How to calculate Madelung's constant for NaCl?

The series $\displaystyle \frac{6}{1}-\frac{12}{\sqrt{2}}+\frac{8}{\sqrt{3}}-\frac{6}{2}+\frac{24}{\sqrt{5}}\pm\ldots$ diverges as has been proven in 1951.

The correct value can be obtained by summation over lattice cubes with edge length $2n+1$ and getting the limit for $n \to \infty$. This has been shown in 1985.

Wikipedia provides more information and further reading on this.


How to determine net charge of a polycation using Wade's rules?

As far as I can tell, your calculations are correct. A closo structure would require $2n+2$ cluster electrons (or $4n+2$ total electrons), which is satisfied by the formula $[\ce{Bi}_n]^{(n-2)+}$. As such, for $n  = 6$, a closo octahedral structure would correspond to the ion $\ce{[Bi_6]^4+}$ as you have outlined.

The contradiction seems to be that the ion $\ce{[Bi_6]^2+}$ is not closo, but rather nido, according to e.g. Hampel & Ruck:[1]

  The $\ce{[Bi_6]^2+}$ cation, a nido cluster with 16 skeletal electrons, has the shape of a distorted octahedron with an opened edge.

These require $2n$ cluster electrons for complete population of the bonding molecular orbitals, consistent with the formula $\ce{[Bi_6]^2+}$. I do not currently have access to Holleman-Wiberg (nor would I understand the German version!), so I cannot check what they have written, but it is possible that it is a simple typo.

Interestingly, some of the higher bismuth polycations also display a tendency to have "extra" electrons. For example, Greenwood and Earnshaw[2] describe the following cations:

$$\begin{array}{cccc}\hline
\text{Cation} & \text{Formal oxidation state} & \text{Cluster structure} & \text{Point group symmetry} \\ \hline
\ce{Bi+} & 1.00 & - & - \\
\ce{Bi3+} & 0.33 & \text{Triangle} & D_\mathrm{3h} \\
\ce{Bi5^3+} & 0.60 & \text{Trigonal bipyramid} & D_\mathrm{3h} \\
\ce{Bi8^2+} & 0.25 & \text{Square antiprism} & D_\mathrm{4h} \\
\ce{Bi9^5+} & 0.56 & \text{Tricapped trigonal prism} & C_\mathrm{3h} (\sim\!D_\mathrm{3h}) \\ \hline
\end{array}$$

The rationalisation which they provide for this is

  [The application of Wade's rules] would account for the stoichiometries $\ce{Bi3+}$ and $\ce{Bi5^3+}$ but would also lead one to expect $\ce{Bi8^6+}$ and $\ce{Bi9^7+}$ for the larger clusters. However, these charges are very large and it seems likely that the lowest-lying nonbonding orbital would also be occupied in $\ce{(Bi+)_n^2-}$. For $\ce{(Bi+)8^2-}$ this is an $\mathrm e_1$ orbital which can accommodate 4 electrons, thereby reducing the charge from $\ce{Bi8^6+}$ to $\ce{Bi8^2+}$ as observed. In $\ce{(Bi+)9^2-}$ the lowest nonbonding orbital is $\mathrm a_2''$ which can accommodate 2 electrons, thus reducing the charge from $\ce{Bi9^7+}$ to $\ce{Bi9^5+}$ as observed.

References:

Hampel, S.; Ruck, M. Synthesen, Eigenschaften und Kristallstrukturen der Cluster-Salze Bi6[PtBi6Cl12] und Bi2/3[PtBi6Cl12]. Z. Anorg. Allg. Chem. 2006, 632 (7), 1150–1156. DOI: 10.1002/zaac.200500379.
Greenwood, N. N.; Earnshaw, A. Chemistry of the Elements, 2nd ed.; Butterworth–Heinemann: Oxford, U.K., 1997, pp 590–591.


Stability of superoxides of alkali metals

According to the Wikipedia page Alkali metal oxides, the superoxide of $\ce{K}$, $\ce{Rb}$ and $\ce{Cs}$ form when these metals are burnt in air, and according to the Bodner Research Group page Active Metal Reactions, they

  react so rapidly with oxygen they form superoxides, in which the alkali metal reacts with $\ce{O2}$ in a 1:1 mole ratio.

Given, according to the Chem-Guide blog article Alkali metals that

  The fact that a small cation can stabilize a small anion and a large cation can stabilize a large anion explains the formation and stability of these oxides.

Which, according to the blog, results in the  $\ce{Li+}$ ion (which is small, possessing a larger positive field around it) only being able to naturally combine with a small anion $\ce{O^{2-}}$ forming $\ce{Li2O}$, whereas the substantially larger $\ce{K+}$, $\ce{Rb+}$ and $\ce{Cs+}$ have weaker positive fields and can stabilise the larger superoxide anion $\ce{O2^{-}}$ forming superoxides.


Are the potentials of the electrolytes equal in Galvanic cell?

There is a potential difference between an electrode and an electrolyte, but it is not directly measurable and the electrolyte potential is rather undefined.
We do not use electrolyte potentials as reference potentials, as reference must be well defined. By mechanical analogy, it would be like defining the relative height wrt a level that is not clearly given. By other words, the conventional electrode potential is NOT the potential difference between the electrode and electrolyte, even if this difference does exist.
By convention, the electrode potential is considered the potential difference between the electrode and the standard hydrogen electrode(SHE) $E - E^{\circ}_{\ce{H2/H+/Pt}}$.
The cell potential is called rather the cell voltage or the cell electrode potential difference. The cell voltage is difference of electrode potentials, which are difference of electrode potentials wrt SHE.
$$U_\mathrm{cell} = E_\mathrm{cathode} - E_\mathrm{anode} \\ = ( E_\mathrm{cathode,absolute} - E^{\circ}_{\ce{H2/H+/Pt}}) - (E_\mathrm{anode,absolute} -  E^{\circ}_{\ce{H2/H+/Pt}})\\
= E_\mathrm{cathode,absolute}  - E_\mathrm{anode,absolute}$$
Note that cathode is the more negative one for electrolytic cells, but more positive one for galvanic cells.
The cell voltage and the potential electrode difference wrt SHE ( = the cell voltage if the other electrode were replaced by SHE ) does not depend on the electrolyte potential.


Reagent notation: meaning of dash(-) and line(|) or slash(\) in reagents

The '|' in $\ce{Pd|BaSO4}$ implies that the reaction is taking place in the presence of $\ce{Pd}$ as a metal catalyst. This metal provides surface area for reaction to occur.
Also '/' in $\ce{Cr2O3-Al2O3/Mn2O3}$ means that $\ce{Cr2O3}$ can be used in combination with either $\ce{Al2O3}$ or $\ce{Mn2O3}$.


Of course it will, provided you have a billion of it. (One molecule does not rotate much, after all.) Also, each should come without the rest of the body, or at least without the other hand, otherwise that would make a racemic mixture. (I'll leave the grisly details to the reader's imagination.) Also, each hand should form the same gesture, otherwise the destructive interference will ruin the effect.
Also, light of suitable wavelength would be microwave radiation, and a cursory search seems to indicate that the methods of polarizing it and measuring its polarization are known.
With all these conditions fulfilled, I don't see any difference from the situation with the molecules and ordinary light.


Equilibrium calculation gives negative molarity

The equation $2.48\times 10^{-3} = \frac{(0.1-x)(0.031-x)}{0.25+x}$ is correct, but it simplifies to a second degree polynomial, so it has two solutions, not one. You just picked the incorrect root (negative concentrations have no physical meaning). Can you find the right answer now?


How can compounds be called reduced if nothing is known about the reaction they formed in?

While we don’t know exactly how minerals on meteorites were formed, we can make very educated guesses on the conditions they were formed in. Most notably, outer space only contains very low concentrations of the compound most associated with oxidation on Earth: gaseous oxygen.

But this is not the main reason why the label ‘remarkably reduced’ is accurate here. We will immediately recognise the mineral in question as an ionic compound and given that it is made up of a metal and a nonmetal, we can equally immediately assign oxidation states as shown below.

$$\ce{\overset{+II}{Ca}\overset{-II}{S}}$$

There is no surprise about calcium which generally is found in its $\mathrm{+II}$ oxidation state. The surprise here is that sulphide is the anion which is a rare occurance. On Earth, most mineralic sulphur is oxidised and mainly to sulphate ($\mathrm{+VI}$). This is basically the extreme opposite and caused among others by the high atmospheric oxygen concentration.

Of course, the label ‘remarkably reduced’ can only be applied to some of the elements of the compound. Language-wise, we are looking at the oxidation state and merely stating that a state this low is uncommon.


Several carbonyl groups, which one does ethylene glycol protect?

Several factors influence the overall rate of a reaction under various conditions. Among the crucial factors are:

structural features of the carbonyl compound;
the role of protons or other Lewis acids in activating the carbonyl group toward nucleophilic attack;
the reactivity of the nucleophilic species and its influence on subsequent steps; and
the stability of the tetrahedral intermediate and the extent to which it proceeds to product rather than reverting to starting material. (Ref.1)

Limiting ourselves to current molecule, aldehydes  are more reactive than ketones. Electron-withdrawing substituents render the carbonyl group more electrophilic.

Image credit: Organic Chemistry, Structure and Function Seventh Ed by Peter Vollhardt, University of California at Berkeley, Neil Schore, University of California at Davis

Because aldehydes form acetals more readily than ketones, we can protect an aldehyde selectively in the presence of a ketone. The following example shows the reduction of a ketone in the presence of a more reactive aldehyde (Ref.2):

With reference to your question, C-2 is more reactive then C-1 carbonyl group for the above reasons.

References

Advanced Organic Chemistry, Fifth Edition, Part A: Structure and Mechanisms by Francis A. Carey and Richard J. Sundberg, University of Virginia
Organic Chemistry, Eighth Edition, by L. G. Wade Jr., Whitman College
Reactivity of Aldehydes & Ketones - Chem LibreTexts


When reactants are in their transition state, can they go back into their reactants form instead of progressing to form the products?

In the gas phase the reactants have kinetic energy as they move towards one another, so you would think that they would just cross the transition state, but molecules also have vibrational energy which means that it is possible to reach the transition state and then turn back. See the figure below which shows a simple LEPS calculation of the O+H2 reaction.

The solid black line shows the position and vibration of the H2 molecule as it approached the O atom and then 'reflects' at the transition state. It has just enough energy to cross, but the vibrational motion and position on the energy surface prevents it.  The axes show the separation of H and H2 (mid point) and of H and OH products.

In solution (or high pressure gas phase) the situation is a little different because collisions with may solvent inhibit or enhance the motion of the reactants towards one another. The species diffuse together and also diffuse along the transition state pathway so that the transition state can be crossed and recrossed many times before reaction occurs or the reactants are reformed. This approach is named after Kramers and is the most important one to understand when thinking about transition states in solution. (Strangely, it is rarely mentioned in Chemistry textbooks even though it was developed more than 70 years ago.)


The question: Why is the electrode potential for a cathode positive?

Since question didn't directed to what kind of cell, the answer is it is not always positive:

As M. Farooq clarified, cathode is where reduction is happening in a cell. Thus, in Galvanic cell it is positive as the reduction reaction, $\ce{Y+ + e- -> Y}$, happens. The electrons needed for this purpose came from the anode. However, in an electrolytic cell, electrons from an outer source (e.g., a battery), flow to the negative end where the reduction reaction, $\ce{M+ + e- -> M}$, happens. Thus, the cathode is negative here.

Edit: I agree with  Poutnik's comment below. Accordingly, it may be better to say cathodes are more negative than anodes in electrolytic cells and more positive than anodes in galvanic cells. As more positive does not mean positive and more negative does not mean negative. The convention of marking positive and negative cell contacts is good enough for every day life, but it says nothing about the sign of electrode potentials either with respect to free electron or SHE (e.g., relevant half-cell reduction reaction's potential may carry positive or negative sign, based on the corresponding reaction).


How would substitution reaction results differ between reacting halides with a tertiary bromoalkane vs reacting it with a primary bromoalkane?

There are (at least) four major factors to consider in predicting the results of these reactions: (a) the nature of the substrate, (b) the strength of the nucleophile, (c) the quality of the leaving group, and (d) solvent effects. Halides are able to function, to varying degrees, as both good leaving groups and strong nucleophiles.

In protic solvents, negative charges are stabilized by the $\ce{\delta^{+}}$ of the solvent hydrogens bonded to some electronegative atom (conventionally limited to $\ce{N, O, F}$). Since nucleophiles carry either a full or partial negative charge, protic solvents tend to reduce the reactivity of nucleophiles. Because smaller nucleophiles are more easily and thoroughly solvated than large and polarizable nucleophiles, the effect on them is more drastic. Consequently, in protic solvents, the relative nucleophilicity of the halides is generally given as $\ce{I- > Br- > Cl- > F-}$. In aprotic solvents, the reactivity is precisely the opposite. Hence, the proper selection of solvent is crucial. In your first example, there is no particular reason I can think of to suppose that the yield of 1-chlorooctane would inevitably be low. Indeed, in an aprotic solvent, the usual choice for SN2 reactions, $\ce{Cl-}$ can be expected to be a stronger nucleophile and poorer leaving group than $\ce{Br-}$. I would expect the yield to be good, especially with an excess of $\ce{KCl}$.

On the matter of the substrate, with a tertiary carbon, SN2 is all but completely impossible. In that situation, elimination reactions are typically favored if some base is present. The halides, however, are extremely weak bases (being the conjugate bases of strong acids), so if a non-basic solvent is selected, substitution by an SN1 mechanism seems most probable. Selection of a relatively non-basic, non-nucleophilic protic solvent would be ideal. (Off the top of my head, DMSO comes to mind, which is only weakly nucleophilic, especially by comparison to the halides.) You're right that the strength of the nucleophile is not the deciding factor in an SN1, since the rate-determining step is the loss of the leaving group and the resultant carbocation formation. In a protic solvent, $\ce{I-}$ should theoretically be more effective as a nucleophile than $\ce{Cl-}$. However, I would still expect the latter reaction to work, particularly if the concentration of $\ce{Cl-}$ is high. I'm fairly confident that the difference in reactivity between, e.g., $\ce{Cl-}$ and $\ce{Br-}$ would be largely negated if the former is available in vastly larger concentration in the solution.

Edit: On a second reading, I just noticed that you labeled the halide solutions as aqueous. I don't know if that's deliberate, but I don't think that would make sense. For one thing, the reagents (namely, the haloalkanes) would be largely insoluble in water. Second, because water is an amphoteric species, it could be potentially reactive.


Predict Entropy based on the Balanced Equation

Aqueous salts will always have less order than solid salts. There is much less order when ions are floating around in solution than when they are in their rigid lattice structures. This is what I think you are failing to consider. The reaction is not so much:

$$\ce{2Na_{(s)} + 2H2O_{(l)} -> 2NaOH_{(aq)} + H2_{(g)}}$$

But really:

$$\ce{2Na_{(s)} + 2H2O_{(l)} -> 2Na^{+}_{(aq)} +2OH^{-}_{(aq)} + H2_{(g)}}$$

What results from the reaction of 2 atoms of a very structured metal and 2 molecules of liquid water, is 4 aqueous ions that are free to move within the solution, and one molecule of gas. Not only are there now more species present, but they are also in a relatively more free states than before.


Do ions in a salt bridge come from the half cells in electrochemistry?

You must think electrically, without thinking first to the chemistry. In the anodic solution first, the reaction produces positive ions $\ce{Mg^2+}$ or $\ce{Zn^2+}$. It does not really matter here. The problem is that this cannot stay like this. Either the positive ions have to move where negative ions are available. Or new negative ions must arrive in the solution. Or both.

The simplest way for solving this difficulty is to mix the two solutions. Because a similar difficulty appears on the cathodic region : positive ions ($\ce{Ag+}$ or $\ce{Cu^2+}$) are here discharged, leaving an excess of positive ions around the cathode. By mixing the solutions the excess negative ions produced around the cathode move to the anode where they are needed. But mixing the solutions is the least thing to do, because now $\ce{Ag+}$ or $\ce{Cu^2+}$ ions can touch the $\ce{Zn}$ or $\ce{Mg}$ anode. And the chemical reaction will go on in the contact between positive ion and metal. The same electrons of the half-reaction are produced but they will not go through the long external circuit form the anode to the cathode. They prefer the shortest circuit direct from metal to ion in solution, and metallic $\ce{Ag}$ will  soon be formed on the $\ce{Mg}$ anode (or $\ce{Cu}$ on the $\ce{Zn}$ plate). No current will be emitted in the outer circuit. It is the end of the cell. So it is absolutely necessary to prevent the $\ce{Cu^2+}$ or $\ce{Ag+}$ ions to touch the anode, so the two compartments must be separated.

That is why we introduce the salt bridge. It must contain ions that do not react with any ion contained in the two compartments. Their nature is not important. Usually it is $\ce{K+}$ and $\ce{NO3-}$.

So let's go back to what's happening electrically in solution when the cell is working. There is an excess of positive ions around the anode, and an excess of positive ions around the cathode. To compensate this situation, the negative ions of the bridge are attracted by the positive ions in excess at the anode, and the positive ions of the bridge are attracted by the cathode. Both quit the bridge and get mixed in the solutions. This allow the cell to work.

After some time, the bridge is loosing all its stuff, so some positive ions of the anode ($\ce{Zn^2+}$or $\ce{Mg^2+}$) may enter the bridge and replace the initial positive ions $\ce{K+}$that have moved to the cathode. And symmetrically some negative ions ($\ce{NO3-}$ or $\ce{SO4^2-}$) from the anode will enter the bridge to replace the initial ions $\ce{NO3-}$.


how: an enzyme is able to effectively make a reaction intramolecular by binding the reactants in the correct conformation in close proximity within the active site this can speed up the rate of a reaction dramatically, intramolecular rates avoid diffusion control and have prearranged conformations which is far more efficient than typical intermolecular reactions,
why: enzymes are outrageous, evolution has pushed them, through trial and error, towards optimal configurations that, almost magically, provide ideal conditions for a reaction to occur


Salt water is an excellent electrolyte, and will accelerate galvanic corrosion reaction between two different metals especially in oxygen rich environments (including dissolved oxygen in water). This may create metal oxide deposits on circuits. The longer the exposure time the more corrosion can occur.

If your device is on, that is, there are circuits with electricity flowing, clearly adding a salt water could also short circuit your device as well.

As to whether any of the above results in irreversible damage is hard to say with certainty as some devices have been dried out and appear to function at least following exposure to common tap water.


Determine the amount of substance of calcium chloride used

Yes, you did it correctly, though there is a small issue with significant figures if that is a concern.

$$\mathrm{Molarity}=\frac{\mathrm{moles}}{\mathrm{liters}}$$

So for your values:

\begin{align}
0.20 &= \frac{n}{0.096}\\
n &= \pu{0.20 M} \times \pu{0.096 L} \\&= \pu{0.0192 mol}\, \ce{CaCl2}
\end{align}

However, both your molarity and volume have only two significant figures and you're multiplying those values, so the result should only be reported to two significant figures as well: $n = \pu{0.019 mol}\,\ce{CaCl2}$.


Short answer is yes, HOMO and LUMO refer to canonical orbitals, since only those have at least some physical meaning.

Two more remarks about that:

Koopmans' Theorem does only work if you add or remove one electron.

If you change two or more electrons of your system (e.g. exciting one electron is first removing one, then adding one in another orbital) you get additional terms in energy difference between both states. Those terms are coming from the electron-electron interaction of the two changed orbitals/electrons.

In case of changing only one orbital, those additional terms do not appear since there is no electron-electron self-interaction.

Furthermore, the approximation of Koopmans' Theorem, that orbitals do not change (do not relax, remain stationary) on removing/adding electrons, gets worse the more electrons change.

Orbitals itself are only an approximation. Factorizing the total electronic wave function into orbitals is mathematically only possible for non-interaction electrons, or like in Hartree-Fock if you assume some mean-field for the electron-electron interaction.
The resulting canonical orbitals do not always match the simplified picture based on our chemical intuition.


The order of aromaticity of benzene, thiophene, pyrrole, furan, and pyridine

The main criteria of aromaticity is resonant or delocalization energy.
It is difference between energy contained in bonds if they were localized like in Kekule's formula for benzene and actual energy.
For benzene if is about 150 kJ/mol.
The more this energy the more stable molecule is.

The approaches for calculating this energy may differ.
As for β-values mentioned in paper by @orthocresol it looks like it is Dewar's resonance energy per one π-electron calculated based on modified Hückel's MO method as mentioned in GSE article.

There is a table in this article with the same β-values:

In English it is

Benzene                            0.065     Pentalene                     -0.018
Naphthalene                        0.055     Azulene                        0.023
Anthracene                         0.047     Fulvene                       -0.002
Tetracene                          0.042     Pyridine                       0.058
Cyclooctadecanonaene([18]annulene) 0.012     Pyrrole                        0.039
Cyclobutadiene                    -0.268     Furan                          0.007
Azete (Azacyclobutadiene)         -0.160     Thiophene                      0.032


Mechanism for the reaction between trisilylamine and hydrogen chloride

I am not sure what back bonding you are referring to. It may be related to the outdated concept of silicon using its vacant, high-energy d orbitals in any meaningful way which is not actually the case.

There is a second effect which consists of overlap of nitrogen’s p orbitals with the σ* orbitals of the $\ce{Si-H}$ bonds (which are silicon-centred due to the low electronegativity of silicon). This indeed reduces nitrogen’s basicity slightly compared to other nitrogen compounds but the effect is not that important overall. If you check page 2 of the Evans $\mathrm pK_\mathrm a$ table, you can see that the difference between $\ce{(Me3Si)2NH}$ and $\ce{(iPr)2NH}$ is about 10 logarithmic units which is noticeable (i.e. the HMDS salts are milder bases than LDA) but it does nothing to counteract their overall behaviour. (I am aware that I am arguing with the wrong $\mathrm pK_\mathrm a$ values but sadly the table does not contain data for $\ce{(Me3Si)2NH2+}$ so this is the closest equivalent.)

It is also important to point out that you are using a very strong acid $(\mathrm pK_\mathrm a \approx 8)$ which will protonate practically any nitrogen lone pair that it can come across.


It's like an asteroid falling toward Earth. Potential energy converts to kinetic, until it crashes, then the energy is found in whatever aftermath there is.  The total energy is still zero. The "falling" from a great distance is a change from high potential energy to low. By convention, that higher level is assigned zero, so the lower energy must therefore be negative.

For an electron and a bare nucleus far apart, and not counting any nuclear or non-chemical phenomena, we assign zero as the total energy.  When the electron is allowed to fall, it falls faster and faster, potential energy converting to kinetic, until "crashes" - it enters the 1s orbital, n=1, releasing a photon of 13.6eV. (In constrast, an asteroid makes a big crater, stirs up debris, and creates tremendous heat.)  The photon carries 13.6eV, and so the newly formed hydrogen atom is -13.6.  This is  called binding energy, and is negative by the convention physicists and chemists have followed since long ago.

BTW, that formula is only for any atom with exactly one electron, and Z not too big.  With two electrons it no longer applies, and for the heavier elements such as uranium, there are relativistic effects and finite nucleus size effects requiring more sophisticated math and at the very least, some sort of correction terms to that formula.


Why absorption/transmission spectroscopic data is normalized?

There is confusion here about absorption at one wavelength and the absorption spectrum. At each wavelength the Beer-Lamber law applies $$I_{trans_{\lambda}}=I_{0_{\lambda}}\exp(-\epsilon_{\lambda} c l)$$ for a compound with extinction coefficient $\epsilon$ at wavelength $\lambda$ and concentration c and path length l.

When an absorption spectrum is produced it is normal to show it as optical density vs wavelength, i.e. $OD_{\lambda}=\log(I/I_0)_{\lambda}=-\epsilon_{\lambda} c l $ and in this case the ratio of peaks for the same compound is constant no matter what the concentration. (Naturally assuming that it does not absorb so much that the instrument fails).


Why is the entropy of deuterium greater than that of hydrogen gas

The entropy can be split into three parts arising from translational, rotational and vibrational energies and using statistical mechanics each of there can be calculated from measure properties. The gases $\ce{H2}$ and $\ce{D2}$ differ in each of these three energies due to their differing mass and nuclear spin. The experimental values are hydrogen $130.7$ and deuterium $144.9 ~\pu{J mol^{-1}K^{-1}}$

The translational entropy is given by the Sakur-Tetrode equation,
$$S_{trans} = R\left(\frac{5}{2}+\ln \left(\frac{V}{N} \right)+ \frac{3}{2}\ln\left( \frac{2\pi mkT}{h^2} \right )\right)$$
where m is the mass, V volume, N Avogadro’s number, h the Planck constant and k the Boltzmann constant. Assuming that the gases behave ideally the difference in entropy is small as it depends only on the mass difference all other terms being the same, so $S_{trans}$ is greater for $\ce{D2}$ than $\ce{H2}$ by $(3R/2)\ln(2) \approx 8.64~ \pu{ J mol^{-1}K^{-1}}$.

The vibrational entropy can be ignored at low temperature, for example at room temperature $kT \approx 210~ \pu{cm^{-1}}$ whereas the vibrational quantum are $4395$ and $3118~ \pu{cm^{-1}}$ for $\ce{H2}$ and $\ce{D2}$ respectively.

The contribution from the rotational motion depends both on the mass, as the moment of inertia determines the rotational constant, and on the nuclear spin which via the symmetry properties of Fermions and Bosons affects the weighting of the rotational energy levels.

To calculate the entropy the partition function $Z$ is required. Once this is obtained the entropy is calculated as
$$ S = k\ln(Z)+\frac{U}{T}$$
where the internal energy $U$ is
$$U= kT^2\left ( \frac{\partial \ln(Z)}{\partial T} \right)_V$$

The partition function for odd mass number, assuming a rigid rotor model, is (see this answer for details Why do spin isomers of hydrogen (ortho and para hydrogen) change their nuclear spin with temperature variance? )
$$Z_{odd}= \frac{g_n(g_n-1)}{2}\sum _{J=0,2,..} (2J+1)\exp(-B_eJ(J+1)/(k_BT)) +\\ \frac{g_n(g_n+1)}{2}\sum_{J=1,3,..} (2J+1)\exp(-B_eJ(J+1)/(k_BT)) $$

where $g_n = 2s_n+1$ where $s_n$ is the nuclear spin, $1/2$ for hydrogen and $1$ for deuterium. The rotational quantum numbers are J and $B_e$ the rotational constant assumed to be the same for all J but which differs between hydrogen ($60.85 \pu{cm^{-1}}$) and deuterium ($30.44 \pu{cm^{-1}}$) by a factor of two. (These rotational constants are not typical, values for most molecules which are heavier and have longer bonds are far smaller, for example, $\ce{F2}, B_e =1 \pu{cm^{-1}}$)

The partition function for even mass number
$$Z_{even}= \frac{g_n(g_n+1)}{2}\sum _{J=0,2,..} (2J+1)\exp(-B_eJ(J+1)/(k_BT)) +\\ \frac{g_n(g_n-1)}{2}\sum_{J=1,3,..} (2J+1)\exp(-B_eJ(J+1)/(k_BT)) $$
where the only difference is that due to the weighting (those terms in $g_n$ ) in from of the summations which run over even and odd values of J. Hydrogen has weightings $1$ and $3$, deuterium $6$  and $2$ respectively in the equations above.
Thus Z for hydrogen with $g_n=2$ is
$$Z=1\cdot\Sigma (even~J) + 3\cdot\Sigma(odd~J)$$
and for deuterium, $g_n=3$
$$Z=6\cdot\Sigma (even~J) + 2\cdot\Sigma(odd~J)$$

At low temperatures the argument in the exponential is large meaning that the exponential is small. e.g. $exp(-large ~number) \rightarrow 0$  for all J terms except $J=0$ where the exponential is $1$. Thus in hydrogen the limiting partition function at low temperatures is $Z_{T\rightarrow 0}= 1 + 3(small~number) \approx 1$ and the limiting entropy is $S_{T\rightarrow 0 } R\ln(1) \rightarrow 0$; (and $U(0)=0$)

In deuterium the limit is $Z_{T\rightarrow 0 } = 6 + 2(small~ number) \rightarrow 6 $ and thus the limiting entropy is $S_{T\rightarrow 0 } R\ln(6) \rightarrow 14.9~ \pu{J mol^{-1}K^{-1}}$.
The change of rotational entropy with temperature is shown in the figure.


Acidic strength of organic acids. Substituted PhCOOH acids

The ortho substituted benzoic acid is most acidic, followed by the meta substituted and finally para substituted benzoic acid.

The methyl group, when at ortho position, causes steric inhibition of resonance, which essentially means that the $\ce{-COOH}$ group goes out of the plane of the benzene ring (due to steric hindrance from the $\ce{-CH3}$ group). Due to this, the benzene's $\pi$ electrons can no longer be transferred to the central carbon of the $\ce{-COOH}$ group by resonance, hence increasing the positive charge on it. This makes the hydrogen of the $\ce{-COOH}$ group more acidic.

When the methyl group is on meta position, it operates by +I, thus reducing its acidity. When on para position, it operates by both +I and +R (actually +H, which means hyperconjugation effect), hence decreasing the acidic strength even more.


Why does fusing benzene rings not produce polycyclic alkynes?

The ideal bond angles for alkynes are $180^{\circ}$, which is necessary for optimal overlap of atomic orbitals combining to produce mutually perpendicular $\pi$-type molecular orbitals. Any significant deviation from this is normally inherently destabilizing. Rings, by the very nature of their geometry, often incur additional strain due to their conformational inflexibility and possible demands for non-ideal bond angles (relative to acyclic molecules, that is). Introducing $\pi$-bonds into rings can potentially compound this problem, since they force at least local planarity and shorter bond lengths. These factors all contribute to the immense instability of cyclic alkynes in general.

When you mention stable cycloalkynes, keep in mind that stability is a relative term in this context. While those with a sufficient number of carbons can be produced and isolated under specific experimental conditions, they are not thermodynamically stable relative to their open-chain equivalents, or to most of their various possible structural isomers. In some cases, they can, however, be made kinetically stable by the addition of flanking $\alpha$-substituents which inhibit reaction by their steric bulk1.

Syntheses of cycloalkynes below cyclooctyne are often elaborate, and frequently proceed through extremely reactive intermediates that decompose by specific reaction pathways so as to carefully avoid other products. Benzene (and the various polycyclic aromatic hydrocarbons, for that matter), on the other hand, is an extremely stable molecule which participates in quite a limited collection of reactions. Whatever particular reaction you're suggesting would almost certainly be hugely uphill in energy (i.e., yields a significantly less stable product), and hence very thermodynamically disfavored; it would also certainly have immense kinetic barriers as well (since any such reaction pathway implies at least temporary loss of aromaticity in the transition state[s] and intermediate[s]).

Relevant reading might include Dehydrobenzene and Cycloalkynes by Reinhard W. Hoffmann. There is extensive discussion of the reactivity of cycloalkynes, and numerous examples of spontaneous dimerizations, isomerizations, etc. Evidently, isomerizations to cycloallenes and cyclobutadiene dimers are common, which is a profound testament to the instability of cycloalkynes.

1. For example, 3,3,7,7-tetramethylcycloheptyne, a good description of which (and synthetic scheme for) is given in Strained Hydrocarbons by Dodziuk.


How long the block starting with element 121 will be?

As you noted, this is a very appropriate question in light of the IUPAC announcement that we have just finished filling Period 7!

The names of the subshells s, p, d, and f are named after the old spectroscopic terms sharp, principal, diffuse, and fundamental. We ran out of fancy names after that, so the subsequent subshells are named in alphabetical order - g, h, and so on - which means that after the 8s block is filled, we would theoretically have a 5g block.

The orbitals in the g subshell would be labelled with the quantum number $l = 4$, so $m_l$ would take integer values between $-4$ and $4$ (inclusive) giving a total of nine g-orbitals. Each g-orbital could hold two electrons with opposite spins, so the g-block would have $18$ electrons.

However, it is worth noting that the electronic configurations may or may not obey the aufbau principle fully. Whether the 5g orbitals will actually be filled or not will certainly not be easy to determine, considering how short the half-lives of those elements are likely to be.

Wikipedia has an article which talks about it.


Is there a relationship between boiling point and van der Waals constants

There will be a relationship, but as the comments on the question indicate, the relationship will not be simple. There are two complications:

The van der Waals equation is an empirical model built to fit experimental pressure/volume/temperature data for real gasses.
The boiling point of a liquid is not a single number but a line on the pressure/temperature phase diagram.

Supposed the van der Waals equation can predict phase behavior (source: 1) However, I made some graphs. The van der Waals constant data are from Wikipedia and the boiling point data looked up from the Wikipedia articles.

  Constant a

The blue line is a cubic polynomial fit, while the grey shaded area is the error. With the exception of the metal outlier (mercury), this fit seems pretty good.

  Constant b

The blue line is a logarithmic fit, while the grey shaded area is the error. Again, with the exception of the metal outlier (mercury), this fit seems pretty good.

  Finally, I decided to plot the van der Waals constants against each other.

The van der Waals constants are fairly co-linear, so any clear relationship between one and boiling points implies a clear relationship between the other and boiling points.

Source:

Terrell L. Hill, 2012 [1960], "An Introduction to Statistical Thermodynamics" [Dover Books on Physics], Chicago, IL, USA:R.R. Donnelly (Courier/Dover), ISBN 0486130908, referenced on the Wikipedia article, but I do not have access.


Why are vinylic and arylic carbocations highly unstable?

Background

S-orbitals are lower in energy than p-orbitals, therefore the more s-character in an orbital, the lower its energy and the more stable any electrons occupying the orbital.  In other words, electrons will be more stable (lower in energy) in an $\ce{sp}$ orbital than an $\ce{sp^2}$ orbital, and electrons in an $\ce{sp^2}$ orbital will be lower in energy than those in an $\ce{sp^3}$ orbital.

This is the basis of the inductive effect.  For example, it explains why the electrons in an $\ce{sp^3}$ methyl group will be drawn towards a connected $\ce{sp^2}$ carbon in an aromatic ring - the methyl group is inductively electron donating to the aromatic ring as its $\ce{sp^3}$ electrons drift towards the lower energy $\ce{sp^2}$ aromatic carbon.

What we're really saying is that an $\ce{sp}$ hybridized carbon atom is more electronegative (it's lower in energy and the electrons prefer to move towards it) than an $\ce{sp^2}$ hybridized carbon atom, which in turn is more electronegative than an $\ce{sp^3}$ hybridized carbon atom.

Answer

Look at the figure below, notice that in the alkyl carbocation on the left the cationic center is attached to an $\ce{sp^3}$ carbon, whereas in the vinylic cation in the middle, the cationic center is attached to a more electronegative $\ce{sp^2}$ carbon.  For a positive center to be attached to a more electronegative group is destabilizing.  Hence the vinylic cation is less stable than a typical alkyl cation.

Things are even worse with the aryl carbocation on the right.  Here the positive carbon is attached to 2 $\ce{sp^2}$ carbons. Destabilizing the aryl cation even further is its geometry.  A vinyl cation prefers to be linear, but due to geometrical constraints imposed by the aromatic ring the aryl cation must be bent and the empty orbital is forced to be $\ce{sp^2}$ rather than $\ce{p}$.  These three factors combine to make the aryl carbocation even higher in energy than the vinyl cation.


Why is a very small peak with larger m/z not considered to be the molecular ion?

Without knowing more details, it is hard to guess, but at this $m/z$, it seems likely that the peak is the result of one $^{12}\ce{C}$ being substituted by one $^{13}\ce{C}$. It is more useful to assume a uniform mass of 12 for carbon when analyzing such a spectrum.

Note that with more carbons in a larger molecule, it becomes more likely that at least one of them is a $^{13}\ce{C}$. Thus, the spectra become more complicated that way.

This whole analysis goes out the window if this is not an organic molecule, though similar patterns may arise for other molecules - it depends on the isotopes and their distribution.


Ca cannot be detected in General Metal Mode. You have to change to "Mining Mode"
In this mode, you can measure Ca properly.


What determines whether an aldol reaction will occur again or eliminate?

Partially yes. The big problem with aldol reactions in general is that there are a lot of species that can react together to form stable compounds resulting in low yield. There are numerous ways to counter this:

Specific enolate equivalants, like lithium enolates, silyl enol ethers, aza enolates, enamines, etc. These are among the most important.
Lack of proton in $\alpha$-position.
Strong thermodynamic favoring of certain products in combination with reversibility of the reaction.
etc.

As for your example the $\beta$-hydroxy ketone is more sterically hindered than acetone and is therefore not very likely to be the major contributor (although polymerization can still happen of course).

In general you cannot predict whether or not elimination of $\ce{H2O}$ occurs after the aldol reaction, but the likeliness can be increased by forcing certain reaction conditions.


How is the dissolution of acetic acid that makes its aqueous solution a poor electrolyte?

There are two questions being asked here:

Why is aqueous acetic acid a weak electrolyte?

When acetic acid is dissolved in water there is an equilibrium reaction:
$$\ce{CH3COOH + H2O <=> CH3COO- + H3O+}$$

Since acetic acid is a weak acid, the equilibrium position lies well to the left, with only a small fraction of the acetic acid molecules reacting to form ethanoate and hydronium ions. The presence of this small amount of ions results in aqueous acetic acid being a weak electrolyte.

Why is acetic acid highly soluble in water?

Since the vast majority of acetic acid molecules do not dissociate when a sample is dissolved in water, the solubility has to do with the interactions between acetic acid molecules and water molecules. Water has a network of hydrogen bonds between molecules in its liquid phase and so when a substance dissolves in water this bonding is disrupted. However, acetic acid is able to form many new hydrogen bonds to water molecules and so this results in a highly favourable interaction, leading to the high solubility of acetic acid in water.


Fractional Bond Order - How to Visualize?

Maybe some examples will help.  In the picture below, I've drawn some resonance structures for butadiene and acrolein.  What do they tell us?  They suggest that the central "single" bond should have some double bond character, it's more than a single bond, but less than a double bond.  By the same token, the "double"

bonds at the ends of the molecules should both have some single bond character.  These are examples of fractional bond order.  The real world consequences of fractional bond order is a higher barrier to rotation about the central "single" bond in these two molecules than we might have expected, and a lower barrier to rotation in the pi bonds than we might have expected.


What makes scorpions glow under UV light?

As mentioned in this comment, which links to this paper, and this comment, which links to this Wired article, there are at least two molecules:

$\beta$-carboline:

4-methyl-7-hydroxycoumarin:

$\tiny\text{Yes I poached those comments, they could have been answers.}$


Would 2 C₂ be 2 moles of diatomic carbon, or 4 moles of carbon?

You can have 1 mole of $\ce{C_2}$ which contains 2 mole of $\ce{C}$ just as you can have for example 1 mole of $\ce{C_2H_5OH}$ which contains 2 mole of $\ce{C}$, 6 mole of $\ce{H}$ and 1 mole of $\ce{O}$.

The point is that you always have to mention whether you are indicating moles of molecule or moles of atoms. That is why a statement like: "I have 1 mole of oxygen" is ambiguous, because it could be referring to oxygen gas (i.e. $\ce{O_2}$) or to atomic oxygen (i.e. $\ce{O}$).


Reactions of diborane, silane, methane, aluminium, boron, sodium, phosphorus trichloride, white phosphorus and iron with water at RT

$$\ce{B2H6 + 6H2O -> 2H3BO3 + 6H2 \\
SiH4 + 4H2O -> Si(OH)4 + 4H2 \\
2Na + 2H2O -> 2NaOH + H2 }$$

The rest do not react with water under ordinary conditions. $\ce{Al, Fe, B}$ react only with steam or when they are red hot. White $\ce{P}$ undergoes easy hydrolysis when coordinated with ruthenium, as illustrated by Vaira et al. [1]. $\ce{CH4}$ does not undergo hydrolysis with water and $\ce{PCl3}$ does not give hydrogen on hydrolysis.

Reference

Vaira, M. D.; Frediani, P.; Seniori Costantini, S.; Peruzzini, M.; Stoppioni, P. Easy Hydrolysis of White Phosphorus Coordinated to Ruthenium. Dalton Transactions 2005, 0 (13), 2234–2236. DOI: 10.1039/B504795A.


If the pH is zero, what would be the normality of a sulphuric acid solution?

$\ce{H2SO4}$ has 2 $\mathrm pK_\mathrm a$s, one that is strong whereby $\ce{H2SO4}$ ionizes to $\ce{HSO4-}$ and $\ce{H+}$ and another that is weak whereby $\ce{HSO4-}$ ionizies to $\ce{SO4^{2-}}$ and $\ce{H+}$

At $\mathrm p[\ce{H+}] = 0, [\ce{H+}] = \pu{1M}$

At $\mathrm {pH} = 0$, sulfuric acid is substantially $\ce{HSO4-}$ and $\ce{H+}$, with little $\ce{H2SO4}$ or  $\ce{SO4^{2-}}$.

Therefore at $\mathrm p[\ce{H+}] = 0$, sulfuric acid concentration is $\pu{1M}$, because each mole of sulfuric acid has released one mole of $\ce{H+}$.

Normality of an acid is the concentration of the acid times the number of acidic protons per molecule.  Sulfuric acid has two acidic protons per molecule.

Normality $= \pu{1M} \times 2 = \pu{2N}$


Why is ionization energy measured in gaseous phase?

You want to determine the energy for a single molecule and avoid stabilization effects through intermolecular interactions.


Is there any method to synthesize salicylaldehyde from salicylic acid?

Salicylic acid can be converted to salicylaldehyde using a triangular or square wave alternating voltage. The process of this conversion is explained in this paper$\ce{^{1}}$but it is behind paywall so here's the abstract:

  The effect of superimposing sinusoidal, triangular, and square‐wave
  alternating voltage (aV) on the potentiostatic reduction of salicylic
  acid to salicylaldehyde has been studied. Polarization curves with aV
  modulation were measured with an amalgamated copper rotating disk
  electrode. Enhanced dc current densities were observed with increasing
  magnitude of superimposed aV. However, aV frequency and type did not
  have significant effect on the resulting dc current densities. Batch
  cell electrolysis experiments were conducted with a H‐type cell. The
  conversion of salicylic acid and yield of salicylaldehyde were
  measured as functions of electrolysis time, aV magnitude, type, and
  frequency. The superimposed aV significantly enhanced the reaction
  rate, conversion of salicylic acid, and the yield of salicylaldehyde.
  The energy consumption of ac electrolysis was higher than that of dc
  electrolysis.

doi: 10.1149/1.2100342 J. Electrochem. Soc. 1987 volume 134,issue 12, 3074-3077


How many electric vehicles can our current supply of lithium power?

You assume there is X lithium and all of it cost Y. And then there would be no lithium to mine. Thats not quite right. As you are ready to pay more, harder to get sources become profitable and become part of available reserves. Sea has almost unlimited amount of lithium, but it is about x5 as expensive to mine.
The total lithium content of seawater is very large and is estimated as 230 billion tonnes
(wiki)


Reaction of substituted furan with methyl vinyl ketone with subsequent rearomatisation

Your Diels–Alder adduct is pretty much correct. Without any further evidence we cannot definitely say whether the exo or the endo adduct is formed (you have drawn the exo adduct, which generally arises via thermodynamic control). However, the stereochemistry of this adduct does not affect the final product, so it does not matter which one is formed.

By the way, you also should think about the regioselectivity of the Diels-Alder reaction, i.e. why do the $\ce{NH2}$ and $\ce{COMe}$ end up ortho to each other and not meta. You need to have a better reason than simply "that's the way it was drawn in the product". Generally, this is rationalised by looking for the largest coefficients in the HOMO of the diene and the LUMO of the dienophile. The HOMO of 2-aminofuran is likely to have a large coefficient on C–5 (opposite the amino group), whereas the LUMO of methyl vinyl ketone has a large coefficient β to the carbonyl group.

The subsequent steps are not anything fancy. Generally, $\mathrm{sp^3}$ carbons that are bonded to two heteroatoms tend to collapse (e.g. tetrahedral intermediates in nucleophilic acyl substitution; hydrates; etc.) In this case the nitrogen kicks out the oxygen. Note that the other option of oxygen kicking out nitrogen is not possible, as that would lead to a bridgehead double bond.

Ordinarily you need some sort of acid or base catalyst, but presumably here, the thermodynamic driving force for the collapse is large (relief of ring strain + product is aromatic), and the high temperature provides the energy required to overcome the activation barrier. The last elimination step probably occurs via an E1 mechanism, since there's no good base present.


Calculating time needed to coat an object using the process of electrolysis

The trick you're missing here is that you don't account for how many electrons are needed to reduce $\ce{CrO3}$ to $\ce{Cr}$.
Cr has an oxidation state of 6 in $\ce{CrO3}$, so it takes 6 electrons to reduce every chromium atom.
The full form of Faraday's law is $$\frac{m}{M} = \frac{q}{zF}$$  where $z$ is the number of electrons transferred.
Therefore, you need 6 times as much charge as you initially calculated, which works out to 8550 seconds or 2:22:30.


Inorganic chemistry handbook with reaction lists

An Intro in Russian edition of Reactivity of Inorganic Substances by Lidin et al. claims its uniqueness:

Настоящее пособие носит информационно-справочный характер, не имеет аналогов в отечественной и зарубежной химической литературе, охватывает все разделы неорганической химии, изучаемые в химических вузах.
(English) This reference handbook has no analogues among domestic and foreign chemical literature, covers all inorganic chemistry sections studied in university chemistry course.

However, Turova's Inorganic Chemistry in Tables comes pretty close.
It doesn't only list reactions per element groups, but also arranges them in visual blocks; available both in Russian and English [1, 2]:

References

Турова, Н. Я. Неорганическая Химия в Таблицах; Высший химический колледж РАН: Москва, 1997.
Turova, N. Inorganic Chemistry in Tables; Springer: Heidelberg ; New York, 2011. ISBN 978-3-642-20486-9.


What is the state of stationery phase in paper chromatography?

It's a bit of both, actually, and also ion-exchange chromatography.
The cellulose and other components of paper do exhibit adsorption and ion exchange properties with some analytes, but as paper chromatography is usually performed, partitioning between the flowing solvent and trapped water is the major mode of separation.
Paper chromatography typically uses water-saturated solvents and enclosed chambers where water vapour can keep water trapped in the paper (the paper is equilibrated with this vapour before development).
Since cellulose is very hydrophilic, immiscible or poorly miscible solvents are usually used, which keeps the water stuck to the cellulose as it has much higher affinity for water than the solvent.

This is an early paper showing the apparatus used for paper chromatography of amino acids and this book details the theory involved starting on page 4.


How to determine concentration of an unknown solution using Beer's Law?

This can be an issue in many calibration experiments, and falls into the broader and extremely important field of "design of experiments" within the field of chemometrics. Clearly, your range of calibration data is not quite sufficient for your application, but you have several options available to you, each with pros and cons, some of which have been mentioned in the comments

Go back to the calibration curve, and take some standard measurements at higher/lower concentrations than you did. Pros: this will be the most accurate method, and will help you to avoid this problem with future measurements. Cons: takes time and requires more measurements.
As commented, extrapolate from your model. Pros: extremely quick. Cons: likely to be inaccurate unless you are 100% sure Beer's Law holds for your sample at all concentrations. There is also no way of estimating the errors.
As commented, dilute your sample with a known quantity of solvent and remeasure. Then multiply your answer by the dilution factor. Pros: Fairly accurate, takes less time than performing more measurements for your model. Cons: Depending upon your lab skills and precision of your equipment, errors could start to propagate and become larger.

Good luck!


Why, when chemical equilibrium is defined as when two opposing reactions are equal in rate, can the products or reactants be favoured?

A chemical reaction can be reversible, and a reversible reaction can be at equilibrium, but they are not the same thing.

Equilibrium is the state where the forward and reverse reactions are the same rate.  This is clearly not true at the start of a reaction because there are no products yet.  As the reaction progresses the forward reaction rate decreases and the reverse reaction rate increases until they are equal.  At that point the concentrations of reactants and products no longer change.

The position of the equilibrium can be far to the left, so that hardly any product is formed before it reaches equilibrium.  This would be because the forward reaction is much slower than the reverse reaction.  If the position of the equilibrium is far to the right, almost all of the reactants have been used up because the forward reaction is much faster than the reverse reaction.

If you do something to change the conditions, like increase the temperature or add more reactants, you break the equilibrium and the reactions are no longer at the same rate.  As above the reaction will progress until the rates are the same again.  Depending on the change, this new equilibrium position may be more to the left or let or more to the right.

It is tempting to think that if you increase temperature, the rate of both reactions will increase the same.  This is not the case.  Although both reactions will increase in rate, because of the usual collision theory, but the same temperature change can have a much bigger effect on the rate of one reaction than the other.


As per the rules of inorganic chemistry nomenclature the element of the compound which is more electronegative is to be written at the end of molecular formula and its name is to be written after all the elements which have lower electronegativity are written hence the given compound is pronounced as silicon-carbide and not carbon-silicide

a more detailed view about electronegativity of elements can be found in this picture


Which element is being reduced in this reaction?

Your reasoning is not entirely right. In those compounds, O is always (-II), K and Na always (+I). However, it's not true for S, which can have different oxidation states. So does Mn.

So, in KMnO4, Mn must be (+VII). In MnO2, it is (+IV). Thus, Mn is reduced.

For sulfur: in NaSO3, it must be be (+V). In NaSO4, it must be (+VII). Thus, sulfur is oxidized.

However (big however, in fact): the sulfate anion is not SO4–, but SO42–. Similarly, sulfite is SO32–, not SO3–. So, your equation should be instead:

$$\ce{2KMnO_4 +3 Na_2SO_3 +H_2O -> 2MnO_2 +3Na_2SO_4 +2KOH}$$

which means that sulfur is oxidized from (+IV) to (+VI).


Why does the position of equilibrium depend only on the relative Gibbs energies of the products and the reactants?

The short answer is that the approximate entropy of mixing is a simple function of the reaction quotient Q, so it is implicitly considered when calculating the equilibrium constant from $\Delta_\mathrm{r} G$.

  Surely one must take into account the entropy of mixing of the products and the reactants, which wouldn't be dependent on the Gibbs energies, to work out the lowest point (energy wise) and find the position of equilibrium.

Yes, the entropy of mixing is important in describing an equilibrium and figuring out the equilibrium constant. Without the entropy of mixing, all reactions would go to completion, see Why does equilibrium exist? and Why can't a reaction go to completion?.

The Gibbs energy of reaction has an enthalpy and an entropy component:

$$\Delta_\mathrm{r} G = \Delta_\mathrm{r} H - T \Delta_\mathrm{r} S$$

The enthalpy does not depend on concentrations (or reaction quotient Q), but the entropy does:

$$\Delta_\mathrm{r} S = \Delta_\mathrm{r} S^\circ - R \ln Q$$

The term $R \ln Q$ is the entropy of mixing (relative to everything being at standard state, i.e. $Q = 1$). Having both reactants and products present results in a higher entropy than having just reactants or just products.

Where is the mixing?

There is no actual mixing, just making reactants or products, but the expression is the same. You could do a thought experiment where the reaction proceeds via an enzyme that takes up reactants from one container and puts them into another (so reactants and products would never be in the same container). In a second step, you would mix the two containers so that reactants and products are in the same container.

Equilibrium and the entropy of mixing

As a result of the concentration-dependent entropy term, the Gibbs energy of reaction changes (increases) as the reaction goes towards equilibrium. When it reaches equilibrium, it is zero. The expression for $\Delta_\mathrm{r} G$ is:

$$\tag{1}\Delta_\mathrm{r} G = \Delta_\mathrm{r} H - T \Delta_\mathrm{r} S$$ $$ = \Delta_\mathrm{r} H^\circ - T (\Delta_\mathrm{r} S^\circ - R \ln Q)$$ $$ = \Delta_\mathrm{r} G^\circ + R T \ln Q$$

At equilibrium, $Q = K$ (that's the special name we give to $Q$ for a set of concentrations that is at equilibrium) and $\Delta_\mathrm{r} G = 0$. If we make these substitutions in equation (1), we get:

$$\tag{2}0 = \Delta_\mathrm{r} G^\circ + R T \ln K$$

Solving for $\Delta_\mathrm{r} G^\circ$, we get the well-known relationship between standard Gibbs energy of reaction and the equilibrium constant:

$$\Delta_\mathrm{r} G^\circ = - R T \ln K$$

The left-hand side does not make any reference to concentration dependence (it refers to a single state, the standard state), whereas the right-hand side contains the concentration-dependent term (entropy of mixing times temperature, again compared to the standard state where Q is one).


Enolate vs hydrate formation

Both the enolate and the hydrate form.

Aldehydes (and ketones) react with water in the presence of acid or base to form hydrates. However, for more aldehydes in aqueous solution (even dilute solution), the hydrate is only a minor component of the mixture (~1%).

The other 99% of the aldehyde molecules are perfectly able to form the enolate, but only a small number of them do because the difference in pKa between most alpha hydrogens on aldehydes and ketones (~21) and water (15.7).

So, in actuality most of the aldehyde molecules at any given time are neither enolates nor hydrates. Now, if a reaction should occur to consume enolate, then Le Châtelier will come out and nudge the system to make more enolate.

  "Favored" resonance contibutors

The two resonance contributors $x$ and $y$ are not separate compounds. It is not possible for one structure or another to be "favored", though we often use that language because we can relate to comparing the structures of two distinct compounds, like the cis and trans isomers of an alkene. What we really mean when we say "favored" is that the resonance contributor is more important for describing the structure and behavior of the ion than the other. The true structure of the enolate ion is some linear combination of resonance contributors $x$ and $y$ (as well as some other minor things). Thus, the true structure might be like a weighted average over the contributors with $x$ having higher weight.

For example:

$$enolate = \frac{2}{3}x + \frac{1}{3}y$$


Since you haven't mentioned the carboxylic acid to be measured, I assume the carboxylic acid to be $\ce{HCOOH}$. It has a $\mathrm pK_\mathrm a$ of $3.75$
Now, as you mentioned, phenol (or carbolic acid) has a $\mathrm pK_\mathrm a$ of $10.0$. There is a difference of about $6$ in the two values. This is because of the effective delocalization that takes place in formic acid compared to that of phenol. The best way to check for this is by computing the charges on the atoms in their respective conjugate bases and then check the stabilities of the two.
So first, we check the delocalization on the phenoxide ion.
The Mulliken partial charges on the atoms ($\ce{C-1}$ as carbon connecting to oxygen and we move counter-clockwise. $\ce{H-2}$ signifies the hydrogen bonded to $\ce{C-2}$)
$$
\begin{array} {c}
\begin{array}{lr}
\text{Atom} & \text{Charge} \\ \hline
\ce{C-1} & 0.307 \\
\ce{C-2} & -0.246 \\
\ce{C-3} & -0.204 \\
\ce{C-4} & -0.243 \\
\ce{C-5} & -0.204\\
\ce{C-6} & -0.246 \\
\end{array}
\begin{array}{lr}
\text{Atom} & \text{Charge} \\ \hline
\ce{H-1} & 0.103 \\
\ce{H-2} & 0.100 \\
\ce{H-3} & 0.089 \\
\ce{H-4} & 0.100 \\
\ce{H-5} & 0.103\\
\ce{O} & -0.659\\
\end{array}
\end{array}
$$
Now, we check the delocalization on the formate ion,
$$
\begin{array}{lr}
\text{Atom} & \text{Charge} \\ \hline
\ce{C} & 0.363 \\
\ce{H} & -0.083 \\
\ce{O-1} & -0.639 \\
\ce{O-2} & -0.639 \\
\end{array}
$$
Comparing the two, we can see that the more effective delocalization has taken place in formic acid since the net charge on an individual oxygen atom is less ($-0.659$ in phenoxide ion and $-0.639$ in the formate ion)
Due to this, we can say that the net effective charge is much better delocalized in formic acid which means that the conjugate base is better stabilized. Even though there were five possible resonance structures for the phenoxide ion, there do not contribute to the final resonance hybrid as well as the two equivalent resonance structures for the formate ion.
Since order of conjugate base stability is directly proportional to acidity. Hence formic acid would be more acidic than phenol.
Note: These values were calculated on WebMO with geometry optimization using Gamess DFT B3LYP/3-21G


How do I find the empirical formula for cobalt oxalate hydrate with mass of oxalate and oxide produced?

I'm not exactly sure - but I guess you are trying to find the amount of water in your sample and to determine the ratio to cobalt oxalate.
Some parts of your calculation are correct, you do however forget that $\ce{CO2}$ is released. This would be my approach:
$$n_\ce{Co3O4} = \frac{\pu{115.8 mg}}{\pu{240.79 g/mol}} = \pu{0.4809 mmol}$$
This means that there was three times as much $\ce{CoC2O4}$ in your sample.
$$n_\ce{CoC2O4} = \pu{1.443 mmol}$$
That equals a weight of $\pu{212.1 mg}$. Your original sample weighed $\pu{328.3 mg}$, so there is $\pu{116.2 mg}$ left that seem to be $\ce{H2O}$ ($\pu{18.015 g/mol}$), giving you.
$$n_\ce{H2O} = \pu{6.45 mmol}$$
The ratio of cobalt oxalate to water then is $\left(\frac{6.450}{1.443} = 4.5\right)$ which implies the empirical formula is $\ce{2CoC2O4.9H2O}$


Just looking at the electronegativities, there seems to be no reason why atoms such as boron, carbon or silicium — which all have similar electronegativities to hydrogen — should not undergo hydrogen-bond–like intermolecular interactions.

There is one key difference that separates hydrogen from all the other nonmetals that readily form 2-electron-2-centre (i.e. standard) bonds at ambient temperature and pressure: the lack of any core electrons.

Atoms like boron and carbon (or worse: silicon) have a low-lying 1s atomic orbital that is generally not considered to take part in bonding. However, its spherical structure surrounds the nucleus and effectively shields its positive charge from the outside world. Hydrogen, on the other hand, has a basically unshielded nucleus consisting only of a positively charged proton.

The positively charged hydrogen nucleus can now interact well with other orbitals on other atoms (typically part of other molecules). Any in-between core electrons will repulse the electrons that the electropositive element is trying to interact with. Hence, the stabilisation is greatest in hydrogen’s case.


If you are using a gas stove with a methane ($\ce{CH4}$) flame, an interesting things occurs if you heat $\ce{CuO}$ in the gas part of the flame. I once  created cupric oxide from heating a penny to glowing red and then shaking in air. Then, I inserted the dark colored penny into the flame zone, and it appeared to be cleaned!

Apparently, the hot cupric oxide is reduced by $\ce{CH4}$ to $\ce{Cu2O + Cu}$. As such, my take on a possible reaction liberating elemental copper, $\ce{Cu2O}$ and carbon:

$$\ce{CH4 + 3 CuO -> 2 H2O + Cu2O + C + Cu}$$

Now, the flame test for Carbon is red-orange and for Copper green/blue, so a mixture of colors could theoretically be observed in this experiment as the methane flame itself participates in element production.

Normally, however, this does not occur.


While calculating the effective nuclear charge, why don't we consider effect of outer electrons on inner ones?

As chemists, we are typically more directly concerned with the effective nuclear charge on outer electrons than inner electrons, because it's only the outer electrons that interact with other atoms and thus determine an atom's chemical properties. [Though the effective nuclear charge on inner electrons will affect their radial probability distributions and thus create a second-order effect on the effective nuclear charge of outer electrons.]
Nevertheless, this is an interesting question.
When we are calculating the effective nuclear charge seen by electrons in an atom (say, for outer electrons) (which directly impacts, for instance, their ionization energies), we start by doing so for isolated atoms, i.e., atoms in the absence of an external field.  Adding an external field (from the presence of other atoms, especially bonding atoms) is an added complication we won't consider here.
And in isolated atoms, the probability distributions of all electrons are spherically symmetrical.
So let's consider an inner electron in an isolated atom.  According to Gauss's Law, any charge that is spherically distributed about the nucleus, and is farther from the nucleus than that electron, "cancels itself out" (it integrates to zero), i.e. it has no net effect on the effective charge felt by the electron.
So an inner electron is only affected by outer electrons to the extent that the outer electrons spend some time closer to the nucleus than the inner electron.  [Again, when they are further from the nucleus than the inner electron, they have no effect on it, because of the spherical symmetry of their probability distributions.]
Based on the relative probability distributions of inner and outer electrons, the percent of the time this happens (outer electrons being closer to the nucleus than inner ones) is low.  Having said that, when it does happen, it will cause the inner electrons to feel a reduced nuclear charge.  I.e., to the extent that outer electrons affect inner electrons (in isolated atoms), they don't push them towards the nucleus, they push them away!
Hence there is no "reverse shielding" effect by outer electrons on inner electrons (again, in isolated atoms).  There is only a (small) shielding effect.


pKa explanation through empirical formula?

Can we explain the above pka with (ONLY) the knowledge of the given
  numbers of atoms of C, H and O using the theory of Usanovich

No, nobody's theory (including Usanovich's) can explain $\ce{pK_{a}}$ differences based on molecular formula coefficients.  This is best illustrated by the observation that even compounds with the same coefficients (same values for "abcdef") can have different $\ce{pK_{a}}$ values.  For example, $\ce{C3H6O}$

\begin{aligned}
\ce{&acetone &&pK_{a}=20}\\
\ce{&allyl~ alcohol &&pK_{a}=$15.5$}\\
\end{aligned}


Each carbon atom (you seem to have identified them) needs four of those stick-like thingies (the lines, sometimes called "bonds"). These include the dashed lines and thick triangles as well as the "double bond" you see at the lower oxygen which looks like this ($=$). That one counts twice. Every time you only see 3 or even 2 for one carbon atom, that means 1 (or 2) hydrogen atoms are implied.


What are the disadvantages of using the InChI format?

You have to evaluate shortcomings as compared to something. You can make a list and say "but InChI doesn't represent X" where X is ranging around

mixtures
Markush groups
polymers from repeat units
reactions
toasters

The point is: so what? All representations have some limitations. But limitations aren't shortcomings unless you have a different representation that doesn't have those limitations and that is not traded off by other limitations or bloat or other disadvantages.

What do you have to compare InChI with? In my world: SMILES and with MOLFILEs. InChI is most closely related to SMILEs. SMILES is easier to just type, but it isn't canonicalized and it doesn't have the layers of specificity that allow you to find approximate equivalences so easily as with InChI. InChI connection layer is very similar to SMILEs in principle, but it is a lot harder to read because you need to replace the numbers with the atoms from the preceding formula layer. But it is easy to convert InChI to SMILES, except where InChI is better. For example, parenthesized groups, so you represent the carboxyl group -C(=O)OH as -C(O,O)H, which isn't a thing in SMILES.

Where InChI shines is that many isomerisms are abstracted or pushed into more detailed layers. There is no other format that does this as well or at all as InChI, which is why it is good. But InChI is conceived as an "identifier" and some of the details are not at all represented, most specifically the atom positions. That's where MOLFILEs are so easy to work with, if you want to just display the molecule. But in that regard, InChI is no worse than SMILES.

So, you should be more specific in terms of what your goals are that you are asking for "disadvantages" and you need to stay very critical of text that berates InChI or any other standard with a bunch of rationalizations for why not to choose standard A (because the author is committed to standard B). I know this sort of superficial argument is being made a lot in any informatics sector, people like their baby and berate the other's baby. Especially be weary of those who say "don't use InChI, InChI is bad" and then "you really should follow the work on XYZ, because XYZ will be able to represent everything including the toaster."

InChI has it's firm place when the purpose is to represent, refer to, identify basic molecular structures. That's it. If you want mixtures, reactions, polymer repeat units, consider using InChI as the core building block and then build higher order structures around it. For example, obviously there is no reason to push reactions into the InChI representation, just make a simple structure that lists the reaction participants on both sides, their stoichiometry and the enthalpy, activation energies, etc. etc., these can be done in higher order data structures much better than shoving that information into a single InChI++ string. Likewise for polymers, represent each repeat unit as an InChI and then combine them. If your goal is such higher order molecules, you would profitably use InChI as the representation of those core building blocks.

Now that said, I give you my personal pet peeve about InChI, i.e., that is in scope of current InChI where I think it has, let's say, pitfalls, not necessarily that it made the wrong choice. And that is, ionic bonds and coordination bonds. Especially ionic bonds are represented as two separate structures, recognizable with the period . in the formula separating the multiple structures. This is actually a rudimentary representation of mixtures. So in my view, it should not even have introduced this dot notation but simply forced people to represent the parts of these ionic mixes as separate InChIs. I come from the biochemical / physiology / pharmaceutical / medical world where such ionic bonds usually do not even really exist in nature, as the ions are all dissociated in solution. But all that can be disputed. Same for metal coordination or chelate bonds, you want to say where the metal ion associates with, and I think that's not in InChI. So in my world it would be better not to have any of this mixing and coordination inside InChI and we would use higher order structures for it instead, but always use InChI as the basic building blocks.


Why does equilibrium constant change when the reaction is multiplied by a number and nothing in the real has been changed?

First of all, you switched the numerator and denominator in the definition of $K$. If $\ce{A <=> B}$ , then $K=\frac{[B]}{[A]}$. Products over reactants.

Notice that, if $K$ is large, then mathematically it is necessary for either the numerator to be large and/or the denominator to be small. Therefore, qualitatively, if $K$ is large, then you either have a lot of products, or little left of reagents, both of which imply a reaction which occurs close to completion.

Now here's the problem - we were speaking qualitatively. Your first sentence ("a big equilibrium constant shows that the reaction wants to get complete") is only a rule of thumb. It happens to work well because in reactions where equilibria descriptions are quantitatively reliable/exist at all, the stoichiometric coefficients are relatively small when written with the lowest integers (using the lowest integer coefficients is almost always done). The rule of thumb is borne out of repeated calculations under these implied conditions. They are not the case for your second equilibrium example, so such a rule should not be expected to still be valid.

Indeed, though $10^{20}$ is an impressive number at first, when you actually go do the calculations to find the relative amounts of reactants and products, you will at some point have to take its twentieth root, promptly turning it back into a much smaller number. Another way to think about it is that with such large stoichiometric coefficients, even a small change in concentrations is amplified enormously by the large exponents and has a huge impact on the value of the reaction quotient; a small increase in products will go a long way to reaching the equilibrium condition.


Symmetry is a very broad topic, and as was mentioned in the comments there are many kinds of symmetry. However, when dealing with molecules you should focus on point group symmetry. A good place to start is this link.

But let's get back to the question:

You mention that you know water is asymmetrical and is therefore polar but at the same time you also note that you CAN draw a line through it to cut it into two equal halves. In that way, you can say water has a mirror plane and is therefore symmetric. This type of symmetry is more precisely defined as $C_{2V}$. Ammonia has  $C_{3V}$, Benzene has  $D_{6h}$, and so on and so forth.

Now, symmetry and polarity do have some relation but you can't tie them together, the case in point being the hydrocarbon $\ce{CHMeEtPr}$. From the point of view of symmetry, this is definitely not a symmetric molecule. And despite its complete asymmetry this is definitely what we would call a non-polar molecule.

So are there any properties related to symmetry? Yes - the ability for a molecule to rotate plane polarized light. We measure this and call it optical activity, and call the asymmetric molecules chiral.

Hope that helps!


What is the formula of manganese oxohydroxide: MnOOH and MnO(OH)2?

First, the difference the substances is the oxidation number of manganese. In $\ce{MnOOH}$, the oxidation number of manganese is 3+. While in $\ce{MnO(OH)_2}$, the oxidation number for manganese is 4+.

However this doesn't mean $\ce{MnOOH}$ is manganese (III) oxyhydroxide and $\ce{MnO(OH)_2}$ is manganese (IV) oxyhydroxide, not for $\ce{MnO(OH)_2}$ at least. If you move the element around, you will get $\ce{MnO(OH)_2}$ is actually $\ce{H_2MnO_3}$. $\ce{H_2MnO_3}$ is dihydroxy(oxo)manganese which is an acidic substance.

For the question regarding the structure for both substances, I'm sorry that I don't haven't the image but I can describe it (Hope that would do). For $\ce{MnOOH}$, manganese is single bonded with $\ce{OH}$ and manganese is double bonded with oxygen. For $\ce{MnO(OH)_2}$, manganese have two single bonds with both of the $\ce{OH}$ and manganese is double bonded with oxygen.

In conclusion, $\ce{MnOOH}$ is manganese (III) oxyhydroxide and $\ce{MnO(OH)_2}$ is dihydroxy(oxo)manganese. In structure wise, manganese is single bonded with $\ce{OH}$ and manganese is double bonded with oxygen for $\ce{MnOOH}$. For $\ce{MnO(OH)_2}$, it likes $\ce{MnOOH}$ but have one more single bond with $\ce{OH}$.


In the condensation reaction of an aldehyde with a hydroxyl amine, why is an oxime formed and not an amide?

I don't understand why it's not possible for the acidic hydrogen of the aldehyde to be abstracted by the $\ce{OH-}$ radical of the amine

There is a lot wrong in this sentence and that’s the key to the answer. For one, there is no hydroxide radical (neither in hydroxyl amines nor anywhere else); if anything it is a hydroxyl radical $\ce{HO^{.}}$ — but it is not present in hydroxyl amine. If you check out the mechanism of the condensation, you will realise that no basic hydroxide species occurs:

If any basic species ever exists in here, it is the oxygen anion in the first step. Note especially that the hydroxyl group on nitrogen never loses its proton and thus never becomes basic in this mechanism.

I can hear you asking ‘so why doesn’t the oxyanion in step two abstract the neighbouring hydrogen?’ And this is really the key:

That hydrogen is not acidic!

Never, never, never ever deprotonate this hydrogen on an aldehyde. The only way to make it moderately acidic (but you would still need very strong bases to do so) is something along the lines of the Corey-Seebach umpolung. It generates a cyclic S,S-acetal out of your aldehyde which is stable. Its $\mathrm{p}K_\mathrm{a}$ is low enough that it is actually accessable. That of an aldehyde is not.

Also, to finally generate an amide, you would need some kind of migration of the hydroxyl oxygen to the aldehyde carbon after hypothetic deprotonation. I’m not saying it’s impossible, but rather unlikely (but that is difficult to judge due to the unlikelyness of the previous step). Note that an amide’s carbon has an oxidation state of $\mathrm{+ III}$ while the oxime’s carbon has $\mathrm{+I}$. You need a redox mechanism to get there.


Why is the boiling point of rectified spirit lower than alcohol absolute?

Ethanol and water have weaker intermolecular interactions than water with water and ethanol with ethanol. For this reason ethanol and water molecules can more easily escape from the liquid mixture, called azeotrope, to the gaseous phase than it is the case for the pure alcohol. Consequently the azeotrope has a lower boiling point.

What is considered as an impurity depends on your use case. E.g. 96% pure ethanol may qualify as extra fine spirit.


Litharge is $\ce{PbO}$. It is probably introduced to save silver from compounds containing oxidized silver like $\ce{AgCl}$. If metallic silver powder is mixed with $\ce{AgCl}$, this compound will react with $\ce{PbO}$ according to $$\ce{2 AgCl + PbO -> PbCl2 + Ag2O}$$ And $\ce{Ag2O}$ is an instable compound being easily transformed into $\ce{Ag + O2}$ at high temperature. So all metallic silver and silver from silver compounds are finally dissolved into lead, because lead dissolves metallic silver and not silver compounds. Without litharge, silver compounds like $\ce{AgCl}$ are lost.


I think you are looking for nuclear transmutation which occurs via nuclear reactions or radioactive decay, as explained in the referenced wikipedia page. As you can imagine then, implementation has some important impediments.


Does "silica colloidal anhydrous" contain aluminium?

It will probably have some aluminium in it, likely to be less than 0.1%, and most probably less than 0.01%. Not only that, also expect Ti and Zr in those quantities.

This level of impurities does not matter for most practical applications. If you do require exceptionally pure silica, you can buy it for a premium.

A note about “anhydrous”: when exposed to air, powdered silica will absorb moisture. Particularly if it’s amorphous. If you want it to be really anhydrous, dry the thing before using. Above 100 C will remove most of it, but completely removing all traces of water will require firing at ~1000 C in a noble metal crucible such as platinum.


Which of the following samples contains the largest number of atoms?

Geeze, NaCl has 1 mole of Na and 1 mole of Cl per mole of NaCl. So that is 2 moles of atoms per mole of NaCl.

$\ce{K2SO4}$ has 2 moles of potassium, 1 mole of S and 4 moles of O per mole of $\ce{K2SO4}$. So that is 7 moles of atoms per mole $\ce{K2SO4}$.

You don't need to convert to atoms using Avogadro's number then back to moles. That complicates the problem needlessly. Just use the stoichiometry and you can do the arithmetic in your head.


The collection of nitrogen $(\ce{N2})$ by downward displacement of water works since nitrogen is poorly soluble in water (at room temperature and standard pressure) and does not chemically react with water.

For the same reason, some impurities (aerosols, water-soluble gases and vapours – for example $\ce{HNO3}$) can be removed by passing the stream of nitrogen through a washing bottle that is filled with water.

Nitric oxide $(\ce{NO})$ is only poorly soluble in water and therefore cannot be removed by this method. However, $\ce{NO}$ can be oxidized. Strong oxidizing agents can convert $\ce{NO}$ into $\ce{NO3-}$:

$$\ce{NO + 2H2O <=> NO3- + 4H+ + 3e-}\quad E^\circ = +0.959\ \mathrm V$$

A suitable strong oxidizing agent is acidic dichromate $(\ce{Cr2O7^2-})$:

$$\ce{2Cr^3+ + 7H2O <=> Cr2O7^2- +14H+ + 6e-}\quad E^\circ = +1.38\ \mathrm V$$

The balanced overall reaction is:

$$\ce{2NO + Cr2O7^2- +6H+ <=> 2NO3- + 2Cr^3+ + 3H2O}$$


First of all charge of atom is not well defined (maybe except AiM). Formal charge of both is zero, but that does not help us to understand chemistry. Above mentioned reaction shows that nitrogen is more negative, and if you do simple QM computation both Mulliken and NBO gives ~-.44 for N and ~+.147 for Cl. If you want to have a bit feeling about it, think that way: both have similar electronegativity but Cl is bigger so it's easier to polarize.


How to simplify these system of equilibrium equations

Since the equilibrium constant for the formation of the complex ion is very large, I assume that $\ce{[Ni(CN)4^{2-}] >> [Ni^{2+}]}$

From the comments:

  [comments:] That assumption is incorrect. The equilibrium constant seems large, but the exponents are high, so it is misleading. In fact, more of the cyanide is in the form of HCN than in complex with nickel, and there is very little free cyanide.

OP replied:

  How may I make a better assumption to solve this problem?

In the absence of nickel, the pH is around 5 and HCN is the major species, with the concentration of cyanide in the micromolar range (standard weak acid problem). When you add nickel, the concentration of cyanide gets lowered further. Because the free cyanide concentration contributes little to the total, we can combine the two reactions to see whether there is more complex or more HCN and to determine the pH:

$$\ce{4HCN + Ni^2+ <=> Ni(CN)4^2- + 4H+}$$

The equilibrium constant is

$$K_\mathrm{combined} = \pu{5.8e−16}$$

So for this reaction, reactants are favored. HCN and nickel concentrations hardly decrease compared to the initial concentrations. If we call the concentration of the complex $x$, we get:

$$K_\mathrm{combined} = \frac{x (4x)^4}{[\ce{HCN}]^4 \cdot [\ce{Ni^2+}]}$$

Solving for $x$, we get $\pu{2.24e−5}$, confirming that reactants are favored. From this we can calculate the pH as

$$\mathrm{pH} = -\log 4x = 4.05 $$

Checking the solution

Upon adding nickel, the pH turned a bit more acidic. That makes sense as a micromolar concentration of nickel complex formed, releasing some more protons from HCN. We can calculate the cyanide concentration from either one of the given equilibrium constants, and verify that it is miniscule. Then, we can verify that all reactions are at equilibrium (at pH 4, hydroxide is a minor species as well, and has little impact on the exact solution).


How exactly do we superimpose molecules while checking optical activity?

You're being confused by the rotation. You take the mirror image. Any rotation or sequence of rotations that superimpose the mirror image over the original means that the molecule is achiral. It's not just a 180-degree rotation.


Liquid crystals are controlled by electric fields and their structure manipulates the polarisation of light
Liquid Crystals are usually made from long rod-shaped molecules which behave in some ways like liquids (they flow) but in other ways like crystals (they form structures with a lot of regularity). The name is appropriate.
The key features that matter are that the structures they form are easy to manipulate with electrical fields and that the rod-like nature of the structure interacts with light in roughly the same way as polaroid: it selectively alters the polarisation of the light passing through.
In LCD devices the liquid crystals are placed between two polaroid filters and in front of a backlight so the the default orientation of the filters plus the liquid crystal layer lets most light through. When a small electric field is applied across the liquid crystal it can be arranged so that it changes the orientation of the molecules in the layer and twists the light so it no longer passes through the polaroid layers. This can be arranged to happen one pixel at a time with suitable electronics. So an electrical field can drive a display.
That's the basic principle. Real displays have evolved to have some very complex arrangements of these effects to improve speed and the twistiness of the LC layer to make faster responding and higher contrast pixels.


How is the mol highway used in this problem?

You are on the right track. Every element has a molar mass that is defined under standard conditions, which is in this case the natural abundance of an element taken into account. Pressure, volume and temperature will not change the mass of a compound. Since every element has a specific molar mass, every compound does too. The molar mass is a measurement of how much mass does one mole of a substance have.

The molar mass is usually referred to with $M$, while the mass is referred to as $m$. The amount of substance is $n$. This gives you the following relationship: $$M=\frac{m}{n}$$

Since you have given $m(\ce{C3H8})=11~\mathrm{g}$ and you already looked up $M(\ce{C3H8})=44.1~\mathrm{g\,mol^{-1}}$, you can use this formula to determine $n(\ce{C3H8})$.

In this question it is quite hard to explain the use of significant figures. Those are used to imply a certain inaccuracy. Not enough information is given by the question, as of how accurate the measurement is. It is a mere exercise of converting one property into another. Here you should not worry about it.


What metals aren't dissolved in/attacked by aqua regia?

Keep in mind

The answer will depend upon the reaction conditions. Most importantly,

physical state of the metal: porosity, degree of comminution;
temperature;
mechanical aggravation of metal surface during reaction.

Often times a chemistry text mentions that no reaction occurs. The reaction might still happen. It is just that for the specified parameters the process is meaningless and negligible.

Short overview of aqua regia

Aqua regia is the $3:1$ volumetric mixture of $\ce{HCl}$ and $\ce{HNO3}$. Its additional reactive power draws from monochlorine created in situ.$^{[1]\ [2]}$

$$\ce{HNO3 + 3HCl -> Cl2 + NOCl + H2O\\ NOCl -> Cl + NO}$$

  Which metals are impervious to $3:1\ \ce{HCl/HNO3}$?

Almost every metal will react with aqua regia provided certain criteria are met.$^{[1]\ [2]}$ The closest you will probably get is ruthenium $\ce{Ru}$, and perhaps osmium $\ce{Os}$. To the best of my knowledge, $\ce{Ru}$ will not react with aqua regia in a meaningful way even if aqua regia is boiling.$^{[2]}$ The difference with $\ce{Os}$ is that powdered osmium is attacked by boiling aqua regia.$^{[1]\ [2]}$

$$\ce{Ru + HNO3 + HCl $\kern.6em\not\kern -.6em \longrightarrow$}$$

$$\ce{\underset{powder}{Os} + $\underbrace{\mathrm{HNO_3}}_{\text{boiling}}$ -> OsO4 + N_xO_y + H2O \\ OsO4 + 2H2O <=> H2[OsO4(OH)2] \\ OsO4 + HCl ->OsO2Cl2 + Cl2 + 2H2O\\ OsO2Cl2 + HCl ->OsCl4 + Cl2 + 2H2O\\ 2OsO2Cl2 + H2O <=> OsO2 + H2[OsO2Cl4] \\ 3OsCl4 + 2H2O <=> OsO2 + 2H2[OsCl6]\\ OsO2 + 6HCl <=> H2[OsCl6] + 2H2O}$$

Brief discussion about the list provided in the comments

Titanium $\ce{Ti}$ does react, and does so at room temperature.

$$\ce{3Ti + $\underbrace{\mathrm{12HCl + 4HNO_3}}_{\text{room temperature}}$ -> 3TiCl4 + 4NO + 8H2O}$$

Rhenium $\ce{Re}$ reacts slowly at room temperature $\ce{->HReO4}$. This will further react with $\ce{HCl -> ReCl4 + Cl2}$.$^{[2]}$

Hafnium $\ce{Hf}$ does react at room temperature. The reaction is slower than with titanium; overall equation is identical.$^{[2]}$

Tantalum $\ce{Ta}$ reacts when aqua regia is heated to $150\ ^{\circ}\mathrm{C}$. Rhodium $\ce{Rh}$ reacts in a grinded state. As a large compact piece, iridium $\ce{Ir}$ is affected over temperatures of $100\ ^{\circ}\mathrm{C}$. Niobium $\ce{Nb}$ is inert at room temperatures.$^{[2]}$

Summary: ruthenium $\ce{Ru}$ is your best bet.

  What makes metals $\ce{Ru}$ and $\ce{Os}$ so stable in aqua regia?

The nobility of these metals is not the best explanation. As you correctly pointed out, $\ce{Pt}$ and $\ce{Au}$ react fine. This is direct evidence that for other metals a protective layer should form. The layer varies from metal to metal, but usually is either an oxide (or oxide hydrate), or a chloride.

Effectiveness of mechanical aggravation also points to stable, non-reactive compound formation on the metal's surface.

For ruthenium, as of now I am unsure what this precipitate could be. If anyone has a reference, please edit or leave a comment.$^\text{[reference needed]}$

  What happens with silver?

Silver and aqua regia react very poorly, and for a short amount of time.$^{[2]}$ The culprit is $\ce{AgCl}$ ($K_s = 1.8 \cdot 10^{-10}$)$^{[2]}$. A slow reaction might still take place due to complexation.$^{[2]}$

Surprisingly, silver reacts with $\ce{HBr}$!$^{[2]}$ Its solubility product is even worse, $K_s = 5.0 \cdot 10^{-13}$.$^{[3]}$ My guess is that this layer is not as dense as $\ce{AgCl}$ but this still needs verifying.$^\text{[citation needed]}$

References

(In progress)

$[1]$ N. N. Ahmetov. Anorgaaniline keemia. (1974)

$[2]$ H. Karik, Kalle Truus. Elementide keemia. (2003)

$[3]$ Skoog, West, Holler, Crouch. Fundamentals of Analytical Chemistry. 9th edition. (2014)


Radiofrequency transmitter in an NMR experiment: Is there an involvement of (electromagnetic) radio wave?

I finally asked an NMR spectroscopist who was the editor of the Journal of Magnetic Resonance and knows the inside out of NMR spectrometers. He sent a scanned copy of Hoult's paper from 1989. The title explains it all "The magnetic resonance myth of radio waves
." Surprisingly very low citations (15). It is amazing how much nonsense we have been fed with radio waves absorption. Sometimes innocent student questions on these forums can lead to introspection.

Here is the article Radio Waves Myth in NMR. Karl will like it.

The conclusion is that there is no involvement of radio waves, neither in absorption nor in emission (of course). The emission process is easier to understand but the transmission process seems to have confused even the experienced spectroscopists and more MRI people.

This is from another paper of Hoult. "The Origins and Present Status of the Radio Wave Controversy in NMR"

  The origins, history, and present status of the controversy
  surrounding a quantum description of the NMR signal as being due to
  radio waves are traced. With the Principle of Relativity and Coulomb’s
  Law as formal starting points and the minimum of mathematics needed
  for understanding, the derivation of a classical electromagnetic
  theory of signal reception is first given. The agreement between that
  classical theory and a recent NMR experiment is then presented,
  leading to proof that, except for the highest field imaging
  experiments, there is no significant contribution of radio waves to
  the signal.


What about the aromaticity of cyclopropenylidene?

This is a problem of counting lone electron pairs as part of the delocalized pi system.  In pyrrole you count the lone pair on the nitrogen, in pyridine you don't.  What's the difference?
In pyrrole you have the lone pair left over after you have given every ring a ligand (the outer hydrogen atoms).  With the sigma bonding requirements thus satisfied, the lone pair goes into the pi system.  In pyridine the nitrogen is missing its ligand, so the lone pair must replace that in the plane of the ring and can't go into the pi system.
Which situation corresponds to your compound depends on what compound is meant, and there seems to be confusion on this point.  Cyclopropenylidene ($\ce{C3H2}$) and cyclopropenyl anion ($\ce{C3H3^-}$) differ only by a carbon-hydrogen bond, and the shorthand method commonly used to render organic structures thus fails to distinguish them.  The problem statement should be checked to ascertain which of these compounds is meant because they give different outcomes, and the answer key is right for only one of them.
If it is cyclopropenylidene
In cyclopropenylidene, the carbon with the lone pair is missing a ligand like pyridine, so the lone pair has to replace the missing ligand in the plane of the ring.  That leaves two pi electrons to be delocalized which makes the ring aromatic.  Cyclopropenylidene is not only aromatic by the electron count, it is a sufficiently preferred three-carbon structure to be seen in interstellar space.
If it is cyclopropenyl anion
Cyclopropenyl anion has all the ligands in place, so the lone pair goes into the pi system like pyrrole.  Unfortunately in this case, adding that lone pair gets you four pi electrons in the ring and that's bad.  In fact it turns out to be really had, even worse than larger (and thus more flexible) formally antiaromatic rings like cyclooctatetraene.
Cyclopropenyl anion goes through some serious contortions to get rid of this antiaromaticity.  See this question.


Is following graph is of a dibasic or monobasic acid?

As I understand it, if the data is absolutely correct, the curve should not change direction of curvature in that manner if it is monibasic. Each time it's double differentiation changes signs 2 times, I think it passes an equivalence point no matter how shallow it is. It should be a dibasic acid. You can get a function of [H+] using equilibrium constant (assume it to be monobasic) and taking x as moles of naoh added in other words, the "initial" moles of salt x is and "initial" moles of acid is a-x, a is the moles of acid we started with.By "initial" I mean the moles before equilibrium. You would get a quadratic equation, see at how many times sign of double differentiation of -log([H+]) wrt x changes for any value of x. Here is oxalic acid's curve.


What factors are influencing a woosh rocket?

For home projects like this I believe one of the most important things is to ensure you have a good fuel to air (oxygen) ratio. It is surprisingly easy to use too much propellant which can take up air space within the combustion chamber. More fuel doesn't necessarily mean bigger bang.


4-Bromo, 2,4-di-methylhexane and 3-Bromo, 3,5-di-methylhexane are the same thing?

Both names, ‘4-bromo-2,4-dimethylhexane’ and ‘3-bromo-3,5-dimethylhexane’ are unambiguous and describe the same compound. However, only ‘4-bromo-2,4-dimethylhexane’ is in accordance with IUPAC recommendations.
First of all, the seniority order of functional classes, which is used to choose the functional class (here: …hexane), does not apply to numbering of locants within the same compound class.
Concerning numbering of locants within the same compound class, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(…)

Furthermore,

P-14.3.5 Lowest set of locants
The lowest set of locants is defined as the set that, when compared term by term with
other locant sets, each cited in order of increasing value, has the lowest term at the first
point of difference; (…)

Therefore, the example is correctly named as 4-bromo-2,4-dimethylhexane rather than 3-bromo-3,5-dimethylhexane since the locant set ‘2,4,4’ is lower than ‘3,3,5’.