Would the concentration of oxygen around fire increase as we get close to the fire (from the outside)

A fire has no way to directly "draw in" oxygen for it to burn. It consumes what oxygen is in the immediate vicinity of the flame, depleting the air in oxygen compared to the concentration of oxygen in air further away. This sets up a concentration gradient such that oxygen will diffuse into the depleted region near the fire.

An even more important process than diffusion of oxygen into depleted air is convection driven by the heat of the fire. Oxygen-poor air that has been in the fire gets hot and rises, drawing fresh air into the base of the fire. Note that this still isn't oxygen being drawn to the fire, but a mass of fresh air flowing into the lower pressure area at the fire resulting from convection.

The bottom line is that a fire consumes oxygen, so the region around the fire will be slightly depleted in oxygen, not enriched as you suggest.


Using the quasi-steady assumption for reaction-diffusion equations

This is an example of a reaction-diffusion system where the participating species both react as well as diffuse. Such systems can be described using partial differential equations.

The mathematical model describes how the system changes with time. So we have $$\frac{\mathrm dC}{\mathrm dt}=f(C)$$

where $C$ is a vector with each element denoting the concentration a species ($C_i$).

At quasi-steady state the rates of some of the species will remain unchanged over time.

$$\frac{\mathrm dC_i}{\mathrm dt}=0$$

The reaction part usually comes from mass action kinetic models (which gives you rates for zeroth/first/second order reactions).

The diffusion part comes from Fick's laws.

$$\frac{\partial C}{\partial t}=D\nabla^2C(x,y,z,t)$$

Here the $\nabla^2$ denotes spatial double derivatives $\large\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}$

which kind of describe how the different species would be distributed in space. The reaction-diffusion equation is a sum of the reaction and the diffusion parts. At quasi-steady state, the concentration of a given species at any given point in the space would remain unchanged over time. This only sets the time derivative to zero, as I previously mentioned.

The time derivative denotes the net accumulation rate of a species. In the absence of diffusion, it will simply be the sum of rates of all reactions leading to formation of the species minus the sum of rates of all reactions causing removal of the species (principle of mass balance).

In the absence of reactions the diffusion terms will describe how the spatial pattern of the species will change over time.

At (quasi) steady state, the time derivative goes to zero which means that all the reaction and diffusion terms balance each other out such that there is no change in the concentration of the corresponding species over time.


(Effective) number of molecules interacting with light to produce optical activity

I think the picture above is self-explaining.

Now you must know the diameter and lenght of the lightpath so you can calculate the illuminated volume. Taking this volume and the concentration c you get the mol quantity (6.022 * 10^23) of molecules which have caused optical activity.


Yep, it has to do with the orbitals.

$\ce{CO2}$ is linear, so even though the $\ce{C-O}$ bonds have individual dipole moments, the overall dipole moment is zero as these cancel out (they point in opposite directions, as shown in the diagram below).

On the other hand, $\ce{H2O}$ is "bent", which means that the individual dipole moments of the bond are at an angle to each other. They add up to give a net dipole moment (shown with grey in the diagram).

The colors indicate electron density, red is more dense/blue is less dense. Dipole moment is from low density to high density.

OK, so why do these molecules have differing shapes? This is where orbitals come in. I'll try to explain as much as I can without going into orbitals.

Carbon has an outer shell electronic configuration as $2s^22p^2$. Out of these four electrons, two are used in $\pi$ bonds, and two in $\sigma$ bonds. If you don't know what those are, just look at it like this for now: A set of bonds between two atoms will have one and only one $\sigma$ bond, with the rest of them $\pi$ bonds. So, any single bond is made up of just a $\sigma$ bond, a double bond is made up of one $\sigma$ and one $\pi$, and a triple bond is made up of one $\sigma$ and two $\pi$ bonds. What these types of bonds actually are can be explained if you know what an orbital is.

Now, what VSEPR says is that the geometry of the molecule is only decided by the $\sigma$ bonds and lone pairs on the central atom. You count up the $\sigma$ bonds and lone pairs (lets say they add up to $x$), and decide the geometry based on that. The geometry is the most stable configuration of $x$ hybrid orbitals. In simple terms, if we took $x$ balloons and tied them together, the directions the balloons point in help us correspond to where the bonds and lone pairs lie:

In $\ce{CO2}$, we have two $\sigma$ bonds and two $\pi$ bonds (as each double bond has one of each type). Each bond takes up one electron from carbon, so we have no leftover electrons for forming any lone pair. Since we have two $\sigma$ bonds and 0 lone pairs, $x=2$, giving us the structure given by the first set of balloons, which is linear. And $\ce{CO2}$ is indeed linear:

Now let's take water. The central atom (Oxygen) has a valence configuration of $2s^22p^4$, that is, 6 electrons. In water, since we have two single bonds, we have one $\sigma$ bond each (and no $\pi$ bonds). So we have total two $\sigma$ bonds.

But this leaves us with $6-2=4$ unpaired valence electrons. These form two "lone pairs" (pairs of electrons which do not bond). With two lone pairs and two $\sigma$ bonds, $x=4$. This gives us a tetrahedral structure (third in the balloon diagram). Two of the four points in the tetrahedron are occupied by the lone pairs, and two by bonds:

(Note that the angle 104.5 is not the same as the angle in perfect tetrahedra, 109.25--this is due to the lone pairs repelling each other)

So finally, we have the following "bent" structure for water:

From the structure, as shown above, it is very easy to check if the molecule has a dipole moment.


Why is the enthalpy change not zero in an adiabatic process?

The Enthalpy $H$ is defined as $H=U+PV$. Therefore,

$$\Delta H=\Delta U + P\Delta V +V\Delta P$$

For an adiabatic process, $q=0$. therefore from the first law of thermodynamics,
$$\Delta U = q +w =q-P\Delta V$$
$$\Delta U=w=-P\Delta V$$
Substituting this in the first equation you get,
$$\Delta H=V\Delta P$$
If $\Delta P$ is zero during the process (Isobaric) then the enthalpy change will indeed be zero, but it will be non-zero if the process is not isobaric.


pH is defined for water solutions; using it elsewhere is either impossible or leads to contradictory results like yours. Who said pH should not be too high? Well, you might argue that high pH implies alkaline solution, does it not? No, it doesn't. Alkaline solutions are (roughly speaking) those with high concentration of $\ce{OH-}$. In water this means the same thing, because $\ce{[H+]\cdot[OH-]=10^{-14}}$, so low $\ce{[H+]}$ means high $\ce{[OH-]}$ and vice versa. In hexane, not really.

To determine pKa values which are way beyond the physically possible range of pH, we extend it with things like Hammet's $H_0$, and thus build a ladder of measures suitable for a much wider range. You may call it "effective pH" or like that, but remember: it has little to do with literal concentration of $\ce{H+}$, and your hypothetical solution of $\ce{10^{-30}M\; HCl}$ in hexane is in fact neither acidic nor basic.


In most Raman experiments, the incident radiation is not near (or at) an absorbing wavelength, and so you will never access a real, honest-to-God excited state (stationary state). (If it was close to an absorbing wavelength, we would be taking about a resonance Raman experiment.)

Now obviously something is going on between the molecule and the incident light. What you are really doing is preparing a superposition state. A superposition of real excited states, or stationary states, of your Hamiltonian. This is what we call a "virtual state," and it is not an eigenstate of your Hamiltonian, so its energy is undefined. (You are free to compute an expectation value, though.) This virtual state doesn't last forever, and, when it decays, scatters radiation away in some direction (it doesn't matter where, unless you are doing an angle-resolved Raman experiment).

(By the way, if you look closely at the equation porphyrin gave, the sum over $v$ virtual states is actually a sum over real $v$ excited stationary states, such that $E_v$ is defined. Otherwise that expression is nonsense. This equation essentially is the sum of overlaps of the initial state into the states that make up your virtual state, and then multiplied by the sum of overlaps of the excited states that make up the virtual state into the final state. This is the Kramers-Heisenberg-Dirac expression for Raman amplitudes.)

If the superposition collapses back to the ground state, the scattered radiation will have the same wavelength as the incident light, and we have Rayleigh scattering. If the superposition collapses to an excited vibrational states, we have Stokes Raman. If your initial state was vibrationally excited and you collapse to a lower vibrational state, you get anti-Stokes Raman. In the actual experiment you will observe all types and can see a Rayleigh line in the spectra. Most spectrometers will filter this wavelength out, however, leaving just the Raman peaks. Usually the virtual state is dominated by the ground state and most of the scattered light is Rayleigh. This is why Raman spectroscopy was too weak to be useful until we had better optical components.

Obviously this isn't the only way to think about Raman. You can use path integrals and/or QED if you want, which is much closer to Wildcat's answer. In fact, I think you must use QED if you want to work out virtual state lifetimes. Here is a good reference if you want to dig a bit deeper:

  Williams, Stewart O., and Dan G. Imre. "Raman spectroscopy:
  time-dependent pictures." The Journal of Physical Chemistry 92, no. 12
  (1988): 3363-3374.


What is the mechanism of action of anaesthetics?

This answer is a general answer, based on some research.

The mechanisms behind anaesthetics is not entirely understood, according to the article Mechanisms of Anesthesia: Towards Integrating Network, Cellular, and Molecular Level Modeling (Arhem et al. 2003), primarily due to, as you alluded to in your question, the fact that anaesthetics come from a very diverse group of chemicals.

In the fragmentary data presented in the article, the conclusions the authors made were that

  general anesthetics modulate the activity of ion channels, the main targets being GABAA and NMDA channels and possibly voltage-gated and background channels, thereby directly or indirectly hyperpolarizing neurons in thalamocortical loops, and thereby disrupting coherent oscillatory activity in the cortex. In our view, it does not seem unreasonable that the ultimate target is NMDA channels.

There is also a suggestion that the unconsciousness that results is in a way like sleep, an adaptive response to the chemical input.

Interesting perspectives are provided in the article Why anesthetic mechanism research
has failed, and what to do about it (Hammeroff), where the author suggests, research is needed in investigating anaesthetic mechanisms by

  Look closely at the quantum nature of anesthetic binding by van der Waals
  London forces.

In terms of local anaesthetics, the slides included in the presentation Local Anesthetics: Overview state that the mechanism here is due to

  Ionized form of weak base
  blocks $\ce{Na^+}$
  channels
  by binding to an internal
  sequence involved in channel inactivation


Making a 10L HCl solution with 8.77% HCl in a 100 mL solution from a 16 M stock solution

I figured out the problem I had. I needed a way to convert grams to milliliters, which was density.

I found the density of HCl for the 8.77% concentration:
$$ HCl \ Density = 1.04 g/mL$$
I then found how much HCl would be in 100g of solution:
$$ 8.77\% \ of \ 100g, \ so \ 8.77g \ HCl \ and \ 91.23g \ solvent$$
I did some dimensional analysis to get to moles divided by liters, so that I could get the molarity:
$$ (\frac{8.77 g \ HCl}{1})*(\frac{1 \ mol}{36.461g})*(\frac{1}{100g \ solution})*(\frac{1.04g}{1 mL})*(\frac{1000 mL}{1 L}) = 2.5 M$$
Then, I did a nice dilution calculation:
$$ 10 L *2.5M = 16M *L_2$$
$$ \frac{25}{16} = 1.5625 L$$

To do the dilution I would place 1.5625 L HCl into a beaker, and then fill with solvent until I reached a total solution volume of 10 L.


Baking powder is baking soda ($\ce{NaHCO3}$) plus a mild acid salt or dry acid, such as $\ce{Ca(H2PO4)2}$ or tartaric acid. These ingredients should pose no hazard in a stove unless there is remaining powder blocking a vent hole or gas nozzle. In that case, just use a paper clip or other tool to open the vent.

Tartaric acid is theoretically flammable, but the amount you might have left is so small as to be innocuous. You've probably cleaned the stove about as well as baking soda would, once you remove all obvious deposits of the white powder. Bon appétit!


Why is there a max radius for an orbit of an electron around a nucleus?

Perhaps you can clarify your question a bit, but if you are referring to the size of the radial distribution of electron density, this increases with increasing principal quantum number, up to the ionisation limit, at which point the spacing between energy levels trends towards 0 and the interaction energy trends towards zero. After this, the electron is for all intents and purposes unbound. This is illustrated for the hydrogen atom in the Lyman series (image borrowed from Wikipedia):

The energy of the largest transition $n=\infty \rightarrow n=1$ is 13.6 eV, or half a Hartree (this is the energy of a photon with a wavelength of 912 Ångström). This is the binding energy of the hydrogen atom. Even the Lyman gamma line - representing a $n=4 \rightarrow n=1$ transition - is quite close to the ionisation limit. As n increases, from a strictly electrostatic point of view the orbital becomes far less stable, and this is reflected in the general negative trend of ionisation energies across the periodic table, however the complete picture is a bit more complex than that because electrons can interact with electrons as well as the nucleus.

The key point is that as n increases, the energy required lose an electron entirely becomes smaller and smaller, and this fixes a practical limit on where we can say the electron is actually bound. This has a caveat though - cold gasses can form unusually long-lived Rydberg states with ridiculous principal quantum numbers with correspondingly enormous orbital radii (up to micrometers!*) and semiclassical electron orbits. The reasons for the stability of these states is outside of my knowledge.

**To paraphrase, Alisa Bokulich points out in 'Re-examining the Quantum-classical Relation' that Rydberg atoms are of comparable size to grains of (fine) sand, which is pretty neat*


What does the charge distribution around a Stone-Wales defect look like?

FWIW, here are some Hückel calculations on the 5775 PAH.

I'm showing the first 7 energy levels. Each molecular graph presents the contribution of each atom to the electron density ($\mathrm{Abs}[c_i]^2$). This is not normalized. I do not use wavefunctions, only the coefficients multiplying them, assuming

$$\Psi_{\mathrm{mol}} = \sum_i^n c_i\psi_i$$

In this approximation (HMO), the first 7 energy levels will be occupied (14 pi electrons).

I will first give you pictures of each energy level, and then a linear combination of each. Take it for what it is, as there is much information lost in the process.

To me, this simply suggests that indeed, the system will behave as if composed of tropylium and cyclopentadienyl ions. But this is not publication quality calculation either =)

Here are compared the linear combination of the first 7 energy levels of the 5775 and 6666 systems. It is clear that electron distribution is not the same in both PAHs, which hints that my calculations are not so far fetched.

5775

6666


What purpose does mercury dichloride have in fireworks?

Most likely, mercury was used as a colorant, specifically red.  This list shows mercury providing a red color in flame tests, and the same would presumably be seen in fireworks.

Several other, less dangerous elements also give red or reddish colors.  This site identifies lithium as providing the color in red fireworks, and also mentions lithium carbonate specifically.

Be safe this 4th, handle fireworks with care if you do so at all!


Materials for growing large quality sugar crystals

The Exploratorium has directions for making rock candy, the common name for sucrose crystals. In short:

Make a saturated solution of sucrose -- 4 cups sugar dissolved in just 2 cups of boiling water.
Immerse the fabric in the solution, and let it cool slowly (to make larger crystals). You might use a covered styrofoam container to slow the cooling. If crystallization doesn't start in a few days, you might need to drop one seed crystal of sugar into the container.
You can experiment with different types of food color in the sugar solution to color the crystals, but in the process of crystallization, foreign molecules of dye may be excluded from the crystals. The larger the crystal, the less colored it may appear.

Enjoy! Be sure to use sanitary equipment and safe colors, if anyone is to eat this. BTW, ordinary cane sugar is one of the purest substances sold in quantity, better than 99.9% sucrose.


Do the salt bridge ions have to match the ions in the half-cells solutions?

In short: a priori no, it need not.

The purpose of the salt brigde is to close the electric circuit (transport of charges) between the two half-cells.  To work as intended, all you need to do is to fill the bridge with any salt whose ions, both cation and anion do not react with either half cell components, nor under the external potential applied, and ideally have the same mobility in the electrolyte of the bridge, expressed by the transfer number.  This is why aqueous solutions of KCl are used so often for this purpose, but you are free to choose differently.

And not to forget, the tube normally ends by a diaphragm at each side, to slow down considerbly the mechanical exchange of matter (ions) between the bridge's content and the half cells.


A palindromic stretch of DNA is a strand whose reverse complement is itself.  So 5'-AAAT-3' is not palindromic.  It's reverse complement is 5'-ATTT-3'.  Those two pieces of DNA are not identical.  However, 5'-GGATCC-3' is palindromic, because the reverse complement is identical.


Why does the trimethylammonium cation have a larger -I effect than the ammonium cation?

$\ce{N(CH3)3+}$ has more -I than $\ce{NH3+}$
, this irony happens because if you consider
$\ce{N-CH3}$ bond and $\ce{N-H}$ bond,  which is more polar?
Of course $\ce{N-H}$ bond will be more polar  due to more difference in electronegativity  value,  so that implies electron density will be more on $\ce{N}$ in $\ce{NH3+}$ than $\ce{N}$ on $\ce{N(CH3)3+}$.
And therefore since electron density is less on $\ce{N}$ of $\ce{N(CH3)3+}$ , it will have greater -I effect.


Does water freely absorb atmospheric CO2?

Er, the quick, simple technical answer is "Yes, water will absorb $\ce{CO2}$". However, there is a "but" coming. Actually, two "buts".

First, plain water will not be able to absorb much $\ce{CO2}$. Henry's Constant for $\ce{CO2}$ in $\ce{H2O}$ is $K_H=29.76 \mathrm{atm\over mol/L}$. Considering the partial pressure of $\ce{CO2}$ in air is roughly $3.5\times10^{-4}~\mathrm{atm}$, one liter of water can absorb, at best $1.18\times10^{-4}~\mathrm{mol}~\ce{CO_2}$ which is only about $0.1$% of a gram of $\ce{CO2}$. Not very much. (Note, I'm ignoring the conversion to carbonate, $\ce{CO3^{2-}}$, and bicarbonate, $\ce{HCO3^{-}}$, because in otherwise neutral water, they don't exist in appreciable quantities (see: Carbonic Acid)

Second, that's the thermodynamic best you can do. However, you have to contend with the kinetics and diffusion limitations as well. Basically, it takes time for the $\ce{CO2}$ to go from the air to the water, and then to get well mixed within the water.

Now, there's a couple things you can do to help $\ce{CO2}$ absorption. First, instead of using neutral water, use and alkaline solution. For example, if you add $\ce{NaOH}$ to your water, the equilibrium will shift toward the carbonate ion, $\ce{CO3^{2-}}$:

$$\ce{2NaOH + CO2->Na2CO3 + H2O}$$

This allows orders of magnitude more $\ce{CO2}$ to be absorbed than would possible, basically, you become limited by the amount of sodium hydroxide you have and the kinetics.

Speaking of kinetics, how can we help that out? Packed bed towers! Imagine having a tall column filled with beads. Blow your air up from the bottom, while allowing the alkaline solution to flow from the top. This will maximize contact between the two phases and enhance the transfer of $\ce{CO2}$ from the gas to liquid.

And that is basically how scrubbers work in chemical plants, and one method of carbon sequestration. However, the very low concentration of $\ce{CO2}$ in air ($\sim\!400~\mathrm{ppm}$) makes pulling appreciable amounts of $\ce{CO2}$ from the air infeasible and an inefficient use of resources.


What does oscillator strength have to do with state character?

This does not seem to make any sense to me either unless I'm missing something very obvious. Suffice it to say that spin changing transitions are generally strongly forbidden, unless paramagnetic electron spin or heavy atoms make them allowed, and that spin 'forbidden-ness' is stronger than symmetry 'forbidden-ness'. Symmetry forbidden transitions can be made allowed by 'intensity borrowing' (Herzberg-Teller) involving a higher electronic state & a vibration of suitable symmetry, as in benzene.  When both spin & symmetry are forbidden transitions are very weak indeed. $n\pi ^*$ transitions are often symmetry forbidden (e.g. in carbonyls) and so weak compared to $\pi \pi ^*$. Other than these general rules you just have to look at the individual molecules to work out what is going on, it might be quicker than trying to understand the diagram.


I don't see any flaws in your argumentation. I agree that $R, T, V = \mathrm{const}$, hence

$$\frac{p(\ce{He})}{n(\ce{He})} = \frac{p(\ce{O2})}{n(\ce{O2})}$$

$$m(\ce{O2}) = \frac{p(\ce{O2}) \cdot M(\ce{O2})}{p(\ce{He}) \cdot M(\ce{He}) } \cdot m(\ce{He})$$

Just one minor thing: you incorrectly converted partial pressure of helium: $\pu{42.5 torr}$ is $\pu{0.056 atm}$, not $\pu{0.559 atm}$, therefore your resulting mass should be 10 times higher.

In fact, there was no need to convert $\pu{torr}$ to $\pu{atm}$ in the first place. Don't do unnecessary steps if you can avoid them. At the same time don't drop any units and always remember to include them in your final calculation.

$$m(\ce{O2}) = \frac{\pu{158 torr} \cdot \pu{31.99 g mol-1}}{\pu{42.5 torr} \cdot \pu{4.00 g mol-1}} \cdot \pu{1.42 g} = \pu{42.22 g}$$


Thermodynamic functions are strictly defined only for macroscopic systems (systems that have an essentially infinite number of atoms). You can't apply them to individual atoms because that would be confusing large-scale averages with individual microscopic values.

Here's an analogy: the average speed of cars on a stretch of highway might be 55 mph, but it's possible that no individual car is traveling at that exact speed.

Another analogy: the average global temperature is rising over long periods of time, but if I look at the thermometer outside my window, the local temperature is falling over a short period of time... I can't say that my local temperature drop disproves global warming, because weather is a different thing from climate.

Suppose that you knew the speed of a single atom. Now, for a macroscopic ideal gas, the average speed of molecules $\overline{v}$ is $$\overline{v} = \sqrt{\frac{8 R T}{\pi M}}$$ It would be wrong for you to plug in the speed of your single atom and solve for its temperature, because that equation was derived by assuming that you had a macroscopic system (with a certain speed distribution over many molecules); it only applies to average velocities, not individual velocities.

This applies to other thermodynamic functions, too. Let's look at a simple chemical example. For a macroscopic crystal, the Gibbs free energy might be written as $$G = Ng(P,T)$$ where $N$ is the number of atoms and $g$ is the Gibbs free energy per atom. Little $g$ is a function of pressure and temperature.

So can we say that little $g$ is the Gibbs free energy of a single atom? No. It's the average Gibbs free energy per atom for a huge number of atoms. To see the difference,  consider a small cluster of atoms from that same crystal, the previous equation will need some correction terms:

$$G = Ng(P,T) + a(P,T)N^{2/3} + b(T)\ln{N} + c(P,T)$$
where the "a" term is a surface free energy and the last couple of terms might come from free energy contributed by things like rotation of the cluster (Source: T. L. Hill, Thermodynamics of Small Systems, Dover, 1962.) You can't say that $g(P,T)$ is the Gibbs free energy of a single atom in the cluster because of those correction terms.

TL;DR: Thermodynamic functions are averages over very large numbers of particles; random fluctuations are tiny compared to the mean values. But with small systems, those fluctuations become quite important, and there are additional effects that cause macroscopic averages to poorly describe microscopic systems.


How do non-polar substances dissolve in non-polar solvents?

The electron density distribution in molecules (including nonpolar ones) is not static. Therefor, as a function of time, the electron density is not uniform. Occasionally, randomly, the electron density in a molecule will shift to produce a spontaneous dipole: part of the molecule now has more electron density ($\delta^-$) and part of the molecule now has less electron density ($\delta^+$). This spontaneous dipole is transient. The electron density will shift back to negate it and then shift again to create a new spontaneous dipole. In a vacuum, this behavior would be a curiosity.

In the presence of other molecules, these spontaneous dipoles have a propagating effect. If molecule A develops a spontaneous dipole, then the electron density in neighboring molecule B will by respond by developing a spontaneous (but opposite) dipole. The $\delta^+$ region of B will be close to the $\delta^-$ region of A. This new dipole in molecule B is an induced dipole. This induced dipole in B will then induce a new induced dipole in another neighboring molecule C. By this time the transient dipole in A is already fading, perhaps to be replaced by a new induced dipole from another molecule.

These random, transient, but continuously propagating dipoles have attractive forces associated with them. These forces are named London dispersion forces after the physicist who proposed them. The magnitude of these forces scales with increasing surface area of the molecule. Thus, larger molecules will overall have stronger London dispersion forces than smaller molecules. Linear rod-like molecules will have stronger forces than spherical molecules (spheres having the smallest surface area to volume ratio of the 3-dimensional solids).

More complex attractive forces also arise from nonpolar molecules. Benzene (and other aromatic molecules) can pi stack, which relies on the strong permanent electric quadrupole. Benzene also has a (weak but) permanent magnetic dipole due to its ring current, so some component of the attraction between benzene molecules may be magnetic and not coulombic. Or, the two attractions may be one and the same, as benzene's electric quadrupole may result from its having a magnetic dipole. As his part of the discussion is devolving into physics, I will end here.


Frontier orbital analysis of zinc-catalysed carbonyl-ene reaction

First of all, a molecule only has one HOMO and LUMO. Don't confuse MOs with localized orbitals from VB theory.
You can basically approximate this mechanism with four MOs:

LUMO+1: Mainly $\pi^*_{C=C}$ with contributions from $\sigma^*_{C-H}$
LUMO: Mainly $\pi^*_{C=O}$
HOMO: Mainly $\pi_{C=C}$ with contributions from $\sigma_{C-H}$
HOMO-1: Mainly $p_O$ with contributions from $\sigma$-orbitals on the carbonyl carbon. We can regard this MO as one of the lone pairs of the oxygen.

The corresponding interactions will roughly look like this:

Hence, the HOMO - LUMO interaction explains the C-C bond formation, while the HOMO-1 - LUMO+1 interaction explains C-H bond breaking and O-H bond formation. Both HOMO and LUMO+1 also explain the shift of the double bond.
Now to your question:

Q1: Yes, $\sigma_{C-H}$ is part of HOMO, but has a lower contribution compared to $\pi_{C=C}$. So I would mainly call the C=C bond as donor. As you can see $\pi_{C=O}$ is not part of the FMOs, so your proposal makes little sense. It mainly contributes to more stable occupied MOs.
Q2: If I understand correctly, you propose an antarafacial [2+2], which is in principle symmetry allowed, but would lead to a very strained transition state with a high barrier. You are free to swap the sings on MOs, but as soon as interact with another MO, you take the one that leads to stabilization of course.

Edit: As reference, I got the FMOs from a HF/6-31G calculation on a slightly simplified system:


The obvious choice for the source of carbonate is sodium bicarbonate (UPDATE: sodium bicarbonate is not soluble enough per other answer, but sodium carbonate is a possibility), which has a positive enthalpy of solution even at high concentrations, so let's call that B.

The acid is a little trickier. If you look up the standard enthalpy of solution of most acids, you'll find negative values (meaning the temp would go up when they are added to water). However, these values refer to infinite dilution in neutral water. Once you've added a little bit and the pH has dropped, the reaction $\ce{H+ + HO- -> H2O}$, which has a large negative enthalpy, essentially stops happening and the enthalpy of solution becomes positive. This is easy to tell by looking at the maximum solubility as a function of temperature. If it increases, then the enthalpy change of solution at saturation is positive. So at the high concentrations in the problem (4 g per 30 mL), many acids that are solid at room temp will cause a temperature drop when dissolved. Citric acid is one example.

A solution of citric acid is certainly acidic enough that it will cause rapidly production of carbon dioxide gas when mixed with a solution of sodium carbonate.


How would 2-(chloromethyl)-1-methylpyrrolidine recyclize into 3-chloro-1-methylpiperidine with only water?

Forget about the water present - its just a solvent.
The best nucleophile present is the N-lone pair; if you think about it in 3-D you will see that it can do a nucleophile attack on the CH2-Cl to displace Cl- and form the aziridinium cation. Cl- can then attack this at either carbon of the [3] ring, one attack merely reforms starting material, the other gives the 3-Cl-piperidine product shown.


Lines on inside of hexagonal shapes of structure diagrams

Noradrenaline/norepinephrine is an aromatic organic compound containing a benzene ring (the hexagon). Essentially, the vertices in the diagram are carbon atoms and the number of lines represents the covalent bond order; 1 line (eg connecting an OH to the benzene ring) means a single covalent bond. Any 'spare' bonds are where a C-H bond is. So it can be rewritten like this:

HOWEVER, when it comes to a benzene ring (the hexagon), you do not simply have alternating single and double bonds. In fact you have what is known as a 'resonance structure', where the electrons/bond is shared around the entire ring. One way of thinking of it is that there are 1.5 bonds between each carbon atom in the ring. Another way of thinking of it is that pi-orbitals from each carbon atom merge to form a ring combined orbital structure, so instead of existing in that figure-8 type area around one carbon nucleus, those 6 electrons can be in a donut shaped area above and below the carbon nuclei:

This is why you often see a hexagon with a circle inside it to represent a benzene ring as it is less misleading than the alternate single and double bonds. In that case, you would redraw the noradrenaline diagram like this:

The advantage, though, of having the alternate double and single bonds is in drawing diagrams when working out reactions.


What do 'sigma' and 'pi' in front of ligands in the formula of a coordination compound mean?

Concise Inorganic Chemistry1 has the following to say about this tetra(cyclopentadienyl) titanium complex:

  Tetra(Cyclopentadienyl) compounds such as $\ce{Ti(C5H5)4}$ may be made from $\ce{TiCl4}$ and $\ce{Na(C5H5)}$. The formula may be written as $\ce{[Ti(\eta_5-C5H5)2(\eta_1-C5H5)2]}$ where two cyclopentadienyl rings are attached by 5 C-atoms ($\pi$ - bonded) and two rings are attached by one C-atom ($\sigma$ - bonded).

  NMR studies on these tetra(cyclopentadienyl) compounds suggest that in the $\eta_1$ rings the C bonded to Ti continually changes, and the $\eta_1$ and $\eta_5$ rings change their roles.

The first paragraph explains quite well what the author intends by the quantifiers $\sigma$ and $\pi$. Two rings are bonded by $\pi$ donation with a hapticity of 5 and two rings are $\sigma$ bonded with a hapticity of 1. This picture should clear things up

References:

Lee, J. D. Concise Inorganic Chemistry. 5th ed, Wiley publishers & Blackwell Science, pp. 696, 1996


The total pressure of the system is:

$\frac{1.0\times0.5+1.0\times0.4+2.0\times0.2}{1.0+1.0+2.0}=1.3/4=0.325$

i.e. just the sum of the $PV$s divided by the sum of the $V$s.  Then $0.325$ is the sum of the partial pressures by the Law of Partial Pressures.  So this at least suggest that the individual partial pressures are near $0.1$.

Can you figure out how to parcel out the individual pressures from here?


Why was rhenium used as an igniter in flash bulbs?

New tungsten alloys containing rhenium were developed as a filament material in flashbulbs for the purpose of improving their reliability. Weak batteries (anything but rare in the carbon-zinc era) could easily result is a slightly-delayed flash or no flash at all, spoiling the photo in either case.
An important characteristic of batteries is their internal resistance. This impedance is small in a fresh dry-cell battery but increases as the battery wears out, eventually reaching a value high enough to make the battery unusable. The rhenium acts to "increase the cold resistance of tungsten wire filament so as to permit sufficient power to be obtained over large variations in impedance."
As the prior post suggested, anti-oxidation properties of the rhenium also helped greatly  reduce the chance of a filament burn-out and consequent flash lamp ignition failure.
Because photography wasn't cheap back in the 50's and 60's, the vast majority of cameras and flash units sat in closets at least nine-tenths of the time. Even unused, their carbon-zinc batteries could go bad in three years. Considering the many millions of flash  pictures attempted with nearly-dead batteries in that era, this reliability feature looks like an excellent idea.
For details, please see United States Patent No. 3,123,993 granted 3/10/1964 to George W. Cressman, Chagrin Falls, and Louis A. Demchock, Jr., Cleveland, Ohio, assignors to General Electric Company.


Assigning R/S configurations to Fischer projections of 3‐chloro‐2‐hydroxybutanoic acid

The compound mentioned is (2S,3R)-3-chloro-2-hydroxybutanoic acid. For Carbon atom #2, the CIP priority order is

$\ce{Cl}$
Carbon #1
$\ce{CH3}$
$\ce{H}$

The methyl group is lower priority because Carbon #1 has a hydroxyl group attached to it, which is higher priority than Hydrogen. This implies that the orientation at carbon atom #2 is R.
For more in-depth information, you might want to check out Wikipedia


As orthocresol commented (vide supra), migratory aptitude in cationic reactions has been given as: $\ce{H}=\mathrm{Ph} \gt \ce{(CH3)3C \gt (CH3)2CH \gt CH3CH2 \gt CH3}$ (Wikipedia and Migratory aptitude in pinacol-pinacolone rearrangement). In your question, the secondary carbocation formed when protonated has three groups, which can migrate to give more stable tertiary carbocation. They, with their migratory aptitude, are:  $\ce{-R^1CH2(CH3)CH\!- \gt R^2CH2CH2\!- \gt H3C\!-}$ (where $\ce{R^1}$ and $\ce{R^2}$ are continuation of rest of cyclopentane ring).

Now, if  $\ce{H3C\!-}$ group migrated, it'd give option (D); if  $\ce{R^2CH2CH2\!-}$ group migrated, it'd give option (A); and if  $\ce{R^1CH2(CH3)CH\!-}$ group migrated, it'd give option (B). If nothing migrated you have option (C). Although all 4 products are possible in the reaction mixture ((C) would be very trace amount), asking is the major product, and hence is is the product given but migration of group with highest migratory aptitude. Therefore, it should be the product given in option (B).


If given a half reaction, how do you determine if an element can exist in a acidic conditions?

Doing so will result in the two half equations

$\ce{2MnO_4^2- -> 2MnO_4^- + 2e^-}$
and

$\ce{MnO_4^2- + 4H^+ +2e^- -> MnO_2 + 2H2O}$

which you then combine into the total equation

$\ce{3MnO_4^2- + 4H+ <=> 2MnO4- + MnO2 + 2 H2O}$.

The question is which way this goes. You can see that you have loads of $\ce{H+}$ on the left-hand side so you can expect that in acidic conditions the equilibrium will be shifted to the right hand side. You can quantify this by calculating the electrochemical potential from the Nernst equation for the two half cells.


What is the @ symbol used for in material science?

The at symbol @ universally means "at the site"  or "at the rate of".
It is routinely used in supramolecular chemistry to denote guest@host relationship because

it follows the original designation of denoting the binding site;
its shape resembles a guest (letter "a") entrapped within a host shell (circle sign "○") providing a visual cue for an intuitive image (e.g. see endohedral fullerene).

In the context of the paper by Guo et al. [1] you are referring to, paraffin is encapsulated inside $\ce{SiO2}$ shell (obtained in situ), and these silicon dioxide particles were then embedded in silicone rubber matrix.
Here, in the notation SR/Pa@SiO2 the forward slash "/" implies composite material (phase separator), and the at sign "@" refers to host-guest interactions as mentioned before.
Here is a brief infographics I roughly created with Inkscape to illustrate the concept:

An obligatory reference quotation [2, p. 203]:

3.6.17 At Symbol (@)
The at symbol ( @ ) is most commonly used in email addresses. Its modern use is primarily in accounting (where it means “at the rate of”) and should not be used in scientific writing as a substitute for “at.”
Two legitimate uses of the at symbol are

In chemical formulae, the @ is used to denote trapped atoms or molecules. For instance, $\ce{La@C60}$ means lanthanum inside a fullerene cage.
In genetics, an at symbol after a gene symbol indicates that it is part of a gene cluster.

References

Guo, Y.; Yang, W.; Jiang, Z.; He, F.; Zhang, K.; He, R.; Wu, J.; Fan, J. Silicone Rubber/Paraffin@silicon Dioxide Form-Stable Phase Change Materials with Thermal Energy Storage and Enhanced Mechanical Property. Solar Energy Materials and Solar Cells 2019, 196, 16–24. DOI: 10.1016/j.solmat.2019.03.034.
The Manual of Scientific Style: A Guide for Authors, Editors, and Researchers, 1st ed.; Rabinowitz, H., Vogel, S., Eds.; Elsevier/Academic Press: Amsterdam; Burlington, MA, 2009. ISBN 978-0-12-373980-3.


If you had included the units in your calculation, you would have noticed why your equation is not correct.

Molar mass $M$ is defined as
$$M=\frac mn\tag1$$
where $m$ is mass and $n$ is amount of substance.
Since the Avogadro constant $N_\mathrm A$ is
$$N_\mathrm A=\frac Nn\tag2$$
where $N$ is the number of particles, the mass $m$ of one atom $(N=1)$ is
$$m=\frac M{N_\mathrm A}\tag3$$

Density $\rho$ is defined as
$$\rho=\frac mV\tag4$$
where $V$ is volume.
Thus, the volume of a sample is
$$V=\frac m\rho\tag5$$
Using Equation $\text{(3)}$, the volume $V$ can be calculated for a single atom:
$$V=\frac M{N_\mathrm A\cdot\rho}\tag6$$

Assuming that a fraction of $88\,\%$ of the volume $V$ is filled with a hard sphere, the volume $V_\text{sphere}$ of the sphere is
$$\begin{align}
V_\text{sphere}&=0.88\times V\tag7\\[6pt]
&=0.88\times\frac M{N_\mathrm A\cdot\rho}\tag8
\end{align}$$

Since the volume of a sphere is
$$V_\text{sphere}=\frac43\pi r^3\tag9$$
where $r$ is the radius of the sphere, the radius $r$ is
$$\begin{align}
r&=\sqrt[3]{\frac{3V_\text{sphere}}{4\pi}}\tag{10}\\[6pt]
&=\sqrt[3]{\frac{3\times0.88\times M}{4\pi\cdot N_\mathrm A\cdot\rho}}\tag{11}\\[6pt]
&=\sqrt[3]{\frac{3\times0.88\times 107.86820\ \mathrm{g\ mol^{-1}}}{4\pi\times
6.02214076\times10^{23}\ \mathrm{mol^{-1}} \times 10.5\ \mathrm{g\ cm^{-3}}}}\\[6pt]
&=1.53\times10^{-8}\ \mathrm{cm}\\[6pt]
&=1.53\times10^{-10}\ \mathrm m\\[6pt]
&=153\times10^{-12}\ \mathrm m\\
\end{align}$$


Why are the mass ratios of O to KClO3 and O2 to KClO3 the same?

Your calculations are correct. The key is that you are measuring percentage by mass: the mass of oxygen in 2.062 g of $\mathrm{K}\mathrm{Cl}\mathrm{O}_3$ is the same whether it is considered the mass of individual oxygen atoms or of $\mathrm{O}_2$ (or of any other allotrope of oxygen, for that matter). You have half as many moles of $\mathrm{O}_2$, but the mass is the same (since the mass of 1 mol $\mathrm{O}_2$ is twice the mass of 1 mol O).


How to get the coordinates or .cif file of β  and χ ?

Jiang et al. 2018 (DOI: 10.1016/j.jechem.2018.01.026) may be relevant literature? Here's a quote:

These include three allotropes named borophene, beta-12(β12) and chi-3(χ3), which
have been successfully synthesized by physical methods [26,27].
[26] A. J. Mannix, X. F. Zhou, B. Kiraly, J. D. Wood, D. Alducin, B. D. Myers, X. L. Liu, B. L. Fisher, U. Santiago, J. R. Guest, M. J. Yacaman, A. Ponce, A. R. Oganv, M. C. Hersam, N. P. Guisinger, Science 350 (2015) 1513–1516.
[27] B. J. Feng. J. Zhang, Q. Zhong, W. B. Li, S. Li, H, Li, P. Cheng, S. Meng, L. Chen, K. H. Wu, Nat. Chem. 8 (2016) 563-568.

I also found this one:
Izadi Vishkayi, S., Bagheri Tagani, M. Edge-Dependent Electronic and Magnetic Characteristics of Freestanding β 12-Borophene Nanoribbons. Nano-Micro Lett. 10, 14 (2018). DOI: 10.1007/s40820-017-0167-z
Perhaps you may find the link to the .cif from all of the above?


Why does only one reaction occur when so many are possible?

Your example is not quite the case when multiple reactions are possible. Also, $\ce{H+}$ not only can, but actually does combine with $\ce{CO3^2-}$ to form a new particle: $\ce{HCO3-}$, the hydrocarbonate anion. This is where it would end if you'd use just one tiny droplet of $\ce{HCl}$. Otherwise, that anion would attract another $\ce{H+}$ and form carbonic acid $\ce{H2CO3}$, which is unstable and decomposes into $\ce{H2O}$ and $\ce{CO2}$, to produce the summary equation which you already know.

As for your question in general, in most cases when so many reactions are possible, they indeed occur all at once. It just so happens that they go at different rates, and one of them outruns the others by a wide margin, so you may consider it "the only one" and safely ignore the rest. Or it does not, and you end up with a bucketful of brown goo. Such examples rarely make it to the textbooks, though.

All this is especially big deal in the field of organic chemistry, where all reactions produce more or less "side products". Read the details on just about any reaction, and you'll see these yield figures to the right: 96%, 92%... or maybe less if the authors were less lucky. That's how much of our reactants went down the desired pathway. The rest went someplace else.


Can you fix rubber on a handle after it has all but turned into a liquid?

I've had the same problem with other types of rubber-coated tool handles. The best solution is to completely remove the rubber coating, which takes a lot of scraping and then scrubbing (outdoors) with paint thinner on paper towels. Invariably, as you have already noticed, everything touched gets coated with hard-to-remove black goop.

Another choice would be to wipe off the loose stuff and then over-wrap the handle with a strong cloth tape, checking carefully that it will not slip when you use the maul.

You might cover the tape with heat-shrink tubing, if you have the right diameter stuff, or with tubing cut from an old bicycle inner-tube (cheap or free, and makes a good grip).


Finding pH using the -log, why does my calculation differ from the book's answer?

Because barium hydroxide is $\ce{Ba(OH)2}$

The book is assuming each mole of barium hydroxide release 2 moles of hydroxide ions, while you are assuming only 1 mole is released.


Is this a possible structural isomer for Butene(C4H8)

Yes, the formula you drew is an example of isobutylene.  Isobutylene has the "official" IUPAC name of 2-methylpropene, and is an important industrial chemical.  Because it:

has a formula of $\ce{C4H8}$, and thus contains four carbon atoms;
contains a carbon-carbon double bond;
and doesn't contain any other functional groups

... it is indeed an isomer of butene (sometimes known as butylene).  You got this one right!$%edit$


Why don't sponges foam up an hour later?

Dish soap traditionally contained foaming agents such as sodium laureth sulfate which is chemically stable over long periods of time. But more recently environmentally conscious soap companies are turning to foaming agents that have a shorter half-life. These chemicals tend to break down sooner - perhaps just by being exposed to air. This helps reduce foaming in the settling basins of wasterwater treatment plants, or storm drains that enter the ocean and it takes less water to rinse away the soap.

Read the ingredients of the soap you are using and research to see which are the foaming agents, and what the stable half-life is.


Does H2O2 break down into H2O + O or H2O + O2?

$\ce{H2O2}$ -> $\ce{H2O +1/2O2}$

This might make it easier to understand.

$\ce{H2O2}$ will homolytically cleave for form two $\ce{.OH}$ radicals. Radicals are very reactive and will start a chain reaction, but ultimately you will end up with water and oxygen products from $\ce{H2O2}$.

(Why does it form two hydroxy radicals? because the O-O single bond is weak and unstable - see MO theory)

How do we know this? Look at a bottle of $\ce{H2O2}$ that has been left out of the fridge for too long. There will be a slight pressure build up of $\ce{O2}$ gas and the % concentration $\ce{H2O2}$ will be down due to formation of $\ce{H2O}$and loss of $\ce{H2O2}$.

For sake of completion,
You can alternatively balance the equation thus:
$\ce{2H2O2 -> 2H2O + O2}$


Why is bromine produced at the anode when aqueous sodium bromide undergoes electrolysis?

This is  how I see it:

In the electrolysis of NaBr, water is reduced at the cathode. This occurs because water is more easily reduced than are sodium ions. This is reflected in their standard reduction potential.

At cathode reduction of water occurs:

$$\ce{2H2O(l) + 2e- -> H2(g) + 2OH-(aq)}$$

And hydrogen gas is produced

At the anode, where oxidation occurs, the standard oxidation potential of water is –1.23 volts, while that for bromide ions is -1.07 volts.

The production of bromine itself has a negative electrode potential.
One of the half-reactions must be reversed to yield an oxidation.

$$\ce{Br^2- -> Br2 + 2e-}~~~~~~~ Eº -1.07$$

Remember that when one reverses a reaction, the sign of Eº (+ or –) for that reaction is also reversed.

This means that bromide ions are more easily oxidized than water.

It is important to note: When current begins to flow, the distribution of ions around the electrodes changes, and the equilibrium electrode potentials no longer accurately apply.

I think other factors also apply;

Concentration
Temperature
Nature of ions


Why potential energy is neglected in this problem for calculating the energy difference between two stationary Bohr Orbits?


How it was observed, that elements in columns 6 and 11 assume abnormal electron configuration?

I assume you to mean

  How it was first observed, that elements in columns 6 and 11 assume abnormal electron configurations?

Chemists (starting with alchemists) have were doing stoichiometric reactions for a very long time.

The point is that if there was only a single "normal electron configuration" for chromium, then there would only be one chromium chloride. However chemists have synthesized three.

Chromium(II) chloride, also known as chromous chloride.
Chromium(III) chloride, also known as chromic chloride or chromium
trichloride
Chromium(IV) chloride

Thus the fact that there must be "abnormal" electron configurations was first discovered by stoichiometry, not spectroscopy.

Other columns in the periodic chart do have elements with multiple electron configurations. So such an "abnormality" isn't unique.

If your question means

  How are the bonding energies measured for the elements in columns 6 and 11 in various electron configurations?

Then ESCA would be one technique.


Why must the total increase in oxidation numbers equal the total decrease of them in redox reactions?

The reaction is with an advantage written in the simplified ion form:

$$\ce{5 Fe^2+ + MnO4- + 8 H+ -> 5 Fe^3+ + Mn^2+ + 4 H2O}$$

Total oxidation numbers $+17=+17$

The oxidation number is formally the net charge an atom within a molecule would have held, if we had broken all bonds and bond electrons  had sticked to atoms according the atom electronegativity.

Therefore, the constant total sum of oxidation numbers across all involved atoms directly relates to the constant total sum of valence electrons and to the law of charge conservation.


Why is an S-S bond stronger than an O-O bond?

TL;DR: The $\ce{O-O}$ and $\ce{S-S}$ bonds, such as those in $\ce{O2^2-}$ and $\ce{S2^2-}$, are derived from $\sigma$-type overlap. However, because the $\pi$ and $\pi^*$ MOs are also filled, the $\pi$-type overlap also affects the strength of the bond, although the bond order is unaffected. Bond strengths normally decrease down the group due to poorer $\sigma$ overlap. The first member of each group is an anomaly because for these elements, the $\pi^*$ orbital is strongly antibonding and population of this orbital weakens the bond.

Setting the stage

The simplest species with an $\ce{O-O}$ bond would be the peroxide anion, $\ce{O2^2-}$, for which we can easily construct an MO diagram. The $\mathrm{1s}$ and $\mathrm{2s}$ orbitals do not contribute to the discussion so they have been neglected. For $\ce{S2^2-}$, the diagram is qualitatively the same, except that $\mathrm{2p}$ needs to be changed to a $\mathrm{3p}$.

The main bonding contribution comes from, of course, the $\sigma$ MO. The greater the $\sigma$ MO is lowered in energy from the constituent $\mathrm{2p}$ AOs, the more the electrons are stabilised, and hence the stronger the bond.

However, even though the $\pi$ bond order is zero, the population of both $\pi$ and $\pi^*$ orbitals does also affect the bond strength. This is because the $\pi^*$ orbital is more antibonding than the $\pi$ orbital is bonding. (See these questions for more details: 1, 2.) So, when both $\pi$ and $\pi^*$ orbitals are fully occupied, there is a net antibonding effect. This doesn't reduce the bond order; the bond order is still 1. The only effect is to just weaken the bond a little.

Comparing the $\sigma$-type overlap

The two AOs that overlap to form the $\sigma$ bond are the two $\mathrm{p}_z$ orbitals. The extent to which the $\sigma$ MO is stabilised depends on an integral, called the overlap, between the two $n\mathrm{p}_z$ orbitals ($n = 2,3$). Formally, this is defined as

$$S^{(\sigma)}_{n\mathrm{p}n\mathrm{p}} = \left\langle n\mathrm{p}_{z,\ce{A}}\middle| n\mathrm{p}_{z,\ce{B}}\right\rangle = \int (\phi_{n\mathrm{p}_{z,\ce{A}}})^*(\phi_{n\mathrm{p}_{z,\ce{B}}})\,\mathrm{d}\tau$$

It turns out that, going down the group, this quantity decreases. This has to do with the $n\mathrm{p}$ orbitals becoming more diffuse down the group, which reduces their overlap.

Therefore, going down the group, the stabilisation of the $\sigma$ MO decreases, and one would expect the $\ce{X-X}$ bond to become weaker. That is indeed observed for the Group 14 elements. However, it certainly doesn't seem to work here. That's because we ignored the other two important orbitals.

Comparing the $\pi$-type overlap

The answer for our question lies in these two orbitals. The larger the splitting of the $\pi$ and $\pi^*$ MOs, the larger the net antibonding effect will be. Conversely, if there is zero splitting, then there will be no net antibonding effect.

The magnitude of splitting of the $\pi$ and $\pi^*$ MOs again depends on the overlap integral between the two $n\mathrm{p}$ AOs, but this time they are $\mathrm{p}_x$ and $\mathrm{p}_y$ orbitals. And as we found out earlier, this quantity decreases down the group; meaning that the net $\pi$-type antibonding effect also weakens going down the group.

Putting it all together

Actually, to look solely at oxygen and sulfur would be doing ourselves a disservice. So let's look at how the trend continues.

$$\begin{array}{|c|c|c|c|}
\hline
\mathbf{X} & \mathbf{BDE(X-X)\ /\ kJ\ mol^{-1}} & \mathbf{X} & \mathbf{BDE(X-X)\ /\ kJ\ mol^{-1}} \\
\hline
\ce{O} & 144 & \ce{F} & 158 \\
\ce{S} & 266 & \ce{Cl} & 243 \\
\ce{Se} & 192 & \ce{Br} & 193 \\
\ce{Te} & 126 & \ce{I} & 151 \\
\hline
\end{array}$$
(Source: Prof. Dermot O'Hare's web page.)

You can see that the trend goes this way: there is an overall decrease going from the second member of each group downwards. However, the first member has an exceptionally weak single bond.

The rationalisation, based on the two factors discussed earlier, is straightforward. The general decrease in bond strength arises due to weakening $\sigma$-type overlap. However, in the first member of each group, the strong $\pi$-type overlap serves to weaken the bond.

I also added the Group 17 elements in the table above. That's because the trend is exactly the same, and it's not a fluke! The MO diagram of $\ce{F2}$ is practically the same as that of $\ce{O2^2-}$, so all of the arguments above also apply to the halogens.

How about the double bonds?

In order to look at the double bond, we want to find a species that has an $\ce{O-O}$ bond order of $2$. That's not hard at all. It's called dioxygen, $\ce{O2}$, and its MO scheme is exactly the same as above except that there are two fewer electrons in the $\pi^*$ orbitals.

Since there are only two electrons in the $\pi^*$ MOs as compared to four in the $\pi$ MOs, overall the $\pi$ and $\pi^*$ orbitals generate a net bonding effect. (After all, this is where the second "bond" comes from.) Since the $\pi$-$\pi^*$ splitting is much larger in $\ce{O2}$ than in $\ce{S2}$, the $\pi$ bond in $\ce{O2}$ is much stronger than the $\pi$ bond in $\ce{S2}$.

So, in this case, both the $\sigma$ and the $\pi$ bonds in $\ce{O2}$ are stronger than in $\ce{S2}$. There should be absolutely no question now as to which of the $\ce{O=O}$ or the $\ce{S=S}$ bonds is stronger!


Products of the reaction between chromium(III) chloride, zinc and sulfuric acid

I know that chromium(III) reduces to chromium(II), but I'm not sure about the reaction. Would it be $$\ce{CrCl3·6H2O + Zn -> Cr(H2O)6Cl2 + ZnCl + H2}?$$

Your concept is right and I am glad nobody taught you the concept of nascent hydrogen. It is an obsolete idea of the 1920-40s. Please don't use it anywhere. Wikipedia has a nice summary https://en.wikipedia.org/wiki/Nascent_hydrogen and of course there are several articles which negate this very concept.

You can just simply write the ionic equation involving Cr(III), Zn, resulting in Cr(II) and Zn(II). Hydrogen is just a side-product, which is not participating in the reaction. Write the conditions over the arrow.

Alternatively, write the aquated forms as a complex. Your current equation has incorrect formula of zinc salt. Check what it should be!


Reason behind conversion of acid to acid bromide in Hell-Volhard-Zelinsky reaction

The $\alpha$ halogenation occurs through the enol form.

The formation of the acyl halide is key because carboxylic acids do not form enols readily since carboxylic acid proton is removed before the $\alpha$ hydrogen. Acyl halides lack the carboxylic acid hydrogen.

The phosphorus halides form the acyl halides readily because phosphorus has a huge affinity for oxygen.
Sometimes, phosphorus tribromide is not enough to form the acyl halide but, since there is also bromine in the media, phosphorus pentabromide is formed which can form acyl halides even of solid acids.

The halogenation is performed by the $\ce{X2}$, either bromine or chlorine.
When $\ce{PBr3}$ and $\ce{Cl2}$ is used, the alpha-chloro acid is obtained and not the bromo acid.


Why are noble gases used in "neon" lamps

Strictly speaking "neon" lamps contain neon but the term is often used colloquially for a whole range of coloured lighting probably because the red neon tubes are one of the commonest.

But it isn't just noble gases that are used, though they are the most common. Carbon dioxide is sometimes used. And there are many discharge lamp that add metals or metal salts, though these are more common in other uses such as high intensity street lighting.

The noble gases and mixtures of them are commonly used in coloured advertising lighting because they offer a range of colours in a simple low-pressure discharge lamp without the added complexity of fluorescent coatings (as in common fluorescent tubes) or high-pressure systems (as is required in high intensity systems used in street or industrial lighting). Advertising signage often consists of long glass tubes in custom shapes and the simplicity of simple gas discharge makes the results cheaper than more sophisticated alternatives.

The colours are derived from the energy levels of the electrons in the gases and it just so happens that the range of those energy levels in noble gases correspond to a useful variety of colours.

There is a good summary of the colours (and other types of discharge lamps) in this Wiki article.


To which block of the periodic table does helium belong?

Blocks in the periodic table should not be mixed up with groups (like noble gases). The reason why Helium is considered a noble gas is because its outermost (and at the same time only) shell is fully occupied by its 2 electrons. When you look at the electron configurations in the PTE you can see that the first element which happens to have an occupied p-orbital is boron. Due to this fact boron also has to be the first p-block element, since the block definition is as follows:

  An element belongs to the block x, if the electron, which is added in comparison to the previous element (by atomic number), is in an orbital of type x in the ground-state electron configuration of that element.

Or the fast explanation: Why should helium be considered a p-block element, if it doesn't even have any electron in a p-orbital or any shell including a p-orbital?


exchanging H2O in buffer for D2O - pH vs pD?

The articles by Glasoe,[1] Covington et. al.,[2] and (much more recently) Krężel and Bal[3] suggest in their intros that pD values can be obtained from a pH calibrated electrode by adding a correction constant of about $0.4$. This was determined by comparing the measured pH when the same amount of acid or base was dissolved in $\ce{H2O}$ and $\ce{D2O}$.
References:

P. F. Glasoe, Citation Classic - use of glass electrodes to measure acidities in deuterium-oxide. Current Contents/ Physical Chemical & Earth Sciences 1979, 16, 12-12. A1979HZ34800001 (pdf) Original paper: P. K. Glasoe, F. A. Long, J. Phys. Chem. 1960, 64,:188-190. Doi:10.1021/j100830a521
A. K. Covington, M. Paabo, R. A. Robinson, R. G. Bates, Anal. Chem. 1968, 40 (4), 700–706. DOI: 10.1021/ac60260a013
A. Krȩżel, W. Bal, J. Inorg. Biochem. 2004, 98 (1), 161-166. DOI: 10.1016/j.jinorgbio.2003.10.001 Available at researchgate.net.


Why is DMSO more polar than other solvents, yet it dissolves ionic compounds worse?

Having large dipole moments and dielectric constants, as DMSO does, does not imply that it has the ability to solvate ionic compounds like $\ce{CuCl2}$.

Although both methanol and DMSO have oxygen atoms with their lone pair of electrons to help solvate cations, there is no corresponding functionality in DMSO to solvate a small, inorganic anion like chloride.  Methanol on the other hand can solvate chloride via hydrogen bonding in the same way that water does.

These properties are not necessarily applicable to a coordination compound, so there is no reason to expect methanol to act as a better solvent than DMSO in this case.


Here's how I would approach it. I would start by making some (over)simplifying assumptions:

1) O $p_x$ orbitals are mostly nonbonding, so I'm going to ignore interactions of those orbitals with carbon orbitals

2) Due to $sp$ mixing, the carbon $p_y$ and $s$ orbitals can be merged into $sp$ orbitals aligned on the $y$ axis.

3) Same for the O $p_y$ and $s$ orbitals. In this case, the mixed $sp$ orbital with its large lobe oriented away from C is mostly nonbonding, so I'll ignore interactions of it with carbon orbitals

With those assumptions in place, I focus on building and ordering the occupied molecular orbitals. The highest energy occupied orbitals will be some of those oxygen-based nonbonding orbitals, so let's start by putting all eight of them at the top of the diagram. The relative energy levels among them will be affected by the number of positive and negative interactions, so the highest energy one will likely be the $b_{1g}$ where all of the $p_x$ orbitals have anti-bonding interactions with their neighbors. I mention this orbital in particular because it is the HOMO, and I think that the $\sigma$ interaction between oxalate and a metal center would come from the interaction of this orbital with the metal $d_{x^2-y^2}$ or $d_{xy}$, for example.

Mixed in with the $p_x$ and $sp_y$ nonbonding orbitals will be the two $\pi_{nb}$ orbitals, also centered on the oxygens. These are the $b_{2g}$ and $a_u$ orbitals in your set above. Since there isn't a carbon $\pi$ orbital with the right symmetry to interact with either of these, the resulting MO's have a nodal plane aligned with the C-C bond. These are the orbitals that will have $\pi$ interactions with a metal center if the oxalate is acting as bidentate ligand.

Below the nonbonding $\pi$ orbitals, we have the two $\pi$ bonding orbitals formed from the $b_{1u}$ and $b_{3g}$ central and oxygen $p_z$ orbital combinations. The $b_{3g}$ is higher energy of the two because of the C=C $\pi^*$ interaction.

At the bottom of the occupied orbitals, we have the five primarily $\sigma$ orbitals. I think the highest energy of the group is the $\sigma_{C-C}$ formed from the mixed $sp_y$ carbon orbitals oriented towards each other, ie a mix of the two $a_g$ central orbitals. Since the density is primarily between the carbons, I'm going to ignore the interactions with the O orbitals.

The four $\sigma_{C-O}$ orbitals come from the pairings of the central $b_{2u}$, $b_{1g}$, $b_{3u}$ and $a_g$ orbitals with their ligand orbital counterparts. In energy, from low to high I would guess $a_g$ ($\sigma_{C-O}$ and $\sigma_{C-C}$), $b_{3u}$ ($\sigma_{C-O}$ and $\pi_{C-C}$), $b_{1g}$ ($\sigma_{C-O}$ and $\pi^*_{C-C}$), then $b_{2u}$ ($\sigma_{C-O}$ and $\sigma^*_{C-C}$).


Why do aromatic hydrogen atoms have a higher chemical shift value compared to aliphatic hydrogen atoms?

The chemical shift gives you information about how well shielded the nuclei are from the magnetic field. A proton at higher chemical shift values is deshielded, so the aromatic protons are obviously less shielded than aliphatic protons.

One effect that causes deshielding is the presense of electronegative atoms that draw electrons away from other atoms and thereby deshield them. But that is not the cause of the aromatic chemical shift.

The reason for that one is the ring current that is induced in a magnetic field. The induced current creates a local magnetic field that has the same direction as the external field outside the aromatic ring where the attached protons are, leading to the deshielding of the nuclei.


How can I mask the Chromium (III) and Chromium (VI) ions in my solution?

If I summarize, you want to make an EDTA titration of $\ce{Al^{3+}}$ in a solution containing also $\ce{Cr^{3+}}$ and $\ce{Cr^{6+}}$. The $\ce{Cr^{6+}}$ ions are usually in the form $\ce{CrO4^{2-}}$ and they don't interfere with $\ce{Al^{3+}}$ in EDTA titrations. But they can be eliminated by precipitation. The best thing to do is to oxidize $\ce{Cr^{3+}}$ into $\ce{CrO4^{2-}}$. In the old Treadwell's book, they propose to take $5$ mL $\ce{Cr^{3+}}$ solution, add $0.5$ mL of a saturated solution of $\ce{Br_2}$ + $0.5$ mL $\ce{KOH}$ $4$ mol/L. The reaction is finished at room temperature in $5$ minutes, and $\ce{Al^{3+}}$ ions do not interfere. $$\ce{2 Cr^{3+} + 3 Br2 + 16 OH^- -> 2 CrO4^{2-} + 6 Br^- + 8 H2O}$$ At the end of the reaction, a solution of barium chloride $\ce{BaCl2}$ or baryum hydroxide $\ce{Ba(OH)2}$ is slowly added to get rid of the chromate ions, by the reaction $$\ce{Ba^{2+} + 2 CrO4^{2-} -> BaCrO4(s)}$$ After filtration of the insoluble $\ce{BaCrO4}$, the obtained solution does not contain any more chromium, it can be acidified and the $\ce{Al^{3+}}$ ions can be titrated by EDTA.


Everyone says something different

Well, actually they don't. A ketone has a nucleophilic and an electrophilic center. The carbonyl carbon is electrophilic and can be attacked by nucleophiles, the oxygen itself can act as a nucleophile and can, for example, be protonated. And yes, if we protonate the oxygen the "ketone", or better the carbonyl carbon, get's more electrophilic.

An example would be the Luche reduction. In this case the carbonyl oxygen coordinates to cerium chloride, thus acting as nucleophile, and makes the carbonyl carbon even more reactive towards nucleophiles.


What is the reason of explosion of batteries?

The most common cause for a battery to explode is the contents of it heating to a point where it exerts a force on the battery too large for it to handle. Basically, if the battery feels very hot, it could be under large pressure and the object that it is powering should be powered off if possible. The recent reports of smartphones catching fire are due to a manufacturing error by Samsung on the Galaxy Note 7. A portion of these phones were not insulated correctly and it caused a short circuit that ended up burning holes through the phones. This was not an explosion, but the phones did catch on fire. In conclusion, you can use your phone while charging if it's not a Galaxy Note 7 (which has been recalled).


Biomolecular energy functions and entropy

I am sure that a more definitive answer could be provided by someone more familiar with the field, but here are a couple of observations.

First, I think it is a bit of an exaggeration to say that protein folding is almost entirely the result of entropy because the protein will have larger entropy in the unfolded state than in the folded state. The main entropic increase on protein folding will be the entropy of the solvent (and the system solvent + protein is what matters thermodynamically). This is the so-called hydrophobic effect which is what you're referring to by saying folding buries the hydrophobic residues.

Notice, however, that an energy function which contains solvent-solvent, solvent-residue, and residue-residue interactions implicitly contains this entropic effect. This comes by maximizing solvent-solvent and solvent-residue interactions. In order for this to happen, the hydrophobic residues will naturally become buried because this will increase the average number of hydrogen bonds between solvent molecules (I'm assuming the solvent is water) and strengthen the solvent-residue interactions at the exterior of the protein.

Second, the reason an energy-only model of protein folding might get "lucky" by ignoring entropy is that even if the net enthalpy change for protein folding is roughly zero, this does not mean that energy did not play a significant role in the folding process. People often talk about how proteins cannot simply be randomly walking the conformation space because it would take an endless amount of time to find the energy minimum by doing this. This is exactly why intermolecular interactions between residues get discussed so frequently.

So, an energy-only approach will very likely identify the residue-residue interactions which are important structural features of the final folded state. It is entirely possible that there are steps which are energetically destabilizing, but when these steps take place (if they are to take place in the real world), there must be a very constrained conformation space for the step to take place by randomly sampling states. Nonetheless, each step in the protein folding process must hit an energy minimum at some point, so just searching the energetics is a perfectly reasonable first approach. Furthermore, when this approach fails, it tells you that whatever is going on must be quite sensitive to the entropy of the system, so it is not like you are flying blind.

Third, it could easily be the case that energy-only approaches sometimes arrive at the correct structures completely accidentally. That is, there is more than one path from the unfolded state (which unfolded state?) to the folded state. The "real" path, if there is a single path that proteins take, might be quite different from the one that a model like this will take. In fact, if the protein is being sort of randomly perturbed and energy minima are being sought out, the pathway is likely complete nonsense, but this doesn't really matter as long the as protein is folding. It would be interesting to see how models like this do with proteins that have very energetically similar but distinct folded states. If what I'm saying in this part is true, one would expect the model randomly ends up in either state when you run the simulation. This would indicate that in the real world, there is a kinetic effect which forces the folding to follow a certain path and the model does not capture this.


Liquid phosphorus is white phosphorus. It melts at 44.1 °C.
Red phosphorus does not melt. It burns at 200 °C and sublimes in an argon atmosphere at 280 °C, without melting.


What are the side effects of adding dry ice to a swimming pool?

Solubility of $\ce{CO2}$ in water is $\pu{1.45 g/L}$ at SATP. Apparent dissociation constant for carbonic acid at the same conditions is $K_\text{a}(\ce{H2CO3}) = \pu{4.47e−7 mol L^{-1}}$.

Now lets speculate how much dry ice could be disposed and what the pool capacity might be, guessing that the water warmed back up. It's hard to tell which pool size your apartment complex (hopefully not this one) might have, so I'm going to propose some values (marked in $\mathrm{\color{red}{red}}$) which look plausible to me.

Assuming $m(\ce{CO2}) = \color{red}{\pu{10 kg}}$, and the pool volume of approx. $V = \color{red}{\pu{5e5 L}}$ (olympic pool is $\pu{2.5e6 L}$, I took $\frac{1}{50}$ of that), one notices that in order to completely dissolve $\pu{10 kg}$ of $\ce{CO2}$, $\pu{6.9e3 L}$ of water is required (based on solubility data), which means that we have more than enough water in the pool to completely dissolve all $\ce{CO2}$.

For the weak acid

$$[\ce{H+}] \approx \sqrt{K_\text{a} C} = \sqrt{\frac{K_\text{a} m}{M V}} = \sqrt{\frac{\pu{4.47e−7 mol L^{-1}} \cdot \pu{e4 g}}{\pu{44 g mol ^{-1} \cdot \pu{5e5 L} }}} = \pu{1.43e-5 mol L^{-1}}$$

so the pool water will indeed be acidified to the level of

$$\mathrm{pH} = -\log[\ce{H+}] = 4.85,$$

assuming the values I've willfully established, and that all carbon dioxide completely dissolved in water. You can put your values into the equation and argue with your landlord further.


Double surface in the pressure temperature concentration curve for a binary mixture

The liquid phase is above the surface. The vapor phase is below the surface.

The planar cut at a constant temperature is easiest to understand. The bottom curve will show the composition of the liquid phase as a function of molar fraction, and the upper curve will show the composition of the vapor phase.

Likewise for the curves of molar fraction versus temperature at a given pressure.


How does Potassium Hexfluoronickelate(IV) exist in a stable state?

To quote from the comments: you can get things into an amazing variety of oxidation state given the right environment.
In the case of maximal oxidation states, particularly above $+2$ or $+3$ in the $3d$ transition series, the right environment (at least in simple species) involves complexation with ligands that are highly electronegative, capable of pulling electrons even from a strongly positive central atom, and resistant to oxidation themselves in an electron-poor setting.  Meaning, in practice, oxide or fluoride ligands.
We can therefore roughly classify high oxidation state complexes into oxo and fluoro complexes.  Oxo complexes are more likely to be seen with early transition metals, where the central metal ion contributes few $d$ electrons and there are a lot of vacant orbitals with which the strongly $\pi$-donating oxygen can form back-bonds.  The best known examples being chromate/dichromate ($\ce{Cr(VI)}$), manganate and permanganate ($\ce{Mn(VI), Mn(VII)}$), and ferrate (most commonly, $\ce{Fe(VI)}$).  Even cobalt reaches its apparent maximum $+5$ oxidation state with an oxide-ligand species, $\ce{[CoO4]^{3-}}$.
Later transition metals, which have lower maximum oxidation states due to increased electronegativity/ionization energy, have more $d$ electrons coming from the cationic shells and thus are no longer so stabilized by $\pi$ back-donation.  Then the fluoro complexes, with a relatively $\pi$-inactive ligand, take over.  In the $3d$ series cobalt, cited above as forming an oxide complex with the formula $\ce{[CoO4]^{3-}}$, also forms a +4 fluoride complex $\ce{[CoF6]^{2-}}$(1).  Fluoride complexes are also seen with both nickel and copper in oxidation states above $+2$, including the $\ce{[NiF6]^{2-}}$ ion mentioned in this question and the copper(III) species $\ce{[CuF6]^{3-}}$.  Zinc in the $3d$ series is limited to oxidation states of $+2$ or less, but if mercury has a tetrafluoride $\ce{HgF4}$ (disputed), it's no coincidence that this too is a fluoride.
Reference

J.W. Quail and G.A. Rivett, "Fluoride Complexes of Tetravalent
Cobalt",  Can J Chem 50:2447 (1972).


I have a theory for the question.

First lets look at what the graph is telling us about the radius and the atomic number. For the Noble gases, as the atomic number is increasing the radius increases. It is because of the addition of more electron shells (simplification for now).

Helium has a full 1s orbital. Whereas Neon has a configuration of $1s^2,2s^2,2p^6$. Therefore the electronstatic force occurring on the electrons is slightly bigger than the electron on the helium atoms. Whereas Argon has a configuration of $[Ne]3s^23p^6$.

Now my theory. In He, we only had $1s^2$ and in Neon we added the extra shell and we went up til the 2p. Now the shells also act as barriers sometimes, therefore we have something called an effective nuclear charge. Now, since He has 1s2 and we added an extra shell, well the $1s2$ does not act much of a barrier and the nucleus is still able to attract the electrons in the second shell ($2s^2 2p^6$) electrons and the radii does not change drastically. But when you add a full shell on top of that, well then we have a problem. Because then the second shell electrons can act as a barrier in the electrostatic force, and the effective nuclear charge decreases and the repulsion between the electrons in the third shell will be greater, hence a greater atomic radii occurs between Ne and Ar, compared to He and Ne.

Now this trend will not be seen in the other ones. It is because the electron configuration is one thing but is not the only thing that is being shown on the graph. All of the other ones are ions. When something is an ion, for example the positive ones, then there are two things to look at, electron configuration and the charge. Meaning in a positive ion, you have more protons than electrons, meaning that the electrostatic force will much more greater in $Li^+$ than the one in $He$, even though they have the same electron configuration.

Whereas the trend in the noble gases is based on the fact that they all have $e^- = p^+$ and are neutral, whereas the others are not.

I hope I make sense here. But like I said, its just my theory for the question and I can be wrong :)


Predicting color of a TMB derivative affected by peroxidase

Your question: How could one predict from the structure of the one product that it is going to have a blue color in that situation?

This is a very broad question. OP has to go to chemistry fundamentals to learn, how one can predict the absorbance wavelength of a compound by analyzing the structure of the given compound. One easy subject for this study for OP is basic fundamentals of UV-visible spectroscopy in organic chemistry. For example, how absorbance wavelength increase with each additional conjugated double bond (Absorption in the ultraviolet and visible regions is related to the transition of electrons). Look following table:

$$
\begin{array}{c|lcr}
\text{Substance} & \text{Absorption Peak }(\lambda_{max}) & \text{Molar Absorption Coefficient }(\epsilon) \\\hline
\ce{CH2=CH2} & \pu{180nm} & 10000  \\
\ce{CH2=CH-CH=CH2} & \pu{217nm} & 21000  \\
\text{Vitamin A (5 conjugate double bonds)} & \pu{328nm} & 51000  \\
\text{Benzene} & \pu{255nm} & 180  \\
\text{Naphthalene} & \pu{286nm} & 360  \\
\text{Anthracene} & \pu{325nm} & 7100  \\
\hline
\end{array}
$$

Thus, one can suggest diimine, 4-(4-imino-3,5-dimethylcyclohexa-2,5-dien-1-yl)-2,6-dimethylcyclohexa-2,5-dien-1-imine, which contains 5 directly conjugated double bonds (two of them are imines), may absorb wavelengths of lights way larger than $\pu{328  nm}$ given for Vitamin A. That's because it involves transition of $n \rightarrow \pi^*$ in addition to $\pi \rightarrow \pi^*$ in Vitamin A (e.g., $\pi \rightarrow \pi^*$ in acetone is $\pu{195 nm}$ in addition to $n \rightarrow \pi^*$ transition at $\pu{274 nm}$, when compared to $\pi \rightarrow \pi^*$ transition of ethylene at $\pu{180 nm}$). Experimental evidence has shown that 4-(4-imino-3,5-dimethylcyclohexa-2,5-dien-1-yl)-2,6-dimethylcyclohexa-2,5-dien-1-imine absorbs light at $\pu{450 nm}$, giving its characteristic complimentary color $\color{blue}{\text{Blue}}$ (Ref.1).

References:

P. D. Josephy, T. Eling, R. P. Mason, “The horseradish peroxidase-catalyzed oxidation of 3,5,3',5'-tetramethylbenzidine. Free radical and charge-transfer complex intermediates,” The Journal of Biological Chemistry 1982, 257, 3669-3675 (http://www.jbc.org/content/257/7/3669).
For basic learning, read: Master Organic Chemistry.


Fajans' Rule is  used to predict whether a chemical bond will be covalent or ionic.

Polarization will be increased by:

1) High charge and small size of the cation

2) Ionic potential

3) High charge and large size of the anion

4) The polarizability of an anion is related to the deformability of its                                            electron cloud (i.e. its "softness")

5) An incomplete valence shell electron configuration

In pentahalides, the oxidation state is +5 and in trihalides, the oxidation state is +3. Since the metal ion with a high charge has more polarizing power, pentahalides are more covalent than trihalides.


There are many types of magnetic properties, including ferromagnetism, paramagnetism, diamagnetism, antiferromagnetism, ferrimagnetism, superparamagnetism, metamagnetism, spin glasses, and helimagnetism.  Many of these are too weak to cause any noticeable interaction with a magnet. The type of everyday magnetism you're thinking of, which nickel has, is ferromagnetism.

While nickel is ferromagnetic,  copper is not. As you said, the American nickel is currently 25% nickel and 75% copper.  According to this paper (from 1931!), in order for a nickel-copper alloy to be ferrogmagnetic, it must contain at least 56% nickel:

  ... 56 percent nickel is required before the alloy shows ferromagnetic properties at ordinary temperatures.

https://journals.aps.org/pr/abstract/10.1103/PhysRev.38.828

E. H. Williams, Magnetic Properties of Copper-Nickel Alloys. Phys. Rev. 38, 828 (1931).


Why doesn't the N-Br bond dissociate quickly in NBS?

Actually, the sequence of reactions leading to a low but constant concentration of $\ce{Br^.}$ radicals is very simple:

$$\begin{align}\ce{R2N-Br &<<=> R2N^. + Br^.}\tag{1}\\[0.4em]
\ce{Br^. + Br-NR2 &<=> Br2 + R2N^.}\tag{2}\\[0.4em]
\ce{Br2 &<<=> 2 Br^.}\tag{3}\end{align}$$

The $\ce{N-Br}$ bond is very weak and can easily dissociate homolytically to give both bromine radicals and succinimide radicals. Bromine radicals can combine with other NBS molecules to form free bromine and another succinimide radical and bromine molecules can dissociate to give bromine radicals.

All of these steps are reversible and the equilibrium favours the non-radicals side. Whether your reaction needs a constant supply of bromine or a constant supply of radicals doesn’t matter as NBS can supply them.

Only if there is already some bromide in solution can a fourth reaction take place:

$$\ce{R2N-Br + Br- -> R2N- + Br2}\tag{4}$$

This is the only reaction that has a dedicated forward-direction. If bromide is present, bromine molecules can be generated much more rapidly than radicals, so the generation of radicals will have to come from bromine dissociation.

While it is true that the NBS radical is somewhat well stabilised, that does not make it that much easier for the $\ce{N-Br}$ bond to dissociate: radicals are still radicals and two-electron-two-centre bonds are still energetically favoured. The $\mathrm{S_N2}$ displacement is much better since an anion is equally well stabilised but has a  lower general energy.


Colour change when excess ammonia added to [Cu(H2O)6]2+

Like the OP mentioned, the replacememt of a weak field ligand ($\ce{H2O}$) by a stronger one ($\ce{NH3}$) certainly increases the $\Delta_o$ of the $\ce{Cu(II)}$ complex.

A chemical sample appears colored when it absorbs a certain wavelength of light. What we see is the sum total of the remaining colors. The colour visible to the human eye would be the complement of the colour(i.e. wavelength) absorbed.

The increase in $\Delta_o$ leads to the absorption of a lower wavelength (or higher frequency) of light, as evident from:
$$E=h\nu= \frac {hc}\lambda $$

Note that a higher frequency would mean a shift towards the blue end.

Quantitatively speaking, the light blue colour of the initial hexaaqua complex was due to the absorption of the orange spectrum range of light.

Back to the color wheel, a shift towards the blue side (in the green direction, please) gives us a lighter shade of orange.

  Why does the solution turn deep blue?

Observe the complement(the colour diametrically opposite) of orange-yellow is definitely a deeper shade of blue.


The mass of electron is never given in question you should remember it. Mass of electron is $9.109 \times 10^{-31} \mathrm{kg}$.

Now read the question Kinetic Energy is given, so from that you can easily find velocity of electron.

$$K.E.=\frac{1}{2}mv^2$$

so You will get $v= 6 \times10^6 \mathrm{m/s}$

Now you can substitute value of m and v in formula $\lambda = \frac{h}{mv}$

You will get wave length of electron as $0.12 \times 10^{-9} \mathrm{m}= 0.12 \mathrm{nm}$


Yes, this is correct. Assuming that you are talking about solubility in water, it is discussed by JF Liebman, Structural Chemistry, 15, 165 (2004) which needs a subscription to read in full, but much of the key information is given on the first couple of pages which are available as a preview. The relevant thermodynamic quantities (in kJ mol$^{-1}$) for dissolution of one mole in water are given as a table, with AgCl compared with NaCl and KCl:

  $$ \begin{array}{cccc}  & \Delta H & \Delta G & T\Delta S \\ \hline \text{NaCl} & +4 & -9 & +13 \\ \text{KCl} & +17 & -5 & +22 \\ \text{AgCl} & +45 & +37 & +8 \end{array} $$

The author got the original data from NBS Tables of Chemical Thermodynamic Properties.
All the entropy terms are favourable.
All the enthalpy terms are unfavourable, but the AgCl has a much more unfavourable enthalpy of solution than the alkali metal chlorides, and the
entropy cannot compensate for this. The Ag$^+$ ion is poorly hydrated in solution
compared with Na$^+$ and K$^+$.


Your question will probably get closed because any answer is going to be an opinion. Instead of recommending a book-I'll shed a little light on what your journey will be like.

The Subject(s)

Your pursuit to describe high level biological functionalities using first principals covers many fields. I don't think you'll find a single book that explains it all. Instead, you're going to have to learn from a number of different sources.

Physical Chemistry - This will give an overview of electronic structure using quantum mechanics
Physical Organic Chemistry - Learn how to apply physical chemistry to organic molecules.
Bio-Chemistry - Apply organic chemistry to biological processes.
Physical Bio-Chemistry Combines physical chemistry with bio chemistry.

You're really after point 4, but it will need to learn a fair amount before getting there.

The Reality

The reality of things is that we don't need to dig down to first principals to describe many biological processes. Chemistry is able to abstract enough of the nitty-gritty away to come up with easier & faster high level concepts.

Take programming for example, web developers don't need to write websites in assembly language because someone has gone ahead and abstracted it all into a nice user friendly language. You could spend a lifetime coding a site like stackexchange in assembly-or you could use a higher level language and do it in a few months.

Just how you won't find a book on website design using assembly language (first principals), you probably won't be finding one describing biology using pure quantum mechanics.

Further Steps

Assess your current knowledge of physical and (general) chemistry. Once you know which level you're at, go to different university sites and check which books the professors are using to teach. If you find a course that is covering the topics that you're interested in-contact the TA or professor and ask for recommendations; I'm sure they'd be happy to help.


Total spin number 0 together with $B_2$ state - confusion

We can see what a $^1B_2$ will look like by consider the MO diagram for water, a $C_{2v}$ molecule with $10$ electrons.

$\hspace{9ex}$

This is the configuration that we might expect for the ground state, which has $^1A_1$ symmetry. We could get a configuration with $^1B_2$ in a couple ways, namely by exciting form $1b_2$ into $4a_1$ or by exciting from either $2a_1$ or $3a_1$ into $2b_2$ without changing the spin of whatever electron we move. These new configurations would still be singlets because we haven't changed any of the spins and would have $B_2$ symmetry since clearly $B_2\otimes A_1=B_2$.

As to your question about spin multiplicity, if we consider a system with a single electron, we account for both the spin up and spin down configuration by determining the spin multiplicity. The spin multiplicity for $S=\frac{1}{2}$ is $2S+1=2$, a doublet, which gives us exactly the number of spins states we would expect.


Let's start with the question: How do we get from your answer to the official answer?  We can transform your answer into the official answer by first adding 3 ammonium ions to each side:

$\begin{aligned}\ce{H2O} + \ce{NH4+} + 2\ce{S^{2-}} + \ce{O2} & \rightarrow \ce{NH3} + 2\ce{S} + 3\ce{OH-} \\
+ 3 \ce{NH3}+3\ce{H^+} & \rightarrow + 3 \ce{NH3}+3\ce{H^+}\\
\ce{H2O} + 4\ce{NH4+} + 2\ce{S^{2-}} + \ce{O2} & \rightarrow 4\ce{NH3} +3\ce{H^+} + 2\ce{S} + 3\ce{OH-}
\end{aligned}$

Then noting that the $\ce{H+}$ and $\ce{OH-}$ will 'neutralize' one another on the product side, producing $\ce{3H2O}$, one of which will 'cancel' the water molecule on the reagent side:

$\begin{aligned}
\ce{H2O} + 4\ce{NH4+} + 2\ce{S^{2-}} + \ce{O2} & \rightarrow 4\ce{NH3} + 2\ce{S} + 3\ce{H2O} \\
-\ce{H2O} & \rightarrow - \ce{H2O}\\
4\ce{NH4+}+2\ce{S^2-}+\ce{O2} & \rightarrow 4\ce{NH3}+2\ce{S}+2\ce{H2O}
\end{aligned}$

Producing what is the 'official' result.  Now we can ask: are the two answers equivalent?  Because $\ce{NH4+}$ is a weak acid in water, it will dissociate and we can write your answer as:

$\ce{H2O + NH3 + H+ +2S^2- + O2} \rightarrow \ce{NH3 + 2S + 3OH-}$

Your answer is a balanced redox reaction somewhere in between acidic and basic conditions, which is a rather unusual formulation.  I'm not going to go so far as to say, this can't happen, but it is unlikely, and certainly not how we typically describe balanced redox reactions.


Why does the gas from dry ice flow down?

The gas flows downwards because it is more dense. It is more dense both because $\ce{CO2}$ is heavier than $\ce{N2}$ and $\ce{O2}$ and because the gas is colder than the ambient air.

The mist is water in the surrounding air condensing into microscopic droplets due to the lowered temperature.  The amount of mist depends on the humidity of the air; in perfectly dry air there will be no mist.

All of the gases involved are colourless. The mist appears white because it is in small particles and so it scatters any light falling on it.  It is the same colour as "natural" mist and for the same reason.


Crossing at critical temperature of a gas

At the critical temperature/pressure, all properties of liquid and gaseous phase  mutually converge to the same values. There is just one phase at these temperatures, neither liquid, neither gas. Or, both at the same time.
Changing  pressure at temperature above the critical temperature causes continuous changes of the supercritical phase properties without the edge of the gas/liquid phase transition.
Changing  temperature at pressure above the critical pressure causes continuous changes of the supercritical phase properties without the edge of the gas/liquid phase transition.


What happens to the gas inside cold drinks as the cold drink is frozen?

Well, you are certainly right in that the solubility of the gas will grow if you decrease the temperature, and so it will continue all the way down to the freezing point. That's when things get hairy.

Ice is quite different from liquid water. It does not dissolve any $\ce{CO2}$ to speak of.(*) So pretty much all $\ce{CO2}$ will leave the solution and go to the gas phase, adding its pressure to that of ice in an effort to tear the bottle open. (Some plastic bottles manage to withstand it, some don't.) The dissolved gas and whatever little salts are there will lower the freezing point of water and in fact make it an interval rather than a point, but that's another story.

After thawing, much of $\ce{CO2}$ will happily go back to the solution (though not instantly), and the drink will be more or less restored to its original state.

(*) For any physical quantity, there are layers upon layers of scientists from different fields, each confined to a certain range of the said quantity and calling everything above it "infinity" and everything below it "zero". The levels of dissolved $\ce{CO2}$ in the ice core samples from lake Vostok are quite a big deal for paleoclimatology. Not my field, though.


Can Substances on Titan be Reacted at a Thermodynamic Gain?

The thermodynamic problem is not so much with the hydrocarbons reacting with O2 as it is with forming O2 from water.  That reaction is energetically uphill.  A big hill.  In fact water is a waste product of your second reaction.  The reason the second reaction is energetically favorable is due to the high stability of CO2 and water relative to the reactants, methane and oxygen.

Would it make sense that you should be able to then convert the water from your second reaction into O2 and H2 in a way that would consume less energy than it took to make the water in the first place?  That is almost equivalent to what you are proposing by starting out with the water already present on Titan.

The prohibitive energetics of your reactions apply on Earth as well as Titan.  You need a process to split the water to burn with hydrocarbons that will consume less energy than the energy you will get back from burning the hydrocarbons.  You could conceive of using solar power as the energy source for converting water into O2 and H2.  Or other energy sources.  But that wasn't your question, and you definitely won't get the energy back that you would require via your second reaction.  And good luck with my solar energy idea on Titan ;)


Can a molecule contain a linear series of conjugated pi bonds twisted through 180 degrees?

The material you imagine has been a target for Natta, best known together with Ziegler for they work on isotactic polymers.

Natta was interested in what you describe - polyacetylene - as the extended conjugation would have likely led to unusual electronic properties.

Calculations and models were around theoretically, before and in  parallel to the actual synthesis.

The material prepared by Natta was however insoluble and basically no work up was possible.  This is not totally surprising if we look at the structure.

After decades, the fortuitous use of thousand time the amount of catalyst led to the formation of nice shiny films of polyacetylene. Still insoluble but so different from commons residues to attract sufficient attention to cause the gathering of brilliant scientists and finally the emergence of an entire field, that of conducting (or better, conjugated) polymers and so called plastic electronics.

https://www.nobelprize.org/nobel_prizes/chemistry/laureates/2000/

https://en.m.wikipedia.org/wiki/Polyacetylene

This is to answer the final part of the question.

Concernig the first part, if I understand correctly your picture, I can tentatively answer in an apparently convoluted way.

As far a chain of alternating single and double bonds is considered, and without accounting for delocalisation, we have basically free rotation around the single bound. As such, any statistical coil can be obtained.

Conversely, if one imagine a "metal like" structure with full electron delocalisation, the chain is than stiff.

The reality is somewhere in the middle. Real conjugated polymers should be regarded as delocalized pi systems (which at the end is the reason of the huge interest on them) inerhently not much flexible and not very soluble.

Still, a long chain is susceptible to adopt a variety of elical or coiled conformations as a result of collective little departures from planarity. Ideally, you can turn very rigid rod on themselves providing they are long enough.

One cannot apply Flory theory as in polyethylene, but adaptations exist to account for rigidity due to order bond > 1 in conjugated polymers.

Also let me note that independent of the extent of delocalisation (also known as effective conjugation) a linear series of sp2 hybridized C atoms can close to a ring - no need for examples here).

Additional notes:

1- Solubility in common solvents is a major challenge towards real  application of conjugated polymers. A common strategy is the side substitution with moderately long alkyl chains. This in generak results in improved solubility via an extra entropy term.

2- polyanilyne far preceded the interest on polyacetylene. This systems is more complicated and in a way interesting, but less as far organic electronics is in scope.

3*- a detailed and chemically exhaustive paper appeared well before the paper mentioned in the linked Nobel prize. It has been published in the Journal of the  Australian chemical society and it details the preparation and properties of polythiophene.

*(Making products and selling them are two different tasks. This is the lecture I got)


Why some reaction have no transition state?

There is nothing special in having an activation barrier between reactants and products although, obviously, very many reactions do. But some reactions, such as electron transfer, are observed experimentally to have no barrier which happens when the product potential crosses that of the reactant at its minimum (i.e. between the 'normal' and 'inverted' regions).  In fact  electron transfer is an interesting case as it is possible to control the barrier height by changing the redox potential. Some atom diatom reactions such as H+F2 also have very very small barriers.  Whether you want to call the crossing point a transition state when it has a zero barrier is a moot point. The behaviour is more like water going over a waterfall, so limited by the transport of the wavepacket along the reaction path, rather than sometimes being reflected at the barrier and sometimes crossing.


Role of the counter/auxiliary electrode in cyclic voltammetry

You are right that the counter electrode must balance the redox reactions happening at the working electrode.
For every electron that goes to reduce an $\ce{Fe^3+}$ ion at the working electrode, an electron must be collected from an oxidation reaction at the counter electrode.

What that reaction actually is depends on a lot of things.
First, we must consider what is actually in the solution to oxidize at the counter electrode.
In an $\ce{Fe^3+}$ solution, unless the working electrode is very close to the counter electrode, the newly formed $\ce{Fe^2+}$ ions will not diffuse efficiently to the counter electrode, so something else must balance the bulk of the current.
In many aqueous solutions, this might be the water itself being oxidized, unless there is a more easily oxidized substance in the solution (a supporting electrolyte, for instance).
If the counter electrode material is chosen poorly, it may even be the counter electrode itself oxidizing.

In practice, electrochemists often design cells such that the reactions at the counter electrode are unimportant, perhaps by placing the counter electrode in a separate vessel connected with a salt bridge, or at the least assuring that potential counter electrode products will not interfere with the reactions at the working electrode.