Relevance of strain energy, total energy and stability of molecule?

Would anybody please highlight the difference between total strain energy of a molecule, total energy of a molecule

Strain energy is usually defined as the sum of all angle and steric strains that a certain molecule possesses, compared to an unstrained analogue molecule.

For example, here is a table that compares various cyclic alkanes to cyclohexane:

wikipedia

In this way, stain energy can be thought of as a positive contribution to potential energy that destabilizes a molecule above the stability that its ideal bonding would suggest.

Total energy of a molecule could be defined in lots of ways, often it is done as a $\Delta H_f$, which could be positive or negative, and compares the potential energy of a molecule (i.e. how strong the bonds are, stronger means lower potential energy) to the elements from which it is formed. However, this will also take into account any intermolecular forces present and these are not inherent to a single molecule, but to a collection of molecules in standard states.

We could also define the energy of a molecule by thinking of the sum of the bond enthalpies, which define how strong the bonds are compared to unbonded atoms (specifically by breaking bonds homolytically, forming radical species, in the gas phase). The larger the value, the stronger the bonds are and the lower the potential energy of the molecule.

In computational chemistry, the total energy of a molecule can theoretically be calculated using quantum mechanical means. This is done using energy minimization software. There is a good post here about the theory in this process. This value will take into account the bond enthalpy/strength and strain within a molecule.

  and what we can infer about the stability of the molecule using the total strain energy of the molecule as well as the scope of using such inferences made.

In general, the greater the strain energy and the higher the total energy, the less stable/more reactive the molecule.


The free electron pairs of the oxirane may act as a Lewis base and coordinate with a Lewis acid like $\ce{Li+}$.  This form of activation facilitates the attack of the alanate $\ce{[AlH4]-}$ as a $\sigma$-nucleophile at the less substituted side of the epoxide.

Once opened, the nucleophilic oxygen attacks at the electron deficient centre of $\ce{AlH3}$ (Lewis acid).  But the $\ce{Al-O}$ bond hydrolyzes upon the workup, yielding the alcohol and the hydroxides of lithium and aluminum.


Relationship between microplastics and wildfire smoke

The term «polymer» only refers to using small molecules (called «monomer», needn't be of one type) multiple times to build a larger molecule.  E.g., if you would use the small molecule thrice to get a larger one, it were a trimer, if four times, a tetramer to be more specific; but these were all examples of «polymers».  The term does not tell if the material were of industrial / artificial (like for a PET bottle), or biological origin (e.g., lignin, chitin, cotton, or kreatin).
Given the small size of molecules, even tiny particles known as microplastic consist of many polymer molecules each.  Like if the band of a tagliatelle would just be cut into smaller and smaller pieces and still count as a noodle.  The fine mechanical division  / chopping into smaller and smaller fragments only increases the accessible area per unit of weight.  Similar to a combustion engine, this material then spread out may yield a combustion more efficient with less smoke and soot than, e.g., a torch, or a candle. Not all polymers burn equally well, even if finely divided.
The «problem» with wood, compared to natural gas - chemically speaking - is that it consists of many different compounds.  Beside the lignin, there are a number of minerals, sugars, fluids, resins, etc. and water present in a tree; each with its own characteristics.  Some will catch fire quickly no matter what, like human hair.  Some need to be heated first so that fire nibbles them away.  Meanwhile others may be heated and then burn as gases, rather than as solid or liquid, sometimes liberated like in an explosion.  Some materials require much air (wind) per unit of mass to burn, or otherwise would just smoulder; so now equally soil, weather and topography of the terrain enter the the set of parameters yielding a typical wild fire with incomplete combustion and smoke.


What is the explanation of the energy gap law in radiationless transitions?

These are a few 'rough and ready notes' and some data on radiationless transitions.

Intersystem crossing $S-T$ and Internal Conversion $S_1-S_0$ are described by the Fermi Golden Rule:

The rate constant decreases exponentially with energy gap between S and T in solution. In contrast in isolated molecules the gas phase rate increases with energy gap, see figures below.

Molecules in condensed phase

Molecules isolated in the vapour phase

The pure singlet and triplet states cannot interconvert because of the strict spin selection rule. An interaction, which is spin-orbit coupling, mixes the states together to make new states that contain a bit of each initial state. The singlet now contains a little triplet and triplet contains a little singlet. As the S1 and S0 states are orthogonal to one another without an interaction; $\int S_1 S_0 d\tau = 0$ so no transition. States must have different symmetries if coupled by a photon similarly  a vibration with the correct symmetry can also couple S1 and S0 to cause internal conversion.

Transitions occur because an interaction mixes basis states together (singlet and triplet, say) to form new states containing part of each. These new eigenstates are shown schematically below, the thick lines on the lower pair show the amount of each base state in the mixed state.

One state can couple to many others, and when there are very many final states (below) we think of them as a quasi-continuum of levels. The radiationless transition rate is the rate from one state (optically excited) to these many final states.

Equation to solve is $\displaystyle (H_0+\Omega)\psi=i\hbar \frac{\partial}{\partial t}\psi$ to get rate constant $W_{kn}$ between levels $k,n$

Robinson & Frosch simplified the calculation of Fermi Golden Rule with approximations;
(a)Assume that the interaction energy is the same to all $n$ levels ,
(b)The (vibrational) Franck-Condon factors take an average value.
(c) The density of states $\rho$ is made to fit data! ( $\rho$ = number of states / wavenumber)

The approximate formula is: $\displaystyle W_E =\frac{2\pi}{\hbar}|V|^2F_c\rho(E)$  where $W$ is rate constant at energy $E$, $V$ is interaction matrix element between $k$ and $n$ , $F_c$ is the average Franck Condon factor and $|V|^2F_C=\int\psi_1\Omega\psi_2d\tau$. The $\Omega$ is spin-orbit operator for intersystem (singlet to triplet) crossing, kinetic energy operator for internal conversion.  The term $\rho(E)$ is the effective density of states at energy $E$.

This formula is not good for accurate calculation but indicates what is important.

The full and detailed calculation shows that :

(a) In the initial state, say singlet, only certain vibrational modes are effective in causing a radiationless transition.

(b) These modes are separated out and called promoting modes, at least one is present, and can only be identified by experiment.

(c) These promoting modes are the only modes which do not have zero electronic coupling terms.

(d) promoting modes have zero displacement between the two electronic states and are non-totally symmetric

(e) In the final state accepting modes receive the vibrational energy, these are modes of largest frequency so that as few quanta as possible of each are excited. Usually these are CH or CC vibrations.

There are two cases molecules in condensed phase and isolated molecules in gas phase.

(a) Condensed phase

The energy gap law  $\displaystyle K_{ST}=Ae^{-\beta\Delta E}$ is observed experimentally and is obtained theoretically from $W_{kn}$ by taking a Boltzmann distribution over initial levels and then the maximum value of resulting expression. It turns out that this term is greater than all the others combined. (btw, this energy gap law has nothing to do with Arrhenius equation.)

The full expression is

$$W=|V|^2\sum_m \left[\prod_n\frac{X_n^{m_n}}{m_n!}\right] \delta (\Delta E -\hbar \omega_k-\hbar\sum\omega_n m_n)$$

where $\displaystyle X_n=\frac{\mu_n\omega_n}{2\hbar}(Q_n-Q_k)^2$ is the displacement between vibration $n$ between energy surfaces $Q_n\to Q_k$. The term in the square bracket is the 'Franck Condon Factor density of states' assuming harmonic oscillators.  $\omega_k$ is the promoting mode and has no displacement between the two states. The $\delta$ term ensures energy conservation, i.e. $\delta(0)=1$, else is zero.

This equation is very difficult to use, but to a good approximation the complicated $\Sigma\Pi$ term can be replaced by its maximum value and a significant simplification is obtained.

We now take $X$ as being the displacement of the vibration of maximum frequency, $\omega$, usually CH or CC vibration. $\Delta E$ is singlet -triplet energy gap. The expression for the energy gap law is

$$W=|V|^2e^{-X}e^{-\gamma\Delta E/\hbar\omega}, \qquad \gamma= \ln\left(\frac{\Delta E}{\hbar\omega X}\right)-1$$

and $\gamma$ has values in the range 3 to 5 typically and is effectively a constant.

Its success is that it describes fall off with energy gap and also D/H difference in rate constants due to difference in frequency, see first figure above.  The difference to the gas phase is the averaging over the Boltzmann distribution.

(b) Gas phase

In the gas phase with a large molecule, say benzene, we are in the statistical limit with a huge number of final states, effectively a continuum, then the calculation is from one initially excited level $k$ to many levels $n$ in the final state.

$$W_{kn} \sim |V|^2\sum_n \big|\int\psi_n\Omega \psi_k d\tau\big|^2 \delta (\Delta E_n-\Delta E_k )$$

the equations to use in this case can be found in Beddard et al Chem. Phys. Letts. 1973, v 18 p 481, if you cannot get to this paper I can post again.


How do you determine which entity gains or loses an H+ in an equation?

Both of these equations are valid, as $\ce{HCO3-}$ is amphoteric (i.e. both an acid and a base).

To answer your question, one has to consider that the $\mathrm{p}K_\text{a}$ value for $\ce{H2CO3}$ is 6.37 and that for $\ce{HCO3-}$ is 10.32. So if we started from "pure" $\ce{HCO3-}$ (let's assume sodium bicarbonate) and add it to water, what would happen?

The $\mathrm{p}K_\text{b}$ for $\ce{HCO3-}$ is of $14 - 6.37 = 7.63$, it is larger than the $\mathrm{p}K_\text{a}$ for the acidic function (10.32). Therefore although both a weak base and a weak acid, it is a stronger base than acid, increasing the pH when added in water.


IUPAC nomenclature for tricarboxylic acids

Systematic names of carboxylic acids are generally formed using the suffix ‘oic acid’ $\ce{{}-(C)OOH}$, or ‘carboxylic acid’ $\ce{{}-COOH}$, or the prefix ‘carboxy’ $\ce{{}-COOH}$. Note that the $\ce{{}-(C)OOH}$ group of ‘oic acid’ formally replaces a terminal $\ce{{}-CH3}$ group whereas the $\ce{{}-COOH}$ group of ‘carboxylic acid’ or ‘carboxy’ is a usual substituent that replaces just a $\ce{{}-H}$.

The relevant rules in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) read as follows.

  P-65.1.2.1 Carboxylic acid groups, $\ce{{}-COOH}$, that conceptually replace a $\ce{{}-CH3}$ group of methane or terminate an unbranched hydrocarbon chain are named by replacing the final ‘e’ of the name of the corresponding hydrocarbon by the suffix ‘oic acid’. No locants are necessary to denote the positions of the carboxylic acid groups in a hydrocarbon chain; locants are used when hydrocarbon chains are modified by skeletal replacement, as shown in P-15.4.3.2.3. Except for formic acid, acetic acid, oxalic acid (see P-65.1.1.1), and oxamic acid (see P-65.1.1.1), systematically formed names are preferred IUPAC names; the names given in P-65 .1.1.2 are retained names for use in general nomenclature.

That’s why the name of the unsubstituted parent structure of the first compound that is given in the question is a heptanedioic acid.

  P-65.1.2.2.1 If an unbranched chain is linked to more than two carboxy groups, all carboxy groups are named from the parent hydride by substitutive use of the suffix ‘carboxylic acid’, preceded by the appropriate numerical prefix ‘tri’, ‘tetra’ etc. and appropriate locants.

That’s why the name of the second compound is pentane-1,3,5-tricarboxylic acid.

  P-65.1.2.2.3 The prefixes ‘carboxy’ and ‘oxalo’
  When another group is present that has priority for citation as suffix, for example, a free valence, or when all carboxylic acid groups cannot be described by a suffix, carboxylic acid groups, $\ce{{}-COOH}$, are indicated by the preferred prefix ‘carboxy’ (also used in general nomenclature). The prefix ‘oxalo’ is recommended as the preferred prefix for $\ce{{}-CO-COOR}$, but cannot be used to lengthen a carbon chain. In general nomenclature, the compound prefix ‘carboxycarbonyl’ may be used, but the compound prefix ‘carboxyformyl’ is not recommended.

That’s why the complete name of the first compound is 4-(carboxymethyl)heptanedioic acid.


Why does hybridization produce a more stable configuration?

What causes orbital hybridization?

Hybridization only occurs in response to bond formation, that is an atom will not just spontaneously hybridize on its own.  Further, an atom will only hybridize if hybridization leads to the formation of a more stable (lower energy) molecule.

Nature is very efficient at exploring the energy surface and finding whatever energy minimum is available.  For example, ammonia ($\ce{NH3}$) is roughly a tetrahedron with $\ce{H-N-H}$ bond angles around 109.5° (the tetrahedral angle) using (roughly) $\mathrm{sp^3}$ hybridized orbitals. For ammonia, mixing (hybridizing) it's $\mathrm{2s}$ and three $\mathrm{2p}$ orbitals to form four $\mathrm{sp^3}$ hybridized orbitals that are used to form three $\ce{N-H}$ bonds and one lone pair orbital produces a more stable molecule than if it were not hybridized.  On the other hand, the analogue phosphine ($\ce{PH3}$) chooses to remain roughly unhybridized, using p-orbitals to form its $\ce{P-H}$ bonds with a resultant $\ce{H-P-H}$ bond angle around 90°.  Bond strengths, steric factors, etc. all play into determining whether a hybridized or unhybridized structure will be the more stable.


Activated charcoal cleans water mainly by adsorbing the impurities.

You can also couple normal charcoal with sand and form alternate layers of charcoal and sand and pass water through it. It's a very effective Indian style of filtering water. Nearly all suspended particles gets blocked by the above mentioned filtering style.

You can also convert normal charcoal into activated charcoal to serve your purpose.

Now irradiated water cannot be treated by these methods alone. Normal carbon filtering is used along with several other methods such as ion exchange methods and other such high level water processing methods to treat irradiated water. High levels of radiation can't be treated fully though.

You can refer to this link to learn about effective radioactive water purification techniques.


Dealing with different units (psi) in ideal gas equation

In the given experiment, there is no indication that the volume of the container changes and/or a reaction takes place. Both the volume and the amount of substance can therefore assumed to be constant.
Let's assume that the enclosed air can be treated as an ideal gas.
Under these conditions, the pressure is proportional to the absolute temperature (in Kelvin):
\begin{align}
p &\propto T & \frac{p_1}{p_2} &= \frac{T_1}{T_2}
\end{align}
This relation is known as Amonton's Law.

I am really confused as to what psi is and how am I supposed to convert it to figure out the final pressure.

Using the equation above, you would only have to convert the temperatures to Kelvin and could give the final pressure as $\pu{psi}$ too.
If you insist on conversion to Pascal ($\pu{Pa}$) as your favourite SI unit for pressure, remember that a pressure is defined as a force acting on a surface (area):
$$\pu{1 Pa} = \pu{1 N//m^2} = \pu{1 kg//m s^2}$$
The $\pu{psi}$ is a still rather common unit and it stands for pounds per square inch.
This is double weird, since both the force and the area are expressed in non-SI units.
Note that the pound is not the German Pfund ($= \pu{0.500 kg}$), but the angloamerican pound with $\pu{1 lb} = \pu{0.453 kg}$.
The area part in $\pu{psi}$ is given in square inches with $\pu{1 in} = \pu{0.0254 m}$.
In summary, $\pu{1 psi} \approx \pu{6895 Pa}$.


What is the difference between the equilibrium position and the equilibrium constant?

Here's a more simplistic response.

Equilibrium position: a particular set of concentrations of reactant and product species.
Equilibrium constant: a number that describes how far "forward" equilibrium lies for a particular system (as described by a particular equation).

Example: for the system $\ce{A <=> B}$, the mass action expression is $\ce{\frac{[B]}{[A]}}$.  Let's say $K = 200$.
This means, at equilibrium, $\ce{[B]}$ will be 200 times greater than $\ce{[A]}$. That's what the equilibrium constant tells you - equilibrium for this system lies to the right, and the K value is greater than 1.

There are infinitely many different equilibrium positions that satisfy K.  We could have $\ce{[B]} = 20~\mathrm{M}$ and $\ce{[A]} = 0.1~\mathrm{M}$.  Or we could have $\ce{[B]} = 1~\mathrm{M}$ and $\ce{[A]} = 0.005~\mathrm{M}$.  Those are two different equilibrium positions that are both at equilibrium.  In both cases, $Q = K$ and the system is at equilibrium.

Some notes:

If we describe the exact same system with a different equation, $\ce{B <=> A}$, then the mass action expression is $\ce{\frac{[A]}{[B]}}$. We would find that $K = 0.005$ (that is, $1/200$) and we would say that equilibrium lies to the left.  Of course, the molecules don't behave any differently, but the equation we use to describe the system is different, and that difference is reflected in a different $K$.
If we have a system with more than one product or reactant or with stoichiometric coefficients other than 1, then it's a bit more complicated than just saying something like "product concentrations are 200x greater than reactant concentrations". Instead, equilibrium may be achieved at many different equilibrium positions. You have to look at the mass action expression to interpret the meaning of $K$.
The equilibrium constant $K$ is just one way of expressing how far "forward" a particular reaction will proceed.  Gibbs free energy change ($\Delta G$) is another way. And, for redox systems, cell potential is yet another way.


A reasonable mechanism for the transformation is shown below. Since this is under basic conditions, I've written a mechanism that avoids cationic intermediates. Deprotonating the hydrazine gives a very nucleophilic nitrogen that can attack the nitro group. Proton transfer gives a nitrogen anion, which can kick out hydroxide. Two additional proton transfer steps lead to the product, HOBt.

A SciFinder search didn't reveal any examples of the intermolecular variant of this reaction, although I didn't find this exact intramolecular reaction either.


Essentially it means that you exclude the possibility of splitting an element even further via nuclear means.


How fast do electrons move around the nucleus?

The ratio of the speed of an electron traveling in the first Bohr orbit to the speed of light is given by the handy equation

$$\mathrm{V_{rel}=\frac{[Z]}{[137]}}$$

where Z is the atomic number of the element under consideration and 137 is the speed of light in atomic units, also known as the fine structure constant.  Consequently a 1s electron in the hydrogen atom will travel at approximately 0.7% the speed of light. In silver (Z=47) the 1s electron will travel around 34% the speed of light, while the 1s electron in gold (Z=79) will travel at about 58% the speed of light.

Once we get up to around silver, the electrons are traveling at relativistic speeds and this can dramatically impact properties of the atom. For example, the relativistic mass of an electron is given by

$$\mathrm{m_{rel}=\frac{m_{e}}{\sqrt{1-(V_{rel}/c)^2}}}$$

where $\ce{m_{e},~ V_{rel}~ and~ c}$ are the electron rest mass, the velocity of the electron, and the speed of light respectively.  The following figure provides a graphical representation of how the electron mass increases as the electron velocity increases.

The following equation relates the ratio of the relativistic radius of the first Bohr orbit $\ce{R_{rel}}$ to the normal radius $\ce{R_{o}}$, to the relativistic velocity of the electron

$$\mathrm{\frac{[R_{rel}]}{[R_{o}]}=\sqrt{1-(V_{rel}/c)^2}}$$

As the relativistic velocity of the electron increases, the orbital radius contracts (the above ratio becomes smaller).  For silver, the first Bohr radius contracts ~6%, while for gold the contraction is ~ 18%.

Look at these earlier Chem SE answers to see the interesting physical effects that atoms can exhibit when their electrons travel at relativistic speeds.

francium
gold
mercury


how to increase Keq(equilibrium) with temperatrue

You can easily see how the equilibrium constant varies with temperature using the Van’t Hoff equation:
$$\frac{\mathrm{d}\ln{K_\text{eq}}}{\mathrm{d}T}=\frac{\Delta H^\circ}{R{T^2}}$$

Rearranging the equation a little bit, we get that:
$$\mathrm{d}\ln K_\text{eq} = \frac{\Delta H^\circ}{RT^2}\mathrm{d}T$$

Now we can integrate each side of this equation:
 $$\int_{K_1}^{K_2} \mathrm{d}\ln K = \int_{T_1}^{T_2} \frac{\Delta H^\circ}{RT^2}\mathrm{d}T$$
Because $\Delta H^\circ$ and $R$ are constants, this will be equivalent to :
$$\int_{K_1}^{K_2} \mathrm{d}\ln K = \frac{\Delta H^\circ}{R} \int_{T_1}^{T_2} \frac{1}{T^2}\mathrm{d}T$$
Solving the integral we will get this :
$$\ln\left(\frac{K_2}{K_1}\right) = \frac{\Delta H^\circ}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$

For $\Delta H^\circ > 0$ we have two cases: when $T_2>T_1$ and $T_2<T_1$.
If $T_2>T_1$ then $\frac{1}{T_2}<\frac{1}{T_1}$ and hence $\frac{1}{T_2} - \frac{1}{T_1} < 0$ . Knowing that $\Delta H^\circ>0$, this means that $\frac{-\Delta H^\circ}{R}$ will also be negative, and the product of these 2 will be positive:
 $$\frac{-\Delta H^\circ}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)>0$$
And from the Van’t Hoff equation we get that : $$\ln\left(\frac{K_2}{K_1}\right)>0 \Leftrightarrow \frac{K_2}{K_1} > 1 \Leftrightarrow K_2 > K_1$$
i.e. The equilibrium constant increases as the temperature is increased.
Using the same principle for $T_2 < T_1$, we get that $K_2 < K_1$


Is ionic bond just formed by "electrostatic" interaction between two oppositely charged ions?

You are absolutely correct that the electron has an angular momentum, which can be thought of like a magnetic field and which interacts with magnetic fields.
In "light" elements, however, the magnetic interaction is almost completely negligible.

The Hyperphysics page on the Zeeman Effect shows that in a hydrogen atom, a magnetic field of one tesla (which is fairly strong) splits energy levels by $5.78 \times 10^{-5} eV$, or roughly one-millionth of the ionization energy of a hydrogen atom.

From just having done calcuations on the anomalous Zeeman effect for homework (you can use the formula in the link I provided and check it yourself), I can tell you that in a potassium atom, the energy levels of a p-electron split by 0.006 Hartree/Tesla, or 1/10,000 of the ionization energy of a hydrogen atom.

In other words, while there is a magnetic field (or what can be treated as one) generated by electron motion, it's fairly safe to ignore it in lighter elements.

As far as heavy elements go...well, you undergo what's called j-j coupling, caused by the fact that the electrons are going so fast that their motion becomes relativistic. To describe that motion requires the Dirac equation, which I have not studied in any level of detail. Instinct says that there are other considerations which are much more significant than magnetic interactions, but I have no proof whatsoever to back it up.


Given advancements of computational power an machine learning, how is it still not possible to predict products from chemical reactions?

It's not an all-or-nothing situation, where either we can reliably prediction reaction products, or we have not tried to.

For instance, drug companies, for many years, have been making extensive use of increasingly powerful, sophisticated, and refined in silico predictive models to choose likely candidates for in vitro testing.  And I would be surprised if they've not also begun to develop, evaluate, and incorporate ML.

At the same time, these models are not perfect.  So the answer is that it is being done, it's been done for a while, we're getting better and better at it, but we still have a long way to go.

It will be interesting to see what happens with quantum computing, which is currently in its infancy.


How do I type a simple chemical equation in Microsoft Word?

If you are using MS Word 2007 or newer, use the equation feature. It is designed for math but works okay for chemistry.

Go to the insert tab. (For shortcut you can press Alt+= sight together)

Click on the equation button on the far right.

Type in your equation. Use the buttons in the ribbon to do superscripts and subscripts. Alternatively you can use _ for subscript and ^ for superscript.  The default is to have letters italicized (as variables), so you will want to fix that.

There are also shortcut commands to render most the common things you want. For example, underscore _ creates a subscript and a caret ^ creates a superscript Shortcut for typing subscript and superscript in MS Word 2007|2010|2013|2016 and office 365  . You have access to a wide range of arrows from a pull-down menu, but -> will give you a simple right arrow (although it is not very long). This feature on Word will also accept some (but not all) tex commands for formatting equations.

To get a long arrow, click on the operator but and choose the arrow with the word "yields" written over it under common operator structures. For up arrow and down arrow showing gas liberation and precipitation use \uparrow or \downarrow followed by space Shortcut for typing arrows of chemical equation in Word 2007 and above.

Click on the word "yields" and replace it with as many spaces as you need to create an arrow of whatever length you want. Shortcut for other types of arrows is.

Finally, finish your equation.

If you need to type above or below arrow just type "\above(text above arrow goes here)[space]".Similarly tying below arrow just type \below(test below goes here)[space]". How to type chemical equation and arrows in Word 2007 and above.

For older versions of MS Word, go to the insert menu and click on the equation, which launches the Equation Editor Program (you can also find this program on your computer by searching for eqnedt.exe), which gives you the same ability to create equations.


Why do the melting and boiling points of the noble gases increase when the atomic number increases?

The melting and boiling points of noble gases are very low in comparison to those of other substances of comparable atomic and molecular masses. This indicates that only weak van der Waals forces or weak London dispersion forces are present between the atoms of the noble gases in the liquid or the solid state.

The van der Waals force increases with the increase in the size of the atom, and therefore, in general, the boiling and melting points increase from $\ce{He}$ to $\ce{Rn}$.

Helium boils at $-269\ \mathrm{^\circ C}$. Argon has larger mass than helium and have larger dispersion forces. Because of larger size the outer electrons are less tightly held in the larger atoms so that instantaneous dipoles are more easily induced resulting in greater interaction between argon atoms. Therefore, its boiling point ($-186\ \mathrm{^\circ C}$) is more than that of $\ce{He}$.

Similarly, because of increased dispersion forces, the boiling and melting points of monoatomic noble gases increase from helium to radon.

For more data of melting and boiling points of noble gas compounds, read this page


not understanding buffering region vs equivalence point

This is a matter of definitions which you must learn. Let's say that the titrant is NaOH and that we have a monoprotic acid.

Equivalence point - The equivalence point occurs when the moles of added base are equal to the moles of the initial acid species.

Notice that this doesn't infer anything about the pH of the solution.

Now for a polyprotic acid there are multiple equivalence points. The first equivalence point occurs when the moles of added base are equal to the moles of the initial acid species. The second occurs when twice as many moles of added base are added as compared to the initial acid species. The third three times as much base as acid and so on.

Again notice that this doesn't infer anything about the pH of the solution at the multiple equivalence points.

When the concentrations of the acid and base species are equal, pH = pKa."

This is a poor statement. Let's let the acid be $\ce{HA}$ so the conjugate base is $\ce{A^-}$. The statement would have been much better expressed as:

When the concentrations of the acid and its conjugate base are equal, pH = pKa."

Now to wrap this up, imagine doing this experiment. We start off with $\ce{HA}$ in solution. As we add more base we get to the pH where $\ce{[HA] = [A^-]}$ and where (1) there is maximum buffering capacity, and (2) the pH = pKa of the acid. We now add more base and we finally get to the equivalence point.

PS - For titrations you should also know the endpoint which is where the chemist gets some sort of signal to stop the titration.


Etymology of saturation degrees (-ane, -ene, -yne) in aliphatic compounds

I found the following information through a website linked to Yale University.

The naming structure seems to have evolved from some of the early nonsystematic names given to hydrocarbons. It started with Dumas who in 1835 came up with the name Methylene for the $\ce{CH2}$ radical. The "-ene" suffix is a Greek feminine patronymic which was used to denote that Methylene was the "daughter of" (derived from) the "spirit of wood" or what we now call methanol (Dumas believed methanol had the formula $\ce{CH2\cdot H2O}$).

The naming structure caught on for ethylene since it was related to ethyl(even though the meth- in Methylene came from a Greek work for wine and not the numerical prefix used in methyl).

When the use of these feminine patronymics (-ene, -ine, -one) became common during the 1860s, Hoffman proposed making a consistent nomenclature that followed the standard progression of vowels: a e i o u. This led to the suffixes -ane, -ene, -ine(-yne), -one, and -une to refer to increasingly unsaturated hydrocarbons.

Since -one was already in use for ketone structures, only the first three suffixes caught on.


Hund's rules do not apply to different configurations that arise from the occupancy of orbitals with different energies.

For example, with a $\mathrm{2p^2}$ configuration (e.g. carbon), the allowed term symbols are $^1D$, $^3P$, and $^1S$. From Hund's first rule, you can predict that the most stable one out of these three is the $^3P$ term.

You cannot compare these three terms with a term arising from a different electronic configuration, e.g. a $\mathrm{2p3p}$ configuration, which could give rise to a $^3D$ term which would be "more stable" on the basis of Hund's second rule.

The aufbau principle is much more important here. A simplified diagram would look something like this (don't judge me, it's not to scale and I have no time to calculate the exact splitting):

Term symbols are associated with specific electronic configurations and while we often write term symbols on their own, we should always remember that a term symbol without an electronic configuration makes no sense.

Just to show you why it doesn't make sense: with your lithium example, if you really wanted to, then you might as well promote the 2s electron to a 3d orbital, since that would give you a $^2D$ state. Higher $L$, wonderful! Why not go further and promote it to a 4f orbital, or a 5g orbital? We might as well ionise it...


How to choose the correct name for calcium sulfate based on a calculations of the crystallisation water included?

From my calculations

\begin{align}
x &= \frac{M(\ce{CaSO4}) \times \left[m(\ce{CaSO4 * xH2O}) - m(\ce{CaSO4})\right]}{M(\ce{H2O}) \times m(\ce{CaSO4})} =\\
&= \frac{\pu{136.14 g mol-1} \times (\pu{1.77 g} - \pu{1.07 g})}{\pu{18.02 g mol-1} \times \pu{1.07 g}} =\\
&= 4.94,
\end{align}

which gives $\ce{CaSO4* 5 H2O}$, calcium(II) sulfate pentahydrate.


What is the mechanism for the reaction of acetyl chloride and aniline?

The mechanism is just a regular nucleophilic attack of an amine on a carbonyl, with the triethylamine just there to pick up any loose protons at the end of the reaction.


chemically inert molecule with piezoelectric properties

The ur-piezoelectric crystal is quartz. Being composed of $\ce{SiO2}$, it more or less epitomizes chemical/biological inertness. It has the additional advantage of being comparatively cheap and common. There are a number of other natural minerals which are piezoelectric (e.g., various perovskites). Most of those minerals are, as far as I know, relatively unreactive and non-toxic. Naturally, if toxicity is a concern, avoid the common lead-based compounds.

I've seen mention of materials that exhibit both ferrimagnetic and piezoelectric properties. If ferrimagnetism is acceptable to you instead of paramagentism, then you may wish to consider that as an avenue for research. If you need paramagnetic materials specifically, then it's probably not worth your time (unless your project is compatible with subjecting those materials to their respective Curie temperatures).


Comparing formula for enthalpy change with bond dissociation energy and formation enthalpy

Yes. The two methods are equivalent.

You can try this with the dimerization of $\ce{NO_2}$ into $\ce{N_2O_4}$. In this reaction we form one $\ce{N-N}$ bond. Every other bond stays the same.

$\ce{2NO_2 <=> N_2O_4}$

If we calculate the enthalpy of the reaction through bond energies, we get the negative of the $\ce{N-N}$ bond energy. Why? Because we only formed one bond in this entire reaction. Every other bond stays the same - we start with four resonance stabilized $\ce{N-O}$ bonds and end with four resonance stabilized $\ce{N-O}$ bonds. The net change in bonds is the formation of a $\ce{N-N}$ bond. Here is a picture of the process (note that this picture doesn't show any resonance stabilization but there is; the "pi" bonding electrons are delocalized despite what the picture implies).

Bond energy is a positive value; the negative of this value implies that the dimerization reaction is exothermic - which is obviously true! We are going from a radical species - nitrogen dioxide - into a species with no unpaired electrons. Radicals are reactive!

We can also calculate this using heats of formation for both nitrogen tetraoxide and nitrogen dioxide and we'd get the same enthalpy of reaction.


Ultimately those structures are wrong.  Most of the community of chemistry educators teaches that formal charge reduction creates hypervalency.  All modern research and models shows this not to be the case.  You can't hybridize a d orbital and even get the proper tetrahedral geometry here.  Reducing formal charges to 0 makes no real world sense, when sulfur has a measured empirical charge of around 1.7 in sulfate.

Sulfate is 4 single bonds.  The bond length difference is due to the fact that the sulfur carries extra charge, as does the oxygens.  This ionic like bond is of course going to be shorter than more neutral forms of S-O.  Further, why would two double and two single hybrid bonds average to be the same length as a double bond?

Essentially any general chemistry solution that invokes d-orbitals as it's solution, outside of transition metal chemistry, is wrong.  Sulfur hexafluoride is not in fact 6 single bonds.

Unfortunately, biochemists are not going to update their picture of phosphate anytime in the next fifty years because no one cares.

The fact that basically every science major in the world learns this wrong is abhorrent.


Not all reactions can or should be represented with a double arrow.  The general-audience example is that you "can't unbake a cake".  You are right that there are instances that we assume that k2 is negligible (relative to k1) and therefore a reaction is treated as if it is reversible.

Here's a sample plot of how the concentrations of A for the irreversible and reversible reactions compare for k1 = k2 = 1.

No surprises here; the reversible reaction results in an equilibrium concentration of A. We can compare the concentration of A in each of the reactions as a function of kr:

Here I've plotted Log[k2] on the x-axis and the ratio of A in the irreversible reaction to the reversible reaction.  The different lines are at various times (in seconds).  I've kept k1 = 1 and the initial concentration 1 as well.  The two equations predict the same concentration of A in the limit of very small k2, and by very small I suspect this means at least a factor of 10^4 or more (not shown on the graph).

While you are correct that the integrated rate laws get much more complicated for reversible reactions, the one you are considering is still solvable in closed form.


Predicting Ka/Kb values using values for solubility in water

Your example, $\ce {HCl}$, is considered to be a strong acid. Thus, it it dissociate $100\%$ in water. Thus, the molarity of the given solution (if it exists):

$$\frac{\pu{720 g}}{\pu{1.0 L}}\times\frac{\pu{1.0 mol}}{\pu{36.5 g}} = \pu{19.7 mol L^-1}$$
That means, since $\ce {HCl (aq) -> H+(aq) + Cl-(aq)}$, $[\ce {H+}]$ of the solution is $19.7$, and $[\ce {Cl-}]=19.7$ as well so you can find $\mathrm{pH}$ of the solution. But, since no $\ce {HCl}$ remains in the solution, $K_\mathrm{a}$ has no meaning.


Equations for reactions dealing with coordination compounds

I found the answer which I'm assuming was just getting rid of an unnecessary spectator ion:

SCN-(aq) + [Fe(H2O)6]3+(aq) -> [Fe(H2O)5(SCN)]2+(aq) + H2O(l)


Trisulfur does exist, but is not circular, it is ozone-like and unstable at common temperatures.

My guess is that while angle strain does exist in ring form, it is not that harsh. Unlike carbon, having natural bond angle ~109 degree, for bivalent sulfur it is closer to 90 degree (sulfur exhibits little to no hybridization).  However, the S-S bond strength is only 40% of C-C bond and in trisulfur ring  all S-S bonds would have dihedral angle equal to zero, which is unfavorable because of nonbonding pairs interaction. This issue is a lot more severe than in tetraphosphorus, and leads to ring opening.

For further information on sulfur allotropes see the Wikipedia article.


Degree of dissociation of weak electrolyte at infinite dilution

There are a number of ways to explain this. In words, the entropic gain with dissociation overwhelms any attraction between counterions as the electrolyte is diluted.

Mathematically: for a m:n electrolyte undergoing dissociation according to the reaction

$$\ce{A_mB_n -> mA^{\nu_A -} + nB^{\nu_B +}}$$

if the initial concentration of electrolyte $\ce{A_mB_n}$ is $c_{T}$, define the degree of dissociation as

$$\alpha = \frac{c_{AB}}{c_{T}}$$

where $c_{AB}$ is the concentration of dissociated $\ce{A_mB_n}$. Assuming ideality and unit molarity as standard concentration, the equation for the equilibrium constant can be written as $$K=\frac{c_A^m c_B^n}{c_{AB}}$$ and solved in terms of the degree of dissociation. For a 1:1 electrolyte this leads to

$$\alpha = \frac{\beta}{2}\left[\left(1 + \frac{4}{\beta} \right)^{1/2} -1\right] \tag{1}$$

where $\beta = K/c_T$. As $c_T \rightarrow 0$ it follows from this last expression that $\alpha  \rightarrow 1$ #QED

PS The following equation (from Wikipedia's page on Ostwald's law of dilution) $$\alpha=\sqrt{\frac{K}{c_T}} \tag{2}$$ does not hold over the entire concentration range $c_T$. It can be obtained from Eq. (1) by assuming $K<<1$ and therefore $\beta<<1$. For increasingly dilute solution Eq. (2) fails and  $\alpha  \rightarrow 1$.


Hydration number: Carbonic acid in water

You may wish to refer to the publication of Saykally et al., The hydration structure of aqueous carbonic acid from X-ray
absorption spectroscopy

Basically, to determine hydration number, X-ray spectroscopy was used. This relates to the short existence of carbonic acid as, since it is so short lived, it is necessary to analyze it in such a time frame. As such, this was "the first measurement of X-ray absorption spectra for aqueous carbonic acid, enabled by the use of rapid-flow liquid microjet technology" which allowed it to be captured.

Then, by analyzing the spectra, it is possible to calculate the approximate hydration number, as "the increasing H-bond strength is proportional to the hydration
number (measured for distances up to $2.5\mathrm{\overset{\circ}A}$): $\ce{Na2CO3}$ (5.55), $\ce{HCO3}$
(4.26), $\ce{H2CO3}$ (3.17)".


How to calculate the pH of a buffered solution with Henderson Hasselbalch?

The main equilibrium in such a solution is the following:

$\ce{CH3COOH +CH3COO^- <=>CH3COOH + CH3COO^-}$.

This should be your ICE chart's equation too. I suspect that the equation you have for your ICE table is the following:

$\ce{CH3COOH +H_2O <=>CH3COO^- + H_3^+O}$

You can certainly solve the second equation with an ICE table by plugging in the initial concentration of the conjugate base, $\ce{CH3COO^-}$.

  However, doesn't this assume that the system is already at equilibrium
  and that the concentrations of $\ce{[CH3COO-]}$ and $\ce{[CH3COOH]}$
  are the concentrations at equilibrium?

Yes. The concentrations (or more accurately, activities) used in the H-H equation are equilibrium concentrations.

In the case of a 0.5 M acetic acid/acetate ion buffer, we can, for most purposes, conclude that the initial molarity of the main system components - acetic acid and acetate ion will remain constant. This is because the main acid-base reaction in the system is the following:

$\ce{CH3COOH +CH3COO^- <=>CH3COOH + CH3COO^-}$.

As we can see, the above equation is symmetric. There is no net change in system composition resulting from the above equation and the given molarities.

We also have these equilibria:

$\ce{CH3COOH + H2O- <=>H_3^+O + CH3COO^-}$
$\ce{H2O + CH3COO^- <=>CH3COOH + HO^-}$

But both acetic acid and its conjugate base, acetate ion, are weak. So these won't do much to change the system composition. Plus these aren't the favored reaction in the system; that would be the one above (the one between the strongest acid and strongest base in the system).

If you wanted to solve this system analytically - i.e. you want an even more accurate measure of the system's pH, then you would have to employ the law of conservation of mass and invoke the concept of solution electro neutrality.

Mass conservation:

$1.0 M$ $\ce{=[CH3COOH] + [CH3COO^{-}]}$

Solution electroneutrality:

$\ce{[HO^-{}] + [CH3COO^{-}] = [H_3^+O]}$.

These two statements, coupled with $\ce{K_a(CH3COOH)}$ and $\ce{K_b(CH3COO^{-})}$, and $\ce{K_w}$, give you a system of equations in which you can solve for $\ce{[H_3^+O]}$.


What would a diagram of the metal lattice structure of CuSO4 look like?

From the WebElements.com page on copper sulphate, the metal lattice structure looks like:

The yellow pyramids are cooper and the other structures are sulphates.

An alternate view comes from the Wikipedia page of copper (II) sulphate:

The pink spheres represent copper, yellow are sulphur and red are oxygen.

Below is a 2D representation of copper sulphate, from the Royal Society of Chemistry

A similar question with some more insight is found in How do copper sulphate and the like change crystal structure on adding water of hydration?


Why is there a lone pair in thionyl fluoride?

When you want to write a Lewis structure, I suggest you start from considering how many valence electrons each atom has. In your case, you would have:

S = 6
F = 7; total = 14
O = 6

The final total would be 14 + 6 + 6 = 26. The structure you guessed for is correct, sulfur is the central atom. You start by drawing a single bond per each atom, getting something like this:

F
|
S–O
|
F

Then, you can draw a double bond for oxygen:

F
|
S=O
|
F

If you do the math, by counting the electrons that you put in the previous structure, you would have 2 + 2 + 4 (two S-F bonds and a S=O) = 8.
Then we will add in the lone pairs: three on each halogen, two for oxygen, and one for sulfur. Therefore we add 3 × 2 × 2 = 12 electrons (three lone pairs on each fluorine) plus 4 electrons on O (the two lone pairs), and we have a total of 24. The final 2 , adding up to 26, come from the lone pair on sulfur.

This makes sense because sulfur is in period 3, so it is possible for it to have more than 8 electrons: in other words, the octet rule applies not only for 8 electrons, if that makes sense.


Products formed on cleavage of epoxides using HI

Cleavage of epoxides can happen in 2 ways, resulting from the cleavage with the 2 carbons to which the oxygen atom is bonded.
If the medium is acidic, it will follow the SN1 pathway, proceeding via the most stable carbocation. This makes sense, since in acidic medium, the alcohol can get protonated and break away from one of the carbon atoms.
If the medium is basic, it will proceed via the SN2 pathway.
In other words, the type of cleavage doesn’t depend on the concentration of $\ce{HI}$, the only thing that matters is the acidity or basicity of the medium.


How do I go about balancing these equations?

I agree with you, the right equilibration of the redox reaction is the following:
$$\ce{4 Co^2+(aq) + 2H2SO3(aq) + 2H^+ (aq)-> 4 Co^3+(aq) + S2O3^2- (aq) + 3H2O(l)}$$

The problem is  in your assumptions about the oxidization number change at the beginning, and more precisely about sulfur.
The oxidation number of  sulfur in $\ce{H2SO3}$ is four. But in thiosulfate ion, you have two inequivalent sulfur atoms:
Please see:
 Oxidation states of the sulfur atoms in the thiosulfate ion
The oxidation number is $\ce{+IV}$ for the central one, while it's zero for the terminal one.

In sum, the variation in the oxidation number for sulfure is four, not two.
(As if you have two sulfur atoms with oxidation number of four in $\ce{2H2SO3}$. One of them changes its oxidation number to zero. while the other one stays intact).


Which is the least stable trihalide of nitrogen?

Nitrogen triiodide $\ce{NI3}$ is an extremely sensitive explosive that explodes with the slightest touch when dry.

Nitrogen trifluoride $\ce{NF3}$ is so stable that it is a greenhouse gas global warming concern, with lifetime in the atmosphere of hundreds of years.

$\ce{NI3}$ is unstable due to steric strain as you are saying.


Steric Effects in Friedel-Crafts Reactions

Based on literature, I'd say the unsuccessful Friedel-Crafts reaction between benzoyl chloride and mesitylene is not due to steric reasoning. My argument is based on at least one report of Friedel-Crafts acylation on following polymethylated aromatic compounds (Ref.1):

The mono- and di-acylation have been taken place between 1,3-substitutions in each occasion as indicated in the abstract:

Diacylation of mesitylene, durene and isodurene by the Friedel–Crafts method, using the acylating agent in excess (as both solvent and reagent), gives derivatives which are of potential use as monomers. The bis(p-fluorobenzoyl) derivatives, for example, can function as the electrophilic components in poly(ether ether ketone)(PEEK)-type Polymerisations, and the bisnitrobenzoyl compounds are potential sources of novel diamines. Tribenzoylation [although not tris(p-fluorobenzoylation)] of mesitylene has been observed, but dibenzoylation of m-xylene is a much less efficient process than previously reported.

The reasoning for the resistance to dibenzoylation on m-xylene must be rationalized due to the deactivation of in-plane carbonyl $\pi$-electrons of benzoyl group on aromatic ring after first benzoylation (highly possible 1-benzoyl-2,4-dimethylbenzene). All other instances (2-benzoyl-1,3-dimethyl substitutions), the benzoyl carbonyl group is in out of plane with the corresponding aromatic nucleus (Ref.2), probably due to steric hindrance:
$$
\begin{array}{c|ccc}
\text{Compound} & \nu_\ce{C=O} \ (\ce{CH3C=O}) & \nu_\ce{C=O} \ (\ce{C6H5C=O}) & \text{ECA} \ (c.a.) \\
\hline
\text{Acetophenone}  & \pu{1688 cm-1} & - & -\\
\text{Benzophenone}  & - & \pu{1664 cm-1} & 42^\circ\\
\text{2,4,6-trimethylacetylbenzene}  & \pu{1703 cm-1} & - & 73^\circ\\
\text{2,4,6-trimethylbenzylbenzene}  & -& \pu{1672 cm-1}  & 90^\circ, 40^\circ\\
\text{2,4,6-trimethyl-1,3-diacetylbenzene}  & \pu{1705 cm-1} & - & -\\
\hline
\end{array}
$$
$^a$ ECA: Effective Conformational Angle (whenever two angles are given, first refer to the angle with mesitylene ring).
A comprehensive study of mono- and di-acylation of mesitylene have been done and in all cases, at least mono-acylation was achieved in $\gt 90\%$ whenever anticipated Ref:2). The abstract states:

The Friedel–Crafts acylation of mesitylene has been studied in detail. In the acetylation, propionylation or benzoylation reactions mono- or di-ketones may be formed. Mesitoylation leads only to the monoketone. A two-stage acylation procedure is described for the synthesis of acetylbenzoylmesitylene and acetylpropionylmesitylene. Reversibility has been shown to be a factor in these acylations, especially where an aliphatic acyl group is involved. The relationship between the stereochemistry of mesityl ketones and their u.v., i.r., and $\ce{^1H}$ n.m.r. spectra is discussed.

Overall, these results suggest that Friedel-Crafts acylation reaction would not effected by steric hindrence, at least created by 1,3-dimethyl substitutions. For example, 2,4,6-trimethylacetylbenzene (acetylmesitylene) was achieved in 92% yield; 2,4,6-trimethylbenzoylbenzene (benzoylmesitylene) in 83% yield; 2,4,6-trimethyl-1,3-diacetylylbenzene (diacetylmesitylene) in 95% yield; and 2,4,6-trimethyl-1,3-dibenzoylbenzene (dibenzoylmesitylene) in 89% yield (Ref.2).
It is also noteworthy that the nitration of mesitylene is not effected by steric effects as well. You can find more details on this aspect in here.

References:

Brian N. Hendy,   Kevin H. Patterson,   David M. Smith,   Susan E. Gardner, Neil J. Nicolson, “Polyalkylated aromatic monomers and polymers. Part 1.—Friedel–Crafts diacylation of polymethylbenzenes as a means of monomer synthesis,” J. Mater. Chem. 1995, 5(2), 199-204 (https://doi.org/10.1039/JM9950500199).
P. H. Gore, J. A. Hoskins, “Friedel–Crafts acylations of aromatic hydrocarbons. Part VIII. Mono- and di-acylation of mesitylene,” J. Chem. Soc. C 1970, (4), 517-522 (https://doi.org/10.1039/J39700000517).


Finding concentrations in a voltaic cell

I think you were on the right track!

You were right that you need the Nernst equation, which must be how you determined the initial $E_\text{cell}$.

$$E_\text{cell}=E_\text{cell}^\circ-\frac{RT}{nF}\ln Q$$

You correctly identified that your $\ce{Co|Co^2+||Ni|Ni^2+}$ cell has the cell standard potential

$$E_\text{cell}=E_\text{cathode}^\circ-E_\text{anode}^\circ=-0.25\ \mathrm V--0.28\ \mathrm V=0.03\ \mathrm V$$

So the question is essentially asking what concentrations are needed for your cell to have the standard potential. Again looking at the Nernst equation, you can solve for the reaction quotient $Q$ when $E_\text{cell}=E_\text{cell}^\circ$

$$E_\text{cell}=E_\text{cell}^\circ-\frac{RT}{nF}\ln Q$$

$$E_\text{cell}-E_\text{cell}^\circ=0=\frac{RT}{nF}\ln Q$$

$$\ln Q=0$$

$$Q=1$$

At this point, you should be able to solve for the concentration of $\ce{Ni^2+}$ with simple algebra.


Intermolecular interaction Van Der Waals

Intermolecular attraction can reduce the volume, in the same way that increasing the pressure can lower the volume (Boyle's law).
One way to see the effect is to take the derivative of the volume, as given by the van der Waals (vdW) equation, with respect to the attractive parameter $a.$
The vdW equation is
$$\left(p+\frac{a}{\bar{V}^2}\right)(\bar{V}-b)=RT\tag{1}$$
Ignore for simplicity the excluded volume term $b,$ so that
$$p\bar{V}=RT-\frac{a}{\bar{V}}\tag{2}$$
Then the derivative of the volume wrt $a$ is
$$\left(\frac{\partial V}{\partial a} \right)_{p,T} = \frac{\bar{V}}{a-p\bar{V}^2}=-(RT)^{-1}\tag{3}$$
In essence what this result says is that the volume decreases as you increase the parameter $a.$


How does the force constant change when a heavier atom is used in vibrational spectroscopy?

In the harmonic oscillator approximation, the wavenumber of vibration for a diatomic molecule is
$$\tilde{\nu} = \frac{1}{2\pi c}\sqrt{\frac{k}{\mu}}$$
Here, $k$ is the force constant of your bond, and $\mu$ is the reduced mass for the two atoms.
$$\mu=\frac{m_1m_2}{m_1+m_2}$$
When you change isotopes, only $\mu$ changes. The force constant depends on the strength of the bond, which depends on the chemical nature of the elements, and can be assumed to stay constant regardless of changing isotopes.
You have to calculate the $\mu$ for the second problem, and use that with the $k$ you got from the first problem to calculate the wavenumber of the fundamental band.


Why are different colors, according to sugar concentration observed in Benedict's test?

Benedict's Reagent is just simply $\ce{Cu^2+}$ ions in the form of copper citrate. When Benedict's Reagent finds an aldose (a sugar with an aldehyde group), it can oxidize the aldose to a carboxylic acid. For example:

Since D-Glucose is oxidized, Cu is reduced to a red precipitate ($\ce{Cu2O}$). The colors range form green to red because the original Copper Citrate ($\ce{C6H8Cu2O7^4+}$) is blue in color.

Source: Quora


How is the acetoxy group ortho/para-directing in electrophilic aromatic substitution?

Phenoxide is more reactive than phenol towards electrophilic substitution. However, the number of lone pairs isn't the factor that controls it, because it is not possible for both lone pairs to be simultaneously in conjugation with the benzene ring. If one lone pair is in conjugation with the benzene ring, the other lone pairs will not be oriented correctly to overlap with the pi system of the ring. So, you can have one, two, or three lone pairs - it does not matter because any lone pairs beyond the first are useless.

Only the availability of the lone pairs matters. That has to be discussed with inductive or resonance effects. With an $\ce{-OH}$ group you have a hydrogen atom fighting for some of the electrons (inductive effect), and with a $\ce{-O^-}$ group there is nothing else.

Moving on to the acetoxy group $\ce{-OC(O)CH3}$, the lone pair on oxygen is delocalised into the electron-withdrawing carbonyl group, via resonance. You can think of it as the $\ce{C=O}$ and the benzene ring both fighting over the oxygen lone pair, which makes it less available for donation to the pi system. Consequently, the acetoxy group is still o/p-directing (because of the presence of at least one lone pair), but is not as strongly activating as the hydroxy group (because the lone pair is less available).


Calculating the standard reduction potential for the oxidation of water

According to Hess’ law, we expect the Gibbs energy change $\Delta G$ of the two partial reactions to add up to the total Gibbs energy change:

$$\Delta G_1+\Delta G_2=\Delta G_3$$

For an electrochemical reaction, the change in Gibbs energy corresponds to an electrochemical potential $E$ according to

$$\Delta G = -zFE$$

where
$F$ is the Faraday constant and
$z$ is the number of electrons transferred.

Since for the two partial reactions the number of electrons is $z_1=z_2=2$, but for the total reaction $z_3=4$, we expect:

$$\begin{align}\Delta G_1+\Delta G_2&=\Delta G_3\\
z_1E_1+z_2E_2&=z_3E_3\\
2\times0.695\ \mathrm V+2\times1.763\ \mathrm V&=4\times1.229\ \mathrm V
\end{align}$$


To see these jelly flakes in an ultrasound bath is a normal thing -- referring to ``it happens regardless if the water was tap-water or deionized water, a just a matter of time''.  Perhaps this occurs faster in the former, than in the later, especially if the bath is used to clean glassware.  But it equally happens with glassware judged as clean (by eye) for example if the bath is deployed to initiate / support a reaction inside a reaction flask.  Ultrasound-assisted generation of Grignard reagents is one example.

The ultrasound bath works by sonication, and the cavitation simply detaches the oily/fatty/grease surface film covering quickly all what was touched once by naked hand (e.g. sweat), and breaks this into even smaller pieces.  Over time -- when the ultrasound bath is switched off --, this thermodynamically unstable emulsion decomposes, splitting into a water soluble phase, and a hydrophobic phase that then tends to agglomerate.  (I did not check the presence of organisms living in or on them since discharging the bath is a chore regularly done, too.)

Side note:  For me, these residues appear to be more a liquid, then a solid -- hence I used the term emulsion.  If they were a solid, the thermodynamically unstable mixture would be an suspension.


Color shifting reaction in black sealcoat

It's not a chemical reaction, AFAIK, but due to the Tyndall effect or Rayleigh scattering from the fine particles of the asphalt-water emulsion as water evaporates and the size and/or dispersion of the asphalt particles change.

It is also possible that interference of light in an adhered water layer plays some part. This is the same effect that gives soap bubble their rainbow colors, which change as the bubbles dry -- note that just before it pops from drying, one area of a bubble  may go colorless due to thinning of the film beyond the range of visible light interference.


Sketching lattice points, primitive axes, primitive cell, and basis of atoms

I think the cell after EDIT is correct. Here I added a few neighboring cells:

Admittedly, the atom in the right bottom corner is missing. Apparently the author didn't consider it important.


Probably the biggest drivers behind using methane as a fuel is that it is abundant in natural gas and is (currently) mostly useless as a chemical feedstock.
Ethane makes up a few percent of natural gas and can also be obtained as byproducts of petroleum refining, but the big difference from methane is that ethane is extremely useful in chemical synthesis (mostly to make polyethylene).
In fact, household natural gas often contains a bit of ethane, which may vary depending on the current demand for ethane as a feedstock.
In short: it burns just as well, but ethane has other important uses.

Propane has some use as a feedstock, but it is an attractive fuel because it is easily stored as a liquid without requiring huge pressures.


Na2CO3 + FeBr3 + H2O: why do we get Fe(OH)3?

You have certainly already recognized $\ce{Na2CO3}$ as the base in your equation.

Did you also consider the following:
When $\ce{FeBr3}$ is dissolved in water, the hexaaqua complex $\ce{[Fe(H2O)6]^3+}$ is formed. This complex is fairly acidic:
$$\ce{[Fe(H2O)6]^3+ + H2O<=> [Fe(H2O)5OH]^2+ + H3O+}$$

Ignoring ligand exchange processes, you might state that the $\ce{OH}$ groups come from the deprotonation of the water ligands of your initial complex.


Which would have a more intense transition: a low spin d6 complex or a high spin d5 complex?

You see that both cases are Laporte-forbidden, now how about spin-forbidden?

Assuming the complex vibrates so it's no longer exactly octahedral, can you make a spin-allowed transition for $\ce{[Fe(CN)_6]^{4-}}$?  How about $\ce{[Fe(H_2O)_6]^{3+}}$?  When you answer this you will answer your question.

   You should find that $\ce{[Fe(H_2O)_6]^{3+}}$, with all orbitals occupied by electrons having the same spin, has only spin-forbidden d-d transitions.  Therefore it will undergo weaker transitions than $\ce{[Fe(CN)_6]^{4-}}$ which has some spin-allowed transitions.


How can cooling gases do not change their physical states?

Well since the lowest temperature that can exist is -273.15 C.

All gases condense to liquid or solids almost all to solids because at this temperature molecular activity ceases.

And the title of your question seems contradicting with what you have written. Probably you might have mistaken interpreting "before".

also only ideal gas can remain gas at this temperature as they have no force of attraction.
http://en.wikipedia.org/wiki/Absolute_zero


The fundamental question is: which is more acidic, 2-pyrrolidone or 2-pyrrolidinethione?

If you took a 50-50 mixture of the two and added enough alkali to neutralize half of the mix, what would be the composition of the solution? The mismatch in p orbitals between C and S leads to more C=N bonding, rather than C=S, with a charge separation, leading to more negative charge on sulfur in Y and more positive charge on N, so this N will be more acidic than the N in W. The charge separation by oxygen, leading to more double bond character between positions 1 and 2 in 2-pyrrolidone is an effect also, but apparently not as large as the p-p mismatch between C and S.

The Lewis diagrams don't show clearly enough the actual charge distribution in the anions.

And I thank Jan for pointing out that I completely flipped my actual explanation when I thought I understood what two effects were happening. Although this is not a trick question, I think it needs more context to be able to be of predictive value. Maybe just a way to get you thinking: what could be going on here, and where else might it explain.


You are technically correct in saying that radioactive sugar is another species in that it is possible to distinguish it from 'normal' sugar, although chemically identical. But think about it this away. Some of the added radioactive sugar solution (which could be micromolar or less) finds its way into the solid sugar at the bottom of the flask, but some remains in solution so that (dynamic) equilibrium is reached with respect to the radioactive sugar. You could do it the other way round, add solid radioactive sugar and it will appear in the solution.  What is so special about the radioactive sugar that it can do this and 'normal' sugar cannot? Well, there is nothing special so it is entirely reasonable to suppose that the normal sugar does this also. By inference, other molecules also do this as we know of no experimental reason why sugar is peculiar among all the other types of molecules known.

If you are still not convinced an experiment could be done with a coloured salt such as copper sulfate, so only one species. Make a saturated solution in which there will be solid in equilibrium with water, which will be blue due to the copper ions. Next, filter off the water and retain the solid and (gently) add more water, initially this will be clear but becomes increasingly blue as the sulfate dissolves. You will see equilibrium being established again as the sulfate dissolves, and provided there is some solid left equilibrium solid to solution will be achieved.

There is an experimental technique called temperature jump that is used specifically to study the time it takes to 'relax' back to equilibria and so obtain rate constants for the process. In this method, a short duration laser pulse is used to rapidly heat a solution (typically by $\ll 1$ C) and change its equilibrium a little. The return to equilibrium is followed by optical/spectroscopic methods so nothing has to be taken from the solution. Many reactions have been studies this way e.g. $\ce{H3O+ + OH^- <=> 2H2O}$.


From here

Back in 1960's Dr.Dayhoff was shortening the names of amino acids so that ancient computers could save computing time.

  By the time Dr. Dayhoff got to lysine, there were not too many letters left, so she used the letter K, explaining that K is at least near L in the alphabet.

As for Glutamine.

  For the remaining 5 amino acids, Dr. Dayhoff was reaching somewhat to find an easy-to-remember connection between the single letter and the amino acid.  She assigned aspartic acid, asparagine, glutamic acid and glutamine the letters D, N, E and Q, respectively, noting that D and N are nearer the beginning of the alphabet than E and Q, and that Asp is smaller than Glu, while Asn is smaller than Gln.


From Supniewski, J. "Inorganic Preparative Chemistry", 1958 (in Polish):

  Melt $\pu{36.4 g}$ of vanadium pentoxide $\ce{(V2O5)}$ and $\pu{63.6 g}$ of sodium carbonate $\ce{(Na2CO3)}$ a nickel crucible until you get a uniform, clear melt with no bubbles of $\ce{CO2}$ coming out. Pour the melt on a cold plate and wait until it solidifies to a colorless mass (during the cooling process it should turn dark green, then yellow, then colorless). Pulverize the melt and dissolve it in a minimum amount of cold water, filter, and layer with several volumes of ethyl alcohol. After several hours, colorless crystals of $\ce{Na3VO4 * 16 H2O}$ should separate. Wash the product with ethanol and dry in a desiccator.

Some clarifications as requested in the comments:

"Minimum amount" means that you need to add only so much water to dissolve the salt completely, no more. Just add water in small portions and stir well - if the salt is still visible add some more water, etc. Suppose it took you about $\pu{50 mL}$ of water to dissolve your substance. Now you need to add twice or thrice (i.e., $\pu{100 .. 150 mL}$) the amount of ethanol ("several volumes"). Add the alcohol slowly so the solutions do not mix immediately but form two separate layers, and leave it undisturbed for several hours (I think it may take a day or two).


Phase abbreviations for non-aqueous solutions

According to IUPAC, general notation for any solution would be $\ce{X(sln)}$ [1, p. 54]:

(vi) States of aggregation
The following one-, two- or three-letter symbols are used to represent the states of aggregation of chemical species [...] The letters are appended to the formula symbol in parentheses, and should be printed in Roman (upright) type without a full stop (period).
$$
\begin{array}{ll}
&\ldots\\
&\text{s} &\text{solid}\\
&\text{sln} &\text{solution}\\
&\text{vit} &\text{vitreous substance}\\
&\ldots\\
\end{array}
$$

Similarly, when denoting relation of a given quantity/symbol, subscript $\mathrm{sln}$ notation is used, e.g. $Δ_\mathrm{sln}G^\circ(T,p^\circ)$ [2, pp. 241, 276]:

4.2 Subscripts and superscripts
4.2.1 Subscripts
$$
\begin{array}{ll}
&\ldots\\
&\text{r} &\text{reference, reduced}\\
&\text{sln} &\text{solution}\\
&\text{s} &\text{saline solution}\\
&\ldots\\
\end{array}
$$

References

IUPAC “Green Book” Quantities, Units, and Symbols in Physical Chemistry, 3rd ed.; Cohen, R. E., Mills, I., Eds.; IUPAC Recommendations; RSC Pub: Cambridge, UK, 2007. ISBN 978-0-85404-433-7. (PDF)
Gamsjäger, H.; Lorimer, J. W.; Scharlin, P.; Shaw, D. G. Glossary of Terms Related to Solubility (IUPAC Recommendations 2008). Pure and Applied Chemistry 2008, 80 (2), 233–276. DOI: 10.1351/pac200880020233. (Free Access)


Why do we write electrode potentials next to chemical reactions?

Your question is a valid question, and ignore downvotes. They don't mean anything.

Your understanding is very good and that you realized that the electrode potential is a property of the electrode and it really does not care how the reaction is written. However, a equation is $needed$ to keep track of the electrons lost or gained in the Nernst equation. So as a tradition and a matter of convenience, electrode potentials are quoted along with a balanced half-cell.

Another very important thing to remember (even some university teachers and PhDs lack this understanding), that the sign of the electrode potential in invariant. It does not care how the reaction is written. It is the electrostatic sign of the electrode. This part takes care of the European and American conventions. American convention is due to Gibbs and Latimer and no longer used. Most international electrochemists have agreed to stick to the European convention i.e., write all half reactions as reduction.


Is there a practical re-use value gained by grinding up activated carbon that is spent/saturated?

Activated carbon is called "activated" when its grains contain a large amount of large cavities. These cavities are big enough to hold large molecules of undesired substances (oils, smokes) which are present in the air. Only small molecules like oxygen and nitrogen from the air can go through without being retained. So the activated carbon removes pollution from the air. After a long time, there are more and more cavities occupied by these unwanted substances. As a consequence, the effectiveness of the activated carbon slowly decreases. It has to be replaced. But it is no use grinding it. If one wants to reuse this carbon again, one should heat it up to such a high temperature that the substances fixed in the cavities are vaporized. Unfortunately some substances like soot will never get vaporized. So the effectivity of the activated carbon will never be entirely restored


The products of a chain reaction beginning with acetone and sodium amide

Acetone and sodamide give acetone sodium enolate A, nothing further happens until acetylene is introduced. The pKas of acetone and acetylene are both around 25 so an exchange occurs reforming acetone and creating sodium acetylide. This seems an unusual way of making sodium acetylide, but I don't think this is a real world example (unless someone can find a reference).

Sodium acetylide adds to acetone giving the sodium alkoxide of tertiary alcohol B

Work up with dilute acid gives the tertiary alcohol C.

Hydrogenation over Pd catalyst reduces the triple bond to a double bond giving D.

Passing D over alumina at 400C dehydrates the tertiary alcohol to give isoprene.


How is fluorine gas contained and transported, why aren't the metal cylinder and tube in this video attacked?

I think it is mentioned somewhat around 4:35 or so in the link you added. The metal cylinder that has to store elemental fluorine will have to be made from a special metal like nickel or monel. Those form metal fluoride layers on the inside but they remain on the metal and allow no further fluorine to attack the rest of the metal which is called passivation.

About the plastic tube, it could be Teflon but I don't know if that works or not. Basically, the problem with fluorine especially about demonstrations was always the container. You could have easily done experiments in metal containers but then no one would be able to see through. Then there were, of course, fluorinated plastics but they tend to be quite rough and it all becomes blurred. Which is probably also the reason why there is no real fluorine ampoule for element collectors. That was at least how I understood it until quite recently when I was listening to a talk by Florian Kraus, who is a fluorine chemist. And they are using a special polymer that can hold fluorine and other highly fluorinating interhalogens but is much more transparent. So I guess by now there is quite a variety of polymers available that will hold fluorine.

Elemental fluorine could be stored in glass at least theoretically. It is not the fluorine that attacks the glass but hydrogen fluoride or I think even more precise something like $\ce{H2F^+}$. You could avoid that to some degree by drying the fluorine but the problem is the glass itself. As glass forms a polymer structure of $\ce{[SiO4]^4-}$-tetrahedra that join all vertices there has to be a sudden end once there is no more repeating unit added. But that would mean that the outside of a glass surface is all negatively charged (ignoring metal cation additives). Therefore there are many $\ce{Si-OH}$ groups on the outside which tend to be acidic. When fluorine reacts with them it will form hydrogen fluoride that will attack the glass.


Why do stabilised ylids lead to trans alkenes in the Wittig reaction?

TL;DR: With stabilised ylids, the trans oxaphosphetane is formed, under kinetic control. This is due to minimisation of dipole-dipole repulsions in the transition state for oxaphosphetane formation:

There is no equilibrium established between cis and trans oxaphosphetanes. The formation of the trans oxaphosphetane is irreversible, and its collapse leads to the formation of the (E)-alkene.

Long answer

Part 1: Oxaphosphetane reversibility?

The mechanism of the Wittig reaction has historically been controversial. Pioneering research, most notably by Edwin Vedejs and his group, has led to our current understanding of the mechanism as being a concerted [2+2] cycloaddition to form the oxaphosphetane directly. For more information, see: Which is the currently accepted mechanism of a Wittig reaction?

Within this framework, the formation of (Z)-alkenes from unstabilised ylids is understood to be because of a minimisation of steric repulsion in the cycloaddition transition state, as was proposed by Vedejs in 1988.1 For more information, see: Why does the unstabilised Wittig reaction selectively form cis alkenes?

For a long time, it was believed that oxaphosphetane formation from stabilised ylids was reversible. Consequently, the trans oxaphosphetane was preferentially formed (due to minimisation of steric repulsions), under thermodynamic control. The collapse of this then led to (E)-alkene formation.

However, this has been proven to be all but incorrect (and is sadly still propagated on widely used websites such as Wikipedia). A major problem with this theory was that oxaphosphetanes formed from stabilised ylids decompose so fast that they are not detectable via NMR spectroscopy.2 It is therefore implausible that the reversal of oxaphosphetane to starting materials can occur before forward reaction to form products.

The nail in the coffin was provided by Vedejs.2 If oxaphosphetane formation (from stabilised ylids) is indeed reversible, then it would mean that if the cis oxaphosphetane is synthesised, it should decompose to form the (E)-alkene via isomerisation to the trans oxaphosphetane. While the oxaphosphetanes themselves cannot be synthesised (they simply decompose to products), the corresponding betaines can be synthesised by deprotonation of β-hydroxyphosphonium salts (their conjugate acids).

The work by Vedejs proved that the syn betaine - which necessarily cyclises to form the cis oxaphosphetane - decomposed stereospecifically to the (Z)-alkene, hence disproving the idea of reversibility in the Wittig reaction. First, the relevant β-hydroxyphosphonium salts were synthesised:

Deprotonation of this formed a betaine, which would immediately cyclise to form a cis oxaphosphetane (this was well-established by then; see the question linked earlier on the mechanism).

If any of the (E)-alkene was formed, then it would mean that there was a mechanism by which either the betaines, or the oxaphosphetanes could interconvert. This would therefore not be conclusive evidence as to whether oxaphosphetane formation was reversible. However, if none of the (E)-alkene was formed, then it would necessarily mean that betaine and oxaphosphetane formation were both irreversible.

In their experiments, an E:Z ratio of > 98:2 was observed for all cases except for R = Ph. This indicated that in fact oxaphosphetane formation was not reversible, and that the stereoselectivity of the Wittig reaction cannot be based on a thermodynamic argument.

For R = Ph, 8% of the (E)-isomer of ethyl cinnamate was observed. Experiments with added 4-chlorobenzaldehyde ruled out the possibility of reversal to the ylid + carbonyl, as any ylid formed would preferentially react with the more electrophilic 4-chlorobenzaldehyde, leading to ethyl 4'-chlorocinnamate - but this was not seen. Furthermore, ethyl cinnamate was shown to be configurationally stable under the experimental conditions. It was finally found out that this occurred because, instead of deprotonating on oxygen, DBU deprotonated the carbon atom α to the ester, which led to loss of stereochemical integrity. This was further supported by deuterium labelling studies.

As a consequence of this, Vedejs concluded that oxaphosphetane formation was irreversible, and noted that for reactions with a stabilised ylid, the Hammond postulate implied a late transition state, which would "feel" stabilising or destabilising influences of the oxaphosphetane. Vedejs therefore proposed an almost-planar transition state, with the stereoselectivity rationalised by steric repulsions between the $\ce{CO2Me}$ and $\ce{R}$ substituents in the TS:

Part 2: Steric repulsions or dipole alignment?

In 2003, Varinder Aggarwal's research group managed to synthesise stabilised ylids with a $\ce{P(OMe)3}$ group instead of the usual $\ce{PPh3}$, and discovered that these ylids led to substantially lower (E)-selectivity.3 Vedejs' model could not be used to rationalise these observations:

To explain these, Aggarwal and Harvey put forth a different model, based on dipole-dipole interactions and supported by DFT studies:4

For the usual $\ce{PPh3}$ ylids, there is a strong dipole pointing from ylid carbon to ester group; in the trans TS, this dipole points opposite to the dipole of the C=O group in the carbonyl component. Hence, there is electrostatic attraction in, and stabilisation of, the trans TS. This explains the usual high (E)-selectivity.

However, for the new $\ce{P(OMe)3}$ ylids, the dipole moment of the ylid is much smaller. This is because the negative charge on the ylid carbon is delocalised into not only the ester group, but also the σ* orbital of the P–O bonds, which are energetically low-lying. This accounts for the lower (E)-selectivity of the phosphite-derived ylid.

As far as I know, this is the currently accepted model for why stabilised ylids lead to (E)-alkenes. Our mechanistic understanding of the Wittig reaction is very nicely summed up by Aggarwal and Harvey,5 in a grand computational study of all types of ylids. In it, they provided a full potential energy surface for the reaction of $\ce{Ph3P=CHCO2Me}$ with benzaldehyde (note that there is a typo in the ylid starting material in the diagram):

Ending with a quote from their paper, which perfectly summarises this answer:

  In the case of the triphenylphosphonium stabilized ylide, TS2 is very low in energy, showing that reversal cannot occur in these reactions, which is in agreement with experiment [...] E/Z selectivity is thereby determined solely by the relative energy of cis and trans TS1s. These energies, as well as the geometry of the respective TS1s, are strongly affected by a dipole-dipole interaction between the two reactants [...] The outcome for stabilized ylides is thus a high E selectivity.

References

Vedejs, E.; Marth, C.F. Mechanism of the Wittig reaction: the role of substituents at phosphorus. J. Am. Chem. Soc. 1988, 110 (12), 3948–3958. DOI: 10.1021/ja00220a037.
Vedejs, E.; Fleck, T. J. Kinetic (not equilibrium) factors are dominant in Wittig reactions of conjugated ylides. J. Am. Chem. Soc. 1989, 111 (15), 5861–5871. DOI: 10.1021/ja00197a055.
Aggarwal, V. K.; Fulton, J. R.; Sheldon, C. G.; de Vicente, J. Generation of Phosphoranes Derived from Phosphites. A New Class of Phosphorus Ylides Leading to High E Selectivity with Semi-stabilizing Groups in Wittig Olefinations. J. Am. Chem. Soc. 2003, 125 (20), 6034–6035. DOI: 10.1021/ja029573x.
Robiette, R.; Richardson, J.; Aggarwal, V. K.; Harvey, J. N. On the Origin of High E Selectivity in the Wittig Reaction of Stabilized Ylides: Importance of Dipole−Dipole Interactions. J. Am. Chem. Soc. 2005, 127 (39), 13468–13469. DOI: 10.1021/ja0539589.
Robiette, R.; Richardson, J.; Aggarwal, V. K.; Harvey, J. N. Reactivity and Selectivity in the Wittig Reaction: A Computational Study. J. Am. Chem. Soc. 2006, 128 (7), 2394–2409. DOI: 10.1021/ja056650q.


You can easily see it: "burn out" the filament of an incandescent lamp from a flashlight (torch) with over-voltage (e.g. put a 1.5 volt lamp across a 12 volt battery). The bluish flash you see is vaporizing tungsten (W), but to the unaided human eye, it looks little different from, say, the "spark" in xenon flash lamps.

Since the filament is so hot, it radiates heat rapidly and cools quickly, so is visible only as a flash. If you were to vaporize a few kg of W, it would be visible longer as a blindingly-bright object.

BTW, notice the film on the "dead" lamp: it is condensed W. If the lamp has a vacuum, the film is silvery, but collision with argon or other inert filler makes it rougher, or grayish.


Unusual oxidation numbers for alkaline earth metals?

This answer is an extension to @Ian Bush answer. Not only magnesium, but every group 2 element has a lower oxidation state of +0(beryllium) and +1(for other metals).

Be(0)

A 2016 paper gives an insight to the existence of a zero valent beryllium complex compound i.e it has been verified through computational method but it is yet to be synthesized.

  [...]The team added a single ligand i.e CAAC = cyclic
  (alkyl)(amino)carbene to $\ce{BeCl2}$ in benzene solution to give
  $\ce{(CAAC)BeCl2}$, and obtained the final product,
  $\ce{[Be(CAAC)2]}$, by reducing the beryllium with KC8 in the presence
  of a second equivalent of CAAC ligand.

Be(I)

  High resolution infrared emission spectra of beryllium monohydride and
  monodeuteride have been recorded. The molecules were generated in a
  furnace-discharge source, at 1500 °C and 333 mA discharge current,
  with beryllium metal and a mixture of helium and hydrogen or deuterium
  gases.(source)

Mg(I)

It has been described in @Ian Bush's answer. For more information, refer to these sources(1 and 2). Some examples are $\ce{Mg2RuH4, Mg3RuH3, and Mg4IrH5}$ having the $\ce{Mg-Mg}$ bond and magnesium diboride containing the metastable $\ce{Mg2^{2+}}$ ion(Credit @Oscar Lanzi).

Ca(I)

A Ca(I) Sandwich Complex $\ce{[(thf)3Ca(μ-C6H3-1,3,5-Ph3)Ca(thf)3]}$ is being described in this 2010 paper.

Sr(I)

  The high-resolution infrared spectrum of gas-phase $\ce{SrF}$ was
  obtained in emission with a Fourier transform spectrometer.[...]
  (Source)

Ba(I)

3 isotopes of barium in $\ce{BaF}$ (Source).  In [2018] Wu and Lerner (source) reported a barium(I) in a complex graphite intercalation compound.

I did not had to google each and every element to search for lower oxidation state. The wikipedia article of oxidation element gave a list of all posible O.S of element. Information regarding unusual O.S of any element can be found in the footnote and thus only clicked the relevant elements for the answer.


Why does an ideal capacitor give rise to a rectangular cyclic voltammogram (CV)?

The kind of analysis truly required is outside the scope of chemistry and would be best handled in the careful hands of an electrical engineer. I'll give a brief attempt that exploits an easy-to-use circuit simulator.

We have to first think of an equivalent circuit that we can use as a model to determine its behavior. I suggest the following:

where my variable is R1's value for its resistance, which I shall set to 0, 10 and 100 ohm. If we were to watch this happen in time then we would see the following:

Quickly converting these to Current vs Voltage and running two other simulations at different resistances and we get:

These results are due to setting up differential equations and solving them appropriately.

You can play with the circuit I made for yourself here.


How to write a reaction that you have never seen?

There is no general method for predicting the products of a chemical reaction, given only the starting materials and no other information.

You will learn certain trends, such as, for example, the increasing reactivity down the Group 1 metals with water, or dilute acids, but this often isn't enough to generalise to more complex situations.

You'll also learn about relative reactivity, that allows you to predict that, for example, although (because zinc is more reactive than copper) zinc reacts with copper sulphate to form zinc sulphate and copper, copper does not react with zinc sulphate in the same way.

Often, even after the products of a reaction are known, it can still take a good deal of research to establish exactly how it took place - how the starting materials interacted to form intermediates, for example, and how those intermediates interacted to form the products.  This is known as the reaction mechanism, and although many reaction mechanisms are known, plenty are not.

You should continue to study, and learn as much as you can about the principles of chemical reactivity, and you'll find that in more and more cases, there are patterns that let you predict what will happen in similar cases.  But there are often exceptions, that are only explainable once you have an in-depth understanding.

There is often no alternative to actually doing the reaction, and seeing if what is predicted actually happens.  This is why chemists do research!


How to get ionic radii for coordination number 12?

The commonly used method of obtaining ionic radii for higher coordination numbers (C.N.) is to extrapolate values from the Shannon's scale [1] using the relationship between ionic radius and coordination number proposed by Zachariasen [2]:

  ... the bond lengths $D(N_1)$ and $D(N_2)$ for cation coordination numbers $N_1$ and $N_2$ were related as follows

  $$D(N_2)=D(N_1)(N_2A_1/N_1A_2)^{1/n} \tag{1}$$

  where $n + 1$ is the exponent in the Born repulsion term of the lattice energy and $A_1/A_2$ is the ratio of the Madelung constants. The quantity $A_1/A_2$ was taken to be $0.972$ for $N_2/N_1 = 12/9 = 8/6 = 4/3$ and $0.927$ for $N_2/N_1 = 12/8 = 9/6 = 6/4 = 3/2$.

Jia [3] used Zachariasen's formula and published effective ionic radii for all lanthanides with C.N. $6$ to $12$.

Using the same approach, selected ionic radii for the 12-coordinated $\ce{A}$-sites in oxo-perovskites $\ce{ABO3}$ [4] as well as halide perovskites $\ce{ABX3}$ [5] have been determined.

References

Shannon, R. D. Revised Effective Ionic Radii and Systematic Studies of Interatomic Distances in Halides and Chalcogenides. Acta Cryst. A 1976, 32 (5), 751–767. https://doi.org/10.1107/S0567739476001551.
Zachariasen, W. H. Bond Lengths in Oxygen and Halogen Compounds of d and f Elements. Journal of the Less Common Metals 1978, 62, 1–7. https://doi.org/10.1016/0022-5088(78)90010-3.
Jia, Y. Q. Crystal Radii and Effective Ionic Radii of the Rare Earth Ions. Journal of Solid State Chemistry 1991, 95 (1), 184–187. https://doi.org/10.1016/0022-4596(91)90388-X.
Li, C.; Soh, K. C. K.; Wu, P. Formability of $\ce{ABO3}$ Perovskites. Journal of Alloys and Compounds 2004, 372 (1), 40–48. https://doi.org/10.1016/j.jallcom.2003.10.017.
Li, C.; Lu, X.; Ding, W.; Feng, L.; Gao, Y.; Guo, Z. Formability of $\ce{ABX3}$ ($\ce{X = F, Cl, Br, I}$) Halide Perovskites. Acta Cryst. B 2008, 64 (6), 702–707. https://doi.org/10.1107/S0108768108032734.


Is there any notion of pH out of solution?

The IUPAC definition of pH is:

The quantity pH is defined in terms of the activity of hydrogen(1+) ions (hydrogen ions) in solution:
$pH = − lg [a(H^+)] = − lg [m(H^+) γ_m (H^+) / m^⦵]$
where a$(\ce{H+})$ is the activity of hydrogen ion (hydrogen 1+) in aqueous solution, $(\ce{H+})$(aq), $γ_m(\ce{H+})$ is the activity coefficient of $\ce{H+}$(aq) (molality basis) at molality m$(\ce{H+})$, and m$^⦵ = 1$ mol kg $^{−1}$ is the standard molality.

So since the definition specifically refers to "aqueous solution", pH is undefined unless an aqueous solution is being considered.
There would be a p[H+] in pure acetic acid based upon the self dissociation of acetic acid, where [H+] is the concentration of hydrogen ions in acetic acid solution, but this is not pH according to the IUPAC definition.
According to Acid-Base Equilibria in Glacial Acetic Acid. III. Acidity Scale. Potentiometric Determination of Dissociation Constants of Acids, Bases and Salts J. Am. Chem. Soc., 1956, vol. 78, pages 2974–2979:

the autoprotolysis constant of acetic acid is calculated to be $3.5 \times 10^{-15}$ (pK = 14.45)

Therefore p[H+] = 7.2 for pure acetic acid.
This value is not a measure of acidity, but simply the concentration of solvated hydrogen ions in pure acetic acid.


Why I can't use a single Slater determinant to describe the ground state of open-shell atoms and molecules?

You are looking for multi-reference cases.$^1$ Another term usually applied is "strong correlation". See What exactly is meant by 'multi-configurational' and 'multireference'?.

Instead of trying to answer myself on what those cases might be, let me quote Frank Jensen (Introduction to Computational Chemistry, 2nd ed., Wiley, 2007, p. 154):

  Consider again the ozone molecule with the two resonance structures shown in Figure 4.9 [which shows a) a zwitterion with a positive charge on the central oxygen b) a triplet configuration with single electrons on terminal oxygens]. Each type of resonance structure essentially translates into a different determinant. If more than one non-equivalent resonance structure is important, this means that the wave function cannot be described even qualitatively correctly at the RHF single-determinant level (benzene, for example, has two equivalent cyclo hexatriene resonance structures, and is adequately described by an RHF wave function). A UHF wave function allows some biradical characters, with the disadvantages discussed [earlier in the text]. Alternatively, a second restricted type [Configurational State Function (CSF)] (consisting of two determinants) with two singly occupied MOs may be included in the wave function. The simplest [multi-configuration SCF] for ozone contains two configurations (often denoted TCSCF), with the optimum MOs and configurational weights determined by the variational principle. The CSFs entering an MCSCF expansion are pure spin states, and MCSCF wave functions therefore do not suffer from the problem of spin contamination.

In some cases, a single determinant may give you an incomplete picture. This is more often the case for triplet or open-shell species, but open-shell singlet cases can also demand a multi-determinant treatment.

Note that multi-reference treatments may also be necessary to obtain accurate quantitative results if the physical ground and excited states are close in energy.

$^1$ The "reference" is due to the use of the Hartree-Fock determinant as the reference state for configuration interaction (CI) post-SCF calculations. There are systems where only bad results can be achieved when using a single reference in a truncated CI. (In full CI, you have every possible reference included.)


Does boiling milk really make the nutrients in it less useful to the body when ingesting it?

The term "denature" is generally used for proteins, and it is true that boiling milk will denature the proteins that are present. Calcium and other such elements would not be "denatured" per se. In fact, I can think of no adverse effect of boiling on the chemical nature of calcium, magnesium, etc. I'm not entirely sure what other minerals and/or nutrients are present in homogenized milk.

As for material being "denatured" during digestion, this is probably an inappropriate term since we generally use the term "metabolized" instead of "denatured". When proteins and other "nutrients" are metabolized, they result in useful products, as opposed to when they are "denatured". So, in summary, I'd say that boiling milk is not the same a metabolizing milk, and I'd say that boiling would likely render milk less "nutritious".


Oxidation state of metal in metal oxo complexes always positive?

Yes. Oxygen is much more electronegative than all of the transition metals, so it is given a formal -2 charge. Keep in mind, though, that these values are not intrinsic properties of these elements/molecules. Their use is intended to help us rationalize some of the chemistry and properties associated with these materials. As an interesting example, consider oxygen difluoride, $\ce{OF_2}$. Fluorine is the most electronegative element (that forms compounds with other elements) and is given a formal charge of -1. To balance the formal charges then, oxygen must have a +2 charge!


What is the name of the "primary" carbon in a benzene derivative?

It's ipso-.

Wikipedia page
IUPAC Gold Book definition of the ipso-attack


Dehydration of carboxylic acids with phosphoric acid

The $\ce{-OH}$ part will not get protonated, since its lone pair is in resonance with the carbonyl part of the $\ce{-COOH}$ group. Instead, the lone pair of the carbonyl part will give its lone pair to $\ce{H+}$.

Something like this :-

It may look insignificant because the product is same, but this is the correct mechanism.


Turns in beta-alpha-beta loops (protein motifs)

So a beta-alpha-beta structure will have two loops that could be turns - between the first beta strand and the alpha helix, and the helix and the second strand. They might be beta turns, but they might not.

Although there are different definitions of beta-turns, one common one is the presence of a hydrogen bond between the carbonyl of residue (i) and the amino of residue (i+3). I'm not actually sure how frequent this is, but you can see this from a search on PDBeMotif. There is also Motivated Proteins, which gives nice 2D (and 3D) images of motifs.

However, you could get other conformations of these loops that are not beta-turns, especially if they are quite long.


What chemical can be used for getting rid of hard water spots on cars?

Maybe the water spots are not limestone, chalk or gypsum, all of which are dissolvable in acid. Maybe they are mud spots, which could contain clays which adhere to the car paint.

If a toilet bowl cleaner based on citric acid (stronger acid than acetic, also a chelator) or Coke (acidic, contains phosphate) do not work quickly (without damage), perhaps an alkaline dispersing agent (like a liquid laundry soap) plus some elbow grease could diminish the visibility of the spots. Dispersing agents for clays include phosphates and anionic polymers like polysilicates (but they are typically quite alkaline - need to check damage potential).

Then, a good waterproofing (waxy) polish could make it easier to resist and remove the spots. Discovering where the spots come from (rain, splash from puddles, lawn sprayer, etc.) could help suggest what materials are present in the spots.


There are a few basic principles that are useful to keep in mind when you want to balance a redox equation.

1) OILRIG - Oxidation Is Loss, Reduction Is Gain (of electrons)

Let's look at the basics of the first equation:  $$\ce{Cu + HNO3-> Cu(NO3)2} + ?$$

On the left hand side we have Cu(0) metal and N(V), on the right hand side we have Cu(II) ion and N(V), so from OILRIG we know that the copper is being oxidised... but we know from OILRIG we have to put the electrons from the Cu(0) somewhere...

2) Hydrogen ions are always +1, and oxygen ions are always (there are a few exceptions but this is a good general rule) -2

... So the only thing that can change is the nitrogen. The nitrogen must be being reduced.

Nitrogen actually has a large number of possible oxidation states, taking any value between -3 and +5, so how do we decide which one it goes to in this case?  This depends on the temperature and the concentration of the acid. We can probably assume room temperature and bench concentration of acid (otherwise conditions need to be specified), and that the nitrogen isn't going to gain 5 electrons and become nitrogen gas, which means that we are going to form an oxide. Of the oxides, the most thermally stable is NO (nitrogen is in oxidation state +2 here).

What does that look like now?

$$\ce{Cu + HNO3-> Cu(NO3)2 + NO}$$

To balance the electron transfers, we need 3 Cu(O) going to Cu(II) to donate 6 electrons for 2 N(V) to go to N(II). Balancing for electrons and nitrogens, we now have:

$$\ce{3Cu + 8HNO3-> 3Cu(NO3)2 + 2NO}$$

Ok, now we need to look at balancing oxygens and hydrogens.

3) Use water if you can to balance oxygen and hydrogen. If not, start looking at hydroxide or hydrogen ions.

Here we need to somehow account for 8 hydrogens and 8x3 - 3x3x2 + 2 = 4 oxygens... That's 4 water molecules, so we finally have the answer you provided at the start!

$$\ce{3Cu + 8HNO3-> 3Cu(NO3)2 + 2NO + 4H2O}$$


How to calculate size of a specific cavity in a protein for every frame in trajectory file?


How does potassium nitrate work as an oxidiser?

If we assume potassium carbonate ($\ce{K2CO3}$) as the only solid reaction product, the overall reaction can be described as:
$$\ce{2 KNO3 + C12H22O11 + 11 O2 -> N2 + 11 CO2 + 11 H2O + K2CO3}$$

when using sucrose for candy. With sorbitol, an equation like

$$\ce{2 KNO3 + C6H14O6 + O2 -> N2 + 5 CO2 + 7 H2O + K2CO3}$$

seems fitting. In both cases, it is assumed that:

the nitrate is completely reduced to nitrogen
additional oxygen is needed

The latter is plausible, the rocket candy is supposed to burn layer-wise at the surface in contact with air. This allows for propulsion rather than one instant explosion.

Whether the nitrite is formed as an intermediate (or which organic intermediates are formed) is beyond my knowledge. However, once this partial reduction (nitrate to nitrite) has happened, I rather doubt that the reaction goes back again (to regenerate the nitrate).

As far as the bubbles are concerned, is it conceivable that the fuse wasn't completely dried and thus water was driven out?


Why are alkyl halides used as solvents for relatively non polar compounds and not polar compounds?

Polarity is relative. Dipole moment (macro scale) and dielectric constant of the solvent (molecular scale)  are probably the most decisive. Water has ε~80, ethanol 25 and DCM is "only" 9. By comparison toluene, benzene or tertachlormethane have ε~2. The last example also illustrates why differences in electronegativity are not enough. Being symmetric (Td) tetrachloromethane has dipole moment of zero, even if there is a big difference between electronegativity of carbon on chlorine.


Effect of temperature on benzenediazonium during diazotization

Diazonium salts are prone to ejection of molecular nitrogen. The process is fairly slow for aromatic diazonium salts, so the salts are usually relatively stable around 0 celsius (alkyldiazonium salsts, though, are a completely different matter). In case of benzenediazonium resulting phenyl-cation is extremely active and will react with first nucleophile present, in many cases it is water, so the product would be phenol.