Temperature measurement using thermometer

I have two samples of water – 'A' and 'B'. My thermometer measured them to be 90 °C. They both have different volumes. So, the kinetic energy of the particles of sample A and B is same or different?

The total kinetic energy is different. The average kinetic energy per particle is the same.

  Through this question, I want to ask whether temperature measurement results depend upon how many particles hit the bulb of my thermometer or it is independent of this?

In your specific example, the same number of particles hit the bulb because it just depends on the area of the bulb, as long as it is covered in water. Not only is water hitting the bulb, the atoms in the bulb are also hitting the water. After they have done this for a while, water and bulb are at the same temperature.

  If it is independent of this, then how thermometer works? As another example, let me consider a gas which has the density of 2 mol/22.4 L. Its temperature recorded by my thermometer is 60 °C. Then its density is reduced to about 10 atoms/22.4 litre. Now, its temperature is the same?

For the example with the two gases at the same temperature but at different pressure (i.e. particle density), the situation is a bit different, but the thermometer still works. There are a different number of "hits" in this case, but if you wait long enough, the gas and the bulb will be at the same temperature.

  Please provide me the simplest answer as I am in class 9 only.

I tried, but if the question addresses a complicated phenomenon, the answer can't be super-simple.

  [From the comments:] So, if i have 10,000000 particles in 1 L and 10 particles in 1L, temperature is going to be same or different?? (I mean the readings)

Ideally, there should be more sample than thermometer. This means that no matter what the temperature of the thermometer was to start, it will have the temperature of the sample afterwards. With only 10 particles in the sample, the 10 particles will take on the temperature of the thermometer (and the enclosure), so you will be unable to measure the initial temperature of the sample.

  [My own question] If the sample keeps hitting the thermometer, why doesn't it get hotter and hotter?

As I said above, the thermometer also keeps hitting the sample. There is some net transfer of energy while the temperatures are different (from hot to cold, warming up the colder and cooling down the hotter). Once the temperatures are the same, sample and thermometer keep "hitting each other", but the temperature of both stays constant (if the entire setup is insulated against the surrounding).


From the Aldrich data:
 $\rho = 1.114\ \frac{\mathrm{g}}{\mathrm{mL}}\ \text{at}\ 25\ \mathrm{^\circ C}$

Converted to a more convenient unit:
 $\rho = 1.114\ \frac{\mathrm{g}}{\mathrm{mL}} \cdot \frac {1000\ \mathrm{mL}}{1\ \mathrm{L}} = 1114\ \frac{\mathrm{g}}{\mathrm{L}}$

Given the molar mass (also from Aldrich) of:
 $M = 78.13\ \frac{\mathrm{g}}{\mathrm{mol}}$

We learn that one liter has:
 $1114\ \frac{\mathrm{g}}{\mathrm{L}} \cdot \frac{\mathrm{mol}}{78.13 \ \mathrm{g}} = \frac{1114}{78.13}\ \frac{\mathrm{mol}}{\mathrm{L}} = 14.26\ \frac{\mathrm{mol}}{\mathrm{L}}$

I can't speak to why your other source has a different number.


Is there a difference between Close Cubic Packing and Face Centered Packing?

The Wikipedia entry for "Cubic crystal systems" says:

...Face-centered cubic (abbreviated cF1 or fcc, and alternatively
called cubic close-packed or ccp)

In another entry it is stated:

...face-centered cubic (fcc) (also called cubic close packed)...

The text of the above references make no statement about one term providing any more or less information than the other.  This does not preclude the possibility that some organization or another may have a preference for one term or the other, but the meaning seems to be the same.


Ethyl acetate disolves PLA - which is typically the ingredient in non-acetone nail polish removers.

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

According to Wikipedia:

  "If the filament is soaked in a small amount of ethyl acetate, it will dissolve and can be used to clean 3D printing extruder heads or remove PLA supports. The boiling point of ethyl acetate is low enough to also smooth PLA in a vapor chamber, similar to ABS and acetone."


Which is the most reactive metal: Caesium or Francium?

The reactivity series consists of elements whose properties are FULLY known and understood. It is on the basis of whether an element can displace another from its salt solution (this is what you've been told at grade 10 Indian schooling level I guess?).
Francium, on the other hand is a radioactive element, with a half life of only 22 minutes for its most stable isotope
(Francium-223). Therefore, we can only make predictions, and not observations, about the nature of its salts and whether they would displace other elements from their salt solution or not.


If you google "why is technetium unstable" you'll see that this question has been around for a while.  Over on physics.stackexchange, it is discussed in Why is technetium unstable?.

I've read some of what others have posted on this topic and the short answer appears to be: because we have such a poor understanding of the atomic nucleus, there is no answer to this question at this time.  Again, the link above would be a good place to start.  One of the answers (3 upvotes) claims to be an answer.  The selected answer (10 upvotes) says there is no answer.


The principal component of wood ash appears to be calcium carbonate. (Wiki entry, "wood ash".) Other components include compounds of potassium and phosphorus. If you heat the calcium carbonate strongly enough, it will decompose into CO2 and calcium oxide. Apparently, strong heating is likely to result in less carbon instead of a larger percentage. The remaining calcium oxide is pretty stable as to temperature although addition of water will readily convert it to calcium hydroxide.

Calcium carbonate is white. White mixed with the black of any carbon present would result in a grey color.


How can I purify Ferric or Ferrous acetate with a very basic lab?

Both the Handbook of Chemistry and the Merck Index state that $\ce{Ferric Acetate}$ exists only in solution. If this solution is evaporated, a basic ferric acetate is obtained that is insoluble in water, with a formula $\ce{Fe(OH)(CH_3COO)_2}$. This basic acetate is soluble into acetic acid.

Nevertheless there has been recent reports stating that solid Trinuclear Ferric Acetate may be obtained in rather concentrated solutions with the following structure : $\ce{[Fe_3O(OAc)_6(H_2O)_3]OAc}$. Its synthesis has been described in A. Laurikenes et al., Lithuan. J. Phys. 56, 1. p.35 (2016).
They mix $11.2$ g Iron powder ($0.200$ mol) and $12$ mL $\ce{CH_3COOH}$  $50$% ($0.100$ mol) at $75°C$. Why this choice ? The contrary would have been a much better choice (interchange $0.1$ and $0.2$)... They obtain a concentrated green solution, which they filtrate, and they add $6$ mL $\ce{H_2O_2}$ $30$%, and filtrate the obtained red mixture. The obtained filtrate is very concentrated. It is then evaporated under vacuum. And they obtain $2.95$ g of the trinuclear compound. The yield is miserable.  But they say that they have been able to recrystallize this compound in ethanol.

It is interesting to know that this substance has three equivalent $\ce{Fe}$ atoms in the apexes of a triangle, with an Oxygen atom in the center of the triangle, and that each $\ce{Fe}$ atom is included in the center of an octahedron, where all apexes are occupied by Oxygen atoms.

See also : K. I. Turte et al. Synthesis and Structure of Trinuclear Iron Acetate, J. Struct. Chem. 43, 108-117 (2002)


Confusing definition of the term precision

In colloquial English, precision is synonymous with accuracy. If you look it up on Google, the first definition is:

  the quality, condition, or fact of being exact and accurate.

In the field of science, however, it takes on another meaning:

  the degree to which repeated measurements under unchanged conditions show the same results

There is also measurement resolution:

  the smallest change [an instrument] can detect in the quantity that it is measuring

Which is similar to the definition precision takes on in numerical analysis:

  the resolution of the representation, typically defined by the number of decimal or binary digit

In short, precision can take on different meanings depending on how you choose to apply it. If you talk about the precision in a set of measured data, you will be using the scientific definition; if you talk about the precision of an instrument, you will be using the numerical analysis definition.


What happens to concrete if you add lye to the mix?

The article referenced in the comment above (Ref 1) is not relevant to concrete. It analyzes solutions with a water to solids (cement plus $NaOH$) ratio of 1, whereas ordinary concrete seldom goes above 0.5. This is a paste, not a structural material. The $NaOH$ speeded up early hydration at 1 M, and retarded it at 2 M. It was apparently affecting the aluminates more than the silicates.
The composition of typical portland cements varies, but is usually considered to be mostly $C_3S$, a faster reactor and $C_2S$, a slower, but eventually stronger component. Ref 2

Many admixtures are used in portland cement concrete (Ref 3), but sodium hydroxide is not one of them. $NaOH$ is used as a promoter with silica-rich additives like flyash in highly modified cements, but portland cement is already silica-poor. When portland cement hydrates, $Ca(OH)_2$ is formed, which is not a really good strengthener, except that over long periods of time, it will become incorporated into the silicate gel, strengthening the concrete somewhat.
However, there are adverse reactions noted with silica-rich aggregates and the high alkalinity of sodium hydroxide (Ref 4), so it is always advisable to keep the sodium level low, whether from an alkali, an admixture, or external applications, like salt.
Ref 1: https://www.sciencedirect.com/science/article/abs/pii/000888468990046X
Ref 2: https://en.wikipedia.org/wiki/Portland_cement
Ref 3: https://www.cement.org/cement-concrete-applications/concrete-materials/chemical-admixtures
Ref 4: https://www.concreteconstruction.net/how-to/repair/chemical-attack-on-hardened-concrete_o


Graphitic acid-Its structure and properties?

Graphitic acid is a compound derived from graphite.

You may find it named "Graphite oxide" as well: as you see, it is obtained by reacting graphite with an oxidizing agent ($\ce{HNO3}$, for instance).

The structure resembles that of graphite, but it carries several oxygen-containing substituents, such as hydroxyl and carbonyl functions, thus having a less regular geometry, compared to that of graphite.

The exact features of the structure depend a lot on the method used for its preparation, and the layers may show some "defects", not described by this ideally planar model, but the key concept is this.

Here is a simple proposed structure for a single layer (source: Wikipedia)

Note that, just like graphite is made of several layers of graphene, graphite oxide is made of several layers that, upon reaction with a base, become graphene oxide.

Here you can find further information, and a nice image of a single graphene oxide film


Suitable safe cleaning solution for aliphatic urethane

Oxalic acid is good at removing rust (iron oxides) and generally not agressive against polymers like polyurethanes.
Chemical resistance of polymers is however a complicated area, because polymers comes in a wide spectrum of formulations, additives etc. Some indications can be found on different manufacturers websides


Where does magnetic inequivalence come from?

Firstly, thank you for asking a very interesting question that certainly pushed my limits of knowledge (and others' as well). I apologise if I have made mistakes, it is a real possibility. All references are to Levitt's Spin Dynamics, 2nd ed. (an excellent book, by the way).

From a quantum mechanical point of view, the complexity in the spectrum of magnetically inequivalent nuclei should be considered the "norm". Fundamentally, in any strongly coupled system where the difference in Larmor frequency between two spins $\Delta \omega$ is much less than the coupling constant $J$, we expect to see complex spectra (see second half of this answer for more discussion).

Magnetic equivalence is a special case which leads to simplifications in the Hamiltonian and hence in the spectrum. As I mentioned in a comment, it can be shown that any coupling between magnetically equivalent nuclei "disappear". The word "disappear" merits further explanation, though. It is not true that the coupling constant $J$ goes to zero, for example. Neither is it true that it is averaged to zero by some sort of physical process, e.g. molecular rotation. It only means that the coupling term in the Hamiltonian has no impact on the expectation value of any quantum mechanical observable, or its evolution. Since an NMR spectra is simply us humans reading out some expectation values, the NMR spectrum itself is therefore unaffected by the coupling term between magnetically equivalent nuclei.

Having established that it is magnetic equivalence, not inequivalence, that is the special case, I would like to rephrase your question to: "Is there a physical explanation for why magnetic equivalence leads to simplification of the spectrum?"

The mathematical reason is given in Appendix A.9 of Levitt. I won't reproduce it here, so I will just mention it has to do with some operators which commute with others, and these only commute under very specific circumstances, i.e. the rules for magnetic equivalence that you have surely been taught (they must be chemically equivalent, plus either they have the same coupling constant to all other nuclei, or there are no other spins in the molecule).

Given the (relative) complexity of the mathematical explanation, I suspected that a physical explanation would not be particularly instructive. Levitt, though, does offer us something (p 461):

  The apparent disappearance of J-couplings between magnetically equivalent spins is quite surprising. A physical explanation runs as follows. For magnetically equivalent spins, the local magnetic fields, whatever their source, are exactly the same on the two spins. It follows that the motion of the two spin polarizations is identical. Whatever happens, the polarizations of the two spins are locked in the same relative orientation. There is, therefore, no need to incorporate a term in the spin Hamiltonian that takes into account the dependence of the energy on this relative spin orientation.

What this means is: it does not matter what the relative orientations of the spins are. For example, in ethene, it does not matter whether two spins are up and two are down, or whether all four are down, or whether all four are somewhere between up and down (i.e. a superposition of $\alpha$ and $\beta$, which is in fact the most likely case by far). The relative orientations can be anything, but the point is that the time evolutions of all four spins are synchronised. Therefore, the coupling terms between them – which are terms, scalar products, that depend on the relative orientation of spin vectors – may be omitted from the Hamiltonian without any impact on the observables of the system.

Returning to a mathematical interpretation, another sign of magnetic equivalence between nuclei $i$ and $j$ is when the labels $i$ and $j$ can be permuted in the spin Hamiltonian without changing it. This implies that $[\hat{\vec{I}}_i,\hat{H}] = [\hat{\vec{I}}_j,\hat{H}]$. From the standard formula

$$\frac{\mathrm d}{\mathrm dt}\langle Q \rangle = -\frac{\mathrm i}{\hbar}\langle[\hat{Q},\hat{H}]\rangle$$

it is immediately obvious that $\mathrm d\langle\vec{I}_i\rangle/\mathrm dt = \mathrm d\langle\vec{I}_j\rangle/\mathrm dt$, thus proving that the spins $i$ and $j$ move in concert.

Regarding ethene: you are correct that the cis and trans (and vicinal) couplings in ethene are different. However, all four of the protons in ethene are chemically equivalent and must be considered together as a group. It is not correct to look at two out of the four protons and call them magnetically inequivalent, without also taking into account the remaining two. In this case, it turns out that if you do the maths, the relevant commutators don't cancel out if you only consider two out of the four nuclei. However, when you consider all four nuclei together, the relevant operators fully commute.

Now, none of the simplifications above apply to magnetically inequivalent nuclei. These experience very tiny instantaneous differences in magnetic fields – hence the name – and therefore all the couplings in the system end up being reflected in the spectrum, as we should have expected a priori. Returning to your analysis ("the coupling constants are the same, so shouldn't they be the same?"), the issue is that the coupling constant $J_\ce{AX}$ is only one part of the coupling term in the Hamiltonian, which is $2\pi J_\ce{AX}(\hat{\vec{I}}_\ce{A}\cdot\hat{\vec{I}}_\ce{X})$. Although $J_\ce{AX}$ is indeed equal to $J_\ce{A'X'}$, the dot product $(\hat{\vec{I}}_\ce{A}\cdot\hat{\vec{I}}_\ce{X})$ is in general not equal to $(\hat{\vec{I}}_\ce{A'}\cdot\hat{\vec{I}}_\ce{X'})$.

Warning: as with the ethene case, one cannot look at individual A and A' nuclei, because they are symmetry-related. You have to look at both of them together as a group. Once you do that, though, the spectra become possible to analyse.

Can magnetic inequivalence be explained without any QM?

This was the original crux of my answer, but I realised it is slightly tangential. I leave it in, however, since there is a possibility it may be useful.

The Hamiltonian of a spin system can be written as:

$$\hat{H} = \sum_i \omega_i \hat{I}_{iz} + \sum_i\sum_{j<i}2\pi J_{ij}(\hat{\vec{I}}_i\cdot\hat{\vec{I}}_j) \tag{1}$$

where $i$ ranges from $1$ to $n$, the number of spins in the system, $\omega_i$ is the Larmor frequency of spin $i$, and $J_{ij}$ is the coupling constant between nuclei $i$ and $j$ in $\pu{Hz}$ (hence the factor of $2\pi$).

Under a particular condition - when $|\omega_i - \omega_j| \gg J_{ij}$, or "weak coupling" - the term $$(\hat{\vec{I}}_i\cdot\hat{\vec{I}}_j) = \hat{I}_{ix}\hat{I}_{jx} + \hat{I}_{iy}\hat{I}_{jy} + \hat{I}_{iz}\hat{I}_{jz} \tag{2}$$ may simply be approximated by $\hat{I}_{iz}\hat{I}_{jz}$. This is called a "secular" approximation, cf. Appendix A.6 of Levitt. The reduced Hamiltonian is then

$$\hat{H} = \sum_i \omega_i \hat{I}_{iz} + \sum_i\sum_{j<i}2\pi J_{ij}(\hat{I}_{iz}\hat{I}_{jz}) \tag{3}$$

Since the Hamiltonian only consists of $\hat{I}_z$-type operators, the eigenstates of the Hamiltonian will simply be products of the eigenstates of $I_{iz}$:

$$\Psi = \psi_1\psi_2\cdots\psi_n \tag{4}$$

where $\psi_i$ is either $\alpha_i$ (up spin) or $\beta_i$ (down spin). The energies of such systems are then found by using $\langle\alpha_i|\hat{I}_{iz}|\alpha_i\rangle = +1/2$ and $\langle\beta_i|\hat{I}_{iz}|\beta_i\rangle = -1/2$. In general, therefore, the first term in the Hamiltonian will give some linear combination of $\omega_i$ (with each coefficient being $\pm 1/2$) and the second term in the Hamiltonian will give another linear combination of $2\pi J_{ij}$ (with each coefficient being $\pm 1/4$).

The allowed transitions correspond to a difference between these energies, and we regain the simple picture that we learn in introductory NMR: peaks at $\omega_i$ (or their equivalent in the chemical shift scale), and splitting between peaks of exactly $2\pi J_{ij}$ (or $J_{ij}$ in the Hz scale). Often, this is taught using a simplified model of considering each spin to be like a bar magnet, i.e. it is either up or down spin. With this model one may achieve some understanding of first-order spectra, and can rationalise the $n+1$ rule, for example.

However, this is an assumption which leads to precisely equation $(4)$. By assigning a distinct "spin" to each nucleus, we discard the possibility of quantum entanglement, i.e. we discard any possible quantum states that are not simple direct products. For example, consider the two-spin wavefunction

$$\Psi = \frac{1}{\sqrt{2}}(\alpha_1\beta_2 + \beta_1\alpha_2) \tag{5}$$

This is a perfectly valid wavefunction, but it is not possible to assign an up or down spin to either of the two nuclei. Equivalently, by assigning up or down spins to each nucleus, we are ruling out the possibility of any such states being important.

Now, this is not a problem as long as we are in the weak coupling regime. It may be helpful to think of the discarded term $(\hat{I}_{ix}\hat{I}_{jx} + \hat{I}_{iy}\hat{I}_{jy})$ as a perturbation to the simplified Hamiltonian in equation $(3)$. The unperturbed eigenfunctions of the simplified Hamiltonian are given by equation $(4)$. If the perturbation is small, i.e. if we are in the weak coupling regime, then the eigenfunctions of the full Hamiltonian are very close to that of the simplified Hamiltonian, and hence our simplified model (with discrete up and down spins) works. However, if the perturbation is large, i.e. strong coupling, then the simplified model fails. The eigenstates are no longer the same, the energy eigenvalues are modified, and hence the transition energies (which depend on the energy eigenvalues) and intensities (which depend on the eigenstates) are totally different.

Having two nuclei which are chemically equivalent, i.e. having the same chemical shift, is a case which simply not play into our simplified picture above. Since they have the same chemical shift, $|\omega_i - \omega_j| \equiv 0$ and hence weak coupling is never true. Note that chemically equivalent nuclei still possess a coupling constant $J_{ij}$ to each other and hence, in fact, chemical equivalence represents the most extreme case of strong coupling.

Fortuitously, when two chemically equivalent nuclei are also magnetically equivalent, it can be shown that the coupling is simply not manifested in the spectra (Appendix A.9, Levitt). Therefore, we can still use our simplified model, but simply with the caveat that "equivalent protons do not couple to each other". Strictly speaking the model is already incorrect, because the eigenstates of the system are no longer products of individual spin functions (see my answer here).

This is, however, not the case for magnetically inequivalent nuclei. Therefore, we have to throw our simplified model out - it just does not work here. It is therefore – to the best of my knowledge – not possible for one to explain the physical origin of magnetic inequivalence based on simplified arguments such as how "spin A sees spin B".


Finite probability from discrete distribution and zero probability from continuous distribution in Maxwell-Boltzmann statistics

The probability of any one specific velocity is not zero, it's just very, very small.  Infinitismal, or tending towards zero, is how we usually put it.  The reason is because the spacing between allowed translational states is inversely proportional to the size of the container, and when you are talking about a macroscopic container (a bottle of gas, say, or beaker of liquid) then from the atomic point of view this is gargantuan and the spacing between allowed translational energies very, very small.  (If you want to get a feel for this, look up the allowed energies for a particle in a box.  You'll see that the size of the box goes in the  denominator and as the box gets bigger the spacing between energy levels gets smaller and smaller.  You can amuse yourself by calculating the exact difference between the 1st and 2nd energy levels for, say, an H atom in a box 1cm x 1 cm x 1cm.  It's really small!)

So small that to do the sum over states in your top formula is a gigantic pain, quite challenging.  But if you convert it to an integral, the integral can be done pretty easily.  So that's what we do.  We replace a huge sum, over 10^20 terms or more, each of size 10^-20 or less, with an integral.  Think of it as the reverse of when you learned integral calculus, and in order to figure out an integral you replaced it with a large sum -- and let the size of each term go to zero while the number of terms went to infinity.

Here, you are taking a sum which has a very large number of terms, each of which is very small, and replacing it with an integral.  It's not exact, but it's very good, and it becomes perfect in the limit of an infinite-sized container, and anyway the errors for ordinary-sized containers are ridiculously small, unmeasureable.


Why work done Is 0 and why not a negative value?

There is no work done on moving a piston against zero external pressure (not if it is ideal, ie lacks mass and there is no friction in its movement). You might consider the definition of $pV$ work
$$w = -\int_{V_\mathrm{ini}}^{V_\mathrm{fin}}p_\mathrm{ext}
\mathrm{d}V$$
Obviously if the external pressure $p_\mathrm{ext}=0$ then $w=0$.

Conceptually, the point is that a collision by a single gas molecule would suffice to drive the piston to its extended position. In addition, that gas molecule would expend no energy, it would not slow down or change course in pushing the piston, it wouldn't feel it at all (this can get complicated, as explained further below). As absurd as that sounds, this is accurate within the constraints imposed by this conceptual question.

The following schematic contrasts the effect of a gas molecule impinging upon an ideal versus a slightly more realistic piston.

In the real case forces during the collision transfer momentum to the piston and reduce or reverse the forward momentum of the gas molecule. In the case of the ideal piston the collision is best modeled as a perfectly inelastic collision between the massless piston and the gas molecule, the resulting united body having the same momentum (in the direction of the piston axis†) as the original momentum of the gas molecule. There are no forces however and therefore no work is done.

† Components of the molecule's momentum along the orthogonal direction are conserved as well, and the molecule must be reflected once it reaches the chamber wall.


Color difference among oxovanadium species

The colour will depend on several factors. You've identified the oxidation state (+5 in both cases) as one.
Other factors are:

Ligands
Complex shape

With $\ce{[VO2]}^{+}$, the actual complex ion is $\ce{[VO2(H2O)4]}^{+}$ (Taken from ChemGuide). So our ligand is water, and the shape is tetrahedral (or square-planar).

The $\ce{[VO4]^{3-}}$ ion usually only exists at high pH in solution but polymerises at lower pHs. These polymeric ions can have colour (e.g. $\ce{[V10O28]^{6-}}$ is orange).(Link for colour / polymerisation)
However, I could find no details as to the ligands (I would assume that they would be hydroxide ions at high pH) nor shape.

If we are talking about the solid, then the same applies... does the solid have any water of crystallisation. (Think copper sulfate... blue when "wet" and white when "dry").


What is difference between H⁺ and proton?

There is no chemical difference, only a psychological one: how do you think about it. They are both the same thing, but many people associate $\ce{H+}$ ions with chemical reactions and protons with particle physics.
A hydrogen atom has one electron and a proton, no neutron. Therefore $\ce{H+}$ is just a proton.

That is why acids are sometimes referred as proton donors as they donate $\ce{H+}$ to a base (also known as proton acceptor).


Among the $4$ equilibria, only $3$ are linearly independent, because we have $(1)+(3)=(2)+(4)$. Let's then choose only three independent reactions

\begin{align}
\ce{NH3+-C5H10-COOH}&\ce{<=> NH3+-C5H10-COO- + H+},\\
\ce{NH3+-C5H10-COO-}&\ce{<=> NH2-C5H10-COO- + H+},\\
\ce{NH3+-C5H10-COO-}&\ce{<=> NH2-C5H10-COOH},
\end{align}

as the basis equations, which are obtained from $(2)$, $(4)$, and $(1)-(2)$. The equilibrium constants of these equations are $K_{a1}$, $K_{a2}$, and $K_{a2}/K_{a1}=10^{-7.24}\ll 1$. Therefore most of the neutral amino acid molecules would exist in the zwitterionic form $\,\ce{NH3+-C5H10-COO-}$. Neglecting the last equilibrium, the equilibrium constants $K_{a1}$ and $K_{a2}$ are those of the $\ce{-COOH}\,$ and $\ce{-NH3+}$ groups in the zwitterions $\,\ce{NH3+-C5H10-COO-}$, which is much more abundant than the original molecules $\,\ce{NH2-C5H10-COOH}\,$ making the latter negligible.


Which element is undergoing oxidation in the combustion of methane?

I believe I have answered my own question - I assumed that H had an oxidation state of -1, when it appears to have an oxidation state of +1.


Redundant out of plane bends in gaussian not recognised

According to Gaussian's Modredund description, there is no such type as "O" anymore in Gaussian16.
Hence, the input deck is rejected. Gaussian16 provides generalized internal coordinates, for which more complex expressions are possible.


How does the 1.5 bond order of ozone contribute to its instability?

It's important to bear in mind stability is always a relative judgment.  Things are stable or not only with respect to that into which they could turn. So we only say ozone is "unstable" because it can and will easily decompose into oxygen (O2) if given a chance.  In some sense it might even be more correct to say what's important here is the stability of O2 (which derives from its strong double bond).  If O2 weren't possible, or were significantly higher in energy itself, then we would say O3 is more "stable" just because it wouldn't have anything into which it could as easily turn, so it would stick around longer.  So "stability" is not just a property of one molecule, but a property of that molecule plus all the other molecules into which it could turn.

A famous example of this is the high stability of N2 (with its very strong triple bond), which means certain compounds (like TNT or ammonium nitrate) with nitrogen atoms in higher oxidation states (e.g. bonded to O atoms) can readily decompose to products including N2 and O2, producing a lot of energy.  That is, they are good explosives, and you can consider that as coming from the strong stability of the O2 and N2 molecules.  (This is somewhat akin to the argument in organic chemistry that the pKa of an acid depends strongly on the stability of its conjugate base -- that into which it can turn.  The more stable the base, the higher the pKa, i.e. less "stable" the acid.)

All that said, fractional bond order does indeed suggest resonance, because it suggests the failure of a single Lewis structure to describe the molecule accurately.  And indeed O3 needs two resonance structures to describe it. It's also correct to say resonance stabilizes a molecule, but, again -- with respect to what?  In general, the answer is "with respect to a similar (perhaps hypothetical) molecule that does not have resonance."  So O3 is more stable (because of resonance) than some hypothetical O3 molecule trapped in one of the O3 resonance structures.  It can still be "unstable" for most purposes, it's just less unstable than this hypothetical non-resonant O3.


Not much is going to happen. Due to low solubility of $\ce{Ca(OH)2}$, further suppressed by the presence of calcium cations from the dissolved $\ce{Ca(NO3)2}$, you get a suspension of calcium hydroxide, which unfiltered looks like white cloudy liquid and is also known as milk of lime.

Regarding the $\mathrm{pH}$ if the solution, $\ce{Ca(NO3)2}$ is prone to hydrolysis and alone would create slightly acidic medium:

Step 1:

$$
\begin{align}
\ce{Ca(NO3)2 + H2O &<=> CaOHNO3 + HNO3} \tag{R1.1}\\
\ce{Ca^2+ + 2 NO3- + H2O &<=> CaOH+ + 2 NO3- + \color{red}{H+}}\tag{R1.2}\\
\ce{Ca^2+ + H2O &<=> CaOH+ + \color{red}{H+}}\tag{R1.3}
\end{align}
$$

Step 2:

$$
\begin{align}
\ce{CaOHNO3 + H2O &<=> Ca(OH)2 + HNO3}\tag{R2.1}\\
\ce{CaOH+ + NO3- + H2O &<=> Ca(OH)2 + NO3- + \color{red}{H+}}\tag{R2.2}\\
\ce{CaOH+ + H2O &<=> Ca(OH)2 + \color{red}{H+}}\tag{R2.3}
\end{align}
$$

(Here .1, .2, .3 stand for molecular, complete ionic, and net ionic equation, correspondingly.)

But, when you add enough $\ce{Ca(OH)2}$ (saturated solution), which is a strong base:

$$\ce{Ca(OH)2 <=> Ca^2+ + 2 \color{blue}{OH-}}\, ,$$

I'd expect the resulting solution to be basic.


How does one synthesise 2-Monoglycerides from Triglycerides?

One way I have learnt is by taking a page out of the human body's book and using pancreatic lipase, found in pancreatic juice, to catalyse the reaction:

$triglyceride+ 2H_2 O \rightleftharpoons   2$-$monoglyceride+2 FAA$

Where FAA stands for, 'Fatty acid anion.'

If someone has a better answer I'll readily accept it over this one, this is just one method.


Does the reaction become more or less favorable as temperature increases?

You should use the van't Hoff equation together with the standard enthalpy of reaction $\Delta H^\circ$ for the process to predict the response to a temperature change. This should provide you with a prediction consistent with Le Châtelier's principle that an endothermic reaction is pushed in the direction of products as $T$ increases.

The general equation for the Gibbs' free energy $$\Delta G = \Delta H - T \Delta S$$ should not be relied upon to determine the response of the reaction equilibrium constant to $T$. The expression is valid at constant $T$, but this is not the problem with using this equation, rather it is its generality: it is not limited to describing reactions at equilibrium. Note that $\Delta G = 0$ for a reaction at equilibrium.

On the other hand, you could consider using $$\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ$$to guide you, and the result should be consistent with analysis based on the van't Hoff equation. It is simpler however to work with the van't Hoff equation directly.  In fact, assuming $\Delta H^\circ$ and $\Delta S^\circ$ are independent of $T$ you can use fairly straightforward mathematical manipulations to arrive at the usual integrated form of the van't Hoff equation.


Did O₄ turn out to explain why liquid oxygen appeared to defy Curie's law? If not, what does?

The observation that oxygen did not seem to strictly follow Curie's law (where $1/\chi$ is linear with temperature)  has been known since Kammerlingh Onnes pointed it out in the 1910's. By the late 1920's, it was known that diluting the liquid with nitrogen preserved Curie's law, suggesting that it took oxygen in close proximity to other oxygen to be the issue.

Fast forward some 80 years, and the computational tools available to us provide some key insight. As one such paper, Ab Initio Molecular Dynamics Investigation of the Structure and the Noncollinear Magnetism in Liquid Oxygen: Occurrence of O$_{4}$ Molecular Units takes a look at structure and properties of liquid oxygen. Quoting from their conclusion:

  In  conclusion, we  performed ab  initio molecular  dynamics  to  investigate  the  noncollinear magnetism  of  a system with an evolving atomic structure. Application to liquid oxygen provides us with a picture in which the large majority of colliding O$_{2}$ molecules assume structural and magnetic configurations which closely resemble those in the O$_{4}$ molecule. Formation of truly long-living molecular O4 units  also  occurs but involves a considerably smaller fraction of O$_{2}$ molecules.

So, it would seem that O$_{4}$ units, whether long lived 'molecules' or temporary short ranged order is the answer to the deviation from Curie's law in liquid oxygen.

Several comments on these O$_{4}$ units - first, as noted in the paper above, 'long lived' means under a picosecond (which doesn't really seem long-lived to me, but...). Further, the authors above point to The Infrared Spectrum of Bound State Oxygen Dimers in the Gas Phase to explain what they mean by O$_{4}$, which they call a "somewhat rigid (O$_{2}$)$_{2}$ molecule".


How is the energy of reactants in a chemical reaction found ?

Energy itself is usually not measured, but rather change in energy ($\Delta E$) is the quantity of practical importance. One may be interested in change in free energy as well, $\Delta G$, changes in enthaly $\Delta H$, and even change in entropy $\Delta S$.

A common way to evaluate the increased energy of transition state ($\Delta E^{\ddagger}$) above the reactant is via the Arhennius Plot.  Here, one evaluates the rates of reaction ($k$) at a set of different temperatures ($T$ in Kelvin).  It is fairly easy to show that a plot of the $\ln(k)$ vs. $1/T$ gives a line whose slope is related to $\Delta H^{\ddagger}$, and where the intercept is related to $\Delta S^{\ddagger}$. (The ideal gas constant is in here too, but it's easy to accomodate.)

(There is an analogous plot for energy change for a reaction, which is called the van't Hoff plot, but it generally requires both reactant and product to be in equilibrium such that the concentrations of both are observable.)

Especially the enthalpy of reaction can be found by methods such as bomb calorimetry.  This measures the heat that is liberated upon reaction.


How to find oxidation state of Co and Cr in [Co(NH₃)₄(NO₂)₂][Cr(NH₃)₃(NO₂)₃]?

Conventionally, first complex ion is positively charged and the latter is negatively charged. Suppose thess charges are $p+$ and $q-$, respectively. Since two ion ratio is 1:1, $|p+|= |q-|$.

Suppose oxidation number of $\ce{Co}$ is $n$ and that of $\ce{Cr}$ is $m$. Hence, for cation complex (Note that the net charges of $\ce{NH3}$ and $\ce{NO2-}$ is $0$ and $-1$, respectively):
  $$n+0\times4+(-1)\times2=+p = n-2$$
and for anion complex:
 $$m+0\times3+(-1)\times3=-q=m-3$$
Since, $|p+|= |q-|$,
$$m-3= +p \; \text{and  }  n-2=-p$$
Or, $m+n=5$.
The only answer agree with this is (A) or (B):

If (A) is correct, then, $\ce{Co^2+}$ and $\ce{Cr^3+}$. Substitute these values in original two equation: $+2+0\times4+(-1)\times2=+0 $ and $+3+0\times4+(-1)\times3=+0 $. This can't be.

If (B) is correct, then, $\ce{Co^3+}$ and $\ce{Cr^2+}$. Substitute these values in original two equation: $+3+0\times4+(-1)\times2=+1 $ and $+2+0\times4+(-1)\times3=-1 $. This is correct.

Therefore, your answer is (B), and the oxidation numbers of $\ce{Co}$ and $\ce{Cr}$ ared $+3$ and $+2$, respectively.


Software or library for creating images of ball and stick models?

For general open source solutions:

Avogadro - Molecular editor GUI. Has a Python API.
Jmol - Chemical structure viewer. Fully scriptable using a custom scripting language. It also has a JavaScript-only version named 'JSmol', allowing models to be displayed in web browsers.
ghemical - Has a GUI for editing molecules, but no API.

For a $\LaTeX$ solution, see this question on the $\TeX$ Stack Exchange site: Help creating solid cylinders along a path in Tikz for crystal lattice diagram

Other solutions:

If you have access to Mathematica (e.g. on a Raspberry Pi running Raspbian [which comes with a free copy of Mathematica]), you can use Mathematica's ChemicalData built-in symbol to create ball and stick models.


Importance of halogens and nitrogen in degree of unsaturation

So here is the method I learned years ago without memorizing an awkward formula. The concept is to reduce the molecular formula to a hydrocarbon and compare it to the most saturated hydrocarbon bearing the same number of carbon atoms.  Consider the infamous free-base, hydroxychloroquine, $\ce{C18H26ClN3O}$.

The following atom replacements may be done in any order. Ignore divalent atoms ($\ce{O,S}$). The formula is now $\ce{C18H26ClN3}$. Replace all halogens with an equal number of hydrogens $\Rightarrow \ce{C18H27N3}$. Substitute $\ce{CH}$ for each nitrogen or trivalent atom $\Rightarrow \ce{C21H30}$. The most saturated $\ce{C21}$ hydrocarbon is $\ce{C21H44}$. Now do the math: $\ce{C21H44}$ - $\ce{C21H30}$ = $\ce{H14}$. Now divide 14 by 2. The Degree of Unsaturation is 7.
As to the question as to why oxygen is dropped from the equation, if the oxygen in hydroxychloroquine weren't there, the degree of unsaturation would still be the same. For example, the Degrees of Unsaturation of tetrahydrofuran and cyclobutane are the same, D.U. = 1.


The effect of concentration on the rate of a homogenous reaction

I've also read online that adding $\ce{H2O2}$ and $\ce{H2SO4}$ creates a "Piranha" solution [...]

This is true - in principle. In reality, Piranha solution is made from concentrated sulfuric acid and 30% hydrogen peroxide.
In the case of your experminent, you will most likely use diluted sulfuric acid and aqueous solutions of hydrogen peroxide with a much lower concentration. Consequently, the Piranha effect will not play a role in your experiment, neither is the oxidizing power of sulfuric acid relevant here.
On a side side note, treating a sample with conc. $\ce{H2SO4}$ is a method for the qualitative determination of iodide: Upon heating, one can observe the violet vapour of iodine.
In your case, you need the sulfuric acid to keep the pH value low. Under these conditions, $\ce{H2O2}$ acts as an oxidant:
$$\ce{H2O2 + 2 H+ + 2 e- -> 2 H2O}$$

Does dissolving into ions also count as a reaction?

No.

Does the Starch (for which I have no given chemical formula) react in solution A?

Starch is a mixture of two natural polymers (amylopectin and amylose), both made from D-glucose. Amylopectin doesn't play a role here, but it doesn't disturb either and it's just cheaper to let you work with the natural mixture.
It's the helical amylose that is relevant here - and it acts as a reagent.
What actually happens in the reaction is the following:
$$\ce{2 I- -> I2 + 2 e-}$$
$$\ce{I2 + I- -> I3-}$$
The linear triodide ion forms a strongly coloured inclusion complex with the amylose in the starch.


The refluxing of methanol or ethanol over magnesium is indeed the method of choice to dry these solvents.

As far as the iodine is concerned, this is a classical trick, which is also used to initiate Grignard rections. Is is assumed that magnesium iodide is formed, which is soluble in the alkanols. As a result, a fresh, activated metal surface is left behind.

This might even work, when the magnesium metal is covered with a very thin oxide layer.


How does selenium protect against mercury poisoning?

Selenium has a very high affinity for mercury. It strongly binds mercury to itself, thus neutralizing the reactivity of mercury. The complex formed Hg - Se is not absorbed by body and excreted out from the body.

In this way mercury is unable to lodge in fatty tissues, resulting in no damage to the body.


What is the reaction rate constant of an effective reaction?

It is not clear that they are related. It is simpler to consider the equivalent scheme $A \rightarrow C \rightarrow E$ with rate constnats $k_1,\, k_2$ then the decay of A is
$A_t/A_0= \exp(-k_1t)$ the amount of  C rises and falls; it is
$$\displaystyle C_t/A_0 = \frac{k_1}{k_2-k_1}(\exp(-k_1t)-\exp(-k_2t))$$ and
E is
$$E/A_0 = 1-A_t-C_t=1-\exp(-k_1t)-\frac{k_1}{k_2-k_1}(\exp(-k_1t)-\exp(-k_2t))$$

When $k_2 $ is very large and much bigger than $k_1$ then $\exp(-k_2t) \rightarrow 0 $ and $\displaystyle E/A_0= 1-(1+\frac{k_1}{k_2})\exp(-k_1t)$ and in the opposite limit $k_1 \gg k_2$ then $E/A_0= 1-\exp(-k_2t)$.

So you can choose which limit is important. In either case the amount of E rises as time proceeds but with different rate constants. In the intermediate case there seems to be no rate constant but only a slope to the curve which will depend on the time at which the measurement in made. You could for example find the half rise time and try to relate that to rate constants, but it looks messy.


No, what is occurring is the mercury is forming a eutectic liquid (also called an amalgam) with the aluminum. Aluminum is quite reactive with air/water, but typically aluminum can form an oxide coating that protects it from a complete reaction with air/water (passivation). When the mercury is added and the surface of the aluminum is damaged (by the drill) the two are free to mix per the phase diagram below now the aluminum in the liquid may react with the air/water, but now on the liquid the oxide formed is easily disrupted, and thus ineffective in protecting the aluminum from oxidation. This allows more aluminum to oxidize and the mercury to dissolve more aluminum as aluminum is depleted from the liquid. This creates a steady state condition where aluminum is continually dissolved and oxidized until the aluminum is completely consumed.

A screw dislocation is a microscopic feature only visible with atomic scale imaging such as with a tunneling electron microscope (TEM).


Balancing the oxidative decomposition of FeS2

I found a solution which works,
$$\ce{4FeS_2 + 15O_2 +14H_2O = 4Fe(OH)_3 + 8H_2SO_4}$$

This is how I found it. Do you know linear algebra ? Because that is very useful. Anyway without to know this is linear algebra you'll be able to understand.

So we are looking about the stoichiometric coefficients $\{\ce{a, b, c, d, e}\}$ of this reaction :

$$\ce{aFeS_2 + bO_2 + cH_2O + d Fe(OH)_3 + eH_2SO_4 \rightarrow 0}$$

Now what you need to do is first to find the number of different elements which are involved in your reaction, here they're four : $\{\ce{\color{\green}{H}, \color{\red}{O}, \color{\purple}{S}, \color{\orange}{Fe}}\}$. I'm not that good in the use of $\LaTeX$ but now the idea is to make column vectors under each of the reagents of your reactions and in these vectors you put in the same order for each the number of each elements there are in and if there is not just put a zero. It will gives you this :

$$a\cdot\begin{pmatrix} \color{\green}{0} \\ \color{\red}{0} \\ \color{\purple}2 \\ \color{\orange}{1} \end{pmatrix}+b\cdot\begin{pmatrix} \color{\green}{0} \\ \color{\red}{2} \\ \color{\purple}{0} \\ \color{\orange}{0} \end{pmatrix}+c\cdot\begin{pmatrix} \color{\green}2 \\ \color{\red}{1} \\ \color{\purple}{0} \\ \color{\orange}{0} \end{pmatrix}+d\cdot\begin{pmatrix} \color{\green}3 \\ \color{\red}{3} \\ \color{\purple}{0} \\ \color{\orange}1 \end{pmatrix}+e\cdot\begin{pmatrix} \color{\green}2 \\  \color{\red}4 \\ \color{\purple}1 \\ \color{\orange}0 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}$$

It can be easily transform into a system like this :

$$\begin{cases}
2c+3d+2e=0 \\
2b+c+3d+4e=0 \\
2a+e=0 \\
a+d=0
\end{cases}$$

So here we are in a good way because we have more unknowns than equation so if the system converge it has an infinity of solutions. And we can multiply by the number we want each stoichiometric coefficients and still have the same reaction.

So here we will determine four unknowns in terms of an other, I chose $e$. And I got :
$$$$\begin{cases}
a=-e/2 \\
b=-15e/8 \\
c=-7e/4 \\
d=e/2 \\
e=e
\end{cases}$$$$

I multiply then by $8$ to have only integers which give me with $e=-1$ :

$$\ce{4FeS_2 + 15O_2 + 14H_2O - 4Fe(OH)_3 - 8H_2SO_4 \rightarrow 0}$$

Then $$\ce{4FeS_2 + 15O_2 + 14H_2O = 4Fe(OH)_3 + 8H_2SO_4}$$


$I$ is the ionic strength of a solution, and is a measurement of the concentration of ions in a solution. $m_i$ is the concentration of the ion i, and $z_i$ is the charge of that ion. The $\frac{1}{2}$ in front of the summation is a result of the fact that both positive and negative ions are considered when calculating the ionic strength. $m_i$ could either be a measurement of Molarity or molality, and your book should state which definition you are using (I assume molality due to the lowercase m).


How do we know the number of atoms in an inorganic compound?

For all compounds, the systematic name of that compound should provide information on the composition and structure. Your case is an example of the nomenclature of binary covalent compounds. Consider the following example:

  dinitrogen tetroxide

The first word in this name tells us which element is first in the formula and how many of them there are. In this example the "di-" prefix tells us we have two of something.

$$\mathrm{dinitrogen = }\ \ce{N2}$$

The second word in the name tells us which element is second in the formula and how many of them there are. In this example the "tetr-" prefix tells us that there are four of something. By convention, we usually use the "ionic" form of the second element in the name even if it is a covalent compound, so "oxide" means "oxygen".

$$\mathrm{tetroxide=}\ \ce{O4}$$

Therefore, the full formula for this compound is $\ce{N2O4}$.

The prefixes are as follows (up to 10, with larger numbers being rare). Note that the trailing "o" or "a" of a prefix is dropped with oxygen to increase ease of pronunciation.

$$\begin{array}{|c|c|} \hline
\mathrm{prefix} & \mathrm{number}\\ \hline
\mathrm{mon(o)} & 1 \\
\mathrm{di} & 2 \\
\mathrm{tri} & 3 \\
\mathrm{tetr(a)} & 4 \\
\mathrm{pent(a)} & 5 \\
\mathrm{hex(a)} & 6 \\
\mathrm{hept(a)} & 7 \\
\mathrm{oct(a)} & 8 \\
\mathrm{non(a)} & 9 \\
\mathrm{dec(a)} & 10 \\
\hline
\end{array}$$

  What about names like iodine monochloride? How many iodine atoms are there in the formula?

The "mono-" prefix is often left out if there is no compound out there that have more than one of those atoms in it. Thus, iodine monochloride is $\ce{ICl}$. Why not iodine chloride? There is another compound out there "iodine trichloride," which is $\ce{ICl3}$.


Why are carbons of Inositol chiral centers?

Consider 2,3-butanediol.

This compound has 2 stereocenters. The meso version of this compound is achiral. However, there are still two stereo centers. As drawn, the left one is (R) and the right one is (S). It is precisely because these two stereocenters are identical but opposite in configuration that leads the molecule to be meso.

A carbon center is asymmetric if it has 4 different substituents regardless of whether or not the whole molecule is chiral. The ring substitution in inositol is such that the centers around the ring are not identical (including when you look at stereochemistry--for example, (R) has higher priority than (S)). This of course isn't enough to help you easily make stereochemical assignments for all of the centers of inositol, but hopefully, this allows you to understand why the centers are still chiral centers.

As an aside, there are a few centers that cannot be chiral because they lie on a mirror plane. Interestingly though, changing the configuration of the center produces a different molecule. These centers are called pseudochiral, and we label them with lower case r and s.


Interesting question.
It looks like this is a rather common notation in geology and refers to the total iron(II) and iron(III) oxide (e.g. ferrous and ferric) content.
For a chemist this is probably somewhat confusing as “total” is normally denoted via subscripted abbreviation “tot”.
Flipping through numerous sources, it appears there is no standardized notation and alongside with $\ce{FeO_T},$ $\ce{FeO_t},$ $\ce{FeOt}$ and $\ce{FeO^\mathrm{t}}$ are being used.

An introductory textbook by Shikazono [1, p. 21] explicitly defines $\ce{FeO^\mathrm{t}}$:

  $\ce{FeO^\mathrm{t}}$: Total iron $(\ce{FeO} + \ce{Fe2O3})$

Reference

Shikazono, N. Introduction to Earth and Planetary System Science: New View of Earth, Planets and Humans; Springer: Tokyo; New York, 2012. ISBN 978-4-431-54058-8.


Rate-limiting in transition state theory

In transition state theory (TST, goldbook) one of the necessary assumptions is that reactants and products are in equilibrium. In principle this gives us
\begin{align}\ce{
A &<=> [AB]^{$\ddagger$} -> B\\
B &<=> [AB]^{$\ddagger$} -> A\\
A &<=> B,
}\end{align}
and in addition to this, also
\begin{align}\ce{
B &<=> [BC]^{$\ddagger$} -> C & B &<=> [BD]^{$\ddagger$} -> D\\
C &<=> [BC]^{$\ddagger$} -> B & D &<=> [BD]^{$\ddagger$} -> B\\
B &<=> C                      & B &<=> D,
}\end{align}
as well as
\begin{align}\ce{
C <=> [BC]^{$\ddagger$} -> & B <=> [BD]^{$\ddagger$} -> D\\
D <=> [BD]^{$\ddagger$} -> & B <=> [BC]^{$\ddagger$} -> C\\
C <=> & B <=> D,
}\end{align}
and
$$\ce{
C <=> A <=> D\\
[AB]^{$\ddagger$} <=> [BC]^{$\ddagger$} <=> [BD]^{$\ddagger$}
}.$$

You can use TST to estimate rate constants.
$$k = \frac{\mathcal{k}_\mathrm{B} T}{\mathrm{h}} \exp\left\{ −\frac{\Delta^\ddagger{}G^\circ}{\mathcal{R} T}\right\}$$

Since all components have to be in equilibrium, you can predict a ratio between them via Boltzmann statistics, hence:
$$\frac{N([BC]^{\ddagger})}{N([BD]^{\ddagger})} =  \exp\left\{ −\frac{\Delta^\ddagger{}G^\circ([BC]^{\ddagger})-\Delta^\ddagger{}G^\circ([BD]^{\ddagger})}{\mathcal{R} T}\right\}=\frac{N(C)}{N(D)} $$

The preceding equilibrium does not need to be considered to determine the ratio of the products. However, it is important to note, that there are many assumptions necessary for TST to work. For example, at higher temperatures, anharmonic corrections have to be considered. More crucial is the assumption, that the transition can be described as a translatory movement, i.e. it is treated with classical mechanics.

For a system with a preceding equilibrium, it is well possible, that the TST approximation breaks down completely.


How to calculate the heat absorbed by a reversible process?

I assume you mean the first law as

$$dU = \delta q + \delta w$$

where $dU$ is the (state) change in internal energy, $\delta q$ is the (path dependent) work, and $\delta w$ is the (path dependent) work.

Saying the walls are diathermal says something about temperature (as you should confirm). Then, you should find a relationship between temperature change and internal energy change for the substance in your system.


According to the Cahn–Ingold–Prelog priority rules, a double bond considered as ($\ce{CCX}$), where $\ce{X}$ depends on the substitution on the double bond. In your case, $\ce{X}$ is hydrogen ($\ce{H}$). Thus, I arranged the priority as depicted in following diagram (chiral carbon in consideration is marked with asterisk):

Hydrogen is the least priority group on the chiral carbon. Among other three groups, the bridgehead carbon is a $\ce{-CH2 -}$ group, hence can be marked as ($\ce{CHH}$) on priority checking. The final two groups can be marked as ($\ce{CCH}$). I used red and blue color on one group (methine on the ring), and black on the other (double bond) to distinguish these two group. Since both has identical substitutions on first carbon, we need to moved to second carbon. On double bond, it is again ($\ce{CCH}$), but bothe red and blue carbon give ($\ce{CHH}$) priority checking. Thus, double bond has the highest priority as indicated. Therefore, assignment for this chiral center is ($S$)- as given in chemspider website.


Group Theory and Chemistry for Undergraduate Thesis

The first 4 chapters of Chemical Applications of Group Theory by F A Cotton should be enough for your student. However there are numerous books available.

Again chemwiki has a good documentation on group theory.


Are there chiral carbons in PVC polymers?

Yes, all polymer chains of the form - (CHR - CH2 -) are in principle chiral, the problem is that in the left and right side of the backbone in  BBL - CHCl - BBR is in effect indistinguishable.

Even in a PVC molecule with perfect tacticity, only the end groups (they usually are different) make the whole thing chiral, and they are too far away. In a structure with a defect tacticity, the next defects in the chain create chirality at a C atom, but they (usually) occur with the same probability distribution on both sides of the local carbon atom and the whole molecule.

That's why in polymers, optical activity is not observed, instead, the relative occurrence of pairs, triplets, etc. of neighbouring C* atoms with identical stereochemistry can be specified (and detected via NMR). That way, you can differentiate e.g. chains which go RRRRRSRRRRRSRRRRRS or RRRRRSSSSSSRRRRRRSSSSS, which will show remarkably different mechanical properties due to their crystalline structure.

A different thing are of course polymers with chiral side groups, e.g. proteins. A solution of those can (or rather does, unless the side group is racemic) show optical activity.


Relative Solubilities of Salts in Acidic Solution

Chloride, Bromide, and Sulfate all correspond to strong acids, while phosphate corresponds to a weak acid.

With 1M H+ available, phosphate will be essentially fully protonated, while the others will still be anions.


Why can alcohol undergo elimination just by sulphuric acid?

In E1, the way I understand it is that the resultant carbocation's protons become reasonably acidic as the conjugate base (an alkene) is very stable. They only require a weak base to be removed, so practically anything can remove them. It looks to me as if the likely weak base candidates are the water produced from the leaving group, or even another alcohol in the reaction mixture.

Primary alcohols will be less susceptible to dehydration by this method, as they follow an E2 mechanism which progresses much more slowly in acidic conditions due to the absence of a strong base (which would usually act to 'shove' the leaving group off, a weaker base gives a weaker 'push').

I have seen a few schemes for E1 and E2 in which the hydrogen sulphate ion acts as the weak base; the usual assumption is that there is a weak base that is able to capture the proton, but we don't necessarily care what it is because anything will do. As long as there isn't another half decent nucleophile in the reaction mixture, elimination will likely prevail.

As you can see from this image I found in Clayden Organic Chemistry, the proton in the reaction mechanism is often drawn as just 'falling off', emphasising that any weak base is capable, and abstraction of the proton is very likely by a large number of species.


What techniques can determine the concentration of ethyl acetate in a solution containing ethyl acetate, acetic acid, ethanol and water?

You might consider taking Aliquots of the reaction (or if you have the equipment directly sampling via a cannula) into a GC-FID or quantitative GC-MS. This way you could monitor the concentrations of all species in solution given the right GC method. Areas under each peak will give you relative amounts, you would need to use an appropriate known amount of an internal standard that remains constant to get absolute numbers.


Two different parts to the question I think:

Are the readings you are getting on your meter meaningful?
If formaldehyde levels are high, what (if anything) can/should you do about it?

Regarding interpreting the reading from the meter, there are a few things you should consider right off:

What does the manufacturer say about accuracy? Typically the results should be what's displayed +/- some amount (say 25%). If this isn't documented, you may be able to contact the manufacturer for more information.
What other substances can cause interference with the sensor. It's very possible that there are other contaminants in the air that are causing the reading to increase. Again, check the documentation and/or contact the vendor.

My experience (especially with non-professional grade equipment) is that direct reading instruments are typically good at helping to understand concentration trends. For example if you check the readings over the course of the day, you may find higher concentrations at certain times and lower concentrations at other times - could be linked to temperature, ventilation rates, or something else. The information is useful for understanding at a high level what's going on and for figuring out why you have a problem.

I have a lot less confidence in the absolute readings that they provide. Professional grade meters for formaldehyde can cost thousands of dollars and require regular calibration. And even then, if I wanted to be confident in my results, I'd prefer to use a wet chemistry method like one of these.

Now, assuming you are still convinced that you have a formaldehyde problem, what can you do about it?

Some people will say that you can attempt a "bake-off". Get some space heaters and run the indoor temperature up to as high as you can get it and keep it there for a day or two. The idea is that heating everything up will cause whatever formaldehyde reservoirs to fully "off-gas", and when the temperature returns to baseline, there won't be any formaldehyde left. I'd be cautious on following this method, however, as some studies have found that any reductions seen are generally short lived.

Three more realistic approaches are:

if you can increase the amount of outdoor air coming into the house, this will go a long way toward diluting concentrations.
NASA has also done research and found that certain types of house-plants are good at absorbing formaldehyde from indoor air.
Now that you're more aware of formaldehyde as a potential indoor air quality problem, keep it in mind the next time you update furniture or your home interior. There's increasing awareness in the market, and you should be able to find low VOC formaldehyde-free alternatives for most relevant products.


I think you are using the first approach that I described.  So, here is the analysis based on the first approach.  Suppose you have the following numbers of moles of A, B, C, and D in the reaction mixture at some time: $n_A$, $n_B$, $n_C$, and $n_D$.

And the total pressure of the mixture (in bars) is held constant at P.  So the partial pressure of species A is $$p_A=\frac{n_A}{n_A+n_B+n_C+n_D}P$$And the partial molar free energy of species A is $$\bar{G}_A=G^0_A+RT\ln{\left(\frac{n_A}{n_A+n_B+n_C+n_D}P\right)}$$where $G^0_A$ is the free energy of pure species A at the reference pressure of 1 bar.  Similar equations can be written for the other species.  The total free energy of the mixture at any point in time during the reaction is:  $$G=n_A\bar{G}_A+n_B\bar{G}_B+n_C\bar{G}_C+n_D\bar{G}_D$$
Now, let the concentrations of the chemical species in the reaction mixture be expressed as follows:
$$n_A=n_{A0}-a\xi$$
$$n_B=n_{B0}-b\xi$$
$$n_C=n_{C0}+c\xi$$
$$n_A=n_{D0}+d\xi$$
where the subscript 0 refers to the initial state and $\xi$ is the conversion.  To find the equilibrium condition, you want to find the value of $ \xi $ that minimizes the free energy of the solution mixture.  To do that, you have a calculus problem to solve.


Well, I might take a different approach at an answer, and try to get to a reasonable Lewis structure with data we already should have had from drawing structural diagrams! It's kinda upside down, since none of what's written below is happening in real-life chemistry.

The structure you have in mind is the diatomic oxygen molecule, with 12 electrons in its system overall.

$\hspace{33ex}$
$\hspace{45ex}$Source

While the peroxide ion overall has 14 electrons in its system, since each oxygen has 6 valence electrons and the system also has two extra electrons indicated by the $2-$ charge of the ion. Now, let's oversimplify this a bit: Imagine that in peroxide also, the oxygen atoms need to fulfill their octet configuration, hence each of them would have 8 electrons in its outer shell.

Let $N_b$ be the number of bonding electrons and $N_{nb}$ the number of non-bonding electrons of each oxygen atom. We know that

A single oxygen atom would have non-bonding electrons and bonding electrons that fill the octet.
All of the electrons in the system, which are the non-bonding electrons for each of the two oxygens, and the bonding electrons, make 14 electrons.

Hence we solve:

$$\Bigg\lbrace\begin{array}\\ N_b + N_{nb} = 8 \\ N_b + 2N_{nb} = 14 \end{array}\Bigg\rbrace \Rightarrow \boxed{N_b = 2},~\boxed{N_{nb} = 6} $$

Hence we could figure out how many bonding and non-bonding electrons there are in compounds we can draw Lewis structures for, by finding out the overall number of electrons (taking into account the number of electrons in the valence of each atom and the number of electrons lost or gained due to the charge of the ion) with a rudimentary approach.

Try to find out how many bonding and non-bonding electrons there are for the azide, permanganate, manganate, sulfate, oxalate, salicylate, and superoxide anions and thionyl, sulfuryl, nitrosyl and methanium cations.


There are (were) three main nomenclature systems used for naming organic compounds:

Common Name (or trivial name). Under this convention this molecule is neopentane. For alkanes, this naming system is based on the positioning of methyl substituents, and gives rise to the iso-, sec-, neo- namiong prefixes.
Derived Name. Who uses this method these days???? This system identifies the central carbon atom which has the most carbon groups attached, and then names the groups as a prefix, and ends the name with 'methane'. The formula are often written with the central carbon unit in square brackets. You may still see derived names on MSDS/SDS under the 'Other Names' section. Under this system this molecule is tetramethylmethane. I can think of more important things to teach you in class if this is what you are learning.
IUPAC Name
This is the accepted naming convention in use today, which allows for retained names, which usually have their origins in the trivial naming system. Its naming system is based on a systematic framework and is used universally. In older texts (pre 1970s), it was known as the Geneva system or congress. Under this convention, this molecule is 2,2-dimethylpropane.

I can't even find any decent references for you at this point, but will update if I find some.


Why don't electrons have the same spin even when in same orbit?

The Pauli Exclusion Principle says no two electrons can have the same quantum numbers, and there are roughly two ways to interpret this.

First, an 'orbital' may host two electrons so long as they have opposite spins.

Second, an 'orbital' may only host one electron, but there is a second orbital that is geometrically and locationally identical which hosts an electron of the opposite spin of the first orbital.

Why?  It is written into the fabric of the universe.  That is how the math works out.  Experiments have strongly supported Quantum Mechanics.


Homemade preservative for natural aloe-vera gel?

I suffer from chronic back pain and have tried just about everything on the market.Now I grow and harvest my own Aloe Vera gel,
put the gel in a 50ml roll-on bottle 1/2 full with gel, a 1/2 tsp vitamin C powder
and 10 drops vitamin E oil then add 98.9% DMSO by syringe
10ml at a time till the strength is good enough for me. I personally use 25ml
DMSO with the Aloe gel for remarkable pain relief


Summarizing from the FDA website (some of this info could vary by country, but I think the general ideas should hold anywhere) commercial use of the term "alcohol" doesn't coincide with its use in chemistry. Cosmetic products use "alcohol" to mean ethanol, so you may find other alcohol compounds in a product that is ostensibly "$100\%$ alcohol free".


How to distribute the Boltzmann energy to internal and external motion?

The translational energy of a mole of molecules is $\tfrac{3}{2}RT$ which corresponds to an average kinetic energy of $\tfrac{1}{2}M\langle v^2\rangle$.  The average reflects a Boltzmann distribution of speeds
$$f(v)=4\pi\left(\frac{M}{2\pi RT} \right)^{3/2}v^2\operatorname{e}^{-\tfrac{1}{2}Mv^2/RT} $$
i.e. some molecules have very low velocity and some have very high velocities.

Kinetic energy is constantly being transferred, so if you could monitor its velocity over time its average kinetic energy would be $\tfrac{1}{2}M\langle v^2\rangle=\tfrac{3}{2}RT$

In addition one mole of gas has $\tfrac{3}{2}RT$ of rotational energy ($RT$ for linear molecules), which also reflects an average rotational kinetic energy.  For individual molecules at any given time, the amount of translational and rotational energy is independent.  For example, a molecule can move very slowly while rotating very rapidly – until it collides with another molecule and things change.  However, on average there is an equal amount of translational and rotational energy (for non-linear molecules).


The reason only five isomers can be formed is the three methyl groups are equivalent by symmetry.

I believe your arbitrary distinction of branch carbon/methyl from lateral carbon/methyl is creating confusion


How to calculate the kinetic order of an enzymatic reaction?

The legal limit in some states of 0.08% BAC corresponds to about 17 mM ethanol in the blood. The $K_M$ value for human liver alcohol dehydrogenase is 0.5 mM according to this paper. So at this particular legal limit, the enzyme works at 97% of its maximal capacity. ($v = v_{max} \frac{[S]}{[S] + K_M}$).

Above or very much above the legal limit, the rate does not change much, making it approximately zero order in ethanol. Because the breakdown of alcohol does not get faster with more intake, it can take past the next morning to metabolize sufficient alcohol to reach values below the legal limit if someone drinks heavily at night.

For ethanol concentrations below the $K_M$, the reaction is approximately first order in ethanol, so it gets slower and slower as the ethanol concentration approaches very small concentration.

While the $K_M$ is the same for different individuals, the amount of enzyme is not (depends on the size and health of the liver, gender, and on genetics). That's where the body mass comes in.


What would explain urine that resists freezing?

Urine is an aqueous solution of urea, uric acid, sodium salts and whatnot. These solutions usually exhibit freezing-point depression.

As Brian has correctly pointed out in a comment, the effect isn't that strong and a quick and dirty look in the literature shows that a freezing point around -1.3 to -2.2 °C.

You might want to try a cooling bath with dry ice in isopropanol or acetone or other drastic methods to achieve shock-freezing.


Acid catalysed dehydration of alcohol with a bicyclo group

Pathway 2 is unlikely. The tert-butyl group stays put, i.e., it remains attached to the same carbon throughout the reaction. It is the rest of the molecule that rearranges. The mechanism of this reaction was demonstrated 70 years ago. Compare this reaction with Possible nonclassical ion from a bicyclic system on ChemSE. There is information at the link on labeling to elaborate the mechanism.


Yes and no, with more electron withdrawing groups at ortho and para positions would certainly increase acidity but other factors do play a role. For example,
In trinitrobenzoic acid

here the two $\ce{NO2}$ at ortho positions are very bulky and due to steric repulsion rotate the $\ce{COOH}$ group thus making the compound non planar due to which there will be no more resonance thus, reducing acidity drastically. This is the most important factor that plays role in reducing the acidity.


Why do rare earth metal oxides vary in color so much?

TL;DR f-f transitions in lanthanide ions give rise to the colours of the ions. The electronic states can be derived from quantum mechanics, and their energies vary from ion to ion, which give rise to different colours.. The colours are pale because these transitions are Laporte-forbidden and therefore weak.

These absorption bands do indeed arise from electronic transitions, but the theory required to describe them is more sophisticated than simple crystal field theory where one speaks of d-d* transitions. Instead one would have to construct the symmetries of the electronic states via direct products. In the case of the transition metals, there is some description here. For example, a $\mathrm{d^2}$ octahedral transition metal complex would have a $\mathrm{^3T_{1g}}$ ground state and there would be three spin-allowed transitions to excited states ($\mathrm{^3T_{2g}}$, $\mathrm{^3T_{1g}}$, $\mathrm{^3A_{2g}}$). Not all of them may fall within the visible region, but often, more than one of them do.

As a concrete example, here's the spectrum of $\ce{[V(H2O)6]^2+}$.1 In wavenumber units the visible spectrum is roughly $14\,000\mathrm{-}25\,000~\mathrm{cm^{-1}}$:

You can hopefully already appreciate that it is not as simple as "d-d* transitions" which would imply that the complex only absorbs at one wavelength.

Now, for the lanthanides, exactly the same principles apply. The lanthanides characteristically do not interact strongly with any bound ligands, and therefore the effect of the ligand field is almost negligible. The relevant electronic terms are therefore the atomic term symbols, which are obtained by Russell-Saunders coupling, which is more fully described in numerous physical chemistry textbooks. For example, consider $\ce{Pr^3+}$ ($\mathrm{4f^2}$). The allowed term symbols are

$$\mathrm{{}^3H, {}^3F, {}^3P, {}^1I, {}^1G, {}^1D, {}^1S}$$

and following spin-orbit coupling (which is quite a significant effect in the lanthanides) one gets the electronic terms

$$\mathrm{{}^3H_4, {}^3H_5, {}^3H_6, {}^3F_2, {}^3F_3, {}^3F_4, {}^3P_0, {}^3P_1, {}^3P_2, {}^1I_6, {}^1G_4, {}^1D_2, {}^1S_0}$$

The ground state as predicted by Hund's rules is $\mathrm{^3H_4}$. There are therefore up to 12 different transitions that must be considered. In some other lanthanide ions, there are even more possible states.

The bottom line is that each lanthanide ion, with its unique electronic configuration, will have different excited states and different excitation energies. Therefore their absorption spectra differ and they display different colours. $\ce{Pr^3+}$ itself has the following spectrum.2 The lower excited states don't fall within the visible region.

As for why they are "bright", that is simply related to the selection rules. For transitions between atomic states, the requirements are

$$\Delta S = 0; \Delta l = \pm 1; \Delta L = 0, \pm 1; \Delta J = 0, \pm 1 \text{ except }J = 0 \not\leftrightarrow 0$$

and for centrosymmetric systems like atoms there is an additional constraint that

$$\mathrm{g} \not\leftrightarrow \mathrm{g}; \mathrm{u} \not\leftrightarrow \mathrm{u}; \mathrm{g} \leftrightarrow \mathrm{u}$$

which is the Laporte selection rule. The bottom line is that these "f-f transitions" (I am using this term to loosely refer to the collective set of electronic transitions) are parity-conserving and therefore Laporte-forbidden. Vibronic coupling, which is the pathway which allows this selection rule can be relaxed, is not of consequence in the lanthanide ions because of the lack of covalency; the colours are not intense, and therefore typically a pale colour is observed. (Compare $\ce{KMnO4}$ which has a fully parity-allowed LMCT transition; its dark purple colour is exactly the opposite of pale.)

Much greater discussion of the electronic spectra of lanthanides can be found in ref 3.

References

Oxford lecture notes.
Aspinall, H. C. Chemistry of the f-Block Elements; p 35.
Cotton, S. Lanthanide and Actinide Chemistry; pp 61 onwards.


How to determine the order of acidity of the following dimethyl nitrophenols

The right answer should be 3 as the $\ce{NO2}$ group has a negative mesomeric effect, which is only possible with 3. In 2, due to steric inhibition of resonance, it exhibits only -I effect, and although 3 contains H-bonding, it slightly affects acidity.


How does a hydroboration mechanism actually work?

I had to write the complete mechanism for you so that you can understand it better. See the image below. I have explained it in the simplest form. Some of the points I would like to discuss before you look at the mechanism are:

In BH3, Boron is less electronegative than H (i.e. hydrogen) , hence in the syn addition I have shown, hydrogen ends up being attached to a more substituted carbon atom.
For every bonding orbital we have an anti-bonding orbital which lies exactly opposite to the bonding orbital and each bonding-antibonding orbital pair can contain a maximum of 2 electrons in them.

Ok now lets head towards the mechanism which I wrote exclusively for you:

If you have any further doubts in the mechanism then you can ask in the comment section below my answer. I hope it helps.


What analytical method(s) to use for finding the contaminations in a fatty base material

I have found that with the ICP-MS method we can at least get a quantitative approach of the elements in these samples.
Unfortunately it is not so easy to figure out the form these contaminations are in, especially the phosphate.


How can I mix methylene chloride, methanol and hydrofluric acid to make a paint stripper?

It is not clear how reliable the concentrations given in the SDS are. It is not uncommon to give a range, but actually manufacture product at one end of the range.
I would take 25 parts of methanol and 70 parts of dichloromethane and add hydrofluoric acid till it becomes cloudy - then back off (add some more CH$_3$OH + CH$_2$Cl$_2$). You may be able to add 4.6 parts of hydrofluoric acid (round up to 5 for the SDS).
The solubility of dichloromethane in water is 1.60% at 20$^o$C and the solubility of water in dichloromethane is 0.24% at 20$^o$C. Ref 1  The solubility of dichloromethane in water increases as the temperature decreases. Ref 2
One unknown is the concentration of hydrofluoric acid and whether higher concentrations of HF are more easily dissolved in this mixture than lower concentrations. The SDS may not give you useful information, but I suggest trying to use the most concentrated HF available - you can always dilute it, and H$_2$O is no friend of this combination.
Methanol and water (and thus, probably also hydrofluoric acid) are miscible and so are methanol and dichloromethane. A recent question on StackExchange (Ref 3) gave a ternary diagram of water-methanol-dichloromethane:

The line separating the one-phase (M) from the two-phase regions (I) does not go all the way to the top, but you can see that there are homogeneous mixtures with a lot of water, methanol and dichloromethane.
The Safety Data Sheet is required to list hazardous ingredients, but not non-hazardous materials. Some stabilizer might be added, hidden in the 15% range of dichloromethane and 10% range of methanol, that could keep the mixture homogeneous. One compound that comes to mind is propylene glycol. If the commercial product is cloudy, it might be stabilized with a thickener like guar or xanthan gum. Wax has been used as an evaporation inhibitor in paint removers containing dichloromethane; it might also act as a thickener and stabilizer.
Ref 1. https://macro.lsu.edu/HowTo/solvents/Dichloromethane.htm
Ref 2. https://en.wikipedia.org/wiki/Dichloromethane
Ref 3. Mixture of water, methanol and dichloromethane


Which textbook is this image on protein tertiary structure from?

The illustration is likely taken from a later edition of Baum's Introduction to organic and biological chemistry.
The first edition [1, p. 292] presents nearly identical illustration under a different number alongside with a note that the image was adapted from Bennett's Graphic Biochemistry [2]:

Unfortunately, at the moment (2020-10-16) [1] is available on The Internet Archive only for a limited amount of time for borrowing, and I was unable to find a full-text source for [2].
References

Stuart J. Baum. Introduction to Organic and Biological Chemistry, 1st ed.; Macmillan: London, 1970.
Bennett T. P. Graphic Biochemistry. The Chemistry of Biological Molecules; Macmillan, 1968; Vol. 1.


The first thing needed for a dative bond is a lone electron pair in one atom and an empty orbital in another atom.
This condition is necessary but enough.

There is also a formula that fails in many cases but may sometimes become handy:
Formal Charge = (Oxidiation State of the centeral atom - 8 + Digit of its group number)/2

e.g.: For SO3 it would be: (6-2)/2 Which means that it has 2 co-ordinate bonds.
However for a compound like NH4+ it won't work.

Of course, safest way is to find out how the compound had formed in the first place.
e.g.: If we know that BF3 + F- made BF4-, we would find out that F- had shared with B two of its electrons, filling the empty orbital belonging to B.


Finding Ka of a monoprotic weak acid from pH and concentration of added sodium salt

I don't know why shouldn't it work, this is my approach, leading to final result, what's the fallacy?

Consider the reaction
$$\ce{A-} + \ce{H2O} \ce{<=>} HA + \ce{OH-}$$
It can be broken into two steps
$$\ce{A-} + \ce{H+} \ce{<=>} HA$$

$K = \dfrac{1}{K_a}$
$$\ce{H2O} \ce{<=>} \ce{H+} + \ce{OH-}$$

$K = K_w = 10^{-14}$ at 25°C

If h is hydrolysis constant, approximating $h\ll 1$, then [$\ce{A-}$] = c(1 - h) $\approx$ c and [$\ce{OH-}$] = [$\ce{HA}$] = ch.

$\therefore$, $K_h = K_b = \dfrac{K_w}{K_a} = ch^2$.

$\implies K_a = \dfrac{K_w}{ch^2} = K_\mathrm{a} = K_\mathrm{w} \frac{[\ce{A-}]}{[\ce{OH-}]^2}$

$h = \sqrt{\dfrac{K_w}{K_{a}.c}}$

$[\ce{OH-}] = ch = \sqrt{\dfrac{K_{w}.c}{K_a}}$

pH = 7 + $\dfrac{1}{2} log {c} + \dfrac{1}{2} log {pK_a}$

Which is a well known result...


How to synthesize t-Boc protected derivative of PVA

Interesting, and i like the part with the photoacid.

I can't read the full details of the cited article (paywall), but since the authors aren't very pleased with the functionalization of the PVA, how about a change in the order of events:

prepare or buy the monomer (vinyl tBOC)
polymerize it according to this article (or a better source)


About the meaning of the names of f orbitals

I would have to say "f" overwhelmingly stands for "fundamental" (not "fine").

An early (1915) reference is "Some Recent Discoveries in Spectrum Series" Astrophysical Journal, vol. 41, p.323

  Within each of these three systems there are, in general, four different types of series, known as "principle", "sharp" (second subordinate), "diffuse" (first subordinate), and "fundamental" (Bergmann).  These names are not always very appropriate; for instance the "sharp" series is often diffuse, the "fundamental" may be composed of the faintest lines in the spectrum, but this is not vital.

The fundamental series was discovered by Arno Bergmann in 1907, and is sometimes called the Bergmann series. It was named "fundamental" by William M. Hicks, I beleive in A Critical Study of Spectral Series. Part I: The Alkalies H and He (1911): "F is the symbol for the new series (fundamental) as P.S.D. stand for those already known" (page 57) and "It was my first impression that this would be the same for all the alkalies and the .987 of the earlier discussion. It would be the basis of a fundamental series for each
element, and the letter F which has been attached to it had its origin in this idea." (page 94).

The names existed before orbital theory.  There is no need to investigate f-block elements to observe excitations to f-orbitals.  Electrons of any elment can be excited to an f-orbital. In fact, Bergmann discovered the series in the sodium spectrum.  What was orginially known as the sharp series, were transition to s-orbitals, the primary series transitions to p-orbitals, etc.

See The Origin of the s, p, d, f Orbital Labels for additional information.


f-electrons in chemistry of heavy transition metals

"f-hole is critical to describe DoS of IrO2 correctly."

[...]

This implies, that f-electrons are somewhat involved into chemistry of Ir.

Perhaps naïvely, but I would dispute that the second statement follows from the first. While f-electrons are generally not critical for the chemical reactivity of molecules or materials, their inclusion indeed is critical for a proper description of the Density of States. The magnetic and optical properties, notably, can be derived from the DoS and depend directly from f electrons.


The steps for hormone level measurement generally follow a similar sequence.  For the most accurate analysis liquid chromatography-mass spectroscopy (lc-ms) is typically used.  The liquid chromatography separates the sample into fractions, conditions are selected such that a clean, separate peak is seen for the hormone being evaluated.  The purified hormone derived from the liquid chromatography step is then introduced into the mass spectrometer for quantification.

The general process involves:

sample collection: a sample of saliva, blood or urine is collected
addition of a known quantity of hormone to act as an internal standard.  Often a $\ce{^{13}C}$ labeled hormone is used (commercially available).  The parent ion for the labeled standard will occur one mass unit higher than the parent ion for naturally occurring $\ce{^{12}C}$-containing collected sample.  If the peak height of the sample in the mass spec is half as large as that of the standard, then the concentration of hormone in the collected sample is half that in the standard.
chemically separate the hormone from any large binding proteins
isolate the fraction containing the hormone using liquid-liquid extraction
run a second liquid-liquid extraction to remove highly polar materials
run lc-ms analysis on sample that has now been purified and enriched in hormone

Here is a link to a detailed procedure for testosterone analysis.

To keep things understandable I have simplified the description of some of the
steps in the process. For example lc-tandem mass spectrometry is often used to improve accuracy.

Research today in hormone quantification is focused on reducing cost and turn-around time.  A significant amount of work is particularly being done on analyzing the raw collected sample without further purification.


Shea Stearin Extraction by Dry Fractionation

Shea butter is a complex combination of esters:

(Wikipedia). The overall ratio of oleic to stearic acids is about 2/1, but statistically, there will a little of the high melting stearic, more of the 2 stearic to 1 oleic; much more of the 2 oleic to 1 stearic, and a lot of all oleic ester. The melting point of shea butter is low (body temperature) and so if it does not precipitate out the tristearin even at room temperature, you will have to do something different.

A fractional crystallization involving higher temperatures would use a poor solvent for the tristearin (mp ~ 72C), like ethanol. The oleic esters will be more soluble. Make a 20-30% solution in hot ethanol (be careful!!) and let it cool. Experimenting with similar alcohols (propyl, isopropyl, and perhaps with a bit of water added (~5-10%) could give better results.

Oiling out is a common problem in trying to crystallize organic compounds.


Here is the usual citation for the toxic products and symptoms created upon mixing NH3 and NaOCl (not in lab conditions, which involves employing very cold dilute aqueous solutions):

When bleach is mixed with ammonia, toxic gases called chloramines are produced. Exposure to chloramine gases can cause the following symptoms:
Coughing.
Nausea.
Shortness of breath.
Watery eyes.
Chest pain.
Irritation to the throat, nose, and eyes.
Wheezing.
Pneumonia and fluid in the lungs.

The chemistry proceeds along the lines:
$\ce{NaOCl ⇌ Na+ + OH- + HOCl}$
$\ce{NH3 + HOCl ⇌ NH2Cl (monochloramine) + H2O}$
$\ce{NH2Cl + HOCl ⇌ NHCl2 (dichloramine) + H2O}$
$\ce{NHCl2 + HOCl ⇌ NCl3 (trichloramine) + H2O}$
Wikipedia also cites possible disproportion paths in acidic conditions starting with monochloramine, to quote:

In an acidic medium at pH values of around 4, chloramine disproportionates to form dichloramine, which in turn disproportionates again at pH values below 3 to form nitrogen trichloride:
$\ce{2 NH2Cl + H+ ⇌ NHCl2 + NH4+}$
$\ce{3 NHCl2 + H+ ⇌ 2 NCl3 + NH4+}$
At low pH values, nitrogen trichloride dominates and at pH 3–5 dichloramine dominates. These equilibria are disturbed by the irreversible decomposition of both compounds:
$\ce{NHCl2 + NCl3 + 2 H2O → N2 + 3 HCl + 2 HOCl}$

Note, with respect to any formed nitrogen trichloride (appears as a yellow oily liquid), extreme caution as it is a very unstable explosive compound upon exposure to shock, heating, sunlight and organics, even while in aqueous solutions. Because of its dangerous sensitivity, $\ce{NCl3}$ is best avoided.
Also, with respect to exposure of swimmers where chloramine can be formed in non-chloramine pools from free chlorine interacting with amine groups present in organic substances, or chloramine treated pools employing NH2Cl as a disinfectant, especially in the case of indoor pools, per the same Wikipedia source:

There is also evidence that exposure to chloramine, can contribute to respiratory problems, including asthma, among swimmers.[11] Respiratory problems related to chloramine exposure are common and prevalent among competitive swimmers.[12]

On the question, can one safely disposed of aqueous NH2Cl? Be careful, as on just adding it to a water source may increase the solubility of heavy metals. This can result, for example, in Lead poisoning for animals and humans (see 'Chloramine Catch: Water Disinfectant Can Raise Lead Exposure).


Why does NaCl dissolve in H2O despite its strong ionic bond?

The electrostatic forces between the positive cations and anions are strong. Coulomb's Law expresses the potential energy of two charges separated by a distance.

$$E_p = \frac{Q_1Q_2}{4\pi \epsilon_0 r} $$

As you may already know, water molecules stick to each other through dipole-dipole forces due to the permanent dipole that water faces due to the high electronegativity of the oxygen atom.

As ionic solids are added to water, water molecules proceed to surround each ion on the surface of the solid, forming a sphere of hydration. In the process, ions are separated from each other.

The $\delta^-$ charge on the oxygen atoms of water are attracted to cations and inversely, repels the $\delta^+$ hydrogen atoms. Thus, for cations, the oxygens of water point inward, and for anions, the hydrogens face inward respectively. The most important thing is that the ion-dipole interactions and separation of ions with little change in energy.

We can relate the potential energy of the ions to the two partial charges of a polar molecule like water:

$$ E_p \propto - \frac{|z|\mu}{r^2}$$

Z is the charge number of the ion and $\mu$ is the dipole moment of the polar molecule. Potential energy is lowered by the interaction between the solvent molecules and the ion. The $r^2$ term indicates that the interaction between ions and dipoles depends more on distance than the charges between two ions.

Thus, for hydration to occur, ion-dipole interactions must occur at the surface of the ion, and thus, ion-dipole interactions are strong for small, highly charged ions such as $\ce{Mg^{2+}}$, $\ce{Li^{2+}}$ etc.

$\ce{AgCl}$ is very slightly soluble in water and will not dissociate into it's ions. $\ce{HF}$ is a weak acid thus it does not deprotonate easily.


Mechanism of Grignard addition to carbonyls

This reaction can be viewed as an example of a pericyclic (concerted) Ene reaction.  The ene reaction is actually fairly common in organic chemistry.   The following diagram details the mechanism as a concerted reaction with a cyclic 6 membered transition state involving 6 electrons (an "aromatic" transition state).

A note on the Ene reaction...

The Ene reaction is really just a variation on the Diels-Alder theme.  The Diels-Alder reaction is a concerted [4 pi + 2 pi] reaction, whereas the Ene reaction is a [(2 pi + 2 sigma) + 2 pi] concerted reaction.


Is there a way to dehydrate hydrated aluminum chloride for use as a Lewis acid?

I think obtaining anhydrous aluminium(III) chloride from hexahydrate $\ce{AlCl6 · 6 H2O}$ isn't worth it, both economically and time-wise.
Thionyl chloride might work, but, as you said, it's an unpleasant reactant to work with, which is also likely more expensive than dry $\ce{AlCl3}$ (or is on par, depending on the purity of aluminium trichloride).
A common way to prepare water-free $\ce{AlCl3}$ in the lab is to heat pure aluminium metal with the excess of dry $\ce{HCl}$ gas at $\pu{150 °C}.$
The product is then sublimated (above $\pu{180 °C}$) and collected in the condenser (air-cooled).
The quirk here is you have to make sure pure aluminium is used, not an alloy; but this is usually evident from the color of the product, which should be white/colorless.
If you already have $\ce{AlCl3}$ labelled as anhydrous, but you suspect it was slightly moisturized, it can be regenerated via sublimation [1, p. 636]:

Aluminium chloride (anhydrous) […] Sublime it several times in an all-glass system under nitrogen at 30-50 mm pressure.  It has also been sublimed in a stream of dry $\ce{HCl}$ and has been subjected to a preliminary sublimation through a section of granular aluminium metal.

An extensive description is provided in Experimental part in [2, p. 1230]):

Aluminum Chloride.—The aluminum chloride (Baker Chemical Company) was purified by subliming three times down a straight Pyrex tube (with constrictions) which contained a small amount of nitrogen gas (50 mm.). In the first sublimation, the aluminum chloride mas sublimed through a section of granular aluminum metal, which was supported with plugs of glass wool. After the final sublimation and before sealing the tube off, aluminum chloride mas sublimed back down the tube until the remaining material formed a loose plug.
In order to remove the aluminum chloride from the tube, the tube is scratched and then the scratch touched with a hot rod in such a manner that the tube cracks but does not open. The tube may then be weighed, the tube tapped slightly to open. and after removing the aluminum chloride, the empty tube weighed. Using these techniques, it is possible to prepare very pure samples of aluminum chloride and weigh and transfer them without bringing them into contact with air.

Remember to store anhydrous $\ce{AlCl3}$ in an air-tight vessel and perform all operations with it in a fume hood.
Also, keep in mind that you can purchase $\ce{AlCl3}$ dissolved in organic solvents such as nitrobenzene, THF, or even $\ce{AlCl3}$-functionalized silica gel.
References

Armarego, W. L. Purification of Laboratory Chemicals; Elsevier: Boston, MA, 2017. ISBN 978-0-12-805457-4.
Jensen, F. R. Substitution of Polynuclear Aromatic Compounds. I. The Friedel-Crafts Benzoylation of Naphthalene. J. Am. Chem. Soc. 1957, 79 (5), 1226–1231. DOI: 10.1021/ja01562a051.


Do conjugated polymers absorb very low energy photons?

In principle yes, longer conjugated regions absorb at longer wavelengths. The simplest model when thinking about this is to consider a particle in a box, or on a ring if considering cyclic conjugation as in the chlorophyll's. However, in very long polymers, such as polyacetylene, the conjugation length is in fact rather small, approx 10 to 20 double bonds. Nanometer thick  films appear brown and absorb below 800 nm ish whereas if the conjugation length were very large, as per the chemical structure you draw, the molecules would absorb to longer wavelengths and appear black.

This limited conjugation is due to two things, first the molecules are  twisted and bent in forming the polymer and these distortions alter the electronic properties just enough to shorten the conjugation length. Second even when this twisting etc. were removed there are two ways of arranging the double bonds depending how you start. At their junction these two forms form what is called a 'solition' or 'topological kink'  and this is  a non-bonding electron between two regions of conjugation. This also limits conjugation.


How can I reduce Crystallization temperature of seed oil to be used in winter as biodiesel?

Waxing and pour point are typical specs for fuel. How cold is your winter? If it is very cold, and your tank is above ground, there is little to be done. Insulate the tank (metal-backed rock wool) and heat it.  Use a more unsaturated feedstock with no saturated components. That lowers its pour point and removes waxing.  Did you dry your fuel?  Unsaturated fuels oxidize and deposit varnish.  "Biofuel" is a good substrate for microbial growth.

Enviro-whinerism, sustainability, social consciousness, are Luddite lies.  They assume thermodynamics is amenable to legislation and good feelings, and weather is always mild and cooperative.  None of that obtains in the real world.

http://en.wikipedia.org/wiki/Pour_point
http://www.infineum.com/Pages/pourpoint.aspx
http://www.aftonchemical.com/Solutions/LubricantComponents/Pages/PourPointDepressants.aspx
http://www.globalp.com/products/product.cfm?productID=32
Chemicals


Reaction of amines with diethyl oxalate (Hofmann amine separation method)

When a 1° amine attacks to a carboxyl group, The structure below (left) is generated which lowers the energy level of the intermediate such that the second amine can easily react with the other carboxyl group.
It is clear that 2° amines (right) don't have such proton for hydrogen bonding with the oxygen. Why the second 2° amine doesn't attack to the other carboxyl at all is maybe due to the decreased partial positive charge of the amide (less electronegativity of nitrogen) toward the carboxyl.
Finally, 3° amines don't have even a leaving proton to substitute with the ethoxide.


Metal vapors do conduct electricity, as in the mercury vapor lamp (also used in fluorescent lamps), but, since the atoms are isolated or in small (molecular) clusters, conduction is ionic, not metallic. Since electrons are far less massive (and therefore more mobile) than the heavy ions, they are responsible for the majority of the conducted current.

Electrons are forcibly ripped away from the mercury atom (well, raised to higher levels) -- the first ionization potential is 10.39V, and its second is 18.65V, so in a mercury vapor lamp or mercury vapor rectifier, the potential drop of at least 20 volts implies more than one of the outer electrons are lost from many of the atoms.

Note that ionized gases are much less conductive than solid or liquid metals. Consider a 70 watt sodium vapor lamp, with an arc length of perhaps 10 cm and a potential drop of ~18 volts. The finest commercially available wire, 40 gauge, would have a potential drop of just ~1.3 volts for metallic copper wire, or ~4 volts for sodium, which has 1/3 the conductivity of metallic copper (if they made fine sodium wire), at the same length and current (~3 A).
See the useful chart explaining sodium-vapor ionization .


The reorganisation is just the extension/contraction of of bonds and most importantly solvent motion /reorientation to equalise the energy at which point the electron tunnels and relaxation occurs into the product. Reorganisation happens before the reaction, it is, as it were, how the 'hill' is climbed to reach the transition state thats all. (This is the insight that Marcus had). Consequently it is the rate limiting process, even though it may take less than a picosecond in some reactions.  After reaction, relaxation (reorganisation) also occurs so that the solvent reaches equilibrium with the new charges distribution.

Figure source; R. Marcus, Biochimica et Biophysica Acta., 811, 1985, 265 - 322)


Does the emf of an electrode potential change over time?

Yes. Remember that the standard electrode potentials are for 298K at 1 atm of pressure and 1N concentration of reagents. As the concentration of ions to be reduced at the cathode decreases the emf will fall as well and is reflected in the Nernst equation.$$E_{cell} = E^\circ_{cell} - \frac{RT}{nF}ln(Q)$$