How to calculate Absolute error under square root sign?

A slightly different approach to @Zhe's answer is to use the 'propagation of errors' formula. Your formula is $y=\sqrt{x}$  with $x=15.13$ and an error $\sigma_x=0.005$ then the formula becomes $\displaystyle \sigma_y=\frac{\mathrm dy}{\mathrm dx}\sigma_x = \frac{\sigma_x}{2\sqrt{x}}= 6.4\cdot 10^{-4}$.

[if you have a formula in more than one variable , say $x$ and $z$ then $y=f(x,z)$ and the equation is more complicated but still quite easy if done in steps and is $\displaystyle \sigma_y^2=\left(\frac{\mathrm dy}{\mathrm dx} \right)^2_z\sigma^2_x+ \left(\frac{\mathrm dy}{\mathrm dz} \right)^2_x\sigma^2_z$. Notice that the square of the final error is calculated in this case.]


What is the sense of electronic configuration of an atom if an electron can be in any superposition of energy eigenfunctions?

The answer that will satisfy you depend on how familiar you are with the mathematics and principles of quantum chemistry.

So, let's start small. Hydrogen atom. Exact expressions for eigenfunctions for the Hydrogen are known, they are the orbitals we know and love(?). Although the electron can be in any combination of eigenfunctions, it can only stay that way when you are not looking. When you do look, that is, take a measurement, it will collapse to a specific eigenfunction. That eigenfunction could be 1s or it could be 6h, but it turns out that 99.999% of the time, when you measure the energy of a Hydrogen atom, it will display the familiar 1s. That is the sense in which we can say that the electron in a Hydrogen atom is is in orbital 1s, what we really mean is that, typically, for the vast majority of time, the electron of the Hydrogen will be occupying the ground eigenstate.

Now that is for Hydrogen. For heavier elements, you no longer have orbitals, strictly speaking, you have more complicated eigenstates, obtained by solving the Schrodinger Equation to each atom separately. Currently, there is no analytic solution to the equation when you include a second electron. We approximate the ground state of each atom as though the electron were occupying different hydrogen-like orbitals (hydrogen-like because they differ from the pure hydrogen orbitals in a few parameters that are added).

Because electrons cannot occupy the same quantum state (considering spin), the way to do that is to assume that each electron is in a spin orbital. The way to distribute the electrons through the infinitely many hydrogen-like orbitals is to distribute them in the way you have learned, following the Pauli diagram. The resulting electron distribution is an approximation to the true ground state, using hydrogen-like orbitals, because the true ground state is not known.

O hope I made myself understood, but if you need clarification, please ask and I shall edit the answer as much as possible. I have fond memories of the time when I was asking the same questions :)

EDIT 1

You asked why is the probability of finding the electron in 1s orbital highest when we measure its energy?

The doubt comes because from your understanding these probability comes from writing the initial state as a linear combination of eigenbasis of our choice (in this case the energy eigenfunctions) and the initial state can be any function (normalisable and continuous)

Now, while in principle this is correct, in practice that is not so. When you have a vast sample of hydrogen atoms and you measure the energy of each one of them, the vast majority of them will return the ground state for most experimental conditions. That is because experimental setups that actually manage a high degree of superposition with excited states are very hard to prepare.

This means that the experimentally expected state for an electron in a Hydrogen atom is 1s for the vast majority of experimental setups. This is the relevant information. The preparation of an arbitrary superimposed state is not trivial, and they do not occur easily in nature.

This leads us to the approximation using hydrogen-like orbitals and the aufbau principle, according to which the ground state (the most experimentally relevant eigenstate) of an atom can be approximated as though each electron occupied one spin orbital around the nucleus, following the Pauli diagram of energies.

It is not so much a theoretical result as much as a way to make theory and practice agree. There is a way to reach the aufbau principle from first principles, but it involves doing statistical thermodynamics using Fermi Statistics. These calculations would tell you how electrons usually distribute through their available energy levels, and it would become clear that the aufbau principle is pretty sensible, from an experimental standpoint.


No special light is required, although you will want to keep the light source distant to avoid melting your ice. You will also want to experiment with light angles to get the best resolution of the different angled facets of the ice crystal.

Without the right equipment and experience it will be difficult to get an image quite as spectacular as the crystal you've shown. The keys to getting the best image are avoiding heat, from a lamp or the surrounding air, which causes sintering of the finest faceted crystals.

A good way to view ice under a microscope is to place crushed dry ice under a dark covering (depending on your equipment, getting something that fits and still focuses may be your greatest challenge). Whether you use glass or plastic or whatever else isn't very important. The dry ice is cold enough that ice will nucleate on the surface of your choice and grow from there. Be prepared to watch the whole show, from freezing to melting!


What is the Lewis structure of the azide ion?

Your proposed structure is wrong. Nitrogen does not exceed the octet in any of its known compounds (and even if $\ce{NF5}$ will be discovered it will not exceed the octet according to everything we know now). However, if you have a formal negative charge that means an additional electron added to the 5 nitrogen usually has; if four of those six electrons are used to build the double bonds, there is still a lone pair on nitrogen for a total of 10 electrons.

If you are having trouble determining Lewis structures, there are four quick calculations that you can perform to help you:

Add up all valence electrons the atoms are bringing into the compound.
Each nitrogen has five electrons plus there is one negative charge (additional electron) so:

$$3\times5+1=16\tag{1}$$
Add up how many valence electrons would be needed so that each atom has an octet (for hydrogen: dublet) of its own.
Each nitrogen would want eight electrons so:

$$3\times8=24\tag{2}$$
Take $(2)-(1)$. This represents the number of electrons the atoms must share, i.e. the number of bonds.

$$24-16=8\tag{3}$$
Take $(1)-(3)$. This represents the number of electrons that do not have to take part in bonds; these must then be distributed as lone pairs.

$$16-8=8\tag{4}$$

Then, start drawing but make sure you have as many lone pairs and bonding electrons as the equations state. Ignoring the lone pairs, we can get the following possible structures for $\ce{N3-}$:

$$\ce{N#N-N}\qquad\qquad\ce{N=N=N}\qquad\qquad\ce{N-N#N}$$

(The exercise of distributing four lone pairs across the three nitrogens so that each ultimately has eight valence electrons is left to the reader because I am too lazy to open ChemDraw to draw the structures.)

After you have done that, you need to take a look at potential formal charges. For that, split each bond homogenously (i.e. give each atom one of the bonding electrons) and count. Compare that count to what an atom should have; the difference corresponds to the atom’s formal charge. (Because electrons are negative, an additional electron corresponds to a charge of $-1$.) When done for those three structures, we arrive at:

$$\ce{N#\overset{+}{N}-\overset{2-}{N}}\qquad\qquad\ce{\overset{-}{N}=\overset{+}{N}=\overset{-}{N}}\qquad\qquad\ce{\overset{2-}{N}-\overset{+}{N}#N}$$

In each of those cases, the formal charges sum up to the overall charge of the molecular ion ($-1$) which is an indication that we have done it correctly. (Again, I have sheepishly left out the lone pairs; you can use my formal charges to determine where they should have been and how many.)

There is no principle of zero formal charges. However, when debating between different structures, a structure with less formal charges is often (not always!) more ‘favourable’. (The actual term should be ‘contributes more to the overall picture’ but that may confuse too much at this stage.)

But which of the three is correct? They all are! In fact, this is what is known as mesomery: we have a number of (resonance) structures that all explain the actual compound a little bit but neither holds the absolute truth. To show this, resonance arrows are usually drawn between the depictions:

$$\ce{N#\overset{+}{N}-\overset{2-}{N} <-> \overset{-}{N}=\overset{+}{N}=\overset{-}{N} <-> \overset{2-}{N}-\overset{+}{N}#N}$$

The key difference between correct structures and your proposition is that the central nitrogen atom can never carry a negative formal charge as it needs to accomodate four bonds to its neighbours which is only possible for $\ce{N+}$.

As for the answer given in the homework response: It is not strictly correct because it is incomplete. All three structures should be marked as correct – until the concept of resonance has been formally introduced at which point only a combination of the three should be.


Why does the opening of an epoxide occur via an Sn2 like mechanism when using a methanoate ion as a nucleophile and methanol as a solvent?

Quotation from Clayden:

  Without protonation, the epoxide oxygen is a poor leaving group, and leaves only if ‘pushed’
  by a strong nucleophile: the reaction becomes pure SN2. Steric hindrance becomes the controlling
  factor and methoxide attacks only the primary end of the epoxide.


Create combinatorial library of biologically plausible chemical compounds based on library of known compounds

I think your best bet would be to go to the KEGG Reaction Database (http://www.kegg.jp/kegg/reaction/) which is a comprehensive database of known biochemical (enzyme-catalyzed) reactions, plus various associated tools. It includes a module that predicts a likely biosynthetic pathway for any pair of structures. All your pheromones are natural products and are formed via enzymatic reactions, so you should be able to explore out from your starting set by exhaustively applying the reactions in the KEGG database to each one of them. There may be a module that does this for you  automatically, but it is not documented on the KEGG web site.


How do I isolate calcium from calcium carbonate?

The answer is you cannot do it almost for sure.

To obtain $\ce{Ca}$ metal from $\ce{Ca (+II)}$, you need to perform a reduction process. For this purpose, you would need a very strong reducing agent, because calcium is very close to the bottom in the Ellingham diagram (used in metallurgy for this same purpose: obtain metals from its salts).

So you will need to perform an electrolysis of molten calcium chloride, which is not cheap, and by any reason not doable at home or at an standard lab (just a quick view of calcium salts melting point will give you a hint about why is this...)


Difference between relaxation and resonance leading to an absorption spectral feature?

The difference can be viewed in two ways, I guess.

Firstly, for relaxation spectroscopy, you apply a forced oscillation on your sample. That can be mechanical (dynamic rheology, mHz to .1kHz, several kHz with very special, not commerically available equipment) or electrical (dielectric spectroscopy, low mHz to GHz range). The mathematical concepts behind both methods are very similar. For vibrational, MR, UV, etc. spectroscopy, the sample does just not interact with frequencies outside of the linewidth of some QM transition. The width of a resonant peak is given by the energy needed for the transition, and the lifetime of the excited state. The width of a feature in a relaxation spectrum is typically at least half an order of magnitude, from "sample can follow the external stimulus and stay close to a (dynamic) equillibrium" to "external stimulus is much too fast for the sample to follow at all".

The other point is that in relaxation spectroscopy, energy is transferred and dissipated into the sample below a specific frequency (invariably as heat), and reflected (elastically) at higher frequencies. The energy is not quantised, because the induced change in the sample is translatory. The peak frequency itself has no real significance, it's inverse is called a "relaxation time" of the sample. A statistical quantity, mostly. For QM transitions, energy is only transmitted at the given frequency/energy, and nothing happens otherwise, except Rayleigh scattering.

Basically, resonant spectroscopy deals with individual quantum mechanical species, and relaxation spectroscopy deals with the classical, unquantised newtonian properties of the whole sample. NMR is a bit of a hybrid: The induced transitions are purest QM, but you can only observe the classic induction of the ensemble, never individual photons.

Technically, relaxation spectra are usually taken by applying the stimulus at discrete frequencies, stepwise, and you want to have a phase-sensitive detector (voltage - current, deformation - torque).


Peculiar deprotonation of an alkenyl proton

Decided to convert this to an answer anyway. Disclaimer: I'm not an organic chemist (anymore), and corrections/suggestions are more than welcome.

One distinct mechanistic possibility here is that you deprotonate γ to the carbonyl, then the extended enolate reacts at the position you see, and then after that there's some tautomerisation which brings the double bond back into conjugation with the carbonyl (maybe during workup).

These are specific conditions under which you might expect deprotonation at the γ-position over the other α-position. In particular, the extended enolate resulting from this is more stable than the cross-conjugated enolate resulting from α-deprotonation, so this is called the "thermodynamic enolate". Paquette et al. use this term in the paper (J. Am. Chem. Soc. 1987, 109 (10), 3017–3024):

  Toward this end, the thermodynamic enolate of 6b was exposed to [...] 1-bromo-2-methylpropane.

I won't comment on the specifics of this reaction, but will mention that it can be very dangerous if scaled up.

If you wanted to deprotonate the other α-position, you would probably have more success with something like LDA. That deprotonation relies on chelation of the carbonyl oxygen to lithium (via a cyclic transition state*), so goes for the proton nearer to the carbonyl. That would be the "kinetic enolate".

* The six-membered chair-like transition state is widely taught, but is actually of questionable accuracy in this specific case. I think the transition state is cyclic, but is eight-membered or something like that. See e.g. slides 50 onwards of this very informative presentation from David Macmillan's group: "Lies My Sophomore Organic
Professor Told Me".


Do all metal salts in aqueous solution contain metal aquo complexes?

[OP] But do all metal salts in aqueous solution contain such complex ions?

This is a good question and in many cases, the exact structure and dynamics of aqueous metal ions is still being studied. In the simplest terms, if something is in aqueous solution, it will be surrounded by water. To start getting some more details, take a look at the Wikipedia article on metal aquo complexes. It says that metals surrounded by a first shell of six waters (with an octahedral geometry) is very common. They also point out that some of these complexes are long-lived while others are not:

  Rates vary over many orders of magnitude. The main factor affecting rates is charge: highly charged metal aquo cations exchange their water more slowly than singly charged species. Thus, the exchange rates for $\ce{[Na(H2O)6]+ and [Al(H2O)6]^3+}$ differ by a factor of $\pu{e9}$. Electron configuration is also a major factor, illustrated by the fact that the rates of water exchange for $\ce{[Al(H2O)6]3+ and [Ir(H2O)6]^3+}$ differ by a factor of $\pu{e9}$ also.[3] Water exchange usually follows a dissociative substitution pathway, so the rate constants indicate first order reactions.

Type of metal

  [OP] Is the metal required to be a transition metal?

For sodium ions in aqueous solutions, just calling them $\ce{Na+(aq)}$ is probably sufficient because the waters change places so rapidly. For a nickel complex with waters in the first shell not exchanging that quickly, it makes more sense to talk of a aquo complex and to study its geometry.

Other coordination numbers and geometries

If you dig deeper, you will encounter other coordination numbers and geometries. For example, according to this report, copper(II) likes to surround itself with five ligands in a square pyramidal fashion. Here is an excerpt of the abstract:

  Cu K-edge extended X-ray absorption fine structure (EXAFS) and Minuit X-ray absorption near-edge structure (MXAN) analyses were combined to evaluate the structure of the copper(II) imidazole complex ion in liquid aqueous solution. [...] This core square-pyramidal motif has persisted through [Cu(H2O)5]2+, [Cu(NH3)4(NH3,H2O)]2+,(1, 2) and now [Cu(Im)4Lax)]2+ and appears to be the geometry preferred by unconstrained aqueous-phase copper(II) complex ions.

I think for geometries that are not the optimal space-filling ones, you need transition metals, but I have not done a careful survey of the available literature.


Does the order of a written redox couple matter?

My understanding is that a redox couple is an unordered pair of two
  conjugate species

This is conceptually perfect and there is no problem when we talk about electrode potentials of half cells because as I had mentioned in your earlier queries, the electrode potential value and its associated sign do not know nor care how you write the half cell.

What is creating confusion in your mind is that you are trying to abbreviate a half cell. Abbreviating a half cell is of no use, because it is creating a problem of ordering.

All we care about the order is when we write a full cell. For that, we must follow the IUPAC recommendation and rules in order to prevent chaos in the published work. There your half-cell is no longer an unordered pair.

I used to remember the IUPAC recommendation with a mnemonic: Write Reduction on the Right.

This has to be parsed as follows: Reduction is on the right, so you must order the copper half cell in such a way that it appears as reduction.

On the left, of course, we have an oxidation half cell and it must be written as such.

Hope that clarifies your confusion.


Acidities of (dimethylamino)acetophenone and (diphenylamino)acetophenone

Yeah, you're thinking in the right way! In the first structure, the hydrogen atom will not have a tendency to leave since the negative charge on the  corresponding conjugate base is destabilized due to lone pair of nitrogen atom(the N atom intensifies the negative charge on the  carbanion formed by donating its electrons in the p orbital of carbanion). Whereas in the second structure the lone pair of N atom is already dispersed in the two phenyl rings so the lone pairs have very less tendency of going into the p orbital of carbanion. Therefore, the second structure is more acidic since its conjugate base is more stable( or less destabilized) than the first structure


Bonding of the complex when both weak and strong field ligand are present

$\ce{NH3}$ and $\ce{H2O}$ are strong and weak field ligands respectively according to spectrochemical series.

No, this is not correct. I keep saying this time and time again. Ammonia is a medium-field ligand at best, and water is the defined ‘zero point’ below which weak field ligands begin.

  I understand that pairing of electrons occur in the case of strong field ligands and it does not occur in the case of weak field ligands.

This is simplified at best. The transition from low spin to high spin depends on both the metal and the ligands (and also the geometry). In fact, some complexes are known both in a high-spin and in a low-spin state with identical central metal and ligands. You always need to consider the big picture to determine whether a high spin or low spin complex will result.

I am also going to completely ignore the stuff you wrote about hybridisation. You should not consider hybridisation in transition metal complexes as I have posted numerous times.

To answer this question you should ask yourself how many ligands you have around the nickel centre and which geometry that implies. The complex being tetraamminediaquanickel(II) results in six ligands and thus an octahedral configuration. The next step should be to consider the molecular orbital scheme of an octahedral complex (although the d-orbitals alone carry all the necessary information so even the crystal field model can answer it).

Figure 1: simplified MO scheme of an octahedral $\ce{[ML6]}$ complex ignoring π $\ce{L\bond{->}M}$ interactions. Originally taken from Professor Klüfers’ web scriptum of his coordination chemistry course at the LMU Munich.

The important bit to note is that the d orbitals split into a $\mathrm{t_{2g}}$ and an $\mathrm{e_g}$ part. $\mathrm{e_g}$ corresponds to the $\mathrm{d}_{z^2}$ and $\mathrm{d}_{x^2- y^2}$ orbitals. Thus, a perfect octahedral geometry will generate two degenerate orbitals which should be filled with parallel spins by Hund’s rule. Even if you start to assume Jahn-Teller distortions here (which are not likely, since a high-spin $\mathrm{d^8}$ octahedral configuration is not susceptible to Jahn Teller because there are no unevenly populated degenerate orbitals), that would not lead to an energy difference large enough to induce spin pairing. Remember that a typical 3d-metal octahedral complex is high-spin even in $\mathrm{d^6}$ cases; the energy difference between $\mathrm{t_{2g}}$ and $\mathrm{e_g}$ is generally much larger than what can be achieved by Jahn-Teller distortion.

Now you may decide to argue using square planar complexes. However, the complex in question cannot be square planar since there are six ligands.

Therefore, a magnetic moment of zero is not an option in the case presented; $\ce{[Ni(H2O)2(NH3)4]^2+}$ cannot reasonably adopt a low-spin configuration. The high-spin complex is paramagnetic and in a triplet state ($2S+1 = 3$).


Which has greater N-O bond length: NO or NOCl?

Also relevant:  the nitrogen-chlorine bond lengths:

$\ce{NOCl}=196\text{ pm}$ Wikipedia

$\ce{NCl3}=176\text{ pm}$ Wikipedia

The oxygen in nitrosyl chloride is donating its lone pair to the antibonding orbital in the nitrogen-chlorine linkage, weakening that bond but adding more pi bonding between nitrogen and oxygen to become like a triple bond.

If we were supposed to compare bond lengths using "simple" principles of bonding, this is a poor choice.  Nitrosyl chloride presents more than meets the eye.


As said by Oscar Lanzi, they are called "ternary metal chalcogenides". However, the name depends on the type of metal present in the compound. From here:

Systems to be studied include the layered intercalates $\ce{AMX2}$ (A=alkali
metal; $\ce{X=S, Se, Te}$) which are potential ionic conductors suitable for
use in solid state batteries: Group 2 ternary chalcogenides of type
$\ce{AMX3}$ and $\ce{AM2X5}$ ($\ce{A=Mg, Ca, Sr, Ba}$ ; $\ce{X=S, Se, Te}$) - some of which are superconducting and ternary mixed transition metal chalcogenides $\ce{MM'X2}$ ($\ce{X=S, Se, Te}$) examples of which have a wide range of magnetic properties including colossal magneto resistance.

You'll find more information on chalogenides of formula $\ce{ABS3}$ in this paper:

The structural diversity of ABS3 compounds with d0 electronic
configuration for the B-cation, John A. Brehma, Joseph W. Bennett,
Michael Rutenberg Schoenberg, Ilya Grinberg, and Andrew M. Rappe, J.
Chem. Phys. 140, 224703 (2014); DOI: 10.1063/1.4879659 (PDF
link)

Also, metal chalcogenides of different composition has also been noted:

($\ce{AB2X4}$): Wold A., Dwight K. (1993) Ternary Transition Metal Chalcogenides AB2X4. In: Solid State Chemistry. Springer, Dordrecht. DOI: 10.1007/978-94-011-1476-9_12

Mixed transition-post transition metal chalogenides of form $\ce{ABS2}$ and $\ce{AB2S4}$

($\ce{CuInS2,AgInS2}$): Ternary Metal Chalcogenides: Into the Exciton
and Biexciton Dynamics, Tushar Debnath and Hirendra N. Ghosh,The Journal of Physical Chemistry Letters 2019 10 (20), 6227-6238, DOI: 10.1021/acs.jpclett.9b01596

($\ce{AgBiS2}$): Effect of different growth temperature on the formation of ternary metal chalcogenides AgBiS2, S.Sugarthia,G.Bakiyaraj, R.Abinayab, M.Navaneethana, J.Archanaa, M.Shimomurad, Materials Science in Semiconductor Processing
Volume 107, 1 March 2020, 104781, DOI: 10.1016/j.mssp.2019.104781

($\ce{ZnIn2S4}$): Ionothermal/hydrothermal synthesis of the ternary metal chalcogenide ZnIn2S4, Cuixia Li, Honghua Lia, Lijun Han, Chunshan Li, Suojiang Zhang, Materials Letters, Volume 65, Issues 15–16, August 2011, Pages 2537-2540, DOI: 10.1016/j.matlet.2011.05.052

Note: if the chalcogen is oxygen i.e $\ce{ABO3}$, then it is called Perovskite.


Is there any connection between "static correlation" and the Born-Oppenheimer approximation?

Credit to Rudi_Birnbaum for coming up in a comment to the question with what I think is the key here.

Static correlation is a difficult to grasp concept of which even a clear definition is beyond me. However, it seems clear that a full Configuration Interaction (FCI) calculation yields all correlation energy, thus, must correctly handle static correlation. From the discussion of the Born-Oppenheimer approximation (BO) by Jensen,$^{[1]}$ I gather that the BO approximation is essentially a modification of the Hamilton operator. This shortcoming cannot be healed by a FCI calculation that only considers the electronic problem. So, in this sense, the two concepts are unrelated.

(And now for the speculative part of my answer.)

However, I cannot speak as to whether they frequently show up together in chemical systems. Though I cannot find a reference for it, I recall that the beryllium atom requires a multi-reference treatment. Being a single atom that is rather heavy, the BO approximation could be acceptable for it. Thus, this would be a case where the two are not related.


What property allows ozone to absorb uv light

Shortwave UV light (100-240nm) has enough energy to knock an electron out of an $\ce{O_2}$ bond creating an excited molecule $\ce{O^{excited}_2}$ which decomposes to two free radicals of oxygen. Each of the free radicals then combines with different oxygen molecules to create a net of two ozone molecules.

The crux here is that the electron is not "knocked out" of the oxygen molecule, the electron is just moved to a higher energy orbital of the same oxygen molecule. Since orbitals are quantized it takes a very specific energy to make the transition happen.

Now there is a second reaction. Longwave UV light (240nm - 350nm) has enough energy to knock an electron out of an $\ce{O_3}$ bond creating an excited molecule $\ce{O^{excited}_3}$ which decomposes to a free radical of oxygen and an oxygen molecule.

Thus oxygen protects the ground from shortwave UV, and ozone from longwave UV. Neither alone would protects us from all UV as effectively.

Now the rest of the story is that all molecules bouncing around in the air are not in the same energy state due to atom-atom vibrations. So the UV absorption occurs as a "band" rather than as a "fine line" like in the absorption spectra of an atom in free space.


What condenser can be used for simple distillation?

That sounds good.  Here's a drawing of the set up.  I'd use a water-cooled condenser.


Clarification about Frost diagram y-axis conventions?

The UC Davis version is incorrect. The free energy of the system ($\Delta G$) is equal to -nFE, so y-axis is reported as $\Delta G / F$ or -nE, according to the original report by Arthur Frost (J. Am. Chem. Soc., 1951, 73 (6), 2680-2682).

Note: In the original report free energy was denoted by $\Delta F$, rather than $\Delta G$. The x-axis designation is insignificant.


Does Combining Solid/Liquid Concentration with $K_c$ Matter when Comparing $K_c$ of Different Reactions?

The equilibrium constant of a reaction containing solids or liquids doesn't contain hidden values for their concentrations.

The issue is that we use the concentration (or pressure) of a substance in the equilibrium expression because it is a shortcut.  What we are actually putting into the expression is the activity of that substance, a more complicated idea.  It very often reduces to the ratio of the concentration of a solution to the concentration at standard state.  Since the concentration of a solution at standard state is 1M, there's no point in dividing through by 1 for each term.

For solids and liquids, the concentration at standard state is its molar density. ($C = \frac{mol}{L}$)  This value doesn't depend on the quantity of substance available, so the ratio between the density at standard state and the density at any point in a reaction is 1.  The values ARE in the equilibrium expression; they're just all 1's.

As to your concern about comparing equilibrium constants, we can only do that directly for reactions with the same number of reactants and products.  Adding (or removing) any term to either side interferes with the comparison.

Compare:
$\ce{A(aq) +B(aq)->AB(aq)}$
$\ce{2A(aq) +B(aq)->A2B(aq)}$

We have two similar reactions but the equilibrium constants can vary by a huge amount due to the presence of a second A atom in the reactants.  Similarly, with a constant solid/liquid concentration (activity = 1) the equilibrium constant effectively ignores that product or reactant.

It's just one of the limitations of the equilibrium constant.

Check out the wiki on Thermodynamic Activity for more information.


Nuclear power: what material is the control rod made from?

There are various designs and materials for control rods, which are more technically correctly called control assemblies. The choice depends on the type and purpose of the reactor, the size of the core, the availability of other means for power control (e.g. boric acid in the reactor coolant, burnable poison rods, $\ce{UO2-Gd2O3}$ fuels, recirculation pumps, or thermal-hydraulic feedback), patent rights, and tradition.

The control assemblies of typical boiling water reactors (BWRs) are inserted from the bottom because, in a BWR, the neutron flux is larger in the lower part of the core and the space above the core is occupied by the water separator and the steam dryer. Therefore, the design should consider saving weight. BWRs usually use cross-shaped control assemblies. The four blades of one cruciform control assembly are inserted between four fuel assemblies. The blades are made of stainless steel and contain tubes that are filled with boron carbide ($\ce{B4C}$) as neutron-absorbing material. In addition, hafnium ($\ce{Hf}$) absorber may be placed in selected locations of selected control assemblies.

The control assemblies of typical pressurized water reactors (PWRs) are inserted from above. Each control assembly has multiple rod fingers, which enter openings at the upper end of a fuel assembly. These control assemblies are very heavy because, when the reactor is tripped, they have to quickly drop into the core against the pressure of the reactor coolant. The neutron absorbing material mainly is an $\ce{Ag-In-Cd}$ alloy (usually 80 % $\ce{Ag}$, 15 % $\ce{In}$, and 5 % $\ce{Cd}$) encapsulated in an austenitic steel cladding. These three neutron absorbers together effectively block the entire energy spectrum of fast, epithermal, and thermal neutrons. In other PWR core designs, the control rods are divided into two groups: so-called black assemblies made up of rods consisting of an $\ce{Ag-In-Cd}$ alloy in the lower section and of $\ce{B4C}$ in the upper part, and so-called grey assemblies containing absorber rods made of $\ce{Ag-In-Cd}$ alloy and steel.


Maximum Pressure of Covalently Bonded Material

With the twist of thumbscrew, a hand-held diamond anvil cell will exert 300 GPa (43,500,000 psi).  No chemical bonding problem there. If you stack stuff - planetary and stellar cores - it gets more interesting, depending on composition and temperature.  Temperature matters.

http://en.wikipedia.org/wiki/Electron_degeneracy_pressure

Magnetic fields exert pressure.  How much magnetic field diddles with chemical bonding to give you a pressure estimate?

http://en.wikipedia.org/wiki/Magnetar
http://en.wikipedia.org/wiki/Magnetic_pressure
  10^10 teslas is 4.0×10^25 J/m^3 E/c^2 mass density more than 10^4 times that of lead.

That'll put some hurt on chemical bonding.  Scale down to 347 kJ/mole-bond multiplied in a molar volume of diamond.  Atomizing diamond requires about 800 kJ/mol per atom against cohesive energy.  The strongest chemical bonding interaction is roughly 3 eV.  Then you can pontificate with a Lennard-Jones potential.

http://en.wikipedia.org/wiki/Lennard-Jones_potential


How to calculate the bond energy of C-Cl?

The bond dissociation enthalpy* $D$ is the enthalpy change for conversion of bonded into dissociated molecules following homolytic cleavage of the bond in question:
$$\ce{A-B (g)->A(g) + B(g)},\qquad D(\ce{A-B})=\Delta H^\circ_f (\ce{A(g)})+\Delta H^\circ_f (\ce{B(g)})-\Delta H^\circ_f (\ce{A-B(g)})$$
This is the property usually discussed as representing the "strength" of a bond, and is positive because the process of breaking a bond requires an input of heat (is endothermic). Breaking stronger bonds requires a greater positive enthalpy change.
The enthalpy change for formation of the bond by the dissociated molecular constituents is negative the dissociation enthalpy:
$$\Delta H^\circ (\ce{A-B})=-D (\ce{A-B})$$
and is a negative quantity because heat is released when bonds are formed (the process is exothermic).
The enthalpy of formation of a $\ce{Cl-Cl}$ bond is the enthalpy change for the following process:
$$\ce{2Cl(g)-> Cl2(g)}$$
and therefore can be expressed as
$$\Delta H^\circ (\ce{Cl-Cl})=\Delta H^\circ_f (\ce{Cl-Cl(g)})-2\Delta H^\circ_f (\ce{Cl(g)})$$
and since $\Delta H^\circ_f (\ce{Cl2(g)})=0$,
$$\Delta H^\circ (\ce{Cl-Cl})=-2\Delta H^\circ_f (\ce{Cl(g)})=-D(\ce{Cl-Cl})$$
In the process
$$\ce{C(g) + 4 Cl (g)  -> CCl4 (g)}$$
four $\ce{C-Cl}$ bonds are formed, so the average heat of formation of one bond is
$$\Delta H^\circ(\ce{C-Cl}) = \frac{1}{4}\left(\Delta H^\circ_\mathrm{f}(\ce{CCl_4(g))}-\Delta H^\circ_\mathrm{f}(\ce{C(g)}) - 4 \Delta H^\circ_\mathrm{f}(\ce{Cl(g)}) \right) $$
which is a negative quantity (exothermic).
Note again that the corresponding bond dissociation enthalpy is then a positive quantity.

*Sources (from IUPAC Gold Book website):

PAC, 1990, 62, 2167. (Glossary of atmospheric chemistry terms (Recommendations 1990)) on page 2176.
PAC, 1994, 66, 1077. (Glossary of terms used in physical organic chemistry (IUPAC Recommendations 1994)) on page 1089.


This is an example of a Thermal syn elimination. The Wikipedia article is a good place to start: here

image from here
From ref 2

Unsymmetrical Acetate and xanthate esters yield a mixture of all the possible alkenes, but there is usually a predominance of the more highly substituted alkene.


Determine most electronegative element based on successive ionization energy data

Let's look at what amounts of energy we get from IEs.

$$\ce{X}: 577 \rightarrow 1815 \rightarrow 2740 \rightarrow 11600 \rightarrow 15000$$

After taking out three electrons, we see a first jump in IEs. Thus, there are three electrons in the valence shell.

$$\ce{Y}: 800 \rightarrow 2427 \rightarrow 3660 \rightarrow 25026 \rightarrow 32827$$

Again, after three electrons out, we have the first jump. The result is the same as X: Three electrons in the valence shell.

$$\ce{Z}: 738 \rightarrow 1451 \rightarrow 7733 \rightarrow 10543 \rightarrow 13630$$

You see the first bump in IEs in between IE2 and IE3. So, the third electron comes out of a lower shell and thus, two electrons in the valence shell.

The more the electrons, the more shells they will fill and the farther they'll be from the nucleus (among other lesser important factors). That means X has more electrons than Y because of relatively lower IEs, and therefore Y is more electronegative as it's closer to the fluorine.

You're correct that from left to right we get more electronegativity for the atoms, but note that
though with exceptions, such as Al and Ga, from top to bottom you get lesser electronegativities.

Even in the case of Al and Ga, the one that's more electronegative will get relatively more IE to lose electrons. So that's not what you should worry about. So let's conclude:

  If you see two elements that are in the same group, the one with a generally higher IE is more electronegative.


What is responsible for the smell of incomplete hydrocarbon burning and what sensor can detect it?

In analytical chemistry, sensors refer to analyte specific devices, e.g., glucose sensors, pH sensor etc. Smell on the other, generated by a burning complex mixtures such as fuels can rarely be associated with one compound. People have spent their life (e.g. Alan Marshall) studying the constituents of fuels. There are literally thousands and thousands of compounds.

However, if you were in a research lab investigating odors, you would be using something called as a olfactory gas chromatography technique. There is a chromatography column which separates all the gaseous/volatile components and a scientist puts a nose at the end of the column and records the "response". Later you can identify that particular odor by a mass spectrometer.

In short, there is no sensor per se. You need a instrumentation which would fit in a kitchen sized area to determine what causes that particular odor after burning an extremely complicated mixture such as fuel oil.

Here is how it looks like:


How can you tell how much actually reacted in an acid base reaction

The procedure is called titration. It consists in adding a pH indicator (that changes color depending on pH) to the solution and adding measured quantities of base or acid until the indicator changes color and determining how much reacted from the amount used.

An alternative is to use a pH meter instead of an indicator specially in automated analysis.


The effect of baking soda on water electrolysis

Definitely do not used stripped copper wire, unless you fancy a nice blue solution and electrodes that need constant replacement. You see, copper oxidizes quite readily, so you will end up with lots of $\ce{CuO}$, and a loss in the yield of $\ce{H2}$ and $\ce{O2}$. The best things would be noble metals like platinum, but stainless steel should do fine for your purposes/likely budget.


Why is logP quite a bit lower in cyclohexane than in hexane?

Think of the hole in the water that the molecules occupies as a bubble. The bigger the bubble the greater the area over which surface tension is acting.

The highest volume to surface ratio is for a sphere. So hexane is like a rod and cyclohexane is like hexagonal pyramid. Guess which has a higher volume to surface ratio?

edit more information...

Not that it is exactly the same, but look at the densities of the liquids. The cyclohexane molecule occupies a small volume in the liquid as compared to hexane, hence it is denser, even though it has 2 hydrogen atoms less per molecule.

Now water has 18 ml/mole so we can figure out roughly how many water molecules are displaced by a molecule of each hydrocarbon.

                                                     # water
                                                     molecules
             density         molecular mass  ml/mole displaced
Cyclohexane  0.7781 g/mL     84.16 g/mol     108.2    6.0
Hexane       0.6548 g/mL     86.18 g/mol     131.6    7.3


Reaction of an alkyl halide with silver nitrite

On treating ethanolic solution of haloalkane with silver nitrite ($\ce{Ag-O-N=O}$), nitroalkane is formed because since the bond between $\ce{Ag-O}$ is covalent, the lone pair on nitrogen acts as an attacking site for nucleophilic substitution.
$$\ce{R-X + AgNO2 -> R-NO2 + AgX}$$
But on the other hand if haloalkane is treated with potassium nitrite ($\ce{KNO2}$), alkyl nitrite is formed as major product because since the bond between $\ce{K-O}$ is ionic in nature, the negative charge on oxygen serves as an attacking site.
$$\ce{R-X + KNO2 -> R-O-N=O + KX}$$


What is the difference between free energy and standard free energy

$\Delta G^0$ is change in free energy change in going reversibly from molar stoichiometric quantities of  pure reactants, each at temperature T and pressure 1 bar, to corresponding stoichiometric quantities of  pure products, each at temperature T and pressure 1 bar.  So $\Delta G^0$, $\Delta H^0$, and $\Delta S^0$ are all functions of T, but not pressure (which is fixed in the initial and final states as 1 bar).  The values of these function changes tabulated in your text book are all for 298 K.  But values at other temperatures can be obtained from integration, for example, in the case of $\Delta G^0(T)$ using the van't Hoff equation.
$\Delta G$ is change in free energy change in going reversibly from molar stoichiometric quantities of  pure reactants, each at temperature T and its own arbitrary pressure, to corresponding stoichiometric quantities of  pure products, each at temperature T and its own arbitrary pressure. If these arbitrary pressures happen to match the partial pressures of the same reactants and products in an equilibrium mixture, then $\Delta G$ will turn out equal to zero.  Otherwise, $$\Delta G=\Delta G^0+RT\ln(Q)=RT\ln{(Q/K)}$$and, for equilibrium, Q=K.


How to prove that the orientation of the atomic orbitals in the superposition $\psi= a\psi_{1} + b\psi_{2}$depends on the coefficients $a$ & $b$?

In quantum mechanics states/orbitals are defined within the mathematical framework of the Hilbert space. A Hilbert space is a vector space that generalizes the notion of Euclidean space -  one example of the latter would be the normal two- or three-dimensional spaces you might be used to from basic geometry. So, orbitals are nothing more but abstract vectors in a Hilbert space and you can draw many analogies between them and the more familiar two- or three dimensional vectors you learn about in high school. So, let's think of quantum mechanical states as vectors for now. If you add two vectors, $\vec{a}$ and $\vec{b}$, you get another vector, $\vec{c} = \vec{a} + \vec{b}$, which if $\vec{a}$ and $\vec{b}$ are linearly independent, will have a direction different from either $\vec{a}$ or $\vec{b}$. You can also choose different coefficients for the superposition, e.g. $1$ for $\vec{a}$ and $-1$ for $\vec{b}$ yielding a vector $\vec{d} = \vec{a} - \vec{b}$ which now has a direction different from either $\vec{a}$, $\vec{b}$ or $\vec{c}$. So, the direction of a sum vector changes with the coefficients.

The same principle holds true for orbitals. If you for example think of the $\mathrm{p}_{x}$ orbital as a vector pointing in the $x$ direction, i.e. in two dimensions $\bigl(\begin{smallmatrix}1 \\ 0 \end{smallmatrix}\bigr)$, and of the $\mathrm{p}_{y}$ orbital as a vector pointing in the $y$ direction, i.e. in two dimensions $\bigl(\begin{smallmatrix}0 \\ 1 \end{smallmatrix}\bigr)$, then a superposition of both (with equal coefficients of $\frac{1}{\sqrt{2}}$ to ensure normalisation) will give you a $\mathrm{p}$ orbital pointing in the $\bigl(\begin{smallmatrix}1 \\ 1 \end{smallmatrix}\bigr)$ direction.

But as with vectors you are by no means confined to simply add them together with equal coefficients. For example, if you add $2$ parts of $\mathrm{p}_{x}$ and $1$ part of $\mathrm{p}_{y}$ you get a $\mathrm{p}$ orbital pointing in the $\bigl(\begin{smallmatrix}2 \\ 1 \end{smallmatrix}\bigr)$ direction, and that is clearly different from $\bigl(\begin{smallmatrix}1 \\ 1 \end{smallmatrix}\bigr)$.


How should one check an instant-equilibration algorithm for conservation?

I would think one of those conditions would always be sufficient. If you have shown that $T'-T=0$ for all $t$ and that $\partial_tT=0$, then  $\partial_tT'=0$ must be true by default. That is $\partial_tT=0$ shows that $T$ is a constant and if $T'-\text{constant}=0$ for all $t$, then $T'$ is also a constant.


Why does the 3rd electron shell start filling up with scandium?

The relative energies of the electronic subshells have been calculated for atoms in the vicinity of $Z=20$ (J. Chem. Educ., 1994, 71 (6), 469), and the result is surprising:

Looking at this graph, by all means the electronic configuration of scandium should in fact be $\ce{1s^2 2s^2 2p^6 3s^2 3p^6}$ $\color{blue}{\ce{3d^3}}$ in order to minimize orbital energies! The graph does not show the energy of the $\ce{4p}$ subshell, but it would lie somewhat above both curves shown. However, there is an important effect not being considered here, which is the destabilizing interelectronic repulsions. Electrons in a same subshell tend to repel each other more (intrasubshell repulsion) than electrons in different subshells (intersubshell repulsion). It turns out (and no one can really qualitatively explain why, see below) the repulsion is such that in an atom with $Z=21$, the configuration $\ce{[Ar] 4s^2 3d^1}$ is lower in energy than other potential candidates, such as $\ce{[Ar] 3d^3}$ or $\ce{[Ar] 4s^1 3d^2}$. Theoretically, the odd configurations $\ce{[Ar] 4s^2 4p^1}$ or $\ce{[Ar] 4s^1 3d^1 4p^1}$ would have lower interelectron repulsion energies than the observed ground state, but in these cases the lowered interelectron repulsions do not compensate the requirement of populating the higher-energy $\ce{4p}$ subshell. The comparison of several possible electron configurations and the joint minimization of orbital energy and electron repulsion energies does not seem to be something that can be done without resorting to heavy calculations or direct experimental validation.
In addition to some material I linked in comments, this article from Eric Scerri's blog (a chemist who focuses on aspects of periodicity, including electronic distribution) states, regarding the electronic configuration of scandium compared to calcium and to the $\ce{Sc^3+}$ ion (emphasis mine):

This amounts to saying that all three of the final electrons enter $\ce{3d}$ but two of them are repelled into an energetically less favourable orbital, the $\ce{4s}$, because the overall result is more advantageous for the atom as a whole.  But this is not something that can be predicted.  Why is it 2 electrons, rather than one or even none?  In cases like chromium and copper just one electron is pushed into the $\ce{4s}$ orbital.  In an analogous case from the second transition series, the palladium atom, the competition occurs between the $\ce{5s}$ and $\ce{4d}$ orbitals. In this case none of the electrons are pushed up into the $\ce{5s}$ orbital and the resulting configuration has an outer shell of $\ce{[Kr] 4d^10}$.
None of this can be predicted in simple terms from a rule of thumb and so it seems almost worth masking this fact by claiming that the overall configuration can be predicted, at least as far as the cases in which two electrons are pushed up into the relevant s orbital.  To those who like to present a rather triumphal image of science it is too much to admit that we cannot make these predictions. The use of the sloppy Aufbau seems to avoid this problem since it gives the correct overall configuration and hardly anybody smells a rat.

So unfortunately, it seems that even though most of the effects which combine to result in the observed electronic configurations are known, there is no qualitative way to predict where the configurations are going to mismatch with the Aufbau principle or the energy levels of the orbitals. I have read the statement that the Aufbau principle is most decidedly wrong for practically every atom with respect to the placement of the orbital energy levels, but incredibly it happens to predict the configuration of the valence shell for most atoms.


Why is potassium phosphate KH2PO4 in this reaction?

I think that’s just another mistake the OP of the other question made. Potassium phosphate is $\ce{K3PO4}$ and any student using any hydrogenphosphate instead would lose marks if I were marking the exam.

If $\ce{KH2PO4}$ is meant to be added, use the correct term: potassium dihydrogenphosphate.

This nomenclature can be found in the current version of IUPAC’s Red Book, sections IR-8.2 and IR-8.4 and the anions in question are explicitly named in table IR-8.1 therein.


Does it matter which end of a chain one sets to be 1 when drawing the structure of organic compounds?

Both of those are the same molecule (flip the one on the left over its horizontal axis, and they are identical).  It does matter where you start counting from, because the numbers chosen for the functional groups should be as low as possible:

Numbering of the chain. This is done by first numbering the chain in both directions (left to right and right to left), and then choosing the numbering which follows these rules, in order of precedence:

Has the lowest-numbered locant (or locants) for the suffix functional group. Locants are the numbers on the carbons to which the substituent is directly attached.
Has the lowest-numbered locants for multiple bonds (The locant of a multiple bond is the number of the adjacent carbon with a lower number).
Has the lowest-numbered locants for double bonds.
Has the lowest-numbered locants for prefixes.

 From IUPAC nomenclature
So, given the choice between 1,3,4-trimethylcyclohexane and 1,2,4-trimethylcyclohexane, you would choose the latter (for both molecules).


Should the tip of the burette of an automatic titrator be immersed in the analyte?

In classical titrimetry you would avoid dipping the burette into the sample. However, for autotitrators, the electronic dispenser tip (electronic burette) should be dipped in the sample. This is to avoid any error due to the a drop clinging to the tip. In manual titration one would wash it. Most autotitrators are like this design including Karl Fischer system. Your concern can be alleviated by the fact that you need pressure to cause a back flow of the sample into the tip of the dispensor.


What does Brightness/ intensity of spectral lines depend on?

The answer "The number of photons of same frequency or wavelength ABSORBED or EMITTED" from your book is a general way of saying when spectral lines are brighter (or darker). Your understanding of this general mechanism is (I think) mostly correct.

From what I understand, the differences in intensity of different spectral lines, however, is related to two (possibly more) things:

1) If a photon hits an atom (or molecule) and that photon has the exact energy needed to excite that atom does not necessarily means that the photon WILL be absorbed. There is an absorption probability associated with each energy transition. For example, one transition might have a 80% probability of absorption, while another has 0% (because the transition is forbidden).

2) Another factor that influences intensity is related to what state the system is in. For example, if two transitions, lets call them 0->2 and 1->2 both have a 100% probability of adsorption, but all the atoms/molecules are in state 0, then the 1->2 transition will not happen. That is, it will be dark/low intensity.

I hope this is helpful.


Why would breathing pure oxygen be a bad idea?

The other answers here, describing oxygen toxicity are telling what can go wrong if you have too much oxygen, but they are not describing two important concepts that should appear with their descriptions. Also, there is a basic safety issue with handling pressure tanks of high oxygen fraction.

An important property of breathed oxygen is its partial pressure. At normal conditions at sea level, the partial pressure of oxygen is about 0.21 atm. This is compatible with the widely known estimate that the atmosphere is about 78% nitrogen, 21% oxygen, and 1% "other". Partial pressures are added to give total pressure; this is Dalton's Law. As long as you don't use toxic gasses, you can replace the nitrogen and "other" with other gasses, like Helium, as long as you keep the partial pressure of oxygen near 0.21, and breathe the resulting mixtures without adverse effects.

There are two hazards that can be understood by considering the partial pressure of oxygen. If the partial pressure drops below about 0.16 atm, a normal person experiences hypoxia. This can happen by entering a room where oxygen has been removed. For instance, entering a room which has a constant source of nitrogen constantly displacing the room air, lowering the concentration -- and partial pressure -- of oxygen. Another way is to go to the tops of tall mountains. The total atmospheric pressure is lowered and the partial pressure of oxygen can be as low as 0.07 atm (summit of Mt. Everest) which is why very high altitude climbing requires carrying additional oxygen. Yet a third way is "horsing around" with Helium tanks -- repeatedly inhaling helium to produce very high pitched voices deprives the body of oxygen and the partial pressure of dissolved oxygen in the body falls, perhaps leading to loss of consciousness.

Alternatively, if the partial pressure rises above about 1.4 atm, a normal person experiences hyperoxia which can lead to oxygen toxicity (described in the other answers). At 1.6 atm the risk of central nervous system oxygen toxicity is very high. So, don't regulate the pressure that high? There's a problem. If you were to make a 10-foot long snorkel and dive to the bottom of a swimming pool to use it, you would fail to inhale. The pressure of air at your mouth would be about 1 atm, because the 10-foot column of air in the snorkel doesn't weigh very much. The pressure of water trying to squeeze the air out of you (like a tube of toothpaste) is about 1.3 atm. Your diaphragm is not strong enough to overcome the squeezing and fill your lungs with air. Divers overcome this problem by using a regulator (specifically, a demand valve), which allows the gas pressure at the outlet to be very near that of the ambient pressure. The principle job of the regulator is to reduce the very high pressure inside the tank to a much lower pressure at the outlet. The demand valve tries to only supply gas when the diver inhales and tries to supply it at very nearly ambient pressure. Notice that at depth the ambient pressure can be much greater than 1 atm, increasing by about 1 atm per 10 m (or 33 feet). If the regulator were to supply normal air at 2 atm pressure, the partial pressure of oxygen would be 0.42 atm. If at 3 atm, 0.63 atm. So as a diver descends, the partial pressure of oxygen automatically increases as a consequence of having to increase the gas pressure to allow the diver to inflate their lungs. Around 65 m (220 ft), the partial pressure of oxygen in an "air mix" would be high enough to risk hyperoxia and other dangerous consequences.

Now imagine a gas cylinder containing 100% oxygen. If we breathe from it at the surface, the partial pressure of oxygen is 1 atm -- high, but not dangerous. At a depth of 10 m, the partial pressure of supplied oxygen is 2 atm -- exceeding acceptable exposure limits. This is a general pattern -- raising the oxygen fraction of diving gasses decreases the maximum diving depth.

And you can't lower the partial pressure much because the lower limit, 0.16 atm, isn't that much lower than the 0.21 atm of sea level atmosphere.

One general category of solutions is to change gas mixes at various depths. This is complicated, requires a great deal of planning, and is outside the scope of your question. But it is certainly not as straightforward as just simplifying the gas mixtures or just raising the partial pressure of oxygen.

Additionally, compressed oxygen is a relatively annoying gas to work with. It is not itself flammable, but it makes every nearby organic thing flammable. For instance using grease or oil on or near an oxygen fitting risks spontaneously igniting the grease or oil. Merely having grease on your hand while handling oxygen refilling gear (with a small leak) can burn your hand.


Sadly, the lemon battery is a bit of a cheat: it's not the lemon that's providing the energy at all. The very first battery ever was very similar to the lemon cell, but used damp cloth, or cardboard soaked in salty water (might be worth giving it a go, I've never tried).

The energy really comes from the metal of the two electrodes which have had lots of energy piled into them (heat etc) when they were refined. When you wonder where the energy comes from, think of the guys in the refinery and their furnaces, some of that is "in" the metal, and you're getting it back.

The purpose of the solution (or briney felt discs) is to give one part of this reaction (the electrons) a good run for their money, going through the electrical circuit, as they can't travel through the electrolyte, while the other part, the ions, can cross the other way, through the solution. They meet up at the other end, and the reaction is complete and the zinc slowly corrodes away.

So what you're testing is the quality of electrolytes. Salts and acids in solution make good electrolytes: salty water, lemons, potatoes, coke. Milk acts as an electrolyte because it contains some ions from salts (which nutritionists helpfully/confusingly call electrolytes).

Pure water is pretty bad as an electrolyte. There's one of two options. First, it's not absolutely useless as one, and maybe the clock requires so little power that water is good enough. The other option is that tap water is hard, or otherwise has salts (minerals etc) in it which increase the ion concentration greatly. Probably a combination of those two factors. It might be worth trying distilled water (if you have some for the car, for example) which is probably much poorer.

More about the "first battery" is here: it's lots of these metal-electrolyte-other-metal cells end-to-end. The zinc is slowly being corroded by the copper the necessary combination of a. ions (which are big, heavy old things that amble through the electrolyte) and b. poor, zippy, lightweight electrons which can't, and have to go "the long way round" through the electrical circuit. https://en.wikipedia.org/wiki/Voltaic_pile


There are bi-cycles that are chiral, the one shown in the question is not. This can be shown, as pointed out in the comments, by superimposing the molecule on its mirror image, which is possible. Another fast way of ruling out chirality is finding the mirror plane within the molecule (in this case. perpendicular to the both cycles).


What is spatially and chemically heterogeneous materials?

Spatially heterogeneous means that the composition varies as a function of position within the material (varies "in space").

Chemically heterogeneous means that the chemical composition just varies in general. In this case, I assume they desire to emphasize that the composition varies among species of widely differing molecular weights—from large polymers down to small molecules.

So, it appears to me that "spatially and chemically heterogeneous" was used to emphasize that the chemical composition varies significantly, in particular as compared to prior assumptions; and that this compositional variability itself varies as a function of position within a given volume of the material.

In general, "spatially and chemically heterogeneous" would be rather a redundant term: a material can't really be spatially heterogeneous without also being chemically heterogeneous: if a substance is chemically pure, there is no possibility of any spatial distribution of compositional variations. So, I think the particular history of this material, which apparently had been assumed to consist only of polymeric material, is what led the authors to use the phrase.


Variation in atomic radii of elements in different blocks?

Valence electrons experience an electrostatic force from the nucleus. The nucleus has its character positive charge, but due to shielding by core electrons the total positive charge is not completely felt by the electrons, so the actual net positive charge felt by the valence electron has its own name and is called the effective nuclear charge $Z_{eff}$. This effective nuclear charge, largely effects the size of the atom. The general trend of the periodic table is that as you go down a group, and go from right to left, the $Z_{eff}$ decreases, and you see an increase in the size of the atom. The easy way to think of it, is that as you go from left to right, the number of valence electrons, and  protons increases but the number core electrons stays the same; so, you see an increase in $Z_{eff}$.

$Z_{eff}$ is calculated using Slater's Rules. Where
$$Z_{eff}=Z-s$$
Where $Z$ is the atomic number and $s$ is the shielding constant.

The rules for calculating are as follows (from wikipedia):

Firstly,(1)(4) the electrons are arranged into a sequence of groups in order of increasing principal quantum number n, and for equal n in order of increasing azimuthal quantum number l, except that s- and p- orbitals are kept together.

$\mathrm{(1s) (2s,2p) (3s,3p) (3d) (4s,4p) (4d) (4f) (5s, 5p) (5d)}$ etc.
Each group is given a different shielding constant which depends upon the number and types of electrons in those groups preceding it.

The shielding constant for each group is formed as the sum of the following contributions:

An amount of 0.35 from each other electron within the same group except for the (1s) group, where the other electron contributes only 0.30.
If the group is of the (s p) type, an amount of 0.85 from each electron with principal quantum number n one less than that of the group, and an amount of 1.00 for each electron with principal quantum number two or more less.

If the group is of the (d) or (f), type, an amount of 1.00 for each electron "closer" to the atom than the group. This includes
i) electrons with a smaller principal quantum number n and
ii) electrons with an equal principal quantum number and a smaller azimuthal quantum number l.


Which statement is correct about electron orbitals and energy levels?

Although @orthocresol got this first, and it should be credited to him, I'm posting it as an answer.

In option (B), the statement is incorrect as it refers to $d$ orbitals, which can (according to Pauli's Exclusion Principle) hold 2 electrons each, rather than the $d$ subshell, which can hold 10 electrons, as it has a total of 5 $d$ orbitals which can hold 2 electrons each, and $5*2=10$.

Thus, only option (D) is correct.


What's the strongest known organic acid?

According to "The Strongest Isolable Acid" Angewandte Chemie International Edition 43 (40): 5352–5355, carborane acids are the strongest, with $\ce {H(CHB11Cl11)}$ being the strongest.

Depending upon ones preferred definition of "organic" this might not be considered organic, but the derivative $\ce{H(CHB11Me5Br6)}$ is strong enough to fully protonate benzene.

$\ce{H(CHB11Me11)}$ is stronger than $\ce{H(CHB11Me5Br6)}$ and only slightly weaker than $\ce {H(CHB11Cl11)}$, so if $\ce {H(CHB11Cl11)}$ is disregarded as no being organic, then $\ce{H(CHB11Me11)}$ is strongest.  If $\ce{H(CHB11Me11)}$ still isn't organic enough, then protonated benzene or a protonated alkane.

(A later article and the article in Nicolau's comment report $\ce {H(CHB11F11)}$ and $\ce {H(CEtB11F11)}$ which may be even stronger than the above)


is there any difference between glycerin and glycerol?

Nope, there is no chemical difference between glycerol, glycerin or glycerine.  All 3 names refer to the same compound, propane-1,2,3-triol.


Chemists who need to control which gases are in their vessels (e.g. those who work with air sensitive compounds) do this routinely.

There are two ways to achieve gas replacement. The more reliable way is to work with vacuum-lines that allow switching the input to a vessel between a vacuum and an inert gas. First the air is removed and then the inert gas introduced.

Often, though, this is overkill. In many cases where liquids are involved the liquid needs to be degassed (to remove air dissolved in the liquid). In many cases it is good enough to bubble the inert gas through the liquid and the vessel for long enough to sweep out the unwanted gases. Bubbling the inert gas through the liquid and allowing it to escape the vessel is good enough if done for long enough.

If it isn't important to remove all traces of the original gas, this is usually an effective technique and gets more effective the longer the inert gas is allowed to flow.


What is the difference between Half Equivalence point and Equivalence point?

The equivalence point is where the amount of moles of acid and base are equal, resulting a solution of only salt and water. If you are titrating an acid against a base, the half equivalence point will be the point at which half the acid has been neutralised by the base. For instance, if you have 1 mole of acid and you add 0.5 mole of base, exactly half of the acid will have been neutralised. The solution remaining will be half salt and half acid.


What would be the unit for the calculation of the rate of reaction?

The formula has been stated and you simply need to look out for the units of the quantities
 from the formula. Mass(g)/Time(s). Which is g/s.


In Perkins reaction, condensation occurs between aromatic aldehydes (which cannot undergo self condensation) with an acid anhydirde in presence of sodium or potassium salt of corresponding acid.

Consider Perkins reaction of benzaldehye with acetic anhydride to form cinnamic acid. The first two steps occur in weakly basic medium (sodium acetate). The last two occur in acidic medium (protonation and subsequent dehydration). Notice that the last step is particularly favoured owing to formation of a conjugated product.

This wouldn't be the case for aliphatic aldehydes (with no alpha Hydrogen , else self condensation may occur). Also, if the anhydride used has only one alpha Hydrogen (here acetic anhydride has 3 alpha Hydrogen), then the reaction barely proceeds to completion.Instead an aldol type product is obtained. This shows the role conjugation plays in making the last step fairly irreversible.

Furthermore, this reaction occurs with vinylogs, hetrocyclic aldehydes and even pthalic anhydrides. Notice that in all cases , final product is conjugated.

It would be worthwhile to mention another alternative mechanism, which has been mentioned in Peter Stykes, Wikipedia as well.

Some Support for this mechanism comes from fact that anhydrides with only 1 alpha hydrogen donot give this reaction. Again, here the base promoted dehydration step would be largely irreversible incase of aromatic aldehydes due to conjugation.

Hope this helps.

References:

Perkin, W. H. J. Chem. Soc. 1868, 21, 181
XXIII.—On the hydride of aceto-salicyl.DOI: 10.1039/js8682100181
Peter Stykes. Organic mechanism. Sixth Edition

3.https://en.wikipedia.org/wiki/Perkin_reaction

SN Sanyal: Reactions,Rearrangements and Reagents.

5.http://www.name-reaction.com/perkin-reaction


To understand this phenomenon first we must point out that there are very different solid substances and solvents. Different combinations of them lead to different interactions and properties.

Anyway for the regular case - ionic or soluble compound in water - there is at least two steps to realize the solution:

Edited Answer:

Break the solid structure (eg. the crystal lattice). This step generally absorbs energy (endothermic, see lattice energy);
Solvate the compound. Because there are interactions between matter, the solvent might organize itself around the compound (solvation shell). Those molecules involved are not any more available, they are tied to the compound, they are not any more free water. This step generally releases energy (exothermic, see solvation energy). You can think about this step as a reaction of the solvent to recover form the intrusion by coating the intruder with one or several layers of its molecules. Then it 'feels' if there were only water. You will find details of the complex solvation process in good Physical Chemistry book.

Succeeding these two unit steps leads to dissolution. Solubility is the thermodynamic limit beyond that you cannot dissolve further molecule of a substance. Off course this is related with every substance present in the solution and available free water. Solubility are generally expressed for pure substance only.

When you realize your solution, speaking of heat exchanged with the environment, you experienced the balance between those two reactions. This is why the heat balance can lead to a global endothermic or exothermic reaction.


Standard Reduction Potentials and Electropositivity

Your reasoning is reasonable. However, although we can make the association between electropositivity and reduction potential qualitatively, relying on the former one to imply about the reduction potential is not always safe. For instance, following the trend of electropositivity on the periodic table, we would imply based on the reasoning that you proposed, that Zn which has a greater electropositivity than Cu (check the table bellow) has also a greater reduction potential, which is not the case. When it comes to transition metals, there are actualy some other effects that play a role on the reduction potential of the atom, ex.: shielding effects, final electron configuration, etc. The bottom line is, although electropositivity could help to justify a certain trends in reduction potential, it should not be used to make rigorous predictions of these trends since it could easly lead to mistakes mainly when considering transition metals.

P.S. The table shows the trend according to electronegativity. You can visualize the electropositivity trend by inverting the color scale at the bottom.


You could call ethyl alcohol "ethyl hydroxide" analogously to "sodium hydroxide". But it is a poor analogy, since (unlike NaOH) ethyl alcohol is not an ionic compound, does not dissociate in water to form ethyl cation and hydroxide anion, and does not raise the pH when dissolved in water.


Why does my sodium acetate look like metalic magma?

I wouldn't bother trying to purify it. As Mithoron said in his legendary comment "Caramel made your product a black unholy abomination of science". I would have distilled the vinegar and diluted it (because concentrated acetic acid is a bit dangerous if you don't know what you're doing) and start again. If you want pure product, make sure your starting materials aren't filled with "unholy abomination of science" causing impurities. If you really want to purify it, maybe recrystallise it in hot water? Or a solvent that sodium acetate can become saturated in.


C-N bond lengths in 2-bromo-1,3,5-trinitrobenzene

By rotating the molecule 180° about the $\mathrm{C_2}$ symmetry axis that passes through the bromine and the nitro group in the 5-position, you can interconvert $\ce{C_1-NO2}$ and $\ce{C_3-NO2}$.  Groups that can be interconverted by a rotational axis must be equivalent.  There is no symmetry element that interconcerts $\ce{C_1-NO2}$ (or $\ce{C_3-NO2}$) and $\ce{C_5-NO2}$, therefore these groups cannot be equivalent. So symmetry tells us that the  $\ce{C_1-NO2}$ and $\ce{C_3-NO2}$ bond lengths are exactly the same, and both are different from the $\ce{C_5-NO2}$ bond length.

Symmetry can only tell us if things are the same or different, it can't tell us the magnitude of any difference.  So I don't know for sure which set of carbon-nitro bonds is longer or shorter, but your argument based on steric strain seems reasonable.  It is also possible that the bulky bromine causes the adjacent nitro groups to twist out of the plane of the aromatic ring, reducing resonance overlap between nitrogen and carbon and also lengthening those adjacent nitro bonds.


I think some of your assignments are not correct. The doublet at $\pu{1.3 ppm}$ is correctly assigned. But close inspect of that signal suggests $\pu{12 Hz}$ coupling constant. Therefore, the signal given by the single proton causing that splitting should be a visual quartet with $\pu{12 Hz}$ coupling constant. But, your assignment for that proton is broad singlet at $\pu{4.75 ppm}$ with 2.7 integration (keep in mind that ChemDraw suggestion is not always true). I think that signal at $\pu{4.75 ppm}$ belongs to $\ce{NH3+}$. The small signal at $\pu{4.1 ppm}$, which is seemingly a quartet with large coupling constant is your $\ce{H-C(Ph)(CH3)NH3+}$ resonance. My best bet for the broad singlet next to it at $\pu{4.2 ppm}$ is crystalline water from  (S)-alpha-acetoxyphenylacetic acid ((+)-O-Acetyl-L-mandelic acid), which is crystalline solid.


Why are low spin tetrahedral complexes so rare?

Hybridisation won't explain anything in transition metal complexes, so please stop using it, at least to the extent where it is possible to avoid using it. Quite literally everything about transition metal complexes is better rationalised using MO theory, and I am not exaggerating.

The reason why low-spin $T_\mathrm d$ complexes are rare is because the splitting parameter, $\Delta_t$, is significantly smaller than the corresponding octahedral parameter $\Delta_o$. In crystal field theory, there is a complicated derivation which leads to the conclusion that (all things being equal)

$$\Delta_t = \frac{4}{9}\Delta_o$$

For more information, please see: Why do tetrahedral complexes have approximately 4/9 the field split of octahedral complexes? and Why do octahedral metal ligand complexes have greater splitting than tetrahedral complexes?. Of course, this relationship is not exact in the real world, because CFT is a very simplified model; ligands are not point charges. However, it is still true in a qualitative sense.

Since the splitting $\Delta_t$ is smaller, it is usually easier to promote an electron to the higher-energy $\mathrm t_2$ orbitals, rather than to pair the electrons up in the lower-energy $\mathrm e$ orbitals. Consequently, most tetrahedral complexes, especially those of the first-row transition metals, are high-spin. Low-spin ones do exist (e.g. J. Chem. Soc., Chem. Commun. 1986, 1491), but aren't common.


How to determine if a mixing process is spontaneous at constant temperature but variable pressure

Volume of H2 = $\frac{nRT}{P}=\frac{(2)R(298)}{2}$

Volume of N2 = $\frac{nRT}{P}=\frac{(4)R(298)}{3}$

Total volume = $\frac{7}{3}(298)R$

Final pressure = $\frac{nRT}{V}=\frac{(6)R(298)}{(\frac{7}{3})R(298)}=\frac{18}{7}\ atm$

Final partial pressure H2 = $\frac{18}{(7)(3)}=\frac{6}{7}\ atm$

Final partial pressure N2 = $\frac{18}{7}\frac{2}{3}=\frac{12}{7}\ atm$

$$\Delta S=-n_{H_2}R\ln{\frac{\frac{6}{7}}{2}}-n_{N_2}R\ln{\frac{\frac{12}{7}}{4}}=-(2)R(-0.8473)-(4)R(-0.8473)=5.08R$$

For ideal gas mixing, $$\Delta H = 0$$

Therefore, $$\Delta G=\Delta H-T\Delta S$$

Another way of doing this is to (a) first allow the divider to float so that the two chambers equilibrate in pressure and then (b) remove the divider and allow the gases to mix.  You calculate $\Delta G$ for each step, and then add them.  In the end, you get the same answer I got.


Which Phenols give the colours blue, green or violet in the neutral FeCl3 test?

The colour may vary from blue, green or even red depending upon the
  nature of the phenol.(Source)

o,m,p-cresols and resorcinol gives violet or blue coloration. (source)
beta-naphthol gives green coloration. (Source)
alpha-naphthol gives pink coloration. (")


Which is the central atom in Lewis Dot Structures?

The video is wrong if it says "the central atom ... is always the least electronegative atom".

There are chlorine oxides with Cl central such as perchlorate, chlorate and $\ce{ClO2}$, even though $\ce{Cl2O}$ has O central.

There are $\ce{SeCSe}$, $\ce{CAt4}$, $\ce{NI3}$ and $\ce{Li6C}$ each with the more electronegative element central.


Henry Rzepa gives a mechanism in which two borane molecules attach to the carboxyl group of a carboxylic acid.  The boron atoms are attached to the electronegative but electron-rich portions of the carboxyl group (meaning the oxygen atoms) and then transfer hydride ion moieties to the carbon while taking away one of the oxygens.  This exchange of hydrogen for oxygen reduces the carboxyl carbon.  Other carbonyl compounds may be assumed to react similarly.


Would trans hydride ligands on a metal complex be beneficial for a catalyst?

In alkene metathesis you need large ligands that direct the stereoselectivity of the reaction, and a hydride ligand is very small.

Also for hydrodefluorination I don't see an advantage of a trans hydride ligand on the catalyst. cis hydride ligands could be beneficial, as they then take the place of the fluorine atom on the substrate.


Formation of Lactone from 5-amino-decanoic acid?

I doubt your book is asking you to convert this molecule into a lactone. This would be very hard to do. It could, however, be easily converted into a lactam. 5 and 6 membered lactams are formed by simple dehydration. For your γ-lactam, this can be achieved by heating the amino carboxylic acid, forming the thermodynamically favourable, stable 6-membered lactam. Similarly, lactones can be be interconverted by adjusting the pH.

IF you really wanted to convert the amine into a lactone, it would involve some pretty uncommon reagents. This blog post mentions that pyrylium salts can be used to make a primary amine into a leaving group, which would, presumably, allow you to substitute a hydroxyl group, thereby giving you access to the lactone.
http://curlyarrow.blogspot.hu/2009/09/how-to-turn-amine-into-leaving-group.html


Electron affinities of the chalcogens and halogens

The amount of screening is the same in the two groups but the effective positive charge that the incoming electron feels is stronger in Group 7.

For instance, oxygen has the electronic structure of
$$\ce{1s^{2} 2s^{2} 2p_{x}^{2} 2p_{y}^{1} 2p_{z}^{1}}$$ and 8 protons. The effective positive charge that pulls the incoming electron (which approaches the second level) is $8 - 2 = 6$.

In fluorine we have,
$$\ce{1s^{2} 2s^{2} 2p_{x}^{2} 2p_{y}^{2} 2p_{z}^{1}}$$ and 9 protons. The effective positive charge is $9 - 2 = 7$.

Therefore, the net pull from Group 7 is stronger than Group 6.


Why is osmotic pressure only for solvent

Osmosis depends on the separation of a solution from its pure solvent by means of a  membrane permeable to solvent but not to solute. The semipermeable membrane can take several forms, cellophane for example is impermeable to high molecular mass compounds, such as polymers, but allows water to pass though. Palladium foil is permeable to hydrogen gas but not nitrogen or oxygen so osmosis can be performed in the gas phase.

The experimental arrangement is for the pure solvent to be separated from the solution by the semipermeable membrane. The natural tendency is for the solvent to flow from the pure solvent chamber into the solution. The tendency is counteracted by applying pressure only to the solution side. The excess pressure is called the osmotic pressure.

Now if the solute as well as solvent could diffuse through the semi-permeable membrane there would soon be a solution of different concentrations on either side of the membrane. Osmosis would now work both ways across the membrane and result in equilibrium when the same solution concentration is on both sides.


stone which turns to black on exposing to atmosphere?

There is a compound called cerussite or 'natural white lead' which turns to black 'lead sulphide' when it reacts with hydrogen sulphide in the atmosphere. It was commonly used as make up by women in medieval times and is sometimes still used as a white pigment in 'lead paint'.

Of course, we all now know lead is toxic to many organs and tissues including the heart, bones, intestines, kidneys, and reproductive and nervous systems, as well as causing potentially permanent learning and behavior disorders, particularly in children.

For this reason, it's use has diminished over time, except perhaps amongst New Age crystal 'healers' who claim it can "infuse your aura with light" and "bring amazing change into your life", not only through a lowering of general intelligence, but, it can also help prevent bacterial infections.

Having said all this, you may be surprised to know that another toxic lead-based white crystalline substance called lead acetate is approved by the FDA for use in 'progressive' hair dyes, where it reacts with the sulphur in hair proteins to slowly turn grey hair black, over several applications of course, so 'no-one' will notice.

image of lead(II) acetate crystals, courtesy of Wikipedia

PS: In case you were wondering why lead-based products are still used, given that there are many alternatives out there, many of which are labelled 'non-toxic', 'natural' and 'organic'...

the National Cancer Institute reports that there are over 5000 different chemicals used in hair dyes, including toxic coal-tars, aromatic amines and phenols. Some of these are also carcinogenic (cancer-causing agents) and can be absorbed through the skin or inhaled. Whilst the industry has been working towards eliminating some of these chemicals since the late 1970's, people who regularly began using hair dyes before 1980 may be at increased risk of developing non-Hodgkin lymphoma (a form of blood cancer).

The International Agency for Research on Cancer (IARC) has said that "whilst workplace exposure as a hairdresser or barber is probably carcinogenic to humans", personal hair dye use is “not classifiable as to its carcinogenicity to humans,” based on "a lack of evidence from studies in people".


What is the hybridization of the central carbon in buta‐1,2‐diene?

You must be clear that the hybridization is decided by the number of σ-electrons, so for $\ce{CH4}$ the number of σ-electrons is four and so its hybridization is $\mathrm{sp^3}.$
In case of $\ce{CH3-CH=\overset{\ast}{C}=CH2}$ the $\ce{\overset{\ast}{C}}$ is having two σ- and two π-bonds, so its hybridization is $\mathrm{sp}.$
While deciding hybridization we need to only care about bonds, and that too only the σ-bonds. Remember that possible hybridization for $\ce{C}$ are $\mathrm{sp^3},$ $\mathrm{sp^2}$ and $\mathrm{sp}.$


How can you tell if a metal ion will hydrolyze to form an acidic solution?

So the ability of a metal cation to hydrolyze water molecules to form acidic solutions is based on the charge of the ion and the size of the atom. The larger atoms will stabilize charge better (i.e. it spreads it out over a larger space). Unstable cations form hydroxides or oxides with water molecules and release hydronium ions.
Estimations of the acidity for metal cations can be found with the equation below,
$pka = 15.14 - 88.16(\frac{Z^2}{r})$
Z being the charge of the ion and r being the atomic radius. I bold the word estimation cause this isn’t a perfect system and I want to make that clear. However, better equations do exist.
Electronegativity does play a role too. The more electronegative a metal cation is, the more acidic it will be.
I would recommend taking a look at this link for more information


Water has a high heat capacity, so I agree with JoeHobbit that a spray of hot water via hose would probably work best. A power washer that is capable of heating the water might be worth getting. A secondary advantage of doing this is that the water spray would remove accumulated road salt, which accelerates corrosion. Short of that, a space heater that blows hot air on the truck would be a less effective but also less messy solution. A moving stream of hot water or air should be far more effective than static immersion.

Small amounts of ice deposits can be removed with an aerosol deicer spray.

See, there is actually a lot of chemistry going on here.


The colour of black tea is given by a complex mixture of theaflavins and thearubigins.

These compounds are formed by enzymatic oxidation and condensation of polyphenols of green tea, typically of gallocatechin and epigallocatechin gallate ( EGCG )

Basic dissociated forms of these compounds have strong absorption of visible light due to delocalization of phenolate electrons. This absorption  decreases a lot when phenolates are protonized in acidic environment.

So yes, in some sense, black tea infusion is a $\mathrm{pH}$ indicator, similar to litmus or juice from red beet roots. Many natural colourful compounds have $\mathrm{pH}$ dependent colour, as their molecules undergo acid-basic reactions.

But synthetic indicators are better due stability, sharper and better defined colour transition.


You run into all sorts of problems when the pH is outside of the range of 1 to 13. For example, in Environ. Sci. Technol. (2000) 342, p. 254-258, they say:

  [...] the former National Bureau of Standards (NBS) established a set of conventions that limits measurements to 1 < pH < 13 and to ionic strength, I < 0.1. The main limitations are the activity coefficient expression, the range of defined standard pH buffers, and interferences with the reversible response of the glass H+-sensitive membrane electrode.

They go on to use standard of concentrated sulfuric acid to define the pH, using a Pitzer model to predict activity coefficients (figure from researchgate) :

  In reality it is completely opposite. So what is the catch?

If you do the math, the extreme pH of -5 shown in the figure corresponds to a hydrogen ion activity of hundred-thousand. The concentration of water for pure water is "only" ~55 mol/L, so this must reflect an activity coefficient far from one.


If I have a chiral substrate, should I always assume that both enantiomers are present?

In this SN1 reaction there would be two products even if only one enantiomer reactions.

Enantiomerically pure (S)-2,3-dimethyl-3-pentanol reacts with HBr to produce both (S)-3-bromo-2,3-dimethylpentane and (R)-3-bromo-2,3-dimethylpentane in equal proportions as a racemic mixture.

ALSO, enantiomerically pure (R)-2,3-dimethyl-3-pentanol reacts with HBr to produce both (S)-3-bromo-2,3-dimethylpentane and (R)-3-bromo-2,3-dimethylpentane in equal proportions as a racemic mixture.

When this kind of phenomenon happens, there must be an achiral intermediate that can react to produce either enantiomer with equal propensity. In this case, what is the structure (geometry and everything) of the carbocation formed when the leaving group leaves?


Difference between Force Field and topology, and other related questions

1 - "Force Field" (FF) means the set of equations that describe the interactions between atoms and the set of numerical parameters that people using those particular functions have come up with to describe some molecules (typically proteins, Nucleic Acids, lipids, and a few others). Many FFs exist because of different choices in functional forms, in the procedures adopted to obtain the numerical parameters etc etc.
A topology is just the description of a molecule in terms of parameters that can be replaced by the numerical ones contained in the FF of your choice, so that the equations of the FF can be worked out numerically.

2 - see 1.

3 - Water is described separately because there are specific models for water (TIP3P and so on) that work reasonably well, so people stick to these models indipendently of their choices for the rest of the system.
What do you mean by "it is not included"? I can see water parameters in all MD programs I have used.
Usually, FFs are always divided by "optimisation" let's say, meaning that a set of parameters have been optimised for a particular task/family of molecules, and those are given together.

4 - if you want to model a molecule for which you don't have the parameters, then you should fit them according to the procedures that have been used to obtain the other parameters belonging to the FF (good luck). If parameters exist for a similar molecule, you can try to change those to represent your molecule.

In general, it is not a good idea to use mixed parameters, meaning e.g. AMBER for the protein, CHARMM for a DNA in the same simulation.
Choose one FF and if you need parameters for some new molecule, you have all my sympathy (you can find tools to obtain these "pseudo-automatically", though).


The answer may be more simple than some may expect, I start with Black powder, which contains, to quote:

Black powder is a granular mixture of

a nitrate, typically potassium nitrate (KNO3), which supplies oxygen for the reaction;

charcoal, which provides carbon and other fuel for the reaction, simplified as carbon (C);

sulfur (S), which, while also serving as a fuel, lowers the temperature required to ignite the mixture, thereby increasing the rate of combustion.

Next, the warming of an organic together with sulfur can produce a volatile sulfur-containing compound (see, for example, 'Reactions of sulfur with organic compounds XXI. The action of sulfur on 2-thienylchloromethane and di-2-thenyl sulfide').
Further, mixing of the new compound with carbon powder could account for the observed product especially after evaporation of all volatiles.
Note, per the comment: "problems of stinking, liquifying or decaying animal greases, and some people tell they used petroleum jelly or lithium grease for decades with no particular problems", the precise sulfur-based product, if any, created could account for the differences reported.
Also, this proposed explanation requires an element of heating, which likely accounts for the development of the reported problem over time.
So, to precisely answer the question: "Are petroleum-based products prone to produce hard residues with black powder? Yes, per my assessment.


How can aluminium oxide be called an acid?

The most basic definition of "acid" is that it is a proton donor (or one which accepts a lone pair)

All of this stuff is done in an aqueous medium, so we can assume that all aqueous ions and molecules are present. With this assumption (in this case, we are assuming that $\ce{OH-}$ is available to react), we get the following equation:

$$\ce{Al2O3 + OH- -> 2AlO2- + H+}$$

Similarly, we get:

$$\ce{Al2O3 + 6H+ -> 2Al^3+ + H2O}$$

where it is acting like a proton acceptor (base).


What is the platinum doing in a Zippo hand warmer?

Platinum is a catalyst and decreases the activation energy, so the process can start at lower temperatures.


How to Find the Mass of an Individual Element in a Compound/Molecule

Step 1: Find the Molar Mass of Adrenaline

To do this, you must need to know its chemical formula which is $\ce{C9H13NO3}$. Now to calculate its molar mass, we add up all of the molar masses of each atom: $$M =9\times12~\mathrm{g~mol^{-1}} + 1\times13~\mathrm{g~mol^{-1}} + 14~\mathrm{g~mol^{-1}} + 3\times16~\mathrm{g~mol^{-1}} = 183~\mathrm{g~mol^{-1}}$$

Step 2: Find the Mass of Oxygen in 1 mol of Adrenaline

In each molecule of adrenaline, there is 3 molecules of oxygen atoms. This means that in $1\ \mathrm{mol}$ of adrenaline, there are $3\ \mathrm{mol}$ of oxygen. Therefore the mass of oxygen in $1\ \mathrm{mol}$ of adrenaline is: $\mathrm{3~mol\times16~g~mol^{-1} = 48~g}$.

Step 3: Calculate Percentage Composition of Oxygen by Mass

We now know that in every mole of adrenaline, there are $48\ \mathrm{g}$ of oxygen. We also know that $1\ \mathrm{mol}$ of adrenaline weighs $183\ \mathrm{g}$. Therefore in $183\ \mathrm{g}$ of adrenaline, there are $48\ \mathrm{g}$ of oxygen. All have to do is simply divide by them by each other and multiply by $100$ to get the percentage composition of oxygen. Therefore the percentage composition is $26.23\,\%$.

Step 4: Calculate Mass of Oxygen in Sample

Now we know the percentage composition of oxygen which is $26.23\,\%$. This means that in every gram of adrenaline, there is $0.2623\ \mathrm{g}$ of oxygen. Therefore to find the mass of oxygen in the sample: $$m_\text{oxygen} = 0.2623 \times 6.8~\mathrm{g} = 1.78~\mathrm{g}$$


Why can you sanitize paper money notes with a Clothes Iron?

In short: This is possible because «paper money» actually is not referring to the paper like paper from the office supply for (your) desktop printer or xerox, but to cotton paper.
Because of the cotton, if folded and forgotten in a pocket of your jeans close to $\pu{40 ^\circ{}C}$, they may physically survive a pass in the washing machine with a mild non-bleaching detergent (including the spinning).  They obviously suffer from the dye if e.g., the jeans blead, or mechanically if you try to unfold them after if passing a dryer.  Of course different to the ones made of polymers as (but not limited to) in Australia.


How can solids undergo combustion if it only occurs in the gas phase?

Combustion is a gaseous phase phenomenon. Oil and gasoline have a high enough vapor pressure at ambient temperatures to produce a gaseous phase of fuel above the liquid. In contrast, hold a lighter up to a piece of wood and try to get it to light. It won't—at least not for quite some time. This is because solid fuels must first undergo endothermic pyrolysis before real combustion can occur. This produces a slew of products which are what truly undergo combustion when solids are burned. Because combustion is exothermic, once a high enough temperature is reached the solid will autopyrolyze, making combustion a self-sustaining reaction.


How to know properties of ion Aluminum -27?

Aluminum-27 refers to an isotope of aluminum. For almost all chemical reactions the isotope makes no difference in the chemistry. You need "exotic" (meaning very unusual) conditions for the isotope to matter.

Unusual conditions would mostly happen in the gas state. For example a lighter gas would diffuse faster. This was the basis for the separation of uranium isotopes at Oak Ridge during WWII to make the first atomic bomb.


Why can't light pass through a gold foil, but alpha particles can?

Light can pass through a gold foil though, it just has to be thin enough.

Pure gold is a very malleable substance and can be beaten with a hammer into foils of around 100 nm thickness. Sources suggest the gold foil used in the Geiger–Marsden experiment (known more commonly as the Rutherford gold foil experiment) was about 86 nm thick. Somewhere around this region of thickness, gold starts letting light through, as well as alpha particles. It is well known that ~20 nm thick gold (or any other metal) deposited onto a surface by thermal evaporation in a vacuum is essentially invisible and can be used as a transparent conducting surface/electrode. As the foil gets thicker, the gold starts to absorb a noticeable amount of photons, especially in the violet/blue part of the electromagnetic spectrum, while scattering away the red light. The light scattered off a semi-transparent gold film acquires a reddish/yellow tinge, while the light that still does make it through is blueish. When the film reaches 100 nm or so, it becomes practically as opaque as the bulk metal.

In truth, comparing the mean free path or collision cross-sections of photons and alpha particles through a gold sheet is somewhat more complicated than just saying alpha particles should penetrate less because they're massive, so it would be conceivable for a film of appropriate thickness to block photons yet still allow alpha particles through. A very coarse, hand-waving explanation is that photons are absorbed in the electrosphere of a gold atom, while an alpha particle is only significantly pushed away if it bangs almost straight into the nucleus, a region composing approximately $10^{-15}$ of the volume of the atom.


Computing the bond spring constant from first principles

tldr, you want the force constant from the Hessian matrix .. ideally in internal coordinates

First off, you're saying 'first principles,' which most computational chemists would take to be some level of quantum chemical method, whether density functional or wavefunction-based. None of that is available in Open Babel - it only implements a very small number of classical force fields.

Even if you have optimal bond lengths and energies, that doesn't directly give you the force constants. For that, you want the second derivative.

In principle, you could start perturbing atoms and evaluating the energy. In that way you could build up a stochastic approximation of the PES and evaluate numerical second derivatives. I don't recommend that.

A more accurate way would be at least to get the forces / gradients at each of those points, which you can use to build up the Hessian matrix. There are some notes about that as part of the psi4numpy notebooks which can be used to drive the Psi4 quantum program.

The problem is that for any molecule over a few atoms, the Cartesian Hessian matrix won't directly give you a bond force constant. Most vibrations include multiple atoms, and are rarely along any particular Cartesian axis.

The psi4numpy notebooks give some indication of how you can evaluate the Hessian in internal coordinates, which should be closer to what you want.


How does hyperconjugation lead to stability?

What you are ignoring is the magnitude of the effect.

While hyperconjugation does decrease the electron density of a localized bond and slightly increases the energy, the overlap with a high-energy, empty orbital is at least very poor. Think that the orbitals are very far from each other and have different symmetries. The total electron density of this bond would remain largely unaffected.

On the other hand, the difference between zero electrons and partial electron density for an empty orbital of high energy is significant, and it increases (to a moderate extent) the stability of, say, a carbocation. Besides, a single orbital often receives electron density by hypercongujation with more than one bond.


Why is double bond stronger than single bond?

A double bond is composed of a π-bond and a σ-bond. In order to break a bond such as this you must provide the molecule with the energy needed to break the σ-bond alone plus the one needed to break the π-bond alone. In other words, you must provide the sum of those two amounts of energy. Thus, it goes without saying that the sum is a greater number than the two energies alone.