How would I find R/S configuration in this case?

While assigning priority in I, $\ce{-CHO}$ is second priority group instead of $\ce{CH2OH}$ as in $\ce{CHO}$ there is a double bond between $\ce{C}$ and $\ce{O}$, so we can consider the duplicate form, i.e. $\ce{C}$ is bonded to two $\ce{O}$ atoms and $\ce{O}$ is bonded to two $\ce{C}$ atoms. So, the rotation is counterclockwise, but $\ce{H}$ is in front of the observer, so we consider the opposite rotation, i.e. clockwise. Thus it is R configuration.

Instead of using the rule that we have to consider the opposite rotation (while moving the fingers from the more priority group to the less priority), when the least priority is in front of the observer. You can visualize it this way, when you see the the compound I, drawn on paper, then $\ce{H}$ and $\ce{OH}$ is towards you and $\ce{CHO}$ and $\ce{CH2OH}$ is away from you and the rotation is counterclockwise. But when you see the same compound below the paper, the $\ce{H}$ and $\ce{OH}$ is away from you and $\ce{CHO}$ and $\ce{CH2OH}$ is towards you, but now the rotation is clockwise. So now we consider this rotation to say that it is in R configuration.

The figures are basically the Fischer projections of the corresponding molecules. The actual representation of Fischer projections is below.

Hence the 'pushing $\ce{H}$ as far as possible' thing is done when $\ce{H}$ is on dashed bond (into the screen). Just use the rules of Fischer projections and change the molecule to the form which you feel most comfortable to determine R/S configurations.


Let me begin my discussion by first introducing the notion, using some intuitive model building, and will try to derive, or perhaps guess the van der Waals equation of state.

The perfect gas equation, $ PV = nRT $ doesn’t allow for any interactions between molecules. The van der Waals equation is a refinement of this model, in the sense that it introduces the possibility of both repulsive and attractive interactions.

Imagine the gas molecules are small, hard, impenetrable spheres. This accounts for repulsive interactions, as now they can’t get arbitrarily close to each other. Consequently, they are no longer free to move about in a volume, $V$, and only have access to a much smaller volume $V-nb$ where $nb$ is roughly the volume taken up by the molecules.

The modified equation now becomes:
$$P = \frac{nRT}{V-nb}$$

Now to calculate the excluded volume:
$$V_\text{molecule} = \frac{4}{3}\pi r^3$$
where $r$ is obviously the radius.

Now note, a particle is surrounded by a sphere of radius $2r$ (two times the original radius) that is forbidden for the centres of the other particles. If the distance between two particle centres were to be smaller than $2r$, it would mean that the two particles penetrate each other, which, by definition, hard spheres are unable to do.

The excluded volume for the two particles (of average diameter $d$ or radius $r$) is: $\frac{4}{3}\pi (2r)^3$ or, $8V_\text{molecule}$. Divide this by two to get the volume excluded per molecule and we get $b \approx 4V_\text{molecule}$

Note: Of course, molecules are not infinitely hard, as van der Waals thought; there exists some “softness”. Thus, the factor four yields an upper bound; empirical values for b maybe lower.

Now turn on an attractive force between the particles.

Assume, notwithstanding the existence of this force, the density of the fluid is homogeneous.

And assume that a large majority of the particles are always surrounded by other particles, which attract each other (unless they get too close, when they repel; thus, the bulk of the particles do not experience a net force pulling them to the right or to the left. They feel that the container is infinitely large without any walls.

Now, the surface particles feel a net force from the bulk particles pulling them into the container, because there are no particles on the side where the wall is (and we assume here that there is no interaction between walls and particles, which is not true in reality).

This net force decreases the force exerted onto the wall by the particles in the surface layer. The net force on a surface particle, pulling it into the container, is proportional to the number density:
$$F = \frac{N_\mathrm a}{V_\mathrm m}$$

Assuming homogeneity, the number of surface particles is also proportional to the number density. So the total force on the walls is decreased by a factor proportional to the square of density.

Since pressure is force per unit area, it's decrease proportional to:
$$a'C^2 = a'\left(\frac{N_\mathrm a}{V_\mathrm m}\right)^2 = \frac{a}{V_\mathrm m^2}$$

Now, the modified equation becomes:
$$ P = \frac{RT}{V_\mathrm m-b}- \frac{a}{V_\mathrm m^2}$$

Derivation using Statistical Mechanics

For a system of $N$ particles, in 3 dimensions

The partition function:
$$ Z = \frac{1}{N!} \int \mathrm dp^{3N}\mathrm dx^{3N}\mathrm e^{-\frac{\beta}{2M} \Sigma p^2_n}\mathrm e^{-\beta u(x)} $$

This integral factorises into two, an integral over $p$, and an integral over $x$, i.e. a kinetic energy contribution and a potential energy contribution.

I write the result as
$$Z = \frac{z^N}{N!}$$
for an ideal gas, this result is
$$\frac{V}{\Lambda^{3N}}$$
where
$$ \Lambda = \frac{h}{(2\pi mk_\mathrm bT)^{1/2}}$$
(thermal wavelength)

For a van der Waals gas,

$$
U(r) =
\begin{cases}
\infty, & \text{when }r<d \\
-\epsilon\left(\frac{d}{r}\right)^6, & \text{when }r \ge d
\end{cases}
$$

basically attractive up to a certain distance, and repulsive when they come too close. r is the distance between the spheres and d is the distance where they touch. Also, each particle moves independently in an average potential field offered by the other particles

Also, for an ideal gas we integrate over all space i.e. the entire volume of the container, however, we cannot do that anymore. Instead our volume of integration is $V - Nb'$ (or $V – nb$, same as what we did before, $n$ is number of moles instead of number of particles).
$$b'= \frac{2\pi d^3}{3}$$

Now, assume that in a shell of thickness $\mathrm dr$ there are $ \frac{N}{V} 4 \pi r^2\mathrm dr$ particles, i.e. the position of the particles is averaged (reality, like always is different)

So for a single particle, the attraction felt is
$$\phi = \int_{d}^{\infty} u(r)\frac{N}{V} 4 \pi r^2\mathrm dr$$
and
$$z = \frac{(V-Nb')\mathrm e^{-\phi/2k_\mathrm bT}}{\Lambda^3}$$

I will omit the computations and quote the results,

$$\phi = -2a' \frac{N}{V}$$
where
$$a' = \epsilon (2/3) \pi d^3$$

going back and putting things together,

$$\ln Z = N\ln(V-Nb') + \frac{N^2 a'}{Vk_\mathrm bT} - N\ln(\Lambda^3) - \ln(N!)$$

Now,
$$ P = k_\mathrm bT \frac{\partial \ln Z}{\partial V}$$
which turns out to be,

$$ P = \frac{Nk_\mathrm bT}{V-Nb'}- \frac{N^2 a'}{V^2}$$

UPDATE: Physical Interpretation

I am writing this section because I get the impression (from the discussion the in the comments) that my answer hasn't adequately addressed this.

In the limit $P \to 0$ (or $V_m \to \infty $) we can expect perfect gas behaviour, i.e $PV = \text{constant}$

However, say at high pressures, we observe that real gases would have larger molar volumes than expected, so we remove a term from the volume.

Similarly, the pressure is lower than ideal gas pressure, so we add a term to it to compensate.

So, instead of saying $PV_m = \text{constant}$, we say $ (P +\text{"something"})(V_m - \text{"something"}) = \text{constant}$

It also accounts for critical behaviour, simply solve the following system to obtain critical constants (since the critical point, is a point of inflection on the PV diagram)
$$ \frac{\partial P}{\partial V_m} = 0 $$
$$ \frac{\partial^2 P}{\partial V_m^2} = 0 $$

Of course, it would be too optimistic to expect a simple, elegant model such as this to truly determine all properties of gases with complete precision; the van der waals equation of state does have its failings, but if nothing else, it improves at least the qualitative discussion of properties/behaviours of real gases


Why does white phosphorous exhibit chemiluminescence?

From here:

  The best-known and first well-documented example of deliberately
  induced, man-made chemiluminescence was the discovery of elemental
  phosphorus by a German alchemist H. Brand. The waxy white phosphorus
  is slowly oxidized at room temperature giving off faint greenish glow.
  The formation of excited reactive intermediate $\ce{(PO)2}$ and
  $\ce{HPO}$ is responsible for the green glow.

That's where the term phosphorescence come from.


How to compare the bond lengths of an peroxide complex before and after oxidation?

The short and unpleasant answer is that we cannot derive the (spectroscopic) oxidation states based on theory alone. We need experiments (preferably crystal structures and some way to determine the oxidation states of cobalt which are accessible by experiment).

Oxidation states are unfortunately only easy for inorganic molecular compounds or simple salts. In your case, you have a bidentate bridging dioxygen ligand. This could be any of the species $\ce{O2, O2^.-, O2^2-}$ and potentially even more (with an even weaker oxygen–oxygen bond). This is what is called a noninnocent ligand: it can undergo redox chemistry with the central atom. The difficulties of finding out what the actual oxidation state is, is outlined in this question concerning the nature of the $\ce{Fe-O2}$ bond in haemoglobin.

For these cases, IUPAC has defined a formal oxidation state which can be derived according to a set of rules so that one may at least name the complex in oxidation nomenclature. In these rules, both the cyanido and the peroxido ligand would be seen as anions, peroxido being $\ce{O2^2-}$ and cyanido obviously $\ce{CN-}$. By considering the total charge of the $\ce{[Co2(CN)10\mu{-}(O2)]^6-}$ we can allocate cobalt an oxidation state of $\mathrm{+III}$ in the reactant where the maths adds up nicely.

Upon oxidation two things can happen:

one of the two cobalt ions supplies the electron, resulting in a decacyanido-μ-peroxidodicobaltate(II,III)
the peroxido bridge supplies the electron resulting in a superoxido complex with again two cobaltate(III) nuclei.

The resulting crystal structures would be indistinguishable except if one is able to synthesise both products and compare the bond lengths. Therefore, one must resort to other methods of determining the oxidation states. The most practicable way in my humble opinion would be to examine the UV/Vis spectra of the cobalt centres. These should remain low-spin due to the ten cyanido ligands. If we happen to have two $\mathrm{d^6}$, then cobalt was not oxidised. If one of the cobalt centres turns out to be $\mathrm{d^7}$ after the reaction, this should result in a noticable change from the original reactant spectrum.

In the absence of experimental evidence we need to resort to hidden clues in the question. You were asked to consider the oxygen–oxygen bond length. If the metal centre was oxidised, this should not change significantly. However, if the peroxido ligand was supplying the electron for oxidation, the bond order would increase when the species transforms from peroxido to superoxido and this should result in a reduced bond length.

Only with this meta-knowledge we can a posteriori deduce that oxygen’s oxidation state must have gone from $\mathrm{-I}$ to $-\frac12$.


In the following reaction is E2 or Sn2 favorable?

This is simply wrong. The person setting this question has assumed that the hydrazine will be completely selective for the ketone over the alkyl bromide and that the strong basic conditions will also cause elimination of the Br- to give the styrene. Selectivity seems most unlikely give the typical forcing conditions used in W-K reductions

Hydrazine is a powerful nucleophile and reacts with alkyl bromides under mild conditions often giving polyalkylation reference here.

I consider it most unlikely there will be any selectivity unless confronted by literature example.


Can we relate approximate Hartree-Fock orbitals to true solutions in the basis-set limit?

In quantum chemistry numerical experimentation is the norm and rigorous mathematical proof the rare exception. As a result, not many rigorous results are know, not even for the good-old Hartree-Fock method. For a review of mathematical results I recommend you have a look at these references:

Claude Le Bris, Computational chemistry from the perspective of numerical analysis, Acta Numerica (2005), pp. 363–444
Pablo Echenique and J. L. Alonso, A mathematical and computational review of Hartree–Fock SCF methods in quantum chemistry, Molecular Physics, Vol. 105, Nos. 23–24, 10 December–20 December 2007, 3057–3098

In particular, section 3.3 of the first reference is relevant, but don't expect to find mathematically strong results. For example, here's a quote:

The choice of an AO basis for solving
a given problem mostly relies upon some practical know-how. The lack of
rigorous understanding is a pity, because the output of the calculations
(typically some molecular property) might be very sensitive to the choice
of the basis set. The only available measures of the quality of the basis
set are obtained, in the chemistry literature, by choosing test cases, i.e.,
reference systems, where the solution of the exact Schrödinger equation may
be computed, mostly through numerical computations and, when possible,
with the help of an analytic calculation.

From a practical point of view I don't see why occupied Hartree-Fock orbitals expanded in the usual Gaussian basis sets should not converge (in the integral norm sense, ie leaving aside the lack of cusp at the nucleus) to the exact solutions of the Hartree-Fock equations. I have little idea as to what happens to the virtual orbitals.


Why mustn't the sample be dipped in the mobile phase in paper chromatography?

Dipping sample to a mobile phase would cancel all the purpose of chromatography as a separation technique.

Imagine properly performed chromatography as the athletic race, where all runners start at the same point and time. The finish order is by their average speed.

Now imagine a crazy version of a New York marathone, where runners start in  long sequence of runner batches. But nobody would  think about registration of runner starts, so there would be no idea which runner was the fastest to be decorated as the winner.

Back to chromatography, instead of chromatographic spots ( PC, TLC) or peaks (HPLC, GC), there would be useless, smeared, overlapping long bands.


Polyurethane epoxy resin chemistry advice please?

The resin is the same variety commonly used to produce fiberglass baths, boats and many other uses.

The resin is unsaturated polyester, using methyl ethyl ketone peroxide (MEKP) as catalyst, cobalt solution for example cobalt octoate (Coct) as an accelerator.

It isn't often sold to consumers because it is too explosive and corrosive, the MEKP reacts explosively with raw cobalt, and it's dangerous to the eyes, it can cause blindness in 15 seconds if splashed in the eye.

The MEKP doseage is 1-2% per volume of polyester, depending on desired drying time. The cobalt is mixed first and dyes the polyester a whiskey color, you can dose it by color, the darker whisky color the faster it will dry.

Industry standards are 1% and 6% cobalt solution, 1% is dosed at 0.5% to 2%, however there is no way to tell without the product data sheet if the cobalt solution 1%, 6%, 10%.

safety:

MEKP and cobalt mix = flames
MEKP in eyes = blindness
wear goggles and rinse with water and go to hospital if you get some MEKP eye contact.
wear protective clothing and avoid skin contact.

"Cobalt octoate is the most reactive and most commonly used paint drier. Cobalt octoate catalyses the oxygen uptake and accelerates peroxide formation causing a surface film to form rapidly and hence known as oxidation catalyst. It basically a metal caroxylate used to accelerate the drying process like linseed oil. It accelerates the decomposition of the organic peroxide initiators (often called catalysts) and promotes the cross-linking of polyester resins.
"

Note that baths are finished with a gloss product called gel coat.


Why is the number 2 needed for 2-methylpropanal?

Well I suppose you are correct in a way- but it does help us see the molecule quicker in our heads...

it should be written methyl-2-propan-1-al or methyl-2-propan-2-al
otherwise it would be a butan-1-al  or butan-2-al

let me explain further>
the new IUPAC nomenclature  advises the use of numbers instead of the iso epithet- in organic chemistryIn old naming -as with propanol and iso-propanol - so with propanal

Isopropyl alcohol

Preferred IUPAC name

Propan-2-ol

-
Isopropyl alcohol (IUPAC name propan-2-ol; commonly called isopropanol) is a compound with the chemical formula C3H8O. -It is a structural isomer of 1-propanol.

wiki  Chemical Names: ISOBUTYRALDEHYDE; 2-Methylpropanal; 78-84-2; Isobutanal; Isobutylaldehyde; Propanal, 2-methyl- More... Molecular Formula: C4H8O or (CH3)2CHCHO.


What defines basis sets in computational chemistry?

In LCAO, it is the set of atomic orbitals (AOs) that is the basis, and the coefficients are the basis expansion coefficients. For example, take the hydrogen molecule, with 1 atomic orbital on each atom:

The lower energy MO, the bonding one, will be (excluding normalization):

$$
\psi_{\mathrm{1s}} = \phi_{\text{left}} + \phi_{\text{right}}
$$

and the higher energy MO, the antibonding one, will be (excluding normalization):

$$
\psi_{\mathrm{1s}^{*}} = \phi_{\text{left}} - \phi_{\text{right}}
$$

This means that for the bonding MO, $c_{\text{left}} = 1$ and $c_{\text{right}} = 1$, while for the antibonding MO, $c_{\text{left}} = 1$ and $c_{\text{right}} = -1$. If your basis was different (say, the atomic orbitals were of different radial extents for the two atoms), then the coefficients would be different to accommodate that. They aren't part of the basis set.

Since you tagged this with computational-chemistry, I assume you also want to know about basis sets in computational chemistry. Conceptually, they are identical to LCAO-MO bases, but what may be confusing is that each atomic orbital itself may be composed of multiple functions (called primitive functions), rather than just the single function for each AO as seen above. This leads to another set of coefficients, called contraction coefficients, describing how the primitive functions are linearly combined to form a contracted function.

An "implementation" detail is that the functional form of AOs is usually Gaussian functions, which have a parameter in the exponent (here, $\alpha$):

$$
\phi_{r}(x) = e^{-\alpha x^2}
$$

This means that to fully define an atomic orbital basis composed of contracted Gaussian-type orbitals (CGTOs), one needs both the contraction coefficients and exponents. They usually look like the following. For each angular momentum type, there may be one or more contracted functions, each composed of one or more primitives. For example, in the hydrogen definition below, there are 2 functions to describe s orbitals, one of which has three primitives and the other is not contracted. The contraction coefficients are in the first column and the exponents in the second column.

****
H     0
S   3   1.00
     13.0107010              0.19682158E-01
      1.9622572              0.13796524
      0.44453796             0.47831935
S   1   1.00
      0.12194962             1.0000000
****
C     0
S   5   1.00
   1238.4016938              0.54568832082E-02
    186.29004992             0.40638409211E-01
     42.251176346            0.18025593888
     11.676557932            0.46315121755
      3.5930506482           0.44087173314
S   1   1.00
      0.40245147363          1.0000000
S   1   1.00
      0.13090182668          1.0000000
P   3   1.00
      9.4680970621           0.38387871728E-01
      2.0103545142           0.21117025112
      0.54771004707          0.51328172114
P   1   1.00
      0.15268613795          1.0000000
D   1   1.00
      0.8000000              1.0000000
****

The last note is that when performing a Hartree-Fock or similar calculation, the contraction coefficients and the exponents don't change. That would mean the basis set changes. The basis set is fixed, it is the MO coefficients that change, the same coefficients that appear in the LCAO-MO equations.

Image taken from here. Basis set is Def2-SV(P).


Number of grams of CO2 released from a carbonated water

This shows part of the problem.

Henry's law as given is

$K_{AW}^{px} = $1.65*10^3$ = \frac{p_{\ce{CO2(gas)atm}}}{[\ce{CO2(aq)}]}$

where:
$K_{AW}^{px}$ atm = aqueous-phase mixing ratio of $\ce{CO2}$ and the units are $atm$

$[\ce{CO2{(aq)}}]$ = the concentration of $\ce{CO2{(aq)}}$ in $mole/L$

$p_{\ce{CO2(gas)atm}}]$ = the partial pressure of $\ce{CO2{(gas)}}$ in $atm$

So rearranging...

$[\ce{CO2(aq)}] =\frac{p_{\ce{CO2(gas)atm}}}{K_{AW}}$

The moles of $\ce{CO2}$, $m_{\ce{CO2}}$, is given by concentration, $[\ce{CO2}]$, times the volume, $v$ in $L$.

$m_{\ce{CO2(aq)}} = [\ce{CO2(aq)}]v =  \frac{p_{\ce{CO2(gas)atm}}}{K_{AW}}v$


When comparing boiling points of two ionic compounds, what bonding factors should you assess?

To compare 2 ionic compounds' melting point, there are generally 3 determinants:

Ionic Radii, smaller the ionic radius, closer the ions are to each other, therefore stronger the electrostatic attraction between them. (analogous to stronger attraction when closer to a magnet) A good example would be melting points of cations from the same group with the same anion (NB: LiF does not follow the trend, potentially due to its more covalent nature):

$\ce{NaF}\ -\  \pu{993 ^\circ C ( 1819 ^\circ F)}$
$\ce{KF}\  -  \pu{858 ^\circ C ( 1576 ^\circ F)}$
$\ce{RbF}\  -\  \pu{795 ^\circ C ( 1463 ^\circ F)}$

Charge of ions, larger the charge an ion has, more positive or negative the ion is, therefore there is a stronger attraction between ions. (analogous to stronger attraction with two stronger magnets in opposite poles) An example would be $\ce{NaCl}$ and $\ce{MgO}$:

$\ce{NaCl}\ -\ \pu{801 ^\circ C ( 1,474 ^\circ F)}$
$\ce{MgO}\ -\ \pu{2852 ^\circ C ( 5166 ^\circ F)}$

Ratio of ions, in most cases, closer the ratio is to 1 : 1 between anions and cations, stronger the attraction is. larger or smaller the ratio is, the more the anions or cations will have to 'share' or 'divide' the electrostatic attraction from the oppositely-charge ion, decreasing the melting point. An example would be $\ce{NaCl}$ and $\ce{MgCl2}$:

$\ce{NaCl}\ -\ \pu{801 ^\circ C ( 1,474 ^\circ F)}\ [1 : 1]$
$\ce{MgCl2}\ -\ \pu{714 ^\circ C ( 1317 ^\circ F)}\ [1 : 2]$

The determinants for melting points would generally also apply to the boiling points of ionic compounds. However, it should be noted that other factors would affect the boiling point as well.


Why does slaked lime water lose alkalinity over time?

If you dissolve slaked lime in water it will absorb carbon dioxide from the atmosphere over time.
Carbon dioxide will react with hydroxide ions from the slaked lime forming mainly hydrocarbonate ions. Hydrocarbonate ions will equilibrate with water and at about pH < 8 carbon dioxide will slowly be liberated from the solution. After some time, no significant amounts of hydroxide ions will be left from the slaked lime and the pH of the solution will close into neutral pH.

Well, what will happen to the calcium ions from the slaked lime? You will probably find them as an insoluble deposit at the end. This is because carbonate ions will also be formed during the process, forming calcium carbonate of very low water solubility.


Actually, the technology isn't that new. It seems however, that the setups are becoming more affordable in the hobby market.

The technology is know as stereolithography and is a form of light-induced polymerization (UV curing).

The liquid monomers used have a substituent that can be activated using UV light. Out of practical considerations, UV is more convenient: polymerization of the resin will not occur under natural light.

In principle, the procedure is comparable to UV curing in dental repair. There, a lamp that irradiates the whole area is used: you want a fast hardening all over the tooth.

In 3D printing, this is obviously different! The light source is a laser with a narrow beam: you only want to irradiate a very small volume element at a time.

You don't want a rapid polymerization either: only the volume element irradiated should react!

The laser beam is focussed at the boundary of the resin and different volume elements are addressed by moving the focus in the $xy$ plane.

Rather than moving the (heavy) laser itself, the beam is "moved" using a mirror optic. This allows for a very fast adressing of reaction points within the plane.

UPDATE
In the example linked, in the question and a comment, however, the setup is different from laser-based STL printers.

Irradiation from the bottom of of the resin tank through an oxygen-permeable membrane indicates indeed that the UV curing proceeds in the presence of oxygen. Anyway, if the light source would be just a lamp, one would nevertheless achieve homogenous curing of the whole layer.
But the manufacturer doesn't use the term lamp or light source. Instead, they write about a projector and that's probably what they do:

They are not illuminating the whole layer, but are projecting masks for every slice of the object to be printed.


One possible way:

As you say
$${\bf{S}^2} = S_x^2+S_y^2+S_z^2$$
Thus
\begin{align}
{\bf{S}^2} &= \left( \frac{\hbar}{2}\right)^2
\left[\begin{pmatrix}
0  & 1 \\
1  & 0 \\
\end{pmatrix}\begin{pmatrix}
0  & 1 \\
1  & 0 \\
\end{pmatrix}
+
\begin{pmatrix}
0  & -i \\
i  & 0 \\
\end{pmatrix}
\begin{pmatrix}
0  & -i \\
i  & 0 \\
\end{pmatrix}
+
\begin{pmatrix}
1  & 0 \\
0  & -1 \\
\end{pmatrix}
\begin{pmatrix}
1  & 0 \\
0  & -1 \\
\end{pmatrix}
\right] \\
&=
\left( \frac{\hbar}{2}\right)^2\left[\begin{pmatrix}
1  & 0 \\
0  & 1 \\
\end{pmatrix}+\begin{pmatrix}
1  & 0 \\
0  & 1 \\
\end{pmatrix}+\begin{pmatrix}
1  & 0 \\
0  & 1 \\
\end{pmatrix} \right]\\
&= \left(\frac{3}{4}\right)\hbar^2
\end{align}


Yes, the size* of the $n$-th shell will differ between elements. The reason is that the nuclear charge will shrink each shell as it goes up. Additional electrons cannot fully compensate this effect because they do not perfectly shield the core.

The difference can be measured (see below) or calculated using quantum chemistry with flexible basis sets. Results from the latter can be validated against X-ray spectroscopy. However, even the Bohr model will yield different radii for the first shell when increasing the nuclear charge.

*The definition of the size is not trivial. One can use the van-der-Waals radius, ionic radius or determine the isosphere that contains e.g. 90% of the electron probability density as long as one is consistent about it.


Real life chemistry :boiling carrot dices

My guess would be natural convection. Hot water is less dense than cold water so when the heat source is at the bottom, the freshly heated water rises and the cooler water at the top sinks. The cooler water can't sink through the rising hot water, so the two flows run along side each other. Depending on the fluid properties, this can result in circular "rolls" of flowing water, known as convection cells. If you use a wide pot and small particles, you may be able to see many, but in a relatively narrow pot, you're more likely to observe a single recirculating flow around the centre, as shown in this simulation. The liquid loses more heat around the sides of the pot, so the descending flow rides down the sides and the ascending flow is in the middle where it can stay the hottest.

I'm guessing that your dice is too heavy to ride the ascending flow, but the flow from the edge to the centre pushes all the carrots towards the middle.


Vaporizing NOx (NO and NO2) with hot metal?

Thin wall channels (this would be the "steel" plates perhaps) in the device are coated with a wash coat of aluminum oxide which is porous and increases the surface area. This allows more reactions to take place since the converters also contain precious metals such as platinum, rhodium, and palladium e.g. converting NOX into nitrogen and oxygen, hydrocarbons into water and carbon dioxide, and so on.

These catalytic converters use oxidation and reduction reactions to convert the "toxic" fumes—I'm not certain they are stating to work any differently with this regard than other catalytic converters do though. Oxidation is the loss of electrons and reduction is the gaining of electrons at the molecular level. This in addition to the precious metals help promote the transfer of electrons into the conversion of "toxic" fumes to "non-toxic" fumes.

Furthermore, these can also contain oxygen sensors on them measuring and reporting back to the Engine Control Module the oxygen levels in the exhaust. This will help tell other sensors to adjust air-to-fuel ratio helping to keep the catalytic converter running at optimum efficiency.

I would say if this device has a patent and has held up to the tests thereof, then it's likely legitimate and accurately effective as advertised by the manufacturer.


Why does beryllium lose two electrons when it’s orbitals are already full?

Electrons exist in the outer spaces of elements, away from the nucleus. Orbitals are mathematical descriptions of places in these outer spaces where electrons seem to be when they are there; they can be mathematically manipulated and reorganized but are a way to help you imagine what is really going on.

Manipulating the electrons is done experimentally, not mathematically. E.g., an element can be ionized and the energy required can be measured. Beryllium can suffer the loss of up to four electrons: the first requires 9.3 eV, the second requires 18.2 eV, the third requires 153.9 eV, and the fourth requires 217.7 eV. The energy required to remove the first two electrons is in the range of other metals that lose two electrons (Mg 7.6, 15.0; Ca 6.1, 11.9 eV). So Be can lose 2 electrons with a relatively small amount of energy (which will be supplied by something like empty orbitals and ionic or covalent bonding energy to other atoms). Losing those last two is a big jump and essentially doesn't happen in chemistry (it happens in physics and mathematics!).

The empty orbitals on Be are quite high in energy, i.e., if you stuck an electron into a p orbital on Be, it would be floating so far out from the nucleus that it would be looking for a better place. The idea of orbitals gives you a mental picture of what might happen - whether it really happens or not depends on whether that mathematical picture is true and is the most stable (lowest energy) condition available.

Completing shells or orbitals is a qualitative way of looking at the issue, but numerically, ionization potentials gives you the data you need to develop the picture, and the orbitals (as calculated for the hydrogen atom) are a reasonable way to begin.


R or S configuration of propane-1,2-diol

Your comment is correct where you assign the priority.

Rather than "rotating" or "flipping" the molecule, consider "looking" at it from the angle which makes the H atom be at the back.  In the graphic it is drawn to the left side, so place yourself on the right side and look down the bond.

The priority groups 1-2-3 run anticlockwise, hence it has the S configuration.


Each prostaglandin is named using the prefix 'PG' followed by a letter A to K depending on the nature and position of the substituents on the ring. PGA to PGE and PGJ have a keto group in respective positions on the ring which are further distinguished by the presence or absence of double bonds or hydroxyl groups in various positions in the ring(indicated by numbers or Greek letters*). PGF has two hydroxyl groups and PGK has two keto substituents on the ring. PGG and PGH are bicyclic endoperoxides. An oxygen bridge between carbons 6 and 9 distinguishes prostacyclin (PGI). Also, all prostaglandins have a hydroxyl group in the S-configuration on carbon 15 and a trans-double bond at carbon 13 of the alkyl substituent (R2).

*Numerical subscripts (1 to 3) are used to denote the total number of double bonds in the alkyl substituents, and Greek subscripts (α or β) are used with prostaglandins of the PGF series to describe the stereochemistry of the hydroxyl group on carbon 9.

So for example, if we write PGF2α, the F will indicate two hydroxyl groups in the cyclopentanone ring (F series), the 2 will indicate two double bonds, and the α will indicate that the hydroxyl grouping at carbon 9 is in the α configuration.

References

https://www.lipidhome.co.uk/lipids/fa-eic/eicprost/index.htm (@user55119's link)
https://www.glowm.com/section_view/heading/The%20Prostaglandins:%20Basic%20Chemistry%20and%20Action/item/312#2133
Norman A. Nelson, Journal of Medicinal Chemistry, 1974, 17 (9), 911-918
DOI: 10.1021/jm00255a001


Finding the concentration of ferrous thiocyanate in calibration solutions

So, to find the concentration of $Fe(SCN)^{2+}$, the problem says "the calibration solutions is essentially the concentration of the $[SCN^−]$ in the solution." So, you simply take the volume of $K(SCN)$ solution divided by the total volume, then multiply by $[K(SCN)]$:

$[Fe(SCN)^{2+}]=[SCN^−]=[K(SCN)]=0.002M*\frac{0.5mL}{10.0mL}=0.0001M$ (repeat for 2, 3, 4)

The reason we do not involve $K_c$ is because the large amount of $[Fe^{3+}]$ drives the reaction to the right. The absorbance measurements are going to be useful for determine the amount of $[Fe(SCN)^{2+}]$ in a system where $[Fe^{3+}]$ is not high. In these cases, the concentrations $[Fe^{3+}]$ and $[SCN^-]$ are known, and you use the absorbances you collected to draw a line relating concentration $[Fe(SCN)^{2+}]$ to absorbance. When you know the concentration of your three parts of $K_c$, you can simply use arithmetic to find $K_c$.


Gluten describes a subset of proteins (glutenins and gliadins) found in wheat grain and making up a large fraction of the protein found in common wheat flour (>70%). Since the protein fraction of wheat flour can represent >10%/wt, the gluten fraction commonly represents >5%/wt of the flour.
Given the requirement for an easily implemented chemical test this eliminates more advanced and regularly used methods such as immunoassays (eg ELISA), and maybe even dye-binding assays such as described by the AOAC Official Method 2011.04.
One frequent chemical method to measure the amount of protein in a food sample is known as the Kjeldahl method, which detects nitrogen (which is absent in the carbohydrate). Other assays for nitrogen would also do the trick.
Büchi manufatures equipment for performing the required assay (Ref. 1), and provides an application note based on standard method AOAC 920.87. Although it is unlikely that you will have a Büchi SpeedDigester K-436, K-439, or Kjeldahl Sampler System K-370/K-371, available in your school lab, you'll still require the chemicals and equipment for performing a high temperature digestion and distillation, which should be similar to that listed in the application note. Note this is not a simple experiment.
Unfortunately the Kjeldahl method is likely to be difficult to implement in some educational laboratories given the required equipment, chemicals and experience. An alternative family of less challenging methods to determine protein are biuret methods (BCA and Lowry assays, which exploit protein-copper chelation and secondary detection of reduced copper), as described for instance on the ThermoFisher Scientific website. These however employ absorbance measurements to determine the amount of chelated or reduced copper.
If you are looking to distinguish gluten against a background of other proteins, you may have to isolate the gluten fraction (see eg Ref 2.). In general your approach would depend on the expected concentration and amount of gluten in the sample. Since this is a question about an educational application, it is assumed that you are not concerned with a non-gluten protein background or with detection of low concentrations.
References

Büchi Application Note 027/2010. Nitrogen and Protein Determination in Flour according to the Kjeldahl Method.

Ravinder Kaushik, Naveen Kumar,Manvesh Kumar Sihag, and Aradhita Ray. Isolation, characterization of wheat gluten and its regeneration properties. J Food Sci Technol. 2015 Sep; 52(9): 5930–5937.
doi: 10.1007/s13197-014-1690-2


Is there any empirical upper limit for order of reaction?

The order of a reaction is an experimentally determined quantity and can be  positive, negative or fractional. The order need not be related to the stoichiometric coefficients, although sometimes it is. In some reactions it is not possible to define an order. There seems to be no reason therefore why the order cannot be greater than three; the iodate-iodine reduction (Dushman) reaction $\ce{IO3^- + 5I^- + 6H^+ -> 3I2 + 3H2O}$ has the rate expression $r \approx \ce{[I^-][IO_3^-][H^+]^2  }$.

If the reaction is $aA+bB \rightarrow cC+dD$ then the rate is

$$ r= - \frac{1}{a}\frac{d[A]}{dt} = \frac{1}{c}\frac{d[C]}{dt}= \cdots =k[A]^\alpha[B]^\beta[C]^\gamma \cdots$$

where the order n is $n=\alpha+\beta+\gamma$ and these need not be the same as a, b and c.

If the rate expression is not of this form then it is hard to define an order; one familiar example is the chain reaction $\ce{H2 + Br2}$ where the experimentally determined rate law for the production of HBr has the form $\displaystyle r=\frac{k_2[\ce{H_2}][\ce{Br_2}]^{1/2}}{1+k_2'[\ce{HBr}]/[\ce{Br}]}$ where an overall order cannot be defined.

The order should not be confused with molecularity which describes the number of species reacting in a  postulated elementary reaction and which is always a small positive number. Effectively this means 1 or 2. A molecularity of 3 is not impossible but the chance of three  reactive species colliding at the same time is vanishingly small. In practice when a molecularity of 3 is suspected it is often found that in  subsequent experiments an intermediate is formed between of the two species which then reacts with the third at a slightly later time.


How to make a sample in ppm? (not a specific compound, only the sample)

In layman's terms, ppm is similar to percentage, i.e., how many parts of X are there in a million. By that standard, 1000 mg tea / 0.1 L is indeed 10000 ppm tea "powder" in water. We are assuming that everything dissolves completely.
Scientifically, or in analytical chemistry, you question is meaningless. Tea powder is mixture so one cannot talk about the ppm of tea powder. This is why you will not find this in journals or books. Concentration is specified for a single component.
Tea powder is a very complex mixture. If we have a mixture of say K2SO4 and NaCl, it is not appropriate to mention the concentration of "salts" in water. It will be an ambiguous statement. A better way would be expressing the concentration of K2SO4 as x% and that of NaCl as y %. It is okay in analytical chemistry to mention "total carbonates" or "total acidity"; here the species being analytes are closely related.


Alcohol vapors can certainly form explosive mixtures with air, so filling the car with ethanol vapors is potentially dangerous. A source of ignition, for example a spark, could under the right circumstances ignite the vapors. If the fan motor of the AC system is a brushed motor it can create small sparks. So I would say there is a fire hazard.
But there are other problems with your suggested setup. The ethanol and water vapors would condense on every surface of your car's interior, the windows, the roof, the doors, etc. This would mean you loose a lot of alcohol, so you would need more. Ethanol might attack and start to dissolve some of the interior, glued parts may become loose. And of course the car would smell very strongly like vodka for a long time.
So I would suggest disassembling the dash cluster and whatever else you need to get to the AC system, and using the vodka in liquid form to clean it properly.


For the case of magnesium ion, at least, it seems a stretch to say it dissolves in ammonia solution.  Assume that the ammonia solution has $\mathrm{pH}$ of $11$ and magnesium hydroxide has a solubility product of $5.61×10^{-12}$. We infer that the magnesium ion solubility at equilibrium with the hydroxide in ammonia solution is only $\pu{5.6μM}$ which is not really all that soluble.

Magnesium ion is appreciably soluble not in an ammonia solution but in a buffered solution of ammonia and an ammonium salt.


Travelling wave ion mobility spectrometry: determination of size?

In terms of practical data analysis standpoint, travelling wave-IMS data can generally be treated similarly as drift tube-IMS for which the technical differences is in resolution (better or worse). However, one point of consideration is that

Low gas pressure in TW IMS means strong fields in E/N terms
and thus substantial heating of ions. This may cause fragmentation
or distortion of macromolecular structures, complicating their
detection and characterization. [1]

Your first instinct to assume that the $\ce{[M + H]+}$ adduct is quite rational, however remember that ion conformation is always in flux and can be influenced by collisions with buffer gas. What you may be experiencing is the proton adduct producing a "looser" structure, resulting in a larger collisional cross section. The $\ce{[M + K]+}$ adduct is heavier in mass but may produce a tighter conformation.
A good example of this is in the fragmentation and mobility of polyethylene glycol (PEG) ions $\ce{[PEG + H]+}$ and $\ce{[PEG + Na]+}$. When carrying out fragmentation studies of PEG, it was found that the sodiated ion produced less extensive fragmentation and required higher collision energy.[2] Ion mobility studies have also found that PEG chelates the $\ce{Na+}$ ion similar to that of crown-ether, producing tightly wound "rigid" structures.[3]
These conditions are highly compound/structure dependent however, it should be something to keep in mind especially in the case of analyzing other molecules such as peptides which respond to adduction in "weird and mysterious" ways.
References

Alexandre A. Shvartsburg and Richard D. Smith, Anal. Chem. 2008, 80 (24), 9689–9699.
Jennifer Gidden, Thomas Wyttenbach, Anthony T. Jackson, James H. Scrivens, and Michael T. Bowers, J. Am. Chem. Soc. 2000, 122 (19), 4692–4699.
Rui Chen, Xinlei Yu, and Liang Li, J. Am. Soc. Mass Spectrom. 2002, 13 (7), 888-897.


Does ethanol react with potassium to form potassium ethanolate and hydrogen gas?

You can view water as the simplest alcohol, and indeed, water reacts with all of the Group 1 metals to form hydrogen gas and the corresponding metal hydroxide according to the following equation

$\ce{2M + 2HOH -> 2MOH + H2}$

where $\ce{M}$ is the metal.

The same reaction occurs with simple alcohols to produce the corresponding metal alkoxide and hydrogen gas

$\ce{2M + 2ROH -> 2MOR + H2}$

As you increase the hydrocarbon portion of the alcohol, or reduce accessibility to the $\ce{OH}$ portion of the alcohol (e.g. make the alcohol less like water), you slow down the rate of the reaction.

Here is a nice link that compares and discusses the relative reactivities of the Group 1 metals with water.


Ideal gas law derivation from kinetic theory

This derivation assumes the container is a cube, but the law holds for containers of any shape, right?

The resulting law holds for any shape of container. However, the details of the derivation depend on the geometry. A cube is easy, so we use it.

  This derivation assumes each particle has the same mass, but the law holds for a mixture of gasses of different masses, like air, right?

Yes, it holds for a mixture of any ideal gases - the key is that only the kinetic energy of the gas molecules matters. Heavier molecules would be moving slower at the same temperature, but they would exert the same force (and therefore pressure) on the walls.

  The derivation assumes that the particle intercepts a give side with frequency 2L/vx. Why can we make this assumption?

This assumption works for only one particular case. If you notice the third sentence at the beginning of the pressure derivation (emphasis is mine):

  When a gas molecule collides with the wall of the container perpendicular to the x coordinate axis and bounces off in the opposite direction with the same speed

They are looking at the special case of a collision at a 90 degree angle, which bounces off the opposite wall. In that case, it will take the particle t = L/V time to reach the opposite wall, and t = L/V time to bounce back - so the total is 2L/V. This illustration may help explain it better:

Now this might sound like an unreasonably strict condition, but the reason that it works is that the component velocity distribution for each particle is symmetrical (it's a normal distribution), which means that for every particle that impacts the wall at 95 degrees (for example) there is another that impacts 85 degrees. As a result, the average angle of impact is 90 degrees for each wall, and therefore the average collision time is given by looking at 90 degree impacts for each component of the velocity vector.


Finding sulfate contamination in crystallized silver nitrate

Barium nitrate has a water solubility of $\pu{10.5g/100mL}$ at $\pu{25^oC}$. It isn't specified in the question what concentration of sulfate you suspect might be present, but given that you are trying to check for sulfate by precipitating with barium, barium nitrate should be the way to go.


Bond length comparison between sulfur tetrafluoride and thionyl tetrafluoride

The simplest way to compare these two compounds is to compare sulfur’s oxidation state. In $\ce{SF4}$, sulfur is in the $(\mathrm{+IV})$ state. In $\ce{SOF4}$, it is $(\mathrm{+VI})$. The higher a compound’s oxidation state, the higher its formal positive charge and thus the more stable (i.e. lower in energy) its orbitals are. Orbitals reduced in energy means contracted orbitals which in turn leads to shorter bonds.

Other than that, the two compounds are pretty identical. Both feature two two-electron-two-centre $\ce{S-F}$ single bonds and one four-electron-three-centre $\ce{F\bond{...}S\bond{...}F}$ bond; the latter can be understood with the following resonance depiction:

$$\ce{F^-\ \ S^+-F <-> F-S^+\ \ F-}$$

Even the structures are comparable although the angles are slightly different.


You're confusing different enthalpies (don't worry, it seems to be rather common). Let's stick with NO as an example. The standard enthalpy of formation, $\Delta_fH^\circ$, is defined as the enthalpy of the reaction $\ce{0.5N2 + 0.5O2 = NO}$ at any given temperature. That is, if you need $\Delta_fH^\circ(298\textrm{ K})$, or $\Delta H^\circ_1$, your reaction is $$\ce{0.5N2(gas, 298) + 0.5O2(gas, 298) = NO(gas, 298)}, (1)$$ and for $\Delta_fH^\circ (T)$, or $\Delta H^\circ_2$, provided that the gases in question do not liquify at $T$, the formation reaction is $$\ce{0.5N2(gas, T) + 0.5O2(gas, T) = NO(gas, T)}. (2)$$

Remember that $\Delta_fH^\circ$ for the elements in their stable states is zero at any temperature.

Now let's say that we have to calculate $\Delta_fH^\circ(T)$ using the known value of $\Delta_fH^\circ(298\textrm{ K})$. For that we need to know the heats of three other processes: $$\ce{N2(gas, 298) = N2(gas, T)}, (3)$$ $$\ce{O2(gas, 298) = O2(gas, T)}, (4)$$ $$\ce{NO(gas, 298) = NO(gas, T)}. (5)$$

The enthalpies of $(3)$, $(4)$ and $(5)$ are basically the enthalpies of heating from $298$ to $T$. They are often called "enthalpy increments" and are defined as $\Delta_{298}^TH^\circ=\int_{298}^T C_p\,dT$ . Note that the enthalpy increments can be written between any two temperatures, not just from 298 K.

It is clear that when we add $(1)$ and $(5)$, and subtract half the $(3)$ and $(4)$, we'll get $(2)$. The same with enthalpies: $\Delta_fH^\circ(T) = \Delta_fH^\circ(298\textrm{ K}) + \int_{298}^T \Delta C_p\,dT$, where $\Delta C_p$ is the difference between $C_p$ for products and reactants of reaction $(2)$ with stoichiometric corresponding coefficients (1.0 and 0.5 in our case) taken into account.

Returning to your examples: in the first one, you are asked to find the enthalpy increment for $\ce{NO}$, which is just the amount of heat required to, well, heat it up. In the second example, you are asked to find the enthalpy of formation of $\ce{CH4}$, and I hope that my example above shows you the way of doing it. Also, if you copied the formulae from the textbook right, there's $\Delta$ missing from under the integral (should be $\int \Delta C_p(\ce{CH4})\,dT$).

A rule of thumb: when in doubt about the enthalpies of any process, write down the process thoroughly, with temperatures and phases of all reactants and products. This way it becomes much clearer. It is also useful to remember that in the real world there is no such thing as "The Enthalpy": what one has to deal with is always some change, some $\Delta$.


What is product obtained when B2O3 is reacted with NaOH?

As @Ivan said, it may form metaborate, borax or a tetrahydroxoborate complex all depending on reaction conditions(temperature, pressure, medium, stoichiometry etc.).

$$\ce{B2O3 + 2NaOH ->[\Delta] 2NaBO2 + H2O}$$

Boron trioxide react with sodium hydroxide to produce sodium metaborate and water. This reaction takes place at a temperature of 400-550°C.

$$\ce{2B2O3 + 2NaOH → Na2B4O7 + H2O}$$

Boron trioxide react with sodium hydroxide to produce sodium tetraborate and water. Sodium hydroxide - diluted solution. Boron oxide is amorphous.

$$\ce{B2O3 + 2NaOH + 3H2O → 2Na[B(OH)4]}$$

Boron trioxide react with sodium hydroxide and water to produce sodium tetrahydroxoborate(III). Boron oxide is amorphous. Sodium hydroxide - concentrated solution. The reaction proceeds at room temperature.


Molarity vs Normality in labeling sulfuric acid

Who labels concentrated sulfuric acid as 18N?  36N sounds right to me and Sigma-Aldrich:

  Concentration: The commercial reagent contains 93-98% H2SO4, the balance being water. Its effective concentration is 18 M (36N).

https://www.sigmaaldrich.com/content/dam/sigma-aldrich/docs/Sigma-Aldrich/Product_Information_Sheet/258105pis.pdf

IUPAC discourages use of normality because it is ambiguous.  The number of equivalents can be ambiguous depending upon the purpose (equivalents of protons, equivalents of electrons, etc.).

See the following IUPAC document: http://old.iupac.org/publications/analytical_compendium/Cha06sec3.pdf


Any idea why tap water goes stale overnight?

Exclude reasons one by one
Surface biological contamination: use two vessels, wipe the one of them with H2O2, hydrogen peroxide solution, leave another untreated. Check if time till smell appears is different by a few days. Bacterias can survive 20% vinegar and 100*C for a short time, some of them, if they are in a sleeping stage, it is not quite reliable method i would say.
Tap water biological contamination: use two vessels with water, add a few tablespoons H2O2 with water in one, and just water in other. Check if time till smell appears is different by a few days.
Tap water chemical contamination: buy some bottled water, use two vessels, one with tap water, another with bottled water. Check if it will be different in a few days.
Bulk material of the vessel releasing chemicals: leave water in different material vessels, one is glass, another one is your preferred one. Check if glass vessel has less smell adter a few days. Glass releases the least amount of chemicals into the water.
Dust accumulation: use two vessels, cover one and leave another one open, check if smell will appear faster in one.
Psychological factor: use two identical vessels, ask your friend to give you one or another, randomly, blindfold yourself, do a test 10 times, check if your score is above 8 correct guesses out of 10 tries.
After completing this tests you will narrow down your search significantly to give you some idea about what is going on.


Vapour absorption in 'normal conditions'

Normal or standard conditions are usually ment pressure of 101.325 kPa and temperature of 298.15 K or 293.15 K. Please see the link for reference.


Size of dispersed phase particles in colloid?

Its between $1-1000nm$ as stated on wikipedia in a paragraph name IUPAC definition.

  Colloid: Short synonym for colloidal system.
  Note: Quotation from refs.[3][4]
  Colloidal: State of subdivision such that the molecules or polymolecular particles dispersed in a medium have at least one dimension between approximately 1 nm and 1 μm, or that in a system discontinuities are found at distances of that order.[5]
  Note: Quotation from refs. [3],[4]

Edit: Since it is mention under IUPAC definition that size lies between $1nm-1000nm$, therefore this must be the correct size.


Free energy change in collagen hydrolysis

tldr: reaction is exothermic because the heat of reaction is -2.5 Kcal/mol.

(1) Peptides do hydrolize in presence of enzymes. It means that Gibbs energy of hydrolysis is negative. However, Gibbs energy consists of $\Delta$ H and $\Delta$ S*T. $\Delta$ S of hydrolysis is favorable (more disorder), but $\Delta$H can be either lower or higher.  $\Delta$ H is what we are discussing here (the enthalpy of the reaction).

(2) Some endothermic reactions are really very endothermic (like dissolving NH4NO3 in water). Other reactions are mildly endothermic.  Hydrolysis of meat doesn't have a large $\Delta$ H change. "HEATS OF HYDROLYSIS OF AMIDE AND PEPTIDE BONDS" paper has exact value for the reaction: -2.5 Kcal/mol of peptide.  A negative value indicates the reactants have greater enthalpy, or that it is an exothermic reaction. So the reaction is slightly exothermic.


How to synthesise N,N,3‐trimethylbutanamide from 1‐chloro‐2‐methylpropane?

Phosgene, $\ce{O=CCl2}$, is a potent toxin and in no way would I consider it a safe or even standard laboratory reagent (at least on an undergraduate level).

Allow me to lead you down the garden path of organic synthesis.

We have a haloalkane. What do we know about such compounds; they have a polarised $\ce{C-X}$ bond ($\ce{x}$ is a halogen) and are prone to nucleophilic attack/substitution. In this case, we have a nice primary, haloalkane, hence, any nucleophilic substitution is likely to be SN2.

Now, the target is an amide, but what do we note is different about the target versus the starting material (SM)? There's an extra carbon present (and yes there's no halogen).

Warning bells should be going off at this time. What nucleophilic, reagent do we know of that could extend a carbon chain?
I can think of one: $\ce{CN-}$ (cyanide ion).
So, first react with $\ce{KCN}$, second react with conc. $\ce{H2SO4}$ to give a carboxylic acid, third react with $\ce{NH(Me)2}$ to give the product.
Now, once you've isolated the carboxylic acid, you can react it 'neat' with $\ce{NH(Me)2}$ and molecular sieves. The molecular sieves simply ensure a dry reaction and very high temperatures are needed (approx $\pu{160 ^\circ C}$). You could convert the carboxylic acid to an acyl chloride and react the amine to ensure a very high yield and 'fast' albeit more 'vigorous' reaction; in this case any ordinary organic solvent should suffice.


Is the change of the Gibbs free energy always zero at constant pressure?

You wrote:
$$\mathrm{d}G = \mathrm{d}H - T\,\mathrm{d}S = \mathrm{d}H - đq$$
There are several issues with this. I don't wish to co-opt other answers and comments, but allow me to mention these here, since my answer stands as the accepted (which I have no power to change):

You haven't assumed constant temperature, which means that there is an additional term $-S\,\mathrm{d}T$ which you did not take into account;
You haven't assumed constant composition, which likewise leads to another term $\sum_i \mu_i\,\mathrm{d}n_i$.

In fact, these considerations (particularly the second) are more important than my original point, which is retained below. See Kexanone's answer for discussion of these points. In the following discussion, one should assume constant temperature and composition.

In the second equality, where you equate $đq$ with $T\,\mathrm{d}S$, this assumes that the process is reversible. Without going into too much detail (it can be found easily elsewhere), entropy is defined by the following equation:
$$\mathrm{d}S = \frac{đq_\mathrm{rev}}{T}$$
where the subscript $\mathrm{rev}$ indicates a reversible process. It can be shown that for an irreversible process,
$$\frac{đq_\mathrm{irrev}}{T} < \frac{đq_\mathrm{rev}}{T} = \mathrm{d}S$$
This result is the second law of thermodynamics. Therefore, for an irreversible process, $T\mathrm{d}S > đq$.
You are right in saying that at constant pressure, $\mathrm{d}H = đq$. Putting it all together, we have:
$$\mathrm{d}G = \mathrm{d}H - T\mathrm{d}S = đq - T\mathrm{d}S < 0 \qquad \text{(irrev. process; closed system; }p, T\text{ const.)}$$
Therefore, the direct answer to the question is: $\mathrm{d}G$ is, in general, not equal to $0$ because $T\mathrm{d}S \neq đq$.
In fact, you may notice that the condition $\mathrm{d}G < 0$ is the condition for spontaneity under constant $p$ and $T$. This is no coincidence. The concepts of spontaneity (in the context of $G$) and irreversibility (in the context of $S$) are extremely similar, in that they both refer to a tendency for a process to proceed.

For a reversible process, $T\mathrm{d}S = đq$. Following the logic that you used in your question, this means that $\mathrm{d}G = 0$, which is correct. If $\mathrm{d}G = 0$, the forward process cannot be spontaneous (that would require $\mathrm{d}G < 0$) and the reverse process cannot be spontaneous (that would require $\mathrm{d}G > 0$).
This can only mean one thing: the system is in equilibrium. Indeed, the condition for equilibrium at constant $p$ and $T$ is $\mathrm{d}G = 0$. And at this point, it should not come as a surprise that a reversible process is, by definition, one that is at equilibrium throughout.


Why alkanes containing even number of carbons can be packed more efficiently than alkanes with odd number of carbon atoms?

A recent article in PubMed (Angew Chem Int Ed Engl. 2016 Nov 2;55(45):14090-14095.) suggests that this odd-even effect also affects translation diffusion in the liquid state.

In the introduction of the PubMed article, it also addresses your particular question and confirms your intuition that it relates to the symmetry of the molecule. Interestingly, this was only experimentally confirmed in the last decade. To quote the article:

  Although this striking odd–even effect was known hundreds of years ago, its molecular origin was only revealed in the last decade. With the help of single-crystal diffraction and controlled crystal growth, the crystalline lattice of n-alkanes was determined.[5] It was confirmed that the intermolecular distances between the end groups, namely, the CH3 groups, are responsible for the alternation in the densities and melting points. In short, n-alkanes with an even number of carbon atoms pack more efficiently into ordered periodic crystalline structures and thus exhibit higher densities and melt at higher temperatures, whereas n-alkanes with an odd number of carbon atoms do not pack as effectively, and thus their densities are lower, and lower temperatures are needed for melting or freezing. A “packing effect” has been established to explain the odd–even effect as well as the related physical properties in the solid state.

Edit: For completeness, this is the article they link to that originally proved the effect experimentally: http://onlinelibrary.wiley.com/doi/10.1002/(SICI)1521-3773(19990401)38:7%3C988::AID-ANIE988%3E3.0.CO%3b2-0/full


I am told that carbon dioxide is IR inactive.

You're right, that's not true.  Since carbon dioxide is linear it has $3n-5 = 4$ vibrations and they are pictured below.

The symmetric stretch does not result in a change (of the initially
zero dipole moment), so it is ir-inactive.
The asymmetric stretch does result in a change in dipole moment so it is ir-active.
The bend also results in a change in dipole moment so it too is ir-active.

We expected 4 vibrations and I've only listed 3. This is because the "bend" (let's start by placing the molecule along the x-axis) can occur in the y direction and the z direction.  But these two motions are the same, just deforming in different directions, the bend is said to be degenerate, accounting for the "fourth" vibration.

To sum up, carbon dioxide has 2 ir-active vibrations.

Edit - response to example added (question d) by OP

Question d is incorrect.  Either the author 1) inadvertently switched the column headings (IR active, IR inactive) or 2) meant to use some molecule other than carbon dioxide.

The first 3 rules you learn for interpreting IR and Raman spectra are

The number of molecular vibrational modes equals 3n-6 (3n-5 for linear molecules), where n is the number of atoms.
An ir active band will be observed if a vibration results in a change of the dipole moment. The initial dipole moment in the molecule's equilibrium geometry can be zero; all you need is a change.  The terms "polar" and "non-polar" can be confusing, they often mean different things to different people.  Leave "polar" out of the criteria for ir activity and stick with dipole moment, it is a much better understood term.
the rule of mutual exclusion, it states that, for centrosymmetric molecules (molecules with a center of symmetry, like carbon dioxide), vibrations that are IR active are Raman inactive, and vice versa.  So for carbon dioxide there is 1 Raman band and two IR bands.

Here's a link to a recent SE Chem question: How can I deduce the linearity of XeF2 from the IR spectrum? where these rules were used to determine the structure of a molecule.


Conditions for free radical bromination using NBS and peroxides?

From practical experience, I can tell you that the answer is "All of the above".

First, your peroxide initiator needs some energy to undergo homolytic cleavage. This can either be achieved through UV photolysis (A) or heating (B).

$$\ce{RO-OR <=>[h\nu][\mathrm{or\ }\Delta] RO. + .OR}$$

The amount of energy needed will correlate to the strength of the O-O bond. The energy available from UV irradiation is dependent on wavelength because $E=hc/ \lambda$. The amount of energy available from heat is dependent on temperature.

I've initiated NBS reactions both ways, but I prefer heat since I can more easily heat a larger vessel.

  What about solvent polarity?

Solvent is the tricky point, and depending where you are in your studies, you may not have covered the other things that NBS can do. Under the right circumstances, NBS is also an electrophilic source of bromine, especially in polar protic solvents.

Here are two examples from my personal experience using the same substrate with different conditions:

Using methanol as a solvent without a radical initiator, I got electrophilic bromination of the arene.

Using 1,2-dichloroethane as a solvent with heat and a radical initiator, I got radical bromination of the benzyl position and electrophilic bromination of the arene.

Using 1,2-dichloroethane as a solvent with UV irradiation and a radical initiator, I got radical bromination of the benzyl position.


How to determine what organic molecule has the highest boiling point?

There are commonly three types of intermolecular forces quoted when it comes to melting and boiling points: dipole-dipole interactions, London interactions and hydrogen bonding. Hydrogen bonding is by far the strongest force of them all. In all cases except for very rare ones, for a hydrogen to take part in hydrogen bonding, it must be bonded to $\ce{O, N, F}$ — one of the three very electronegative atoms.

In difluoromethane, there are no hydrogens bound to $\ce{O, N, F}$ — they are both bound to carbon. It is mainly carbon that experiences fluorine’s electronegativity and by the time the inductive effect reaches the hydrogens it has been substantially lowered. On the other hand, methanol and methyl amine both have hydrogens bonded to the electronegative atom — nitrogen or oxygen — so these compounds actually take part in hydrogen bonding. Since we have two compounds that participate in hydrogen bond networks it is safe to assume that these will be the top runners.

As you mentioned, methanol’s hydrogen bonds should be slightly stronger since oxygen has a higher electronegativity. I think your reasoning as to why methanol is stronger than methyl amine is sound enough.

Difluoromethane’s principal intermolecular forces will be dipole-dipole interactions. However, you should also note that the presence of fluorine atoms strongly reduces the underlying London interactions, since $\ce{C-F}$ bonds are more localised and fluorine’s lone pairs are much less disperse than most other carbon-heteroatom bonds — this is due to fluorine’s very high electronegativity. Thus, while we have significant dipole-dipole interactions, we are somewhat lacking London interactions in spite of the rather large molecular weight of $52$.

This is important when comparing it to both methyl amine $M_\mathrm{r} = 31$ and butane $M_\mathrm{r} = 58$. Butane, given its larger electron count ($26$ bonding electrons versus $14$ bonding and lone pair electrons in methyl amine) and its very non-spherical shape can give rise to much stronger London interactions, especially since all its bonds are almost or entirely unpolar $\ce{C-H}$ or $\ce{C-C}$ bonds. A general rule of thumb is the heavier a molecule, the stronger the London interactions and they quickly catch up with what dipoles or hydrogen bonds can do.

So putting it all together:

Difluoromethane has very weak London interactions but some dipole-dipole interactions. It is expected to have a rather low boiling point.
Methyl amine can form hydrogen bonds but is also rather small and thus lacking London interactions. It is less polar than methanol giving rise to lesser dipole-dipole interactions. Put it somewhere in the middle.
Butane is significantly larger in volume than all the others and also has much more electrons that can take part in London interactions. Thus, London interactions are predicted to be rather strong but nothing else is there.
Finally, methanol of course forms strong hydrogen bonds, is a good dipole and can take part in London interactions well. A clear winner.

I wouldn’t know how exactly to rank butane and methyl amine but I would put difluoromethane at lowest and methanol at clearly highest boiling point. This is consistent with the experimental data.


Why isn't 4-hydroxybenzoic acid formed along with salicylic acid in Kolbe's reaction?

I don't understand why two different products are formed in the same reaction using very similar bases.

If the bases make a a difference, there must be a difference between them.
If the anion is the same, the cation must make the difference.

The difference between $\ce{Na+}$ and $\ce{K+}$ lies in the ionic radii of 0.95 and 1.33 $\mathring{A}$, respectively.
It is believed that $\ce{Na+}$ serves as an anchor and guides $\ce{CO2}$ into the ortho position during the formation of the Wheland intermediate. In the latter, $\ce{Na+}$ is chelated between the oxygen atoms of carboxylate and the phenolate. Apparently, this is very effective.
This does not mean that the phenolate doesn't attack $\ce{CO2}$ via C-4 at all. However, the formation of a Wheland intermediate (= the rate-determining step) is normally (and definitely in the case of the para attack) reversible.
It is open to question if full reversibility can be claimed for the $\ce{Na+}$ directed ortho attack under modest reaction conditions.
It is however likely that the back reaction (= cleavage Wheland intermediate) will stronger come into play at elevated temperatures. Consequently, some 4-hydroxybenzoic acid will be formed then.
With the bigger $\ce{K+}$, the anchoring and chelating effect is seemingly not possible, $\ce{K+}$ doesn't fit and the para direction effect of the phenolate prevails.


How exactly do alcohol molecules dissolve in seawater?

Here is my calculation.

The ocean has a volume of 1.3 billion cubic kilometers or 1.3e+18 cubic meters or 1.3e+21 liters. 0.8/46=0.0173 moles of ethyl alcohol x 6.02e+23 molecules per mole = 0.1047e+23 molecules, which if dissolved evenly in 1.3e+21 liters would give 10.47/1.3 or approx. 8 molecules of ethanol per liter.


How do Shaving Creams, Foams, Soaps ensure cut Hair Fragments that stick & clump between Safety Razor blades are easily washed away?

Active chemicals are called surfactants. It is well explained in colloid chemistry books and kind of in wikipedia (https://en.wikipedia.org/wiki/Surfactant)

And it is possible to find alternatives for such chemicals in nature, but they won't be as effective as artificial.


Harmless & pungent volatile liquid for use in a stage show

I would be very, very very cautious in choosing the odorant here. And speed of odour spreading would be quite low unless specific adjustments are employed to fasten air circulation. And even then, it is quite easy to screw because of minor differences in air circulation.

In practice I would drop any unpleasant odours like mercaptans, sticking to powerful, but generally not unpleasant odours. An example would be benzaldehyde. It has quite strong odour, but autocatalitically oxidizes on air, so the odour should vanish reasonably fast. Some food odourant or parfume would be even better if possible.

Another possible option would be acetic aldehyde. It has strong apple-like odour, and can be easily produced by pouring ethyl alcohol into cromium (VI) oxide, immidiately producing strong odour.

Though I must say, most odoured compound are slightly toxic, so it is not impossible to overdo with troublesome consequences.

Still, I strongly recommend to move from odour to light and/or smoke effect, preferably using premanufactured and commercially available devices. It is too easy to screw up with it.


Acidic strength of para substituted benzoic acids

You are somewhat on the right track. However, consider that acidity depends on the relative stabilization of the acid vs. the conjugate base, since the expression for $\mathrm{p}K_\mathrm{a}$ is as follows: $$\mathrm{p}K_\mathrm{a} = -\log\left(\frac{[\ce{A-}][\ce{H+}]}{[\ce{HA}]}\right)$$

Consider this figure for a visualization:

Electron-withdrawing groups like p-$\ce{NO2}$ (nitro group in the para position) increase acidity by making the carboxylate ion stable relative to the protonated form. However, electron-donating groups like p-methoxy decrease acidity relative to standard benzoic acid by destabilizing the (already negatively charged) carboxylate anion!

TL; DR: An electron-withdrawing group like p-$\ce{NO2}$ will better stabilize the negatively charged $\ce{COO-}$ group than an electron-donating group like p-$\ce{OCH3}$.

I found a neat picture from UC Davis that summarizes what I have said:


Predicting the Product of a Substitution/Elimination Reaction

Remember that in a $S_{N}2$ reaction, the centre bearing the $\ce{Cl}$ atom would have to be attacked by $\ce{OH-}$ from the back side. This, however, is prevented by the by two bulky tert.-butyl substituents

As far as the alternative $E2$ pathway is concerned: There are no $\alpha$ hydrogen atoms that could be abstracted by $\ce{OH-}$.

I've uploaded a CML file of the structure to pastebin. I suggest to get yourself a copy of Avogadro and inspect the geometry on your own PC.


How to I neutralize the smell of (VERY) spoiled milk

The disgusting smell is probably butyric acid. Try adding a base: For example, washing soda (sodium carbonate), or perhaps drain cleaner (sodium hydroxide, be careful with this one, this is very corrosive to people/eyes), or even ammonia. This should turn the butyric acid into butyrate ion, which should be much less disagreeable. Then flush down the drain.


Why do amorphous substances have different values of physical properties in different directions?

Yes, you’re exactly right that we are talking about an average. If we could somehow look at a small enough portion of an amorphous solid, we would see differences between the properties in different directions. Indeed, a small enough portion of, say, amorphous ice is a single water molecule, which is certainly not isotropic: we can talk about the direction perpendicular to the molecular plane or whatever.

But when we talk about real, physical materials, as usual in chemistry, we are talking about Avogadro’s number (plus or minus a few orders of magnitude) of molecules. So taking the spatial average of an amorphous structure makes perfect sense, and the law of large numbers means that the tiny differences between properties in different directions will be utterly undetectable.

(As an aside, it’s worth noting that the perfect crystalline structure we imagine is also an average over every unit cell in the material – it’s just that because each unit cell is oriented in more or less the same way, different directions are not “mixed” as part of this average, so crystals can have anisotropic properties.)


What causes energy peaks during a molecular dynamics energy minimization?

At the top of your config file, you imply that you want to perform an NVE calculation, but you’re actually doing NVT, because you have the Langevin thermostat turned on. Langevin works by giving the system a random “kick”, which is probably spiking the energy.


Combustion and pyrolysis of sulfides and organosulfurs

I'd expect $\ce{H2O, CO, CO2}$, and $\ce{SO2}$ to be created during the combustion of these compounds. I'd wager a decent amount of $\ce{SO2}$ was created, but was diluted in the atmosphere.

Regarding $\ce{H2S}$, maybe it was produced, but then reacted according to the following:

$$\ce{H2S + O2 -> SO2 + H2O}$$

Additionally, there were probably a wide variety of odd sulfur compounds, but their existence was transient as they reacted to form $\ce{SO2}$ for instance. Furthermore, the authorities may not even be looking for these more complex molecules.


Expression for fugacity coefficient derived from a pressure-explicite EOS

Mindful of the guideline "don't give answers in comments", I've converted my comment to an answer, trivial though it is.

When $v\rightarrow \infty$ you get the ideal gas eos, so $z\rightarrow 1$. This gives you the result, i.e. $1$ at the lower limit (subtracted) and $z-\ln z$ at the upper one.

When you write down the integrated form, $z-\ln z$, you need to remember that it means that you are evaluating the function $z(v)$ at the two limits $v=\infty$ and $v=v$, and using those values in that expression; you should not be trying to set $z=v$ at the limits, which seems to be (partly) what you have written at the end of your question.

Edit following OP comment.

More details. At constant $T$, $z=z(v)$ is a function of molar volume $v$,
determined by the equation of state, $P(v)$ at the given value of $T$. This equation is unknown, but we can be sure that,
in the ideal gas limit $v\rightarrow\infty$, $z(v)\rightarrow 1$.

Integrating both sides of the equation from the ideal gas limit to
the desired volume $v$, and using $v'$ as the integration variable, gives
eqn (2) of the question in its full form:
$$
\left| \ln \phi(v')\right|_{v'=\infty}^{v'=v}
= \left|  z(v') - \ln z(v') \right|_{v'=\infty}^{v'=v}
+ \frac{1}{RT}\int_{\infty}^v \left(\frac{RT}{v'}-P(v')\right)\mathrm{d}v'
$$
I think it's clearer to distinguish between the integration variable $v'$
and the "upper" limit of integration $v$,
but many people would be happy just to use $v$ instead of $v'$.

Anyway,
on the left,
we know that, in the ideal gas limit,
the fugacity coefficient $\phi(\infty)=1$,
so $\ln\phi(\infty)=0$ and we are just left with $\ln\phi(v)$.
On the right,
similarly,
we substitute in the upper and lower limits for $v'$.
We know $z(\infty)=1$, so
the function being evaluated is $z(\infty)-\ln z(\infty)=1$ at the lower limit,
and $z(v)-\ln z(v)$ at the upper limit.
So the final answer is
$$
\ln \phi(v)
= z(v) - \ln z(v) - 1
+ \frac{1}{RT}\int_{\infty}^v \left(\frac{RT}{v'}-P(v')\right)\mathrm{d}v'
$$
An important point is that the integration variable
is the molar volume $v'$ (or $v$ if you prefer), not $z$.
In evaluating the result, the integration limits $v$ and $\infty$
are substituted for $v'$, the argument of the function being evaluated.
It is incorrect to set $z=v$, or $z=\infty$.
(This should be even more clear in this case,
since $z$ is a dimensionless quantity, whereas $v$ is not).


Why are Kc and Kp interchangeable in the free energy-equilibrium equation?

The equilibrium constant is defined by activities:

The activity of a species in a liquid solution can be approximated to: $a=c/c^0$ where c is the concentration of the species and $c^o$ is the standard concentration ($1M$). This approximation leads to $K_{eq}=K_c$. The activity of gasses is called fugacity and it can be approximated to: $f=p/p^0$ where p is the pressure of the gas and $p^0$ is the standard pressure ($1$ $bar$). This leads to $K_{eq}=K_p$. Therefore, if a reaction takes place in e.g. water, one should use $K_c$, but if the reaction takes place in the gaseous state, $K_p$ should be used when calculating $dG$.


Do various soaps work better when dissolved/used in different water temperatures?

To start, soaps and detergents have two different chemical makeups, meaning that they act different under different conditions.

For general soaps, here is an explanation of how it works:
http://chemistry.about.com/od/cleanerchemistry/a/how-soap-cleans.htm

As a general chemistry note, as the temperature of a solution increases, the ease of a chemical to change (whether it is to react or change shape) increases.  This means that colder temperatures will allow a chemical to remain in the same state.  If the soap depends on a specific conformation/shape to act then colder temperatures may be preferable.  However, if a reaction needs to occur in order for the soap to act some heat may be required.

For Tide Pods, it states that it "dissolves quickly in hot & cold."
http://www.tide.com/en-US/product/tide-laundry-detergent-pacs.jspx

The chemical makeup of Tide Pods is composed primarily of salt compounds and enzymes.  Both salts and enzymes tend to operate better at higher temperatures.  So as long as you don't pass the denaturing temperature of the enzyme, higher temperatures would be preferred.


I think you should think this question a little through:

The easy answer is that atoms are not spherical in shape. Sphere is a solid object, atoms to not have well defined boundaries with "solid cutoffs". Atoms are more like clouds consisting distributions of electrons, neutrons and protons.
Atoms, per definition, are electrons + a single nucleus, ie. the huge positive charge of a point-like object (nucleus) pulls together the whole thing. That result is that electrons approximately move in a central field. Such central field generates spherical symmetry of the electronic states, and that is why you have e.g. spherical harmonics when you describe the atomic orbitals. Check the solution of the Schrodinger equation how it leads to states, and how wave-functions with only radial dependence (i.e. sphere-like) appears.

+++++ Incorrect part of the answer was deleted based on the comments of DavePhD.


Pattern to determine the maximum ionic charge for transition elements?

The pattern you observe for main group elements is due to the number of valence electrons present in the s and p subshells of the outermost shell. The most you can do is either lose all those s and p electrons (giving a ‘cation’[1] charged exactly according to the old main group number — new group number minus 10 for groups 13 through 18) or you can fill these up to a total of eight.

For transition metals, a d subshell is in the process of being filled. Since the d subshell of a lower shell is between the s and p subshells of a higher shell according to the aufbau principle, we can imagine that the highest number of ‘usable electrons’ would be 18. This is not typically the case, however.

For the first few transition metals, you can remove as many electrons as are present in the s and d valence subshells (e.g. 4s and 3d). In the 3d series, manganese is the last one that can do this to give manganese(VII), in the 4d series ruthenium can reach ruthenium(VIII) and in the 5d series iridium can reach iridium(IX).[2] For the remaining elements of a certain transition metal period, the highest accessable oxidation state again decreases. You can find an overview of the maximum oxidation states known for certain elements on Wikipedia.

The only elements that again follow a somewhat more simple scheme are the zinc group (zinc, cadmium and mercury). Since these have a completely filled d subshell in ground state, they are reluctant to give away any d electrons; hence their highest observed charge is $2+$, as if they were in group 2. (Indeed, they behave like group 2 metals in a lot of other respects, too; except mercury which is slightly different again.)

When it comes to the highest possible negative charge or oxidation state, things are more complicated yet again. Since all transition metals are metals, they are not commonly found as anions or anionic particles although some are more likely to than others. For some of these — typically the iron group and the two following groups — this corresponds to $12 - \text{group number}$. This small trend is more of an exception than of a rule, though. Note, for example, how zinc, cadmium and mercury are all known as in the $\mathrm{-II}$ state which does not make sense immediately just looking at the respective shells and subshells.

Many of the actually observable oxidation states can be rationalised by doing a complex molecular orbital analysis of the corresponding compounds. However, for most of these there is no simple and immediately obvious pattern that one can use — with the notable exception of the small bit of pattern outlined in the third paragraph.

Notes:

[1]: A better term here would be oxidation state. For many elements the highest possible free cation is not known but a compound with a corresponding oxidation state is.

[2]: Again, none of these are observable as free cations in standard chemistry; rather they are in compounds. Hence I have stopped using the term charge and started using oxidation state.


D6h symmetry clusters- orbital classification confusion!

The symmetry operations are usually defined with respect to the origin of the coordinate system. For single atoms one naturally chooses $(0,0,0)$, which coincides with the symmetry center.

For example the $\sigma_{xy}$ mirror plane in matrix representation (with $(0,0,0)$ being the symmetry center) would be:

$\sigma_{xy}=\pmatrix{1&0&0\\0&1&0\\0&0&-1}$

which means: invert every $z$ component. If applied to some object centered on $(0,0,0)$ (e.g. the AO of an atom on this position), the result would still be centered on $(0,0,0)$.

In case of molecules we can have at most one atom sitting on the symmetry center. Sometimes none, e.g. in $\ce{C2}$ where the symmetry center is in the middle between both atoms. If you apply the above matrix here (assuming the molecule axis is aligned with the $z$ axis), then it would map the AO from one to the other atom. Therefore it is not symmetric (with respect to the above defined symmetry operation) anymore.

Thus you cannot relate the IRREP of an atomic orbital in the single atom, with its IRREP in the molecule. The symmetry center changed! For example take an $s$ atomic orbital of any of your $\ce{Si}$ atoms. The position of the atom does not coincide with the center of the molecule. Therefore any symmetry operation (except for $E$) would map the $s$ orbital to some other $\ce{Si}$. The AO is no longer characterized by an irreducible representation.

What the code does instead, is first generating Symmetry Adapted Linear Combinations (SALC) of the atomic orbitals. Those are then again characterized by IRREPs (of the molecule) and used for the calculation. Now each obtained MO is indeed only constructed from SALC of the same IRREP. In other words: the Hamiltonian, and therefore the MO coefficient matrix as well, become block diagonal. However, in the output the MO coefficients are transformed back from the SALC basis set to the AO basis set, which is not block diagonal and may contain contributions from any AO for each IRREP of the MOs (an exception would be the AOs of an atom located in the center of the molecule).


Effective Bohr Radius in Transition Metal Complexes?

If you inspect Table II in Ref. 1, which you will find referenced in the link you provide to the WebElements website, you'll see that AU units are used (also in the linked page), not pm, so the radius of the 3d Cu orbital in Ref. 1 is actually

$$ r_{\textrm{Mann}} = 0.613 \times 0.529 Å = 0.324 Å $$

or 32.4 pm.

As for Slaters approach, it is not enough to simply divide $a_0$ by the effective nuclear charge. You can only do this for an unshielded hydrogenic 1s orbital, otherwise it is generally necessary to account for the principal quantum number (in Slater's method, this may be an effective value not equal to the principal quantum number, see Slater's original article$^2$). The location of the maximum follows from the functional form of a Slater wave function:

$$\psi \propto r^{n^*-1}\exp\left({\frac{Z_{\textrm{eff}}}{n^*a_0}r}\right)$$

In general, if you change the nuclear charge of a hydrogenic atom you can perform the transformation $a_0 \rightarrow a_0/Z_{\textrm{eff}}$ to obtain the new orbitals$^3$.

The effective quantum number is n*=3 for a 3rd shell electron so that

$$r_{\textrm{Slater}} = \frac{n^{*2}}{Z_{\textrm{eff}}}=\frac{9}{13.2}\times0.529 Å = 0.361 Å $$

or 36.1 pm.

The difference in $r_{\textrm{max}}$ obtained by the two approaches is therefore not that dramatically different.

Of course there are two other things that seem to be important here: (1) whether using an accurate value of $r_{\textrm{max}}$ can be expected to provide you with an accurate value of the property you seek and (2) how to obtain an accurate estimate of $r_{\textrm{max}}$for a d-orbital of Cu in the transition metal complex.

References

J.B. Mann, Atomic Structure Calculations II. Hartree-Fock wave functions and radial expectation values: hydrogen to lawrencium, LA-3691, Los Alamos Scientific Laboratory, USA, 1968.
J.C. Slater. Atomic Shielding Constants. Phys. Rev., 36, 57, 1930.
W. Greiner. Quantum Mechanics: An Introduction. Springer 1994.


Clarifications on the calculation of the $\Delta\mathrm{G}$ of a reaction

$$ \Delta G^\circ  = \Delta H^\circ - T\Delta S^\circ$$

Gibbs Free energy does not only relate to the spontaneity of a reaction, the magnitude of Gibbs Free energy describes how much "free energy" is available, or how much work is possible.

$\Delta G^\circ$ can either be negative or positive even if $\Delta H^\circ$ is $+$. Therefore,
we have to look at all three terms, $\Delta H^\circ$, $T$, and $\Delta S^\circ$

The Gibbs Free Energy, enthalpy, and entropy are state functions. Although I won't derive it here, we can find $\Delta G$, or $\Delta H$, or $\Delta S$, by using Hess's Law.

The information provided above gives us the $\Delta H_{rxn}$ which is $99.49 \text{kJ/mol}$ and thus endothermic.

We calculate entropy from the a list of experimental tabulated values:

$$\Delta S^\circ_{rxn} = \Sigma \Delta S^\circ_{products} - \Sigma \Delta S^\circ_{reactants} $$

$$\Delta S^\circ_{rxn} = (129.0 \text{J}\cdot\text{mol}^{-1}\text{K} + 4*130.6 \text{J}\cdot\text{mol}^{-1}\text{K}) - (126.8 \text{J}\cdot \text{mol}^{-1}\text{K} + 69.91 \text{J}\cdot\text{mol}^{-1}\text{K})$$

$$\Delta S^\circ_{rxn} = 454.69 \text{J}\cdot\text{mol}^{-1}\text{K} = 0.45469 \text{kJ}\cdot\text{mol}^{-1}\text{K} $$

While solving for entropy, you forgot to multiply the entropy of hydrogen gas with the number of moles. Also, you should always include the states of your reactants and products, as thermodynamic data for compounds differs for different states.

Another thing that is incorrect is standard temperature for thermodynamic reactions is at $25^\circ C$ not $20^\circ C$ and thus your temperature should be $298.15 K$

  I'm also a bit puzzled, 'cause this is a endoergonic reaction right? But the ΔG is negative, so the reaction would be spontaneous if my calculations are correct and if the activation energy is reached right?

This is where a lot of people make mistakes. It is true, that in general, endothermic reactions with a positive enthalpy usually have a positive $\Delta G$ and are thus not spontaneous.

However, if we look at our equation:

$$ \Delta G^\circ  = \Delta H^\circ - T\Delta S^\circ$$

We have to consider our other terms and not only enthalpy, namely that of temperature and entropy.

A reaction is spontaneous if $ \Delta G < 0 $ such that:

If $ \Delta H < 0$ and (-) and $ \Delta S > 0$ (+), the reaction is spontaneous at any temperature.

If $ \Delta H > 0$ and (+) and $ \Delta S < 0$ (-), the reaction is NOT spontaneous at any temperature.

If $ \Delta H < 0$ and (-) and $ \Delta S < 0$ (-), the reaction is spontaneous at low enough temperatures.

If $ \Delta H > 0$ and (+) and $ \Delta S > 0$ (+), the reaction is spontaneous at high enough temperatures.


What is heterogeneous catalysis in gas phase?

"Gas phase" means that the reactants are gases
The basic answer is that heterogeneous catalysis "in the gas phase" just means that the materials being reacted are gases. They catalysts can be liquids or solids (more common).
A simple, but very, very important example is the reaction used to convert ammonia into nitric acid (vital for fertiliser production, important for many organic chemistry building blocks and explosives). In this reaction gaseous mixtures of air and ammonia are passed over platinum gauze and, initially, produce NO (nitric oxide) which is further oxidised homogeneously to dinitrogen tetroxide which ultimately reacts with water to give nitric acid. A good summary of the reaction and its history is given in this Johnson Matthey article.
So gas-phase reactants + solid catalyst -> product. Gas phase heterogeneous catalysis.


Is chirality a physical or a chemical property?

I would not consider chirality as a property--either physical or chemical--it is part of the definition of a molecule's structure, which in turn gives rise to the properties it exhibits (rotation of polarized light, melting point, reactivity with X, etc.). In your example, the fact that hydrogen and deuterium are almost (but not entirely) chemically equivalent will make the substance behave almost the same as its dihydrogen analog, and the two mirror image isomers nearly indistinguishable by physical or chemical means.


How can I predict the product of the reaction of 2-methyltetrahydrofuran with HI?

Initially, the oxygen in the ring gets protonated. However, there is no formation of a carbocation. The ring opening proceeds by an SN2 pathway, where iodide ion attacks at the less hindered carbon:


The popular procedure to experimentally isolate caffeine is sublimation.

Slowly heat a teaspoon of coffee or tea from 150°C to 250°C. Let the sublimated caffeine precipitate at a watch glass cooled with ice. The resublimated caffeine forms white needles that can be identified under a microscope.

The sublimation temperature of caffeine is 178°C.


Condition For NaHCO3 to react to form CO2 with organic acids

$\ce{NaHCO3}$ is the basic sodium salt of carbonic acid, a diprotic acid which exists only in small amounts in aqueous solution, because the equilibrium

$$\ce{CO2 + H2O <=> H2CO3}$$

lies mainly on the educt side (this is also the reason why $\ce{CO2}$ evolution is observed). When this fact is taken into account, a $\ce{pK_{a}}$ value of 6.3 at 25 °C is obtained for the first protolysis step (source). Benzoic acid has a lower $\ce{pK_{a}}$ value of 4.19 and is therefore able to liberate carbonic acid from its salts, like any other acid that is significantly stronger than $\ce{H2CO3}$.

$$\ce{C6H5CO2H + NaHCO3 -> C6H5CO2Na + CO2 + H2O}$$

Just in case that you mean 3,4-Dihydroxycyclobut-3-ene-1,2-dione, which is also known as squaric acid, then a reaction with $\ce{CO2}$ evolution is likely. Squaric acid is a fairly strong diprotic acid, with $\ce{pK_{a}}$ values of 1.5 and 3.4, respectively. The acidity can be attributed to the stabilization of the dianion by resonance:

If carbon dioxide evolution with 1,2-dihydroxy cyclobutanedione is possible depends on the $\ce{pK_{a}}$ value of this compound. However, I would expect it to be less acidic than squaric acid.


Does acetamide respond positively to iodoform test?

From wikipedia article of haloform reaction:

  Acetyl chloride and acetamide don't give this (iodoform or in general haloform) test.

Why?

  Acetamide is not a ketone which is substantially more acidic than the
  amide.  It's about a million times too weak an acid at the alpha
  protons. Hydroxide is too weak a base. (Quora)


Difference in the functioning of alkaline and acidified KMnO4

Permanganate as oxidizing agent works most efficiently in acidic solution, because it is reduced to the greatest extent in this medium, from oxidation state +VII in $\ce{MnO4-}$ to +II in $\ce{Mn^2+}$.

$$\ce{8H+ + MnO4- + 5e- ->~ Mn^2+ + 4H2O}$$

Therefore, the number of electrons transferred from oxidized species per mole $\ce{KMnO4}$ (5 electrons) is the largest, and the least amount of $\ce{KMnO4}$ is needed for the reaction with a certain amount of alcohol. This makes oxidation in acidic medium the best choice for economic reasons.

The oxidation in neutral medium is less efficient, as manganese(VII) is only reduced to manganese(IV), yielding insoluble $\ce{MnO2}$ as a byproduct which needs to be removed from the reaction mixture during purification of the desired product (carboxylic acid).

$$\ce{2H2O + MnO4- + 3e- ->~ MnO2\downarrow + 4OH-}$$

Oxidation in strongly alkaline medium is least efficient in terms of transferred electrons per mole permanganate, as the latter is only reduced to manganate(VI).

$$\ce{MnO4- +e- ->~ MnO4^2-}$$

However, when the reaction is done under these conditions, a higher reaction speed and selectivity can be achieved, presumably because the alcohol will react more readily with $\ce{MnO4-}$ when deprotonated. Nevertheless, oxidation in acidic medium can be preferred for reasons stated above and substrates which are sensitive to strong bases.


What is the structure of a lipid and why is this example non-polar?

Lipids usually consist of 3 fatty acids and 1 glycerol.

That's wrong. You're giving the definition of fats, which are just a subclass of lipids.

The structure in your question is the skeleton of cholesterol - without the stereochemistry.


Which bonds need to be broken to form a liquid?

It could be either chemical bonds or weaker intermolecular forces that change during a change of state. Water is held together by hydrogen bonds and make its boiling point higher than for analogous compounds.

Sulfur changes both ways with temperature:

  Sulfur forms polyatomic molecules with different chemical formulas, the best-known allotrope being octasulfur, cyclo-S8... At 95.2 °C (203.4 °F), below its melting temperature, cyclo-octasulfur changes from α-octasulfur to the β-polymorph. The structure of the S8 ring is virtually unchanged by this phase change, which affects the intermolecular interactions. Between its melting and boiling temperatures, octasulfur changes its allotrope again, turning from β-octasulfur to γ-sulfur, again accompanied by a lower density but increased viscosity due to the formation of polymers. At higher temperatures, the viscosity decreases as depolymerization occurs. Source

Metallic bonds are delocalized, so in melting, there is no real change in bonding (though on boiling, a metallic vapor is quite different).

One could say, though, that this is simply a matter of definition: chemical bonds act over shorter distances, and intermolecular forces over longer, but they are not intrinsically different, the polar or hydrogen bond being an example of one that bridges these realms.