Why do people assume 28 oxygens in chlorite mineral formula?

It doesn't mean that there are 28 oxygen atoms.  It means a charge equivalent of 28 oxide anions, considering that each oxide anion has a -2 charge and each hydroxide anion has a -1 charge.

This is based on the chlorite having an formula including $\ce{O20(OH)16}$

For example see $\ce{Mg10Al2(Si6Al2)O20(OH)16}$ in The effect of Cr on chlorite stability and the appearance of a new Al-rich silicate

In otherwords, as explained in OCTAHEDRAL OCCUPANCY AND THE CHEMICAL
COMPOSITION OF DIAGENETIC (LOW-TEMPERATURE) CHLORITES

  All chlorite analyses were re-calculated as chlorite structural formulae on the basis of 28 oxygen equivalents (56 negative charges) i.e. assuming an ideal anion framework of $\ce{O20(OH)16}$


Is a hydrogen bond considered to be a van der Waals force?

According to the IUPAC gold book a van der Waals force is:

  The attractive or repulsive forces between molecular entities (or between groups within the same molecular entity) other than those due to bond formation or to the electrostatic interaction of ions or of ionic groups with one another or with neutral molecules. The term includes: dipole–dipole, dipole-induced dipole and London (instantaneous induced dipole-induced dipole) forces.

Hydrogen bonding is a type of dipole-dipole interaction, so it would fit the definition of a van der Waals force.

The way I think of it is: van der Waals forces are anything that make a gas non-ideal, since that's how they were originally discovered and defined.


Order of the subshells in an electron configuration

Whilst the said concept is correct, as orbitals increase in number the less they are attracted to the centre of nuclei, consequently the further they move away the higher the energy they possess.

The 2s electrons spent more time closer to the nucleus (when compared to 2p electrons) and this reduces the energy of electrons in s orbitals, the nearer they get, the lower the energy. This is why the s orbital always has a slightly lower energy than p orbital at the same energy level.

The 2s, 3s .... orbitals get progressively away from the nucleus.


Why does magnesium have an exceptionally low melting point?

Following up on my comment, even though I'm not convinced that this is a 'real' answer to the question.  Consulting the SGTE database of elemental free energies [1] (with various updates over the years), one can obtain the melting temperatures for the hcp, fcc, and bcc phases of beryllium, magnesium, calcium, strontium, and barium (rough guesses looking at plots — consider good to approx. $\pm\pu{5 K}):$

$$
\begin{array}{lrrrcc}
\hline
\text{Element} & T_\mathrm{m}(\text{hcp})/\pu{K} & T_\mathrm{m}(\text{bcc})/\pu{K} & T_\mathrm{m}(\text{fcc})/\pu{K} & \text{Phase at RT} & \text{Melt phase} \\
\hline
\ce{Be} & 1544 & 1560 & 1018 & \text{hcp} & \text{bcc} \\
\ce{Mg} & 922 & 755 & 705 & \text{hcp} & \text{hcp} \\
\ce{Ca} & 940 & 1114 & 1063 & \text{fcc} & \text{bcc} \\
\ce{Sr} & 900 & 1050 & 1030 & \text{fcc} & \text{bcc} \\
\ce{Ba} & 635 & 1000 & 709 & \text{bcc} & \text{bcc}\\
\hline
\end{array}
$$

Several things to note, amongst them the variety of which crystal structure is most stable for the different elements at different temperatures. Also, the hcp phase's melting point does almost(!) monotonically decrease going down the table.  The biggest anomaly would actually seem to be the relative instability of the bcc and fcc phases of magnesium (it is the only one that remains hcp up to the melting point).

Overall, it would appear that the group 2 elements display a wide variety of preferences for crystal structure, and hence bonding energy configurations.  Is $\ce{Mg}$ unusual? Not really…

References

Dinsdale, A. T. SGTE Data for Pure Elements. Calphad 1991, 15 (4), 317–425. DOI: 10.1016/0364-5916(91)90030-N.


I've seen a couple of companies making flasks entirely from PTFE or PFA, which might be suitable. They may not hold up to heating with internal vacuum though, for example, check out the notes here.

And yes, the typical safety coatings are on the outside to hold the glass together on fracture.


Intutive understanding of internal energy

For an arbitrary fluid (not necessarily an ideal gas), a combination of the first- and second laws of thermodynamics tells us that, $$\left(\frac{\partial u}{\partial v}\right)_T=-\left[P-T\left(\frac{\partial P}{\partial T}\right)_v\right]$$where U is the molar internal energy and v is the molar volume.  For a fluid which obeys the ideal gas law Pv=RT, the right hand side of this equation is zero.  So the internal energy of a substance which obeys the ideal gas law is independent of volume.  An ideal gas is a substance which obeys the ideal gas law because the interactive forces between its molecules are negligible.  So, yes, it is because ideal gas molecules do not have attractive forces.


Physical significance of chemical potential and fugacity

The chemical potential is the partial derivative of the Gibbs free energy with respect to the number of moles of the specified species at constant T, P, and numbers of moles of all other species.  The equation you wrote is the what this partial derivative reduces to for a species in an ideal gas mixture, where P is the partial pressure of the species in the mixture.  For a real gas, the fugacity of the species replaces the partial pressure of the species in the same equation for the chemical potential.


Approach to model solvent in a lattice: e.g. H2O in NaCl

This is a bit outside my realm of expertise - you'd want to look at molecular dynamics (MD) simulations for more details. Let me take your question in a few parts.

1) A water molecule in a lattice is likely bound by electrostatic (and hydrogen-bond) interactions. So your general assumption is a good one.

2) MD simulations usually have a "thermostat" depending on the type of ensemble used. Your thermal energy will be distributed into different degrees of freedom, such as rotations, translations, and vibrations. So I'd work out the amount of translational free energy from statistical mechanics, and then use the electrostatic interactions to determine the path for jumps. (Your main question here seems like the activation barrier, since each lattice site should have the same energy, right?)

3) If the moves are rare, you might want to look into "rare event sampling" methods, which preserve the correct statistics, but allow you to more efficiently run your model.

Many of these techniques are already implemented in simulation codes. You might want to look at something like LAMMPS which can handle different levels of particles (e.g., atoms, molecules, crystal domains, etc.)

Hope that helps - I can only talk in broad generalities here.


Steric effects play an important role in determining the planarity, aromaticity of a molecule. Cyclobutadiene is sterically unstable(90 degrees angle between bonds), hence having high energy, and thus  anti-aromatic.
A lot of compounds change their conformations for reducing their overall energy. Check for the conformations of glucose, for instance. A compound will always prefer its low energy conformation, which occurs when Cyclooctatetraene is non-planar depicted in the right hand perspective drawing

(reference)
This compound looks like the big brother of benzene, which may incorrectly induce the notion of it being aromatic, or, by Huckel rule, anti-aromatic. But the inner angle of a regular octane shape is 135 degress, while a much stable angle nearby it is 120 degrees(an sp2 config.), which only occurs in the tub config. of Cyclooctatetraen. 120 degrees is not achieved, but ~126 degrees is achieved.
Frozen (at 129 K) cyclooctatetraene's structure was characterized by single crystal diffraction analysis by Claus and Rüger. You find the corresponding interactive model here.


Short answer: there's some twisting of the conjugated structure, particularly in the diene

I ran initial calculations for these using B3LYP density functional theory. If anything this method has a slight tendency for extra delocalization.

I'll follow up later with other methods, but I suspect the general answer will be the same.

Cycloocta-1,3,5-triene

Honestly, I expected the delocalization to be retained in the central conjugated section (albeit not perfectly flat) with some twisting on the outer double bonds.

Indeed, the dihedral angle around that central C=C is ~6.9°, but it's not that different around the other CC=CC bonds (~3° and 5°, respectively).

The whole C=CC=CC=C is fairly co-planar, with the remaining two carbons profoundly out of plane, likely relieving ring strain.

It's clear there's some delocalization because the "single" bonds between the double bonds have bond lengths ~1.46Å, smaller than usual.

Thus, in the triene, there's slight non-planarity to the conjugated section, but it's fairly delocalized.

Cycloocta-1,3-diene

Interestingly, the result is a bit different for the diene. While the C=C bonds are planar (i.e., dihedral angle ~2°) the middle C-C bond is strongly non-planar (i.e., dihedral angle ~38.4°). Moreover, it's 1.475Å in length, longer than the comparable bonds in the triene.

Here, I think the delocalization of two double bonds isn't enough to balance the ring strain.


What do ε₀ and V mean in a Pourbaix diagram?

This is a Pourbaix Diagram (I assume this is where you got this figure). While it's not a common notation as far as I'm aware, $\epsilon_0$ is being used to represent the reduction potential in volts (V). As mentioned at the top of the Wikipedia page, this is basically a phase diagram, that tells you the ranges of voltage/pH for which a particular species of (in this case) iron is most thermodynamically favorable. The dividing lines coincide with an even mixture of the surrounding species.


Dalton's law of partial pressures means that the total pressure in the final volume $V=V_1+V_2$ is equal to the sum of the partial pressures P(N2) + P(Ar) in this volume. However, the partial pressure of N2 in the total volume $V$ is not the same as the pressure of N2 in its initial volume $V_1$. You have to account for the expansion. This probably means you also need to use the ideal gas law for that.


Is an antiaromatic compound more stable than a non-aromatic compound because of more conjugation?

Sometimes, only the delocalisation of electrons is not sufficient to conclude that the compound is more stable. We should also look at which molecular orbital the electron pair is delocalising. But a clear picture of these kind of $\pi$-conjugated systems can be got when it is approached quantum-mechanically, with the help of Huckel-$\pi$ Molecular Orbital Theory.  According to this theory, to get the $\pi$-electronic energy of the system, we have to solve the Secular Determinant, which is $$\ce{det(H_{ij}- ES_{ij}) =0  ,where i,j varies from 1 to n}$$ in a $n$ Carbon System.  In Huckel Theory, for solving the determinants two approximations are taken,$1$. $\ce{H_{ii} = $\alpha$ for all i, and H_{ij} = $\beta$ for |i -j|=1.}$ This approximation tells that the effective Hamiltonian for the whole conjugation is negligible if the carbons are not close to each other(i.e. adjacent).The two different values are obvious because of the in Hamiltonian, the Columb integral, the Exchange integral interferes. $2$. $\ce{S_{ii} =1; and S_{ij}= 0 else}$. This tells the Overlap integral(Note that, this is an integral, and clear overlap in the picture doesn't mean integral should be large) are negligible for any two different carbons.  Taking these two approximations, if you solve the determinant, you will get energy levels for $5$ carbon systems,$$\ce{E = $\alpha$ + 2$\beta$;  $\alpha$; $\alpha$ - $\beta$}$$Typically, the values of $\beta$ are negative and as you increase in $\alpha$ or decrease in $\beta$, the energy is increased and system gets destabilisation. .So, the lowest energy level is $\alpha$ + $2$$\beta$, and then the second and after that the third one.Your first compound has four $\pi$ electrons. So, two electrons will be in the level $\alpha$ +$2$$\beta$, and other two is in $\alpha$ only. So,net energy, is $4$$\alpha$ + $4$$\beta$ in mostly conjugated system $1$st compound.  But in the second case there are only two $\pi$ electrons. That two will go into $\alpha$ + $2$$\beta$.So, net energy is $2$$\alpha$ + $4$$\beta$ in second one.  So, second system has lower energy than the  first as coefficient of $\alpha$ is lesser in second one, though second compound is less conjugated. So, extent of delocalisation doesn't always determine stability.


Why don't substances have a condensation point?

A "condensation point" is sort of a misdirection as to the physical process involved. "Point" is really a stand-in for "temperature."

Consider water in air. Let's say that the temp is 20 degrees C, but the "dew point" is 10 degrees C. So if the temperature drops to 10 degrees C then the air is saturated with water and and a further drop in temperature will result in the formation of dew. So being past the dew point is dependent on the amount of water in the air and the temperature, not just the temperature alone.


You can look it up on James Huheeys book, gold's few properties have been qualitatively explained by relativistic theory. Due to that and high nuclear charge acting on the electron, the single electron in $6s$ is actually attracted to the nucleus strongly, thus the electron is not available to react that easily and thus gold does not corrode.


After reading some more stuff I actually think that crystallization of starch might be the answer to this question. There is a lot of literature about this and it might very well be that it can be sufficiently exothermic to warm the small volume of air in a bag.


Is matter really divided in two categories?

Matter....

Matter is generally categorized as either mixtures or pure substances.  Pure substances are elements or compounds.  Mixtures are either heterogeneous or homogeneous.  For heterogeneous mixtures one should be easily able to discern the different components, while if the mixture is homogeneous it appears uniform.  Solutions and alloys are homogeneous mixtures.

A concept map which I give to my students: https://c7chemistry.wikispaces.com/file/view/CW%20matter%20concept%20map%20new%20rev%20091211.pdf/300804074/CW%20matter%20concept%20map%20new%20rev%20091211.pdf


Could you use a reference color chart? Here is one originally made for skin colors:

Wikipedia has other examples.
Paint stores might give you a nicely printed color chart for free.

There are also online versions of color selectors, for example the one from Pantone, but then you have to be very careful about color calibrating your screen, or at least not changing it between comparisons.

A third option, more realistic if you are only measuring color along one scale, would be to make your own comparison material. Start with a light honey colored liquid and mix in a darker color till it matches your honey. The amount of dark color used gives you the darkness of the honey. You probably have all supplies available in the art department.

Edit:

MaxW found a much better solution, a special color scale for honey. It is called the Pfund scale:


Could a safe water ocean exist on a world with predominantly chlorine atmosphere?

Chlorine gas is dissolved in drinking water at low concentrations for disinfection purposes.  Chlorine gas has also be dissolved in somewhat higher concentrations in swimming pools.

The amount of chlorine dissolved will strongly depend upon pressure.  If the atmospheric pressure on the planet is low the concentration of chlorine gas in the water will be low.

Yes, most of the dissolved chlorine will reversibly react to form HCl and HOCl.

For quantitative information about the equilibrium constant and solubility as a function of pressure and temperature see Solubility of Chlorine in Water Industrial and Engineering Chemistry vol. 33, pages 741–744.


Relationship between surface area and vapor pressure

The rate of evaporation is proportional to surface area. The rate of condensation is proportional to surface area. The vapor pressure is the equilibrium pressure where the rate of evaporation is equal to the rate of condensation. Since the scaling factor is the same, the vapor pressure is independent of the surface area.


Diisocyanates are used in some coatings and polymerize in contact with moisture. Flexithane contains a non-reactive polyurethane (already reacted diisocyanate), and so is simply a solution of rubber in a mixture of solvents (SDS: http://www.hmgcoatings.co.uk/msdspdfs/Flexithane.pdf).
The literature claims dry to touch in 5 minutes, ready for a second coat in 20 minutes at 20C. One way to thicken the paint without adding anything that would destroy its performance would be to evaporate some of the solvent. In a can about half-filled with the paint, stir well, with plenty of ventilation (as advised by the SDS) for a few minutes. At £49.66 to £75 per 500 mL, it could be be an expensive experiment if you don't stir well enough and you get lumps or skin in the paint when you try to apply it.


Is the number of tetrahedral voids always equal to double the number of octahedral voids in any crystal?

Note that, in case of tetrahedral voids, each sphere is in contact with three tetrahedral sites above, and three tetrahedral sites below, hence, there are two tetrahedral sites associated with each sphere. Thus,

Number of tetrahedral voids $(N_{th}) = 2 \times$ (Rank of unit cell)

Or alternatively, if you divide a FCC lattice into 8 octants, each octant will have a tetrahedral site at it's body centre.

For octahedral sites, each sphere is associated with one octahedral sites, thus,

Number of octahedral voids $(N_o) =$ Rank of unit cell

For FCC, besides the body center, there is one site at the center of each of the 12 edges. It is surrounded by 6 atoms, three belonging to the same cell unit cell, and the three belonging to two adjacent unit cells.

For clarity see:

For HCP,


The author of the presentation is talking about large, but not yet very large/infinite distance here. Note the annotations:
$$
\frac{e^2}{4 \pi \epsilon R} = k/R
$$
when $R \rightarrow \infty$: slowly to zero

$$
S_{AB}
$$
when $R \rightarrow \infty$: to zero

$$
\left< 1s_A\left| k/r_A \right| 1s_B\right>
$$

when $R \rightarrow \infty$: rapidly to zero

So there is some regime somewhere, in which the change of the entire expression is dominated by the change in $S_{AB}$. The verification of the correctness of the different notations is apparently left to the reader ...


Why did ammonium chloride fumes turned pink in this synthesis?

$\ce{NH3}$ and $\ce{SCl2}$ form $\ce{NH4Cl}$ which evolves as white fumes. $\ce{NH4Cl}$, when passed through $\ce{SCl2}$ (cherry-red liquid), gives the appearance of dense-pink fumes
$$\ce{\underset{(cherry red)}{6SCl2} + 16 NH3 -> S4N4 +2S + \underset{(white)}{12NH4Cl}}$$
That is why the fumes of $\ce{NH4Cl}$ turned pink


Why does the colour of gold sol change with particle size?

The color of conductive nanoparticles depends on plasmonics. In brief, quantum behavior predominates in the few electrons of a nanoparticle, rather than as a continuous conduction band. Surface plasmons strongly influence the color and polarization of light in sols and colloids. For example, for hundreds of years, stained glass has been made with gold or selenium to get pinks and reds.


As one might expect for an element with no long-lived isotopes, little is known about organopolonium compounds. What is known may be broadly divided into two categories [1,2; cited by Wikipedia]:

Compounds where polonium takes the place of a lighter chalcogen, acting more nearly parallel to sulfur, selenium and tellurium than to oxygen. Examples include polonoethers $\ce{R2Po}$ and arylpolonium halides $\ce{Ar2PoX2,Ar3PoX}$.

Complexes with chelating agents similar to metal complexes.  WP cites complexes with 2,3-butsnediol and theoretical from Ref. [1].

All such compounds are known only in tracer amounts because of both chemical instability and rapid radioactive decay.
References
1.
Zingaro, Ralph A. (2011). "Polonium: Organometallic Chemistry". Encyclopedia of Inorganic and Bioinorganic Chemistry. John Wiley & Sons. p. 1–3. doi:10.1002/9781119951438.eibc0182.
2.
Murin, A. N.; Nefedov, V. D.; Zaitsev, V. M.; Grachev, S. A. (1960). "Production of organopolonium compounds by using chemical alterations taking place during the β-decay of RaE" (PDF). Dokl. Akad. Nauk SSSR (in Russian). 133 (1): 123–125. Retrieved 12 April 2020.


How is the hydrolysis of amides done in a lab?

The answer, as with so much chemistry, is it depends. What substrate are you hydrolysing, is it a primary, secondary or tertiary amide? What else is in the molecule that you want to preserve?

Both acid and base hydrolysis require concentrated conditions and elevated temperature. There is a lab procedure here that uses sodium hydroxide in water at reflux to hydrolyse benzamide to benzoic acid. The standard cleavage of an acetamide protecting group to give amine plus acetic acid is overnight reflux in 2N HCl or KOH in MeOH at reflux ref here, for a benzamide protecting group it is 48h in 6N HCl or HBr in Acetic Acid  at rt.

In my own experience a particularly resistant acetamide required 3 equivalents of KOH in refluxing ethylene glycol to hydrolyse.


As it was mentioned in the comments, finished rubber is a cross-linked polymer, and this is the main reason why getting vulcanized rubber liquefied without destroying it seems to be a nearly impossible task.
Also, there might be other additives, e.g. tires are most likely going to contain powdered charcoal, so that even though one can dissolve the rubber by destroying the cross-linking $\ce{(-S-S-)_n}$ bonds, charcoal is likely going to form another phase, and who knows what might happen to other additives during this process.
Finally, at the end you have to somehow recreate cross-linking for the rubber to retain it's elasticity, and this doesn't happen on its own by just evaporating the solvent (see vulcanisation).

In general, the majority or rubber types can be affected by:

strong oxidizing agents (to the extreme of being irreversibly destroyed);
aromatic and chlorinated hydrocarbons (both have a tendency to soften rubber).
Visually it appears as if rubber is swelling; this can happen if one keeps rubber in chloroform or xylene for an extended period of time – for instance, this property is sometimes used to fit rubber tubing in place.

What I suggest is to use a so-called rubber mix or rubber compound: a semi-finished mix of [natural] rubbers, vulcanizing agents, plasticizers, fillers and additives (carbon, chalk, kaolinite etc.).
Rubber mix is used ubiquitously in the industry to produce all sorts of rubber products, including composite materials such as rubberized fabric.
Rubber mix is a versatile material which is highly viscous at the room temperature, but softens upon heating to approx. 100 °C and can be dissolved in volatile organic solvents (even benzene and gasoline).
The drawback is that at the end you still have to use vulcanization to crosslink the rubber.


For chemical reactions occurring in the interstellar medium, which involve metal-ions, will the metal ions be solvated or not?

Per Wikipedia, the upper bound on the particle density in the interstellar medium is approximately $\pu{E6}$ molecules per cubic centimeter, in "cool, dense regions."

In the context of terrestrial aqueous solvation, where the concentration of water is approximately $\pu{55M}$, the particle density is on the order of $\pu{3.3E22}$.

So, even if every single particle in the vicinity of a metal ion in the ISM were a water molecule, the ISM would be roughly $\pu{E18}$-fold more dilute in neighboring water molecules as compared to aqueous solution.

No, metal ions in the interstellar medium will never be solvated.

Note that this does not imply that it is impossible for aquo-complexes of transition metal ions to occur in the ISM, given their significant affinity for ligands in the first coordination shell and for the appreciable organizing influence (thanks, MaxW!) such ions have on more distant coordination shells. It's just highly improbable that one would find a metal ion associated with a sufficiently statistically large number of water molecules to be considered solvated.

Further, to address the portion of the question about thermodynamic equilibrium: I don't think that's a particularly significant factor, no. On Earth, it's frequently the case that ions are solvated but not in thermodynamic equilibrium: the proton gradient that is sustained in the aqueous milieu of biological organisms to drive ATP synthase is one of the most significant examples that comes to mind. Thus, an absence of thermodynamic equilibrium does not necessarily entail a lack of solvation.


Is it possible to protect molten magnesium with Butane

if it might pull the oxygen out of sand , try using magnesium oxide, surely it won't pull the oxygen from its own self. Or using burnt lime, surely it won't pull oxygen away from calcium.
It might react with butane to form magnesium hydride and carbide ....which would be fun to play with . Also magnesium easily reacts with nitrogen to make the nitride...which is an excellent way to make ammonia


Molecular similarity using Tanimoto coefficient

I was wondering if this means that the measure of distance is not for
  the molecule itself but for its representation?

Inherently, you must decide on a representation of a molecule in cheminformatics. In most cases, the representation is a "2D" valence-bond connection table (e.g., SMILES, SD file, ChemDraw, etc.) familiar to anyone who has taken an organic chemistry class.

So yes, a similarity measure depends on the representation. Tanimoto coefficients are usually based on comparing binary fingerprints.

  If so, is there a measure that compares two molecules based off of
  their inherent structure rather than the particular representation of
  choice?

I'm not sure how to answer this question since a representation forces choices about an "inherent structure" of a molecule. If I pick SMILES, I have no coordinates, implied hydrogen atoms, and choices about implied bond orders aromaticity, etc.

I think perhaps your question is about fingerprints - that "Is is there a similarity measure between a molecular graph - without generating fingerprints as an intermediate step?"

In that case, I think graph similarity measures are what you want, e.g. from Peter Willett, "RASCAL: Calculation of Graph Similarity using Maximum Common Edge Subgraphs" The Computer Journal (2002) pp 631–644

The reason people use fingerprint similarity measures is that they're easy to implement, and have worked fairly well for a wide variety of tasks.


The atomic mass unit (amu) was not renamed to dalton (Da). These are different, albeit related, units.

The atomic mass unit (amu) referred to the relative atomic mass of oxygen, which was taken as 16. However, physicists used the atomic mass of the nuclide $\ce{^16O}$ whereas chemists used the average atomic mass of natural oxygen. This unit became obsolete when IUPAP (1960), IUPAC (1961), ISO, CIPM (1967) and CGPM (1971) agreed to assign the value 12 to the relative atomic mass of the nuclide $\ce{^12C}$.

The unit dalton (Da) and the unified atomic mass unit (u) are non-SI units that are accepted for use with the SI. Actually, ‘dalton’ and ‘unified atomic mass unit’ are alternative names for the same unit, equal to 1/12 times the mass of a free carbon-12 atom, at rest and in its ground state, i.e.

$$1\ \mathrm{Da} = 1\ \mathrm{u} = 1.660\,538\,921(73) \times 10^{-27}\ \mathrm{kg}$$

Therefore, using the old definition of the atomic mass unit (amu) in physics

$$1\ \mathrm{Da} = 1\ \mathrm{u} \approx 1.0003\ \mathrm{amu}$$

and using the old definition of the atomic mass unit (amu) in chemistry

$$1\ \mathrm{Da} = 1\ \mathrm{u} \approx 1.00006\ \mathrm{amu}$$


Heating of Ethers with HI and Cleavage of Ethers

Your ether has $\ce{CH2}$ groups at both sides of the oxygen atom.

I would exclude a $S_N1$ reaction here! Instead, cleavage of the ether proceeds via $S_N2$. The oxonium ion resulting from addition of a proton is attacked by iodide at the least hindered carbon atom. Iodide is quite big ;-)


With which group does the middle oxygen atom of an anhydride remain, on hydrolysis?

What I think is, the marked oxygen atom will stay with acetic acid because the carbon on the right side of the oxygen isotope has a slightly more δ+ charge than the carbon on the left due to the +I effect of methyl group attached to it. So, the marked oxygen should go with acetic acid instead of formic acid in the major product.

  For the reaction the mechanism is usually tetrahedral. Only under acid catalysis does the SN1 mechanism occur and seldom even then.

The tetrahedral reaction mechanism of hydrolysis of anhydrides is similar to that for the hydrolysis of carboxylic esters which usually also proceeds through the tetrahedral mechanism.

  Of the eight mechanisms given by Ingold for ester hydrolysis, seven have actually been observed in hydrolysis of carboxylic esters. The one that has not been observed is the BAC1 mechanism. The most common mechanisms are the BAC2 for basic catalysis and the AAC2 for acid catalysis, that is, the two tetrahedral mechanisms. Both involve acyl-oxygen cleavage.

So considering the mechanisms for ester hydrolysis and anhydride hydrolysis to be similar, it can be proved that the extent of δ+ charge on the carbon being attacked does play a major role in deciding the rate of reaction.

For instance, the rate of hydrolysis for the following esters is in the order shown:

So I think this should be enough to prove what I stated in the very beginning of my answer.

Source: Jerry March Organic Chemistry 6th edition, Organic chemistry by Paula Yurkanis Bruice.


Out of the given reagents, find the ones to be used to convert benzene to p-aminophenol

Steps:

Use $\ce{Br2/FeBr3}$ to make bromobenzene.
Nitrate with $\ce{HNO3/H2SO4}$ to make 4-bromo-nitrobenzene.
Treat with $\ce{NaOH}$ to displace the $\ce{Br}$ making 4-nitrophenol.
Reduce with $\ce{Sn/HCl}$ to make the required p-aminophenol.


What do the molecular orbitals of a single water molecule look like?

First things first, orbitals are not real: neither atomic, nor molecular. Except for, probably, one-electron atoms (also known as hydrogen-like atoms), where one might still argue that to get the orbitals you need to treat the nucleus as a classical point particle that just creates a spherically symmetric electric potential which can be already considered as an approximation. For many-electron systems, one have to necessarily make the so-called mean-field approximation (plus the Born-Oppenheimer approximation prior to that for the case of molecules).

Second, in principle, no further approximation is done in LCAO-MO procedure; it is a purely algebraic trick of expanding a function over a complete set. Such expansion is in principle exact, but the problem is that a complete set is infinite, so in real-world calculations we have to truncate the linear combination at some point which indeed is yet another approximation.

Thirdly, analytical expressions for orbitals are known for one-electron systems only (such as $\ce{H}$ atom, or $\ce{H2+}$ molecular ion) since the electronic Schrödinger equation was solved analytically only for such systems.

And finally, strictly speaking, it is electron density that was imaged with STM, not orbitals. The fact that it closely resembles the density corresponding to orbitals we get in our calculations justifies the quantum chemistry as a whole: at the end of the day, with all the above mentioned approximations we still get something that is close the reality!


How do I figure out how many hydrogens my compound actually has using a mass and NMR spectrum?

Being an NMR fan myself I would inspect that NMR spectrum:

The integrals suggest you have 11 $\ce{^1H}$ or a multiple thereof (the number under each peak is the normalized integral, which is proportional to the number of protons represented by the multiplet). That leaves you with $\pu{87 Da -11 Da}=\pu{76 Da}$ to explain. If you throw in an oxygen you get $\pu{60 Da}$, which is neatly divisible by 12, for a formula $\ce{C5H11OCl}$. Next use the NMR spectrum again to infer the connectivity. The rightmost integrated multiplets (multiplet and triplet) are typical of $\ce{-CH2CH3}$ with $\ce{CH2}$ split by methyl and something else. From the integrals we see clearly there is only one methyl group, and there are no lone H which seems to eliminate the possibility of carbonyl or hydroxyl groups, suggesting an ether.

In the end I just "cheated" (it's 2019 after all) and used the online NMR simulator predictor:

This nice piece of software lets you tinker quickly with different structures and inspect assignments interactively, helping you develop your intuition.

Next you need to verify the MS and IR spectra, but the NMR spectrum is a giveaway that this is the correct hit.


Why is polymer B more dense than polymer A?

First polymerisation proceeds through free radical mechanism, which may cause intermediate radicals to abstract H from other parts of growing chain, thus creating new branches in the process.
Refer Wikipedia for detailed mechanism.

On the other hand, second polymerisation proceeds through condensation reaction, where no such thing is possible (condensation can occur only at terminals).

Furthermore, the chains of B interact through H-bonding, which is much stronger than van der Waal interaction of polymer A. This causes chains to come closer and increases the overall density of polymer.


Which compound reacts faster in the Cannizzaro Reaction?

The mechanism of the Cannizzaro reaction is illustrated below.

The first step involves attack by the nucleophilic hydroxide ion on the positively polarized carbonyl carbon to form a tetrahedral intermediate.  Once the tetrahedral intermediate is formed, substituents on the aromatic ring can have little resonance interaction with the former carbonyl carbon because it is now $\ce{sp^3}$ hybridized. So we need to look at the step leading to the tetrahedral intermediate - the first step.

However, if we examine some of the resonance structures for the starting carbonyl compounds (where the substituents on the aromatic ring can interact with the carbonyl through resonance) we see that the nitro compound places positive charge adjacent to the already positive polarized carbonyl carbon - a destabilizing situation (remember, destabilizing something means making it higher energy, more reactive).  On the other hand, the methoxy compound places negative charge adjacent to the already positive polarized carbonyl carbon - a stabilizing situation.

The destabilized carbonyl in the nitro compound will be more reactive towards nucleophilic attack.


How does Palladium dissociate H₂ so easily?

$\ce{Pd}$ can dissociate $\ce{H2}$ because the resulting $\ce{Pd-H}$ bonds are more stable than the starting $\ce{H2}$.
But the reason why $\ce{Pd}$ is so good at dissociating $\ce{H2}$ is related to the energy barrier to bond formation. The dissociation of $\ce{H2}$ on a $\ce{Pd}$ surface (and on $\ce{Pt}$ and maybe several other metals) has no barrier. So you don't need to put the $\ce{H2}$ in an excited state to go over a barrier and create $\ce{Pd-H}$ bonds.
On $\ce{Cu}$ for example, the bonding is possible but there is a barrier, you need to excite $\ce{H2}$ to dissociate it.

Here is a good theoretical article (with a great title):

It shows the different barriers to dissociation of $\ce{H2}$ on $\ce{Pt}$, $\ce{Ni}$, $\ce{Cu}$ and $\ce{Au}$. It also gives an explanation for such differences (more physics than chemistry).


Why do some gases have lower value of Z for a particular pressure?

The trend you've noticed here is a very interesting way of studying the interactions of a gas with itself.

You are certainly right about how to find the minimum of these curves, but the important thing here is what that minimum actually means.

$$Z=\frac{PV}{RT}$$By analyzing this, we expect that $Z$ will tell us something useful about the way the volume of a gas changes as pressure changes (holding $T$ constant just gives better information on a graph). Specifically, we see that if pressure increases but volume decreases more, we will have a number less than 1 and if volume decreases less than pressure increases, we will have a number greater than 1. So what does that number tell us?

Well, if pressure increases and the volume gets even smaller than we would expect for a real gas, these gases must be attracted to each other. Then, we can interpret the minimum of this graph as being the pressure (for a given temperature) at which a gas molecule is attracted to itself most strongly. Beyond this point, repulsive interactions become more and more important. Notice that there is also no maximum. Or, there is no point where attractive interactions dominate repulsive interactions again which is what we would expect.

Now, to explain why we see the behaviors we do from certain molecules, we have to imagine what is causing there to be attractive and repulsive interactions at all. This is due to polarizability and the amount of van der waal's forces.

Indeed, if we look up value of the parameter $a$ for the molecules present (I have them in a textbook) we find:$$a_{\ce{H2}}=.24646$$$$a_{N_2}=1.3661$$$$a_{\ce{CH4}}=2.3026$$$$a_{\ce{CO2}}=3.6551$$where $a$ is a measure of how attracted the molecules are to themselves (in the horrible unit of $\frac{dm^6*bar}{mol^2}$).

These values do indeed follow the trend we see in the graph above. That is, larger value for $a$ has a larger attractive deviation initially. $\ce{H2}$ is something of a special case because it is effectivly unpolarizable, so we see a linear increase in the compressibility factor with pressure. This basically means that hydrogen will deviate from the ideal gas precisely the same at all pressures.

My best guess at why the slopes then increase at different rates has to do with the size of the actual molecule. Again, looking up the values for $b$ which represents the excluded volume of a molecule, we have:$$b_{\ce{H2}}=.026665$$$$b_{N_2}=.038577$$$$b_{\ce{CH4}}=.043067$$$$b_{\ce{CO2}}=.042816$$with $b$ in units of $\frac{dm^3}{mol}$.

So, this predicts methane will have a slope increasing more quickly than both nitrogen and carbon dioxide which is what we see and that hydrogen will have the shallowest slope which is again what we see.

Hope that helps.


Reaction of CpFe(CO)2Cl with phenyllithium

http://voh.chem.ucla.edu/vohtar/fall06/classes/277/pdf/18-Electron%20Rule.pdf
Summing ligand electron contribution.
DOI:10.1006/jmsp.2001.8470
http://www.ccp14.ac.uk/ccp/web-mirrors/farrugia/~louis/chem2x/Organomet-2.pdf
 Document p. 34.  Note error in drawing the graphic.

Direct displacement gives $\ce{CpFe(CO)2Ph}$, with the phenyl being a simple anion sigma-bonded to the metal. If there are two carbons plus two oxygens in the product, implying the CO ligands are intact, your are done.  If some fancy stiff happened with those ligands and the phenyl anion, you must work it out or look it up.


Commercial liquid chlorine bleaches, used in the laundry, are stabilized with sodium hydroxide at high pH. When you dilute the bleach in water, the alkali concentration also decreases. As a result, the active ion (hypochlorite $\ce{OCl-}$) slowly begins to decompose to the chloride ion (the common salt).


Which of the following is an appropriate set of reactants for the preparation of 1-methoxy-4-nitrobenzene and why?

In principle, both reactions will work.

(a) is a nucleophilic aromatic substitution, in which is $\ce{Br-}$ is displaced by $\ce{CH3O-}$

(b) is a methylation of a phenoxide

I'd probably go for (a) for two reasons:

The nitro group in para position decreases the charge density on the oxygen atom of the phenol. As a result, this anion will be less reactive than an unsubstituted phenoxide.
Methyl bromide with a boiling point of just 4 °C is difficult to handle and a poor choice for a methylation reagent. I'd rather try dimethyl sulfate if I had to.


Computing pH of a buffer, and concentrations

We have to compute the new concentrations after dilution:
$$\ce{[HCOOH]}=\frac{0.02\times 50}{200}=0.005\,\, \mathrm{mol/L}$$
$$\ce{[HCOO^- ]}=\frac{0.05\times 150}{200}=0.0375\,\, \mathrm{mol/L}$$

Now, we can compute the $\mathrm{pH}$ of the buffer solution:$$\mathrm{pH}=\mathrm{p}K_a +\log \frac{\ce{[HCOO^- ]}}{\ce{[HCOOH]}}$$

$$\mathrm{pH}=3.74+\log \frac{0.0375}{0.005}=4.62$$


What factors affect the chemical stability of a compound?

Any change which is spontaneous, takes place to lower the usable energy present in that system (This principle is so important, that even universe obeys it), which depends upon the conditions in which change in taking place. To demonstrate this, lets take an example of a stone at height h, one can do a lot of work with that configuration, it can do work by hitting someone on their head. But in contrast, when it is at ground, it does not do anything much on its own, the reason being it has used up some of its energy stored in the form of gravitational potential energy. Now of course, if conditions change, the stone can again do some work, in fact it is always loosing some of its usable energy energy, in the form of increasing entropy since it is breaking down by the natural forces of wind.
Coming back to your question, why does substitution takes place in case of halo alkane with hydroxide anion. Now, we can see that haloalkane has transformed into an alcohol. If one calculates the usable energy (Gibbs free energy, this encompasses several other factors  which include enthalpy, entropy and thermal energy) stored in haloalkane and corresponding alcohol, it will be clear that usable energy has decreased going from haloalkane to alcohol. To understand more about that, you can look up gibbs energy which quantifies that usable form of energy. Will fluorine bond with fluorine or Cesium- The answer, which I think is obvious to you, is surely cesium. why? Looking at their atoms, imagine cesium has no or very little pull on its valence electrons, while fluorine nucleus has so much pull, that electrostatic force permeates farther than atom. Hence, when two fluorine atoms come close to each other, there is some attraction between both atoms due to their individual pulls, but because both pulls are of same strength, it results in a covalent bond, but in case of cesium, the electrostatic force of fluorine nuclei pulls the valence electrons of cesium taking them away from the cesium nucleus, and thus fluorine's thirst for electrons is satisfied (the rock is at surface now!), while cesium is totally cool about it, since it had very little attraction for those electrons. After this process fluorine has negative and cesium has a positive charge, which attracts both of them and also it decreases the energy stored in fluorine-cesium systemIs CsF so stable that it cannot be dissociated into its components? No its not, put in water and water will dissociate the compound easily and energy is supplied by hydration and increasing entropy.


What is an easy way to reduce a boronic acid to borane?

LAH will attack the boron-oxygen bonds without going after the boron-carbon bond, see here (Eq. Z, p. 142).

But it also deprotonates the hydroxyl groups as a strong base like a Grignard reagent. You need to prevent that.  Convert your acid to an ester (OH -> OR), remove the water and then apply the LAH.


You are quite correct in that it appears at first sight that $\ce{Cu+}$ should be more stable than $\ce{Cu^2+}$, but in aqueous media it isn’t.

Stability in aqueous conditions depends on the hydration energy of the ions when they bond to the water molecules (an exothermic process). The $\ce{Cu^2+}$ ion has a greater charge density than the $\ce{Cu+}$ ion and so forms much stronger bonds releasing more energy.

The extra energy needed for the second ionisation of the copper is more than compensated for by the hydration, so much so that the $\ce{Cu+}$ ion loses an electron to become $\ce{Cu^2+}$ which can then release this hydration energy.  A nearby $\ce{Cu+}$ ion is the most facile reduction target for the removed electron, which is why $\ce{Cu(s)}$ is also formed.

Hence, $\ce{Cu^2+}$ is more stable than $\ce{Cu+}$ in aqueous medium.


Synthesis of oxalyl chloride that doesn't involve phosphorus

Most amide coupling agents are suitable for esters. Feast your eyes on: Amide bond formation - beyond the myth of coupling reagents.

This lists old favourites such as CDI, carbodiimides, etc. A lot of it is overkill. The water soluble carbodiimides might be preferred for ease of urea removal later.

The likes of CDI or $\ce{Ac2O/DMAP}$ can react with your phenol when over-stoichiometric amounts are used.

Alternatives could be: boronic acid derivatives e.g.  phenylboronic acid catalysis; scandium or yttrium triflate catalysts (all requiring water removal, but a Dean-Stark could be used if a water immiscible solvent is used); dimethyl oxalate (already suggested)  but in conjunction with Takasago's ZnTAC24 catalyst. MeOH removal by distillation is required, which is quite trivial for such a volatile component.


The amino group in $\alpha$-amino acids is electron-withdrawing via the inductive effect. Mesomeric (resonance) effects cannot occur here because there is no conjugation of the nitrogen lone pair with the carboxyl group. There are two equivalent ways of looking at how the EWG affects the acidity:

It stabilises the conjugate base via withdrawal of electron density from the negatively charged oxygens;
It polarises the $\ce{O-H}$ bond further than in the unsubstituted carboxylic acid, making the proton more prone to dissociation.

Therefore, the $\mathrm{p}K_\mathrm{a}$ of glycine is $2.34$, significantly lower than the unsubstituted acid (acetic acid, $\mathrm{p}K_\mathrm{a} = 4.76$). Similar effects are seen in 2-chloroacetic acid ($\mathrm{p}K_\mathrm{a} = 2.86$).

As for the comparison between glycine and 2-chloroacetic acid, one thing to bear in mind is that glycine exists as a zwitterion at neutral pH. Therefore, the relevant acid-base equilibria are:

For $\mathrm{p}K_\mathrm{a1}$, which you are interested in, note that the substituent is no longer merely a $\ce{-NH2}$ group but is now a $\ce{-NH3+}$ group. Despite nitrogen and chlorine having extremely similar Pauling electronegativities, the positively charged nitrogen is now more electron-withdrawing than a simple uncharged chloro substituent.

One could draw some parallels with aromatic chemistry: aniline is rather unreactive (compared to chlorobenzene) in electrophilic substitution under acidic conditions, where it is protonated. Of course, the reactivity of aromatic rings is a kinetic effect whereas $\mathrm{p}K_\mathrm{a}$ is a thermodynamic effect, but the underlying rationale is the same: $\ce{-NH3+}$ is more electron-withdrawing than $\ce{-Cl}$.


How to calculate the amount of sodium sulfate needed to lower the silver ion concentration of a saturated silver sulfate solution?

Your calculations are true to a certain extend. However, you neglected the amount of $\ce{SO4^{2-}}$ ions precipitated as $\ce{Ag2SO4}$. And also, since we are calculating in the means of concentration terms, we have to assume that addition of $\ce{Na2SO4}$ solid did not change the original $\pu{0.5 L}$-volume.

You calculated correctly the concentration of $\ce{[SO4^2-]}$ ions ($s$) in the original saturated solution, which is equal to the solubility of $\ce{Ag2SO4}$ ($s$):

$$K_\mathrm{sp} = [\ce{Ag+}]^2[\ce{SO4^2-}]= (2s)^2 \times s= 4s^3= 1.4 \times 10^{-5} $$
$$\therefore \; s = \left(\frac{1.4 \times 10^{-5}}{4}\right)^\frac{1}{3} = 1.52 \times 10^{-2}$$

You also calculated correctly the concentration of $\ce{[SO4^2-]}$ ions in the solution after certain amount of $\ce{Na2SO4}$ solid was added (suppose it is $m$ mols) to make  the concentration of $\ce{Ag+}$ ions is equal to $\pu{4.0 \times 10^{-3} M}$:

$$K_\mathrm{sp} = [\ce{Ag+}]^2[\ce{SO4^2-}]= (4.0 \times 10^{-3})^2 \times [\ce{SO4^2-}] = 1.4 \times 10^{-5}$$
$$\therefore \; [\ce{SO4^2-}] = \left(\frac{1.4 \times 10^{-5}}{16.0 \times 10^{-6}}\right) = \left(\frac{1.4}{1.6}\right) = \pu{0.875 M}$$

Thus, you have $\pu{0.875 M}$ of $\ce{SO4^2-}$ ions in this solution. However, note that some of $\ce{SO4^{2-}}$ ions would be precipitated as solid $\ce{Ag2SO4}$ when you keep adding $\ce{Na2SO4}$ solid (vide supra) in order to maintain the equilibrium (Le Chatelier's principle). This amount is equal to the half of $[\ce{Ag+}]$ ions precipitated with it:

$$\therefore \; [\ce{SO4^2-}]_{ppt} = \frac{1}{2}\left(2 \times 1.52 \times 10^{-2} - 4.0 \times 10^{-3} \right)= \pu{1.32 \times 10^{-2} M}$$

Suppose you add $m$ amount (in moles) of $\ce{Na2SO4}$ to the original solution. which already had $\pu{1.52 \times 10^{-2} M}$ of $\ce{SO4^{2-}}$ ions. Thus, $[\ce{SO4^{2-}}]$ remain in the solution:

$$[\ce{SO4^{2-}}]_{remain} = \pu{0.875 M} = m + 1.52 \times 10^{-2} - 1.32 \times 10^{-2} $$

$$\therefore \; m = 0.875 - 0.20 \times 10^{-2} \approx \pu{0.873 M}$$

Thus, you need to add $\pu{0.873 mol/L}\times \pu{0.5 L} \approx \pu{0.437 mol}$ of $\ce{Na2SO4}$ to the original solution.


How to simulate chemical reactions in GAMESS?

As in geometry optimization, you are searching for stationary point on the potential energy surface (PES). Not for local minimum, but for saddle point, therefore in GAMESS, you specify RUNTYP=SADPOINT. You would also need the correct (non-guess) Hessian matrix, which you can calculate separately with RUNTYP=HESSIAN. In transition state (TS), you should have zero gradients (stationary point) and one imaginary eigenvalue of Hessian (the reaction coordinate in TS).

For working example, see e.g. SN2 transition state. In principle you need very good guess at TS geometry, which you obtain by scanning along some chosen coordinate. The point highest in energy should be reasonably similar to the TS, and you optimize to saddle point starting from there. Already at the guess geometry you must have one imaginary Hessian eigenvalue (resembling the reaction you are looking for, not methyl rotation somewhere else).

But beware, finding the transition states is one of the most difficult topics in computational chemistry and you are not guaranteed to find one, no matter how hard you try. One of the biggest problem is solvation, as you often wish to find reaction barrier in solvent (where majority of chemistry happens), but you can calculate only in vacuum. Just think of how carefully the workbench chemist finds the correct solvent for given reaction, in computer you have only very crude approximations to it (and most often just vacuum).

That said, it is great fun and rewarding to find the transition states. For best results, use the simplest theory available, as it is better to have smooth and well behaving PES, even though it is slightly wrong, so it is no harm to start with HF or BLYP (or B3LYP). To get some ideas, start with simple reactions, i.e. where no charged species are involved, so electrocyclic reactions are probably best choice, e.g. Diels-Alder.


How is (SO4)^2- Charged if S and O all are Neutral Themselves?

The ion is not one S atom and 4 O atoms together. The ion on the whole is having two negative charges, Thus it is somewhat like 1Sulphur + 4 Oxygen+2 electrons. It is not as though there was one sulphur and four neutral oxygen atoms, and they became negatively charged during bond formation. The bond must have formed in some reaction, and in that reaction the charge would have been conserved.


How are t-butyl substituents named according to IUPAC nomenclature?

According to Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013 (Blue Book) (P-57.1.2), the preferred prefix for $\ce{-C(CH3)3}$ is "tert-butyl".

As for where it comes in the name:

  P-14.5 Alphanumerical order
  [...] Alphanumerical order is applied as follows in organic nomenclature. Nonitalic Roman letters are considered first, unless used as locants or part of a compound or composite locant [...]

so the prefix tert is not considered (except in more complicated cases, cf. P-14.5.3). Therefore, the preferred IUPAC name for the molecule in question is 1-tert-butyl-5-heptyl-2-propylcyclononane. Note that there are no parentheses around tert-butyl.

Be warned: tert-butyl has received special treatment from IUPAC on this front. Other non-systematic prefixes have fallen from favour, most notably isopropyl, sec-butyl, and isobutyl. The preferred prefixes are respectively propan-2-yl, butan-2-yl, and 2-methylpropyl.


Deflection of Alpha Particles in Rutherford's model of atom

See, what the Geiger-Marsden-Rutherford experiment achieved was the following: by bombarding (with alpha particles) a one-atom thick gold sheet and counting how many alpha particles passed through, they were able to relate the already known atomic radius with the actual area that could get collided by alpha particles.
Figure 1 shows a sketch of the apparatus.

Figure 1: depiction of the apparatus (courtesy of Wikipedia).
Now follows a simplified rationalisation.
Assume you have an atom-thick sheet of some material.
By knowing the element's molar mass and the mass of the sheet you can calculate how many atoms $n$ are there, and knowing the sheet's surface area $A$ allows you to calculate the atomic surface density $\sigma = n/A$.
Now if the atoms have radius $\rho$, an area $A_\rho = n \pi \rho^2$ of the sheet should be covered by atoms.
This means a ratio
$$r = \frac{A_\rho}{A} = \pi \rho^2 \sigma$$
is actually covered by atoms and $f = 1 - r$ is the ratio of "free space".
Rutherford actually tested if this made sense by throwing particles at the sheet: those scattered simply hit something, the rest passed through.
Imagine $M$ particles were thrown and $m$ of those passed through.
If $M$ is big enough we can set $f_R = m/M$ as Rutherford's "free space" estimate.
Going backwards we may set the coverage ratio
$$r_R = 1 - f_R = 1 - \frac{m}{M} = \frac{A_R}{A} = \pi \rho_R^2 \sigma$$
were $\rho_R$ above is the measured radius of the collision cross section of an atom.
The experiment has shown that $r_R \ll r$ and thus $\rho_R \ll \rho$, which implies that most of the atomic mass is buried inside the atom.
In fact (Wikipedia),

The diameter of the nucleus is in the range of $1.75$ fm ($1.75 \times 10^{-15}$ m) for hydrogen (the diameter of a single proton) to about $15$ fm for the heaviest atoms, such as uranium. These dimensions are much smaller than the diameter of the atom itself (nucleus + electron cloud), by a factor of about $23,000$ (uranium) to about $145,000$ (hydrogen).

In Rutherford's own words (p. 68),

It was quite the most incredible event that has ever happened to me in my life. It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you. On consideration, I realized that this scattering backward must be the result of a single collision, and when I made calculations I saw that it was impossible to get anything of that order of magnitude unless you took a system in which the greater part of the mass of the atom was concentrated in a minute nucleus. It was then that I had the idea of an atom with a minute massive centre, carrying a charge.
--- Ernest Rutherford

Thus, nuclei are simply very small, but very dense, as stated by Enigma and Ilmari Karonen.


Having trouble naming hydrocarbons with the IUPAC nomenclature rules

The first name has the correct alphabetical order of groups: ethyl before dimethyl. The "d" of dimethyl is not involved in alphabetizing. The numbering is correct because the one-to-one comparisons of locants is {1,2,4 < 1,3,4}. The correct name is 4-ethyl-1,2-dimethylcyclohexane. For a brief set of IUPAC rules applicable to this question see, https://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/nomen1.htm


Vinegar was used as one of the main household cleaning and odor removal agents for a good reason. I think in general you can get numerous iron (III) and chromium (III) (which are less pleasant) salts and complexes aside from acetate, e.g. chloride/iodide if you use iodised salt or hypochlorite from chlorinated water.
Diluted acetic acid (vinegar) etches majority of steels quicker than the glacial one, dissolving its passivating $\ce{Cr2O3}$ layer. Aside from concentration, reactivity also greatly depends on the steel type and acetic acid purity [1]:

Carbon and Alloy Steels. Steel is attacked quite rapidly by all concentrations of acetic acid, even at room temperature. Glacial $\ce{CH3COOH}$ at room temperature is less aggressive than aqueous solutions of the acid, but still gives a rate of attack of $\pu{0.8 to 1.3 mm/yr}$ ... Therefore, steel is normally unacceptable for use in $\ce{CH3COOH}$ service.
Stainless Steels. The chromium stainless steels of the 400 series occasionally exhibit low corrosion rates in laboratory tests in dilute acetic
acid. However, because field experience with these materials indicates high corrosion rates and pitting attack, they are rarely used for $\ce{CH3COOH}$ production equipment.
Impurities present in the manufacture of acetic acid, such as acetaldehyde, formic acid, chlorides, and propionic acid, are expected to increase the attack of stainless steels. Contamination with chloride can cause pitting,rapid stress-corrosion cracking, and accelerated corrosion of type 316 stainless steel. Up to $\pu{20 ppm}$ of chloride can be tolerated, but higher concentrations are likely to cause rapid equipment failure.
Transferring heat through a metal wall, as in heat exchangers, can drastically alter the corrosion characteristics of the metal.

If you know the exact steel type used, you can check its stability against various acidic media in several corrosion handbooks, e.g. [1] contains cumulative tables for corrosion behavior of manufactured steels tested under various conditions.

Craig, B. D.; Anderson, D. S. Handbook of Corrosion Data; ASM International, 1994. ISBN 978-0-87170-518-1.


Finding the value of stoichiometric coefficient x

The relationship between $K_p$ and $K_c$ is

$$K_p=K_c(RT)^{\Delta n}$$

where $\Delta n$ is the change in the number of moles of gas molecules over the course of the reaction. The only way you could know this change is from the stoichiometric coefficients. The $\Delta n$ term derives from the laws of mass action in which all partial pressures and/or concentrations are raised to the stoichiometric coefficients.

Consider:

$$\ce{2NO(g) + O2(g) <=> 2NO2(g)}$$
$$K_c = \dfrac{[\ce{NO2}]^2}{[\ce{NO}]^2[\ce{O2}]}$$
$$K_p = \dfrac{P^2_{\ce{NO2}}}{P^2_{\ce{NO}}P_{\ce{O2}}}$$
$$P_x = \dfrac{n_x}{V_x}RT=[\ce{X}]RT$$
$$K_p = \dfrac{([\ce{NO2}]RT)^2}{([\ce{NO}]RT)^2([\ce{O2}]RT)}$$
$$K_p = \dfrac{[\ce{NO2}]^2}{[\ce{NO}]^2[\ce{O2}]}\dfrac{(RT)^2}{(RT)^3}=K_c(RT)^{-1}$$
$$\Delta n = -1 =2-(2+1)$$


Correct equation for Ionic Conductivity (λ) in Solutions?

Short answer: $\lambda = z \mu F$ is correct. The link  provided by @Loong confirms this (page 73 of the document). To add to that I have provided you with a short derivation.

Prerequisite reading (so to speak) for definitions of a few terms/notation would be my answers to this question, and this question

Let me establish an important distinction (or definition) which can (possibly) be a source of confusion. The two relevant equations from my previous answer  to the first question are:

  $$\Lambda _{m}=\Lambda _{m}^{0}-K{\sqrt  {c}}$$

and,

  $$ \Lambda_{m}^{0}= \nu_{+}\lambda_{+}^{0}+\nu _{-}\lambda _{-}^{0}$$

Before I begin my discussion, I would like to note that "$\Lambda_m$" (uppercase ) is the molar conductivity of the solution, and "$\lambda$" is the molar conductivity of the ion (ionic conductivity). The second equation has a "special name", that I did not use in my previous answer-"Kohlrausch's law of independent migration of ions".

In general for dilute solutions of strong electrolytes, molar conductivity can be decomposed into contributions from the different ions in solution (this is the spirit behind the previously quoted equation):

$$ \Lambda_m^0 = \Sigma \nu_i \lambda^0_i$$

Now, let's attempt to relate the ionic conductivity to the ionic mobility (it clearly makes sense to do so). So I am assuming we have a dilute solution of completely disassociated strong electrolyte with molar concentration $c$. Each formula unit of this electrolyte gives rise to $\nu_+$ cations of charge $z_+e$ and $\nu_-$ anions of charge $z_-e$.

The concentration of each type of ion is therefore, $\nu c$ ($\nu$ could either mean $\nu_+$, or $\nu_-$ to avoid unnecessary notational complexity).

the number density of each type of ions is:

$$\text{No. density} = \nu c N_\mathrm{A}$$

Now picture a small window of area $A$ in your solution. The number of ions that can pass through said window in a time interval $\Delta t$ is equal to the number of ions within the distance $s \Delta t$ (where $s$ is the drift speed of the ions)

$s \Delta t A$, happens to define a volume element and the number of ions in this interval is given by: $ V \times \text{no. density}$, i.e

$$ s \Delta t A \nu c N_\mathrm{A}$$

Thus, the flux of ions is

$$J_\mathrm{(ions)} = \frac{s \Delta t A \nu c N_\mathrm{A}}{ \Delta t A} = s\nu c N_\mathrm{A}$$

remember flux always has dimensions of "something" per area per time

Flux of charge is $$J_\mathrm{(charge)} = ze s\nu c N_A = z s \nu c F $$

again, I am omitting the +/- signs for the sake of clarity. Also, note $F := e \times N_\mathrm{A}$

Now, drift speed is $s = \mu E$, so the flux of charge (henceforth, simply J) is

$$J = z \mu \nu c F E$$

The current is charge flux times area, so

$$I = (z \mu \nu c F E)A$$

Additionally, $E = \frac{\Delta \phi}{l}$ (i.e the potential gradient)

$$I = (z \mu \nu c FA) \frac{\Delta \phi}{l} \tag{I}$$

Using Ohm's Law (i.e $\Delta \phi = IR$}, one obtains

$$I = \frac{\Delta \phi}{R} = G \Delta \phi = \frac{\kappa A \Delta \phi}{l} \tag{II}$$

Comparing the (I) and (II) one is lead to $$ \kappa = z \mu \nu cF$$

This is the ionic conductivity. We need to divide by concentration of ions to the get the molar quantity. The concentration of ions however is not $c$, but rather $\nu c$ and our final result is

$$\lambda = z \mu F$$


How can I determine/quantify the presence of boron nitride on a metal surface?

I'd suggest looking into the following:
scanning electron microscopy (should be able to see the film and how intact it is on the surface); Inductively Coupled Plasma tests (destructive).  I believe after your sample is prepared and presented ICP can find B and N emission lines


Why is the silver salt ammoniated to form the Tollen's reagent?

Yes, it is important to convert your silver ion to diamminesilver(I) complex. If you do not add ammonia to silver nitrate, the silver ion is reduced so quickly that colloidal silver metal would appear and the solution would become a black, cloudy liquid and you won't get the silver mirror when it reacts with aldehyde.
It is also important that you maintain the pH of the reaction. During the preparation of Tollen reagent, you will see the use of sodium hydroxide. This is very important for maintaining the pH. During the whole reaction, the pH is between 8-10 and it goes even faster if the pH is beyond 10. The reaction also becomes more sensitive. This has been explained in details in this journal by going through the exact mechanism of the reaction.


It can't happen.  Neutrality in aqueous solution means equal quantities H+ and OH-; since your solution is unlikely to have a net electric charge, so too must your anion and cation quantities be equal.  Buffering from acetic acid being a weak acid is meaningless if you don't have other anions around to make up that balance.

Solution is electrically neutral:
[H+] + [Na+] = [OH-] + [AcO-]

Solution is pH neutral:
[H+] = [OH-]

Substitute the latter in the former, subtract, you can only get:
[Na+] = [AcO-]

If the concentrations of both solutions were the same, and you're titrating by adding both to the same volume, you can only have equal concentrations if you add equal volumes.

If this was not the case (as you describe) we have to violate one of our assumptions:

Solution actually had a net electric charge (you're doing the titration in space and you're bombarded by beta radiation, i.e. unlikely)
Solution was not truly neutral.  Maybe someone calibrated your pH meters wrong, maybe everyone titrated with a faucet while texting and over-shot by a factor of 3.
Contamination with a strong acid added a variable we don't know about to the ion balance.
Most likely, the solutions were not truly the same concentration.  See if your teacher made her NaOH solution from solid NaOH that she let sit around in open air.  It can pick up a lot of moisture from the air, i.e. it's hygroscopic, and before you know it the weight you're measuring on a balance is half water.


Frequency region and "uncoupled oscillators" in vibrational spectroscopy

$\ce{H}$ is just hydrogen and $\ce{X}$, $\ce{Y}$, and $\ce{Z}$ are just other elements, in this case likely carbon, nitrogen, or oxygen. They give an example in the table of $\ce{-X=Y}$ where $\ce{-C=N}$ matches this template of two (generally) different elements double bonded together.
They use coupled here in a slightly different way than most literature. I would may switch to delocalized/localized rather than coupled/uncoupled. The modes they mention are mostly localized to a single bond (e.g. an $\ce{-X-H}$ stretch) whereas in general vibrational modes are delocalized/coupled over the whole molecule (e.g. an aromatic $\ce{CH}$ stretch where all the $\ce{C-H}$ bonds of an aromatic ring move in unison).


Mechanistic explanation of salt lowering temperature of ice slush

Water loses kinetic energy when it detaches from the surface of the ice

Water molecule in the solid phase have more hydrogen bonds than in the liquid phase. It takes energy to break these bonds, so only molecules with sufficient kinetic energy will detach (maybe because they just collided with a water molecule that transferred some kinetic energy to it). So the molecule that detaches will have a low kinetic energy as it goes into the liquid phase. Further collisions will quickly ensure that over time, this particular molecule has an average kinetic energy, but overall the temperature will drop from that detaching event.

Water gains kinetic energy when it attaches to the surface of the ice

This is just the opposite process, so the opposite happens. As additional hydrogen bonds are formed, the molecule has greater kinetic energy (in the form of vibrational energy). Either the molecule had less than average kinetic energy and now has average energy again, or it had average kinetic energy and will have higher kinetic energy, but quickly dissipate this to other molecules.

When you add salt to the liquid water, less molecules attach to the surface of the ice

Compared to pure water, less collisions of particles in the solution with the surface of the ice will lead to attachment of particles because now, some of these particles are not water molecules but sodium or chloride ions instead (they are not incorporated into the crystal lattice of ice but stay solvated by liquid water). As a consequence, there are less attachment events while the detachment events stay the same. With a net melting of ice, the average kinetic energy of particles decreases (the temperature drops). As a secondary effect, this also affects the rates of attachment and detachment, so once the temperature has sufficiently dropped, the two rates will be the same again.

  What energies (kinetic/temperature vs. latent/chemical) are involved, and where?

In my explanation, I talked about kinetic energy (reflecting the temperature) and energy necessary to break hydrogen bonds (could be generalized as intermolecular interactions, and they underlie the observed bulk latent heat).

  How exactly (atomically or subatomically) is one type of energy converted into the other

There is no chemistry or nuclear physics happening, so it is sufficient to talk about the molecular level. Kinetic energy is converted into potential energy through inelastic collisions. Kinetic energy is redistributed between molecules through elastic collisions.

  Why is each step energetically favourable? Or is it not, and it's reversible? How reversible, statistically?

According to microscopic reversibility, all molecular events are reversible. Processes go in a certain direction because it is more probable than in the other direction (2nd law of thermodynamics). The 2nd law of thermodynamics is inherently linked to large numbers of particles, so you need to look at the forest rather than at individual trees.

  How does the concentration of salt change the evolution of the system (again, at a microscopic level).

The higher the concentration of the salt, the lower the concentration of the water in the liquid phase, i.e the fewer events of liquid water attaching to the ice.

  How does adding e.g. ethanol instead of salt change things?

1 mmol/L NaCl has twice the dissolved particles compared to 1 mmol/L ethanol. Otherwise, the effect is the same (colligative property).

  Is liquid water required at all? Is salt able to melt and cool completely frozen ice, or is external force required (stirring, driving over it, etc.) to cause some initial melting for anything to happen? (I've read conflicting answers to this in the past.)

Ice near the freezing point exhibit surface melting. Once there is a tiny amount of water, more ice will melt. On the other hand, if it is cold enough that even saturated solutions will freeze, a mixture of dry salt and ice will not interact.


Can I obtain a different density of a recycled elemental material?

I think it is risky. The density observed lower than the expected density of pure Al is indicative for the presence of something of even lower density than Al itself.

Similar in appearance to aluminium (density $\pu{2.7 g/cm^3}$, melting point $\pu{660 ^\circ{}C}$, source) is magnesium (density $\pu{1.7 g/cm^3}$, melting point $\pu{650 ^\circ{}C}$, source).  Both are used, pure as well as in mixture of other metals (alloys), widely.  Even to the point that they are mixed to form alloys with each other, for example in the frames of laptops

(source)

or items that are used near the sea / in contact with sea water (source).


How to distinguish between CsCl and NH4Cl by a chemical test?

Correct you can distinguish them by their reaction with a base. You should use a base, such as $\ce{NaOH}$. $\ce{Cs^+}$ wouldn't react with $\ce{NaOH}$, while $\ce{NH4+}$ will undergo reaction $$\ce{NH4+ + OH- -> NH3 + H2O}$$ $\ce{NH3}$ is a gas and can be detected with a moist litmus paper.

If you will, you can actually detect $\ce{Cs+}$ using a procedure from this paper:

"Selective Detection of Cs+ in Water Solutions via One-Step Formation of a New Type of Struvite-Like Phosphate". However, this is an exotic reaction (in practice you would use the flame color test) and you don't have to do it since you only need to distinguish samples.


Why is the uncertainty of a digital instrument always the smallest scale division?

Migrating my comments to an answer:

The big issue with your line of reasoning is that the transducer in the scale does not produce decimal output. Like all digital devices, the signal is digital (i.e., binary), so the representation of the weight will also be a binary number.

Then you're in a scenario where the uncertainty is something like $\pm 0.0001_{2}$, which is $\pm 0.0625$. In this case, I would say that the error is in the first decimal place after the decimal point.

You can't give an error of $\pm 0.005$, partly for the reason I gave, but also because the error in measurement is not necessary an error in rounding. It's possible that the transducer is temperature sensitive, so within a reasonable range, it might over or under measure by $0.01$ units.


Can fully-hydrogenated oils be mixed with unsaturated oils for use in food (margarines, etc)?

The purpose of hydrogenation is to convert liquid oils into semi-solids at room temperature.  Palm "oil" is unusual among natural vegetable oils/fats in that it is naturally fairly unsaturated and is thus a semi-solid at room temperature.

Economics is probably the driving factor preventing significant research into the solution (pun intended) you propose with mixing saturated and unsaturated fats to get the desired consistency.  Hydrogenation techniques are cheap, well established, agency approved, etc.  Same idea with palm oil.

I hope I hit the points of your question, comment if I missed anything.


Length of C-H bonds in hybridized bonding orbitals

Yes, the bond length decreases.  One way to think about it is that a sp3 center with 109.5 degree angles is more crowded than an sp2 center with 120 degree angles.  More crowding = more repulsion = longer bond length.

Another way to think about it is that the e-pair in an s orbital is sitting on the nucleus.  Adding a p-orbital moves the electron pair further away from the nucleus.  More p-orbital character = longer bond lengths.

You can test it for yourself using http://molcalc.org/


Is the total entropy change of all isothermal processes 0?

No. Only for reversible processes.
However your equation is correct.
For an irreversible process in which an ideal gas expands isothermally against a constant external pressure
$$\Delta S_\mathrm{total}=\frac{q_\mathrm{rev}}{T}+\frac{q_\mathrm{surroundings}}{T}$$
becomes
$$\Delta S_\mathrm{total}=\frac{q_\mathrm{rev}}{T}+\frac{w}{T}$$
where $w=-p \Delta V$, since $q_\mathrm{surroundings}=-q_\mathrm{system}$ and for the isothermal process the first law stipulates that $w=-q_\mathrm{system}$.
For a free expansion $p_\textrm{ext}=0$ and no work is done. No heat is exchanged and the entropy of the surroundings does not change. The change in entropy of the system then drives the process.


Cleaning with Vinegar - what surfaces does it react badly with? Metal window sills ok?

Avoid aluminum alloys when working with vinegar solutions!

With time, the pH effect of acetic acid can penetrate the protective oxide coating. The exposed Al will even react with water:

2 Al (s) + 6 H2O (l) --> 2 Al(OH)3 + 3 H2 (g)

and with the presence of alloy metals, galvanic corrosion (especially if the 'dirt' is rich in salts that can serve as a good electrolyte) which apparently accelerates the attack on the Aluminum.

To quote a source:

  corrosivity environment for aluminium in food industry are foodstuffs with pH 3 – 5, such as fruit juices, jams and acidic canned fruits or hot gravies, sauces as well as dressings, vegetables and fish pickled in brines with 1 –3 % salt [1]. In particular, acetic acid needs to be taken into corrosion consideration, due to its wide usage in food industry (vegetable and fish pickling), and its representative properties among acids and juices in fruits, vegetables and other organic materials that can corrode metals. Although aluminium has a good resistance to acetic acid solution at room temperature, aluminium can corrode in almost any concentration of acetic acid at any temperature if the acid is contaminated with the proper species [2].

Further comments related to the presence of select metals:

  Although aluminium has a good resistance to almost all the concentrations of acetic acid at room temperature, care must be taken that the metal is free of other impurities such as iron, copper, tin and lead even in traces [2]. With increasing purity of aluminium, its resistance to acetic acid solution increases, and 99.5% aluminium can be used for the majority of engineering purposes, but components added to its alloy can increase the corrosion of aluminium in acetic acid solution [14]. Since the metal corrosion occurs via electrochemical reactions at the interface between the metal and an electrolyte solution, electrochemical techniques are ideal for the study of corrosion processes.

And importantly, the effect of pH:

  It is well known that aluminium resistance is related to the thin and compact layer of naturally formed oxide on aluminium surface, but this oxide layer is stable only in pH range 4-8. Lower or higher pH values caused prominent destroying of protective layer and thus the significant metal dissolution [15].


What other intermoelcular forces can exist between neutral molecules?

This can be a bit tricky when you are first starting your journey into the world of intermolecular forces. What I like to look for first are what (if any) heteroatoms are present when the backbone of the molecule is composed of carbon atoms.

In your case there is a nitrogen present and it is attached to 2 hydrogen atoms. This is a perfect example of a molecule that will exhibit hydrogen bonding as an intermolecular force. Hydrogen bonding, at least in my opinion, is the easiest to pick out. N, O, and F atoms bonded to Hydrogen are the only species in which this attractive force between molecules is observed. Now if you were to remove that $\ce{NH2}$ group and just had $\ce{CH3CH3}$ you would be correct. The only intermolecular force between the molecules would be London forces.


Using drip extraction, how much caffeine is extracted from coffee as a function of the amount of water used?

Caffeine has a solubility in water of 20% at 80 $^\circ$C, while the coffee has between 1 and 4% caffeine.
That means that you have more than enough water to extract all the caffeine in the coffee.
With the quantities you have given, I estimate that you are using from 12 to 24 mL of water per gram of coffee (bulk density 0.25 g/mL).

So, it is not very likely that a change in volume will affect the total quantity of extracted caffeine much. Only the concentration per unit volume will be lower if more water is used.
But then, solid-liquid extraction is not that perfect and possibly you will extract some more caffeine with more water. The main property that influences the extraction is the size of the milled coffee. The smaller the better is the extraction.

Many other components of coffee are extracted fairly fast.
Therefore, here we do a percolation (expresso) which extracts everything with very less water. The concentration of caffeine per unit volume seems to be higher, but the dose of coffee is much smaller.


Why is the yield of substitution reactions decreased at higher temperatures?

In short, this is because both elimination and substitution occur parallely and are competing; amount of elimination product + substitution product = constant. If the elimination rate increases more than the increase in substitution rate, then the ratio of elimination product to substitution product will increase. This is what happens on heating, though the rate of both elimination and substitution increase with temperature.

Broadly, there are two types of reactions: those that are kinetically controlled, and those that are thermodynamically controlled. Most elimination-substitution reactions are exothermic and are kinetically controlled. That means that the proportion of products (elimination or substitution) depends only on the rate of elimination and substitution, and not on the thermodynamic favour for the reaction.

Given that the reaction proceeds to completion, the ratio of elimination product to substitution product is equal to the ratio of elimination rate to substitution rate. Suppose elimination and substitution rates are equal, then the products form in a 50:50 proportion.

The ratio of elimination rate to substitution rate is

$$
\frac{e^{\frac{-\Delta G_{e}^{*}}{RT}}}{e^{\frac{-\Delta G_{s}^{*}}{RT}}} = e^{\frac{\Delta G_{s}^{*}-\Delta G_{e}^{*}}{RT}}
$$

where $ \frac{-\Delta G_{e}^{*}}{RT}$  and  $\frac{-\Delta G_{s}^{*}}{RT}$ are the energy barriers for elimination and substitution respectively.

Since the barrier for elimination is more than the barrier for substitution, the above ratio increases with increase in temperature, and hence more elimination product is formed at higher temperatures.


Flavonoid Xanthine Oxydase Inhibitors vs Hydroxychloroquine

HCQ does show antiviral properties in vitro, but, in vivo (in humans), when taken after an initial dose of 800mg for 4 days at 600 mg/day, it was not found to prevent COVID-19 in high-risk exposure settings.  Thus, it didn't stop people from getting SARS-CoV-2 infection followed by onset of symptoms/severity.
I suspect this is exactly what would happen with flavonoids, many of which have shown antiviral properties in vitro (laboratory settings).  The information on flavonoids and SARS-CoV (2003) is very sparse, but several flavonoids did inhibit SARS-CoV in vitro (see last paragraph below).  I do know there was a clinical trial approved in 2020 by NIH in which the investigators received an IND (From FDA)  to use diosmin as a supplement to treat patients with COVID-19 (i.e. hospitalized).  However, I don't know if was completed.  Diosmin is available OTC as a supplement, and used via prescription under the brand names Dafflon® and Vasculera® for chronic venous insufficiency (CVI), which is caused by venous hypertension (VH). VH, in turn, aggravates metabolic imbalances leading to a self perpetuating cycle of further metabolic changes, including venous acidosis. These changes promote further inflammation in vascular tissue leading to edema, skin damage and possible ulceration and deep vein thrombosis (DVT). Varicose veins and hemorrhoids are also part of the spectrum of CVI disorders.
Interestingly, it took 16 years before a group published the most comprehensive paper on in vitro inhibition if SARS-CoV (2003) by many flavonoids that were considered.  I review these in a pre-print on flavonoids that was just accepted for publication.


Cooking is potentially lethal.
Just like how going out for a walk is potentially lethal (You could get run over by a car, maybe develop a stroke, or get hit by a meteorite).

Cooking or cookery is the art, technology, science and craft of preparing food for consumption.
 From this Wikipedia article on Cooking

The basic idea behind cooking, as a whole, (regardless of the actual physical process) is to either make certain inedible (or indigestible) items  edible (digestible), or otherwise unsavory items more palatable.
Sometimes we have toxic substances that are rendered "safe", through the reactions that the cooking process affords. Other times, the opposite happens.

Now, does cooking (boiling, frying, baking) something generate toxicants?
Almost certainly yes. But the amount and type of toxicant(s) produced is what determines the overall lethality of the item you've cooked.
The potato chips/fries we all hold so dear almost invariably contain acrolein and acrylamides, however in trace amounts. If, you chose to let the potatoes simmer for a little too long, then you'll better be able to taste the wrath of those substances.
It is the dose, that maketh the poison (as I've discussed in a previous answer).

The cooking process does, as you point out, initiate a gamut of reactions: Some simply break down the food into products you can better digest, absorb and assimilate... while others lead to products that aren't as useful.


Aluminum can and is used as both anodes and cathodes in electrochemical cells, but there are some peculiarities to using it as an anode in aqueous solutions.
As you note, aluminum forms a passivating oxide layer quite readily, even by exposure to atmosphere.
In an aqueous solution, if the potential is high enough, $\ce{OH-}$ and $\ce{O^2-}$ are generated at the anode, which can then react with the aluminum to produce aluminum oxide.
$\ce{Al^3+}$ can also be generated directly.
The electric field will draw the anions through the growing aluminum oxide layer towards the aluminum surface and the $\ce{Al^3+}$ towards the solution, making the oxide layer grow both away from the electrode surface and into the surface of the electrode.
In this way, coatings thicker than the normal passivation in air can be produced.
However, aluminum oxide is a good electrical insulator, thus if a dense non-porous layer is grown, it will become impossible to pass current through it and growth will stop, leaving a relatively thin oxide layer (this is how the dielectric layers in electrolytic capacitors are made).
This is the normal behaviour in aqueous solutions at near-neutral pH (5–7). [1]

However, if a thick aluminum oxide layer is desired (e.g. to produce coatings on aluminum parts for dying or durability), maintaining porosity is necessary to avoid completely blocking access to the surface.
One technique that is commonly used is using a low pH solution, which tends to redissolve some of the oxide and neutralize some of the formed $\ce{OH-}$, leaving pores in the oxide layer through which the ions can travel and continue to react.
These pores also give a good structure to retain dyes or lubricants, but generally need to be sealed after to protect against corrosion.


Nomenclature of Organic Compound from Condensed Formula

Are you looking for this? 1,4-dibenzylbenzene

http://www.chemspider.com/Chemical-Structure.205991.html?rid=80e8a67d-0782-4794-bf27-db26ccd1ce39&page_num=0

What I did was go to their Structure tab and input what I thought to be the compound you typed. If this isn't it, you can probably draw what you think it is.

Hope this helped!


Are ortho, meta and para used in IUPAC nomenclature?

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book),

The locants o, m, p are no longer recommended; the numerical locants '1,2-', '1,3-', and '1,4-' must be used in substitutive names. However, as an exception, the three isomers of xylene are still recognized as o-, m-, and p-xylene in general IUPAC nomenclature (see P-22.1.3).

and

The letters o, m, and p have been used in place of ortho, meta, and para, respectively, to designate the 1,2-, 1,3-, and 1,4- isomers of disubstituted benzene. This usage is strongly discouraged and is not used in preferred IUPAC names.

The preferred IUPAC name for the compound that is given in the question is 1-chloro-2-methylbenzene.

The name 2-chlorotoluene may be used in general nomenclature.


How can I calculate the percentage of dissociated dinitrogen tetroxide?

$\ce{N2O4 (g)}$ say initially is $P\ \mathrm{atm}$.

$$PV = nRT$$

$$P(1)= (36.8/92)( 0.082 )(300 K)$$

$$P= 9.84\ \mathrm{atm}$$

Now we have:

$$\ce{N2O4(g) <=>2NO2(g)}$$

Considering initial pressure of N2O4 as p, at equilibrium, assuming 'x' as the degree of dissociation, we will have pressure of N2O4 to be p(1-x) and pressure of NO2 as 2px.

So, $$K_p = \frac{[\ce{NO2}]^2}{[\ce{N2O4}]}$$

$$K_p= \frac{(2px)^2}{p(1-x)}$$

$$K_p= \frac{4px^2}{(1-x)}$$

assuming $1-x$ approximately as $1$,

$$K_p= 4px^2$$

substituting $p$ as $9.84\ \mathrm{atm}$ and $K_p$ as $0.1642$:

$$0.1642= 4(9.84)x^2$$

$$x^2= \frac{0.1642}{4(9.84)}$$

$$x^2= 0.1642/39.36$$

$$x^2= 0.00417$$

$$x= 0.0645$$

degree of dissociation is 0.0645

Hence % of dissociation is $0.0645 \times (100) = 6.45\ \%$ (near to your answer)


(P1,T1) -> to (P2,T2) then what will be the change in enthalpy for ideal gas?

The enthalpy of an ideal gas does not depend on pressure.  It is a function only of temperature.  So $C_pdT$ is always equal to dH for an ideal gas of constant composition.  The subscript p on $C_p$ merely indicates that $C_p$ can be measured directly by determining the heat Q absorbed by the gas in a constant pressure experiment.


Having an electron in an antibonding orbital means that the electron exists in a high energy orbital.  If anything, this means the electron will be "motivated" to react and form a bond, particularly if bond formation lowers the energy of the electron.  An electron in an antibonding orbital can form a bonding (as opposed to antibonding) bond with positive overlap.


Very nice question. I also had the same question. These orbitals are nothing more than functions (wavefunctions). They are solutions to Schrodinger's equation, a second order differential equation. So you solve the equation and the solutions you get are determined by $4$ numbers also called quantum numbers $n, l, m_l, m_s$. A shell consists of all orbitals with same value of $n$ whereas a subshell consists of all orbitals with same values of $n$ and $l$. As others have pointed out the solutions are exactly only for the hydrogen and hydrogen like atoms.


Functional groups are a unifying idea in organic chemistry. There are tens of millions of different organic chemicals but many of their key properties are determined by the functional groups they contain. For example, amines (compounds with the $\ce{-NH2}$ group) are basic and often smelly; carboxylic acids (containing the $\ce{-COOH}$ group) are acidic.

The bar in the symbol represents the bond to the rest of the molecule. Sometimes chemists write this slightly differently by including a specific symbol for the rest of the molecule like this: R-OH where R is the rest of the alcohol. So a chemist might talk about a series of similar alcohols by defining R as $\ce{CH3}$ (methanol) $\ce{CH3CH2}$  (ethanol) or $\ce{(CH3)2CH}$ (isopropanol).

This simplifying terminology allows chemists to talk about large groups of related compounds without having to be too verbose and writing out all the details of every structure.

This is an organic chemistry concept so the functional group attaches covalently to a carbon. Some groups can also occur in inorganic compounds but the terminology would be different. For example, Sodium Hydroxide contains an isolated OH but this isn't an alcohol and wouldn't be written or described using terminology related to functional groups.


When put in water a mixture such as $\ce{HCl}$ and $\ce{Pb(NO3)4}$ goes to complete ionic dissociation. Lead nitrate is soluble in water and hydrochloric acid is a very strong acid. Nitric acid is also strong, which means those dissolved nitrate ions are not reversing back to the molecular form.

Their ions will be fairly stable in solution and will only combine in case a more stable compound is formed. You could come up with reactions in which some compounds will go out of solution (like $\ce{Cl2}$), such as $\ce{ 2HCl + HNO3 \to HNO2 + Cl2 + H2O} $. Lead could also act as an oxidizing agent. But I would argue that if you consider the change in free energy or the net redox potential of those reactions, you'd find that they are quite unfavourable. If you want to calculate it itself, you could start from a reduction potential table like this one.


Spectrometer vs Spectrometry vs Spectroscopy

Long time ago, I wrote mass spectroscopy, by mistake, in an undergraduate exam and the instructor told me "You were the only one in my class who wrote mass spectroscopy." The explanation was that MS is mass spectrometry because it does not deal with light or electromagnetic radiation and that's what the textbook said. At that time internet was not that common. Now I searched the term "mass spectroscopy" in papers using Google Scholar and there are 1,100,000 results with this exact phrase. There are 2,330,000 results for mass spectrometry.

It seems, given the usage, and the difference is not that strict. Consulting the controller of all  chemical terminologies, IUPAC says

https://goldbook.iupac.org/terms/view/S05848

  The study of physical systems by the electromagnetic radiation with
  which they interact or that thay [spelling mistake] produce. Spectrometry is the
  measurement of such radiations as a means of obtaining information
  about the systems and their components. In certain types of optical
  spectroscopy, the radiation originates from an external source and is
  modified by the system, whereas in other types, the radiation
  originates within the system itself.

You will see famous books, especially by Griffiths with the title "Fourier Transform Infrared Spectrometry" and nobody raised an eyebrow.


If a bond dissociation energy is listed, does it refer to homolytic or heterolytic fission?

This Wikipedia article summarizes the definition of bond-dissociation well:

  Bond-dissociation energy (BDE or D0) is one measure of the strength of a chemical bond. It can be defined as the standard enthalpy change when a bond is cleaved by homolysis, with reactants and products of the homolysis reaction at $\pu{0 K}$ (absolute zero). For instance, the bond-dissociation energy for one of the $\ce{C–H}$ bonds in ethane ($\ce{C2H6}$) is defined by the process: $\ce{CH3CH2–H → CH3CH2• + H•}$

Thus the definition of bond-dissociation refers to homolytic fission, and in your example the bond energy of $\ce{A-B}$ is the energy required to make "A radical + B radical".


Does clay absorb water because of polarity?

I am no expert regarding clay (which is a surprisingly varied field) and will therefore give a more general answer. I suspect that there is also quite complex interaction with ions on the surfaces of the layers and such things.

Materials can be generally categorized on a rather continuous scale, the extremes of which are called hydrophilic and hydrophobic (or lipophobic/lipophilic which are synonyms).

If something is hydrophilic, it "likes" to be wet, i.e. it is energetically favourable for the surface to be wet because this minimized surface energy of the system. The opposite applies to hydrophobic substances.

As you have observed yourself, clay is quite hydrophilic.

One can indeed estimate whether a material is hydrophilic based on its chemistry: water likes to form hydrogen bonds and water complexes (e.g. around Na+). Therefore things are more hydrophilic the more of said things they have. A good example is that sugars and carbohydrates are very soluble in water despite being large, whereas even small alkanes are barely soluble.

The explanation involving air pressure makes no sense, because to move matter there has to be a pressure difference. But if you imagine clay as a porous substance, the pores might as well be filled with air and there is an equilibrium. In fact (as easily observed when burning earthware) clay gets more voluminous with higher water content.


I am not a professional or anything close, but I think the answer to your question is simply that $C$ is jointly proportional to $A$ and $B$. The following link explains joint-proportion (https://www.mathwords.com/jk/joint_variation.htm). Just in case, the link gets broken in the future:


In a review (PDF, CalTech) of the properties of elemental sulfur, the author notes that the most stable form of sulfur at STP is in fact cycloocta-sulfur, $\ce{S8}$, with other polyatomic species also found. Elemental sulfur is commonly formed as a molecular solid, with arrangements of $\ce{S_n}$ rings or chains, and the S-S bonds can be Raman active.

The paper offers depictions of some of those structures, and also tabulates the principal Raman bands for selected allotropes of elemental sulfur.


Is carbon–sodium bond mainly ionic or covalent?

Electronegativity of Carbon is $2.55$ and that of Sodium is $0.93$ (from Wikipedia) so by the Haney-Smith equation (from J.D. Lee's Inorganic chemistry, third edition)
$$\%\text{Ionic Character} = 16 |\Delta\chi| =3.5 (\Delta\chi)^2,$$
where $\Delta \chi$ is the difference in electronegativity of the two atoms. Carrying out the calculation gives us
$$\% \textrm{Ionic Character} = 35.1054$$
So, it shows that the bond $\ce{C-Na}$ has a less ionic character.

In Organic Chemistry by Morrison and Boyd we find

  The carbon-magnesium bond is covalent but highly polar, with carbon pulling the electrons from electropositive magnesium [...].
  But, whatever the metal, it is less electronegative than carbon, and the carbon-metal bond - like the one in the Grignard reagent - is highly polar. Although, the organic group is not full-fledged carbanion [...] it nevertheless has considerable carbanion character.

Note: No bond is completely ionic or covalent, what matters is which character is dominant and hence the bond will behave like that.


Why is it not possible to determine the rate constant this way?

The average rate $\Delta[A]/\Delta t$ over time interval $\Delta t$ is only approximately equal to the true instantaneous rate $\rm d[A]/dt$ at time $t$. The difference between the two can be significant if the instantaneous rate changes over the chosen time interval.

Here's an analogy: suppose you're driving your car on the highway at a steady 55 mph. Your average and instantaneous rates of travel will be much the same even over a long time interval. On the other hand, if you're driving through a city with a lot of starts and stops at intersections, your instantaneous rate of travel won't match your average rate of travel unless you're looking at averages over very short time intervals.

To see how far off we'll be if we don't use a small time interval, let's compare the average and instantaneous rates for your reaction with time intervals of 0.1, 0.5, and 1.0 s:

The larger the time interval, the larger the difference between $\Delta[A]/\Delta t$ and $\rm d[A]/dt$ and the larger the error in the rate constants estimated using your method.

So why not just use very small time intervals? Well, there's a limit to how small you can make $\Delta t$ if you're computing $\Delta[A]/\Delta t$ from experimental measurements. You're taking differences between concentrations that are almost the same size. Any error in experimental measurements of $\rm [A]$ may well swamp your rate estimates as you go to smaller and smaller time intervals.