Imagine that you need to titrate the sample S with a solution of the reagent R.

Unfortunately, R isn't available as a stock solution and maybe not even in solid form. All you have is it's precursor P and a recipe for the reaction $\ce{P + X + Y -> R}$. You never know how good that works out.

At this point, titration of S with R doesn't make any sense - unless you can determine the concentration of R!

Here, a stock solution with an exact concentration of the helper H, maybe in a sealed vial from a commerical source comes into play.

The titration of R with H to give the exact concentration of the solution of R is the standardization of the titrant.


Calculating Standard Gibbs Energy of Dissolution

You are correct that you need $\Delta G^\circ =-RT\ln{K}$ and that the equilibrium constant in this case is $K_{sp}$ the solubility product constant, which is defined for a process like this one to be:

$$\ce{A_{m} B_{n} (s) <=> mA^{n+}(aq) + nB^{m-}(aq)}$$
$$K_{sp} = \ce{[A^{n+}]^{m} [B^{m-}]^{n}}$$

In your case $K_{sp}=[\ce{Hg2^{2+}}][\ce{I-}]^2$

You are given the solubility (maximum obtainable concentration) of $\ce{Hg2I2}$ in water at $25 \ ^\circ \text{C}$. How can you use that information to find $[\ce{Hg2^{2+}}]$ and $[\ce{I-}]$ when the solution is saturated?


Is there a least reactive chemical species?

I think a good argument can be made for either helium or neon, the most noble of the noble gasses. Those are the two prototypical unreactive elements. They are the only two stable elements for which no more complex compounds (i.e., other than the single atoms themselves) have yet been isolated, at any temperature. The slightly more reactive element argon will admit formation of compounds such as argon hydrofluoride ($\ce{HArF}$). This compound is only stable up to $\mathrm{17\ K}$, because any hotter and the frail bonds are overcome by random thermal collisions which break the compound apart into $\ce{Ar}$ and $\ce{HF}$.

Picking which of helium or neon is less reactive is a bit more difficult. A naive analysis of periodic trends would point to helium as the most inert, but more detailed computational studies suggest that at least in some cases neon may be less reactive. For example, the extremely Lewis acidic compound beryllium monoxide ($\ce{BeO}$) may potentially form an isolable, if very weakly bound, compound with helium, $\ce{HeBeO}$, but neon is not thought to form the analogous $\ce{NeBeO}$. $\ce{HHeF}$ may also be just barely stable, whereas $\ce{HNeF}$ is not thought to form. None of these have yet been observed in the laboratory, but there is definitely ongoing research into coaxing helium and neon to make isolable compounds.

All that said, we can get helium and neon to react, if we drop the requirement that the product must be isolated (that is, "put into a bottle"). Chemical species such as $\ce{He2^+}$, $\ce{Ne2^+}$ and $\ce{HeNe^+}$ have long been known from mass spectrometry experiments, they just can't be isolated because that would require the presence of a counterbalancing negative ion, which would immediately proceed to react with the positive ion and cause decomposition with release of the noble gas. You can also bring what is arguably the most reactive species in chemistry, the hydrogen cation ($\ce{H^+}$) into the fray. Helium and neon will both easily react with $\ce{H^+}$ to form $\ce{HeH+}$ and $\ce{NeH+}$ , as shown by the exothermic proton affinities of $\ce{He}$ and $\ce{Ne}$. Again, these composite particles cannot be isolated, as they are amongst the strongest Brønsted-Lowry acids in existence, and will protonate anything they come into contact with in order to release the neutral noble gas atom. Some other possible non-isolable relevant noble gas ions are $\ce{FHeO^{-}}$, $\ce{HeCCH+}$ and $\ce{PbHe15^{2+}}$ (!), among others.

Edit: I forgot to mention that there are a few other cases of helium or neon binding to other atoms. By exciting one of the electrons in the atom, it is possible to coax it to bond with another one, forming an excimer or exciplex. See for example, the dihelium excimer $\mathrm{He_2^{*}}$. This is a short-lived species that still can't be isolated, however, because in a matter of microseconds it releases a photon and de-excites, promptly separating into the two free atoms.


TL;DR version: Chirality is a basic property of a molecule, while optical activity is a bulk physical property possessed by samples of chiral compounds. Specific rotation is an intrinsic property of chiral compounds and a measure of optical activity under standardized conditions. Optical rotation is not responsible for the disparate behavior of enantiomers of biologically active compounds.

Chirality is an intrinsic property of a molecule (or, indeed, of certain geometric structures, more generally). A chiral molecule is one that is not superposable on its mirror image molecule (its enantiomer), meaning that no series of rotations or translations will make the two geometrically identical in three-dimensional space. One of the consequences of chirality is that chiral molecules in bulk will measurably rotate plane-polarized light by a specific amount and in a specific direction. (The distinction here is crucial, since any molecule[s] will in general cause some rotation, but the random distribution and spatial orientation of the large number of molecules in any given sample will produce no observable net rotation if the molecules are achiral. The rotation caused by one molecule is likely to be negated by another identical molecule in an opposite spatial orientation.) The key point is that chirality is an intrinsic property of a molecule, while optical activity is a bulk physical property of said molecules gathered en masse. Specific rotation, on the other hand, can be viewed as an intrinsic property, since it is defined for a specific sample under an exact set of conditions (light wavelength, path length, concentration [for samples in solution], density [for pure liquid samples], temperature).

Optical activity is not the cause of any changes in the properties of chiral compounds, nor is it responsible for the sometimes wildly different biological effects of certain enantiomers (which I presume may be what you're actually referring to, forgive me if I'm wrong). The enantiomers of chiral molecules can often differ widely in their biological effects due to the basic nature of proteins, which are the constituent biological molecules that most if not all biologically active substances target in one way or another. Proteins are constructed from amino acids, which often themselves contain multiple chiral centers, and these proteins go on to fold into very specific and complex conformations. Enzyme active sites and cell receptors (among various other biological structures) are mostly comprised of proteins with very specific three-dimensional structures. As such, they interact significantly only with molecules that have the appropriate shape. Hence, enantiomers of chiral substances, which differ in their three-dimensional shape, will exhibit variable degrees and types of interactions with proteins, and can hence impact biological targets in disparate ways. This is largely what makes chirality so important to chemists, and partly explains why so much of the most respected (not to mention economically important) cutting-edge work recently (at least in organic chemistry) has revolved around enantioselective or enantiospecific synthesis techniques.


How can I check if this mystery compound is strontium bromide?

Relying on the fact that the compound is almost certainly a strontium compound, I'll look at the reactions with sulfuric acid of the specific anions:

$\ce{SrBr2}$

$$\ce{SrBr2 + H2SO4 ->  2HBr + SrSO4}$$
Both compounds $\ce{HBr}$ and $\ce{SrSO4}$ are white or colourless. BUT! concentrated sulfuric acid oxidizes $\ce{HBr}$ to bromine gas:
$$\ce{2 HBr + H2SO4 → Br2 + SO2 + 2H2O}$$
And bromine gas has a yellow/red colour. So your compound is strontium bromide.

$\ce{Sr(NO2)2}$

This is a yellow compound. As you mentioned, the compound was white, so it isn't strontium nitrite.

$\ce{Sr(NO3)2}$

$$\ce{Sr(NO3)2 + H2SO4 ->   2HNO3 + SrSO4}$$
Both compounds $\ce{HNO3}$ and $\ce{SrSO4}$ are white or colorless. After looking some more stuff up, it turns out that sulfuric acid and nitric acid don't react (at room temperature) to produce any yellow/brown $\ce{N2O4}$ gas.


In part, your vision already exists and uses electrons as "ink". In a process kown as electron capture, a proton "swallows" an electron of sufficient energy and is thus converted to a neutron.

$$\ce{^{A}_{Z}X + e- -> ^{A}_{Z-1}Y + \nu_{e}}$$

As a consequence, the atomic mass number A remains the same, but the atomic number Z is decreased by one: Element X is converted to element Y.

A typical example would be the conversion of aluminium to magnesium:

$$\ce{^{26}_{13}Al + e- -> ^{26}_{12}Mg + \nu_{e}}$$


Better understanding the pyrolysis reaction

The presumption here is that the biomass is purely sugars: glucose- and xylose-based starches and cellulose.  The essential components of lignin, which make up about 20% of the biomass, is phenols and propane. You may need to rebalance based on those additional components.


The statement is correct.

The most common method for the esterification of amino acids is with anhydrous acid in alcohol. The most convenient way of doing this is with thionyl chloride example procedure or chlorotrimethylsilane example procedure as a source of anhydrous HCl.

The original procedures with HCl gas or conc. sulfuric acid are slower and lower yielding recent example


When a solution is diluted, the number of moles of the solute decreases. what would happen to the molarity?

The number of moles of solute does not decrease on dilution.
For example suppose you add 2 moles of NaOH to 100cm3.
Now if you dilute this solution to 1000cm3 you  are still going to have 2 moles of NaOH.
But now the total volume has changed ,hence the concentration also changes.


You are right: 1 cubic meter of water contains $\pu{1E-4}$ moles of $\ce{H+}$ ions. But one mole is $\pu{6E23}$ atoms or ions.

So, $\pu{1E-4 mol}$ contains $\pu{1E-4}\times\pu{6E23} = \pu{6E19 ions}.$
Here we are.


Significant figure rules when plugging a value into an equation or formula

If the conversion is exact, all uncertainties carry over. Arguing the same for significant digits is very tricky and can lead you on erroneous paths. It is thus easier for you to convert the uncertainty that is present in the originally measured value into a new uncertainty for the converted value and then infer from that the number of significant digits.

Let's assume that your measurement is $\vartheta = 14.0 \pm 0.1 ~ \mathrm{^\circ C}$.

The conversion to Fahrenheit is
$$ \phi(\vartheta) = \left(\frac{9}{5} \frac{\vartheta}{\mathrm{^\circ C}} + 32 \right) ~ \mathrm{^\circ F} \,,$$
which we can straightforwardly apply to the mean of the measurement which gives us $57.2~\mathrm{^\circ F}$.

To do the error propagation under the assumption that the errors are independent, we do the following:
$$\sigma_\phi = \sqrt{\left( \frac{\partial\phi}{\partial\vartheta} \right)^2 \sigma_\vartheta^2} = \frac{\partial\phi}{\partial\vartheta} \sigma_\vartheta = \frac{9}{5}\frac{\mathrm{^\circ F}}{\mathrm{^\circ C}} \sigma_\vartheta = \frac{9}{5}\frac{\mathrm{^\circ F}}{\mathrm{^\circ C}} \cdot 0.1~\mathrm{^\circ C} = 0.18~\mathrm{^\circ F}$$
The term under the root would in principle be a sum, but here we only have one variable with an uncertainty associated with it. For more details, look up this Wikipedia article on error propagation.

Anyway, our newly converted value is correctly expressed as
$$ \phi = 57.20 \pm 0.18 ~ \mathrm{^\circ F} \,.$$


The composition of petrol, also called 'gasoline' or 'gas' in the USA, despite it being a liquid at room temperature, contains a complex mixture of hydrocarbon compounds selected for use in internal combustion engines.

The term "95% octane" refers to the 'octane rating', 'octane number' or 'anti-knock index' of the fuel, not the amount of 'octane' in the fuel (since 'octane' is a specific organic compound in the group of alkanes).  The octane number / rating is an 'anti-knock index', based on an arbitrary scale indexed to a liquid mixture of iso-octane (2,2,4-trimethylpentane and n-heptane). Whilst different countries have different standards for anti-knock index, in the USA, the American Society for Testing and Materials (ASTM) defines two popular indices called the 'research octane number' (RON) and motor octane number (MON). Regardless if the specific stanard, the general principle is that the higher the octane number, the more the fuel can be 'compressed' before 'auto-igniting'. Auto-ignition is not desirable in petrol engines as it causes 'knocking' of the pistons which greatly shortens their life.

What petrol (gasoline) actually contains is a blend of refinery products, including reformate (high in aromatic hydrocarbons, low in alkenes), olefinic hydrocarbons, alkylate (high in paraffinic hydrocarbons) and isomerate (high in branched isomers of pentane and hexane). Specific blends vary from manufacturer. Manufacturerers even offer a range of blends at the bowser (gas station) including 10% ethanol and 'premium' blends for high performance engines.

Refer to 'Development of a Detailed Gasoline Composition-Based Octane Model' for more information.


Mechanism of oxidation by dichlorotriazenetrione

Dichlorotriazenetrione, also known as dichloroisocyanuric acid, is slowly hydrolyzed in water to isocyanuric acid and hypochlorous acid, which serves as the oxidation agent. The N-Cl bond is easily hydrolyzed because of its relative weakness, which is also reflected in the stability of nitrogen halides. The slow hydrolysis releases a low but constant amount of oxidizers. $\ce{HOCl}$ itself oxidizes organic compounds, being reduced to $\ce{HCl}$ in the process.

In aqueous solution, $\ce{HOCl}$ is in equilibrium with its anhydride $\ce{Cl2O}$ and elemental chlorine, which also serve as oxidizers. Furthermore, it slowly decomposes to $\ce{HCl}$ and oxygen, so that a mixture of different oxidation agents is present in its solution.

$$\ce{HCl + HOCl \rightleftharpoons Cl2 + H2O}$$

$$\ce{2HOCl \rightleftharpoons Cl2O + H2O}$$

$$\ce{2HOCl \rightarrow 2HCl + O2}$$


Is the following rate law possible for the given reaction?

You should use steady state approximation which leads to the mentioned answer:

$$\frac{\mathrm{d}[\ce{CD}]}{\mathrm{d}t} = 0$$

$$
\begin{align}
\ce{C + D &->[$k_1$] CD}\\
\ce{C + CD &->[$k_2$] E}
\end{align}
$$

$$\ce{->[QSSA]}\quad k_1[\ce{C}][\ce{D}] = k_2[\ce{C}][\ce{CD}] \quad\to\quad k_1[\ce{D}] = k_2[\ce{CD}]$$

$$\frac{\mathrm{d}[\ce{E}]}{\mathrm{d}t} = k_2[\ce{C}][\ce{CD}] = k_1[\ce{C}][\ce{D}]$$


Can I use gasoline as a non-polar solvent?

Gasoline is a mixture of several compounds, paraffins, cycloalkanes, and olefins (alkenes), the purification can be a challenge.


Theoretical basis behind orbital correlation diagram for pericyclic reactions

Introduction

Instead of the usual cycloadditions (which has some extra complications in terms of the reacting orbitals), let's consider the concerted reaction $\ce{H2 + D2 -> 2HD}$. We'll come back to the cycloadditions at the end.

occurring via a square planar transition state. This can be analysed with the W–H rules and you'll find that it's a thermally forbidden process $(_\sigma 2_\mathrm{s} + {}_\sigma 2_\mathrm{s})$. So, hopefully that's what we find at the end of this discussion.

Ignoring the isotope labels,1 there is $D_\mathrm{2h}$ symmetry throughout the reaction, so we will categorise the orbitals by their symmetry in this $D_\mathrm{2h}$ point group.2 I've defined the coordinate axes such that the reaction takes place in the xy-plane, with the z-axis perpendicular to the plane of the paper.

There is one simplification which can be made: the symmetry of an orbital under the three mirror planes uniquely specifies the irreducible representation under which an orbital transforms. Hence, in order to identify the point group, we don't need to bother with the symmetry under the five other operations (identity, three rotations, and inversion); we simply need to find the symmetry under the three reflections.3

Reactant and product orbitals

The relevant MOs will be constructed from linear combinations of the 1s orbitals on each atom. First, we look at the orbitals of the reactants, labelled $\psi_1$ through $\psi_4$. We will overlap the 1s orbitals on adjacent atoms to produce $\sigma$ and $\sigma^*$ orbitals for both $\ce{H2}$ and $\ce{D2}$. However, since these MOs do not have definite symmetry under all the mirror planes, we need to take linear combinations of these MOs.

Hopefully the energy ordering of these MOs is clear (I actually listed them in increasing order of energy). It can be deduced from the extent of antibonding/bonding character between each pair of 1s orbitals.

The product MOs, labelled with $\phi_1$ through $\phi_4$, have a very similar form. It should not really be surprising, as the system is exactly the same, just rotated 90 degrees.

Orbital correlation diagrams

We can define a reaction coordinate $x$, such that $x = 0$ before the reaction (i.e. reactants) and $x = 1$ after the reaction (i.e. products). Over the course of the reaction, there will always be four MOs; and these four MOs will always transform as $\mathrm{A_g + B_{1g} + B_{2u} + B_{3u}}$, regardless of what the value of $x$ is. This has to be the case because the basis set (four 1s orbitals) transforms as a representation which can be reduced into that sum.

As the reaction proceeds, the positions of the nuclei vary continuously. Therefore, the Hamiltonian, and also the energies of these MOs $\varepsilon_i$ $(i = 1,2,3,4)$, will vary as the reaction proceeds, so $\varepsilon_i$ is a continuous function of $x$. The orbital correlation diagram simply plots how $\varepsilon_i$ varies with $x$. There will be four curves on the diagram: one for the MO with symmetry $\mathrm{A_g}$, one for the MO with symmetry $\mathrm{B_{1g}}$, and so on.

In theory, one could find the MOs and their energies at every value of $x$, and then you would find four curves, one for each $i$. However, for a qualitative understanding, this isn't necessary. We already know the approximate values of $\varepsilon_i(0)$ and $\varepsilon_i(1)$, and we also know the MOs to which these values correspond. So all we need to do is to figure out which reactant MO $\psi_i$ correlates with which product MO $\phi_j$.

From the symmetry labels, this mapping is quite self-evident, as the $\mathrm{A_g}$ reactant MO must become the $\mathrm{A_g}$ product MO, and so on.

In this example, it also makes physical sense that as the reaction proceeds, the reactant MO $\psi_2$ will simply become the product MO $\phi_3$ (pictured above). For other correlation diagrams, there is a subtlety, but it is a point we will return to later on.

State correlation diagrams

Orbital correlation diagrams are all fine and dandy, but they can be put on a much more solid footing by constructing the relevant state correlation diagrams. I have deliberately avoided using the word "state" up until now. Here, "state" refers to an electronic state of the system as a whole.4 The simplest expression for an electronic state is given by a Slater determinant, and we will simply abbreviate it to the usual form that the organic chemists do. For example, in the ground state of the reactants, we can write

$$\Psi_1 = \psi_1^2\psi_2^2$$

where the antisymmetrisation is implicit. The use of capital $\Psi$ instead of small $\psi$ emphasises that this is a total electronic state of a system. The subscript 1 indicates that it is the ground state.

The electronic states of the reactants are well-described by single Slater determinants. For a system with 4 MOs and 4 electrons, there are a total of $8!/4!4! = 70$ states, and so theoretically we need to look at $\Psi_1$ through $\Psi_{70}$. However, it suffices to consider the first three singlet states (the triplet states can be ignored5). These are

$$\Psi_1 = \psi_1^2\psi_2^2 \qquad \Psi_2 = \psi_1^2\psi_2\psi_3 \qquad \Psi_3 = \psi_1^2\psi_3^2$$

which transform as $\mathrm{A_g}$, $\mathrm{B_{1g}}$, and $\mathrm{A_g}$ respectively.6 Likewise we will consider the three product states

$$\Phi_1 = \phi_1^2\phi_2^2 \qquad \Phi_2 = \phi_1^2\phi_2\phi_3 \qquad \Phi_3 = \phi_1^2\phi_3^2$$

which transform as $\mathrm{A_g}$, $\mathrm{B_{1g}}$, and $\mathrm{A_g}$ respectively.

The state correlation diagram plots the energies of these states $E_i$ as they vary with the reaction coordinate $x$, so the question is again: which reactant state $\Psi_i$ correlates with which product state $\Phi_j$.

Let's first make the assumption that a single Slater determinant is sufficient to describe an electronic state, throughout the entire reaction coordinate. In this case, the assignment is very straightforward. Since $\psi_1$ and $\psi_2$ in the reactants smoothly become $\phi_1$ and $\phi_3$, it means that the reactant state $\Psi_1 = \psi_1^2\psi_2^2$ must smoothly become the product state $\Phi_3 = \phi_1^2\phi_3^2$ as the reaction coordinate increases. A similar consideration for the other states leads to the diagram below.

Let's pause for a moment and figure out what this diagram means. It means that this molecular Hamiltonian has three approximate solutions under consideration; we label these solutions as $S_1$, $S_2$, and $S_3$. The first approximate solution, $S_1$, can be described as a Slater determinant:

$$S_1(x) = f(x)^2g(x)^2$$

where $f$ and $g$ are MOs which vary with $x$. At the point $x = 0$, we have $f = \psi_1$, $g = \psi_2$, and hence $S_1 = \psi_1^2\psi_2^2 = \Psi_1$. As $x$ increases, $\psi_1$ continuously distorts towards $\phi_1$, eventually becoming $\phi_1$ when $x = 1$, and likewise $\psi_2$ continuously distorts towards $\phi_3$. Therefore the state $S_1$ continuously distorts from $\Psi_1$ to $\Phi_3$. Similar statements can be made of $S_2$ and $S_3$.

Now, we introduce the key assumption that the reaction must proceed adiabatically. This can be formalised very thoroughly in quantum mechanics, but essentially, it means that the state of the system cannot jump from one curve $S_i$ to another curve $S_{j \neq i}$. I'm not entirely sure about the justification for this, but Patterson has written about it in an article:7

  A W–H allowed reaction pathway conserves R (reactant) and P (product) nodal geometries, which in turn guarantees continuous covalent bonding along an adiabatic reaction pathway that takes the reactants through a transition state and on into products. A W–H forbidden pathway can either maintain this continuous bonding (interaction energy between R and P) along an adiabatic pathway and generate products in excited electronic states, or disrupt reactant nodal geometry or bonding to reach diabatically (or nonadiabatically) a lower-energy pathway to products. [...] Continuous bonding is energetically more favorable than either nonbonding or antibonding interactions.

Therefore, if we start the system in the state $\Psi_1$ and allow the reaction to proceed, it has to either end up in the product excited state $\Phi_3$ (as far as this treatment tells us); if it ends up in the product ground state $\Phi_1$, then there must be some disruption of continuous bonding. Both of these possibilities are energetically costly, and therefore the reaction is forbidden.

Configuration interaction

That gave us the correct result, but not for the correct reason. The problem is that this assumes that there is no interaction between $S_1$ and other states $S_n$. Mathematically speaking, this means that in the basis $\{S_1, S_2, S_3\}$ the Hamiltonian is assumed to be diagonal,8 such that

$$\langle S_i | \hat{H} | S_j \rangle = \delta_{ij}\langle S_i | \hat{H} | S_i \rangle = \delta_{ij}E_i$$

The question is, is this true? It turns out that, because the Hamiltonian transforms as the totally symmetric irreducible representation (TSIR), the matrix elements $H_{ij} = \langle S_i | \hat{H} | S_j \rangle$ generally only vanish if $S_i$ and $S_j$ are of different symmetry. Therefore, if two states $S_i$ and $S_j$ have the same symmetry, there will often be an off-diagonal element $H_{ij}$. This is the case with $S_1$ and $S_3$ in our example above.

As $x$ increases, the energies of $S_1$ and $S_3$ approach each other, and these two states mix into each other. The new electronic states are therefore linear combinations of our original solutions, $S_1$ and $S_3$. Since $S_1$ and $S_3$ are Slater determinants, the new electronic states are linear combinations of Slater determinants. This process of mixing different electronic states into the wavefunction is commonly known as configuration interaction (CI).

So much for the states; we're not really all that interested in them. The energies are more valuable. It turns out that the presence of this off-diagonal element leads to the non-crossing rule:9 two curves corresponding to states of the same symmetry cannot cross each other. Instead, they are repelled, and there is said to be an avoided crossing. The repulsion is most pronounced when the energies are close to each other, i.e. near the middle of the diagram. The state correlation diagram must be modified to reflect this. The coefficients $a$ and $b$ in the linear combinations must fulfil the criterion $|a|^2 + |b|^2 = 1$.

Within this framework, we can refine our argument for why the reaction is forbidden. In fact, the transition from $\Psi_1$ (reactant ground state) to $\Phi_1$ (product ground state) is adiabatic; however, there is an extremely large electronic activation energy to it, on the order of an electronic excitation energy, which cannot be provided by thermal means.

Real pericyclic reactions

You might notice that the orbital correlation diagram that we drew above is actually similar to that for the disrotatory ring-opening of cyclobutene. The principles are exactly the same, and the state correlation diagram can be constructed exactly analogously (and in fact, the state correlation diagram for this process is found in Fig. 2 of the question), and of course we find that this reaction is thermally forbidden. But, as I alluded to earlier, there is a slight subtlety. Here is the orbital correlation diagram:

Unlike in our first example, there is no obvious one-to-one correspondence between the form of the reactant MOs and the form of the product MOs. For example, it's not obvious why $\pi$ has to transform into $\phi_3$ and not $\phi_1$. It turns out that both $\phi_1$ and $\phi_3$ are actually linear combinations of what $\pi$ and $\sigma$ become in the new geometry.

In the reactant, cyclobutene, the $\pi$ and $\sigma$ MOs cannot mix by symmetry: there is an additional mirror plane (the plane in which the molecule lies), and $\pi$ is antisymmetric and $\sigma$ symmetric with respect to that mirror plane. However, when the reaction proceeds, the symmetry of the system is decreased, and these orbitals are allowed to mix and form linear combinations.

In our earlier example of $\ce{H2 + D2 -> HD + HD}$, the reactant MO $\psi_2$ does physically become the product MO $\phi_3$, as was fairly obvious from the nodal pattern. However, in this situation where there is orbital mixing, it cannot be said that the $\pi$ orbital physically becomes $\phi_3$ as the reaction proceeds. This point was emphasised by Patterson7 and was also described by Woodward and Hoffmann10 in their review of their rules (pp 797 onwards).

So, the correspondence should be determined using symmetry and the non-crossing rule (which also applies to MOs), without recourse to the form of the MOs.

Another example is that of the Diels–Alder reaction. If you look at the $\pi$ MO of ethylene (the second-lowest reactant MO), you find that it becomes the $\pi$ MO of the product cyclohexene. It makes no physical sense to claim that somehow the MO migrated through several carbons over the course of the reaction. Instead, it is more accurate to say that the three symmetric product MOs are linear combinations of what the three symmetric reactant MOs become over the course of the reaction.

Notes and references

If you're really bothered by it, you can change the reaction to $\ce{\color{red}{H}-\color{red}{H} + \color{blue}{H}-\color{blue}{H} -> \color{red}{H}-\color{blue}{H} + \color{red}{H}-\color{blue}{H}}$. But that's just confusing.
The square planar transition state has a higher symmetry of $D_\mathrm{4h}$. However, since $D_\mathrm{2h}$ is a subgroup of $D_\mathrm{4h}$, all the symmetry elements in $D_\mathrm{2h}$ will be preserved in $D_\mathrm{4h}$ symmetry, but not vice versa. For example, the transition state possesses a $C_4$ rotation axis but neither the reactants nor products do. Therefore, we cannot categorise the reactant or product MOs by their symmetry under a $C_4$ rotation; it only makes sense to categorise the orbitals using the more general $D_\mathrm{2h}$ point group. In organic chemistry parlance, this is what is meant by "identify the symmetry elements that are preserved throughout the reaction".
In organic chemistry, it is "conventional to be less formal with the  notation" (Atkins's words; Molecular Quantum Mechanics 4ed, p 400), and so the irrep symbols are replaced with a series of S's (for a character of +1) and A's (for a character of -1). For example, the $\mathrm{B_{1u}}$ irrep would be labelled as ASS since it has characters of $-1, +1, +1$ under the three mirror planes.
The reason why I have never referred to an MO as a "state of an electron" is because of quantum indistinguishability of the electrons, which necessitates that $\Psi$ be properly antisymmetrised. If the Slater determinant (eq. $(1)$) is expanded, it will become clear that the state of electron 1 is entangled with the state of all the other electrons. The wavefunction $\Psi$ cannot be factorised into a product of composite wavefunctions, and so, it is physically incorrect to speak of an electron being in an orbital.
The triplet states can be ignored because spin states are orthogonal. Therefore, for a singlet reactant to end up as a triplet product, there must be a non-adiabatic transition somewhere, and a key assumption is that the process is adiabatic.
The symmetry of a state is obtained by taking direct products of the populated MOs.
Patterson, R. T. An Improved Interpretation of the Woodward–Hoffmann Rules. J. Chem. Educ. 1999, 76 (7), 1002–1007. DOI: 10.1021/ed076p1002.
The diagonal entries of $\hat{H}$ are dependent on $x$, and hence the (approximate) eigenstates $S_i$ and (approximate) eigenvalues $E_i$ also vary with $x$. The basis set $\{S_1, S_2, S_3\}$ is also dependent on $x$, but we don't really need to worry about how it depends on $x$ are we are only looking for a qualitative description of the process.
In the case of two states mixing, the proof is simple: if you use the basis states $\{|a\rangle,|b\rangle\}$ the Hamiltonian is
$$\mathbf{H} = \begin{pmatrix}H_{aa} & H_{ab} \\ H_{ba} & H_{bb}\end{pmatrix}$$
and the eigenvalues are
$$E_\pm = \frac{(H_{aa} + H_{bb}) \pm \sqrt{(H_{aa} - H_{bb} + 4|H_{ab}|^2)}}{2}$$ since $H_{ba} = H_{ab}^*$. Then $E_+$ can only be equal to $E_-$ if $H_{ab} = 0$.

In general I don't want to prove it, because I don't know how to, but a useful reference for this is: Mead, C. A. The "noncrossing" rule for electronic potential energy surfaces: The role of time‐reversal invariance. J. Chem. Phys. 1979, 70 (5), 2276–2283. DOI: 10.1063/1.437733.
Woodward, R. B.; Hoffmann, R. The Conservation of Orbital Symmetry. Angew. Chem. Int. Ed. 1969, 8 (11), 781–853. DOI: 10.1002/anie.196907811


(1) It is true that the inductive effect is weak across several bonds, BUT conjugated bonds (i.e. alternating simple and multiple bonds) propagate this effect much farther, even more for mesomeric effects. You can think of "conjugated bonds" like a tunnel for electrons.

(2) In your case, the electronic effects will stabilize the salt more/less than the acid. As they are in equilibrium, the salt is more/less favored than its acid counterpart.

I think that the most important thing here is that the more electric charges are "diluted" (i.e distributed among more atoms), the more stable a molecule is. This is an effect of entropy, which is too often poorly described as disorder.


Removing water/moisture from car exhaust

Certainly devices exist to remove water vapour from gas streams. Many labs pass their nitrogen stream through a conc. sulfuric acid bubbler to ensure it is dry. Solid absorbents such as silica gel or Drierite can also be used. The question is what volume of exhaust gas do you want to dry? These absorbents need to be changed frequently to maintain the dryness of the gas stream. If you want to keep a lot of exhaust dry you'll need a lot of drying agent.


Reaction of Sulfur dioxide with Potassium

$\ce{KO2}$ is definitely out of question. True, it does form when $\ce{K}$ reacts with $\ce{O2}$, but that's another story. It is a peroxide (in a broader sense), so it contains oxygen in oxidation states other than -2. $\ce{SO2}$ contains oxygen in -2. Who is going to oxidize oxygen?

Turning down peroxides, we resort to the more realistic option of $\ce{K2O}$. Now, that's a basic oxide, and $\ce{SO2}$ is an acidic oxide. Surely some chemistry is going to happen between them.

As to where the reduced sulfur would go, it is not quite clear and depends on many things. With excess of potassium one might expect ending up with $\ce{K2O}$ and $\ce{K2S}$. Obviously that's not our case. Then we must consider a reaction between $\ce S$ and $\ce{K2SO3}$. Such reaction is known (though probably under different conditions than we have here) and indeed produces thiosulfate.

That being said, partially reduced compounds of sulfur are numerous and I see no reason to favor one above the other without some deep investigation. Have you heard about dithionites, for example? They are known to appear when reducing $\ce{SO2}$ with certain metals; why not here?


What about sodium triphosphate in drinking water?

It appears to be safe, provided that you are putting it into one of the delivery systems that it’s made for which I found through the link you provided.
Here is one of the delivery systems:
https://www.pineco.com/eng/products/dosing/polyphosphate-dosers/spillo-polyphosphates-doser-dm
Per the specs, it delivers 3ppm, which means that you’re only introducing 1ml per 333L of water, which is approximately what the average person uses per day.
Contributing to phosphate load shouldn’t be a concern as someone (with “normal” diet &renal function) has a concentration of 10-300ppm of phosphorus in their urine!
This company also sells water filters for regionally specific water issues like high arsenic, and water softeners with different ion exchange resins based on the ions present in the water. They appear to have a solid understanding of drinking water chemistry, but it’s your domestic water, not mine, so caveat emptor


Let us assume that the initial and final temperatures in both the isobaric and isochoric processes are same. We will limit this discussion to ideal gases only. To calculate heat transfered we will use the expression $$q = nC\Delta T$$ Here C is molar heat capacity of the gas. For isobaric process, this will be $$q_p = nC_p\Delta T$$ For the isochoric process $$q_v=nC_v\Delta T$$
$\Delta T$ for both these processes same. Substracting these two equations we get $$q_p - q_v = n(C_p - C_v)\Delta T$$ The value of $C_p - C_v$ is $R$. Therefore $$q_p - q_v = nR\Delta T$$


Why is a rearrangement observed for the secondary alcohol but not the primary in these examples?

You need to consider the possible and plausible reaction mechanisms for both substrates. They are both nucleophilic substitutions in first approximation, so we need to decide whether $\mathrm{S_N1}$ or $\mathrm{S_N2}$ mechanisms are more plausible.

Primary electrophilic carbons are very likely to undergo $\mathrm{S_N2}$ reactions while $\mathrm{S_N1}$ is strongly disfavoured since primary carbocations are less stable than secondary or tertiary ones.

Secondary electrophilic carbons can react according to both mechanisms, technically. However, both mechanisms are not ideal: the steric strain is rather large so $\mathrm{S_N2}$ reactions are slower and secondary carbocations are still rather unstable compared to tertiary ones, so $\mathrm{S_N1}$ reactions are slower.

In this case, we have the possibility to stabilise the less stable second intermediate in the scheme shown below by a Wagner-Meerwein rearrangement. We can let both hydrogen or methyl migrate, but the hydrogen migration gives us the most stable carbocation (a tertiary one) so it will prevail. Multiple other migration paths will likely lead to minor products.

The primary carbon will not, as stated above, form a primary carbocation since it is too unstable. Hence only the bimolecular substitution will occur. If no primary carbocation is generated, no Wagner-Meerwein rearrangement can occur. And the mechanism of an $\mathrm{S_N2}$ reaction does not allow bromide’s (or anything else’s) migration.

The secondary carbon could undergo bimolecular substitution, too. The reaction rate will probably be slower than that of fragmentation and rearrangement, though, since the latter are unimolecular processes. And other shifts (e.g. methyl followed by hydride) are possible but again slower since you need to shift twice. All in all, a higher proportion of side-products may be expected.

In no case will the bromide shift after substitution; any shift would need to occur before.


What determines humidity limit / dew point of the air? - Why can air only hold a certain amount of water?

One way to understand this is if you think of air as bunch of atoms and molecules moving around in random fashion. The speed at which they move is measured by the temperature of air. (This is true even if we are not talking about air, but we don't need to bother about that here.) So when we say higher temperature we mean that the speed at which these particles move is higher. Furthermore the faster they move the less likely they are to stick together.

Now water molecules like to stick to each other and when too many of them are together they "condense". But if you were to increase the temperature of the air (and hence the water molecules), they end up moving around faster and less likely to stick together. This way you can add more water molecules for the same amount of air.

The other thing to think about is pressure. In some sense it describes how crowded the place is. The higher the pressure, the more crowded it is. So when you increase the pressure, you increase the chances that the water molecules will stick together and hence "condense".

Another way to think about this is through Le Çhatelier's principle, which states that when a system in equilibrium (especially chemical equilibrium) is disturbed it tries to adjust in a way which minimizes the disturbance.

So now think of a box with some water and air. If you leave it long enough the system will reach an equilibrium, wherein no more water will evaporate or condense. But say now you increase the temperature of the air, the system will try to readjust so as to reduce the change in temperature. To do this it will cause the water to evaporate (as evaporation requires/consumes heat) and thereby reducing the change in temperature.
Similarly when you increase the pressure which is similar to decreasing the volume of the box, the air tries to account for this by condensing out some of the water (since liquid water occupies less volume than gaseous water), thereby reducing the effective decrease in volume seen by the air.

You may wonder why other particles in air do not "condense." They do, but only at much lower temperatures and/or higher pressures.


What is the function of nitrobenzene as a solvent in Friedel Crafts alkylation reaction?

Without any more information, it's just that, a solvent. Since it is a poor substrate for Friedel Crafts alkylations, it can be considered inert in these conditions. The choice of nitrobenzene would probably be to carry out the reaction at a relatively high temperature, since it boils at 210.9 °C.


Asymmetry in trigonal bipyramidal geometry

Yes, 5-coordinate compounds are always a bit confusing.

The best way I've answered this question about VSEPR for students is to do a little demo with balloons. Blow up balloons (i.e., electron pairs) and tie them around a central knot. It will definitely give a trigonal bipyramidal shape - it has nothing to do with chemistry, and everything to do with "fitting 5 things around a central point."

I disagree with @Martin that there's no theoretical basis to the shapes. The theory is purely geometric.

There are multiple possible shapes, including square pyramidal, but these are higher in energy because you have more ~90-degree angles. As discussed in the comments, if all 5 atoms are in a plane, the angles are 72 degrees, which is clearly bad. You could in principal have the top of an octahedron, in which case all angles are 90 degrees, but that's also clearly worse. (In practice, square pyramidal structures have the central atom slightly above the plane, increasing some angles slightly beyond 90.)

So I do the demo with the balloons, talk about square pyramidal and how this alternative is clearly slightly higher in energy.

The Berry pseudorotation, as mentioned above, does convert between the shapes, and it's known that at room temperature the axial and equatorial positions scramble quickly.

Update:

I decided to do some quick calculations using Avogadro and MOPAC on $\ce{PF5}$ in square planar versus trigonal bipyramidal using the PM7 semiempirical method. This took longer to write than to run the calculations (seconds).

Here's the optimized square pyramidal geometry:

The phosphorous isn't in the plane, it's a bit above, to increase four of the bond angles and minimize electron-pair repulsion. Note that the largest angle is ~103 degrees, but the F-P-F angles in the basal plane are all ~87 degrees.

Here's the trigonal bipyramidal geometry:

Now the angles are exactly what we expect from VSEPR. We see 90 degree angles between the equatorial plane and the axial F atoms. And the F-P-F in the plane are all 120 degrees.

Just from the angles we can guess that trigonal bipyramidal should be lower in energy.

For reference, the difference in energy between these two geometries using PM7 is ~2.76 kcal/mol. So there's not a huge difference in energy, but it's there.


Is it not safe for health to finish a bottle of wine during a week

The bad taste left after drinking wine can be attributed a number of agents found in wine, particularly alcoholic wines.

Alcoholic wines typically contain 8-15% ethanol. When ethanol mixes with oxygen in air it gets oxidised, gradually producing acetic acid (vinegar).

Furthermore, the reaction of ethanol with urea (naturally present in wine) to form ethyl carbomate (urethane) can produce a saline, bitter taste, although probably not toxic.

The ethanol in wine is produced through the fermentation of carbohydrates by yeast. This process also results in the production of acetaldehydes which contribute to its complex set of flavours.

Ethanol itself however is tasteless. It is however toxic and a class 1 carinogen according to the International Agency for Research on Cancer (IARC).

When consumed, ethanol is oxidised by alcohol dehydrogenase (ADH) enzymes in the body to produce acetaldehyde, which is 10-30 times more toxic than alcohol and possibly carcinogenic in humans.

Recent scientific research on cancer seems to indicate even low levels of alcohol consumption (1 cup per day) 'increases the risk of cancer of oral cavity and pharynx, esophagus and female breast' Source: http://annonc.oxfordjournals.org/content/early/2012/08/21/annonc.mds337.full

So a bad taste may not be the worst thing alcohol leaves in one's mouth.

With regard to the health effects of alcohol, please refer to the World Health Organisation report which is based on data collected scientifically from around the world over many years.


Why do impacting electrons have a low probability of causing ionization?

Ionization has a low probability of occurring if the energy of the incoming electron is exactly equal to the ionization energy. Using a classical analogy, if  ionization occurred the incoming electron would be left with no energy, ie, stationary. The incoming, now stationary, electron would not scatter away but rather remain trapped by the nuclear attraction, such that no net ionization would have occurred. It would also imply (classically) a colinear collision with the departing electron, a seemingly unlikely event. On the other hand, if the incoming electron has a significant energy surplus, such that it has sufficient energy to escape the nuclear attraction (scatter), then ionization is more likely to occur.

There are obviously more accurate non-classical ways to describe the process, but the analogy illustrates the problem. There are also intermediate scenarios, such as those exploited in Auger spectroscopy.


Why is chromate coloured but tungstate colourless?

Both complexes (and many more such as permanganate, sulphate, …) share the same general MO scheme which I am going to shamelessly copy from my older answer:

Figure 1: Qualitative MO scheme of a tetrahedric complex with σ and π bonding between metal and ligands. Double vertical lines represent electron pairs.

The intense colour of permanganate and chromate is due to LMCT transitions: ligand-to-metal charge transfer. Essentially, an electron from the HOMO (a $\mathrm{t_1}$ orbital that can be thought of a ligand-centred p-type orbital) is excited to the LUMO (an $\mathrm{e}$-type orbital that can be thought of one of the metal’s empty d orbitals) by absorbing a photon from the visible range; we then see the complementary colour, meaning that chromate absorbs blue light and looks yellow to our eyes.

The energies of LUMO and HOMO – or rather, the energy difference between the two – determines the wavelength of the absorbed photon and thus the colour that we see. For both permanganate and chromate, the wavelength is well within the visible spectrum resulting in the strongly coloured complexes. In cases such as sulphate or tungstate, the absorption wavelength is in the ultraviolet range which cannot be perceived by human eyes; therefore, the solution appears uncoloured.

Without computer calculations, it is probably impossible to accurately predict the ultimate wavelength but note that such an absorption is practically always present if the corresponding orbitals are available although usually in the ultraviolet range. An obvious counterexample would be any zinc complex in which the metal has a $\mathrm d^{10}$ configuration and the LUMO is thus the much further removed $\mathrm a_1$ orbital; therefore, zinc complexes are typically colourless and do not display such an LMCT band.


Step 1: Find equilibrium concentration for Ag+

As you correctly showed in the first step, we need to find the maximum solubility equilibrium concentration of $\ce{[Ag+]_{eq}}$ that is still allowed. Your calculation is correct, the result is
$$ \ce{[Ag+]_{eq}} = \frac{K_\text{sp}}{\ce{[Cl- ]}} = 1.8\cdot 10^{-9}$$

Step 2: Find equilibrium concentration for [Ag(NH3)2]+

We now need to know how much of the $\ce{AgNO3}$ we put into the solution actually ends up in the complex $\ce{[Ag(NH3)2]+}$, which is not in danger of precipitating. The formation equilibrium is according to the reaction equation
$$ \ce{Ag+ + 2NH3 <=> [Ag(NH3)2]+} \,,$$
and reads
$$ K_\text{f} = \frac{\ce{[Ag(NH3)2+]}}{\ce{[Ag+]_{eq}}\ce{[NH3]^2}} \,.$$

Substituting $x=\ce{[Ag(NH3)2+]}$ we quickly reach
$$ K_\text{f} = \frac{x}{\ce{[Ag+]_{eq}}(1-2x)^2} \,.$$
This is the step where you went wrong: You also need to consider that the concentration of $\ce{NH3}$ is changed by the formation of the complex. In principle we should also have taken this into account in the section above, but there the error is many more magnitudes smaller than here.

Solving the above gives $x = \ce{[Ag(NH3)2+]_{eq}} = 0.0258942$.

Step 3: Calculate the total amount of Ag+ in the solution

Armed with the knowledge from the two previous sections, this step is pretty trivial. As the contribution of free silver ions is many orders of magnitude smaller than the concentration of the complex, we can safely ignore it and approximate:
$$ \ce{[Ag+]_{tot}} = \ce{[Ag+]_{eq}} + \ce{[Ag(NH3)2+]_{eq}} \approx 0.02589 $$

Step 4: Calculate the amount of AgNO3 to add

If we assume that the activity coefficient is 1, we can directly infer the concentration
$$ c_{\ce{Ag^+_{tot}}} = \ce{[Ag+]_{tot}} ~ \mathrm{mol\, L^{-1}} \,.$$

From the concentration we can calculate the amount of $\ce{AgNO3}$ to add, as you did correctly,
$$ \begin{align}
m_{\ce{AgNO3}} &= c_{\ce{Ag^+_{tot}}} V M_{\ce{AgNO3}} \\
&= 0.02589 ~ \mathrm{mol\, L^{-1}} \cdot 1~\mathrm{L} \cdot 169.87 ~ \mathrm{g\, mol^{-1}} \\
&= 4.398~\mathrm{g}\,,
\end{align}$$
which, after rounding to two significant digits, leads directly to the result your solution says you should have gotten:

$$ m_{\ce{AgNO3}} = 4.4~\mathrm{g} $$


Decrease of volume by half in equilibrium - with specific numbers- how to solve?

I solved the quadratic equation by Wolfram Alpha

The solution indicates that $\ce{[Cl2]}=\pu{1.26926M}<\pu{2M}$,but$[\ce{CO}]=\pu{1.26926M}<\pu{2M}$, if we have
1 mol of $\ce{CO}$, 2 Mols of $\ce{Cl2}$,and 9.765 Mols of $\ce{COCl2}$
contained in 1 L.


Appropriate container for storing/keeping chromic acid?

Chromic acid can well be stored in glass bottles with stopper/lid but should be kept away from any organic compounds. Chromic acid is a dangerous oxidising agent and incompatible with various organic compounds. Also, it should be well marked with proper label and hazard symbol to distinguish it from other mineral acids. It has a decent shelf life and can be kept for long period without decomposition if kept well.

Also make sure it remains separated from mineral acid. Chromic acid is known to form solution with sulfuric acid forming chromosulfuric acid which is ironically used to clean laboratory glasswares.

Chromic acid is not a glass etchant.


Are there proteins with permanently bound magnesium?

From the look of things you may actually be referring to a lot of different proteins, however I managed to
grab two that I found fascinating:

Enolase

A glycolytic enzyme which catalyzes the reversible dehydration of 2-phosphoglycerate to phosphoenolpyruvate:

Credits: Lehninger Principles of Biochemistry page 213 ("Enzymes")

In the enzyme active site, 2-phosphoglycerate undergoes strong ionic interactions with two bound Mg2+ ions causing the resulting intermediate to be easily abstracted.

GTPase specifically  ARF1 (ARF for ADP-ribosylation factor). These are important in priming phase in the formation of Clathrin-coated vesicles (CCVs).

Clathrin itself is  a protein that forms a polyhedral framework around vesicles that transport TM, GPIlinked, and secreted proteins from the Golgi to the plasma
membrane.

X-ray structures of ARF1.GDP and ARF1.GTP, were determined by Dagmar Ringe and by Jonathan Goldberg:

Credits Voet and Voet Biochemistry 4th ed page 435 ("Membrane Assembly and Protein Targeting")

X-ray structures of (a) ARF1.GDP (left) and (b)ARF1.GMPPNP (right) (GMPPNP is a nonhydrolyzable GTP analog in which the O atom linking GTP’s beta- and gamma-phosphorus atoms is replaced by an NH group.)

The bound nucleotides are drawn in stick form in white with their phosphorus atoms magenta and their bound $\ce{Mg^2+}$ ions shown as lavender spheres.

In ARF1.GDP, the protein’s N-terminal helix (red) together with its
covalently linked myristoyl group (not present in the X-ray structures) are bound in a shallow hydrophobic groove on the
surface of the protein formed in part by the residues of loop gamma 3.


Covalent bonds other then 2-center, 2-electron bonds

One example to consider is benzene, where each carbon contributes 1 electron to a pi-bonding system of 6 electrons in 3 orbitals. The C-C bond lengths in benzene are intermediate between standard single and double bond lengths. Also take a look at the various boranes such as diborane (B2H6) which contain three-centered (or higher?) bonds involving boron and hydrogen.


Calculate the transition dipole moments between the 1s inner shell and the high-lying states of a molecule

Might have found the option you are looking for. GAMESS controls MO printing with the NPREO keyword. If you can specify ranges of orbitals either by their number (e.g. NPREO(1)=1,10 prints the first 10 orbitals) or relative to the HOMO (e.g. NPREO(1)=-3,-5 prints from HOMO-3 to LUMO+4). You can also give 3rd and 4th numbers to list the range of orbital energies to print.

These print options are described on p.23 of the input descriptions document on the Gordon research group page.


Why is nitrosobenzene electrophilic on O in organocatalytic oxygenations?

Using proline as the catalyst provides enantioselectivity to the process, but also allows N vs O regioselectivity to be controlled (though not always perfectly).
Houk (and subsequently others) applied computational methods to the reaction in an attempt to understand why the nitrosobenzene was acting as an electrophile at oxygen rather than nitrogen, as would traditionally be expected (and as is often the case when no proline is used). The computational studies showed a clear preference for the transition state in which the basic nitrogen of nitrosobenzene was able to interact with the acidic proton of the proline:

Proline catalysis occurs via a proline enamine attack on the oxygen of the nitrosobenzene monomer with a simultaneous proton transfer from the carboxylic acid. Computational evidence shows that the observed heteroatom selectivity materializes through the difference in basicities of the oxygen and the nitrogen atoms.
Source: Cheong, P. H.-Y.; Houk, K. N. J. Am. Chem. Soc. 2004, 126 (43), 13912–13913. DOI: 10.1021/ja0464746.

This study further indicated that the transition state for attack at N (which you drew) is disfavoured by 2.5 kcal/mol:

Ref: Enantioselective Organocatalysis, 2007, Dalco (Wiley)


Why don't alloys self-galvanically corrode?

They certainly can. Most cases where the alloying compound is more noble this is a hazard. Nickel in Iron is a popular example, you can get corrosion acceleration on the iron by reduction happening on the Ni-surfaces.

Where the alloying element is less noble you usually have superficial corrosion to begin with but in the end the resulting surface is protected by the main element - which is common for bronze(s). If you just have lumps, then you do not have a bronze and slowly but certainly, nature will reclaim the tin.

One has to care about the homogeneity of the alloy.


Mechanism for Diels–Alder of alkene with substituted thiophene sulfone with loss of sulfur dioxide

A few references into the rabbit hole and we have an answer. The major reference was given in the link mentioned by OP which was Chem. Ber. 1994, 127 (2), 409–425[1], in which the reaction mechanism of the synthesis is given as follows:

[...] Reaction of 33 with tetrathiophene-S,S-dioxide is performed by heating the reactants in highly concentrated toluene solution to $\pu{120 °C}$ for a total of
24 h. The cycloadduct 34 is obtained in high yield.

Now the question that has arisen in the step from 33 to 34 is the following, under normal conditions of Diels-Alder, there would only be one double bond in the final product. Here there are two. This can be answered going one down into the rabbit hole and using J. Org. Chem. 1980, 45 (5), 856–867.[2]
This paper deals with the specific reaction that causes confusion - the annulation using tetrachlorothiophene 1,1-dioxide. According to this paper:

The compound has a combination of stability and reactivity that makes it a useful synthetic reagent. It undergoes Diels-Alder addition with double bonds to form an adduct which spontaneously loses sulfur dioxide to leave the double bond annelated with a tetrachlorobutadienediyl group.

Furthermore, it talks about the reactivity to olefinic bonds and states the following:

Example 18, from l-hexen-5-yne, reveals the greater reactivity of alkene vs. alkyne. Reaction is exclusively with the alkene bond as far as NMR reveals. Thus, the kinetically favored product is formed rather than the thermodynamically favored one.

There is a minor error in the reaction mechanism as given in the link of OP as the first paper shows that the tetra-chloro derivative needs to be oxidised in order to be converted into cyclohexaindane
References:

Kuck, D. Benzoanellated Centropolyquinanes, 15. Benzoanellated Fenestranes with [5.5.5], [5.5.5.6], and [5.5.5.5] Frameworks: The Route from 1,3-Indandione to Fenestrindan. Chem. Ber. 1994, 127 (2), 409–425. DOI: 10.1002/cber.19941270218.
Raasch, M. S. Annulations with tetrachlorothiophene 1,1-dioxide. J. Org. Chem. 1980, 45 (5), 856–867. DOI: 10.1021/jo01293a019.


How much hydrogen or oxygen will be produced in the electrolysis of water?

I'm not sure it's appropriate to explain the relationship between all the variables above, here, as there are a lot of concepts that would need to be explored. However, fundamentally (assuming there are no other reactions apart from the electrolysis) the charge passed through the cell is proportional to the amount of products formed, described by Faraday's laws of electrolysis. Here is a useful rearrangement:
$$n = \frac{Q}{Fz}$$
Where $Q$ is the charge passed, $F$ is the Faraday constant and $z$ is the number of electrons transferred per redox reaction. For example, $z = 2$ for hydrogen because it takes two electrons to convert $\ce{H+}$ to $\ce{H2}$ as $\ce{2H+ + 2e- <=> H2}$. Therefore, for every coulomb of charge, we get: $$\begin{align}
  n &= \frac{1\ \mathrm{C}}{(96.485\ \mathrm{kC/mol})(2)}\\
    &= 5.18\ \mathrm{µmol}
\end{align}$$ of hydrogen gas produced. So if we can measure the charge (or current integrated over time), we can calculate the amount of product produced. This is true no matter what voltage is applied or what the electrolyte content of the solution is, however if more than one reaction is occurring at an electrode, you won't be able to separate the contributions from each reaction with only the charge as information. There will also be a minor contribution from charging the double layer capacitance when the potential is first applied.

Where things become less straightforward is understanding how changing the potential or the solution resistance affects the amount of charge that flows, but the charge itself is directly related to how much product is being produced.


How to determine the valence electrons of D-block elements in their mono atomic form?

Based on the Aufbau principle, they would all be filling the (x-1)d sublevel, so they would ALL have 2 valence electons in the xs sublevel.

Since the Aufbau priniciple fails to account for the fact that some elements will be more stable with incomplete valence xs sublevels and additional electrons in the (x-1)d sublevel, you just need to memorize the transition metals that are more stable with 1 valence s electron:

Chromium, Possibly Nickel, Copper, Niobium, Molybdenum, Ruthenium, Rhodium, Silver, Platinum and Gold,

And the two extra specials:

Palladium which prefers to fill the x-1d sublevel and not put any in the xs sublevel.  It has 18 in its highest energy level, level 4, but only the 10 4d electrons are considered to be valance electons.

And Lawrencium which has 7s2 (5f14) xp1, or 3 valence electrons, putting 1 electron in the xp sublevel rather than in the (x-1)d sublevel.  It turns out to be a d-block element that is not filling the d-block.

Notes
This will give you the number of electrons in the highest energy level for transition metals.  Transition metals in the highest d sublevel (but not in the highest primary energy level) can provide valence.  Many sources will refer to all of the electrons above the configuration of the prior noble gas to be valence electrons.

On one hand, most of the oxidation states for transition metals have been observed, up to the point where they would have lost all of those d electrons.

On the other hand, the d electrons in transition metals are not de-localized in metallic bonding of transition metals (to the best of my knowledge).


The step in question here is the so-called Wolff Rearrangement.

In this rearrangement, silver in the form of for example $\ce{Ag2O}$ is used as a catalyst. A catalyst facilitates a reaction (lowering the activation energy), but it comes out of the reaction unchanged (which doesn't mean it's not changed during the reaction, it's just changed back at the end ;))

So why silver? Well, transition metals are good catalysts most of the time for different reasons:

They have vacant d-orbitals which leads to
An easy tendency to exhibit variable oxidation states.
They can form reaction intermediates with reactants.
The presence of defects in their crystal lattices.

(I conveniently copied this list from a quora post, because im lazy)

What this essentially means is that due to their electronic structure, transition metals are often able to bind reactants to their surface and activate them for example. There are quite some catalyst reactions that we don't even understand today, but sometimes it's as easy as this

And why silver? Well probably Wolff just tried a couple of transition metal catalysts and chose the one with the best effect/cost ratio! Theoretical expectations don't always meet reality so often you just have to try and find the right reactants/catalysts.


Why is the molar entropy of hexane lower than that of butane?

There are two factors to consider. Certainly, the dominant factor is that the degrees of freedom available for the molecule to disperse energy into increases as the carbon chain extends. Thinking along these lines, one would expect a monotonic increase of entropy with carbon chain length.
However, we must also consider phase. As the carbon chain increases in length, the equilibrium phase at RTP changes from gas to liquid. This occurs between butane and pentane. At this point, there is a drop in the standard entropy, which then proceeds to increase monotonically from this point (pentane). The cause for this drop in entropy upon phase-change is due to the motional degrees of freedom being more constrained by proximity to other molecules in liquids compared to gasses.
See this link, where you can see some nice plots of standard entropy vs. carbon chain length.


The caffeine is in a free base form unless prepared otherwise . Caffeine is an alkaloid and a base , slightly alkaline. If you add base like naoh to it the solution will become more alkaline.


Can ionic compounds can ever "dissolve" as molecular compounds do?

Can we refer to these free-floating ionic compounds as "dissolved" solute?

No. Dissolved implies that the ions are solvated. If the ions are in a solid ionic lattice then they are not solvated by the solvent and so would not be described as part of the dissolved solute.

  Does "XY(aq)" (the aqueous solution of an ionic compound) seem like a good description of this state?

Again no. $\ce{XY(aq)}$, in the context of ionic compounds, refers to $\ce{X^{n+}}$ and $\ce{Y^{n-}}$ ions solvated by water molecules. The dissociation is generally regarded as implicit. For example, $\ce{NaCl(aq)}$ implies an aqueous solution containing $\ce{Na+}$ and $\ce{Cl-}$ ions. A better description of the state you are thinking of would be $\ce{XY(s)}$, using state symbols, or even better you could describe it as a suspension.

From the IUPAC Gold Book:

  Suspension:
  A liquid in which solid particles are dispersed.

This accurately describes your system.


Evaluating (∂U/∂T)P - (∂U/∂T)V for an ideal and van der Waals gas

Here I am going to use $U$ to refer to the total internal energy. If you want the molar internal energy ($U_\mathrm m \equiv U/n$) just divide your final result by $n$.

In another of my answers I have proven (equation $(14)$) that

$$\left(\frac{\partial U}{\partial T}\right)_{\!p} = \left(\frac{\partial U}{\partial V}\right)_{\!T}\left(\frac{\partial V}{\partial T}\right)_{\!p} + \left(\frac{\partial U}{\partial T}\right)_{\!V}$$

The proof up until there is completely general, so this applies for both an ideal and a van der Waals gas.

Furthermore here I have shown that (combining equations $(6)$ and $(8)$)

$$\left(\frac{\partial U}{\partial V}\right)_{\!T} = T\left(\frac{\partial p}{\partial T}\right)_{\!V} - p$$

Putting these together gives you

$$\begin{align}
\left(\frac{\partial U}{\partial T}\right)_{\!p} - \left(\frac{\partial U}{\partial T}\right)_{\!V} &= \left(\frac{\partial V}{\partial T}\right)_{\!p}\left[ T\left(\frac{\partial p}{\partial T}\right)_{\!V} - p\right]
\end{align}$$

All that's left is to evaluate the partial derivatives. The ideal gas case is very easy; since $p = nRT/V$, $(\partial p/\partial T)_V = nR/V$ and the term in square brackets goes to zero.

For the van der Waals gas you need to work at those partial derivatives which are a bit tedious, but should be doable, especially if you employ implicit differentiation. For example, if you take the van der Waals equation of state

$$\left(p + \frac{an^2}{V^2}\right)(V - nb) = nRT$$

and differentiate with respect to $T$ throughout, keeping $V$ constant, one gets

$$\begin{align}
\left[\left(\frac{\partial p}{\partial T}\right)_{\!V} + 0\right](V - nb) + \left(p + \frac{an^2}{V^2}\right)(0) &= nR \\
\left(\frac{\partial p}{\partial T}\right)_{\!V} &= \frac{nR}{V - nb}
\end{align}$$

and likewise, barring any careless mistakes, I obtained

$$\left(\frac{\partial V}{\partial T}\right)_{\!p} = nR\left(p - \frac{an^2}{V^2} + \frac{2abn^3}{V^3}\right)^{-1}$$

Maybe somebody has a less tedious way (and I hope so), but this is all I can offer you myself.


Why do shampoo ingredient labels feature the term "Aqua"?

In most countries, cosmetic product labels use the International Nomenclature of Cosmetic Ingredients (INCI) for listing ingredients.

The INCI name “AQUA” indeed just describes water (which is used as a solvent).


Predicting the products of reaction between ammonia and hydrogen peroxide

Majority of answer content came from this link and this link.

Strong solutions of $\ce{H2O2}$ with a few drops of $\ce{NH4OH}$ (aq. ammonia) is left for 24 hours without any nitrite formation occurring. But upon longer standing, even with a small amount of hydroxide then nitrite forms. Nitrite also forms when a dilute solution of $\ce{H2O2}$ is mixed with $\ce{NH4OH}$  and is evaporated over pure conc. $\ce{H2SO4}$ with a bell jar. $\ce{H2O2}$ forms (even in very dilute solutions) nitrite very rapidly, if the $\ce{H2O2}$ solution is mixed with a few drops of $\ce{NH4OH}$ and a little $\ce{NaOH}$  and this then boiled in a retort to a very small volume.

$\ce{2NO2- +  2H+(aq) <=> H2O + NO + NO2}$

$\ce{NO2- + NO3- + 2H+(aq) <=> H2O + NO2}$

Nitrogen dioxide is then able to attack ammonia.

$\ce{2NO2 + 2NO + 4NH3 -> 2NH4NO2 + 2H2O + 2N2~~~~~~~~~~~~~~~~~~~~(1)}$

In any case, solutions of ammonium nitrite decompose on heating to 60 to 70C.

Perhaps the very slow spontaneous reaction of aqueous ammonia with $\ce{H2O2}$ takes place the small equilibrium with amide anions, $\ce{NH2-}$.

$\ce{NH2- + H2O2 -> NH2- + 2OH-}$

Even in $\ce{NH4OH}$, there is a very slight equilibrium with amide ions.

$\ce{2NH3 <=> NH2- + NH4+}$

Amide rapidly and vigorously reacts with water, so the equilibrium of amide existence in water must be very low.

$\ce{NH2-+ H2O -> NH3 + OH-}$

In the base catalysed decomposition of hydrogen peroxide, the mechanisms is presumably:-

$\ce{H2O2 + OH- -> HOO- + H2O}$

$\ce{HOO- + H2O2 -> HOOOH + OH-}$

$\ce{HOOOH --> HOOO- + H+(aq)}$

$\ce{HOOO- -> OH- + O2}$

Also, ammonium nitrite decomposes into nitrogen and water.

$\ce{NH4NO2 → N2 + 2 H2O~~~~~~~~~~~~~~~~~~~~~~~~~~(2)}$

A combination of ammonia and hydrogen peroxide is often used in the process of bleaching hair. Hydrogen peroxide has the necessary properties to remove the color from hair - it oxidizes one of hair’s melanin pigments to a colorless substance. This results in the hair becoming lighter. However, it is not recommended by hair professionals (those who don't understand chemistry but want your cash), and can cause extensive damage to hair (even causing it to fall out).

Therefore, the overall reaction is:-

$\ce{2(NH3•H2O) + 3H2O2 ->[\Delta] N2 + 8H2O~~~~~~~~~~~~~~~~~~~~~~~(1) + (2)}$


Are common names of substituents accepted by IUPAC?

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) various prefixes are retained for use in general (IUPAC) nomenclature. However, many prefixes are no longer recommended.

  Trivial, common, and traditional prefixes have always been an integral part of organic nomenclature. However, as systematic nomenclature develops and becomes widely used, many of these prefixes fall by the wayside. Accordingly, each set of IUPAC recommendations contains fewer of these traditional prefixes.

The prefix “isopropyl” for $\ce{(CH3)2CH-{}}$ is retained only for use in general (IUPAC) nomenclature.
For the preferred IUPAC name (PIN), the preferred prefix is “propan-2-yl”.
The prefix “1-methylethyl” may be used in general (IUPAC) nomenclature.

The prefix “sec-butyl” for $\ce{CH3-CH2-CH(CH3)-{}}$ was still contained in the 1993 recommendations but is no longer recommended as approved prefix.
For the preferred IUPAC name (PIN), the preferred prefix is “butan-2-yl”.
The prefix “1-methylpropyl” may be used in general (IUPAC) nomenclature.


You can answer such questions using D.U. (Degree of unsaturation)

The formula is $\mathrm{C + 1 - }\frac{H + X - N}{2}$

C = Carbon.
H = Hydrogen.
X = Halogen.
N = Nitrogen.

If you get D.U. to be one, then in the structure there could be:

1 double bond.
1 ring.

Let's take an example of $\ce{C4H8}$ which has D.U. equal to one.

It can have three structures of 1 double bond:

It can have two structures of 1 ring:

Now for D.U. equal to two, the possibilities are:

2 double bond.
2 ring.
1 double bond and 1 ring.
1 triple bond.

Let's take your example of $\ce{C4H6}$ which has D.U. equal to two:

It can have two structures of 2 double bond:

It can have one structure of 2 ring:

It can have four structures of 1 double bond + 1 ring:

It can have two structures of 1 triple bond:

So, the answer is 5 cyclic isomers as you can see above.


Why is the boiling point of CH3COOH higher than that of C2H5OH?

Yes you are correct. The molecules are polar in nature and are bound by intermolecular hydrogen bonding. But you must pay attention to the extent of polarization in both the molecules. Consider the alcohol. Here the carbon bearing the $\ce{-OH}$ group is the only polarizing group present. But for $\ce{CH3COOH}$ , the carbonyl carbon is polarized by an $\ce{-OH}$ group as well an $\ce{=O}$ group attached to it, thus increasing its effective polarization more than the alcohol. Therefore $\ce{CH3COOH}$ has greater boiling point.


Is there a formula or way to determine whether an acid is a strong one or a weak one without conducting experiments in real?

Yes, there are ways to determine whether an acid is strong or weak without going into the lab yourself. That being said, someone must still go into the lab.

The most straightforward way would be calculating ΔG for an acid-base reaction. Remember, a negative delta G means that a reaction is spontaneous. Since strong acids have a Ka > 1, their ΔG values should be negative.


Relation between ionic resonance energy and polarity

I'm not too familiar with resonance theory, but I'll try:

According to modern VBT, the general idea is that the state of the molecule is not described by a single state wave function, but rather a linear combination ("superposition" if you will) of multiple different state wave functions. In your particular case, the resonance is between a "dominant" covalent structure (involving a completely equal sharing of electron density) and a "minor contributor" ionic structure (with unequal electron density).

Since resonance energy is defined as the amount of energy needed to convert the true delocalized structure into that of the most stable contributing structure (which can be interpreted as the amount of stabilisation provided by the linear combination of the state wave functions), a higher resonance energy would mean a greater contribution (and hence weighing coefficient) from the ionic structure. This means that there is a greater ionic character to the bond (and it is more polar).

Hope this helped.


Most of the ore will be oxidized if thermally treated under air. That's why it's usual to heat it under controled atmosphere or even in vacuum.


Mechanism of formation of 2-naphthol red dye (aka Sudan 1)

The most important thing to point out here is that your arrows are backwards. The diazonium is not the nucleophile. It is the electrophile. The diazonium ion has one lone pair. If it were a nucleophile, there's no way that you'd end up with the diazo compounds drawn, where each nitrogen has a lone pair.

This also means that drawing out all of your resonance structures doesn't mean anything because that is not the intermediate. The real question to ask here is which site on the naphthol ring is more nucleophilic.

Consider the naphthol as an enol for the purposes of the addition. Your two starting structures are exactly the right ones for focus one. The unsaturation in the enol in the top example is such that the nucleophilic site is benzylic while, in the other case, it is far from the ring.

The intermediates after attack by an electrophile are shown (technically, the carbonyls should be protonated still, but I'm too busy to fix that now). Only in the top case, where we substitute adjacent to the other ring, is aromaticity still preserved in the other ring. This is why the substitution is strongly preferred there as opposed to the other position, where we proceed via an intermediate that is not aromatic.


Conventional heating compared to microwave heating of chemical reactions

Your question about the 'magic' microwave effect is one that I think most chemists who've used a microwave have asked. Anecdotally, some reactions do often just work better when carried out in a microwave. The whole debate is quite nicely summarised in this quote from a (fascinating...) book on the use of microwaves in chemical synthesis:

  Since the early days of microwave synthesis, the observed rate accelerations and sometimes altered product distributions compared to oil-bath experiments have led to speculation on the existence of so-called “specific” or “non-thermal” microwave effects. Historically, such effects were claimed when the outcome of a synthesis performed under microwave conditions was different from that of the convention- ally heated counterpart at the same apparent temperature. Reviewing the present literature, it appears that today most scientists agree that in the majority of cases the reason for the observed rate enhancements is a purely thermal/kinetic effect, i.e., a consequence of the high reaction temperatures that can rapidly be attained when irradiating polar materials in a microwave field

  Source: Microwaves in Organic and Medicinal Chemistry

Microwave radiation:

Source: Microwaves in Organic and Medicinal Chemistry

From the table above, its clear that microwave radiation is not strong enough to cause any actual chemical change itself (other than via the temperature it creates), for microwave radiation used by common household and lab microwaves (2450 MHz, which is important as it avoids interfering with communication equipment), the energy is sufficiently low that it couldn't in theory even over-come a hydrogen bond.

Thermal/pressure effects:

There are several reasons why the use of a microwave may be beneficial compared to a standard reflux setup in a PEG bath (or any other conventional heating method):

the microwave is able to quickly reach a given temperature, with only a short induction period. The result of this is that the reaction reaches the desired temperature quickly, rather than slowly warming from room temperature. This means that the reaction gets going immediately, which can be especially beneficial in cases where the catalyst/reagents being used aren't terribly stable (your metathesis is one such example where the catalyst can quickly degrade in solution).
the microwave heats the solution directly and uniformly, compared to heating conveniently where the glass is being heated and in turn the glass heats the solution.

Source: Biotage. Left image = microwave heating, right image = oil bath, both taken after 1 min of heating

the vials used in microwave reactions are sealed (they're closed systems), which means in addition to temperature, they're able to carry out reactions under increased pressure, allowing the reaction to be heated above the boiling point of the solvent being used.

Source: Biotage

Overall, the easiest way to show all of this is by looking at the read-out from a microwave reaction, showing how quickly the desired temperature can be reached and how this remains constant, along with the pressure, throughout the entire reaction run:

Non-thermal effects:

Non-thermal effects, or 'specific microwave effects' are defined as accelerations of chemical transformations in a microwave field that cannot be achieved or duplicated by conventional heating, but essentially are still thermal effects. These have been considered by many groups and indeed a wikipedia page exists discussing them, however they're difficult to show/prove experimentally.

For solution phase organic synthesis, its generally accepted that there is no specific microwave effect. Only in solid phase systems (ionic lattices for example) is there anything to suggest something unique going on that cannot be achieved using standard heating methods alone.


Confusion about the relation between gibbs free energy and electromotive force

What that means is that $\Delta G$ is the maximum work available and therefore is only a thermodynamic limit. That's why $E$ is measured with a voltmeter where no current is flowing (actually extremely low current, i.e. the voltmeter has very high impedance). An analogy would be measuring gas pressure in a pressurized vessel, only in e-chem this is "electron pressure". Once you do work with an e-chem cell, the current will encounter many resistances which translate into voltage drop and in turn are energy losses and E will change.


How can I prepare a PBS (Phosphate Buffer Solution)? pH=7.4

Google is your friend!

The preparation of pH 7.4 PBS (with KCl and KH2PO4) is described on this site   here. Preparation of many other buffer systems are also described.

Here are the conditions (lifted from the linked site)

Prepare 800 mL of distilled water in a suitable container.
Add 8 g of NaCl to the solution.
Add 200 mg of KCl to the solution.
Add 1.44 g of Na2HPO4 to the solution.
Add 240 mg of KH2PO4 to the solution.
Adjust solution to desired pH (typically pH ≈ 7.4).
Add distilled water until volume is 1 L.


Can ethanol be oxidized by hydrogen peroxide?

First things first: Don't mess with higher concentrations of hydrogen peroxide unless you are a trained and well-equipped chemist. We're talking "steel-reinforced gauntlets" here. This chemical is as volatile as nitroglycerine.

At 3% you have mostly de-ionized water and for good reason. Hydrogen peroxide is extremely, violently unstable, and is just as likely to oxidize anything else than what you actually intend to react it with.

Anything above a concentration of 30% is very dangerous. Above 60% is suicidal in ill-equipped scenarios. 100% is a hypothetical, and is in essence a potential component of rocket fuel.

So, yes it will work. You may lose a hand, but it will work.


How do you calculate molar specific heat?

Specific heat has the units of $\mathrm{J/(K\cdot kg)}$.

Molar specific heat is in units of $\mathrm{J/(K\cdot mol)}$, and is the amount of heat needed (in joules) to raise the temperature of $1$ mole of something, by $1$ kelvin (assuming no phase changes).

So, the conversion factor you need, from dimension analysis, will have unit $\pu{kg/mol}$.

$\pu{kg/mol}$ is the SI unit for molar mass.

Multiply the specific heat by the molar mass to get the molar specific heat.

For example, the molar mass of water is $\approx \pu{0.018 kg/mol}$.

The specific heat of water is $\approx 4186\ \mathrm{J/(K\cdot kg)}$.

So the molar specific heat of water is $4186\ \mathrm{J/(K\cdot kg)} \times \pu{0.018 kg/mol} \approx 75\ \mathrm{J/(K\cdot mol)}$


How do zeolites catalyse the cracking of hydrocarbons?

The video you reference to does not tell the whole story.  Yes, zeolites are minerals with open pores, similar to a filter, occurring in nature and are produced commercially, too. Equally to their structure, their composition is interesting -- as both are used to tailor the reactivity and selectivity of this material.  ZSM-5 is one example of designed remarkably high Brønsted acidity.  Hydrocarbons in contact with these sites split into fragments and rearrange because of the intermediate generation of carbocations.  It is advantageous, as these reactions are heterogeneously catalysed by these "solid state acids".


A review on $\ce{Ag+}$ quotes the following situations, which may be provoked by the Lewis acidic behavior of a metal cation.

  Silver treated cells exhibited electron-light regions in
  their cytoplasm thus forcing DNA to condense. Authors state that
  condensed DNA molecules lose their ability to replicate.

I presume that $\ce{Ag+}$, acting as a Lewis acid, will reduce the negative charge on DNA, thus inducing DNA aggregation.

  Electron-dense region were also observed around the cell wall inside
  the cytoplasm, and the cytoplasmic membrane experienced  shrinkage
  and  detachment  from  the  cell wall.

  There was also indication  that  silver  ions  interact  with  thiol
  groups leading to the deactivation of enzymatic proteins.

Here one sees the action of $\ce{Ag+}$ on thiols which may be interpreted as an acid-base interaction of the Lewis type.

  Also the effect of $\ce{Ag+}$ on the respiratory electron chain of E. coli was stressed, since $\ce{Ag+}$ is toxic to microorganisms due to inhibition of the
  respiratory chain at multiple sites.

It appears that in many cases, one can link the antibacterial function of metal ions to their acidic properties. And certainly, those discussed above are not the only ones.


Electroplating a spoon (electrolytic cell) using a electrochemical voltaic cell

Let's start with the last question: The points A and C have the same potential, because we  generally neglect the resistance of the wire. The same argument is valid for points B and D. But according to your scheme, you have a short circuit (as C and D are connected).

Now, to electroplate the spoon, you want the silver to be oxidized at the silver electrode. So, silver electrode is the anode, and must be connected to pole (+) of the voltaic cell (point B):
$$\ce{Ag -> Ag+ + e- }$$
The spoon is the electrode where ion silver will be reduced. So, it's the cathode and must be connected to pole (-) of of the voltaic cell (point A). The electrons enter the cathode and reduced the silver ion according to the equation:
$$\ce{Ag+ + e- -> Ag}$$

So, you're right!


Finding the equilibrium temperature between two substances in a constant pressure calorimeter

Assuming the heat absorption of the calorimeter is negligible (so if it's made of an insulating material, the only changes in heat will be of the contents themselves).

Substance 1 (the one at the higher initial temperature) has a heat = $q_1$

Substance 2 has a heat of $q_2$

Since substance 1 will lose heat until it comes to equilibrium with substance 2, the heat lost by 1 will be absorbed by 2, so $q_1 = -q_2$

If $q_1 = m_1c_{p1}(T_{{1}_{initial}} - T_{{1}_{final}})$ and $q_2 = m_2c_{p2}(T_{{2}_{initial}} - T_{{2}_{final}})$ and $T_{1_{final}} = T_{2_{final}} = T_f$ you can solve for $T_f$.

If the heat absorption of the calorimeter was not negligible, then $q_1 + q_2 + q_{cal} = 0$


I like your "why?" question. So I dig down to find out why Post-It notes stick to any surfaces, yet, when removed, it's able to post again in another place. This is my finding and it makes sense Adhesives:

  Look at the back of a sticky note using an electron microscope and you'll see not a continuous film of adhesive but lots of microscopic glue bubbles, known as microcapsules, which are about 10–100 times bigger and much weaker than the glue particles you'd find lazing around on normal sticky tape. When you push a Post-it® onto a table, some of these relatively large sticky capsules cling to the surface, providing just enough adhesive force to hold the weight of the paper in the little note. Every time you attach and peel off the note, dust and dirt attach to the adhesive capsules, so they progressively lose their stickiness. But since there are so many capsules of all different sizes, a Post-it® note does go on sticking for quite a while.

I believe, when you dipped your sticky note in water, these microcapsules filled with water, which would be a thicker layer than the water in other areas. So, gum layer area contains more water, thus, that part is actually more damp than the rest of the paper.


The purpose of the salt bridge is to prevent the two half-cell solutions from mixing. It is possible to make a really bad galvanic cell by putting both half-cells in a single solution, but they rapidly self discharge as the oxidizing agent ($\ce{Cu^+}$ in your example) can diffuse through the solution and react directly with the other electrode (Zn in the example), with the electrons being transferred directly instead of going through the external circuit.

Having a "bridge" of just water soaked paper is similar to having both half-cells in the same container but poses some issues. First, the diffusion of ions through the bridge is going to be very slow, increasing the cell resistance, second, the self-discharge problem above will occur, and third, because different ions diffuse at different rates, a junction potential develops as the two containers diffuse into one another.

So a salt bridge's job is to not only maintain electroneutrality between the two containers, but to do so without allowing the solutions to mix. An inert salt is used to prevent any reactions from occurring between the salt and the solutions, and things like $\ce{KCl}$ are commonly used for analytical cells as potassium and chloride ions have very similar diffusion coefficients, minimizing junction potential.


If sulphuric acid barely evaporates, why we can smell it?

There are at least two plausible explanations for why an odorless compound with very low vapor pressure could nevertheless generate an odor:

Sulfuric acid is highly reactive.  If you are pouring or dispensing sulfuric acid, you are probably generating some mist. This is a concern in industrial settings where sulfuric acid is used:

https://www.cdc.gov/niosh/docs/81-123/pdfs/0577.pdf

Thus, even if the sulfuric acid itself has neither a significant vapor pressure nor an odor, as the mist enters your nose it could react with substances in your nasal passages to create compounds that have an odor.  Some studies have reported the perception of odor from sulfuric acid mists:
https://www.epa.gov/sites/production/files/2014-09/documents/sulfuric_acid_interim_dec_2008_v1.pdf

And even without misting, particulates in the air, as they fall into an open container of sulfuric acid, could react with sulfuric acid to generate compounds that create a smell.  This effect is seen with metals -- the metallic "smell" is not from the metals themselves (which have an extremely small vapor pressure at room temperature), but from reaction products of the metals:

Why can we smell copper?


$\ce{Na_2CO_3}$ forms several hydrates with decomposition point close to room temperature (30-40 Celsium). Moreover, the solubility of decahydrate decreases with temperature decrease, and solubility of anhydrous salt almost does not depend on temperature and is much higher than solubility of decahydrate.  The relevant pictures are googleable by words 'sodium carbonate solubility curve', see 'images' google service, not general search.

So, I believe you have prepared a solution of the salt at the higher temperature at day, but the salt mostly precipitated in form of decahydrate overnight, because the solution cooled significantly at night. Bonus points for this version if you have significant changes in temperature at your work during the nychthemeron.

Second version is that you indeed prepared seemingly oversaturated solution because it is not saturated for anhydrous $\ce{Na_2CO_3}$, but overnight it crystallized, because it is saturated for decahydrate. I'm a bit sceptical about such possibility, though, because significantly oversaturated solution crystallize even on slightest imperfections, and the still not-dissolved salt is a great collection of such imperfections.

In addition, if the solution stayed overnight open, it indeed may consumed some amount of atmospheric $\ce{CO_2}$ and partially transform into sodium hydrocarbonate with much lesser solubility.


Comparing Acidic Strength of H3PO4 and H3AsO4

The stability of an acid is judged by the strength of it's conjugate base. If the conjugate base is stable, it implies that it is a weak base and hence the corresponding acid is stronger.
Now consider $\ce{H3PO4}$. The structure has one $\ce{P=O}$ bond and $\ce{3 P-OH}$ bonds.

The conjugate base would be one Hydrogen short from an $\ce{OH}$ group. Similarly with $\ce{H3AsO4}$. However, the conjugation (resonance) will be better in the case of $\ce{P=O}$ bond, since it is an orbital overlap of oxygen's $\mathrm{2p}$ and phosphorus' $\mathrm{3p}$ as against arsenic's $\mathrm{4p}$. $\mathrm{2p}$ and $\mathrm{3p}$ are of comparable size and account for a better overlap and better resonance, distribution of electronic charge and hence a more stable conjugate base.
Consequently, $\ce{H3PO4}$ is better than $\ce{H3AsO4}$ as an acid.


Is there a simple field test for heavy water?

There's a very simple test for $\ce{D_{2}O}$ that springs to mind - ice cubes made with heavy water sink in light water. I assume you're talking about differentiating a glass of essentially pure heavy water from, say, tap water, in which case this test should work rather well and requires equipment no more sophisticated than a freezer, ice cube tray and some glasses.


What are the possible products in allylic bromination of 3-methylcyclopentene?

Your comment shows that you are on the right track.

You have chosen A and C because they are formed when the initial allylic radical instantaneously reacts with $\ce{Br2}$ in the propagation step of the reaction, isn't it? (Hint: Just say YES!)

Now give the allylic radicals corresponding to A and C a bit of time, just enough to flip the dot.

How do these other radicals look like? There you have the other two products.

For an extra point: Which isomer of methylcyclopentene could lead to G?


Do Lewis acids and bases turn change colour of litmus

You should keep in mind that the concept of acid-bases is not restricted to water. Lewis acid definition is applicable to gaseous, aqueous and non-aqueous systems. Consequently, all Bronsted-Lowry acids are Lewis acids. The idea applies to bases. Note that the converse is not true just like an apple is fruit but not all fruits are apples. The idea of litmus test is rather restricted to aqueous systems. So in short, the all relatively strong Lewis acids and bases in water can affect the litmus.
If you put a blue litmus is very dry HCl (gas), it would still turn it red because paper is never dry. It always has moisture absorbed in it.


You do not need the water vapour for this effect, the air that is contained in the bottle is sufficient.

You may use the ideal gas law to estimate the amount of air in the volume of the bottle of about $V=1\ \mathrm l=0.001\ \mathrm{m^3}$ when the air is heated to about $T=60\ \mathrm{^\circ C}=333.15\ \mathrm K$ (using hot water) at normal pressure $p=1\ \mathrm{bar}=100\,000\ \mathrm{Pa}$ (since the bottle is still open):

$$\begin{align}
pV&=nRT\\[3pt]
n&=\frac{pV}{RT}\\[3pt]
&=\frac{100\,000\ \mathrm{Pa}\times0.001\ \mathrm{m^3}}{8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}}\times333.15\ \mathrm K}\\[3pt]
&=0.036\ \mathrm{mol}
\end{align}$$

If you now close the bottle, the amount of air $n$ is trapped.

When this air is cooled to about $T=0\ \mathrm{^\circ C}=273.15\ \mathrm K$ (using ice water) at an approximately constant volume of $V=1\ \mathrm l=0.001\ \mathrm{m^3}$, the pressure in the bottle decreases according to the ideal gas law:

$$\begin{align}
pV&=nRT\\[3pt]
p&=\frac{nRT}{V}\\[3pt]
&=\frac{0.036\ \mathrm{mol}\times8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}}\times273.15\ \mathrm K}{0.001\ \mathrm{m^3}}\\[3pt]
&=8.2\times10^4\ \mathrm{Pa}=0.82\ \mathrm{bar}
\end{align}$$

Therefore, the ambient air pressure of $p=1\ \mathrm{bar}=100\,000\ \mathrm{Pa}$ compresses the bottle until the pressures are equalized (or the bottle fails). The volume in the bottle decreases according to the ideal gas law:

$$\begin{align}
pV&=nRT\\[3pt]
V&=\frac{nRT}{p}\\[3pt]
&=\frac{0.036\ \mathrm{mol}\times8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}}\times273.15\ \mathrm K}{100\,000\ \mathrm{Pa}}\\[3pt]
&=8.2\times10^{-4}\ \mathrm{m^3}=0.82\ \mathrm l
\end{align}$$

Nevertheless, any condensation of water vapour would help to achieve this effect. The resulting changes in pressure and volume due to condensation of water can be even more extreme than the contraction of air. For example, if a volume of $V=1\ \mathrm l$ is filled with steam at $T=120\ \mathrm{^\circ C}$ and $p=1\ \mathrm{bar}$ (without any air) and then cooled until the steam is almost completely condensed to liquid water (at about $T=99.6\ \mathrm{^\circ C}$), the resulting liquid volume is only $V=0.00058\ \mathrm l=0.58\ \mathrm{ml}$.


Every different compound has a different CAS number, even if they are enanthiomers when isolated and published (their racemic mixture has another CAS number). No exception! See following examples:

L-(−)-Glyceraldehyde: CAS Number [497-09-6].

D-(+)-Glyceraldehyde: CAS Number [453-17-8].

DL-(+/-)-Glyceraldehyde: CAS Number [56-82-6].

For your question, I found the CAS Number of D-(+)-glucosamine hydrochloride, which is [66-84-2]. The CAS Number [3416-24-8] is for glucosamine (free base).

The CAS Number of D-glucosamine sulfate is [14999-43-0]. Each molecule contains two molecules of glucosamines and one molecule of $\ce{H2SO4}$, the two acedic hydrogens on each of amine groups. Its molecular formula is $\ce{[C6H14NO5]2.SO4}$, giving the formula weight: $\pu{456.42 g/mol}$.

The CAS Number of D-glucosamine sulfate potassium chloride is [216699-44-4]. Each molecule contains two molecules of glucosamines, one molecule of $\ce{H2SO4}$, and two molecules of $\ce{KCl}$. Its molecular formula is $\ce{[C6H14NO5]2.SO4.2KCl}$, giving the formula weight: $\pu{605.52 g/mol}$. However, NIH website listed [1296149-08-0] as the CAS Number of glucosamine sulfate Potassium chloride, which must be issued a fairly recent time. Spectrum Chemical Co. sells the compound as a dietary supplement under the same CAS number.


Your approach isn't bad: it seems to me that you're considering 3 half-lives to get from 400 mm $\ce{Hg}$ to 50 mm $\ce{Hg}$, and that works for the purpose of answering the question. That said, I think the following steps are going to serve you better when approaching problems like this one, as we need to explicitly solve for $k$ and then find $t$ where some concentration of reactants is specified.

What you've been given is simply the half-life ($t_{1/2}$) of reactant $\ce{A}$ and the change in pressure of the reactant at some time after the reaction begins. That change in pressure is a fine proxy for concentration, since we are in the gas phase and are considering the change of one reactant only. I'm using the concentrations in the expressions here and not the partial pressures, given that the conversion factors of one to the other cancel: we're interested in ratios.

For a first-order reaction, the integrated expression describing the disappearance of $\ce{A}$ is

$$\ln\left({\ce{[A]}\over\ce{[A]_0}}\right) = -kt$$

and the usual expression for half-life (where I've rearranged to solve for $k$) is

$$k = -{\ln(1/2)\over 8.8\;\mathrm{min}} = 0.0787\;\mathrm{min}^{-1}$$

So now, we have $k$. We also have $\ce{[A]_{0}} = 400$ mm $\ce{Hg}$ and $\ce{[A]} = 50$ mm $\ce{Hg}$ (from the problem statement). Rearranging the first equation I've written above, we solve for $t$:

$$-\left({1\over 0.0787\;\mathrm{min}^{-1}}\right)\cdot\ln\left({50\over 400}\right) = t = 26.42\;\mathrm{min}$$

where I have omitted the units of mm $\ce{Hg}$ in the natural logarithm argument because they cancel.

As you properly note, the time is slightly longer than the 26.4 minutes given in your choice of answers. I believe that the answers are just rounded to the tenths place, which obviously has created a situation where one can calculate the "right" answer using the wrong methodology.


Does ice melt faster when submerged in water?

Assuming the water and air are the same temperature (and that the air is mostly stationary) then the ice will melt quicker in the water. This is due to water having a higher thermal conductivity and heat capacity. In fact, this is easy to test: place a cocktail stick through an ice cube and lay it on top of a glass filled right to the top with (room temperature) water. The submerged half will melt quicker than that on top.

Regarding the second question: A warm object will cool quicker in icy water than in just ice for the same reason above but also because of increased surface area contact. However, icy water will be warmer than just ice. So the item will potentially be able to achieve a lower temperature in just ice (slow and steady wins the race).


What are potential sources of low recovery in this acid-base extraction experiment?

I can only assume that you we following some recipe for the purification of the benzoic acid. Frankly if I were doing this I would have done it differently.

Wikipedia shows that the solubility of benzoic acid is 1.7 g/L (0 °C) and 56.31 g/L (100 °C) in water. So I'd added benzoic acid to a test tube and put in a boiling water bath. I'd added just enough water to get the benzoic acid to dissolve and then let the liquid cool in air. When down to room temp, I'd put it into an ice bath. After 10 minutes in ice bath I'd pour off water. I'd wash with a few drops of ice cold water. I'd test for purity and if not good enough I'd repeat.

0.34 g of crude benzoic acid should dissolve in about 6 ml of boiling water.

Each water cycle should have an efficiency of close to (56.31 - 1.7)/56.31 = 97%