Lone pairs can sometimes be delocalized. An acceptable rule of thumb is that if an atom possesses lone pairs and is connected to an sp2 atom, then the two atoms participate in resonance. So consider phenol. Oxygen has lone pairs. It's connected to an sp2 carbon. And so yes, the oxygen can participate in resonance with the sp2 carbon it's connected to, and also the other sp2 carbons in the aromatic ring. We can ascertain the fact that every atom is phenol is sp2 by noting that phenol is planar.

In general, a negative formal charge makes delocalization a (more) favorable move as this decreases the electron (charge) density on one atom. Spreading out electron density between or among multiple atoms generally lessens reactivity.

Now, your question is why would oxygen exhibit a +I or electron-donating effect despite its high electronegativity. Well, it can participate in resonance donation as explained above; see phenol and its conjugate base, the phenoxide ion, for examples. Examining both in terms of electrophilic aromatic substitution also reveals that the oxygen atom in both is strongly electron donating. True, oxygen can withdraw electron density via sigma bonds (inductive effect) but doing so is unfavorable especially in the phenoxide ion in which the oxygen already bears a negative formal charge. As your teacher explained, bearing charge isn't something that is very favorable.

And in general, you're going to find that resonance effects outweigh inductive effects. Hence the reason why an sp2 oxygen connected to another sp2 atom generally exhibits a +I effect. Consider dimethyl malonate and methyl acetoacetate . Which is more acidic? From inductive effects alone we'd say dimethyl malonate because the oxygen is highly electronegative as opposed to an sp3 carbon and should withdraw more electron density via the inductive effect. From a resonance-standpoint we'd say the opposite; the oxygen in the ester donates electron density via resonance more than a methyl group can inductively donate electron density. And you'd have to realize that the resonance-standpoint is more important in evaluating the acidities of the two weak acids.

A common exception to resonance>inductive effect in organic chemistry is an sp2 atom connected to a halogen. Yes, the halogen has lone pairs and these can be delocalized into the adjacent p-orbital. But the halogens (except possibly fluorine) are much bigger than many other atoms, and therefore the magnitude of resonance donation is limited. So the great electronegativity of the halogens trumps their magnitude resonance donation, and we can see this in the context of electrophilic aromatic substitution again; halogens are ortho-para directors (delocalize the lone pairs in the Lewis structures to see why) but are also mild deactivators.


The Conductivity of Electrolytes based on structure of Ionic Crystal

It turns out, according to this paper, stated:

  The solvent-berg model which put forward the suggestion that smaller ions are strongly interacting with the nearest neighbour shell of solvent molecules. This was considered to be particularly true of cations since these are generally smaller in size than the corresponding anions and have a higher charge density. The ion essentially carries this shell of solvent molecules long enough that this leads to a larger effective diameter which lowers its conductivity to a value smaller than the conductivity of larger ions which have no strongly attached shells of solvent.

And in terms of drag from the polarity of the solvent, namely water, they showed a formula for dielectric friction:

ζDF = 3qi2(ε0 − ε∞)τD/(cri3(2ε0 + 1)ε0)

Where:
ζDF is the dielectric friction,
ri and qi are the radius and charge of the ion,
ε0 and ε∞ are the static and high frequency dielectric constants of the solvent,
c is the product of the limiting ionic conductivity and viscosity of the solvent,

So, I assume that the ionic radius will play a role in its drag by the surrounding polar solvents which will affect its mobility, and therefore the conductance.


Why do we neglect the protons from water's autoprotolysis?

You could consider them, but what would be the result?

$0.01 M$ of $\ce{HCl}$ with $1\times10^{-7} M$ of $\ce{H^+}$ from water gives you $0.0100001 M$ of $H^+$. Find the pH of this concentration of $\ce{H^+}$ and compare with the pH you would obtain if you used $0.01 M$ directly.

You shouldn't see that the resulting pH differ significantly.


CH2 group in alkyl glucoside gives two separate HNMR signals

Without the sugar, those two hydrogen atoms are enantiotopic, since substitution of those protons would lead to different enantiomers. Achiral measurements still will appear identical though.

Once the sugar is attached, the two hydrogens are diastereotopic, and are chemically and magnetically different. Especially so close to the chiral environment of the sugar, these two hydrogen see enough of a difference in their local environments that their shifts are different.


What is the hybridization of chromium in chromate and dichromate ions?

The d orbitals are commonly grouped into two groups because that is how they transform under both octahedral (including a g subscript) and tetrahedral (no g subscript) symmetry:

$\mathrm{t_{2(g)}}: \mathrm d_{xy}, \mathrm d_{xz}, \mathrm d_{yz}$
$\mathrm{e_{(g)}}: \mathrm d_{z^2}$ and $\mathrm d_{x^2-y^2}$

In tetrahedral environments, the p orbitals also transform as $\mathrm t_2$ while in octahedral ones they transform as $\mathrm{t_{1u}}$. By analogy to main group tetrahedral molecules, it would be easy to assume that just the three $\mathrm t_2$ d orbitals take part in the formation of tetrahedral complexes (throw in one s orbital for completion of the four).

In reality, all orbitals that transform as $\mathrm t_2$ will be represented in the final complex in some way or another, so that whatever bonding and antibonding orbitals of $\mathrm t_2$ symmetry you end up with will always include some contribution from the original p orbitals. Furthermore, outside of hydride complexes the ligands will also have available orbitals for π-type bonds which will complicate the picture further. The final MO for chromate will look something like the following:

Figure 1: tetrahedral $\ce{[ML4]}$ complex with ligand π contributions.

The metal orbitals on the left are d, s, p from bottom to top. The ligand orbitals on the right are the oxygens’ p orbitals (all 12 of them).

Don’t worry, these schemes will come to haunt you at one point when studying chemistry.


How to find the concentration of an of ammonia solution by titration with nitric acid?

Yes, although your presentation is a bit confusing.
To be clear,

You determined the quantity of $\ce{HNO3}$ consumed during the titration
$$0.151\ \mathrm{M} \times 26.2\ \mathrm{mL} = 3.956 \  \mathrm{mmol}$$
Then given this titration equation:
$$\ce{HNO3 + NH3 -> NO3- + NH4+}$$
You deduced that the $25~\mathrm{ml}$ sample contained $3.956 \  \mathrm{mmol}$ of $\ce{NH3}$, hence a concentration of
$$\frac{3.956\ \mathrm{mmol}}{25\ \mathrm{mL}}=0.1582\ \mathrm{M}$$
From which you calculated the amount of $\ce{NH3}$ in the $500\ \mathrm{mL}$ sample:
$$0.1582\ \mathrm{M} \times 500\ \mathrm{mL} = 79.12 \  \mathrm{mmol}$$
And finally since the original $20\ \mathrm{mL}$ sample also contains $79.12\ \mathrm{mmol}$ of $\ce{NH3}$, that gives you the original concentration:
$$\frac{79.12\ \mathrm{mmol}}{20\ \mathrm{mL}}=3.956\ \mathrm{M}$$

What I think you should have done in place of 4 and 5 is:

4'. Since the $25\ \mathrm{mL}$ sample is taken from a solution diluted from $20\ \mathrm{mL}$ to $500\ \mathrm{mL}$, the original concentration is:
$$0.1582 \ \mathrm{M} \times \frac{500\ \mathrm{mL}}{20\ \mathrm{mL}}=3.956\ \mathrm{M}$$

Which spares you one calculation's work and potential errors.


Use of basis set in DFT (Density Functional Theory)

This answer only deals with the most common variety of Density Functional Theory, namely, Kohn-Sham DFT. This is what most people mean by "DFT", but, as noted in the comments, things such as orbital-free DFT exist.

Kohn-Sham DFT was created to solve a historical problem of DFT: The electronic kinetic energy term (as used in the Thomas-Fermi model) was not accurate enough. Kohn and Sham proposed to make use of what is typically called a "fictitious, non-interacting reference", that is, a wave-function whose main purpose would be to yield a density to be used in all other (nucleus-electron attraction, Coulomb, exchange, and correlation energy/potential) and whose secondary function is to provide the kinetic energy, which is accurate from wave-function theory. This wave-function works just like a Slater determinant and is called the Kohn-Sham determinant. See this question for pointers on where to read more: Equivalent of Szabo and Ostlund book for DFT.

As often in DFT, practicality showed a way that was justified later. When implementing this idea, one learns that one can basically use a HF code and replace the exchange term by the exchange-correlation (XC) potential (which is projected back on to the AO basis set). The potential needs to evaluated numerically on a grid, a procedure that yields the XC energy for almost all choices of density functionals. Thus one obtains the energy and the KS operator (the equivalent of the Fock operator) and can perform a SCF procedure in a given basis set. For the mathematical details, see e.g. JA Pople, PMW Gill, BG Johnson Chem Phys Lett 199, 557 (1992).

To answer your questions in this context:
1) The initial density depends on the guess, which is a whole different can of worms, but the same is true for HF calculations. (Procedures based on tabulated atomic densities exist, but the initial guess business is a bit of a dark art, and I don't think this is what you meant by your question.) The basis set and its MO-like coefficients (defining the KS determinant) are used on every iteration to yield the density.

2) Naturally, yes. The MO-coefficients of basis sets change on geometry change for a given system, just like in HF. The procedure and workable algorithms are very similar to HF (because it all is).


How can mass spectrometry be used to determine concentration?

As indicated in the Wikipedia page, Inductively Coupled Plasma Mass Spectrometry (ICP-MS) is the technique of choice to quantify metal ions in a sample. In summary, this technique is based on the atomization and ionization of the sample metal atoms in a plasma torch, and the ions are then separated based on their mass, usually with a quadrupole mass filter, and then directly quantified on the detector. External calibration with standard solutions of known metal concentration or isotopic dilution then affords access to absolute concentrations.


Comparing first, second, and third ionization energies between atoms

If you know the trends for $1^\text{st}$ ionization energies, a good first approximation for the trends of $n^\text{th}$ ionization energies at any given point is the $1^\text{st}$ ionization trend $n$ elements earlier.

As an example, look at the first and second ionization energies for the second row elements (Table).
From $\ce{Li}$ to $\ce{Be}$, the $1^\text{st}$ ionization energy increases due to the greater effective charge on the outermost electron. From $\ce{Be}$ to $\ce{B}$, the $1^\text{st}$ ionization energy slightly decreases, as the electron we removing is in the p-subshell and is shielded by the inner s-subshell electrons. From $\ce{B}$ to $\ce{C}$, it increases again.

What about the $2^\text{nd}$ ionizations? Well from $\ce{Be}$ to $\ce{B}$, it increase, as the $2^\text{nd}$ electron from each is pulled from the s-subshell, but $\ce{B^+}$ has a greater effective charge than $\ce{Be^+}$. From $\ce{B}$ to $\ce{C}$, it decreases, as the $2^\text{nd}$ electron pulled from $\ce{C}$ is still in the less shielded p-subshell.

The $\ce{Be}$ to $\ce{B}$ trend in $2^\text{nd}$ ionization matches up with the $\ce{Li}$ to $\ce{Be}$ trend in $1^\text{st}$ ionization. The same is true for $\ce{B}$ to $\ce{C}$ (2nd) and $\ce{Be}$ to $\ce{B}$ (1st).


For future titration practices, you should never use the buret until the end. Most high quality burets (Class A) are limited to 50 mL (max) or even 25 mL. If it is a made up exercise by a teacher, they should not promote this approach either. You will have far more error from the incorrect volume measurements rather than the choice of your indicator. Also be careful with reporting concentrations: 0.1 M NaOH is not the same as 0.10 M NaOH.

Anyway, one cannot determine the indicator error from the given data. Practically, you would plot a pH titration curve on a graph paper. Determine the equivalence point (= the true point at which the titration is stoichiometrically complete) from the graph. Lets say, you get 40.56 mL (last digit estimated). With an indicator you get an end-point at 40.60 mL. The end-point is the point where the indicator "hints" you the end of the titration. Now the excess concentration of acetic acid calculated via the end-point is your indicator error.


Why is 1-ethylidene-4-methylcyclohexane chiral?

The strict criterion for a compound to display chirality is that it must not be superimposable upon its mirror image. Let's ignore the chair conformation of the ring for a while, and assume it adopts a planar conformation. You could draw a side-on view of the ring like this:

We can see that if we reflect the molecule (in the plane of the ring + double bond), its mirror image is not superimposable on itself, which makes it chiral:

In practice, the cyclohexane ring does adopt a chair conformation. That does not fundamentally affect the fact that the compound is chiral, as the presence of a chair conformation cannot "remove" this existing source of chirality (for example, there is no way for the hydrogen and methyl groups to swap places by virtue of a cyclohexane ring flip or a similar process).
Actually, this is very similar to the case of an allene, which you may or may not be familiar with. Notice how the conformation drawn above looks almost like this allene:

with the planar ring taking the place of the second double bond. These are examples of axial chirality (Wikipedia; IUPAC), where the chirality stems from the disposition of groups about an axis (in both cases, the axis in question is the C=C bond axis). Further explanation of axial chirality in allenes can be found in this question.

A note on nomenclature
The tetrahedral chiral centre can be named with the usual (R) and (S) stereodescriptors, as described in user55119's answer.
When it comes to the double bond, ChemDraw (and likely some other software) suggest that they can be named with the familiar stereodescriptors (E) and (Z) (the process is also described in user55119's answer). However, this is not fully appropriate.

The exact rules are complicated, but it boils down to the fact that the double bond in question is enantiomorphic: that means that when it is reflected, the configuration of the double bond is reversed. We can see that from the names generated above: the two mirror images have supposedly different configurations at the C=C double bond.
The descriptors (E) and (Z) are supposed to be used only for diastereomorphic double bonds, where the configuration is invariant (i.e. does not change) upon reflection. The large majority of stereogenic double bonds fall into this category, but this one does not.
In place of (E) and (Z), the preferred descriptors are respectively seqTrans and seqCis (see mykhal's answer, and P-92.1.1 (f) of the 2013 IUPAC Blue Book) .


Determining the group number of an element from its successive ionization energies

There are other little jumps to look out for, like the removal of the first electron from  a filled subshell. Consider the ionization energies (in eV) for the third period elements:

$$
\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline
\text{electron}&\ce{Ar}&\ce{Cl}&\ce{S}&\ce{P}&\ce{Si}&\ce{Al}&\ce{Mg}&\ce{Na}\\\hline
\ce{3p}&1521&-&-&-&-&-&-&-\\\hline
\ce{3p}&2666&1251&-&-&-&-&-&-\\\hline
\ce{3p}&3931&2298&1000&-&-&-&-&-\\\hline
\ce{3p}&5771&3822&2252&1012&-&-&-&-\\\hline
\ce{3p}&7238&5159&3357&1907&787&-&-&-\\\hline
\ce{3p}&8781&6542&4556&2914&1577&578&-&-\\\hline
\ce{3s}&11995&9362&7004&4964&3233&1817&738&-\\\hline
\ce{3s}&13842&11018&8496&6374&4356&2745&1451&496\\\hline
\ce{2p}&40760&33604&27107&21267&16091&11577&7733&4562\\\hline
\end{array}
$$

Notice how, in addition to the big jump between $\ce{3s}$ and $\ce{2p}$, there is a small jump between $\ce{3p}$ and $\ce{3s}$ for Ar, Cl, S, P, and Si. If you have five ionization energies, and you don't see the large jump, look for the small jump. If you don't see the small jump, what group is X in?

There is another even smaller jump to look for. Half-filled subshells are also slightly more stable than other configurations (except filled subshells). Thus, it should require a slightly higher increase in ionization energy to remove an electron form any configuration with $\ce{np^3}$ than from $\ce{np^2}$ or $\ce{np^4}$. Let's look at the step wise increase in ionization energy (in other words, how much higher is the ionization energy for the second electron than the ionization energy for the first electron and so on?).

$$
\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline
\text{electron}&\ce{Ar}&\ce{Cl}&\ce{S}&\ce{P}&\ce{Si}&\ce{Al}&\ce{Mg}&\ce{Na}\\\hline
\ce{3p}&-&-&-&-&-&-&-&-\\\hline
\ce{3p}&1145&-&-&-&-&-&-&-\\\hline
\ce{3p}&1265&1047&-&-&-&-&-&-\\\hline
\ce{3p}&1840&1524&1252&-&-&-&-&-\\\hline
\ce{3p}&1467&1337&1105&895&-&-&-&-\\\hline
\ce{3p}&1543&1383&1199&1007&790&-&-&-\\\hline
\ce{3s}&3214&2820&2448&2050&1656&1239&-&-\\\hline
\ce{3s}&1847&1656&1492&1410&1123&928&713&-\\\hline
\ce{2p}&26918&22586&18611&14893&11735&8832&6252&4066\\\hline
\end{array}
$$

You can see that the increase in ionization energy for the third electron to be removed from argon is larger than the increase for the second or fourth electrons. Your numbers don't have this little increase in them at all. Have you considered that X might be a transition metal?


Error in calculation of rate of reaction through graphs

Am I missing something?

Yes. You are missing the rate law and everything that goes along with it. It is only easy to fit zeroth order raw rate data. To produce linear fits, you need to consider the possibility that your reaction has a first or second order rate.

In addition to rate being defined as the (microscopic) change in concentration respect to time, you can also derive a rate law representing this change since the rate of the reaction is going to be proportional to things like the number of collisions between particles (which increases as there are more particles, i.e. higher concentration). So, we get a rate law that is something like the following:

$$\mathrm{rate}=kC^x$$

where $k$ is the rate constant, which is dependent on things like temperature and the activation barrier for the reaction, and $x$ defines the order of reaction, which is dependent on the mechanism and can be loosely interpreted as being related to the number of particles the must collide in the fundamental step(s). The value of $x$ is determined empirically (one such way is the experiment you did), though common values for simple reactions are integers, with $x=0,1,2$ being most common.

With both equations for rate in hand, we can do some to obtain integrated rate laws, which take the form $C(t)$: just the sort of equations that you are looking for. Note the minus signs in my equations. Rate is usually expressed based on consumption of reactants.

  $x=0$, zeroth order

$$\begin{align}-\dfrac{dC}{dt}&=kC^0 = k\\
-dC &= kdt \\
-\int_{C_0}^{C} dC &= \int_0^t kdt \\
C - C_0 &= -kt \\
C &= -kt + C_0
\end{align}$$

A reaction with a zeroth order rate will produce a graph of $C(t)$ that is linear.

  $x=1$, first order

$$\begin{align}-\dfrac{dC}{dt}&=kC^1 = kC\\
-\dfrac{dC}{C} &= kdt \\
-\int_{C_0}^{C} \dfrac{1}{C} dC &= \int_0^t kdt \\
\ln C - \ln C_0 &= -kt\\
\ln C &= -kt + ln C_0 \\
C &= C_0 e^{-kt}
\end{align}$$

A reaction with a first order rate will produce a graph of $C(t)$ that shows exponential decay. However, my derivation provides a linear equation: $\ln C = -kt + \ln C_0$. If you determine the natural log of each concentration and plot against $t$, you will get a line (if the reaction is first order).

  $x=2$, second order

$$\begin{align}-\dfrac{dC}{dt}&=kC^2\\
-\dfrac{dC}{C^2} &= kdt \\
-\int_{C_0}^{C} \dfrac{1}{C^2}dC &= \int_0^t kdt \\
\dfrac{1}{C} - \dfrac{1}{C_0} &= -kt \\
\dfrac{1}{C} &= -kt + \dfrac{1}{C_0} \\
C &= -\dfrac{C_0}{1-C_0 kt}
\end{align}$$

A reaction with a second order rate will produce a graph of $C(t)$ that shows power decay. However, my derivation provides a linear equation: $\dfrac{1}{C} = -kt + \dfrac{1}{C_0}$. If you determine the inverse of each concentration and plot against $t$, you will get a line (if the reaction is second order).


Net ionic equation for a synthesis reaction?

You are right, you can't write ionic equation for that reaction. Quick search showed me that even the mechanism doesn't really involve any kind of ions, which, I'd say, is completely logical, considering reaction conditions, which favor homolytic, not heterolytic  bond cleavage.

You can only write ionic equations for reactions where ions are in charge, and they are preferably formed in polar medium.

P. S. The mechanism of S+O2 reaction isn't researched much, so here is the study of H2S+O2 reaction .


Is benzocyclobutadiene aromatic or anti-aromatic?

Benzocyclobutadiene has an aromatic benzene ring as well as an anti-aromatic cyclobutadiene ring. This gives it the characteristics of both aromatic and anti-aromatic compounds.
Now, according to Cyclobutadiene and its related compounds, [1,p 180]:

Benzocyclobutadiene, the monobenzo derivative of cyclobutadiene, is intermediate in structure between the unstable parent compound, cyclobutadiene and it stable dibenzo analog biphenylene. Benzocyclobutadiene and its derivatives resemble cyclobutadiene rather than biphenylene in being known as only transient reaction intermediates.

Reading this, we can say that benzocyclobutadiene resembles cyclobutadiene more than biphenylene and so has a dominant anti-aromatic character. This means that like cyclobutadiene, the cyclobutadiene part of the ring undergoes dimerisation to form a more stable compound.
Note: The preview of the book referenced has an entire chapter based on just benzocyclobutadiene. If interested, you can go through it for more information
References:

Cyclobutadiene and Related Compounds; M.P. Cava, M.J Mitchell; Chemistry Department, Wayne State University, Detroit 1967. ISBN 978-01-2164-450-5. [Preview]


Why doesn't nitric oxide react with water?

Although nitric oxide is thermodynamically unstable, it is kinetically stable as its decomposition rate is very slow[ref] and so it is considered stable and won't normally react to anything.

So, nitric oxide wouldn't normally hydrolyze easily. But in the presence of air, it forms nitrous acid $(\ce{HONO})$:

$$\ce{4NO + O2 + 2 H2O -> 4 HNO2}$$

However, nitrous acid is very unstable and quickly disproportionates:

$$
\begin{align}
\ce{2 HNO2 &-> NO2 + NO + H2O &(cold dil. soln.)}\\
\ce{3 HNO2 &-> HNO3 + 2 NO + H2O &(warm conc. soln.)}
\end{align}$$

Since $\ce{HNO2}$ is required to prepare diazonium salts (a versatile compound in organic chemistry), it is prepared in-situ by reaction of $\ce{NaNO2}$ and $\ce{HCl}$. Due to its instability, it is prepared for one-time use and is not stored for further preparations.

Reference: J. Phys. Chem., 1958, 62 (3), pp 360–361 https://doi.org/10.1021/j150561a030


Is heat an extensive or intensive property?

In thermodynamics, heat is not a function of state but a process function. So for the question

  Is heat a property of matter?

the answer is no.

  ... is it extensive or intensive?

Since the concept of intensive and extensive properties only applies to physical properties that are a function of state, heat is neither intensive nor extensive.

As Chester said in the comments, it is worth noting that heat (as well as work) applied to a system can cause changes in the thermodynamic equilibrium state of the system.


The Z isomer is destabilized by comparison to the E isomer due to steric strain. This is a consequence of the isopropyl groups being on the same side of the double bond in the Z isomer, resulting in electron-electron repulsion between the groups. This also, at least partially, accounts for the increase in the bond angle that should be observed in the Z isomer: increasing that angle also increases the distance between the isopropyl groups, minimizing the mutual repulsion between them (admittedly, this is at the expense of the other angles, but the bulkiness of the isopropyls by comparison to hydrogens necessitates that distortion from idealized geometry). A similar effect occurs with the geminal isopropyl groups in the 1,1- isomer.

Additionally, the degree of substitution of an alkene has a dramatic effect on stability. The presence of alkyl groups on the carbons of an alkene is generally stabilizing (as can be empirically confirmed by examining thermodynamic data, i.e., heats of hydrogenation and heats of combustion). Evidently, calculations grounded in molecular orbital theory indicate that the mixing of $\sigma$ type molecular orbitals of alkanes with the $\pi$ orbitals of less-substituted alkenes is typically energy-lowering overall, and that the $\pi^{*}$ antibonding orbitals are sufficiently close in energy to the $\sigma$ orbitals of alkanes that hyperconjugative stabilization occurs (albeit relatively small in magnitude, since the energies are not very close). In the 1,1- isomer, only one carbon of the alkene can be stabilized by hyperconjugation, while both carbons should be able to benefit from it in the 1,2- isomers. (Speculatively, I would also think that the extent of orbital overlap should be poorer in the 1,1- isomer, since the geminal isopropyl groups would likely be staggered with respect to one another, which might prevent proper alignment of the alkyl group orbitals with those of the alkene $\pi$ system.)


Briefly, the cause for surface tension is the difference in the energetic state between a molecule at the surface compared to a molecule inside the liquid phase: the molecule inside has lower energetic state as it has more inermolecular interactions (or, to get to the surface, a molecule needs to break intermolecular bonds, which needs energy). This is more pronounced with higher polarity or when ions (charges) are around as the interactions are then stronger.

You can increase the polarity of a (polar) solvent to get higher surface tension, by dissolving salts.


What does a molecules color have to do with its bond/orbital energies?

Rather than give answers to your specific question I try to explain what happens in general as there is some confusion in your question. You should really consult a book on spectroscopy, of which there are many, for a thorough explanation.

In a molecule the arrangement of electron in orbitals is called a state, hence ground state, first exited state etc. When all the electrons a molecule has fill up all the orbitals with no gaps left, this is the ground state. They will fill up half of all the orbitals, two to an orbital.  When a photon is absorbed and it has sufficient energy (usually in the visible or UV part of the spectrum) an electron is promoted from one of the bonding orbital to a anti-bonding orbital which has higher energy. This arrangement of electrons is called an excited state. This transition between states is what gives molecules and atoms their colour.

In iodine the difference between the lowest excited state and the ground state occurs in the visible part of the spectrum. (for simplicity, I'm ignoring any vibrational transitions accompanying the electronic excitation). Benzene has a larger energy gap between ground and excited state than I$_2$ and so absorbs only in the UV and looks transparent to us.

The same basic idea applies to all atoms and molecules. Where the molecules or atoms absorb is a function of the exact electronic energy difference between ground state and excited state of the particular species. An understanding of the nature of the bonding and so energetics allows us to predict trends in absorption spectra, but exact values can only be obtained from quantum calculations.


Is there a reliable way to determine if a chemical system is more suited to be calculated using a multireference method instead of DFT?

You should have a look at the so called FOD (Fermi Occupied Density) analysis by Grimme et al. That is accessible by ORCA 4 very easy and straight forward by specifying "FOD" as a keyword. [1-3]

From ORCA Input Library [1]:

  The FOD plot above (plotted with Chemcraft) indicates a significant and delocalized FOD. Multiconfigurational methods should be used to study this molecule and any single-reference result (HF, DFT, MP2, CC should be regarded with suspicion)

Via Multiwfn you can access the "local total/dynamic/nondynamic electron correlation function" and "total/dynamic/nondynamic electron correlation index" [4,5].

Visual abstract from [5]. You can see the dynamic and non-dynamic correlation from which you also can see if single reference methods are acceptable or not.

[1] ORCA Input Library, FOD analysis
[2] S. Grimme, A. Hansen, Angew. Chem. Int. Ed. 2015, 54 (42), 12308-12313
[3] C. A. Bauer, A. Hansen, S. Grimme, Chem. Eur. J. 2017, 23 (25), 6150-6164
[4] E. Ramos-Cordoba, P. Salvadorc, E. Matito, Phys. Chem. Chem. Phys. 2016, 18, 24015-24023
[5] E. Ramos-Cordoba, E. Matito, J. Chem. Theory Comput. 2017, 13 (6), 2705–2711


Why should 2,6-Dimethyl-4-nitrophenol be more acidic than 3,5-Dimethyl-4-nitrophenol

In the 3,5-dimethyl isomer, steric effects come into play and influence the acidity.

Resonance structures like B in the figure below, that stabilize the phenoxide anion explain why 4-nitrophenol ($\ce{R=H}$) is more acidic than phenol.  Resonance structure B requires that the nitro group be planar with the aromatic ring.  If methyl groups are placed adjacent to the nitro group ($\ce{R=CH3}$), the steric interactions between the two methyl groups and the nitro group becomes large and forces the nitro group to rotate out of the aromatic plane,  Hence resonance structure B can no longer contribute and the compound is less acidic.


Gibbs Free Energy and Equlibrium Constant

Q is supposed to be calculated in terms of partial pressures, not moles.  $\Delta G^0$ is based on a reversible process you have devised to take the pure reactants in stoichiometric proportions and 1 atm (actually, 1 bar) and 298 K and convert them to pure products in stoichiometric proportions, also at 1 atm. and 298 K.


What is the difference between Molecularity and Order Of Reaction?

In fact, molecularity and order of reaction are identical if the reaction is elementary. They may become different when the reaction is complex and has a mechanism.

The molecularity of a reaction is defined as the number of molecules or ions that participate in the rate determining step. A mechanism in which two reacting species combine in the transition state of the rate-determining step is called bimolecular. If a single species makes up the transition state, the reaction would be called unimolecular. The relatively improbable case of three independent species coming together in the transition state would be called termolecular.

The order of a reaction is an experimentally derived number. Let's consider the general reaction: $\ce{A + B-> C + D}$
 The reaction rate$=\ce{k[A]^n.[B]^m}$ where the rate constant $k$ is a proportionality constant that reflects the nature of the reaction, $\ce{[A]}$ is the concentration of reactant $\ce{A}$, $\ce{[B]}$ is the concentration of reactant $\ce{B}$, and $n$ & $m$ are the exponents used to fit the rate equation to the experimental data. Chemists refer to the sum $n + m $ as the kinetic order of a reaction.

In a simple bimolecular reaction $n$ and $m$ would both be $1$, and the reaction would be termed second order, supporting a mechanism in which a molecule of reactant $\ce{A}$ and one of $\ce{B}$ are incorporated in the transition state of the rate-determining step. A bimolecular reaction in which two molecules of reactant $\ce{A}$ (and no $\ce{B}$) are present in the transition state would be expected to give a kinetic equation in which $n=2$ and $m=0$ (also second order).


Just heat it up: Isonitriles can be thermally rearranged to nitriles.

$$\ce{R-NC ->[\Delta] R-CN}$$

Please have a look at some references, such as:

Michael J. S. Dewar, M. C. Kohn, Ground states of $\sigma$-bonded molecules. XVI. Rearrangement of methyl isocyanide to acetonitrile, J. Am. Chem. Soc., 1972, 94, 2704-2706 DOI

Michael Meier, Barbara Mueller, Christoph Ruechardt, The isonitrile-nitrile rearrangement. A reaction without a structure-reactivity relationship, J. Org. Chem., 1987, 52, 648-652
DOI

UPDATE

As far as the sequence outlined in the question is concerned:

The synthesis of methyl formamide from methyl isonitrile is pretty much the opposite of how things are done in real life. Isonitriles, used in the Passerini and the Ugi reaction, have shown to be available by dehydration of formamides, e.g. using $\ce{POCl3}$. For an example with cyclohexyl isonitrile, see Org. Synth., 1961, 41, 13.
N-Methyl formamide is far from being unstable! Release of methyl amine, which, by the way, has a BP around -6 °C might be possible under alkaline conditions.
Making methanol from methylamine, again, is a rather pointless mind game. In real life, methylamine is produced by reacting methanol with ammonia at hight temperatures in the presence of acidic catalysts.
You definitely want to make methyl chloride in the lab! With a BP around -24 °C it's very easy to handle, and the GHS classification with H351 (Suspected of causing cancer) and H373 (May cause damage to organs through prolonged or repeated exposure), this is a real gem. Yes, I'm kidding! Don't do it!
For a nucleophilic substitution with cyanide, the high volatility of methyl chloride again probably doesn't make your life easier, but just shorter.

In summary, my suggestion is simple: Forget about the proposed synthesis! It is laborious, difficult to handle in the lab, and an economical nightmare.


The assessment of nodal planes is not as trivial as it might seem at first, because they are usually not planes at all.

Orbitals can be explained in terms of symmetry operations. There are three different operations and each of them has a unique element (in cartesian coordinates):

Rotation: Axis
Mirror: Plane
Inversion: Point

When we talk about hydrogen type atomic orbitals, we usually refer to their cartesian form. It is already fairly obvious for the $\ce{2s}$, that there is no nodal plane, while there are certainly nodes.

The representations of $\ce{s}$ orbitals are always trivial with respect to the total symmetry of the external field (= Point group). That means any symmetry operation, which is given through the external field will be matched by the $\ce{s}$ orbital. Another term for this is totally symmetric.

The representation of $\ce{p}$ orbitals are always antisymmetric with respect to inversion in any given external field.

The representation of $\ce{d}$ orbitals is always antisymmetric to two mirror planes. That excludes of course the $\ce{d_{z^2}}$ orbital, which is not a cartesian, but a spherical representation. (Note: In cartesian space there are 6 $\ce{d}$ orbitals. As quantum numbers only allow 5 spherical orbitals, they have to be transformed. See for example at chemissian.com or Schlegel and Frisch.) Early basis sets for quantum chemical calculations used the cartesian representation, because it wa easier to code and calculate.

Due to the geometric arrangement of nuclei in a molecule, the external field is given through the point group of the molecule. Every molecular orbital can be expressed through symmetry operations of that point group.

Cyclopropane has $D_\mathrm{3h}$ symmetry. And every molecular orbital has to respect that. I have already written about orbitals these orbitals in Why does cyclopropane give bromine water test?
However, I would like to shed some more light on this, because I believe the picture you found is not quite correct.
The following picture shows the depicted orbitals from your source, as well as the LUMO.

Where W1 corresponds to HOMO-2, W2 and W3 correspond to both HOMO, W4 corresponds to LUMO+3, finally W5 and W6 correspond to LUMO+2.
The ordering of these orbitals is taken from a BP86/cc-pVTZ calculation of Gaussian09. A methodology that I believe gives fairly accurate results for molecular orbitals. (It is however noteworthy, that unoccupied orbitals are constructed from occupied orbitals. That means, that their physical meaning is limited.)

Let's have a closer look at the orbitals. The HOMO-2 is of $A_1'$ symmetry and therefore behaves like an $\ce{s}$ orbital. There is no antibonding nodal surface in this orbital. That means at any bonding axis of the ring structure the sign of the wavefunction does not change. There is however a nodal surface that does not affect bonding, as was already pointed out in some comments.

The HOMOs are of $E'$ symmetry and therefore behave like $\ce{p_x}$ and $\ce{p_y}$ orbitals. These orbitals have one antibonding nodal surface. In a result of symmetry restrictions these nodal surfaces are perpendicular to each other.

This is also true for the LUMO+2s. In addition to this there is another antibonding nodal surfaces with respect to the $\ce{C-H}$ bonds.

The LUMO again is of $A_1'$ symmetry and behaves like a $\ce{s}$ orbital. There is again no antibonding nodal surface with respect to the $\ce{C-C}$ bonds, but the main electron density would be outside of any bond and there is an antibonding nodal surfaces with respect to the $\ce{C-H}$ bonds.

The LUMO+3 is of $A_2'$ symmetry and does therefore not behave like any other atomic orbital. Here we find three antibonding nodal surfaces with respect to the $\ce{C-C}$ bonds.

The bonding situation in cyclopropane is very complicated and can be explained in many different ways. It is not as obvious, that the molecule is a $\sigma$ aromate, having in plane $\pi$ orbitals (HOMO). The explanation of its molecular orbitals were first composed in terms of a Walsh like behaviour on the basis of an extended Hückel calculation with a minimum basis.

When you disregard the $\ce{C-H}$ bonds and only allow three carbon $\ce{sp^2}$ and three carbon $\ce{p}$ orbitals, then you can only arrive at the boning picture that was depicted in your source (and many other publications), as you can only have six molecular orbitals formed from these. There are, however, many more atomic orbitals that mix into the final wavefunction. First of all it is necessary to state, that the used $\ce{sp^2}$ orbitals are not of the ideal $\frac13\ce{s}+\frac23\ce{p}$ composition (see more about that in the linked question and the answer of ron therein).

Now finally for the discussion about the correlation of nodal planes and increasing energy. It is a very common statement, that is missing it's most important limitation. It is only completely true for orbitals of the same symmetry. (This is also a necessary requirement.)

For example, the energy of the $\ce{s}$ orbital series increase with the main quantum number, as you add one more nodal surface per one increment in $n$: $\ce{1s->0; 2s->1; 3s->2;...}$. The same applies to the $\ce{p}$ orbital series: $\ce{2p->1; 3p->2; ...}$.

Now this statement became popular, when talking about aromatic systems. In the framework of Hückel's molecular orbital method, this must be true, since all regarded orbitals are of the same symmetry.

However, it is usually true, that an orbital with fewer nodal planes is more stable than another with more, but this is more a gut feeling, than actual, factual science.


Synthesise the CBZ derivative of glycine from glycine and benzyl chloroformate

I agree with Ron's answer except for d. Using an extra equivalent of NaOH is the right procedure since you need to "mop up" the HCl that is produced when $\ce{CbzCl}$ reacts with the amino group. What is not correct is this:
When you wash the reaction mixture with ether you get rid of any impurities and your product should stay in the aqueous phase. So you have to acidify the aqueous layer and not the ether one. pH 1 should be fine for the Cbz unless you leave it there and go away for the weekend.


How can a molecular orbital with a nodal plane still be considered "bonding"?

The number of nodal planes isn't the determining factor. Yes, having more nodal planes increase the energy of the orbital, but you need to consider the fact that some MOs (specifically the anti bonding ones) are raised in energy. Overall, the total energy of the orbitals is the same as before, so you stabilize some and destabilize others. Take a look at the MOs and you can see that the there's two bonding MOs with one nodal plane perpendicular to the plane of the benzene with $E_{1}$ symmetry. Likewise, there's a pair of MOs with two nodal planes that with $E_{1}$ symmetry that are anti-bonding. These pairs happened to be correlated in that the increase in energy of the anti-bonding orbitals is essentially matched by the stabilization of the bonding orbitals.


Confused with 'discharge potential',preferential discharge theory

I can answer your first two questions.

  I am confused by 'require', do ions upon discharging gain energy or lose energy?

Ions lose free energy during discharging, as you are referring to the spontaneous process of transferring charge at the electrodes with conversion of some of the ions into neutral species (in general you are speaking about the process described in the next paragraph).

For a process to proceed spontaneously, the free energy change for the system (combined half-cells, say) has to be negative. The Nernst equation tells us that the free energy difference is given by $$\Delta G = -nFE$$ where E is the total cell potential. This value must be negative for the cell to discharge, which means E is required to be positive. This is what is meant by "require" (language is used in funny ways, ions don't really require anything). You have more choices than just discharging the cell: you can charge it (creating more of some of the ions). Unless the potential is such to cause discharge you may actually produce more of the ions that are disfavored under spontaneous (discharging) conditions.

  are ions in general not less stable than neutral counterparts?

You should not attempt to generalize, as this depends on the context (details of the reaction in question). An ion may be the most stable state of an atom in one context (an oxidizing environment, say) whereas it will be unstable in another (a corresponding reducing environment). Moreover, you can apply a charging potential and thereby invert the sense of stability, rendering one species more stable than the other.


What type of a reaction is it?

I would describe it first and foremost as a redox reaction. Iron is oxidised and hydrogen is reduced; the half equations would be:
$$\ce{2Fe -> Fe^2+ + Fe^3+ + 5e-}$$ $$\ce{H2O + 2e- -> H2 + O^2-}$$

  Why have we combined $\ce{FeO}$ and $\ce{Fe2O3}$ resulting in the formation of $\ce{Fe3O4}$?

This is a good observation. $\ce{Fe3O4}$ contains both iron(II) and iron(III) ions and is sometimes written as $\ce{FeO \cdot Fe2O3}$. The dot in the middle signifies that there is some form of bonding between the two compounds (another common example of this is water of crystallisation). $\ce{Fe3O4}$ is a compound and not a mixture because it does not consist of two separate $\ce{FeO}$ and $\ce{Fe2O3}$ phases but rather it is a single crystal structure containing $\ce{Fe^2+}$, $\ce{Fe^3+}$ and $\ce{O^2-}$ ions.

  $\ce{Fe3O4}$ has a cubic inverse spinel structure which consists of a cubic close packed array of oxide ions where all of the $\ce{Fe^2+}$ ions occupy half of the octahedral sites and the $\ce{Fe^3+}$ are split evenly across the remaining octahedral sites and the tetrahedral sites.


Final temperature after caustic dilution: Interpreting the graph

The labels on the curves are the initial temperatures of the 50% caustic soda solutions.


I suggest to stay with commercially available, read-made solutions. These products are thoroughly tested and certified.

They typically contain around 5% of various benzalkonium chlorides, such as N-Myristyl-N-benzyl-N,N-dimethylammonium chloride, as biocides.

UPDATE As far as the mildew is concerned: The orange ladybird (16 white spots) and the 22-spot ladybird apparently feed on mildew. If these two are naturally found in your region, get some. This is as eco-friendly as it can get to keep mildew in the garden close to a minumum.


Why Acetone does not behave like its computational values?

It seems that the results of the calculations are more or less fine and the OP just misinterpreted the NIST data. As I said in my comment above, NIST does not claim that $\lambda_{\mathrm{max}}=276 \, \mathrm{nm}$. Clearly only a small region of wavelength is shown on the graph and in the paper[1] referenced on the NIST page it is said (emphasis mine):

  There are two diffuse ultraviolet bands; the first at 2800 Å. is very
  weak with its oscillator strength $f \sim 0.0004$ and the second at
  about 1900 Å  is moderately intense with a maximum extinction
  coefficient $\epsilon_{\mathrm{m}} \sim 1000$. McMurry has identified
  the 2800 Å. band as a forbidden $\pi^* \leftarrow n$ transition
  involving excitation of a non-bonding $\ce{O}$ electron to an
  anti-bonding $\pi$ orbital between the $\ce{C}$ and $\ce{O}$ of the
  carbonyl group.

A high-resolution photoabsorption spectrum in the energy range 3.7-10.8 eV can be found, for instance, in this recent study[2] and is more or less consistent with the result of the OP's calculations.

Noel S. Bayliss, Eion G. McRae, J. Phys. Chem., 1954, 58 (11), 1006–1011.
M. Nobre, A. Fernandes, F. Ferreira da Silva, R. Antunes, D. Almeida, V. Kokhan, S. V. Hoffmann, N. J. Mason, S. Eden, P. Limão-Vieira, Phys. Chem. Chem. Phys., 2008, 10, 550-560. (available at researchgate.net)


Why do electron donating groups decrease the acidity of aromatic carboxylic acids?

You seem to pay too much importance to these individual structures. However, note that there are no individual resonance structure in existence, but only the resonance hybrid - a combination of all these resonance structures in proportion to their stability.

The final three structures that you've drawn have a really negligible contribution to the final resonance hybrid of the benzoate anion. Why, you ask? Because they exhibit charge separation: creation of an additional positive and a negative charge.

Hence, your conclusion is wrong, because it's based on the invalid premise that the third (and by extension, the fifth) resonance structures have a major role in the resonance hybrid. The resonance hybrid in fact has no appreciable positive charge on the ring carbons.

The correct way to explain this is that you should look at the resonance hybrid as a whole, which is negatively charged. Adding an electron donating group to the ring would increase the negative charge further, thereby destabilizing the anion.


In aromatic compounds, the enol form is more stable than the keto form due to the aromatic character. In normal conditions, aromatic compounds exist in enol form but in basic conditions, it tautomerises to keto form.  Phloroglucinol predominantly exist in enol form under normal condition.

iodoform test

Compounds containing ketomethyl group($\ce{-COCH3}$) will give iodoform test. e.g.- propanone, acetophenone etc. Thus phloroglucinol will not give iodoform test because it does not contain ketomethyl group.

ceric ammonium nitrate

Generally, CAN test is done for alcohols which give pink or red colour. But for phenols and phenolic compounds, it gives brown or black test. So, this test helps to differentiate phenols from alcohols. See the detailed question on Chem.SE

2,4 DNP test

Phloroglucinol predominantly exist in enol form under normal condition and thus not give  2,4 DNP test. However, in basic condition (pH- 11-13) it forms 3,5-dihydroxy-2,5-cyclohexadienone dianion,  the deprotonated intermediate in the interconversion of the two forms, referred to as the enolate anion which might give the 2,4 -DNP test. See here

One final note

Don't take tautomerism into account while experimentally determining the functional group in organic compounds . If you consider tautomerism, there will be two functional group for the compound and you will get confused as of which functional group should be considered. For ex- propanone tautomerises to form prop-1-en-2-ol which contains an alcoholic group and then you will get confused which test should be used- Ester or CAN test(for alcohol) or 2,4 DNP test (for ketone).


Can salt and water be obtained from a reaction between a gas and an acid?

If you want to get salt, then you should look for a volatile base.
$\ce{NH3}$ does our job here.
$$\ce{2NH3 + H2SO4 -> (NH4)2SO4}$$

However, I am not sure if water could be formed. If there exists something like $\ce{NH4OH}$ vapours, then sure, water will be formed.


How to derive the Gibbs free energy for an ideal gas?

The fundamental relation for an Ideal Gas, in the entropy representation is

$$S(U,V,N) =
   NS_0 + NR\ln\left[ \left( \frac{U}{U_0} \right)^c
                      \left( \frac{V}{V_0} \right)
                      \left( \frac{N_0}{N} \right)^{c+1} \right]$$

The Gibbs Free Energy, is the Legendre Transform $G \equiv U[T,p]$

so, $G = U + pV -TS$

First, let's get the equation of state $S(T,V,N) $ from the the fundamental relation. We eliminate $U$ using $U = cNRT$

$$S(T,V,N) =
  NS_0 + NR\ln\left[ \left( \frac{T}{T_0} \right)^c
                     \left( \frac{V}{V_0} \right)
                     \left( \frac{N_0}{N} \right) \right].$$

Also, for an Ideal Gas, $p =\frac{NRT}{V}$

$$G(T,p,N) =
  cNRT - TNS_0
  - NRT\ln\left[ \left( \frac{T}{T_0} \right)^c
                 \left( \frac{p_0}{p} \right)
                 \left( \frac{N_0}{N} \right) \right] + NRT $$

Exercise: Find $G_0 = G(T_0,p_0, N_0)$ and substitute into the fundamental relation derived above to get rid of $S_0$


How does one analytically quantify whether or not a compound is a SIRT1 activator?

It's important to distinguish SIRT1, which is a gene, from the protein it encodes, sirtuin 1 (some papers use SIRT1 for the protein too, but to avoid confusion, I'll use Sirt1). When looking at compounds that affect gene expression, one method of measuring the effects is measuring the change in mRNA levels for that gene when the compound is administered, usually with northern blotting or qPCR. In this case, because the gene encodes a protein, quantifying the protein levels also tells us about the expression of the gene.

The SIRT1 gene doesn't have any regulatory role itself, so in this case, people are more interested in the Sirt1 protein. While we can measure protein concentrations, Sirt1 is an enzyme that deacetylates proteins and resveratrol has been proposed to directly alter the activity of the enzyme (it's also known to increase SIRT1 expression, but this isn't the controversial aspect), so it's important to examine the effects of resveratrol on Sirt1's ability to deacetylate certain proteins. The way this is done is to expose various proteins/peptides to Sirt1 and look at whether they deacetylate. One method of measuring this is to use a special peptide as the substrate to which a fluorophore can be attached if the acetyl group is cleaved.

The reason resveratrol's effects have been debated is that the early studies that showed large activation effects were later found to only occur because of the fluorophore on the substrate and not resveratrol alone. Some work since then using other detection methods have shown activation effects: This paper tried many different substrates on a microarray using fluorescently labelled antibodies as well as kinetic measurements by mass spectrometry and found that resveratrol increased Sirt1's activity for some substrates, decreased it for others, and left many the same, though they didn't offer much in the way of structural insights that might be responsible. I haven't done a thorough literature search, but it seems like the debate is less contentious now that people are using other assays to examine activity.

Just to summarize your questions, it's not hard to quantify whether a molecule increases the expression of SIRT1, but because Sirt1 is an enzyme, any kind of label you add to measure activity could have its own effects on that activity. Any detection method that can measure the deacetylation of substrates could potentially be used for studying Sirt1 activity, and there are many, but label-free techniques like mass spectrometry are especially useful because they reduce complications in simulating biological conditions.


Yes, this is how decaffeinated coffee is made. If I recall correctly, this is done commercially by dissolving the caffeine in dichloromethane (DCM), or more recently in supercritical carbon dioxide. This switch is probably over concerns regarding chronic industrial exposure to DCM, which is mildly toxic.

From personal experience, I have extracted caffeine from instant coffee using a separatory funnel with a system of water and dichloromethane. Purification by sublimation finished the job.


Mathematical explanation behind dq = TdS

$\def\d{\mathrm{d}}\def\c{_\mathrm{C}}$Firstly, note that a more correct version of your equation is

$$\delta q_\mathrm{rev}=T\d S.\tag1$$

We use a Greek delta to signify that heat is not, in general, a state function. Its value during a change does not simply depend on the beginning and end points but also on the way which the change is traversed. Subscript '$\mathrm{rev}$' is for a reversible process.

Secondly, I should warn that the proofs I supply are in no way rigorous but should suffice for a standard thermodynamics course.

Thirdly, I have used a small $q$ for heat, reserving capital $Q$ for a partition function.

Classical thermodynamics

The classical definition of entropy in thermodynamics is due to German scientist Rudolf Clausius. He showed that, for a cyclic process, the following inequality holds:

$$\oint\frac{\delta q}{T} \leq 0.\tag{2}$$

  Lemma. For a Carnot cycle, the following line integral over a circular path equals zero: $$\oint\c \frac{\delta q_\mathrm{rev}}{T} = 0.\tag{L}$$

Proof

Recall and remember that in a Carnot cycle efficacy $\eta_\mathrm{C}$ of a system may be expressed in two equivalent ways, by a heat quotient and a temperature quotient.

$$\eta\c=1+\frac{q_2}{q_1}\tag{3a}$$
$$\eta\c=1-\frac{T_2}{T_1}\tag{3b}$$

Substrating one from the other yields

$$\frac{q_2}{q_1} + \frac{T_2}{T_1}=0\tag{4}$$

which, through algebraic manipulation, is the same as

$$\frac{q_1}{T_1} + \frac{q_2}{T_2}=0\tag{5}.$$ Hence for a Carnot cycle, reversible by definition, we optain

$$\oint\c \frac{\delta q_\mathrm{rev}}{T} = 0.\tag{L}$$

I will omit the subscript $_C$ for Carnot in the integral when I mention in the text or it is obvious by the same subscript on some other quantity.

Proof of Clausius integral inequality

Start with the first law of thermodynamics.

$$\d U = \delta q + \delta w \tag6$$

Therefore, for a Carnot engine,

$$w\c = -\oint \delta q_\mathrm{rev}.\tag7$$

Let us take some other engine, called engine $2$. Again from the first law of thermodynamics,

$$w_2 = -\oint \delta q_2.\tag8$$

Assume that, contrary to the theorem,

$$\oint \frac{\delta q_2}{T}>0.\tag9$$

Let us join these two cycles to form a composite engine. The work done by this system will be

$$w_\mathrm{total} = w\c + w_2 = -\oint(\delta q_\mathrm{rev} + \delta q_2) = -\oint\delta q_\mathrm{total}. \tag{10}$$

By our lemma $\mathrm{(L)}$, assumption $(9)$ and result $(10)$, we see that

$$\oint\frac{\delta q_\mathrm{rev} + \delta q_2}{T} \stackrel{(10)}{=} \oint \frac{\delta q_\mathrm{total}}{T} \stackrel{\mathrm{(L)},(9)}{>} 0 \tag{11}.$$

Now change the sign or direction of the Carnot engine, and its size so that no work is done in the composite cycle with engine $2$. (This should always be possible via Carnot's theorem which I will not prove here.) Then

$$w_\mathrm{total} = \oint\delta q_\mathrm{total} = 0. \tag{12}$$

If we break cyclic line integrals apart into $y-1$ smaller integrals that add up to the cyclic integral, we achieve

$$W_\mathrm{total} = \sum_{i<y} \int_{b_i}^{b_{i+1}} \delta q_i = \sum_{i<y} q_i(b_j) \stackrel{(12)}{=} 0,\tag{13a}$$
$$\oint \frac{\delta q_\mathrm{total}}{T} = \sum_{i<y} \int_{b_i}^{b_{i+1}} \frac{\delta q_i}{T_i} = \sum_{i<y} \frac{q_i(b_j)}{T_i} \stackrel{(11)}{>} 0.\tag{13b}$$

The terms $q_i(b_j)$ may be both positive and negative but in equation $(13a)$ cancel out to give zero. In equation $(13b)$ the terms are weighted by temperatures in such a way that positivity wins. Therefore, on the whole, negative $q_i(b_j)$ terms have been divided by larger temperatures whereas positive terms possess smaller temperature denominators. Physically, this implies that heat is transferred from a lower temperature reservoir to a higher temperature bath without doing any net work. This is in violation of the second law of thermodynamics (Clausius statement).

Therefore, our assumption $(9)$ that

$$\oint \frac{\delta q_2}{T}>0$$

is actually false. The result

$$\oint \frac{\delta q_2}{T}\leq 0$$

immediately follows.

  Corollary. For any reversible process, not just a Carnot cycle, $$\oint \frac{\delta q_\mathrm{rev}}{T} = 0.$$ In other words, temperature becomes an integrating factor in a reversible process.

Proof

By Clausius integral inequality $(1)$, we know immediately that

$$\oint \frac{\delta q_\mathrm{rev}}{T} \leq 0. \tag{14}$$

Assume again to the contrary that

$$\oint \frac{\delta q_\mathrm{rev}}{T} < 0. \tag{15}$$

Since this is a reversible process, we may change the direction of the engine. Only the sign of the integral will alter.

$$\oint \frac{\delta q_\mathrm{rev}}{T} > 0. \tag{16}$$

If one proceeds analagously to the proof of the Clausius integral inequality, this will lead to a contradiction. Thus, assumption $(15)$ is not valid, and by $(14)$ we are left with an equality.

Why bother with all of this?

We have now proven that for any reversible process,

$$\oint \frac{\delta q_\mathrm{rev}}{T} = 0. \tag{17}$$

This means that we may define a change in a new state function $S$ which we call entropy such that

$$\Delta S= \int_{S_1}^{S_2} \frac{\delta q_\mathrm{rev}}{T}\tag{18},$$

or in differential form,

$$\d S = \frac{\delta q_\mathrm{rev}}{T}. \tag{19}$$

This is the definition of a differential of entropy. There is nothing else to derive; what we showed above is that we may indeed define such a state function. To achieve your equation, multiply both sides of equation $(19)$ by temperature.

$$T\d S = \delta q_\mathrm{rev}. \tag{CT}$$

Statistical thermodynamics

Another way to go would be to define entropy in statistical thermodynamics. Then the formula $\mathrm{(CT)}$ may be derived.

Statistics in general

For a discrete random variable $X$, entropy is defined in statistics to be the function $\mathrm{H}$ such that

$$\mathrm{H}(X)= -\sum_k p_k\ln p_k \tag{GS}$$

where small $p$ denotes probability.

Boltzmann statistical thermodynamics

Say we have a large number $\mathcal{A}$ of isolated systems which we together call an ensemble. Suppose the system may be described as

\begin{align}\begin{split} \text{energies}\ \ &E_1\ \ &E_2\ \ldots &E_n\ \ldots\\ \text{number of such systems}\ \ \ \  &a_1\ \ &a_2\ \ldots\ &a_n\ \ldots\end{split}\end{align}

where

$$\mathcal{A} = \sum_i a_i.\tag{20}$$

The systems of different energies in this ensemble are distinguishable. The systems that have the same energy, i.e., are degenerate, are indistinguishable. If we supplied each system regardless of energy with an index of energy, the number of ways to order these indices would be

$$\mathcal{A}!.\tag{21a}$$

But as we said, the systems of equal energy are indistinguishable. The number of ways of ordering each of these is

$$a_j!.\tag{21b}$$

So to avoid overcounting we must divide:

$$K(a_1,a_2,\ldots,a_n,\ldots) = \frac{\mathcal{A!}}{\prod_i a_i!}.\tag{21c}$$

You may think of this $K$ as the total number of anagrams of words with $\mathcal{A}$ letters where each letter repeats itself $a_j$ times. Boltzmann used the big letter $W$ instead of $K$ and defined entropy to be

$$S = k_\mathrm{B}\ln W.\tag{BD}$$

The natural logarithm guarantees, among other things, that entropy is not a huge quantity, and that entropy for different (independent) systems is additive. In other words, if

$$W_\mathrm{total} = W_\mathrm{A}W_\mathrm{B}$$

then

$$S_\mathrm{total} = S_\mathrm{A} + S_\mathrm{B}.$$

Boltzmann entropy as a special case of statistical entropy

Begin with the statistical definition.

$$\mathrm{H}(X)= -\sum_k p_k\ln p_k \tag{GS}$$

Say each and every state in $W$ has equal probability. Then, since the sum of probabilities must equal $1$, we optain

$$p_j = \frac{1}{W}.\tag{22}$$

(You may prove that this constraint actually maximises entropy. This is what we expect from a measure of 'information' in a system where 'knowledge of the system' is smallest in the sense that each state has equal probability. See 'Intuitive explanation of entropy?' on math.stackexchange.com by An.Ditlev.)

Substituting

$$\mathrm{H}(X)= -\sum_k \frac{1}{W}\ln \frac{1}{W} = -W \frac{1}{W}\ln \frac{1}{W} = -\ln \frac{1}{W} = \ln W.\tag{23}$$

This shows that Boltzmann constant is actually unnecessary and a figment of historical definitions. We shall see this in one other way down below. We conclude that Boltzmann's definition of entropy is a special case of a more general entropy when probabilities are equal. In thermodynamics, this more general entropy is called Gibbs entropy.

$$S\equiv -k_\mathrm{B}\sum_k p_k\ln p_k\tag{GD}$$

This is our new definition of entropy. Here we have an absolute quantity; in classical thermodynamics (without the third law) we defined the differential of entropy.

Equivalence of Gibbs entropy and Clausius entropy

A partition function $Q$

Intuitively, we assume that the relative populations of system $a_m$ and $a_n$ are given by some function depending on energy.

$$\frac{a_m}{a_n} = f(E_m - E_n)\tag{24}$$

The difference in energy is a natural choice because any arbitrary zero for an energy state should cancel. Furthermore, the function $f$ ought to be transitive.

$$\frac{a_x}{a_n} = \frac{a_m}{a_n}\frac{a_x}{a_m} \iff f(E_x - E_n) = f(E_m - E_n)f(E_x - E_m)\tag{25}$$

The most obvious function with these properties is the exponential function. Thus we define

$$a_j = C\mathrm{e}^{-\beta E_j}, \ \ C\in\mathbb{R},\tag{26}$$

where $\beta$ is positive everywhere by convention. Its form is to be determined. We start by tackling the constant $C$. Sum both sides of equation $(26)$.

$$\mathcal{A}=\sum_i a_i = C\sum_i\mathrm{e}^{-\beta E_i} \tag{27}$$

Divide equation $(26)$ by equation $(27)$.

$$\frac{a_j}{\mathcal{A}} = \frac{\mathrm{e}^{-\beta E_j}}{\sum_i\mathrm{e}^{-\beta E_i}} \tag{28a}$$

In the limit, the left-hand side becomes probability of being in a state $a_j$.

$$p_j = \frac{\mathrm{e}^{-\beta E_j}}{\sum_i\mathrm{e}^{-\beta E_i}} \tag{28b}$$

The denominator is defined to be a partition function $Q$. Assumuming that $E=E(N,V)$ (number of particles and volume, respectively) we see immidiately

$$p_j(N,V,\beta) = \frac{\mathrm{e}^{-\beta E_j(N,V)}}{Q(N,V,\beta)}. \tag{29}$$

The first law of thermodynamics with partition functions

From the definition of expectation value for a discrete variable $X$,

$$\mathrm{E}X = \sum_ip_ix_i.$$

Let us identify internal energy with an expectation value of energy. Then

$$U = \sum_i p_i(N,V,\beta)E_i(N;V).\tag{30}$$

Take a differential of both sides.

$$\d U = \sum_i p_i(N,V,\beta)\d E_i(N;V) + \sum_i E_i(N;V)\d p_i(N,V,\beta)\tag{31}$$

This expression for probability and energy holds for a reversible process. If we associate the first term with work, the second term will be heat. Doing so will allow us to prove the equivalence but it will also lead to correct results for ideal gases if you do the maths. For brevity (ha ha), I will gloss over such discussions but in my opinion the rigorous explanation (if there is one) deserves its own question. (I am not aware of the rigorous proof.) Hence,

$$\delta q_\mathrm{rev} = \sum_i E_i(N;V)\d p_i(N,V,\beta).\tag{32}$$

Proof of equivalence

Recall the definition of Gibbs entropy.

$$S\equiv -k_\mathrm{B}\sum_k p_k\ln p_k\tag{GD}$$

Take the differential of both sides.

$$\d S = -k_\mathrm{B}\sum_k (1 +\ln p_k)\d p_k\tag{33}$$

The term $\sum_k \d p_k$ will equal zero by linearity and because probabilities sum to $1$.

$$\sum_k \d p_k = \d\sum_k p_k = 0\tag{34}$$

Thus,

$$\d S = -k_\mathrm{B}\sum_k \ln p_k\d p_k.\tag{35}$$

Substituting equation $(29)$ into the natural logarithm term in equation $(35)$ yields

$$\d S = k_\mathrm{B}\sum_k \left[\beta E_k(N,V) + \ln Q(N,V,\beta)\right]\d p_k.\tag{36}$$

Similarly to equation $(34)$ the final logarithm term cancels.

$$\sum_k \ln Q(N,V,\beta) \d p_k = \ln Q(N,V,\beta)\d \sum_k p_k = 0\tag{37}$$

We are left with

$$\d S = k_\mathrm{B}\beta\sum_k E_k(N,V) \d p_k(N,V,\beta)\tag{38}$$

which by equation $(32)$ translates to

$$\d S = k_\mathrm{B}\beta\delta q_\mathrm{rev}.\tag{39}$$

Note that $k_\mathrm{B}\beta$ is also an intgrating factor. From our corollary before and to complete the equivalence, we are forced to choose

$$\beta = \frac{1}{k_\mathrm{B}T},\tag{40}$$

and finally,

$$\d S = \frac{\delta q_\mathrm{rev}}{T} \implies T\d S = q_\mathrm{rev}.\tag{19,CT}$$

In conclusion, we have shown that Gibbs entropy is equivalent to Clausius entropy up to an additive constant. This is because a differential of a constant is zero. The choice of $\beta$ again demonstrates how superfluous Boltzmann's constant really is. We introduce it in the Gibbs entropy definiton, only to be forced to get rid of it. For more information on Boltzmann's constant, and how it relates to temperature and the zeroth law, see 'Is the Boltzmann constant really that important?' on physics.stackexchange.com by DanielSank. And so,

$$S_\mathrm{Gibbs} = S_\mathrm{Clausius} + r,\ \ r\in \mathbb{R}.\tag{41}$$

If the most classic form (zero entropy at absolute zero) of the third law of thermodynamics holds, and it always does not,

$$r = 0. \tag{42}$$

TL; DR

The reason why

$$\delta q_\mathrm{rev}=T\d S\tag1$$

holds depends on your starting point.

There are different forms of the first and second laws. Proving that a state function $S$ exists varies based on one's axioms.
Entropy may be defined in various ways. If a classical definition is used, there is only to prove that we may indeed define such a state function. If a statistical definition is applied, we must first prove the classical definition as a special case.
Note that the second law of thermodynamics only defines a difference or differential in entropy. Together with the third law, one may establish an absolute scale. However, the statistical definition is already absolute to begin with.
Boltzmann entropy is a special case of Gibbs entropy, not the other way around. Gibbs entropy agrees with Clausius entropy (in reversible thermodynamics) up to an additive constant.

$$S_\mathrm{Gibbs} = S_\mathrm{Clausius} + r,\ \ r\in \mathbb{R}\tag{41}$$

If the most classical form of the third law holds, $r = 0$.


Why aren't methanal and ethanal also ketones?

Aldehydes deserve a different name because they have different properties, for example being easily oxidized to acids, as in the Benedict test which distinguishes "reducing" sugars like glucose (an aldehyde) from non-reducing ones like fructose (a ketone).  The relation between aldehydes and ketones is like the relation between alcohols and ethers, both of which can be described as R-O-R, but the properties are significantly different when one of the R's is a hydrogen.


Salicylic acid - polarity and solubility

Well it has polar groups - carboxylic and phenolic but benzene ring reduces its polarity. One could say carbon atoms win 7:3 with oxygens, but it's still not bad :) PubChem say it's solubility is about 2g/L, but this site gives more info - it dissolves much better in hot water and its solubility in methanol, which is quite polar, is good, and in benzene (non-polar) bad. Generally it seems to be somewhere in the middle between strongly polar and strongly non-polar compounds.


Why does phenolphthalein form in this reaction?

In the Org. Syn. link you provided phthalic anhydride 1 and p-chlorophenol 2 in 95% H2SO4 and boric acid at 200oC undergo condensation and cyclization. Phthalic anhydride 1 under acid catalysis reacts with p-chlorophenol 2 at the ortho position as shown below to form ketoacid 3. Chlorine is a weak o,p-director and boric acid, I believe, is complexing with the phenolic hydroxyl and the ketone to reduce bond rotation and aid in cyclization to form anthroquinone 4. There is an aqueous work-up at this point followed by treatment with hot aqueous KOH which forms quinizarin 5 by nucleophilic aromatic substitution which is aided by the adjacent keto group.
In the reaction of phthalic anhydride 1 with phenol 7 in the presence of sulfuric acid, (presumably no boric acid), cyclization is repressed because of a lack of electron density at the phenol meta-position in 6. In addition, protonation of the ketone in 6 gives extended conjugation with the phenolic hydroxyl. Preferrable addition of phenol to the ketone of 6 leads to 8 and on to phenolphthalein 9.


How to quantify the difference between two xyz files for the same molecule?

What you describe is to calculate the RMSD after superposition. It is not necessary to do a minimization, there is a classic algorithm to get the best superposition directly: Kabsch algorithm. Sometimes, only a subset of coordinates will be superimposed, either because coordinates are missing in one of the structure, or because some parts of the structure show larger differences and interfere with superimposing the similar parts of the structures.

In some cases, where the structures are very different from one another but there are domains of conserved structures, you could also compute differences of pairwise distances and plot them as a distance difference matrix. Distance differences within the conserved domains will be small, while distance differences from domain to domain will be large.

When you do a superposition and RMSD calculation for the first time, it is nice to be able to check the superposition. Many structure viewers have the capability to visualize the superposition while also calculating the RMSD, for example the "compare" command in jmol.


Calculate mass concentration of Ca(OH)2 solution

There is a general law about the solubility of salts containing doubly or triply charged ions, which is not often presented in textbooks. The law is the following : The measured solubility of such a salt or such a hydroxide is ALWAYS greater than the calculated values derived from solubility products, or from pH measurements. This is due to the fact that the solubility product and the pH data do not take into account the existence of intermediate ions.

In the case of $\ce{Ca(OH)2}$, there are of course ions like $\ce{Ca^2+}$ and $\ce{OH-}$ in its solution. But there are also ions like $\ce{Ca(OH)+}$, and non dissociated molecules $\ce{Ca(OH)2}$. Alas these ions and molecules are rarely mentioned in usual textbooks. But this explains why the solubility of $\ce{Ca(OH)2}$ is bigger than the values obtained from pH measurements. I know that the average reader is reluctant to believe me. I may be more persuasive if I use another example.

Let's speak about the solubility of $\ce{H3PO4}$. Any student in chemistry knows that the solubility is not related to the solubility product $\ce{[H+]^3[PO4^3-]}$, because of the existence of intermediated ions, like $\ce{H2PO4^-}$ and $\ce{HPO4^2-}$. These ions are accepted in the large chemistry family. Everybody accepts that the solubility of $\ce{H3PO4}$ is much bigger that the value derived from the solubility products, or from pH measurements.

That is the same for $\ce{Ca(OH)2}$ and all substances containing a doubly charged ion like $\ce{CaSO4}$, etc. And here the activity is not sufficient to explain this discrepancy. I know that some chemists think that the activity may be introduced to explain this discrepancy. But in the case of $\ce{CaSO4}$ for example, the concentration is so low that the activity factor may be considered as equal to 1.

For $\ce{CaSO4}$, the effect is more pronounced, because of the presence of two doubly charged ions. The solubility calculated from the solubility product is 0.0051 M. The measured solubility, obtained from direct titration is 0.015 M. Three times more !


Naming a given organic compound according to IUPAC naming rules

Yes, your naming would be correct. For saturated hydrocarbons, the first point of difference rule takes precedence after the longest carbon chain. To apply this rule, the locants of each set are ordered numerically in a list from lowest to highest, and the series that has the lower number at the first occasion of difference is determined as the lower locant set, regardless of what comes after this first point of difference. In this case, "3,3,4,7,7" would be the lower locant set than "3,3,6,7,7", because 4 is lower than 6.


This is more a proposition than an answer but after a lot of search on the net, in publications and also with the help of @orthocresol, @Jan and @jerepierre, I propose a reaction that will work. Also the question ask is really unusual, because it is obviously awfull to want to synthetize formaldehyde from urea.

You will find first in this "answer" the best way we found to make it. When I'll have time, I'll add all ideas I had and explain why it was not a good point.

A good question raised by this question is, "is it possible to "cut" the $\ce{C-N}$ bond to replace it by a $\ce{C-H}$ bond?" and if possible in one step. For the moment nobody knows.

The best way we found :

For the moment I can't explain you why I chose to make the N-alkylation but I know I got it in many of my organic chemistry lessons that this groupement will leave the molecule involving LAH so I wanted to do that. Also we are not at one more step from urea to get formaldehyde.

For the other part formaldehyde is much more reactive than the amide function because on the "diamide" both lone pair of nitrogen atoms are involve in mesomeric forms with the oxygen. Then the carbon is really not electrophilic and LAH will react then one the most reactive function in the reactor which will be the aldehyde.

And then the Formox process because it is use in the industry so I'm sure it will work.

After you will find ideas I had but that won't work too. I'll explain as much as as possible when it will be possible. Sometime it will be only because by experiments nobody have ever succeeded to do it.

This one won't work, one of us, I don't remember who didn't find anything in SciFinder about the possibility to make Eschweiler-Clarke reaction from an amide and especially urea.

This is really awfully awkward because you need formaldehyde itself to make the synthesis of formaldehyde. I was not thinking at that when I start but few minutes later it appears to me that I haven't realize before ^^

At the begining I thought it was great to use $\ce{HCl}$ in excess with $\ce{NaNO_2}$ and then get the carbonic acid $\ce{HO(CO)OH}$, because there is nice ways to reduce this acid and its derivatives but just one time. You can find informations in this publication. But when you have $\ce{H(CO)OH}$ I have really no idea of what to do, even $\ce{LiAlH_4}$ won't work. So at the moment I just have the first possible reaction.

Well also in terms of atoms economy using urea to synthetise formaldehyde is absolutely awkward too.

I hope you find some fun here, I keep thinking about it but it looks to be tricky. Have a good day.


A Lugol's iodine solution is an aqueous solution of elemental iodine and potassium iodide. This solution may be prepared in different strengths, typically $2{-}15~\%$, while the strength is determined by the content of elemental iodine (see this table from the Wikipedia article, but note that the first column's title is a bit misleading — the leftmost column lists the different strengths of Lugol's solution, and the other columns list the amount of the each constituent (iodine, iodide, total) in a solution of that strength).

Now, this is in essence an iodine solution, but it is prepared from iodine and potassium iodide due to solubility issues of elemental iodine. See this answer on Chemistry.SE to the question Preparation of iodine solution for a simple and concise explanation.

For the calculation of the amount of iodine and potassium iodide in solution, let us take $5~\%$ Lugol's solution, which is prepared from 5 grams of iodine and 7.5 grams of potassium iodide in $\pu{100 mL}$.

If we have $\pu{5 g}\ (\pu{5000 mg})$ of iodine and $\pu{7.5 g}\ (\pu{7500 mg})$ of potassium iodide in $\pu{100 mL}$, then after dividing by $100$, we have $\pu{50 mg}$ of iodine and $\pu{75 mg}$ of potassium iodide in $\pu{1 mL}$.

Since the commonly accepted volume of one drop is $1/20\ (5~\%)$ of $\pu{1 mL}$, then by multiplying $\pu{50 mg}$ and $\pu{75 mg}$ by $0.05$, we get $\pu{2.5 mg}$ of iodine and $\pu{3.75 mg}$ of potassium iodide in one drop of the solution.

For better clarity, here is the calculation again for I2 content:

$$\frac{\pu{5 g}}{\pu{100 ml}}\cdot \frac{\pu{1000 mg}}{\pu{1 g}} = \frac{\pu{50 mg}}{\pu{1 mL}}  \\ \Longrightarrow  \frac{\pu{50 mg}}{\pu{1 mL}}\cdot \frac{\pu{1  mL}}{\pu{20 drops}} = \frac{\pu{2.5  mg}}{\pu{1 drop}}$$

Do the same for $\pu{7.5 g}$ of potassium iodide to get $\pu{3.75 mg/drop}$.

A word about the actual iodine content of the solution following Jan's comment:

In principle, masses can always be added up (unlike volumes, which should not be added up, due to possible interactions between certain solvents — see here).
Therefore, we can say that the total mass of solutes in one drop is $$\pu{2.5mg}+\pu{3.75 mg} = \pu{6.25 mg}$$ ($\pu{12.5 mg}$ in two drops, per OP question). However, we cannot say that two drops contain $\pu{12.5 mg}$ of iodine, because the solution contains both elemental iodine (I2) and potassium iodide (KI), which are two different constituents. Since KI's role in the solution is to facilitate dissolution of elemental iodine molecules (see description of chemical properties, including equations, here), iodide content is irrelevant and the solution's strength reflects only iodine's content.

In other words, $\pu{12.5 mg}$ is the total quantity of solutes in two drops of Lugol's solution, but this quantity has no practical meaning, since the active ingredient is iodine, and its amount in 2 drops of $5~\%$ Lugol's solution is $\pu{5 mg}$.


Formation of MnO2 vs [Mn(H2O)6]4+ in aqueous solution

Solid $\ce{MnO2}$ can be hydrated, but it's the solid which precipitates from solutions as a product of redox reactions of other $\ce{Mn}$ compounds. There is also for example crystalline $\ce{[MnF4(py)(H2O)]}$ What about aquo cations you mentioned?

One thing is that $\ce{[Mn(H2O)_n]^4+}$ cations should be indeed quite strongly acidic, but that's no problem as another strong acid in solution would prevent precipitation (similar thing like with $\ce{Fe(OH)3}$). The problem is oxidative power, what may be surprising knowing that Mn has higher oxidation states.

Why do I think so? Well, let's see what happens if we try this trick with acid to make a solution. Concentrated $\ce{H2SO4}$ and $\ce{HCl}$ react with $\ce{MnO2}$, but lets see the reactions:

$$\ce{2 MnO2 + 2 H2SO4 → 2 MnSO4 + O2 + 2 H2O}$$
$$\ce{MnO2 + 4 HCl → MnCl2 + Cl2 + 2 H2O}$$

In first case oxygen gets oxidised and in second chlorine!

Now what if we used $\ce{HF}$? Concentrated solutions are really powerful stuff! Unfortunately the problem persists, as you could see earlier $\ce{Mn^{4+}}$ can oxidise even water if acidity is high enough. With difficulty $\ce{MnF4}$ was synthesised, but guess what, Mn can gradually oxidise fluorine in this compound, making $\ce{F2}$ gas! That was first chemical reaction known to do that.

So, generally not much chance for simple aquo complexes.

Check out these articles on Wikipedia for more info and citations to articles: about $\ce{MnF4}$ and about $\ce{MnO2}$.


Nitrosonium ion in the nitration of phenols

Possibly the electrophile is molecular $\ce{HNO3}$. Nitric acid is strong but not super-strong ($\mathrm{p}K_\mathrm{a} = -1.4$), so solutions on the order of a mole per liter have a significant amount of molecular $\ce{HNO3}$. As $\ce{NO2(OH)}$ this can transfer its nitryl ($\ce{NO2+}$) ion moiety to the aromatic ring in exchange for a proton to make the nitrated ring and water. Obviously molecular nitric acid is less electrophilic than the straight-out nitryl ion, so an activated aromatic ring is needed for this to work.


Preparation of ammonium acetate buffer for HPLC

to prepare a 100 mM ammonium acetate solution at pH 9, you’d first prepare a 100 mM ammonium acetate, and then add ammonium hydroxide dropwise until the desired pH was achieved.

Well, this procedure is good enough for a rough screening solution- call it a quick and dirty method. The main issue is that one would not prepare a solution of 100 mM ammonium acetate beforehand and add drops of ammonia to that. Whenever you need to prepare a buffer solution, you should know what volume is desired and what concentration is needed.
For HPLC, let us say you need 1000 mL of buffer containing 100 mM ammonium acetate. One would dissolve the required amount of ammonium acetate in about 900 mL water in a 1 L beaker, add ammonium hydroxide dropwise till you reach the desired pH, add more water till you are close to 950 mL. Check the pH again, and adjust as desired, and then quantitatively transfer the solution to a 1 L volumetric flask and then make up the volume to 1 L. This will be your analytically correct 100 mM ammonium acetate buffer.
Now, coming to the buffer equation. Again, for that you need to define the concentration and volume. What would ask yourself what should be concentration of ammonia should be present in 1 L of 0.1 M ammonium acetate solution such that the final pH is 9 solution.
You would use the pK$_a$ of ammonium ion and from the equation you would figure out the concentration of ammonia.
See example no. 2 here: Buffer calculation
And also study example 3. Ammonia buffers
Now the problem is that conc. ammonia is sold as w/w % solution. Using the density and percentage, you need to determine the molarity of ammonia solution as well to actually prepare a buffer.


Is NaCl crystal always going to have even number of atoms?

Tl;dr: Theoretically yes, practically no.

Needless to say this is a very mathematical question that almost calls for a philosophical answer. The key point to consider is usually the crystal’s surface. Here, there are most defects and the least ideal structure since there is at least one row of atoms which are lacking partners on one side. But I’ll keep that out of my reasoning for a second.

Consider an ideal crystal and a Subtle Knife. You can use the knife’s almost infinitely thin blade to cut exactly between two sheets of atoms. The initial crystal had a $1:1$ distribution, every single sheet has a $1:1$ distribution and thus so will your two bits.

Then consider cutting one of the fragments again, but three atom rows away from where you cut it before. If our ideal crystal is infinitely large, we still get a $1:1$ distribution. Now continue to consider cutting the fragments into smaller fragments until you arrive at a theoretical body of $3 \cdot 3$ size. This cube has to violate the $1:1$ condition. It will be charged depending on which atom type is present in the corners. (The same is true for any other cube of an odd side length when measured in atoms.) However, we have a charged particle (consisting of nine atoms) now. It will attrackt particles of the other charge and pull them towards it. In the long run, one of the following may happen:

An ion of the opposite charge is drawn towards the crystal and attaches to it or inserts into it.
An ion of the extraneous charge is removed.

Both will make us regain our $1:1$ crystal — but it is no longer ideal. However, see below for the evolution of a more realistic view.

Now let’s consider a slightly more realistic case. We no longer have an infinitely large ideal crystal but one with a finite size. Let’s have a cube of even atom length just for simplicity ($1:1$ ratio). And because we want to see our cube, let’s assume a macroscopic size, e.g. a cubic centimetre. Now cut it with your Subtle Knife so that the even side length is split into to cuboids with one odd length. One must have a $+1$ charge, the other must have $-1$. Thus, neither have $1:1$.

One could now say that maybe a chloride ion ‘jumps’ to the other side or something, again creating a non-ideal crystal. Or maybe moisture from the air supplies one side with a proton and the other with a hydroxide. But we could also say that we have a charge of $+1$ and one of $-1$ each delocalised over some $10^{20}$ atoms — no big deal if you think of it.

And let’s go a step further and consider a real crystal. It’s going to have defects, impurities, non-aligned domains and a surface. If it ever had a mathematically exact $1:1$ ratio, then that was purely by chance. We can still delocalise the effective charge over billions of billions of atoms. We would need an extensive charge in a very small area to cause a problem at all.

Even worse if our crystal is put into a saturated $\ce{NaCl}$ solution. (Saturated, because I don’t want it to dissolve.) All the time ions will be entering and leaving solvation in a non-coordinated manner. Both random in time and random in location. The numeric ratio of atoms will change all the time. If we consider a very sub-saturated solution, of course there will be more ions dissolving into solution than recrystallising. But also, those that end up in solution also dissociate from each other. Thus, there is nothing actually keeping an atom of a charge of one (or two, or three, see other metals) close to its counter-ion — charge dissociation is not as bad as it seems on the paper.

So real world crystals will only have the ideal stoichiometry in a short-lived transition state between being too positive and being too negative — and being philosophical I could say that that almost be like a person’s mood.


Why is glycine hydrochloride more acidic than chloroacetic acid?

Glycine hydrochloride is the amino acid glycine adsorving a molecule of Hydrochloric acid, wich is a strong acid. I think this is the main reason of the acidicity difference, since the pKa's are similar (comparing glycine and chloroacetic acid).

Update: glycine is amphoteric: it can either act as a base or an acid.
In water the hydrochloric acid molecule of the glycine hydrochloride will dissociate: $$\ce{H2O +HCl -> H3O+ +Cl-}$$
Then:
$$\ce{Gly +H+ -> GlyH+}$$
$$\ce{GlyH+ +Cl- <-> Gly + H+ + Cl-}$$
You can conclude, therefore, that Glycine hydrochloride is more acid than glycine (and also the chloroacetic acid). Hope it helps, and sorry for the english mistakes!


Ring contraction in pinacol rearrangement of cyclohexane-1,2-diol

Background

You don't specify whether we are starting with cis- or trans-cyclohexane-1,2-diol, but it doesn't matter since both isomers will interconvert under the reaction conditions.  Therefore we will need to consider the reaction products from both isomers.
These reactions run in strong acid are never clean.  I'd bet that some cyclohexanone was also formed, but that it is the minor product.  Here is a link (see section 2. Pinacol Rearrangement, example #3) showing that 1,2-dimethylcyclohexan-1,2-diol produces >90% of the corresponding cyclopentyl analogue and around 6% of the cylcohexanone analogue.
Predominant formation of the cyclopentyl carbaldehyde indicates that our reaction must be kinetically driven. 5-membered rings have ~6 kcal/m more strain than 6-membered rings.  Therefore, if the reaction were thermodynamically controlled the 6-membered ring would predominate.
In the case of cyclohexane-1,2-diol, if a protonated hydroxyl group departed, a seconday carbocation would form.  Secondary hydrocarbon carbocations are not particularly stable, so it is likely that rather than generate a secondary carbocation there is some migrating group migration at the same time the protonated hydroxyl is departing. Such migrations have a strong preference for occurring by way of an anti-periplanar geometry.

Solution

In the figure below I've drawn the 3 isomers of cyclohexane-1,2-diol. The isomer on the left is the axial, axial trans isomer, the isomer in the middle is the equatorial, equatorial trans isomer, and then on the right is the axial, equatorial cis isomer.  I've also labeled the hydroxyls (A and B) for clarity.  In an equilibrium mixture of these isomers we would expect the equatorial, equatorial isomer to predominate because both hydroxyl groups are in the sterically less encumbered equatorial position and both hydroxyls can hydrogen bond with each other in this conformation.

Examining the isomer on the left, neither the $\ce{C-H}$ bond nor the $\ce{C-C}$ bond is aligned anti-periplanar to either hydroxyl A or B, consequently this isomer should be relatively unreactive.

Examining the middle isomer only $\ce{C-C}$ bonds are aligned anti-periplanar to the hydroxyl groups.  Rearrangement of this isomer can only produce the cyclopentyl derivative.

Examining the isomer on the right we see that a $\ce{C-H}$ bond is aligned anti-periplanar to hydroxyl A, while a $\ce{C-C}$ bond is aligned anti-periplanar to hydroxyl B. Hence, this isomer can produce both the cyclopently carbaldehyde and cyclohexanone.

We see that there are 3 pathways leading to the 5-membered ring product and just one leading to the 6-membered ring product. This suggests that the 5-membered ring product should predominate.  If we further consider that, as explained above, the middle isomer should predominate at equilibrium, then our product mixture should be even further enriched in the 5-membered ring product just as you reported.


How can the hydrogen bonds in solid HF be best represented?

At Wikipedia you find this structure (drawn by Benjah-bmm27 on Wikimedia Commons):

So, the answer (c) is correct. Yomen Atassi correctly stated that in such a hydrogen bond the two electronegative partners and the hydrogen prefer a linear arrangement, as this maximizes the orbital overlap for the hydrogen bond. That the configuration (c) is preferred over (a) can basically be explained via the VSEPR theory: a fluoride ion in HF is surrounded by 3 electron pairs and 1 H–F bond: those 4 "ligands" should roughly be arranged tetrahedrally – approximately, not exactly, because electron pairs need more space than bonding electrons – around the F atom, and this leads to the zigzag chains from answer (c).


What equipment to use to make liquid oxygen?

Simply put an open flask in the LN2 and it will fill with liquid oxygen from the air.

Or better: DON'T DO THAT!! What the heck do you want with it?!?

LO2 is highly dangerous. If it comes in contact with anything combustible, it will blow up vigorously. Together with lab solvents, it will likely explode. Make sure the flask or dewar is dead clean.
Chemistry is science, and science is about prediction. If you cannot predict, from knowledge, not hearsay, what your experiment will likely do, don't do it.


Trouble to predict the effect of pressure and temperature for a reaction in equilibrium

When I posted this question I had no idea how changing each factor bring change in equilibrium. Then I bombarded my brain with everything I know about equilibrium and I found a fairly easy way to understand this.

Change in concentration

Imagine two vessels A and B connected to each other through a thin hollow pipe through the middle of the vessel. Let vessel A be reactant and vessel B be product. The vessels are filled with water a little above the pipe. If I add water to vessel A  water flow from A to B. Thus increasing concentrated of A caused the water in vessel A to flow to B; which in other words is the forward reaction.

Change in temperature

Think of a reaction which is in equilibrium.

1)Decreasing the temperature of reaction in equilibrium.

We know that decreasing the temperature favour exothermic reaction
 and exothermic reaction are favoured by low temperature. Exothermic mean giving out heat which remove the deficiency of temperature to bring it back to normal in equilibrium.

2) Increasing the the temperature of reaction in equilibrium

We also know that increasing the temperature favour endothermic reaction or endothermic reaction are favoured by high temperature. Endothermic means to absorb heat energy which causes the excess heat energy ( from high temperature) to bring it back to normal in equilibrium

Effect of pressure

Changing pressure play a significant role in gaseous reaction which are accompanied by change in moles. On increasing pressure of the system equilibrium will shift to the direction where number of moles decrease or where pressure decrease (because pressure is directly proportional to the number of moles). If pressure on the system is increased equilibrium shift in the direction in which decrease in number of moles take place.

This is the second time I answered on stack exchange so please let me know my mistake if any. If I had known how to create the diagrams I usually find in other answers I would have added them.


Oxidation of guanine with singlet oxygen

You're right that ground state $\ce{^3O2}$ doesn't undergo this reaction.

There is no direct excitation of oxygen either.

The key in these processes is the electronic excitation of tetraphenyl porphyrine (TPP). The excited singlet state of the dye undergoes intersystem crossing to a triplet state. The latter interacts with ground state oxygen and generates singlet oxygen.

$$\ce{TPP ->[h\nu] ^1TPP^*}$$
$$\ce{^1TPP^* ->[ISC] ^3TPP^*}$$
$$\ce{^3TPP^* + ^3O2->[sens] TPP + ^1O2}$$

Singlet oxygen then acts like the typical dienophile in a Diels-Alder reaction.


Thermodynamics of a spontaneous isothermal chemical reaction.

The internal energy and enthalpy of a pure ideal gas depends only on temperature.  But, the internal energy and enthalpy of an ideal gas mixture depends both on temperature and on the amount of each chemical species that is present.  To be more precise, it is equal to the internal energy and enthalpy per mole of each component as if it were a pure species times the number of moles of that species present, summed over all the species in the mixture:  $$U=\sum{n_iu_i}$$and$$H=\sum{n_ih_i}$$So, even at constant temperature, $$\Delta U=\sum{(\Delta n_i)u_i}$$and$$\Delta H=\sum{(\Delta n_i)h_i}$$You should have been taught this when they were covering the thermodynamics of mixtures and chemical reactions.


Does calcium hydroxide react with citric acid in orange juice?

Don't know if this is the brand but here is such a label

Therefore calcium hydroxide is an actual additive. In this case advertised along with vitamin D (calcium fortification).

As for what gets to your stomach, there are reasons why ingredients, not compounds produced via cooking are listed. Cooking unlike chemical synthesis isn't supposed to create specific compounds, but food in which all compounds are mixed and every possible equilibrium happens at once.

In the juice there's just more calcium cations and pH may be slightly affected, if it was significantly changed you'd find it by your taste.


Chemical reactions with a room-scale cooling effect

One way to do it is, as user41631 says, with an endothermic reaction.

Another is to use phase-change materials, from which you remove the thermal insulation, to allow the phase-change to take place. Now, all materials change their phases, at some combinations of temperature and pressure. The phrase "phase-change material" is used not to classify materials, but to identify the paricular use of a material - that is, the material is being used because one of its phase changes happens at a temperature and pressure that makes it useful for a specific process.

Such materials are being looked at for the temperature regulation of buildings.

Let's take a material with a melting point of, say, 18°C, and a high latent heat of fusion. Freeze it, insulate it, and take it into a room that's at 22°C. While the material melts, it will absorb heat from the air, cooling the room.

This principle has been applied in building material design, to provide smoothing of temperature variations. See, for example, Schossig et al's "Micro-encapsulated phase-change materials integrated into construction materials", [DOI], from which the following graph is taken, showing the smoothing of temperatures over a week in August , when a phase-change material (PCM) is used (T_Wall_PCM) - this is compared to an identical room without phase-change materials (T_Wall_REF) - as you can see, the PCM reduces the amount of time that the room spends above 26°C, providing more thermal comfort for people in the room.


Is graphite a better lubricant on the Moon than on the Earth?

Graphite, like most solid lubricants, has a layered structure which makes it a good lubricant.
The atoms of a layer are strongly bonded to each other, while the interlayer (van der Waals) forces are much weaker.
The layers align themselves between the sliding surfaces and the layers slide one over the other providing lubrication.

Since it are the shear forces that come into play, gravity has no influence on the system.
From that point of view, there is no difference between its use here or on the Moon.
But, there are also extrinsic factors which influence its performance.

Graphite works best in a humid environment.
This is probably due to the weakening effect of the water molecules on the interlayer forces.
So, graphite should not work well in vacuum.

Another solid lubricant, $\ce{MoS2}$, which has also a layered structure like graphite, works much better in vacuum and should be the lubricant of choice on the Moon.


Pseudochirality with more than two groups on one carbon atom

Ah, now I think I understand what you're asking.  You're saying that we have a central carbon to which is attached a hydrogen and 3 groups with the same structure. Each of these 3 groups contains 2 chiral carbons.  One group is R,R one is R,S and one is S,S.

So yes, the chirality of the central carbon (it is chiral, it is not meso or prochiral - there is no plane of symmetry bisecting the central carbon) can be assigned using the Cahn–Ingold–Prelog priority rules where "R" has higher priority than "S".  Therefore the priority ranking around the central carbon in your molecule would be R,R > R,S > S,S > H.