Is it possible to dope Ca ions in an NaCl crystal to produce a vacancy defect?

A peek at the NaCl - CaCl2 phase diagram (say at FactSage) will show you that NaCl can hold quite a reasonable fraction of CaCl2 in the NaCl crystal. In contrast, CaCl2 has very limited to no solubility of NaCl.


Formulas for non-bonded interaction energies

You've got the Lennard-Jones potential parameters for zinc and oxygen from a certain version of the CHARMM force field. The commonly-used notation is $\varepsilon$ for potential well depth and $\sigma$ is the distance where the pair potential is zero, and $r_{min}$ is the distance which minimizes the L-J potential energy function. The linked Wikipedia article gives the potential as:

$$E_{\rm L-J}(r) = 4\varepsilon\left[\left({\sigma\over r}\right)^{12} - \left({\sigma\over r}\right)^{6}\right] = \varepsilon\left[\left({r_{min}\over r}\right)^{12} - 2\left({r_{min}\over r}\right)^{6}\right]
$$

At $r = r_{min}$, $E_{\rm L-J} = -\varepsilon$ and we can map $r_{min} = 2^{1/6}\sigma$ to work with either form of the L-J potential function.

Accurate Calculation of Hydration Free Energies using Pair-Specific Lennard-Jones Parameters in the CHARMM Drude Polarizable Force Field. J. Chem. Theory Comput. 6(4) 1181–1198 (2010) describes (among many other things) how the L-J potential term is calculated for 2 interacting species (as in your case). See equations (1), (2), and (3) in that paper, reproduced below (where I have used $r_{min}$ instead of $R_{min}$ for consistency).

In the paper, the Lennard-Jones potential is given as

$$E_{\rm L-J}(r) = \varepsilon\left[\left({r_{min}\over r}\right)^{12} - 2\left({r_{min}\over r}\right)^{6}\right]$$

For a pair of interacting species $i,j$:

$$r_{min} = {r_{min,i} + r_{min,j}\over 2}$$

and

$$\varepsilon = (\varepsilon_{i}\cdot\varepsilon_{j})^{1/2}$$

For your system, we compute $r_{min}$ from the given values of $\sigma$ (using the mean of the two values of $\sigma$, as specified in the manuscript):

$$r_{min} = 2^{1/6}\sigma = 2^{1/6}\cdot {1.942 + 3.029\over 2} = 2.789\;\unicode{xc5}$$

and

$$\varepsilon = (0.25\cdot 0.12)^{1/2} = 0.1732\;\mathrm{kcal\;mol^{-1}}$$

Now, at some distance $r$, you can compute $E_{\rm L-J}(r)$.

In the linked publication, the authors also go through the process of how the L-J parameters are derived/fit, which may be of interest to you.

Finally, see this vintage web page authored by a familiar guy which goes through a procedure for taking parameters for computed hydration free energies of cations and deriving L-J parameters for use in the AMBER force field. The procedure is very similar to that used for CHARMM.


If viewing these as charged objects, we're using a physics/electrostatics viewpoint, not usually done in electrochem.  Or at best, it's a topic for postdoc courses in surface nano-chemistry (yet I think if it were discussed in undergrad classes, it would clear up much student misconception!)  The metal/electrolyte interface forms a capacitor.  For this reason we must deal with actual potentials, NOT the shifted values shown in tables of Standard Electrode Potentials.

You're right to be confused.  As with the zinc rod, the copper rod will also acquire a negative charge.

From a physics standpoint, the two electrodes in contact are 1) the electrolyte and 2) the metal.  A half-cell then has two conductive electrodes, not one.  A complete electrolysis cell has four.  In a half-cell, a genuine voltage appears between the metal and the electrolyte.  The voltage is not undefined or somehow abstract or "ghostly."  It's just difficult to measure.  A simple voltmeter cannot do it.

These concepts go back over a century, to when chemistry and physics weren't fully separate.  Alessandro Volta and Michael Faraday studied electrolysis from this viewpoint.  I've seen "single" electrode potentials discussed in 1930s undergrad textbooks, but not in the modern ones.  (Well, Bockris electrochem has it.)

Search: Galvani Potential, also Volta Potential, both found at surfaces and junctions between differing materials.  Perhaps also look for Helmholtz layer as well as Helmholtz Double Layer.

.

Whenever metal first contacts electrolyte, a brief oxidation reaction takes place which establishes a static potential appearing across the double-layer located at the metal surface.  The net reaction halts once these potentials appear.  The copper is then left with a negative charge, same as the zinc, with the electrolyte having net positive charge.

However, if either a salt-bridge or a common electrolyte is connecting the half-cells, the zinc and copper rods will always acquire two different amounts of negative surface-charge, leading to a well-known potential-difference measured between them.

For working with these physics/nanochem concepts, it's useful examine the SHE reference electrode.  The relative potential-difference of the Standard Hydrogen Electrode is roughly 4.44V, not zero.  This is the Galvani potential between metal and electrolyte, also called "Absolute electrode potential," although that term comes from chemistry, and a physicist would insist that this is a genuine potential-difference, not "absolute."  (The two electrodes after all are the conductive electrolyte on one side, and conductive metal on the other!)  If the SHE potential-difference is 4.44V, then all the values in the table of Electrodes need to be shifted by that amount.  In that case, the polarity of every electrode is negative.  Zinc and copper will both end up with the same polarity.  Copper is not positive-charged.

However, once these concepts are grasped, there's no major need for them, since measurements in electrochemistry are always biased by the inherent Galvani potential of various Reference Electrodes (SHE, or SCE, or Ag/AgCl.)  If we ignore electrostatics considerations, and deny that the electrolyte is itself a conductive electrode, then we can take the reference electrode potential-difference to be zero.  In that case, voltage measurements become trivially easy.  No vibrating non-contact probes needed, no quadrant electrometers or Faraday-cup charge-detectors.

But in that case, surface nanochemistry questions such as @Sharad 's cannot be answered unless we first discuss the zero SHE-potential being a "convenient lie," and the table of Standard Electrode Potentials must be shifted by 4.44V to reflect electrostatic reality.  (All the potentials in that table have the same polarity.  Oxidizing of solid metal always removes positive ions, even for the metals listed on the table "below" the SHE's zero-voltage entry.)

See also: we already learned all this in undergrad classes, right?


Number of possible isomers (including optical/geometric ones) with C4H8O molecular formula

It would have been nicer if you included the isomers what you have been counting. However, based on the isomer counts, I assumed you are in correct track but originally thought you might have counted meso-isomer twice (as two stereoisomers) for the total count. Thus I drew all possible isomers and listed accordingly:

Consequently, I also counted 32 isomers (same as yours). When compared to your count, I realized your count of chain alcohols (6), ethers (5), and carbonyl compounds (3) are all correct.

However, your count of three-membered ring alcohols (7) and three-membered ring ethers (6) are incorrect. There are only 6 possible three-membered ring alcohols (one less than your count) while 7 three-membered ring ethers are possible  (one more than your count; see the chart). Therefore, it might not be the miscount of meso-isomer (if this is the reason, you should count 8 possible three-membered ring ethers), but be an error on structure drawing.

Your final count of for-membered ring (4) and five-membered ring (1) ethers are also correct. Precisely, only 1 four-membered ring alcohol is possible.

Conclusively, you and I both have the same structure count but wrong counts on different kind of structures. You can compare your drawing with attached chart here and find out where you made the mistake.


Why do NMR coupling constants change with temperature?

Let’s use vicinal 3-bond $\ce{H-C-C-H}$ proton coupling constants as an example.  Coupling is a through bond phenomenon and as such the coupling constant will change depending if we have better (the coupling constant will increase) or worse (the coupling constant will decrease) overlap between the two $\ce{C-H}$ bonds involved.  Overlap is maximized when the two, vicinal $\ce{C-H}$ bonds are aligned in an eclipsed conformation ($\ce{H-C-C-H }$ dihedral angle =0°) or anti ($\ce{H-C-C-H }$ dihedral angle =180°)  conformation.  When the dihedral angle is ~90° the overlap of the two vicinal $\ce{C-H}$ bonds is minimized.  This thinking is embodied in the Karplus equation which relates vicinal coupling constants to dihedral angle.  The following graph illustrates this dependence of the vicinal coupling constant on the dihedral angle.

image source

All of this tells us that an experimentally observed vicinal coupling constant will be dependent upon the relative populations of the eclipsed, staggered and anti conformations. Said differently, the vicinal coupling constant will be an average of each of the individual coupling constants for each conformation, weighted by the relative population of that conformation. Since the relative conformer populations change with temperature, the coupling constant will also change with temperature.


Why are beta-ketoacids better at decarboxylation than gamma or delta?

It has to do with the stability of the electrons that are ejected from the bond that breaks. In B-ketoacids, the electrons can resonate onto the oxygen and form an enol/enolate intermediate, which is much more stable than forming a carbanion on a gamma or delta ketoacid (because there is nowhere for the electrons to go other than go on the carbon atom that participates in the bond-breaking event).


Are there some similarities between J-coupling and j,j-coupling?

I don't believe there are any commonalities, and the terms have been derived independently.

The term J coupling in NMR was first proposed in 1951 by Erwin Hahn in his paper  describing the spin echo experiment, in which two frequencies are observed; one at the chemical shift (which they called delta, as was convention, even back then) as expected, and a second frequency which they labelled as frequency J. There is no explanation for why the choice of J is given. Gutowsky and McCall  who observed the same phenomenon independently did not give it a label, and the term J splitting stuck. J coupling describes the relationship between nuclear spin, the intrinsic angular momentum of atomic nuclei.

The term jj coupling relates orbital angular momentum of electrons. I'm not going to pretend to know anything about the origins of the term, nor anything about its application in physics and astronomy. However, the term J is sometimes used to represent total angular momentum, which is made up of individual components ji. It is the interaction between these individual j terms that give rise to jj coupling.

Other than being types of interactions between entities of angular momentum, the two couplings are quite unrelated, as are the terms used to describe them.


Potassium permanganate, $\ce{KMnO4}$, is the most common and probably the most applicable oxidizing agent discussed in organic chemistry teaching classes. $\ce{KMnO4}$ is a very strong oxidizing agent, and is capable of oxidizing a wide range of organic molecules. Usually, it will oxidize organic compounds to appropriate carboxylic acids, but, the final products can vary on the conditions used. If you googled enough to find reduction potentials of $\ce{KMnO4}$, you may find following conditions, which are the most used reduction conditions of $\ce{KMnO4}$ (Wikipedia):
\begin{align}
\ce{MnO4- + H+ + e- &-> HMnO4^-} & \mathrm{E^o} &= \pu{0.90V}\tag{1}\label{1}\\
\ce{MnO4- + 8H+ + 5e- &-> Mn^2+ + 4H2O} & \mathrm{E^o} &= \pu{1.51V}\tag{2}\label{2}\\ \ce{MnO4- + 2H2O + 3e- &-> MnO2(s) + 4OH-} & \mathrm{E^o} &= \pu{1.69V}\tag{3}\label{3}\\
\ce{MnO4- + 4H+ + 3e- &-> MnO2(s) + 2H2O} & \mathrm{E^o} &=\pu{1.7V}\tag{4}\label{4}
\end{align}
Other manganese containing compounds other than $\ce{KMnO4}$, which can also be used as oxidizing reagents are listed below:
\begin{align}
\ce{MnO2(s) + 4H+ + e- &-> Mn^3+ + 2H2O} & \mathrm{E^o} &= \pu{0.95V}\tag{5}\label{5}\\
\ce{MnO2(s) + 4H+ + 2e- &-> Mn^2+ + 2H2O} & \mathrm{E^o} &= \pu{1.23V}\tag{6}\label{6}\\
\ce{HMnO4- + 3H+ + 2e- &-> MnO2(s) + 2H2O} & \mathrm{E^o} &= \pu{2.09V}\tag{7}\label{7}\\
\end{align}

Yet, as most of the viewers already commented, I also believe $\ce{KMnO4}$ does not oxidize alkanes, except in following situations:

$\ce{KMnO4}$ oxidizes alkane carbon atoms, which contain sufficiently weak $\ce{C-H}$ bonds such as those in benzylic positions ($\ce{C-H}$ bonds in the $\alpha$-positions of substituted aromatic rings). These oxidizable alkyl benzene substituents can be -$\ce{CH3}$, or any $1^\mathrm{o}$-alkyl groups such as -$\ce{CH2CH3}$, or $2^\mathrm{o}$-alkyl groups such as -$\ce{CH(CH3)2}$. All of them will be oxidized to carboxylic acid group, -$\ce{CO2H}$ (See Scheme 1 below):

It is also important to note that when the benzylic carbon is a $3^\mathrm{o}$-$\ce{C}$ (e.g., corresponding alkyl groups such as $-\ce{C(CH3)3}$ group in tert-butylbenzene, see the example given in RHS of Scheme 1), the oxidation reaction does not happen.

In addition to benzylic carbons, $\ce{KMnO4}$ also oxidizes weak $\ce{C-C}$ bonds such as $\pi$-bonds in alkenes and alkynes (Scheme 2):

$\ce{KMnO4}$ also oxidizes $\sigma~\ce{C-C}$ bonds such as that in glycol-moiety (e.g., $\ce{HO-C-C-OH}$) and $\ce{C^{\mathrm{sp^2}}-O}$ bonds such as those in phenolic compounds (e.g., it oxidizes phenol to 1,4-benzoquinone, commonly known as para-quinone).

At last, as mentioned above, conditions of the reaction can be adjusted such a way that desired compound can be achied in high yields. For example, see following example (Ref.1):

Note that no benzylic carbon or -$\ce{OCH2}$- was oxidized here.

Late Addition:

Although tertiary $\ce{C-H}$ bond in isobutane (preferred IUPAC name: 2-methylpropane; $\ce{CH3CH(CH3)CH3}$) is relatively weaker than othe alkane $\ce{C-H}$ bonds, I’m highly skeptical of the statement of “Oxidizing agents such as potassium permanganate readily oxidize a tertiary hydrogen atom to a hydroxyl group, e.g., isobutane is oxidized to t-butanol.” (vide infra). To my knowledge, the literature evidence for that type of oxidation reactions does not exist. Nevertheless, 2-methylpropane (isobutane) can be oxidized to 2-methylprop-2-ol (t-butanol) by reacting supercritical-phase isobutane with air in the presence of a catalyst such as $\ce{Pd/C}$ at high temperature, Instead of using $\ce{KMnO4}$ (Ref.2):

  Abstract: tert-2-Butyl alcohol can be synthesized efficiently on $\ce{SiO2 –TiO2}$ or $\ce{Pd–C}$ catalyst if air is directly introduced to the supercritical-phase isobutane.

tert-Butanol (TBA) is the initial material in methyl tert-butyl ether (MTBE) production and methyl methacrylate (MMA) synthesis.  MTBE is high-octane-number gasoline additive and MMA is a resin material for plastic synthesis (e.g., PMMA or poly-methyl methacrylate). Thus, one would have wondered why scientists did not attempted to synthesize TBA in a such an easy way like using $\ce{KMnO4}$.

However, in a different note, almost exclusive $\ce{C-H}$ bond activation has been observed in the reaction of linear and branched alkanes with $\ce{CeO2+}$ (Ref.3)

Reference:

A. Abiko, J. C Roberts, T. Takemasa, S. Masamune, “$\ce{KMnO4}$ revisited: Oxidation of aldehydes to carboxylic acids in the tert-butyl alcohol - aqueous $\ce{NaH2PO4}$ system,” Tetrahedron Letters 1986, 27(38), 4537-4540 (https://doi.org/10.1016/S0040-4039(00)84997-7)
L. Fan, Y. Nakayama, K. Fujimoto, “Air oxidation of supercritical phase isobutane to tert-butyl alcohol,”  Chem. Commun. 1997, (13), 1179-1180 (DOI:10.1039/A702308A).
A. Gunay, K. H. Theopold, “$\ce{C-H}$ Bond Activations by Metal Oxo Compounds,”  Chem. Rev. 2010, 110(2), 1060–1081 (DOI: 10.1021/cr900269x).


Quite a fun one to work out. The first thing to know is that there is a stereoelectronic requirement for the migrating group to be anti to the leaving group ($\ce{N2+}$), for orbital overlap reasons.

(apologies for hand-drawing - doing this on my phone)

You need to consider the chair conformations of both stereoisomers, in both possible ring-flip forms. For the cis stereoisomer, you would expect roughly a 1:1 mixture of ring-flip forms. When the $\ce{OH}$ is axial, the group anti to the LG is H, and the cyclohexanone is formed. Similarly, when $\ce{OH}$ is equatorial, the $\ce{C-C}$ bond is anti and the cyclopentane product is formed.

Now, for the trans-isomer there are two factors favouring the cyclopentane. First, the chair form in which both groups are equatorial will predominate, and this is the stereoisomer that leads to the cyclopentane. Secondly, in the (minor) conformer where the $\ce{OH}$ and $\ce{N2}$ are axial, there is no anti group to migrate according to the normal mechanism. The OH itself is anti to the $\ce{N2}$, and I'm not aware (off the top of my head) of any examples of 1,2-hydroxy shifts. However, I expect some epoxide might be formed via an SN2 pathway.


The relevant document including the rules of formal IUPAC nomenclature is the Nomenclature of Inorganic Chemistry — red book from 2005. Section IR-5.4.2.2 (Use of charge and oxidation numbers) is relevant for the cases you discuss. It reads:

[…]
[…]
The oxidation number (see Sections IR-4.6.1 and IR-9.1.2.8) of an element is indicated by a Roman numeral placed in parentheses immediately following the name (modified by the ending ‘ate’ if necessary) of the element to which it refers. The oxidation number may be positive, negative or zero (represented by the numeral $0$). An oxidation number is always non-negative unless the minus sign is explicitly used (the positive sign is never used). Non-integral oxidation numbers are not used for nomenclature purposes.
Several conventions are observed for inferring oxidation numbers, the use of which is particularly common in the names of compounds of transition elements. Hydrogen is considered positive (oxidation number $\mathrm I$) in combination with non-metallic elements and negative (oxidation number $\mathrm{-I}$) in combination with metallic elements. Organic groups combined with metal atoms are treated sometimes as anions (for example, a methyl ligand is usually considered to be a methanide ion, $\ce{CH3}$), sometimes as neutral (e.g. carbon monooxide). Bonds between atoms of the same species make no contribution to oxidation number.
Note that oxidation numbers are no longer recommended when naming homopolyatomic ions. This is to avoid ambiguity. Oxidation numbers refer to the individual atoms of the element in question, even if they are appended to a name containing a multiplicative prefix, cf. Example 12 above. To conform to this practice, dimercury($2+$) (see Section IR-5.3.2.3) would have to be named dimercury($\mathrm{I}$); dioxide($2-$) (see Section IR-5.3.3.3) would be dioxide($\mathrm{-I}$); and ions such as pentabismuth($4+$) (see Section IR-5.3.2.3) and dioxide($1-$) (see Section IR-5.3.3.3), with fractional formal oxidation numbers, could not be named at all.

While the section in itself does not mention the possibility of omitting oxidation numbers, the examples obviously do:

$\ce{Na2[Fe(CO)4]}$   sodium tetracarbonylferrate($\mathrm{-II}$), or sodium tetracarbonylferrate($2-$)

The generally accepted rule is that oxidation numbers can be left out if the element in question is commonly known to exist in only one oxidation state. That, for example, includes groups 1 and 2 but also elements such as zinc of which only one state is common.


Which atom bears the negative charge in cyanide? I am confused by the MO diagram

As commenters have noted, charge isn't the best choice of a word here, but you're right that the HOMO is more centered on the C, and that's really your question.

The answer is a bit complicated. It relates to those dashed lines on the MO diagram. If we naively construct an MO diagram of cyanide ion, we have two favorable sigma interactions. First, the $2s$ orbitals can interact positively, which I'll call $\sigma_s$. That would be the lowest energy MO on our diagram. Second, we can have an end-to-end interaction of $p$ orbitals, which I'll call $\sigma_p$. Since that $\sigma$ interaction is stronger than a side-by-side $\pi$ interaction, we would expect the $\sigma_p$ MO to be lower in energy than the $\pi$ bonding MOs.

If we look at your MO diagram, though, we see that $\sigma_p$ is actually higher than the $\pi$ MOs, not lower as we predicted. It's the HOMO in the diagram. The reason it is higher than we naively expected is because there is mixing between it and the $\sigma_s$ (and also between their anti-bonding counterparts, but ignore that for now). These two MOs can interact positively or negatively. In the positive interaction, we get the (normalized) sum of the two MOs, which is an $sp$ orbital on C with its large lobe toward N and an $sp$ orbital on N with its large lobe towards C (see the lowest MO in the drawings below). Both large lobes have the same phase, so this is a strong bonding interaction that is the lowest MO on your diagram. Both of the starting bonding MOs are biased in favor of contribution from N, since both the $s$ and $p$ AOs of N are lower in energy than their counterparts on C. Thus, this sum orbital has a stronger N contribution as well.

In the negative interaction, we subtract $\sigma_s$ from $\sigma_p$ and again get $sp$-type orbitals on both atoms, but now the large lobes are facing away from each other. That makes the bonding interaction much weaker, so much so that it moves higher in energy than the $\pi$ bonding orbitals. Furthermore, subtraction takes away more N-character than it does C, and you end up with a greater contribution to the new orbital from C than N. As you get more comfortable with MOs, you can verify this mathematically. Another way to think about it is that the final MO is higher than the either the average of a $Cs$ and a $Cp$ orbital or the average of an $Ns$ and an $Np$ orbital, but it's closer to the C $sp$ average (which is higher than the N $sp$ average), so it has more C character by the same logic you used to predict that it would have more N character (which it would if it were just made up of $p$ orbitals).

Below is a drawing of the five occupied orbitals where you can see that the $\sigma$ ones have become $sp$-like and that the carbon contribution is greater in the HOMO. C is on the left and N on the right.


How can methane absorb infrared light within wide range of frequency but CO2 can't?

Comparing the vibrational spectra of two different molecules is just like wondering why two people do not have the same fingerprints. The simplest answer is that these two molecules are different molecules with different atoms hence their vibration modes are supposed to be different.

IR Spectra of Methane and Carbon dioxide are available in NIST. This is how real spectroscopy looked like before the focus shifted from true science to fancier presentations and instruments became black-boxes. Anyway, all you have to note both methane and carbon dioxide absorb quite strongly in the typical range of infrared light 400 to 3800 $cm^{-1}$. The question of wide range does not arise as you can see from the spectra, their absorption band positions are different. Methane's spectra is more complex just because it has more active vibrations (dipole moment change is non-zero) and one can see the fine rotational structure of methane but carbon's dioxide's rotational lines are not resolved.


Mixing isopropyl alcohol and propylene glycol

The two chemicals can be mixed without any reaction between them. However, you should add enough water to avoid fire hazards after the winter. Take a look at this table to find out, what mixture should suffice.

You should never pour solvents onto your soil or down the drain. Contact local authorities to find out, how you can safely dispose of that mixture. You can, of course, reuse the mixture and store it in some plastic cans if you need to empty the basketball hoop.


Why do we say "approximation" in the dipole approximation in spectroscopy?

As per your linked reference, it's an approximation because only the first term in the expansion (here, unity) is considered.

The issue with transitions that are allowed or forbidden arises as a result of using such an approximation: "Forbidden" transitions are not strictly forbidden, it's just that they are weak and, in the linear approximation of the transition dipole operator, are usually not observed.


What is there in an atom between the nucleus and electrons?

It is not necessarily empty space.

For example the flux of neutrinos on Earth is about $3 \times 10^{12}/cm^2/s$.  At a given moment, a neutrino may or may not be within the atom or even within the nucleus of the atom. Similarly, other particles may be present within an atom at a given time.

Also, an electron may be inside the nucleus in accordance with the Fermi Contact Interaction.  So if you have a hydrogen atom, having only one electron, and the electron is inside the nucleus at a given moment, there will be no "between" for there to be space, empty or not.


What causes KNO3 to decompose into KNO2 and Oxygen in gunpowder?

KNO3 (Potassium Nitrate) is a molecule that is used with pyrotechnics and is typically mixed with Carbon and Sulfur as fuels.

When Potassium Nitrate is thermally decomposed it forms Potassium Nitrite and Oxygen.
$$\ce{2KNO3 -> 2KNO2 + O2}$$

Potassium Nitrate is known as an oxidizer for this reason (it provides Oxygen to the reaction). This free Oxygen then reacts with the fuel sources in combustion to create an exothermic reaction. Carbon is typically the main fuel source and Sulfur is added to reduce the ignition temperature required to begin combustion of the Carbon-based fuel.

If you had a mixture of Potassium Nitrate and Sucrose (table sugar), you would see something like this:
$$\ce{5C12H22O11 + 48KNO3 -> 24K2CO3 + 36CO2 + 55H2O +24N2}$$

Where you can see that you have a combustion of the two reactants (after the Potassium Nitrate thermally decomposes, it provides the necessary Oxygen to begin combustion) resulting in the products of combustion, Carbon Dioxide and water, along with the byproducts of Nitrogen gas and Potassium Carbonate.


What is the empirical formula for kerosene?

On p. 28 of the NIST document linked in the question, there is a surrogate mixture formulation of 14 hydrocarbons which closely approximates RP-1. Its empirical formula is $\mathrm {C_{11.8}H_{23.0}}$, and this should be close enough for any sort of kerosene. The properties of this surrogate are given on p. 7.

This means that burning $\pu {1 kg}$ of kerosene in air produces about $\pu {3.15 kg}$ of $\ce {CO2}$ and $\pu {1.26 kg}$ of $\ce {H2O}$ with the extra mass, $\pu {3.41 kg}$, coming from oxygen. Alternately, if you have $\pu {1 kg}$ of a perfect stoichiometric mix of RP-1 and liquid oxygen, burning it would produce $\pu {715 g}$ of $\ce {CO2}$ and $\pu {285 g}$ of $\ce {H2O}$.


how can electrons exchange from 3d orbital to 4s if their energy levels are not equal?

A sort of simplified answer but here goes...

Without some other excitation, electrons only "fall in" towards the nucleus. This is sort of like the fact that rocks roll down hill not up hill. So once an electron "falls in" it has excess energy that must be shed. It can either give the energy off as a photon or transfer the energy to one or more other electron(s) which then go into some excited state (Auger effect).

The transitions occur  according to selection rules. If the selection rules don't allow a particular transition then it is forbidden. If the atom is in free space with no other perturbations, then forbidden transitions would not occur. However for molecules, in liquids and gases, because of vibrational and rotational microstates such a transition may actually occur but it will be relatively weak in intensity to the transitions allowed by the selection rules. Likewise forbidden transitions may occur in solids because of the vibrations of the atoms within the crystal which break the symmetry of an atom, or its excited state.

Now if an atom or molecule is excited by an external energy source, then an electron can move from a lower energy orbital to a empty higher energy orbital. Again these transitions obey the selection rules.


Why is there a bulge at the top of my volumetric flask?

From ASTM manual [1, p. 204], emphasis mine:

Volumetric flasks should be diluted to just below the line, mixed thoroughly, then allowed to stand for 4 to 24 h to allow any heat from an acid-water reaction to dissipate. [...] Thorough mixing consists of at least 20 complete inversions of the volumetric flask, allowing enough time on each for the air bubble to completely traverse the length of the vessel. Some borosilicate volumetric flasks contain a bulge in the neck above the line as an aid to proper mixing.

Volumetric pipetts also have a large bulge in the middle, but it serves a different purpose – to keep the $\rm length : volume$ ratio minimal and still maintain quick liquid collection with high precision.
Reference

Dulski, T. R. A Manual for the Chemical Analysis of Metals; ASTM manual series; ASTM: West Conshohocken, PA, 1996. ISBN 978-0-8031-2066-2.


What do dashes mean in chemical formulas?

This nomenclature refers to a core-shell arrangement of nanoparticles. This is where you have a nanoparticle core consisting of one material that has been coated/decorated/wrapped/overlaid with another material. There are a few accepted forms of nomenclature for describing these types of nanoparticles:

core@shell In your case, it would be Fe2O3@Au
core-shell In your case, it would be Fe2O3-Au
core/shell In your case, it would be Fe2O3/Au

There are plenty of examples of each syntax in the literature, and has been used for eons.

Others may answers for the additional questions you have here.


There is yet no consensus on what occurs when gold(III) hydroxide is heated. Part of the problem is that the exact structure of $\ce{Au(OH)3}$ is unknown, or dubious at best. It has been inconsistently formulated as "truly" $\ce{Au^{III}-(OH)3}$$^{[0]\ [1]\ [2]\ [\text{a}]}$, oxide-hydroxide hydrate $\ce{AuOOH*H2O}$$^{[3]}$, oxide hydrate $\ce{Au2O3*xH2O}$$^{[0]\ [4]}$, even an acid-like hydrate $\ce{H[AuO2]*H2O}$$^{[5]\ [6]}$. I am going to use $\ce{[AuOOH*H2O]^{*}}$ with an asterisk in equations to keep this controversy in mind.

At temperatures below $130\ ^\circ \mathrm{C}$, the decomposition is probably best described by$^{[3]\ [7]}$

$$\ce{[AuOOH*H2O]^{*} ->[below $130\ ^\circ \mathrm{C}$] AuOOH + H2O}.$$

Around $140\ ^\circ \mathrm{C}$ the oxide-hydroxide decomposes further.$^{[3]\ [7]}$

$$\ce{2AuOOH ->[$140-150\ ^\circ \mathrm{C}$] Au2O3 + H2O}$$

Gold(III) oxide is already considerably unstable from $155\ ^\circ \mathrm{C}$ .$^{[1]\ [3]}$ It is not definitively shown what reaction follows. Some posit $\ce{AuO}$$^{[3]}$, which might not exist or actually  be better described as $\ce{Au[AuO2]}$$^{[3]}$, others propose a combined product of elemental gold $\ce{Au}$ and $\ce{Au2O}$$^{[1]}$. There are studies that rule out these products, at least as isolable intermediates, and instead propose $\ce{-> Au + O2}$$^{[0]\ [2]\ [7]\ [8]}$.

$$\ce{2Au2O3 ->[$150-165\ ^\circ \mathrm{C}$] 4AuO + O2}$$

$$\ce{4Au2O3 ->[$150-165\ ^\circ \mathrm{C}$] 4Au + 2Au2O + 5O2}$$

$$\ce{2Au2O3 ->[$150-165;\ 300\ ^\circ \mathrm{C}$] 4Au + 3O2}$$

The proponents of the bottom reaction, a more direct $\ce{-> Au + O2}$ claim that the oxides were mistaken for a complex mixture of $\ce{Au}$, $\ce{Au2O3}$, and interstitial or dissolved $\ce{O2}$.$^{[9]}$

The authors$^{[9]}$, nevertheless, still note

  It  is  not  yet  clear  how  this  “dissolved”  oxygen  should  be  described.  Is it really dissolved in interstitial metal lattice positions,  are there trapped $\ce{Au2O3}$ oxide  clusters in a densely sintered metal  matrix, or encapsulated oxygen gas in lattice imperfections of the gold?$^{[9]}$

If the intermediates $\ce{AuO}$ and $\ce{Au2O}$ exist, then not for too long.  The decomposition of these oxides would most likely consist of multiple interconversions between $\ce{AuO}$ and $\ce{Au2O}$. For example$^{[1]\ [3]\ [\text{b}]}$,

$$\ce{2Au2O + O2 <=> 4AuO},$$
$$\ce{Au + AuO <=> Au2O},$$
$$\ce{Au2O -> Au + O2}.$$

Over $300\ ^\circ \mathrm{C}$ only gold metal $\ce{Au}$ and dioxygen $\ce{O2}$ remain.$^{[9]}$ Another experiment$^{[10]}$ supports this conclusion but differentiates between surface and subsurface oxides. The latter are found to be stable well over $300\ ^\circ \mathrm{C}$$^{[10]}$. More research is required to shed light into what is actually the main decomposition pathway. An additional way of expressing the process at lower temperatures is$^{[9]}$

$$\ce{6AuOOH -> 4Au + Au2O3 + 3O2 + 3H2O.}$$

$[\text{a}]$ $\ce{Au(OH)3}$ is presumed to assume a square planar coordination with four $\ce{-OH}$-groups.$^{[2]}$ A most recent study provides evidence for a four-coordinated linear polymer.$^{[0]}$

$[\text{b}]$ Equations are written on the analogy with copper.$^{[1]\ [3]}$

$[0]$ Daisuke Kawamoto, Hiroaki Ando, Hironori Ohashi, Yasuhiro Kobayashi, Tetsuo Honma, Tamao Ishida, Makoto Tokunaga, Yoshihiro Okaue, Satoshi Utsunomiya, Takushi Yokoyama. 'Structure of a Gold(III) Hydroxide and Determination of Its Solubility'. Bulletin of the Chemical Society of Japan, ($2016$), 89(11), 1385$-$1390. DOI: 10.1246/bcsj.20160228.

$[1]$ H. Karik, K. Truus. ($2003$). Elementide keemia. (p 313)

$[2]$ C. Klanner, N. Weiher, E. A. Willneff, S. L. M. Schroeder. 'In Situ Quick XAFS Studies of the Thermal Decomposition of Gold(III) Oxide'. Hasyweb Annual Report $2000$, ($2000$), part 1, 46, 2986.

$[3]$ R. Ripan, I. Ceteanu. ($1969$). Chimia metalelor. II volume. (p 768$-$769)

$[4]$ I. V. Mironov. 'Properties of Gold(III) Hydroxide and Aquahydroxogold(III) Complexes in Aqueous Solution'. Russian Journal of Inorganic Chemistry, ($2005$), 50(7), 1115

$[5]$ Κ. V. Astakhov, A. G. Elitsur, K. M. Nickoluev. Zhurnal obshcheĭ khimii, ($1951$), 21, 1753.

$[6]$ A. G. Massey, N. R. Thompson, B. F. G. Johnson, R. Davis. ($1973$). The Chemistry of Copper, Silver, Gold. (1975 reprint)

$[7]$ O. Diaz-Morales, F. Calle-Vallejo, C. de Munck, M. T. M. Koper. 'Electrochemical water splitting by gold: evidence for an oxide decomposition mechanism'. Chemical Science,  ($2013$), 4, 2334$-$2343. DOI: 10.1039/C3SC50301A.

$[8]$ C. Klanner, N. Weiher, E. A. Willneff, S. L. M. Schroeder, C. Figulla-Kroschel, M. Jansen. 'Thermal Stability and Decomposition Kinetics of Crystalline  Gold(III) Oxide and Disordered Gold(III) Oxides/Hydroxides'. Hasyweb Annual Report $2001$, ($2001$), part 1, 46, 5440.

$[9]$ M. Peuckert,  F. P. Coenen, H. P. Bonzel. 'On the Surface  Oxidation of a Gold Electrode in $1\ \mathrm{N} \ce{H2SO4}$ Electrolyte'. Surface  Science, ($1984$), 141, 2$-$3, 515$-$532.

$[10]$ L. K. Ono, B. R. Cuenya. 'Formation and Thermal Stability of $\ce{Au2O3}$ on Gold Nanoparticles: Size and Support Effects'. Journal of Physical Chemistry C, ($2008$), 112(12), 4676$–$4686. DOI: 10.1021/jp711277u.


Gaussian calculation problem - Maxcycle values for Opt and SCF

You are mixing two independent, but slightly related topics.

First is the SCF convergence, i.e. obtaining the wavefunction. This step is necessary for any calculation, without it the Gaussian crashes. For well behaved molecules, as the prostaglandines (organics) most probably are, the solution should be found fast (<30 SCF cycles). It also does not matter which converger you choose, i.e. QC or DIIS. How "tight" you decide for the SCF affects the final energy, but only the number of digits (precision). The SCF does not have any "verytight" option, according to manual, only the Conver=N.

On the other hand, the integral and grid settings do have effect on the energies, so all calculations have to be done with the same grid. I won't recommend changing the acc2e settings, the error is negligible.

Regarding the geometry convergence, you should obtain an optimized geometry in less than 100 steps for reasonable organic systems. If this does not happen, something is rotten in your setup and you should look deeper into the problem. In principle, you cannot converge the geometry to "tight" without sufficient accuracy of the SCF, namely the ultrafine grid, which you now do use.

How good the optimization was can be guessed by the frequencies which should be zero (translations and rotations). If they are significantly off, optimize the geometry more tightly. But as said above, you cannot optimize on noisy surface, where the noise is produced by the grid and SCF convergence.


Nanoscopy Nobel Prize - why are all the tools in the visible light range?

As far as fluorophores go, if the excitation wavelength is too short (< 350 nm), the incident light will get stuck in the microscope.  Near-infrared fluorophores, on the other hand, suffer from several drawbacks, especially low quantum yields from competing non-radiative processes.


The fumes will contain trace amounts of NaOH. The fumes are microscopic water droplets, and these droplets will still be caustic, as they are essentially a solution of sodium hydroxide.

There is no real chemical reaction going on here, but rather a dissociation of the sodium and hydroxide ions, which is an exothermic process. Solid sodium hydroxide is 'hazardous', the resulting solution is 'hazardous' and the fumes above the dissolving lye are 'hazardous'.


Why isn't the energy that is used to overcome the activation energy of magnesium metal atoms to react with oxygen, not taken into account?

Typically, the activation energy is much higher than the difference in energy between reactants and products. The activation energy gets used when you go from reactants to activated complex, and more or less becomes available again when you go from activated complex to product. As Waylander stated in the comments, exo- or endothermic is based on the difference in energy between reactants and products, not the size of the activation energy (otherwise, all reactions would be endothermic).

  If magnesium reacts with oxygen to produce magnesium oxide only on the application of heat, then why isn't it categorised as an endothermic reaction?

Just because you are heating something to get it to react does not mean there is a net flow of heat from the surrounding into the reaction. It would be more accurate to say that you are increasing the temperature to allow the reaction to proceed.

  [from the comments]: You just need to heat up a 1/4 inch and get that burning. Then the burning itself supplies enough heat to keep the ribbon burning.

Again, this is about the temperature, not about exothermic or endothermic. There are multiple reasons why a reaction would happen at high temperatures but not at low temperatures:

Change in equilibrium constant: For example, water vapor forms dew at lower temperature, and dew evaporates again at higher temperature. The activation energy is sufficiently low that evaporation and condensation take place at either temperature, but the net direction changes.
Reaction rates: Some reactions have a high activation energy and are too slow at low temperature, so you have to raise the temperature to observe them. This does not mean they are endothermic (check out the thermite reaction).
Allowing the reactants to come together: Magnesium is covered in a layer of magnesium oxide, preventing oxygen in air to react with the magnesium. At higher temperature, magnesium metal melts and comes in contact with air.
A slow step in a multi-step chain reaction: In radical reactions, the initial formation of a radical has the highest activation energy, whereas reaction of a radical with other reactants to form another radical (chain propagation) has a lower activation energy. In this case, the first step would require higher temperature (like a spark) or a catalyst (radical starter) and then the chain reaction can proceed at low temperature.

For all but the first, it does not matter whether the reaction is endothermic or exothermic - higher temperature helps or is required to observe a reaction. You can have a endothermic reaction even at lower temperatures. The heat transfer can happen even under low temperature conditions because the molecules have thermal energy nonetheless.

Using day-to-day language, that sounds weird because we associate heat with temperatures higher than body temperature, so we might doubt that there is heat in cold water. However, reactions that cool down the water even more demonstrate that the water can transfer heat to these processes (e.g. cold water can melt ice).


The hydroxyl groups in phenols are ortho-/para-directing. So, if you want to introduce a group in the meta position with respect to a hydroxyl group you will have to "overwrite" the ortho-/para-preference. This can be done by starting out with meta-directing group which can later be transformed into $\ce{OH}$ in the position where the hydroxyl group shall be. Then you only need to introduce the group you want to be in the meta-position by a simple electrophilic aromatic substitution reaction and subsequently convert the meta-directing group into $\ce{OH}$.

An example for this procedure would be to start out with a carbonyl group at the position of the $\ce{OH}$ group, then nitrate the aromatic ring in the meta position and employ the Baeyer-Villiger oxidation to introduce a $\ce{OH}$ group:

Or you start out with a nitro group at the position of the $\ce{OH}$ group, then introduce chlorine in the meta position of the aromatic ring, reduce the nitro group to an amine, which is then coverted into an diazonium salt which in turn is then transformed into a $\ce{OH}$ group using a Schiemann-type reaction:


Nomenclature confusion regarding carboxylate and ester

There is not supposed to be a space between the substituent prefixes (e.g. methyl) and the parent structure (e.g. propanoate), so your second structure is methylpropanoate, or more correctly 2-methylpropanoate.

In the ester there is a space between methyl and propanoate, i.e. methyl propanoate.


Reaction rate coefficient constant multiplier for equation with multiple atoms consumed?

In general if the reaction is $\ce{aA + bB \to cC +dD}$ the the rate is written as $$r=\displaystyle -\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$$.

If the reaction is indeed second order in oxygen atoms as stated by the OP, then the first equation given by the OP is correct, but unusual (you could just leave the factor two in the denominator on the left hand side instead of multiplying by two on the right hand side).


About dative covalent bonding and expanded octet rule

To answer your first question: the concept of octet expansion for main group metals is no longer generally considered correct and only upheld by the more traditional organic chemists that don’t deal with inorganic bonding situations on a day-to-day basis — or anybody else who didn’t need to recheck what they were taught decades ago. Nowadays, these seemingly octet-expanding structures such as $\ce{XeO4}$ or $\ce{XeF2}$ are explained by dative bonds and/or four-electron-three-centre bonds; both have the beauty of not requiring d orbitals (whose energy would be comparable to the next shell’s s orbital; remember the aufbau principle!).

To answer your second question: no. A dative bond always requires two electrons that are given to one atom upon supply of the bond dissociation energy. It doesn’t matter how a certain bond is created, it only matters how it dissociates; and if it dissociates and was a bond, it will have to have had an electron pair.


How is titanium metal obtained from natural resources?

Titanium is found as an ore, according to the Titanium Occurrences Wikipedia page, specifically, the 2 most economical ores are:

Ores of titanium

Rutile $\ce{TiO2}$ and Ilmenite $\ce{Fe^{2+}TiO3}$

Extraction of titanium metal from ore

According to the chemguide page Titanium,

  Titanium can't be extracted by reducing the ore using carbon as a cheap reducing agent. The problem is that titanium forms a carbide, TiC, if it is heated with carbon, so you don't get the pure metal that you need. The presence of the carbide makes the metal very brittle.

Rather, the ore is chlorinated, then reduced as shown in the schematic from the webpage Titanium Exposed (the chemical processes are included):

Chemguide states that either $Mg$ or $Na$ can be used in the reduction stage, which adds to the cost of production as they need to be extracted themselves.

Asides from the links used in this answer, another good resource is Chemistry of Titanium (Professor John Lancashire - University of the West Indies)


calculate the final concentration of this equation

[…] he should had used the moles of potassium iodide since it is the limiting reactant.

Did any reaction take place upon mixing solutions of $\ce{Pb(NO3)2}$ and $\ce{KI}$?
If so, did the reaction change the number of moles of $\ce{NO3-}$ in solution by precipitation, oxidation, reduction, etc.?
Would the result of the calculation be different if just 300 mL of water were added to the $\ce{Pb(NO3)2}$ solution?

Update I

Now for the rocket science (solubilities at 20 °C):

\begin{array}{cr}
\mathrm{compound} & \mathrm{solubility\ [g\cdot L^{-1}]} \\ \hline
\ce{Pb(NO3)2} & 522\\
\ce{KI} & 1430\\
\ce{KNO3} & 316\\
\ce{PbI2} & 0.76\\
\end{array}

The rather low solubility of $\ce{PbI2}$ suggests that on combination of the two solutions, the following may happen:

$$\ce{Pb^{2+} + 2 NO3- + 2 K+ + 2 I- -> 2 K+ + 2 NO3- + PbI2 v}$$

So again:

Does this in any way change the absolute amount of  $\ce{NO3-}$ in solution?
What does change for the concentration of $\ce{NO3-}$ when 300 mL of another solution is added to 260ml of 1.2 M $\ce{Pb(NO3)2}$ solution?

Update II

Initial $\ce{Pb(NO3)2}$ solution

   260 mL of a 1.2M solution of $\ce{Pb(NO3)2}$ contain $1.2\ \mathrm{\frac{mol}{L}}\ \cdot\ 2\cdot\ 0.26\ \mathrm{L} = 0.624\ \mathrm{mol}\ \ce{NO3-}$

After mixing

   Addition of 300 mL of the $\ce{KI}$ solution results in the precipitation of $\ce{PbI2}$, but does not change the absolute amount of  $\ce{NO3-}$ in solution. However, it changes the total volume and hence the concentration, which calculates to $\frac{0.624\ \mathrm{mol}}{(0.26 + 0.3) \mathrm{L}} = 1.11 \ M$


Monosodium Diphosphate (NaH3P2O7) Preparation

$\ce{NaH3P2O7}$ is an intermediate acid-salt formed during the neutralization of pyrophosphoric acid ($\ce{H4P2O7}$) which becomes unstable as pH rises and leads to formation of trisodium diphosphate, $\ce{Na3HP2O7}$:
$$\ce{H4P2O7 + NaOH ->[pH < 3] NaH3P2O7 + H2O}$$
$$\ce{NaH3P2O7 + NaOH ->[pH = 3-9] Na2H2P2O7 + H2O}$$
$$\ce{Na2H2P2O7 + NaOH ->[pH > 9] Na3HP2O7 + H2O}$$
This acid-salt can also be produced by the reaction of trisodium phosphate and pyrophosphoric acid:
$$\ce{Na3PO4 + H4P2O7 -> NaH3P2O7 + Na2HPO4}$$
Notes and references:

Study of crystallization processes in some rare earth vanadate, tungstate and phosphate systems under hydrothermal conditions, K Byrappa and B Nirmala, Indian J. Phyx. 73A (5), 621-632 (1999) (PDF)
Its CAS number is 13847-74-0
The first dissociation of pyrophosphoric acid gives the anion, $\ce{H3P2O7^-}$. The dissociation constant is pretty large compared to 2nd, 3rd, and 4th dissociation of the acid.

$$\ce{H4P2O7 + H2O <=> H3O+ + H3P2O7¯ ~~~~ K_{a1} \sim 10^{–1}}$$


Why is Libethenite (Cu2(OH)PO4) color green?

First of all, why do you think your copper is not oxidized? Each copper has a charge of +II while $\ce{OH^-}$ and $\ce{PO_4^3-}$ add up to -4.

For all further discussions we need to look deeper into the crystal structure. The compound crystallizes with two crystallographically destinguishable copper ions. One forms a $\ce{[Cu(PO_4)_4(OH)_2]^12-}$ octahedron while the other one is, at least to me only fivefold coordinated to form a trigonal bipyramid $\ce{[Cu(PO_4)_4(OH)]^11-}$.
Of course those are no simple $\ce{OH}$ units as there is further bridging and connections happening but for a simple view this should be okay. So for the color there are, among other effects, still $\ce{d-d}$ transitions happening.

EDIT:

To add something to the color change in the reaction itself. In your copper phosphate you have copper polyhedra that only feature phosphate as ligands, for the $\ce{[Cu_3(PO_4)_2]}$ these would be trigonal bipyramid and square plane.So you have the addition of $\ce{OH^-}$ to these polyhedra as well as some substitution. And introducing a different ligand will also change your splitting energy and therefore change the color.


Is there a method to recover camphene from sublimation at ambient conditions?

MIT has a brilliant technique for sublimating and condensing:

https://www.youtube.com/watch?v=dBNELFi5XiY

The sample is placed in a culture dish sitting on a hotplate. The lid is placed on top.  Finally a beaker with ice is placed on top of the dish to condense the sublimate. After a while crystals form on the top of the lid. These pure crystals are periodically scraped with a policeman and collected.


What is the pKa of 2-methyl-2,4-pentanediol?

Do you mean 2-methyl-2,4-pentanediol?

The $pK_a$ values of 2-propanol (isopropanol) and 2-methylpropan-2-ol (tert. butanol) are 16.5 and 17, respectively and I'd expect your diol to be in this range range, unless deprotonation is featured by intramolecular hydrogen bonding. In that case, the $pK_a$ might be somewhat smaller, around 15.


Could a 'polymer' be formed by hydrogen bonding?

Since a polymer is a

"molecule of high relative molecular mass, the structure of which essentially comprises the multiple repetition of units derived, actually or conceptually, from molecules of low relative molecular mass." (IUPAC Gold Book)

joining two monomers into a dimer is just a specific wording for the more general term polymer like it is for trimer, tetramer, pentamer, hexamer, etc.  While the term polymers often is used about plastics, cellulose, starch, or chitin equally are polymers, too.  Note, this definition does constraint the binding of the monomers to covalent bonds.
The depictions shown as examples by you are valid representations when describing, e.g., the hydrogen-bond pattern in organic crystals.  Here, hydrogen bond donors and hydrogen bond acceptors may contribute significantly to the overall stabilization of the supramolecule an organic crystal may be considered.  Magaret C. Etter and Joel Bernstein popularized  formalisms to identify and systematically describe these pattern.
The one described by you matches for example the situation for $p$-terephthalic acid yielding (in principle) infinite bands hold by the motif $R^2_2(8)$ (a ring of eight atoms in total, comprising 2 donor, and 2 acceptors):

(1990AccChemRes120, includes a guide how to define these graph sets)
This approach equally may be extended to describe the interactions if the monomers are chemically different:

($R^2_4(8)$ motif about the interaction of phosphine / arsine oxides and  nitroanilines in co-crystals, reviewed in 1995ACIE1555)


B-strain

Both B- and F-strains are the terms I found were originally used to describe steric hindrance in complexes [1, pp. 66-67]:

  The concept that steric hindrance in a monodentate ligand can affect the affinity of that ligand for a metal was first introduced by Brown and Bartholomay (1944) who suggested that co-ordination of a tertiary amine with bulky substituents would create strain within the amine. This strain, termed B-strain on account of it occurring behind the ligand atom with respect to the metal, would arise because although in the unco-ordinated amine any steric interaction between the bulky substituents could be relieved to some extent by an increase in the $\ce{C-N-C}$ angles.

Interestingly, further it is pointed out that at the same time there is no other solid evidences that B strain exists. Although modern organic chemistry textbooks are widely using this term for tertiary compounds.

From [2, p. 417]:

  Steric effects of another kind become important in highly branched substrates,
  and ionization can be facilitated by relief of steric crowding in going from the tetrahedral ground state to the TS for ionization. The relative hydrolysis rates in 80% aqueous acetone of $t$-butyl $p$-nitrobenzoate and 2,3,3-trimethyl-2-butyl $p$-nitrobenzoate are 1:4.4.

  This effect has been called B-strain (back strain), and in this example only a modest rate enhancement is observed. As the size of the groups is increased, the effect on rate becomes larger. When all three of the groups in the above example are t-butyl, the solvolysis occurs 13,500 times faster than in t-butyl p-nitrobenzoate.

According to [3, p. 350]:

  Not all steric effects decrease reaction rates. In the hydrolysis of $\ce{RCl}$ by an $\mathrm{S_N1}$ mechanism ..., the ﬁrst step, which is rate determining, involves ionization of the alkyl chloride to a carbocation:

  The central carbon in the alkyl chloride is $\mathrm{sp^3}$ hybridized, with angles of $\sim 109.5^\circ$, but when it is converted to the carbocation, the hybridization becomes $\mathrm{sp^2}$ and the preferred angle is $120^\circ$. If the halide is tertiary and the three alkyl groups are large enough, they will
  be pushed together by the enforced tetrahedral angle, resulting in strain. This type of strain is called B strain (for back strain), and can be relieved by ionization to the carbocation.

  [...]

  Substitution of ethyl groups for the methyl groups of tert-butyl chloride
  does not cause B strain; the increase in rate is relatively small, and the rate smoothly rises with the increasing number of ethyl groups. The increase is caused by normal ﬁeld and resonance (hyperconjugation) effects. Substitution by one isopropyl group is not greatly different. But with the second isopropyl group the crowding is now great enough to cause B strain, and the rate is increased 10-fold. Substitution of a third isopropyl group increases the rate still more.

Further in [3, p. 419]:

  For some tertiary substrates, the rate of $S_N1$ reactions is greatly increased by the relief of B strain in the formation of the carbocation. Except where B strain is involved, $\beta$ branching has little effect on the $S_N1$ mechanism, except that carbocations with $\beta$ branching undergo rearrangements readily.

There is also a Baeyer strain aka angle strain most noticeable for three-membered rings ($60^\circ$ is far from expected tetrahedral angles), which can also probably be abbreviated alike. [3, p. 193]

It also seems like earlier sources used B-strain as a reference to H. C. Brown [4, p. 291]:

  Hammond has proposed a steric "B-strain" (see H. C. Brown, J. Chem. Soc. (London) p. 1248 (1956)) argument to explain the greater stability of primary over secondary or tertiary carbanions. From the viewpoint of hybridization, structure A with $90^\circ$ geometry is preferred in carbanions. But this conformation brings the substituents too close together, and a compromise structure with larger bond angles results. Methide ion, $\ce{^-CH3}$, with the smallest substiuent groups, should be the most stable simple carbanion, since it can more nearly adopt the small angled conformation A preferred on electronic grounds. The more alkyl groups are added, the less able sterically are
  the anions to adopt the favorable conformations with smaller angles, and
  the less stable and more reactive the intermediates become.

F-strain

[1, p. 68] illustrates the origins of the term F-strain:

  The boron trifluoride and boron trichloride complexes of diethyl ether are less stable than those of dimethyl ether, and the same order of stability is observed for the complexes of diethyl and dimethyl sulphides. As expected, the steric interaction is decreased as the distance between the metal and ligand atom is increased. Thus it is decreased when the metal atom is changed from boron to aluminium, and when the ligand atom is changed from oxygen to sulphur. It is largely eliminated, of course, if the substituents on the ligand atom are held back away from the metal, and thus very stable complexes are formed by the tertiary amine quinuclidine, while complexes of tetrahydrofuran and tetrahydropyran (and also the analogous sulphur heterocycles) are far more stable than those of their acyclic analogues.

  Strain which arises as the result of this type of steric interaction has been termed F-strain by H. C. Brown -- who has carried out extensive investigations in this area -- on the grounds that the interactions responsible for the strain occurs mainly in front of the ligand atom with respect to the metal. For a critical but now rather dated review of H. C. Brown’s work, see Hammond (1956).

References

Houghton, R. P. Metal Complexes-Organic Chemistry, 1st ed.; Cambridge University Press: Cambridge; New York, 1979.
Carey, F. A.; Sundberg, R. J. Advanced organic chemistry Part A. Structure and Mechanisms, 5th ed.; Springer: New York, 2007.
Smith, M. March’s advanced organic chemistry: reactions, mechanisms, and structure, 7th ed.; Wiley: Hoboken, New Jersey, 2013.
Hartley, H.; Karabatsos, G. J. Advances in alicyclic chemistry; 1966; Vol. 1.


Metallic aluminum cannot be obtained from a solution of its salts. It is because pure aluminum reacts with water, a little bit like sodium. Of course a piece or a sheet of aluminum does not usually react with water, because the metal is covered by a thin, colorless and waterproof layer of aluminum oxide, which prevents the contact between aluminum and air. And if you remove this layer with a knife, the oxide layer will immediately be remade by air.
The only way of showing that aluminum reacts with water is to dip the aluminum piece in a solution of Mercury chloride. There will be a surface reaction like $$\ce{2 Al + 3 HgCl2 ->  3 Hg + 2 AlCl3}$$ Then the mercury Hg will make an alloy with aluminum. And this alloy cannot be covered by the alumina shield. So such a piece of amalgamated aluminum will be quickly oxidized in air, and will produce long white filaments of $\ce{Al2O3}$ or $\ce{Al(OH)3}$ if abandoned in air for a couple of minutes due to one of the following reactions $$\ce{4Al + 3 O2  -> 2 Al2O3}$$  $$\ce{4 Al + 3 O2 + 6 H2O ->  4 Al(OH)3}$$ This production of alumina filaments is very interessant to observe. Of course it is not allowed any more today, because of the toxicity of mercury.


Answer to your question is within the same Wikipedia page, which states following:

  Mechanism of action: Nitrofurantoin is concentrated in the urine, leading to higher and more effective levels in the urinary tract than in other tissues or compartments. With a $\pu{100 mg}$ oral dose, plasma levels are typically less than $\pu{1 \mu g/ml}$ while in the urine it reaches $\pu{200 \mu g/ml}$. The mechanism of action is unique and complex. The drug works by damaging bacterial DNA, since its reduced form is highly reactive. This is made possible by the rapid reduction of nitrofurantoin inside the bacterial cell by flavoproteins (nitrofuran reductase) to multiple reactive intermediates that attack ribosomal proteins, DNA,[38] respiration, pyruvate metabolism and other macromolecules within the cell. Nitrofurantoin exerts greater effects on bacterial cells than mammalian cells because bacterial cells activate the drug more rapidly. It is not known which of the actions of nitrofurantoin is primarily responsible for its bactericidal activity. The broad mechanism of action for this drug likely is responsible for the low development of resistance to its effects, as the drug affects many different processes important to the bacterial cell.

High availability of nitrofurantoin in the urine ($\pu{200 \mu g/ml}$) compared to that in plasma ($\pu{1 \mu g/ml}$) is due to its high solubility in the urine. At this concentration, it exceeds the requirement for minimum inhibitory concentration for bacteria ($\pu{32 \mu g/ml}$), and as a result, nitrofurantoin becomes bactericidal in urine at therapeutical doses ($\pu{100 mg/capsule}$). Keep in mind that reduced nitrofurantoin (its metabolite by nitrofuran reductase) is much more reactive than nitrofurantoin, thus shorter half-life make it highly lethal to bacteria.


Can somebody explain the low temperature Schottky anomaly for two level systems?

The Schottky anomaly is a general feature of two levels systems. The heat capacity is defined as $\displaystyle C= \left( \frac{\partial U}{\partial T} \right)_\Delta$ where $U$ is the internal energy and the energy gap between the two levels is $k_B\Delta$, $k_B$ being the Boltzmann constant.

The gradient of the internal energy is thus important. At constant energy gap and at very low temperatures  only the lowest level is populated and the internal energy is small, and heat capacity small as the internal energy increase is small, i.e gradient $U$ vs $T$ is small. However, as the temperature is increased the upper level begins to be populated and so the internal energy rapidly increases, then the gradient $U$ vs $T$ is large and so is heat capacity.

When both levels are almost populated equally, which occurs at very high temperature, no more energy can be absorbed and the gradient $dU/dT$ effectively becomes constant and so the heat capacity becomes small and tends to zero.

Similar behaviour is observed when there are three or four etc i.e. a small  number of levels. In the 'normal' case of a molecule there is effectively no upper limit on the number of levels that can be populated as temperature increases, and the slope of $U$ vs $T$ has a constant finite value which leads to $3R$ as the heat capacity per mole at high temperatures.

The figure shows the internal energy (in units of $k_B\Delta $)  and heat capacity (in units of $k_B$) vs $T/\Delta$. To make the plots more general $T/\Delta$ is plotted as this ratio occurs in the final equation.

The equations are:

partition function $ Z=1+e^{-E/k_BT} = 1+e^{-\Delta/T} $ when energy gap $E=k_B\Delta $.

The Internal energy is $\displaystyle U=k_BT^2\left( \frac{\partial \ln(Z)}{\partial T} \right)_\Delta$.

The heat capacity is $C\displaystyle=\left( \frac{\partial U}{\partial T} \right)_\Delta= k_B\frac{ \Delta^2}{T^2}\frac{e^{\Delta/T}  }{(1+e^{\Delta/T})^2}$


Chemistry of toilet bowl mineral stains on porcelain

Many places have water with dissolved iron and manganese, which precipitates out as a brown or black stain when exposed to air.

Massachusetts has areas of podzol, where the fine sandy glacial till had been covered by pine forests. This makes for acid soil, which in anaerobic conditions readily dissolves iron in the ferrous form, $\ce{Fe++}$ from the ground, which is a pale green color, almost unnoticeable. If you dig into a podzol, exposing it to air, the soil turns brownish or orange in a matter of minutes as the iron is oxidized to its ferric, $\ce{Fe+++}$, form. That's what is happening on your porcelain fixtures.

It's hard to dissolve the brown ferric iron stain, but there are two ways to do so:

Reduce the iron (and manganese) to the lower-valence state. Sodium metabisulfite $\ce{Na2S2O5}$ is sold for this purpose under names such as Red-B-Gone. It works well on clothing and light stains on porcelain. Use this with caution, since it can generate noxious sulfur dioxide fumes if it comes in contact with an acid.
Dissolve the oxides in a strong acid, such as $\ce{HCl}$, hydrochloric acid. Though more dangerous than reducing agents, 10%-30% $\ce{HCl}$ can remove thicker layers of stain. Use with great caution, including eye protection and rubber gloves. Rinse immediately with plenty of water if any gets on you or your clothing! $\ce{HCl}$ is available in hardware stores and in plumbing supply shops.

You can also treat the water to remove iron and manganese or to sequester it with a polyphosphate additive.

Do not use alkaline drain cleaners (e.g. $\ce{NaOH}$ and Drano®) on vitreous china, as it etches the glaze and allows stains to soak into the porcelain, making the stains difficult or impossible to remove.


Dihedral angle of gaseous and crystalline HOOH

You've already answered your own question - hydrogen bonding.
First, don't think of it as to why is the gas state different than solid-state.  Rather, think about why is the solid-state different than the gas state. In the gas state, there is very little intermolecular interaction (very, very few species do not follow the ideal gas law to many significant digits). The gas state geometry is close to what is expected for $\mathrm{sp^3}$ tetrahedral geometry and may be considered normal. The link below is a paper discussing the gas phase microwave spectroscopy of $\ce{H2O2}$ in far greater detail than I can hope to here. The description of the torsional rotation energies was not trivial.
Internal‐Rotation in Hydrogen Peroxide: The Far‐Infrared Spectrum and the Determination of the Hindering Potential by
Robert H. Hunt, Robert A. Leacock, C. Wilbur Peters, Karl T. Hecht, The Journal of Chemical Physics 1965, 42, 1931-1946, https://doi.org/10.1063/1.1696228,
mirror
When condensed to the solid-state, molecules get close enough to allow intermolecular forces (in this case hydrogen bonding to a much greater extent than van der Waals). Additionally, the energies of the crystal lattice must be considered.  In this case, th


Complete combustion is, as the name suggests, complete. That on its own doesn’t mean much, but I can tell you that the completeness implies all atoms that were put into combustion leave it fully oxidised (or fully reduced in the case of oxygen, fluorine, chlorine; or unchanged in the case of nitrogen).

An incomplete combustion is one where molecules remain that may still be oxidised, i.e. can react with further oxygen. Examples would be carbon monoxide or elemental carbon (colloquially ‘soot’). Incomplete combustion often results in a yellow flame stemming from carbon particles, while complete combustion typically yields hotter, colourless flames.

In your average Bunsen burner, you can choose between complete and incomplete combustion by regulating the amount of air (meaning: oxygen) that reaches the flame. Complete combustion would follow the equation $\ce{CH4 + 2 O2 -> CO2 + 2 H2O}$. There is no unique equation for incomplete combustion, but one possibility would be $\ce{CH4 + O2 -> C + 2 H2O}$.

By the way, the way living beings burn carbohydrates in cellular respiration is, by definition, a complete combustion since every carbon atom ends up in $\ce{CO2}$ and a set of two hydrogen atoms always ends up as $\ce{H2O}$. The overall equation is $\ce{C6H12O2 + 6O2 -> 6CO2 + 6H2O}$. If glucose is combusted with less than 6 equivalents of $\ce{O2}$, then incomplete combustion results.


Among HClO4 and HClO2 which is the stronger acid if we do not know Ka values?

$\ce{HClO4}$  is stronger acid due to the formation of more stable conjugate base. Also there are more number electronegative oxygen atoms in case of $\ce{HClO4}$ , so acidity should be more.


Why doesn't the oxidation of cellulose oxidise all the alcohol groups, and how does it cause discolouration?

Well Noah,
Primary alcohols, in general can be oxidised more easily, if we are using suitable reagents like acidified dichromate/permanganate etc.

By looking at the reaction it looks like there has been some reagent used like lead teraacetate (LTA) or NaIO4, which attack syn, vicinal diols/(alpha)-hydroxy aldehydes/glyoxal to cleave the C-C bond and produce the oxidized product:

It has been mentioned elsewhere, that cellulose can be oxidised by NaIO4 into some carbonyl derivatives:

This reaction has been responsible for yellowing [at times referred to as discolouration] of paper:

These carbonyl groups arranged in some definite fashion give rise to yellow colour, sometimes called chromophore, which are any group responsible for proving colour to an otherwise colourless compound.


Using Nanodrop for UV/Vis analysis of compounds at limited amounts

In short, i would say that you cant use Nanodrop for such work. Here is why:

From a quick google search it appears that a nanodrop is a UV-Vis spectrometer that can measure very small amounts of liquid. Looks like that  bio-sciences people (I am not a biologist or biochemist) use it to quantify nucleic acids by knowing the approximate molar absorptivity for DNA, RNA etc. at a specific wavelenth, measuring the sample's absorption at that wavelength, and applying the Beer-Lambert law assuming that the only compounds that absorb in the wavelenght that measurement takes place in their sample are the nucleic acids (which i guess is a fair assumption for such samples).

However, I am not sure from your question if you want to measure nucleic acid content in a sample also containing alkaloids as part of the sample matrix or you want to quantify the various alkaloids in your sample.

I think you must mean the latter since you mention GC-MS in which case the nanodrop seems to be a poor choice: UV-Vis spectroscopy in general is a technique with very low resolution in terms of compound separation due to the broadness of the peaks in the absorption spectra. If you had only a few alkaloids in your mixture (which i dont think you do) maybe you could use nanodrop or a normal UV-Vis spectrometer followed by a lot of calibration and absorption band decomvolution in order to identify and quantify each component. If on the other hand you have a complex mixture (which is usually the case with natural extracts) any kind of UV-Vis spectroscopy will fall short. Even NMR spectroscopy which has a much higher resolution would struggle.

This is why GC-MS seems to be the best way for such work: the GC separates the mixture in simpler mixtures (or in the pure compounds if your sample is not too complicated) and then the MS identifies and quantifies the compounds present in each fraction. So i think that you will either need to find more sample or further optimize your GC-MS method to increase sensitivity.

I hope this helps.


How can an Organic chemist know the chemical formula of a natural product she isolated from an organism?

There are two levels of answers. One is historical and one is modern. Historically, determining the chemical formula for had been a trivial job for most small or medium sized molecules. You do a combustion analysis, determine how much carbon dioxide, and water were produced. By doing other classical element detection tests, you would determine if it had halogens, nitrogen and sulfur. These were the most common approaches. The determination of molecular weights was done by classical physical chemistry experiments.

Today you would still do combustion analysis and elemental analysis as a first step and determine the CHNS ratios. Then you run a mass spectrum of the compound, and assess the molecular weight of the compound.

Determining a structural formula may take a long time and others have already mentioned in the comments. Today advanced NMR techniques, high resolution mass spectrometry, X-ray of crystals (if they exist), all help in structure determination. These things have become routine now.

Nobody routinely determines the molecular structure by AFM or STM as far as I know.


Nitrobenzene - Aromatic Substitution Reactions

Of course, there must be some boundary as to whether or not nitrobenzene undergoes a particular electrophilic aromatic substitution (EAS) since nitro-group is very deactivating. I leave it to you to decide that fact on the yield obtained in following reactions, because there are some data you may fail to find but, nonetheless, exist in literature.

Chlorination of nitrobenzene at $\pu{35\!-\! 45 ^{\circ}C}$ in the presence of iron (III) chloride ($\ce{FeCl3}$) gives an isomer mixture, which contains 86% 3-chloronitrobenzene, 10% 2-chloronitrobenzene and 4% 4-chloronitrobenzene (Ref.1). Even if nitro group is $m$-directing, this reaction has given significant amount of $o$-prodouct. You may compare this reaction with nitration reaction of chlorobenzene at $\pu{40\!-\! 70 ^{\circ}C}$ with mixed acids (30% $\ce{HNO3}$ : 56% $\ce{H2SO4}$ : 14% $\ce{H2O}$), which typically gives a 98% yield of an isomer mixture consisting of 34-36% 2-chloronitrobenzene, 63-65% 4-chloronitrobenzene, and about 1% 3-chloronitrobenzene (Ref.1).

A simple and efficient method for the sulfonation of nitrobenzene using liquid $\ce{SO3}$ as the sulfonating agent in a microreactor at $\pu{40\!-\! 110 ^{\circ}C}$ was developed and conducted recently, and 88% yield of 3-nitrobenzenesulfonic acid (m-NBSA) has been obtained (Ref.2).  The main by-product is bis(3-nitrophenyl)sulfone (BNPS) and proposed mechanism was depicted in the following Figure:

On the other hand, the sulfonation of toluene has given $o:m:p$ ratio of  21.2% : 2.1% : 76.7% in 77.8% $\ce{H2SO4}$ (Ref.3). The $o:m:p$ ratio has become 38.8% : 2.6% : 58.6% when solvent has changed to 84.3% $\ce{H2SO4}$ (significant decrease of $p$-product in increasing acidity; Ref.3). Also, note that the sulfonation by sulfuric acid of nitrobenzene itself and of various aromatic compounds in nitrobenzene solution has been studied at a series of temperatures (Ref.4). Yet, no concomitant sulfonation of nitrobenzene in the presence of other activated substrates such as toluene and xylene has been reported.

Using $\ce{NO2+PF6-}$ as the nitrating reagent, Olah had studied nitration of nitrobenzene in two different solvents: He achieved $o:m:p$ ratio of  7.1% : 91.5% : 7.4% in 96% $\ce{H2SO4}$  and that of 10% : 88.5% : 1.5% in $\ce{CH3NO2}$ (Ref.3).

For comparison, same nitrating agent ($\ce{NO2+PF6-}$) has given $o:m:p$ ratio of  68.2% : 2% : 29.8% in $\ce{CH3NO2}$ when subjected to nitrate toluene (Ref.3).

To my knowledge, Friedel-Crafts reaction of nitrobenzene (alkylation or acylation) is yet to be known. I'd love to hear if somebody say otherwise.

References:

International Agency for Research on Cancer (World Health Organization), lARC Monographs on the Evaluation of the Carcinogenic Risk of Chemicals to Humans, Vol. 65, 1996, “2-Chloronitrobenzene, 3-chloronitrobenzene and 4-chloronitrobenzene,” pp. 263-296.
Y. Chen, Y. Su, F. Jiao, G. Chen, “A simple and efficient synthesis protocol for sulfonation of nitrobenzene under solvent-free conditions via a microreactor,” RSC Adv. 2012, 2, 5637–5644 (DOI: 10.1039/C2RA20406A).
George A. Olah, “Aromatic substitution. XXVIII. Mechanism of electrophilic aromatic substitutions,” Acc. Chem. Res. 1971, 4(7), 240–248 (DOI: 10.1021/ar50043a002).
F. J. Stubbs, C. D. Williams, C. N. Hinshelwood, “The Kinetics of Aromatic Sulphonation Reactions. Sulphonation by Sulphuric Acid in Nitrobenzene Solution,” J. Chem. Soc. 1948, 1065–1078 (DOI: 10.1039/JR9480001065).


reaction between copper metal and aqueous sodium bicarbonate

First, reduction potentials.

Theory page
https://en.wikipedia.org/wiki/Reduction_potential?wprov=sfla1

Table of standard electrode potentials
https://en.wikipedia.org/wiki/Standard_electrode_potential_%28data_page%29?wprov=sfla1

Standard electrode potentials of sodium is -2.70V while copper (I) is 5.20V, single electron reduction of copper (II) has 1.59V, full reduction potential is 3.37V

Higher the reduction potential, more the tendency for the species itself to reduce. Reduction potential of Copper is much higher than Sodium.

So conclusion is no.

Lets say if copper can replace sodium. By how much? Under room conditions, RT/F*lne is approximately 0.06V, so the amount of copper ions that will replace sodium will be orders of magnitude 71.5 to 115 times dilute. In other words, non existant.


Iodine is a metalloid. Metalloid elements have one or more allotropes with properties intermediate between those of a metal and a nonmetal.

Metals are lustrous, ductile, and conductive of heat and electricity.

Nonmetals are matte, brittle, and insulators.

Metalloids may be semiconductors, like silicon, germanium, arsenic, and carbon (graphite). They may be lustrous, but brittle, like iodine and tellurium.


First off, hybridization is a concept chemists developed to help explain reality (their observations).  Just like resonance theory and Huckel MO theory, it is often (but not always) a useful way to explain the world around us. A "rule" on hybridization: hybridization occurs in response to a bonding interaction.  Further, hybridization involves the mixing of filled (or partially filled) atomic orbitals to produce new atomic orbitals that can be used to form "directed" bonds (molecular orbitals) that produce more stable molecular systems than an unhybridized atom might produce.  Finally, energetically speaking, hybridization is a net neutral process.  When low energy AOs are mixed with higher energy AOs, the resultant new AOs have the average energy of the initial AOs.  So if some electrons are "promoted", then some are lowered in energy as well

For example, in the formation of methane, carbon mixes the electrons in the 2s and three 2p atomic orbitals to produce 4 $\ce{sp^3}$ hybridized atomic orbitals.  The energy of these 4 equivalent $\ce{sp^3}$ orbitals is in between that of the starting 2s and 2p orbitals. In the case of methane this use of $\ce{sp^3}$ orbitals to produce a tetrahedrally shaped molecule, produces a molecule that is lower in energy than the alternative molecule formed from unhybridized s and p orbitals on carbon.  In the case of phosphine ($\ce{PH_3}$) the situation is reversed.  The $\ce{P-H}$ bonds in phosphine are directed along the x, y and z-axes.  The central phosphorous atom remains unhybridized.  In the case of phosphine, hybridization does not produce a lower energy molecule, so the central phosphorous atom in phosphine remains unhybridized.

Personally, I like the concept of hybridization and find it quite useful.  But you have to be careful where to apply it.  Hybridization is best when it is used to explain reality after the fact.


Could chromatography induce a conformational change?

While I do not have numbers at hand to support it, I think it is possible, chromatography on silica indeed may influence, may restrain the conformational flexibilty of molecules in the mobile phase.

Normal phase chromatography depends on adsorption.  A complete rotation around a $\sigma$-bond, even in absence of other molecules around, needs some energy to pass from one stagggered to an other staggered conformtion.

(figure source)

If there were now -- instead of methyl groups separated by three bonds, as depicted in the diagram -- two groups with a high affinity to the stationary phase, this said energetical barrier would increase.  Assuming lowered temperature and affinity of these groups "just intense enough" (e.g., deportontated sulfonic acids, not per-alkylated ammonium groups), the propability to find one molecule in such $\ce{s-cis}$ conformation will increase.  I just see a potentially relevent publication Langmuir, 2004, 20, 10639–10647 about proteins decoiling on silica.  I imagine, amphiphlic structures like reverse micelles may be equally influenced and will "swap their inner to the outer" as if they were (again) exposed to aqueous systems when facing silica.

Yet I am still hesitant to state if this may by become a constraint, leading to a situation where once conformation is looked, as in the instance of atropisomers.


Is it practical to extract zymase from yeast for ethanol production?

So, do people use such a method, and if not, why not?

No, people do not use such a method.  There are several reasons:

In fact, conversion of glucose to ethanol requires many enzymes, not just one.  The crude extract originally known as zymase was not characterized until later to consist of so many different molecules.  The molecules that comprise zymase include all the enzymes of glycolysis (e.g. glucose-6-phosphate isomerase, triose phosphate isomerase, phosphofructokinase, triose phosphate isomerase, etc.)
The other component of "zymase" as it was understood historically included dialyzable components.  Later these were identified as coenzymes, and are now understood to consist of molecules such as NAD+, ATP, and coenzyme A.
Live yeasts can synthesize all the enzymes and co-enzymes they need to convert glucose to ethanol.  But yeast extracts such as zymase can't. As they convert sugar to ethanol, eventually cofactors are used up, lost to processes such as dilution, oxidative damage, or enzyme promiscuity.  When a zymase extract loses its activity because of these reasons, it can never get them back.  But live yeasts are always synthesizing more coenzymes (and enzymes) in a programmed response to any sensed shortage.
Even zymase would not go to 100% ethanol concentrations.  Enzymes denature and lose their effectiveness at high alcohol concentrations.  The concentration where this happens is probably higher than the concentration that is lethal for yeast, but I don't think it is close to 100%.


Imposing particle number subspace on second quantized Hamiltonians

$\newcommand{\Ket}[1]{\left|#1\right>}$
$\newcommand{\Bra}[1]{\left<#1\right|}$
$\newcommand{\BraKet}[2]{{\left<#1}\left|#2\right>}$

It's an older question, but one worth answering!

For the uninitiated, second quantization is a bookkeeping technique that describes many-particle systems as excited states of a field, usually taken to be the reference field or vacuum state. I found the most clear description of it in atomic_rabbit's answer to this reddit post. The second-quantized operator is useful because this form allows it to be applied to an indeterminately sized system (the number of particles is not explicitly stated in the second-quantized operator).

Now, onto the question:

The "first-quantized" (electronic) Hamiltonian has the form (in atomic units):

$\hat{H} = \sum_i\frac{1}{2}\nabla^2_i - \sum_i\sum_A\frac{Z_A}{r_{iA}}+\frac{1}{2}\sum_i\sum_j\frac{1}{r_{ij}} = \sum_i\hat{h}(i)+\frac{1}{2}\sum_i\sum_j\frac{1}{r_{ij}}$

which, upon second quantization, goes to:

$\hat{H} = \sum_{p,q}\Bra{p}\hat{h}\Ket{q}a^+_pa_q + \frac{1}{4}\sum_{p,q,r,s}\Bra{pq}\Ket{rs}a^+_pa^+_qa_sa_r$

In the case of "first quantization", we construct the matrix form of the operator by projecting it onto whatever basis functions we're using. We then find the eigenvalues of the operator by diagonalizing its matrix form. Your question is an excellent one, because in second quantization, this projection seems to be already done within the operator, and so we should be able to diagonalize the operator outright. Right?

The catch is that the second-quantized operator still has the creation and annihilation operators that have not yet been projected onto any basis. This is equivalent to us not knowing whether there is a particle in the p$^{th}$ state. And what are the states? Those are properties of the field, which we must find out by projecting the second-quantized operator onto said field.


Chemical consideration of biodegradeability

Biodegradability is not a chemical property, much like a price for which you can buy a compound is not its chemical property. Both are the properties caused by some external circumstances, which are not necessarily related to chemistry.

As for the processes, there is hardly anything they have in common. Life is pretty versatile and can consume a wide range of chemical compounds. Wide as it is, though, this range does not constitute all chemistry. There are a great deal of compounds that just never happened to appear in nature before humans made them, so no life form had the chance to develop the ability to consume them. Those are not biodegradable.


Why is H⁺ ion considered the source of acidity, and OH⁻ considered the source of basicity?

Yours is a historical question. Note that in modern chemistry, it is not required that acids have H+ or bases have OH-. The words acidity or basicity referred to certain common behaviors such as acids will liberate hydrogen when come in contact with metals, or they will decompose carbonates to carbon dioxide. Similarly, bases had other common characters. Indicators behaved the same way with acids or bases. No one knows why bases were called bases. Acids we understand, that they tasted sour (=Latin). The term alkali (Group I bases) makes more sense (from Arabic for soda ash or potash).

Definitions are made by humans and I am glad you are not blindly accepting the definitions and at least thinking about them. Arrhenius was among the first ones to think why solutions conduct electricity. When Arrhenius proposed this idea in his PhD, but he knew that his mediocre committee members will not like it. How can anything ionize in water? It was a stupid idea to them. It turned out that most common acids and bases had very high conductivity better than anything else known, even better than salt solutions. Say 0.1 M HCl had way higher electrical conductivity than 0.1 M NaCl? Why? The "proton" in aqueous solution has the highest conductivity known today. What comes next? The hydroxide ion. Also in Arrhenius' time electrical decomposition was known. Faraday did a lot of electrochemical experiments then.

When you electrolyze an acid solution, what do you get at the electrode which has a negative electrostatic charge. Hydrogen gas! What appears at the positive electrode, chlorine gas. How can "H" and "Cl" reach the electrode, this is possible when they have a charge on them as well. So, "H" in acids must exist as "H+"; as a positively charged entity. I have not done justice to Arrhenius history or the history of acid bases.

If you are really (or better word genuinely) interested then search the Theory of Ionization+ Arrhenius History in Google Scholar. Partington's History of Chemistry will have it too but it is meant for serious scholars of chem. history (multiple volumes). It is available online from the Internet Archive.


When using a pH probe, instead of aiming to reach the equivalence point of the titration, you should be titrating beyond it to create a complete titration curve. With an indicator, the end point is clearly indicated by a color change; with a pH probe, however, the end point is indicated by a steep change in pH in a plot of the data, thus requiring the collection of data beyond the end point.

While in theory the pH at the equivalence point of the titration can be calculated (and, for the limiting case of a strong acid and strong base, is exactly the $K_w$ that corresponds to the temperature of your solution), calculating the pH at the equivalence point and aiming to reach that pH is (i) bad lab practice, since if you're able to calculate the pH at the equivalence point, you're presumably trying to experimentally verify theoretical calculations, and (ii) not applicable to your current experiment, because the organic acids present in fruit juice are weak acids, whose conjugate bases are weak bases that will raise the pH at the equivalence point (and render the pH impossible to calculate for your purposes).


Typesetting oxidation state and charge simultaneously

My personal preference in this situation would be to put square brackets around the whole ion and put the charge outside.

$$\ce{[Cr^{VI}2O^{-II}7]^{2-}}$$ gives:
$$\ce{[Cr^{VI}2O^{-II}7]^{2-}}$$

This is sometimes done with polyatomic species anyway (particularly metal-ligand complexes) to emphasise that the charge is on the whole ion, not on any one particular atom.


I would go with your approach but your final product is unstable due to high repulsion's by $\pi$ electrons.

Hence your final product will react further to form a more thermodynamically feasible product.

NOTES:
In second structure two anti-periplanar migrations are possible and the one which leads to the more stable carbonation is favored (or consider the better migrating group based on electronegativity). Third structure rotates about spiral carbon due to repulsion's.


What is the naming convention for Sn and Pb in inter-metallic compounds?

After doing some digging I've found evidence to support that tin based inter-metallic compounds should be named 'stannide' such as in the paper by Fickenscher et al. [1].

Reference

Fickenscher, T.; Rodewald, U. C.; Niehaus, O.; Gerke, B.; Haverkamp, S.; Eckert, H.; Pöttgen, R. The Stannides $\ce{RE3Au6Sn5}$ $(\ce{RE} = \ce{La}, \ce{Ce}, \ce{Pr}, \ce{Nd}, \ce{Sm})$ – Synthesis, Structure, Magnetic Properties and $\ce{^{119}Sn}$ Mössbauer Spectroscopy. Zeitschrift für Naturforschung B 2015, 70 (6), 425–434. DOI: 10.1515/znb-2015-0050.


How can I keep homemade laundry detergent from disassociating?

You should use about 0.25 - 0.35% xanthan gum (thickening agent) per the total weight of the mixture. Using xanthan gum is kind of tricky. You want to thoroughly mix the gum with the baking soda and washing soda (maybe in a bag), and slowly add the sodas to the water while mixing. If you don't do this, the xanthan gum will form lumps that are difficult to manage.

Hydroxyethyl cellulose might also work as a thickener, but it could increase the sudsing effect. Also, I would try adding a dash of dish soap; only use about 1% of the total weight of the mixture though, or else the sudsing effect will cause the washing machine to foam over.


Why does phosphorus tribromide act as a Lewis acid (electron acceptor)?

The why of a mechanism is always the best approach.

The reasons for this attack are:

$\ce{P}$ can expand its octet due to presence of vacant $d$-orbitals.
The $\ce{P-Br}$ bond is polar due to the electronegativity difference between the atoms and hence P develops partial positive charge favouring a nucleophilic attack on it.
Looking at the molecule as a whole, $\ce{Br}$ is a good leaving group so $\ce{P}$ doesn't mind a nucleophilic attack on it by the lone pair of oxygen.
"Nature loves stability" and  $\ce{P-O}$ bond is stronger than $\ce{P-Br}$ so that might be the driving force for this reaction mechanism.


Both ends of the alkene would be converted into a carboxylic acid functional group.

This would be a 'ring-opening' or 'permanganate cleavage' reaction and form adipic acid (hexanedioic).


How would I synthesize ethyl 2‐oxocyclohexane‐1‐carboxylate from cyclohexanone?

As you will see if you look at the reference mentioned by @Mathew-Mahindaratne,   the standard conditions for this transformation is NaH/KH in THF at reflux with diethyl carbonate (EtOCOOEt) with an acetic acid workup.

The Organic Syntheses procedure here uses dimethyl carbonate, but will be equally applicable for diethyl carbonate.


What are the spectroscopic signatures that differentiate triple state water from water at 1 atm?

The triple point of water is point of dynamic equilibrium between three phases of water. Any small change in Pressure or Temperature will lead to one phase regions or two phase lines on the phase diagram.

The triple point is not a state of matter rather a combination/mixture of three phases . So i very much doubt we can get any unique spectroscopic signatures . We would get a distorted result as this point has three different phases.
Also Emission and Absorption spectra depends on  temperature and pressure.


What is the product of a reaction between an acyl chloride and a secondary alcohol?

It's a bit tricky because there's probably a reagent that's missing. You might add DMAP or triethylamine to this reaction which would further activate the acyl chloride. But the alcohol oxygen is the most nucleophilic moiety in this reaction. The most electrophilic moiety is the carbonyl group, especially since it's activated by the electron withdrawing chlorine.

Thus we would predict that a standard acylation (a carbonyl reaction) will take place where we add a nucleophile to the carbon of a carbonyl group. After which, reformation of the carbonyl group will result in kicking out a chloride anion. Finally, a very fast proton transfer will take place that breaks the O-H bond you are referring to. Notice that this is really an after thought and not really the key point of this reaction.

Take a look at the mechanism for the reaction in your textbook (or here).


No, $\ce{CCl4}$ is not an electrolye.  An electrolye is a material that causes ions (charged entities) to form in the solvent.  Carbon tetrachloride neither dissociates into ions nor induces ion formation in the solvent.

The reason is that the $\ce{C-Cl}$ bond is rather strong and won't break under
normal solution conditions.  Nor does the molecule have any net dipole moment.


Calculation of entropy for an isothermal irreversible expansion

Here are the steps to determining the change in entropy for an irreversible process on a closed system:

Use the first law of thermodynamics to determine the final thermodynamic equilibrium state of the system for the irreversible path.
Totally forget about the irreversible path.  It is of no further use.  Focus only on the initial equilibrium state of the system and the final equilibrium state.
Devise a reversible path for the system that takes it from the initial equilibrium state to the final equilibrium state.  This reversible path does not have to bear any resemblance whatsoever to the real irreversible path, other than it must pass through the same initial and final end points.  (Entropy is a function of state).
Calculate the integral of dq/T for the reversible path that you have devised.  This will give you the change in entropy of the system for the irreversible path as well as for the reversible path.

For your problem, this procedure will give you the equation that you have written.  Note that, for the irreversible path, the system temperature matches the surroundings temperature (isothermal) only at the interface with the surroundings, but not throughout the interior of the system.  For an irreversible expansion, this is what we define as "isothermal."