiodide is also very red.  Ordinarily saturated carbon compounds are white or colorless because the electronic transitions involving single bonds to carbon require ultraviolet radiation to go, but the relatively weak bonds in carbon tetraiodide enable transitions in the visible light region.  Iodoform is more lightly colored for a similar reason.

Lead tetraiodide, meanwhile, does not seem to be definitely known to exist.


What concentration of small molecule binding is considered weak or strong?

The threshold for activity varies based on the target, but hit-level activity is typically 1 µM or lower. Targets are normally enzymes or receptors, so the term activity refers to an IC50 or EC50 value.

Smaller molecules have fewer potential sites for intermolecular binding than larger molecules. Therefore, small molecules rarely bind as strongly as larger compounds. For example, a hit from a typical combinatorial library may show activity (KD, IC50, Ki, EC50) at concentrations of 1 µM or lower. In contrast, a hit in a fragment library may be selected with activities of around 1 mM, which is a 1,000- fold difference in activity (a larger IC50 value implies weaker enzyme inhibitor binding).

  In 1999, Teague distinguished between lead-like and drug-like hits,
  and combinatorial library collections. 20 Lead-like hits are
  characterized as having lower molecular weights (<350), activity (>0.1
  µM), and clog P values (<3). The lower values give lead-like compounds
  room to grow into an optimized, high-affinity drug that still
  satisfies Lipinski’s rules. Drug-like hits have higher molecular
  weights (>350) and clog P values (>3) but still modest affinity (~0.1
  µM). The definition of lead-like has since been used as a preliminary
  filter for selecting more promising hits from a screen.

The cut-off for the required activity of a hit is somewhat arbitrary. The threshold may also be based on the performance of the entire library. For example, the discovery group may count all compounds that are two or three standard deviations more active than the average of the full library.

While hits are often selected at an activity level of 1 µM (KD, IC50, etc.), structural changes may provide a lead with activity at concentrations of 0.1 µM (100 nM). Eventually, potency will typically be improved down to the 1–10 nM level during the lead optimization stage.

References:

Medicinal Chemistry E. Stevens


Why are metalloids found in a zig-zag line on the periodic table

In addition to Pritt Balagopal's answer, in which he has clarified the trends in metallic character across the Periodic Table, I would also like to explain why the metalloids occur in a zig-zag fashion.

Firstly, I would like to again clarify that it is possibly a coincidence that the metalloids start from boron, not carbon or any other non-metal in that period. My guess could be the start of electron occupation of the p subshells. Note that boron has one electron in its p subshell. Thus, its chemical properties start to change to become more non-metallic in nature.

Diagonal relationships is a common occurrence in the Periodic Table. However, it is often used to describe the relationship between two elements which are diagonal to each other, not a whole line of them. Thus, I would like to clarify again that this is just my conjecture, not what is the true reason. But I believe this should be part of the correct explanation as it is logically sound.

This site offers a more in-depth explanation as to why diagonal relationships occur:
https://chem.libretexts.org/LibreTexts/University_of_Denver/Chem_2132%3A_Chemistry_of_the_Elements/Unit_3%3A_Descriptive_Chemistry/09%3A_Building_a_Network_of_Ideas_to_Make_Sense_of_the_Periodic_Table/9.3%3A_The_Diagonal_Effect

In short, the diagonal relationships are based on two effects: the increasing effective nuclear charge across a period and the decreasing effective nuclear charge down a group. Effective nuclear charge increases across a period because the number of shielding electrons in the inner shells is constant while there are additional protons in the nucleus of the atoms of the elements across the period. Effective nuclear charge decreases down the group as there is an addition shell of shielding electrons from one element to the next in the group. (These are similar to the trends in metallic character pointed out in Pritt Balagopal's answer.)

These two competing effects cancel each other out and thus, chemists believe that the chemical properties of elements in a diagonal should be similar. The image shows some examples.


Why do some elements with high electronegativity exhibit metallic properties?

Whether an element adopts a metallic or nonmetallic structure does not depend on electronegativity alone.  Aside from electronegativity at least two factors promote formation of a nonmetallic structure:

1)  If the atoms are close to filling their valence shells they are more likely to do so through covalent bonds.  Iodine needs only one more valence electron per atom and can get it by forming covalently bound diatomic molecules; gold needs seven valence electron per atoms to complete the valence shell and goes the metallic bonding route instead.

2)  elements in earlier periods tend to form stronger covalent bonds favoring the nonmetallic structure.  Silicon form the familiar tetrahedral semiconducting structure but lead, below it with similar electronegativity and valence electron configuration, is metallic.

Other factors may enter, too, and Group 14 is an excellent place to look for subtle effects.  Above I said that silicon has the tetrahedral structure we all know and love, but that is under ambient conditions. The tetrahedral structure of silicon is energetically more stable than the metallic one, but the latter has more entropy and thus silicon may be converted to a metal upon heating.  In the case of silicon the temperature for this switchover is high enough so that the metal is molten, so we have simultaneous metallization with melting -- and, therefore, contraction with melting.  Germanium does the same thing, whereas with lead the metallic structure is more stable all the time.  Tin is a transitional case where it's a semiconductor when cold but switches to a metal below the melting point.

So there is indeed more to what makes a metallic element or a nonmetallic one than just electronegativity.


Background

$\ce{O2}$ exists as a paramagnetic, triplet since the two electrons in its two (degenerate) HOMO orbitals are unpaired. There are 6 known phases of solid oxygen with color ranging from pale blue to red to black.  In the liquid phase it has a light blue color.  This color is due to light absorption by the ground state triplet according to the following equation
$$\ce{2 O2(^3\Sigma_{g}) ->C[{h\nu}]\   2 O2(^1\Delta_{g})}$$
This absorption requires light from the red region of the spectrum (~630 nm).  If red light is absorbed , then blue light is transmitted or reflected and this gives rise to the blue color associated with liquid oxygen.

Answer

Note that the light absorption requires 2 molecules of $\ce{O2}$ and a photon, a 3-body process. If we now consider the probability of this 3-body process occurring in the gas phase, we can see that, since a gas is much more dilute than a liquid, the probability for of these 3 items coming together at the same time will be much smaller.  Hence, the probability of photon absorption in the gas phase is much reduced.  To the human eye oxygen gas will appear colorless (at normal and reduced pressures).  However, if you place the gas in a cell and record its visible spectrum you will still be able to detect this absorption.


One possible explanation involves atomic size.  You need a large enough atom in the middle to stabilize all the occupied orbitals in the octahedral structure; sulfur makes it but oxygen does not.
This reference gives the molecular orbitals for sulfur hexafluoride.  A key feature of these molecular orbitals is a doubly degenerate pair, labeled $\mathrm{2e_g}$ in the above source, which effectively contains the "extra" valence electrons in fluorine-based orbitals.
The picture below, taken from a different source due to limited downloading options from the free one, includes one of these combinations (here labeled $\mathrm{3e_g}$, as different nomenclature is used) among others.

If we look carefully we see that the ostensibly nonbonding orbital has antibonding overlap between adjacent lobes and so is technically antibonding.  The same is true of the other component of this degenerate pair.  So to make the orbital effectively nonbonding and keep the octahedral structure from blowing apart, we need a large central atom to act as a "spacer" between orbital lobes.  Apparently sulfur is large enough, oxygen is not.


Reaction of orthocresol with butyrolactone in presence of AlCl3 and heat

I can take a guess since no one else has provided an accepted answer.
First, I don't think lactones will take part in FC-acylation on an arene, but rather prefer alkylation as mentioned in this paper. In the case of o-cresol and heat, this could be a possible explanation for your question why the O-acylation takes place instead of the C-acylation.
Second, when the Fries rearrangement occurs, and the acylium ion is released in the intermediate, the phenolic oxygen is complexed with the aluminum chloride (second to last step here). If you imagine an ortho-methyl group, as well as the complexed aluminum chloride, it's plausible substitution to the remaining ortho-site is not sterically favorable. This would force it to substitute elsewhere on the ring.
As for why "elsewhere" prefers para to the methyl group and not the oxygen—going by what you wrote in your comment—I'm not sure. The para alumino-oxy substitution should produce the more stabilizing resonance structure for the arenium intermediate. At least, in one other reaction of electrophilic aromatic substitution with ortho-cresol, namely its nitration, the 3/5-nitro products are not found. Maybe it's a solvent matter. Wikipedia states a non-polar solvent prefers ortho in the Fries rearrangement; perhaps if the ortho site is blocked, it will settle for para to a methyl substituent.
Edit
In a summary of the comments, the m-hydroxy-ketone was called into question as the correct product. I mentioned none of the papers I found of similar reactions furnished anything but the para or ortho product. Here are the papers I looked through:
paper 1 - first paragraph: o-tolyl acetates yield predominantly p-hydroxy-ketones.
paper 2 - tables 2 and 3
paper 3 - table 1 (page 378)
paper 4 - 5th paragraph
A few others: book p 223, paper, paper


Why is methyl α-D-glucopyranoside preferentially formed from glucopyranose in acidic methanol?

In general, if there is overlap between a filled orbital and an unfilled orbital, a stabilising interaction will take place between the two. A simple example is the resonance present in esters, where the ester oxygen lone pair (a filled orbital) overlaps with the C=O π* orbital (an unfilled orbital):

In this context, anomer refers to a diastereomer differing only in the configuration at C-1. So, the two diastereomeric products you have drawn above are often called the α and β anomers.

In the α anomer, there is one lone pair on the ring oxygen (a filled orbital) which can overlap with the C–OMe σ* orbital (an unfilled orbital):

This leads to a stabilisation of roughly 6 to 7 kJ/mol, and is called the anomeric effect.

In the β anomer, neither of the two lone pairs on the ring oxygen overlap well with the C–OMe σ*.

Of course, steric effects would disfavour the α anomer, which possesses an axial methoxy group. However, this factor only leads to a destabilisation of roughly 3 to 4 kJ/mol, and is therefore outweighed by the stereoelectronic stabilisation provided by the anomeric effect.

The rough numbers are taken from Kirby, A. J. Stereoelectronic Effects (OUP), p 16.

While the lone pair -> σ* explanation is the most commonly invoked, there are also likely to be other factors that are also present: for example, in the α anomer, the dipole moments of the ring oxygen lone pair and the exocyclic C–O bond oppose each other, which is a favourable electrostatic interaction:


Questions regarding the preparation and recrystallisation of methyl 3-nitrobenzoate

Answer 1 - you want to keep the whole reaction cool, preferably below 6 °C, because the nitration reaction is exothermic. If it gets too hot you will lower your yield due to side-reactions. There is also the possibility of runaway reaction degrading the starting material and product. This is why you don't heat it.

Answer 2  - someone somewhere has done the comparison crystallisations and determined that EtOH is a good solvent for the recrystallisation. It is also cheap and safe for students to handle.


Why does 3-hydroxy-butan-2-one give positive Tollens' test?

From the Wikipedia entry on Tollens 1

  Tollens' reagent is a chemical reagent used to determine the presence of an aldehyde, aromatic aldehyde and alpha-hydroxy ketone functional groups

The substrate you have drawn is an alpha-hydroxy ketone.


How do organic chemistry mechanisms become accepted?

Great question!
When I was teaching, Anslyn and Dougherty was a decent text for this. Here are some general comments:

First, please note that you cannot be sure about a mechanism. That's the real killer. You can devise experiments that are consistent with the mechanism but because you cannot devise and run all possible experiments, you can never be sure that your mechanism is correct.

It only takes one good experiment to refute a mechanism. If it's inconsistent with your proposed mechanism, and you're unable to reconcile the differences, then your mechanism is wrong (or incomplete at best).

Writing mechanisms for new reactions is hard. Good thing we have a whole slew of existing reactions that people already have established (highly probable, but not 100% guaranteed) mechanisms for.

Computational chemistry is pretty awesome now and provides some really good insights into how a specific reaction takes place. It doesn't always capture all relevant factors so you need to be careful. Like any tool, it can be used incorrectly.

The types of reactions you run really depend heavily on the kind of reaction you're studying. Here are some typical ones:

Labeling -- very good for complex rearrangements
Kinetics (including kinetic isotope effects) -- good for figuring out rate-determining steps
Stereochemistry -- Good for figuring out if steps are concerted (see this example mechanism I wrote for a different question)
Capturing intermediates -- This can be pretty useful but some species that you capture aren't involved in the reaction, so be careful.
Substitution effects and LFER studies -- Great for determining if charge build-up is accounted for in your mechanism

For named reactions, the Kurti-Czako book generally has seminal references if you want to actually dig through the literature for experiments.
For your specific reaction, what do we think the rate-determining step is? Probably addition into the acylium? You could try to capture the acylium intermediate.
You could run the reaction with reactants that have two labelled oxygens and reactants that have no labelled oxygens. Do they mix? If not, it's fully intramolecular. Otherwise, there's an intermolecular component and the mechanism as written is incomplete.
A quick Google search suggests that the boron trichloride mediated version has been studied via proton, deuterium, and boron NMR. I didn't follow up on this, but there's clearly some depth here.
When I was T.A.ing for Greg Fu, he really liked to use an example with the von Richter reaction. I might be able to find those references...


Material to change its color by temperature condition meets over time

Compounds that irreversibly record the thermal history of the surface they are attached do do exist and others are being developed.

One class, used in harsh industrial environments is described on the Sensor Coating Systems site.

There are chemical systems used to detect high temperatures in refrigerated food, but most don't seem to be single chemicals. There are certainly many devices in development, but how many involve single chemicals is unclear.


What chemical that when cured chemically cross-links to already vulcanized rubber

Various elastomers, such as ethylene propylene diene terpolymer (EPDM), can be cured (crosslinked) using ionizing radiation.

The general radiation effects on polymers are

formation of gas,
cross-linking of polymer chains, and
scission of polymer chains.

However, EPDM rubber contains side chains, ring systems and π electrons, which results in a delocalisation of excitation energy and causes a greater stability of the excited states. Therefore, radiation chemical yields for all dissociative radiation chemical reactions on EPDM are decreased.

The mechanism of cross-linking involves the cleavage of a $\ce{C-H}$ bond on one polymer chain to form a hydrogen atom. The free radical released in the initial radiolysis reaction has enough kinetic energy to abstract a second hydrogen atom from a neighbouring chain to produce molecular hydrogen. Different recombination pathways follow the irreversible loss of hydrogen. Double bond formation results from two events occurring on the same chain; the recombination of two separate macroradicals leads to cross-linking. Cross-linking can also result from the reaction of free radicals with double bonds (comparable to free radical addition polymerization). The radiochemical ageing of EPDM shows the formation of unsaturated bonds (trans-vinylene, vinylidene, and vinyl) as well as the consumption of double bonds by radical addition reactions. Since EPDM already contains double bonds which originate in the diene, the last-mentioned reaction predominates.

The overall effect of cross-linking is that the molecular mass of the polymer steadily increases with radiation dose, leading to branched chains until ultimately a three-dimensional polymer network is formed when each polymer chain is linked to another chain. In contrast, scission is the opposite process to cross-linking in which the rupture of $\ce{C-C}$ bonds occurs. Under irradiation in absence of oxygen, EPDM shows some increased cross-linking with a plateau at roughly $10^5\ \mathrm{Gy}$.

However, further exposure to ionizing radiation makes elastomers hard and brittle until disintegration occurs.


Greg E.'s comments sum up the flaws with your version of the mechanism. While yours contains the same steps, it combines the nucleophilic attach and the proton transfer in a single step. Unless you have experimental evidence for a bimolecular step, the unimolecular step is much more likely. In this case, the unimolecular step is a 5-endo-dig cyclization, which is particularly fast. The alkene nucleophile is already present when the carbocation forms. The rate of the deprotonation is diffusion controlled, as we have to wait for a water molecule to find it.

Also, your logic relating the minority of the product to the unlikeliness of the mechanism is inherently flawed. A minor product does not occur at a lower proportion because the mechanism contains less valid steps. Instead, the distribution of products is related to the relative rates of different mechanisms and the relative energies of the products. That product is a minor product because some other product either forms faster or is more stable or both.

If this was on my exam, and I like the problem so I might use something like it, I would dock you a few points for combining steps that are not likely concerted into a concerted step. For example, if I decided this mechanism is worth 10 points, then you might receive 7 or 8.

  Final note: Remember that experimental evidence trumps all else in science.

Regardless of how intuitively and aesthetically unusual your mechanism appears, if experiments support it, then it becomes a more valid mechanism. As an example of some of the non-intuitive things that are in "accepted" mechanisms, check out halonium ions and sigmatropic rearrangements, especially the one in the Baeyer-Villager Oxidation. Evidence in support of your mechanism might include a failure to observe or trap that final carbocation, a demonstration that the rate depends on the amount of water (since final deprotonation steps tend to be very fast), and others.


How can I synthesize 2‐amino‐3‐(chlorosulfonyl)benzoic acid from toluene or benzene?

Here is a proposed route starting from aniline.
First protect the -NH2 with t-BOC
Directed lithiation of t-BOC-aniline (sBuLi/Et2O, -75C) and quench with MeSSMe to install -SMe in the 2 position.
Directed lithiation under the same conditions quenching with Ethyl Chloroformate to install  -CO2Et in the 6 position.
Oxidise the -SMe group to sulfoxide (MCPBA, DCM)
Pummerer rearrangement of the sulfoxide (Ac2O, aq. base work up) gives the free thiol.
Oxidation of the thiol to the sulfonic acid, one example of possible conditions hydrogen peroxide with Rh catalysis
Form the sulfonyl chloride (conditions using trichlorotriazene here. Depending on your choice of conditions you can remove the -BOC group or preserve it. I think preserving it will give a flexible late stage intermediate that is reasonably easily handled. It is possible to directly form the sulfonyl chloride from the -SMe with chlorine in aq acetic acid conditions here but the conditions are harsh and risk removing the -BOC group


Resolution of enantiomers? How to determine chiral compounds apart from each other?

If the compounds have never been made before, then the enantiomers have to be separated prior to analysis. Then analytical and spectroscopic techniques must be used. The R and S configurations cannot even be assigned until the relative stereochemistry had been elucidated. This is usually done by specific NMR experiments. Chiral HPLC can be used to give enantiomeric purity.


How to identify the reduction in a given redox equation of iron chloride and chlorine?

A redox reaction consists of 2 parts, oxidation and reduction These two parts can be written up as half-equations where one half-equation shows oxidation and the other shows reduction.

It seems that you have mixed up the oxidation and reduction half-equations. Oxidation is losing electrons while reduction is gaining electrons. This can be remembered by the mnemonic:

The question is asking for the reduction half-equation. To do this, you must first recognise what is getting reduced (in other words, what is the oxidant). This can be done by looking at the oxidation states for each element before and after the reaction. Since it is getting reduced, it is gaining electrons and therefore its oxidation state should decrease after the reaction. So by looking at the oxidation states, it should be clear that chlorine is getting reduced as its oxidation state goes from $0$ to $-1$.

Therefore the answer is B

Note: The answer definitely can't be a or c since they are oxidation half-equations


Planarity of given molecules in all conformations

To answer this you need to consider the hybridization and geometry around each atom but also think about all of the single bonds present. Single bonds will (generally) allow rotation along their axes as this can happen to a sigma bond without breaking the orbital overlap. If rotation about a single bond moves other atoms in space then you must consider conformational changes that may take atoms out of the plane.

You are correct with A and D. Option C is a ketene and the central C is sp hybridized, it is very similar to the allene in D. However, the terminal oxgyen has no other atoms bonded to it so it doesn't suffer from atoms out of the plane like D. So, C is completely planar (so far as atoms are concerned!), but not for the reason you stated.

For B, which you have shown, it is planar as drawn. Now consider what would happen if you rotated any of the single bonds. Would you break planarity? If not, it is also completely planar. Interestingly, if you rotate the C2-C3 bond, you rotate the alkyne group compared to the alkene but, as the alkyne group has infinite rotational symmetry along its axis, you can't tell the difference, all atoms remain on the plane!


Look at the examples given here:

(Ignore the stuff written with pen; I have a habit of taking notes when reading)

PCC is taking two hydrogen atoms – one from the carbon atom attached to the -OH group and the other one from the -OH group itself, and then making a C=O double bond. Methanol has both of these. So, PCC should be able to oxidise it to formaldehyde.


Neither hydrogen chloride ($\ce{HCl}(g)$) nor hydrochloric acid ($\ce{HCl}(aq)$) are capable of oxidising iron to the ferric state ($\ce{Fe^3+}$) but in the presence of oxygen (air) the ferrous ions initially produced can then be further oxidised to ferric ions:

$$\ce{4 Fe^2+ + O2 + 2 H2O -> 4 Fe^3+ +4 OH-}$$

For reasons explained here, that process is slow in acidic conditions, though.

Dry chlorine does oxidise iron to $\ce{Fe(III)}$ but at room temperature this is likely to be a very slow process.

However, due to the equilibrium:

$$\ce{Cl2 + H2O <=> HCl + HOCl}$$

moist chlorine is likely to be far more aggressive towards most metals including iron, because $\ce{HOCl}$ is a very strong oxidiser, easily capable of oxidising iron (or $\ce{Fe(II)}$) to $\ce{Fe(III)}$. Due to $\ce{Fe(OH)3}$'s extreme insolubility, the hydrated ferric oxide, i.e. rust, would then precipitate.


Can't we just distill the aniline off?

The boiling point of aniline is around 180 °C. Remember that water boils at 100 °C.

Do water and aniline mix?

Yes, but the solubility of aniline in water is rather low (around $\pu{30 g/l}$).

What's the advantage of the steam distillation?

Aniline and water form an azeotrope with a boiling point around 98 °C.


Convert Concentration (particles/ml) to (mg/ml)

First lets see how to get from particles $N$ per volume to mass $m$ per volume.

Dividing the particle number by the Avogadro constant gives moles.

$$
\frac{N}{V}\cdot \frac{1}{N_\mathrm A}=\frac{n}{V}
$$

Multiplying moles with molar mass gives mass.
$$
\frac{n}{V}\cdot M=\frac{m}{V}
$$
Concluding:
$$
\frac{N}{V}\cdot \frac{M}{N_\mathrm A}=\frac{m}{V}
$$

Now your problem arises because you do not know the molar mass. Because you probably have some distribution function of particle size and not just one size. I just assume now that you made an emulsion polymerization in order to get the latex particles or bought them commercially. Then this distribution is very sharp and you can easily take the averaged molar mass as the molar mass of the most probable particle size.


Why does a hydride shift take place in the cannizzaro reaction?

You are correct that a hydride ion is a terrible leaving group and that it can under no circumstances be eliminated in e.g. an $\mathrm{S_N2}$ reaction. However, there are quite a few reactions where hydride does leave one atom and subsequently attaches to another; Wagner-Meerwein rearrangements jump to mind as one of the most prominent examples.

The feature of a Cannizzaro reaction is that it is not actually a leaving hydride. Rather, the reactivity can be explained in a cyclic transition state as shown in scheme 1 below.

Scheme 1: Transition state and products of a Cannizzaro reaction.

Instead of seeing the reaction as a displaced hydride that subsequently attacks another atom, we can see it as a type of pseudo-electrocyclic reaction. Breaking and forming the bonds will happen almost simultaneously. At no point in the mechanism do we actually have a ‘free’ hydride anywhere.

Note, by the way, that the Cannizzaro reaction is not the only one that features a hydride shift in a cyclic transition state; you may, for example, also be interested in the Evans-Tishchenko asymmetric reduction.


How do I write the complete and net ionic equations for Insoluble compounds?

While lead(II) iodate is indeed sparingly soluble in water, this is not true for sodium hydroxide (111 g/100 mL at 20 °C). The reaction that occurs is a precipitation:

$$\ce{Pb(IO3)2_{(aq)} + 2NaOH_{(aq)} -> Pb(OH)2\downarrow + 2NaIO3_{(aq)}}$$

Both lead salts are not readily soluble in water; however, $\ce{Pb(OH)2}$ has a lower solubility ($K_{sp}=1.43 \times 10^{-20}$) than $\ce{Pb(IO3)2}$ with $K_{sp}=3.69 \times 10^{-13}$. Adding $\ce{NaOH}$ to an aqueous solution of lead iodate should therefore precipitate lead hydroxide, which dissolves again with excess $\ce{NaOH}$ under formation of hydroxoplumbates.


Why is boiling point of Ammonia greater than Arsine?

There is the general trend in the groups 15, 16, 17 of raising of boiling points for the binary compounds with hydrogen, going down the groups.
But the first members of each group - $\ce{NH3, H2O, HF}$ - have anomally with their boiling points being exceptionally high, due hydrogen bonds. The strength of these bonds and the boiling point decreases in the order $\ce{H2O > HF > NH3}$.
See the figure 2 on wordpress page for an illustrative picture:

Hydrogen bonds play role in (not limited to):

evaporation and boiling, as molecules need to gain an extra energy to overcome these bonds and to escape liquid.
condensation, acting as cohesive force for molecules to stick to each other or to a condensation center.
gas pressure, as the inter-molecular cohesive force decreases the gas pressure, compared to the pressure of an ideal gas at the same conditions.

Note that the fact that both NH3 and AsH3 are gases at room temperature is irrelevant for the above.


Why is 2s more stable than 2p, if 2p experiences greater effective nuclear charge?

One big issue here is that we generally don't talk about different ionization energies of anything but the outermost orbitals. Once electrons are in "core," we usually don't care since they generally do not participate in any kind of everyday chemistry.

We generally consider the $2s$ orbital to be lower in energy than the $2p$ because the $2s$ has some density inside the radial node that we think of as "nuclear penetration," i.e., the $2s$ electrons are able to get closer to the nucleus through the $1s$ core electrons. Now, once you add a $3s$ electron, I wonder if the innermost radial probability regions of the $3s$ actually repel the $2s$ electrons more than the $2p$ electrons which are generally farther away. The differential repulsion would have to happen in such a way so that it more than compensates for any nuclear penetration effects of the $2s$ electrons. This would certainly align with the observation that the $2p$ electrons are more stabilized only when we add a $3s$ electron.


"Energy gained needs to be greater than the sum of lost energy in all steps"

http://www.lhup.edu/~dsimanek/museum/unwork.htm
 Perpetual motion machines
http://news.sciencemag.org/chemistry/2014/01/rhubarb-battery-could-store-energy-future
 Flow battery, which works.

1) You cannot win (First Law of thermodynamics from Noether's theorems and the isotropy of time)
2) You can only break even on a very cold day (Second Law of thermodynamics from Carnot cycle and the law of large numbers).
3) It never gets that cold (Third Law of thermodynamics.   If there were an entropy difference at absolute zero, T = 0 K could be reached in a finite number of steps. However, at T = 0 K there is no entropy difference, so an infinite number of steps are needed).


What effect does diluting a buffer have on its buffer capacity?

The buffer capacity β is a direct function of the concentrations of all protolytes in the solution. Thus, if the concentration of the protolytes decreases you should then expect a decrease in the buffer capacity.


First AlCl3 converts 1-chloropropane into propane with a positive charge on terminal carbon

Since secondary carbocation is more stable than primary carbocation it undergoes rearrangement and positive charge shifts to secondary carbonation atom.

Hence we get isopropyl benzene


How much steric hindrance does a phenyl group offer?

despite being what I thought was a rather large substituent

I agree, generally speaking the phenyl group is considered a rather large substituent.

One metric commonly used to compare "steric size" is the cyclohexane A-Value.  In the cyclohexane ring system, bulky substituents prefer the equatorial position rather than the axial position. This is due to steric (this can also be described in electronic terms since it is due to the repulsion between interpenetrating electron clouds) destabilization. For example, in methylcyclohexane, the methyl group can exist in an axial or equatorial position. In the axial position, it is destabilized by the repulsive steric (electronic) interactions it has with the axial hydrogens on carbons 3 and 5.

The larger the A-Value, the larger the substituent and the greater its preference for the equatorial location ($\mathrm{\Delta G = -RTlnK_{eq}}$ where $\mathrm{K_{eq}}$ represents the the axial/equatorial ratio at equilibrium).

Here is a link to a Wikipedia article on A-Values.  Look at the table of A-Values in this article.  A methyl group has an A-Value of 1.7 kcal/mol. a phenyl group has an A-value of 3 kcal/mol.  The only substituent in the table with an A-Value larger than phenyl is the t-butyl substituent with an A-Value > 4 kcal/mol. The phenyl group is a relatively large substituent.


Number of spatial arrangements of tetrasubstituted biphenyl

There are three independent stereogenic centers or elements in this molecule,

chiral carbon atom of the chloroethyl group,
double bond of the propenyl group,
axis of the substituted biphenyl with bulky substituents at 2,2′,6,6′ positions.

Each of them has two configurations. It means that the total number of stereoisomers is 23 = 8. (No one of them is symmetrical and cannot be superimposed with its mirror image, no one can be superimposed with any other one.)

(1M)-2-bromo-2′-[(1R)-1-chloroethyl]-6′-iodo-6-[(1Z)-prop-1-en-1-yl]-1,1′-biphenyl
(1M)-2-bromo-2′-[(1S)-1-chloroethyl]-6′-iodo-6-[(1Z)-prop-1-en-1-yl]-1,1′-biphenyl
(1M)-2-bromo-2′-[(1R)-1-chloroethyl]-6′-iodo-6-[(1E)-prop-1-en-1-yl]-1,1′-biphenyl
(1M)-2-bromo-2′-[(1S)-1-chloroethyl]-6′-iodo-6-[(1E)-prop-1-en-1-yl]-1,1′-biphenyl
(1P)-2-bromo-2′-[(1R)-1-chloroethyl]-6′-iodo-6-[(1Z)-prop-1-en-1-yl]-1,1′-biphenyl
(1P)-2-bromo-2′-[(1S)-1-chloroethyl]-6′-iodo-6-[(1Z)-prop-1-en-1-yl]-1,1′-biphenyl
(1P)-2-bromo-2′-[(1R)-1-chloroethyl]-6′-iodo-6-[(1E)-prop-1-en-1-yl]-1,1′-biphenyl
(1P)-2-bromo-2′-[(1S)-1-chloroethyl]-6′-iodo-6-[(1E)-prop-1-en-1-yl]-1,1′-biphenyl


Reaction between sodium carbonate and water

The three reasons you propose are technically true, but they are not the reason why this is not the process used to produce sodium hydroxide.
But let me first clarify some points (and rant about a bit).

Hydrolysis is a reaction where a compound is split whilst incorporating a molecule of water. The reaction :
$$\mathrm{Na_2CO_3 + H_2O \rightarrow 2 NaOH + CO_2}$$
is not a hydrolysis. It is classified as an acido-basic reaction ($\mathrm{H^{+}}$ exchange) followed by a dehydration reaction. A dehydration is kind of the opposite of an hydrolysis (the true opposite would be called hydration), because is involves the loss of water. The transformation decomposes according to (forgetting about sodium ions):
$$\mathrm{{CO_3}^{2-} + H_2O \rightarrow {HCO_3}^- + HO^-} \\
\mathrm{{HCO_3}^{-} + H_2O \rightarrow {H_2CO_3} + HO^-} \\
\mathrm{H_2CO_3 \rightarrow CO_2 + H_2O }$$
$\mathrm{H_2CO_3}$ is carbonic acid. Carbon dioxide is its anhydride.
The overall transformation actually consumes two molecules of water as acid, and one molecule of water is produced upon dehydration of carbonic acid. But this does not appear in the simplified equation above.

This reaction is actually not thermodynamically favorable ; the reverse transformation is the one that is spontaneous. What drives it here is the fact that the carbon dioxide is gaseous and is driven off the water, into the atmosphere. Unless you work under a $\mathrm{CO_2}$-free atmosphere, it would not be possible to produce sodium hydroxide not contaminated with carbonate.

Now, why is this not used industrially ? The answer lies in the cost of the processes : investments, availability of reactants, price of energy, etc.

Calcinating sodium carbonate at more than 100°C actually gives off sodium oxide ($\mathrm{Na_2O}$), but it also releases $\mathrm{CO_2}$ in the atmosphere, and you have to obtain sodium carbonate first. This is no commonplace mineral. Most of it is actually produced by the Solvay process, which converts two plentiful materials, namely calcium carbonate and sodium chloride into calcium chloride and sodium carbonate :
$$\mathrm{2\,NaCl + CaCO_3 \rightarrow Na_2CO_3 + CaCl_2}$$

The chloralkali process produces sodium hydroxide (and chlorine) directly from sea salt, and the  products are hardly contaminated.


Mechanism for cyclic enamine formation after N-Boc deprotection

Those conditions just give the amine. One you have an amine and an aldehyde or ketone, you should form an imine or iminium ion. Imines are relatively stable, but secondary amines, such as the one in this question, can only form the iminium ion. But these iminium ions frequently generate enamines instead by tautomerization.

You should consult your study materials on imine and iminium ion formation. The mechanisms there are very good to know in general because they reinforce the general reactivity pattern for carbonyl addition and elimination from the tetrahedral intermediate.


How to distinguish between carbon-carbon double bond lengths?

You can draw hyperconjugation structures to compare bond length. The more α-hydrogens the alkene has, the more hyperconjugation structures it has, and the higher is the degree of delocalization of the π-bond and therefore the longer is the bond. When double bond character decreases, bond length increases. Therefore the answer should be (d).


Finding specific activity of an enzyme from Km values for a reversible enzyme reaction

If the enzyme is staturated with substrate, the reaction rate will be $k_2$.

1450 moles per second, per mole of enzyme.

87,000 moles per minute, per mole of enzyme.

87,000,000,000 micromoles per minute, per mole of enzyme.

Based upon "50,000 kDa" (which looks like a typo, 50 kDa sounds more reasonable), one mole of enzyme is 50,000,000 grams or 50,000,000,000 milligrams.

So 87/50 = 1.74 units per milligram.

Then for when product is staturated, use $k_{-1}$ instead of $k_2$.


Calderon & Mohazzabi give an excellent summation of the various theories proposed through the years to explain why ice is so slippery in their 2018 paper "Premelting, Pressure Melting, and Regelation of ice revisited" in the Journal of Applied Mathematics  and Physics.
They  offer both theoretical and experimental evidence that neither pressure melting nor friction melting explain the phenomenon and conclude from atomic force microscopy, among other evidence, that there is a pre-melting Quasi-Liquid surface layer with special properties that make ice skating possible.
References

Calderon, C. and Mohazzabi, P. (2018) Premelting, Pressure Melting, and Regelation of Ice Revisited. Journal of Applied Mathematics and Physics, 6, 2181-2191. doi: [10.4236/jamp.2018.611183.][1]

Preview/read online at :  https://www.researchgate.net/publication/328766489_Premelting_Pressure_Melting_and_Regelation_of_Ice_Revisited


Where to buy/How to build a Devanathan-Stachurski cell?

I would recommend you to do a research to find scientific articles on using DS cells. In this case if you have any specific questions you can try to contact authors directly, hopefully they will reply and be able to help you.

You can start from this article (Mike J. Danielson, Corrosion Science 2002 44 (4), pp. 829-840) http://www.sciencedirect.com/science/article/pii/S0010938X01001032.
The author made his own DS cells as described follow


How to calculate molecular volume from the wavefunction?

the volume seems to be already in the output:

$\text{Molar volume}=305.994~\mathrm{bohr^3/mol}\ (27.304~\mathrm{cm^3/mol})$

But of course, this is defined with the volume inside a density contour of $0.001~\mathrm{e/bohr^3}$.  That may be different from what you measure, because volume is not a quantity with well defined meaning, and what you find in an experiment depends on how you measure it.

Mathematically any atom or molecule have a volume of infinity since the electron spreads to the entire space, but this is clearly not the physically meaning of volume as we usually know it.


Your result is surprising, for the following reasons : 1) Nickel compounds and ions are mainly green. And you say your solution is blue. 2) When treated by hypochlorite ions, Nickel will produce a black precipitate of $\ce{Ni_2O_3}$, which is not what you got . 3) You obtain a solution without any precipitate, although the hypochlorite solution is always basic, which must produce a Nickel Hydroxyde precipitate. 4) The only possibility to have a blue solution with Nickel, is adding ammonia $\ce{NH_3}$, which produces a blue Nickel-Ammonia complex in solution. So my question : Are you sure that you have taken pure Nickel in the beginning ? Other question : With this method you will obtain a lot of calcium chloride with a small amount of nickel (or copper ?) compound.


How do restricted open shell calculations mathematically proceed in the context of the self consistent field iterations?

The paper by Tsuchimochi and Scuseria describes it quite well.

Briefly, Roothan-Hall equation is the same but the ROHF matrix is constructed from the $\alpha$ and $\beta$ Fock matrices from UHF.

$$\mathbf{F}_\mathrm{ROHF} =
\begin{bmatrix}
    \mathbf{R}_{cc}       & \mathbf{F}_{co}^\beta & \mathbf{F}_{cv}^{cs} \\
    \mathbf{F}_{oc}^\beta & \mathbf{R}_{oo} & \mathbf{F}_{ov}^\alpha \\
    \mathbf{F}_{vc}^{cs}  & \mathbf{F}_{vo}^\alpha & \mathbf{R}_{cc} \\
\end{bmatrix}$$

where $c$ is doubly occupied, $o$ is singly occupied, $v$ is virtual and

$$\begin{align}
\mathbf{R}_{cc} &= A_{cc}\mathbf{F}_{cc}^\alpha + B_{cc}\mathbf{F}_{cc}^\beta\\
\mathbf{R}_{oo} &= A_{oo}\mathbf{F}_{oo}^\alpha + B_{oo}\mathbf{F}_{oo}^\beta\\
\mathbf{R}_{vv} &= A_{vv}\mathbf{F}_{vv}^\alpha + B_{vv}\mathbf{F}_{vv}^\beta
\end{align}$$

  Different values for these parameters have been suggested in the literature. Although they do not affect the ROHF wave function and energy, they affect orbital energies whose physical meaning is obscured because of this dependence.

Takashi Tsuchimochi and Gustavo E. Scuseria, J. Chem. Phys. 2010, 133, 141102. DOI: 10.1063/1.3503173; available as ePrint: arXiv:1008.1607 [cond-mat.str-el].


What value of temperature does T represent in Gibbs Free Energy Equation?

We can not calculate the exact amount of G (Gibbs free energy), only we can calculate change of that,so we must put temperature change on this equation:
$$\Delta G = \Delta H - \Delta TS $$


For (1), I think you made the mistake of using a different value for overall cell potential when you worked backwards. Because if you use the correct overall cell potential, you get $K = 31.52$, your $\log K$ would be different if you used $0.685$ (which would make $\log K = 69.43$). Using the correct $E_\mathrm{cell}$ will get you the correct value of $n$ that you're looking for.

To solve for (2), think of what $K$ stands for. It's the ratio of something. Rearrange the ratio to solve for the unknown. It might seem confusing since you might think of an electrochemical solution, but that isn't always the case.


Is benzenesulfonic acid or cyclohexanesulfonic acid more acidic?

Draw out the full line-bond or Lewis structures for each and you will see that there is no way for the benzene ring to contribute to resonance structures that involve the sulfonic acid group, or the sulfonate conjugate base. Instead, the important contribution is the inductive effect. Benzene rings are more electron-withdrawing than alkyl groups, so benzenesulfonic acid should be more acidic than cyclohexanesulfonic acid. A comparison of pKa values for $\ce{PhSO3H}$ ($-2.8$) vs $\ce{MeSO3H}$ ($-1.9$) supports this assertion (Can. J. Chem. 1978, 56 (17), 2342–2354).


Dead trees or charcoal is generally used as a reductant in the production of pure silicon for use in industry what other reductants could be used?

On a small scale, hydrogen gas is certainly an option. Here is a paper: Reduction Kinetics of Metal Oxides by Hydrogen. High temperatures induce the formation of monoatomic hydrogen as the active agent.
However, on a practical note, explosive mixtures of hydrogen with air exist over a wide range (Wikipedia has a good discussion on this topic), so this path is clearly not without danger.
[EDIT] To address a question on trials, not that am aware of, however, here is an interesting experiment one may conduct at a seemingly much lower temperature that is likely safe. For reducing, it is known that in situ creation of the hydrogen atom radical (cited as the major reducing species by Buxton, see Eq 3 on page 2 here). Suggestively, the chemisorption of the hydrogen radical on the surface of metals like Zn, Al and Mg (albeit, effective for a limited lifespan) may be an interesting trial path.
So, as an experiment, proceed to imbue the surface of Mg or Al metal with the hydrogen atom radical (from the traditional nascent hydrogen generation methods based on say Al/NaOH). Place the SiO2 on the imbued metal surface and heat.
As to the reaction mechanics, the •H radical functional behaves based on the equation:
$\ce{ e- + H+ <=> •H }$
where this seemingly reversible formation reaction apparently acts as (e-,H+) pair on ions. For an example, citing a work in Hydrometallurgy 2008: Proceedings of the Sixth International Symposium on p. 818, a suggested commercial reductive leaching equation, to quote:
$\ce{PbS + 2 •H = Pb + H2S (5) }$
Now, could a related reaction path exist with SiO2 at a much lower temperature? To answer, this suggested trial could be quite interesting.


Deuterated solvents vs. regular solvents

In general, it is safe to assume that solubilities in non-deuterated and deuterated solvents are much the same, and this would be the logical solvent of choice if looking for a NMR solvent.

However, there is a small difference in relative solubilities between H/D solvents, and you can gauge potential for differences based on the dielectric properties of the two. By memory, I think H2O/D2O have the largest difference (which is still pretty small).

Another factor that can cause differences in solubilities comes from the relative purity of the two solvents. For instance, many prepackaged deuterated solvents are often drier than bulk bench solvents, and this can sometimes cause significant solubility differences, especially in hygroscopic solvents like DMSO. I've seen countless examples of people claiming a compound was readily soluble in DMSO, only to have it sit like a brick when trying to dissolve in a freshly cracked vial of DMSO-d6.


How can chlorine have an oxidation number of 3 in chlorite?

It is indeed in the +3 oxidation state. Keep in mind, though, that that does not mean chlorine lost three electrons. That polyatomic ion is not made up of smaller ions.


Why are silyl chlorides more readily hydrolysed than alkyl chlorides?

Using $\ce{CCl4}$ and $\ce{SiCl4}$ as the basis for comparison, there are several reasons.

The greater electronegativity difference between Si and Cl leads to a larger $\delta^+$ on silicon, making it more electrophilic and susceptible to attack by water.
Larger size of Si means that there is less steric hindrance towards the approach of water.
It is possible that the hydrolysis of $\ce{SiCl4}$ proceeds via an alternative, faster pathway that involves a five-coordinate intermediate (as opposed to $\ce{CCl4}$, in which five-coordinate carbon does not occur).


Why f is multiplied in solubility of salt constant

Notice that $f$ is equal to the maximum concentration of $\ce{M}$, which is unchanged as part of the acid-base equilibrium. On the other hand, the pH changes the concentration of the counterion in proportion to $f$. Therefore, the second term in the solubility product is scaled accordingly.


Arrange the following in increasing order of acidity: water, ammonia, ethyne and ethane

The order of acidic nature would be: water > ethyne > ammonia > ethane.

Since oxygen has highest electronegativity amongst the atoms of the compounds, so its hydrogen will be most acidic and hence will be best acid among them.

Ethyne contains $\mathrm{sp}$-carbon and as a result the electronegativity of the carbon in ethyne is more than $\mathrm{sp^3}$ due to increased $\mathrm{s}$-character (50%) in the bonds between the carbon atoms.

The electronegativity of the carbon increases to a great extent when  and turns out to be 3.2, which is greater than the electronegativity of nitrogen which is 3.04. Hence the hydrogen of ethyne is more acidic than that of $\ce{NH3}.$

Ethane does not possess any acidic hydrogen hence it is least acidic.


Synthesis of Haloperidol with 4-chlorobutanoyl chloride and piperidin-4-one

The reaction mechanism for the step in question simply involves initial formation of a Grignard reagent followed by nucleophilic attack into the ketone.

When forming Grignard reagents from halides, there is a general trend that the rates of formation are in the order  I > Br >> Cl, making it possible to get selectivity where multiple halides exist.


What does this dashed and curved line mean in pyruvate molecular graph?

It actually hybrid form of the contributing resonance structures of the ion. The significance of the dashed line is that it shows delocalisation of negative charge over the two oxygen atoms.

Note that

Neither of the above two forms represent the actual compound
Actual compound is represented by the resonance hybrid :


Calculating equilibrium constants: adding two or more equations

This is a direct result from the law of mass action.

When you have an equilibrium

$$\ce{A <=> B}$$

with the equilibrium constant

$$K_1=\frac {[\ce{B}]}{[\ce{A}]}\tag1$$

and a second equilibrium

$$\ce{B <=> C}$$

with the equilibrium constant

$$K_2=\frac {[\ce{C}]}{[\ce{B}]}\tag2$$

you can combine both chemical equations:

$$\begin{align}
\ce{A \;&<=> B}\\
\ce{B \;&<=> C}\\
\hline
\ce{A \;&<=> C}
\end{align}$$

Then you can rearrange Equation $\text(1)$

$$\begin{align}
K_1&=\frac {[\ce{B}]}{[\ce{A}]}\tag1\\[6pt]
\Leftrightarrow [\ce{B}]&=K_1\cdot[\ce{A}]
\end{align}$$

and substitute $[\ce{B}]$ in Equation $\text(2)$

$$\begin{align}
K_2&=\frac {[\ce{C}]}{[\ce{B}]}\tag2\\[6pt]
K_2&=\frac {[\ce{C}]}{K_1\cdot[\ce{A}]}\\[6pt]
\Leftrightarrow K_1\cdot K_2&=\frac {[\ce{C}]}{[\ce{A}]}
\end{align}$$

Thus the equilibrium constant for the combined equilibrium is the product $K_2\cdot K_1$.


Why can entropy be measured in joules per kelvin?

The definition of entropy in 'Classical Thermodynamics' is $$\Delta S = \int\frac{\delta Q}{T}$$

The quantity on the left is the entropy change associated with a physical process. $\delta Q$ is an inexact differential, made exact by an integrating factor $\frac{1}{T}$ This notion was devolped by Kelvin and Cartheodory (I believe). This comes out of the analysis of certain kinds of differential equations called Pfaffian forms.

One notes the dimensions ("units") of this quantity, and observers them to be $\mathrm{J\ K^{-1}}$

It was only later, that the statistical definition of entropy (developed by Boltzman) came to be (and was found to be equivalent to the thermodynamic definition).

$$S = -k_B \sum_{i}p_i\ln(p_i)$$

In this framework entropy is a logarithmic measure of the number of states with significant probability of being occupied. In this expression, the the quantity is changed from dimensionless to dimensionful, by the Boltzmann Constat, which has the units, (you guesed it)  $\mathrm{J\ K^{-1}}$

The measure of "disorder" is in the term $\sum_{i}p_i\ln(p_i)$, which has no units. The units are appended to it by the Boltzmann Constant.


Your approach is correct.

The given compound includes the characteristic groups $\ce{-CN}$, which are expressed as suffix ‘carbonitrile’. Since there is only one kind of suffix, the order of seniority of suffixes is not relevant in this case.

Functional suffixes are always attached to the name of the parent structure, which is modified by ‘ene’ or ‘yne’ endings if necessary.

When there is a choice among acyclic parent structures, a principal chain must be chosen. The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) stipulates various criteria for the selection of the principal chain. These criteria are applied, in the order listed, until a decision is reached.

For the given compound, there is a choice between two different principal chains:

    and

The first relevant criterion in this case provides that the principal chain has the maximum number of substituents corresponding to the principal characteristic group (suffix).

  P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).

Therefore, the principal chain of the given structure has all four substituents $\ce{-CN}$ corresponding to the suffix ‘carbonitrile’.

Furthermore, the current IUPAC recommendations stipulate various criteria for the assignment of locants when several structural features appear in a compound.

In particular, the order of seniority provides that low locants are given first to principal characteristic groups (here: carbonitrile) before low locants are given to multiple bonds (here: the double bond) and before low locants are given to detachable alphabetized prefixes (here: methyl).

  P-14.4 NUMBERING

  When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:

  (…)

  (c) principal characteristic groups and free valences (suffixes);

  (…)

  (e) saturation/unsaturation:

    (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;

    (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);

  (f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;

  (…)

Therefore, the resulting name is (1E)-2-methylhept-1-ene-1,3,4,5-tetracarbonitrile (not (6E)-6-methylhept-6-ene-3,4,5,7-tetracarbonitrile) since the locant set ‘1,3,4,5’ for the principal characteristic groups is lower than ‘3,4,5,7’.


I'm not sure if there's definitive answer to not-yet-asked question, whether this compound can be considered organic or inorganic compound.

While organic nomenclature books do not mention it (acording to the current rules it might be called sulfanylidenemethanone, which is somewhat absurd), official IUPAC inorganic chemistry nomenclature (Red Book 2005, table IX, p. 292) lists two names for this compound:

carbonyl sulfide
oxidosulfidocarbon

Your mentioned (organic) systematic name thioxomethanone is just a variant of already mentioned sulfanylidenemethanone, as S= group's preferred name ("preselected prefix") is now sulfanylidene-* instead of thioxo- (rule P-15.5.3.3)

*) see analogy H2C=/H3C− methylidene-/methyl-, and
S=/HS− sulfanylidene-/sulfanyl- (older name: mercapto-)


Do two Miller planes need to have similar arrangement of atoms?

In answer to you first question, no if hkl differ, yes if they do not. I try to explain below.

In a crystal the unit cell defines the repeating unit. Inside the unit cell the atoms are arranged as they are in the molecular structure with the molecule being at the same angle and position within each of the unit cells. Each atom scatters the x-ray radiation depending on how many electrons the atom has and on the position the atom has in the unit cell. The many repeating identical unit cell structures cause constructive interference and so spots are observed on the x-ray detector. Thus the structure is determined from the intensity of the spots on the detector not their position.

The hkl planes are a construction we devise to allow determination of the x-ray amplitudes because the constructive interference that causes scattering has to occur from atoms in the repeated positions as determined by the unit cell. (There are thousands to millions of unit cells in a crystal). Thus we define planes starting at  hkl = (0 0 0) and continue to larger values such as  (1 2 0 ) etc. etc. (100 50 1) etc. (Negative values have a line above the number $\bar 3$ for example.)

Suppose we choose the plane (1 2 3), this is  parallel to (2 4 6) to (4 8 12) and so on and these planes are separated by a distance we will call $d_{hkl}$. We can reduce all these  planes to (1 2 3) and the reason for this is that the phase of the x-ray wave from plane (2 4 6) is $2\pi$ from that of (1 2 3) and that of (4 8 12) is $4\pi$ shifted. But as the phase repeats itself over $2\pi$ all these larger phases are simply the same as $2\pi$. A phase change of $2\pi$ is equivalent to a path difference of $\lambda$ or a whole wavelength, which might be easier to envisage as then waves add together exactly.

Now, if an atom is found bang on the plane (1 2 3) it will have a phase of $2\pi$ (or path-length change of $\lambda$) with respect to the origin so has the maximum scattering intensity possible for that plane. However, consider that the atom in not on the (1 2 3) plane, but at a position $x$ away (and so between two (1 2 3) planes) then the phase it has is now $\displaystyle \frac{x}{d_{hkl}}2\pi $ and so the scattered x-ray waves are out of phase by this amount and so the scattered amplitude is reduced. Now, of course, every the atom in the molecule will be at some distance from one of the parallel (1 2 3) planes and so will have its own particular scattering amplitude to add to the total for that plane.

The same reasoning applies to each of the unique hkl planes and to each atom. It is the un-ravelling of these amplitudes that produces the x-ray diffraction structure and it is made more complicated because the x-ray detectors measure the intensity of the scattered light (which is  amplitude squared (as $I^*I$)) which means that the phase information is lost. This causes great difficulties in solving the x-ray structure.


Work done in expanding a gas reversibly and irreversibly

I will call your processes (1), (2), and (3). I assume (1) is reversible isothermal expansion, (2) is irreversible isothermal expansion, and (3) is unspecified.
The key idea is that the inequality $w_\text{rev} \geq w_\text{irrev}$ can only be applied to reversible and irreversible analogues of the same process. (1) and (2) satisfy this requirement because (1) is the reversible analogue of (2), but (1) is not the reversible analogue of (3).
Physically, (1) is like having a pile of sand on top of a container (containing an ideal gas) whose lid is a piston, and then slowly blowing away the sand. (2) is the same scenario, except you sweep off the sand in one grand gesture. But (3) is a different scenario entirely.
If we refer to the first law, $\Delta U = q + w$, it's clear that, given an initial and a final state, we can make $w$ as large as we want as long as we have a commensurate value of $q$. Just because processes (1), (2) and (3) all have the same initial and final states does not mean that it makes sense to apply the inequality $w_\text{rev} \geq w_\text{irrev}$ to any two of them.


Even though the pressure remains constant, heat can get in (or out) of the system. From the cup calorimeter, we are trying to find $\Delta H$ which is equal to the heat evolved from the system at constant pressure. We measure this heat and our job is done.

$$\Delta H=q_p$$

If we want to calculate $\Delta U$, we just to find the heat evolved from the bomb calorimeter, which can be done by measuring the change in calorimeter's final and initial temperature given that the heat capacity of the calorimeter is known.

$$\Delta U=q_v=nS\Delta T$$


Chiral centers in albuterol-like compounds

You can surely count that nitrogen atom as a "chiral center" because it has four different groups attached to it.

However, the net optical activity due to that chiral center is zero because the two optical isomers from that nitrogen atom are rapidly interconverting with each other ("Amine inversion"). They always exist as a racemic mixture, and hence are optically inactive.


Is cyclopentadiene or cycloheptatriene more acidic?

Essentially it's a case of aromaticity vs number of resonance
  structures.

Well put!

Aromaticity is a very strong driving force so aromaticity wins out; Huckel's rule is more important than the number of resonance structures.  Therefore cyclopentadiene is more acidic than cycloheptatriene.  Even though we can draw 7 resonance structures (one with a negative charge on each carbon atom) for the tropylium anion and only 5 for the cyclopentadienyl anion, Huckel's rule tells us that the cyclopentadiene anion ($4n+2$ π electrons, $n=1$) is aromatic while the tropylium anion with 8 π electrons (doesn't fit $4n+2$) is not.

Another example is cyclooctatetraene.

Like benzene, we could draw resonance structures by shifting the double bonds in this molecule too.  But this molecule is not aromatic, it does not fit the $4n+2$ rule.  In fact, Huckel says with 8 π electrons it is antiaromatic.  To avoid this destabilization cyclooctatetraene adopts a tub-like conformation.

(image source)

Again aromaticity trumps resonance structures.

Edit: Huckel's Rule: Aromaticity - Antiaromaticity

If we look at the energetic positioning of the molecular orbitals (MO's) in a cyclic, conjugated polyene, we can quickly understand the basis for Huckel's rule.

We can use Frost diagrams to determine the energetic positioning of these MO's.  Here is the diagram for benzene, we see that all of the pi electrons are in bonding MO's, the molecule is a stable aromatic that fits the $4n+2$ rule ($n=1$).

Here is the diagram for cyclooctatetraene, and we see that not all of the electrons are in bonding MO's, two π electrons are in non-bonding MO's.  They don't contribute to bonding or stabilization.

This is a $4n$ π electron compound ($n=2$) and accordingly would be classified as antiaromatic by Huckel's rule.  As mentioned above, the compound adopts a non-planar conformation to avoid this destabilization.


When dealing with an aqueous solution, you are correct that the $\ce{H+}$ ion is equivalent to $\ce{H3O+}$ for all intents and purposes. Due to the abundance of water in solution, molecules of $\ce{H2O}$ will readily pick up the hydrogen ions, meaning that most of the $\ce{H+}$ in an aqueous solution is actually of the form $\ce{H3O+}$. While this is the case, using $\ce{H+}$ over $\ce{H3O+}$ is more correct in general contexts. If you are using something like ethanol ($\ce{CH3CH2OH}$) as your solvent, then instead of having hydronium ions, you would have $\ce{CH3CH2OH2+}$ as your measure of concentration of acid.

As for your other questions, the $\ce{H+}$ would be considered an ion rather than an atom. This is because it is a charged species. Either $\ce{H+}$ can be referred to as a proton or as a hydrogen ion, since $\ce{H-}$ is too unstable a species to be formed. While some of your $\ce{H+}$ in solution might be deuterium or tritium, most people will understand what you are talking about if you just call the hydrogen ion a proton.


Calculate voltage, cost and time of nickel coating

The volume of nickel plated is surface area times the thickness.
That's $2m^2 \times \pu{0.03 cm} $.

We'll use cubic cm as the units so it is: $2m \times 1m \times \pu{0.03 cm} = \pu{200 cm} \times \pu{100 cm} \times \pu{0.03 cm} = \pu{600 cm3} $

This helps to find the mass as the density is given as $\pu{8.9 g/cm3}$.

The mass = density $\times$ volume = $ \pu{8.9 g/cm3} \times \pu{600 cm3} = \pu{5.34 kg}$

Convert this to moles of Nickel: $\displaystyle \frac{5.34 \times \pu{10^3 g}}{\pu{58.71 g/mol}} = \pu{90.96 mol} $.
We'll come back to this.

Now for the current, $\pu{0.3 A/cm2}$ applied to a $5 \times \pu{5 cm}$ square plate, meaning that the total current on the plate at a given time is $\pu{0.3 A} \times \pu{25 cm2} = \pu{7.5 A}$ or $\pu{7.5 C/s}$. However, only a fraction of that actually reduces the nickel ($75\%$) so $\pu{5.625 A}$.

The voltage is then $ V = IR = 5.625 \times 0.4 $

The time calculation involves the actual nickel.
$I = \displaystyle \frac{q}{t}$
Where $q =$ total charge

We have $\pu{90.96 mol}$ of Nickel reduced.

$\ce{Ni^{2+} + 2e^- -> Ni}$

That's $\pu{2 mol}$ of electrons per mole of Nickel. So we need $\pu{181.92 mol}$ of electrons.

The charge per mol of electrons is Faraday's constant of $\pu{96485.3 C/mol}$
The total charge is just $\pu{181.92 mol} \times \pu{96485.3 C/mol}$.

Then we rearrange the formula of current for time $\displaystyle t = \frac{q}{I} = \frac{\pu{181.92 mol} \times \pu{96485.3 C/mol}}{\pu{5.625 C/sec}}$
$\text{time} = \pu{3120463 seconds}$ or $\pu{866 hours}$.

That's sort of the basic idea I'd imagine. I may have made a mistake somewhere in between. The time though while seems long is in line with this:
https://sciencing.com/calculate-electroplating-7597391.html

There, they plated just one mole of Cu and with a higher current. That took 2 hours. We have a lot more of $\ce{Ni}$ and smaller current.

Then $\text{Energy} = \text{power} \times \text{time} = VIt$ and you can calculate the cost from there.


Yes, when ice is present at equilibrium with liquid water (at 1 atm), the system is at 0 °C.

If you continue to remove heat from the water (such as in your freezer), the temperature of the system will remain at 0 °C until all the liquid (water) becomes solid (ice) after which, continuing to remove heat will result in a drop of temperature below 0 °C. Conversely, if heat is added to the system, the temperature of the system will remain at 0 °C until all of the solid (ice) has melted into liquid, after which, the temperature will increase above 0 °C. The amount of (latent) heat required to melt ice is called its enthalpy of fusion (334 J/g). This energy is converted into intermolecular bonds (a change in internal or potential energy of the system) rather than a change in kinetic energy (that is, a temperature change).

Note that at a much lower pressure (0.006 atm) and slightly higher temperature (0.01 °C), pure water will be in three states or phases (gas, liquid and ice). This is known as the triple point of water, as illustrated in the phase diagram of water, shown below:


Catalytic energy profiles (is this Wikipedia image misleading?)

Of course, in a strict sense the reaction path using a catalyst should be split into four sub-reactions (all going by the exact image) and each of those sub-reactions should get their own activation energy assigned. Two of the four steps would be endothermic, two exothermic. (Note that it is not necessary for the endothermic steps to occur first; it could be the other way around, mixed or anything.)

However, there is still some truth in labelling ‘$E_\mathrm{a}$ with catalyst’ as such. Even though you have much smaller steps you will still need to add that labelled energy in net terms to overcome the barrier. The principal difference is that it need not be added all at once but can be added stepwise. You could even say that the total activation energy (i.e. adding up all the individual $E_\mathrm{a}$) be even higher, but you regain some of that from the intermediate reactions.

If you decided to supply only enough energy so that a molecule could jump the barrier from intermediate 2 to intermediate 3, that would not make the reaction go. For one, the barrier to get to intermediate to is larger by itself. For two, you would only get that much above the baseline which might be enough to cross the barrier to intermediate one but no further.


Any method for testing Nitrate and Nitrite in biological fluids like aqueous humour?

There is 1982 publication describe nitrate, nitrite, and $\ce{[^{15}N]}$-nitrate in biological fluids. It's abstract describe the procedure briefly:

  Abstract: A new automated system for the analysis of nitrate via reduction with a high-pressure cadmium column is described. Samples of urine, saliva, deproteinized plasma, gastric juice, and milk can be analyzed for nitrate, nitrite, or both with a lower limit of detection of $\pu{1.0 nmol} \; \ce{NO3−}$ or $\ce{NO2−/ml}$. The system allows quantitative reduction of nitrate and automatically eliminates interference from other compounds normally present in urine and other biological fluids. Analysis rate is 30 samples per hour, with preparation for most samples limited to simple dilution with distilled water. The application of gas chromatography/mass spectrometry for the analysis of $\ce{^{15}NO3−}$ in urine after derivatization to $\ce{^{15}NO2}$-benzene is also described.

Thousands of samples of human and rat urine, human saliva, and plasma have been analyzed for nitrate and nitrite using this system. Typical ranges of nitrate and nitrite levels in biological samples are shown in the Table.

$$
\begin{array}{ccc}
\hline
\text{Biological fluid} & \text{$\ce{NO3-}$ Concentration, $\pu{\mu M}$} & \text{$\ce{NO2-}$ Concentration, $\pu{\mu M}$}  \\ \hline
\text{Urine} & 250-2000 & \text{ND}^a \\
\text{Saliva} & 200-600 & 30-210 \\
\text{Plasma} & 15-60 & \text{ND}^a \\
\text{Gastric juice} & 50-85 & 0.4-60^b \\
\text{Milk} & 20-30 & \text{ND}^a \\\hline
\end{array}
$$
$^a$Not determined. $\; \; ^b$Nitrite concentrations at the high end of this range are from people with gastric pathologies.

In general, for methods of the detection and determination of nitrite and nitrate, read Ref.2  (a review) and 3. However, Ref.1 claimed that the method in Ref.2 did not work for biological fluids.

References:

Laura C. Green, David A. Wagner, Joseph Glogowski, Paul L. Skipper, John S. Wishnok, Steven R. Tannenbaum, “Analysis of nitrate, nitrite, and $\ce{[^{15}N]}$nitrate in biological fluids,” Analytical Biochemistry 1982, 126(1), 131–138 (https://doi.org/10.1016/0003-2697(82)90118-X).
Qiu-Hua Wang, Li-Ju Yu, Yang Liu, Lan Lin, Ri-gang Lu, Jian-ping Zhu, Lan He, Zhong-Lin Lu, “Methods for the detection and determination of nitrite and nitrate: A review,” Talanta 2017, 165, 709–720 (https://doi.org/10.1016/j.talanta.2016.12.044).
M. P. Stainton, “Simple, efficient reduction column for use in the automated determination of nitrate in water,” Analytical Chemistry 1974, 46(11), 1616–1616 (https://doi.org/10.1021/ac60347a051).


Restrictions on transitions in Spectroscopy

If the energy is to be transferred from the photon to your atom one energy gap must match the energy of the photon. Thus absorption may occur from the first to second excited state in your example as the energy in both is 2E. However, there has to be some population in the first excited state for this to happen, zero population zero absorption. In fact it is a bit more subtle in that there has to be a difference in population between the two levels involved. It is possible for the photon to stimulate a transition from the upper level to the lower one, just as from the lower one to the upper one. The reason why this seems odd is that in most cases the population in the upper level is vastly smaller than in lower ones (via Boltzmann distribution) and the population in the upper level can often be set to zero without serious error. This is particularly true in electronic spectroscopy, as in the case of atoms, but far less so for pure rotational transitions in molecules which occur in the microwave region of the spectrum. You will find details in any good spectroscopy book.

Thus neither of your cases occurs but something different; transitions between the first to second excited state. How strong this transition is depends on the population in the two levels and on selection rules involving angular momentum and electron spin changes.


How are solvents chosen in organic reactions?

There is no list per se, but there is certainly a process to determine appropriate solvents for a given reaction. For the specific reaction you mention, here are some criteria you would consider to select a solvent:

Because this is an oxidation reaction, choose a solvent that cannot be oxidized, known as an aprotic solvent.
The solvent needs to dissolve the reagents in the reaction: hypochlorite and 4-tert-butylcyclohexanol. Hypochlorite is a polar molecule, and 4-tert-butylcyclohexanol is a little polar thanks to the alcohol group. The alcohol group will be oxidized to a ketone by the reaction, so the solvent should be able to dissolve that reasonably well. Conclusion: choose a moderately polar solvent.

So a polar aprotic solvent is needed for this reaction. There are two more considerations: cost and safety

Obviously, the cheapest solvent with the lowest potential for harm should be selected.

A list of some common polar aprotic solvents for consideration might include:

tetrahydrofuran
N,N-dimethylformamide
acetonitrile
dimethyl sulfoxide
acetone

Tetrahydrofuran could be a reasonable choice, but it forms peroxides over time, which are quite explosive.

N,N-dimethylformamide has some health concerns: GHS07 and GHS08

Acetonitrile also falls into the GHS07 category: it causes cyanide to form in the body at high concentrations.

Dimethyl sulfoxide does not have health or safety concerns.

Acetone also falls into the GHS07 category: it can cause nausea and irritation at high exposure levels.

The health concerns of acetonitrile, dimethyl sulfoxide, and acetone are largely not going to pose a hazard for a laboratory experiment. All three of these can likely be found in any organic teaching lab.

Cost comparison, from Sigma Aldrich:

Acetonitrile: 2 L = $172.50

Dimethyl sulfoxide: 2 L = $208.50

Acetone: 2 L = $94.80

Therefore, choose acetone for this reaction!


Phenol is both a molecule and a functional group (or at least a class of molecules). IUPAC has the definitive answer on the matter:

Phenols:

  Compounds having one or more hydroxy groups attached to a benzene or other arene ring, e.g. 2-naphthol


I read your question a few times, and I think you're confusing occupancy with spatial probability distribution.
Orbitals have a lot of spatial overlap. But if the orbital has two electrons “in it,” then it cannot describe the probability behavior of any other electrons in that atom, as other electrons must be in another orbital.
A 2s electron absolutely can go into a 1s orbital provided there is space. This is exactly what happens with atomic emission spectra when an electron in energy level $n=2$ transitions to $n=1$ and releases energy.


Deciding the major product in this Wurtz reaction

I think that there should be a 'statistical product distribution', if I'm correct the products should be in the ratios 2:1:1.

You get to this conclusion as follows, call 2-chlorobutane A and 2-chloropropane B.

A can react with A to give AA, and
A can react with B to give AB, while
B can react with A to give BA, while
B can react with B to give BB.

When you realise that AB and BA are the same molecule you get AA:2AB:BB; that is, 1:2:1.


What is the reaction of pentane-2,4-dione with chlorine in presence of NaOH?

I assume that there is sufficient base and chlorine and that the conditions are vigorous enough to get beyond dichloride 4, which, being a non-enolic β-diketone at C3, is labile toward Haller-Bauer-type cleavage. Acetylacetone 1 is highly enolic and readily deprotonated. The first two chlorinations lead to dichloride 4. Cleavage of dichloride 4 with base gives enolate 6 which can be protonated by water, or to save space, chlorinated to provide the trichloroketone 8, which is an intermediate in the haloform reaction of acetone. Cleavage with base leads to chloroform (10) and a second equivalent of sodium acetate (7).