Proper definitions of resistivity of conductivity

The specific modifier simply means that stated property per a given quantity, usually mass, usually a gram. In other words resistance or conductance per gram of resisting or conducting material.  But the fundamental definition of the properties for the different word forms of resistance and conductance are not changed by this wording.


Reaction of potassium cyanide with 2-(chloromethyl)furan

Although the answer was already appropriately provided, for anyone interested in further reading, the reaction is covered in detail by Divald and coworkers:1

Essentially, the proposed mechanism starts from the resonance structures of 2-chloromethyl furan, where the chlorine bond is treated as ionic and the electron-donating property of oxygen stabilizes the shared positive charge similar to a benzyl cation. The 5' position is therefore electrophilic (and is presumably further activated by the inductive effects of a neighboring oxygen), resulting in a 60/40 yield of 5/2-methyl substitution.

It is however mentioned early in the paper that 5' substitution is only seen in protic solvents. This could be explained in the proposed mechanism, where after addition to the 5' position, elimination of the geminal hydrogen and tautomerization proceed through exchange with a protic solvent.

Reference

Divald, S.; Chun, M. C.; Joullié, M. M. Chemistry of 2-(chloromethyl)furans. Reaction of 2-(chloromethyl)furans with aqueous potassium cyanide and other nucleophiles. J. Org. Chem. 1976, 41 (17), 2835–2846. DOI: 10.1021/jo00879a010.


Why does lower concentration cause the metal atoms to dissipate more?

It is not primarily that the process $\ce{M(s) -> M^n+(aq) + n e-}$ is faster for more diluted solutions.
The main reason is that the opposite process $\ce{M^n+(aq) + n e- -> M(s)}$ is slower for more diluted solutions.
As the net effect, there is dissolution of $\ce{M^n+(aq)}$ ions and the electrode gaining more negative potential, until the rate of both processes gets equal and the electrode reaches its equilibrium potential.
Because the rate of ion dissolution grows with increased potential,
while the rate of ion depositions grows with decreased potential.
The final effect is, the electrode in then more diluted ion solution has the more negative potential.


Alumina absorbs water to give aluminium oxide hydroxide, a solid which removes the water from the reaction environment.

According to the paper by Pethrick et al. [1]:

  Aluminium oxide exists in several forms which on addition of water give a complex range of oxide-hydroxide; boehmite, bayerite and gibbsite.

References

Pethrick, R. A.; Hayward, D.; Jeffrey, K.; Affrossman, S.; Wilford, P. Investigation of the Hydration and Dehydration of Aluminium Oxide-Hydroxide Using High Frequency Dielectric Measurements between 300 KHz–3 GHz. J Mater Sci 1996, 31 (10), 2623–2629. https://doi.org/10/chcqs8.


Is reverse osmosis a fancy term for filtration?

Ultrafiltration is ongoing at any pressure, even if slowly and liquid always goes to the side with lower pressure.
Reverse osmosis is ongoing only at pressure higher then osmotic pressure. At pressure lower than osmotic pressure, normal osmosis occurs and liquid flows to the side of higher pressure.


Does direction matter in breaking a bond?

A possible real-world example of directional force use may be apparent from the commercial application of so-called magnetizers (see discussion here).
I once experimented with a small inexpensive one and it appears to accelerate select reactions.
A commercial example for reducing the amount of chlorinating agent required in public swimming pools in England demonstrated success to reduce scale build-up, moderate pH and increase available chlorine in a pool, employing magnetic field treatments here.
Some discussion details on how magnetic fields may potentially impact chemical reactions per comments posted previously on StackExchange.
An important point concerns radical based reactions, which per chlorine-based chemistry in sunlight treated swimming pools, is likely extensive and may be potentially impacted. This could particularly alter pool chemistry especially surrounding the fate of the chlorine radical, where an alteration of spins along with a possible decrease in the reverse recombination reaction to Cl2 via the normal reactions:
$\ce{Cl2 + hv -> •Cl + •Cl}$
$\ce{ •Cl + •Cl -> Cl2}$
where the above reaction is impaired due to spin changes by magnetic fields, resulting in more chlorine radicals. The latter can slowly react with water, for example, importantly producing the powerful hydroxyl radical:
$\ce{•Cl + H2O <=> HCl + •OH}$
Source: See Table l in "Photochemistry of HCl and Other Minor Constituents in the Atmosphere of Venus" by Prinn, R.G. as an example.
More detail on mechanics is provided by this source, to quote:

Radical pairs in chemical reactions are known to respond to magnetic fields. This involves pairs of transient radicals being created simultaneously, such that the two electron spins, one on each radical, are correlated. Their chemical fate is largely dictated by weak magnetic interactions, so that you get radicals with unpaired electron spins either antiparallel (singlet state) or parallel (a triplet state).
Even though the interactions of electron spins with an applied field is very weak, there is enough time to affect the balance between singlet and triplet radical pair states and therefore to change the yields of the products formed from them,...


Regioselectivity in addition of hydrogen iodide to vinyl chloride

As far as I can see, that reaction follows Markovnikov's rule. The hydrogen goes where there is more hydrogen and the halide goes where there is less hydrogen.
If you are confused as to why the product of that reaction is 1-chloro-1-iodoethane, it's because the electromeric effect the chloride atom exhibits trumps the inductive effect. We need to take the electromeric effects into consideration because the chloride atom is attached to a $\mathrm{sp^2}$ carbon. The inductive effect makes it attract electrons to itself, but the electromeric effect makes it push them away, like a $\ce{CH3-}$ would.

It's the reason why $\ce{CH3-}$ and $\ce{Cl-}$, when attached to an aromatic nucleus, orient new substituents in the same positions, ortho and para (though $\ce{CH3-}$ has a stronger effect than $\ce{Cl-}$).


What is the effect of adding salt to ice?

From http://antoine.frostburg.edu/chem/senese/101/solutions/faq/why-salt-melts-ice.shtml:

Consider that ice, a crystalline solid, forms via weak partial-polar forces between water molecules, known as hydrogen bonding. As liquid water cools, it loses the energy that keeps these forces from attracting and capturing neighboring water molecules, and so these bonds form in a crystalline structure. Exactly what that structure is depends on the temperature and pressure at which the ice forms; your everyday freezer ice as well as most of the stuff outside in January is Ice Ih, which is a very spacious hexagonal arrangement that makes ice less dense than liquid water.

Also consider that even within what looks like solid ice, individual molecules "melt" and "freeze" all the time as they transiently gain enough energy to break free of the lattice, and then lose it and are recaptured. When the rate at which molecules melt and freeze are equal, the ice is stable. This happens at 0.01*C; below this temperature, more water will freeze, while above this temperature, more water will melt. Right at 0.01*C, water remains water and ice remains ice (the other half of this coexistence, besides temperature, is that water like most substances requires more energy gain or loss to change its phase, known as the Latent Heat of Vaporization, and for water that's 334J/mL, 80 times what's needed to cool liquid water from 2*C to 1*C).

Introduce salt, and these dissolved ions interfere with this hydrogen bonding within the ice, preventing it from happening at the normal freezing point of water. Now the melting molecules don't refreeze, because their partial polar charges are more strongly attracted to the full charges of the sodium and chloride ions than to other water molecules. That upsets the balance; more water is melting than freezing at the normal freezing temperature, and so the ice melts. It takes more energy loss within this system before the water is too unenergetic to stay liquid and settles into crystal, releasing the salt, which then finds other liquid water at the boundary layer between liquid and ice.

This creates a vicious cycle; as the salt concentrates within the remaining liquid water, its freezing point is lowered even further as the sodium and chlorine, which won't resolidify as salt crystals until there's just too little water to remain dissolved, will become even more attracted to the remaining water, even further lowering the remaining liquid's freezing point as the water must free itself from even more ions before it can be captured by the ice crystal.

So, an aqueous salt solution, in nearly any concentration, will lower the boiling point of the solution below 0*C, but the minimum freezing point is at -21.1*C, which is the freezing point of a fully-saturated salt solution. Below this temperature, salt that loses its water to freezing can't stay dissolved, and precipitates out as a separate crystal solid suspended within the ice.


How is solid PVA filament (for 3D printing) synthesized?

All the reactions I've seen that make PVA (also called PVOH, PVAl), create a gel-like substance that could be used as a glue or perhaps a film.

Yes, the reaction you describe is done in a solvent (water, ethanol, methanol, etc...) so that no acetate groups are unaccessable to reaction in a crystalline phase. A gel is formed because when poly(vinylacetate) is fully hydrolyzed, it is less soluble, typical white glue has about $12\%$ of the acetate groups unhydrolyzed to reduce crystallinity and keep solubility high.  The polyvinyl alcohol in filament is just dried out solid PVA, whereas the gel has water in a polymer matrix. All they have to do is melt and extrude the dry PVA powder into filament.

  I don't actually need/want it in the typical extruded filament form that would be needed for a 3D printer.  A simple solid blob of the stuff would suffice.

Side note: you can buy PVA powder online with less mess than self synthesis or depending on your application, you might be able to just use dried out white glue.


Might as well make an answer of the comments.

Aldehydes are not only able to be oxidized to acids, they are quite reactive towards this oxidation.  Exposure to air would be enough to trigger enough oxidation and acid formation to get litmus to react, especially if the litmus is moist.  Aldehydes are not alone in this respect; for example iron(ii) salts can be oxidized by air too, especially if you try to store them as a solution or after breaking the original seal on the bottle.

Moist litmus paper is a good qualitative test for acidity or basicity because it is highly sensitive to small amounts of acid or base.  Reaction with sodium carbonate might not be as sensitive if you are looking for enough carbon dioxide to visibly form bubbles, especially as carbon dioxide has some solubility in most solvents.

Yes, you need to follow good practices inspecting, testing and if necessary, purifying sensitive chemicals if you want high precision or accuracy.


Calculating the UV/Vis spectrum of perylene

I won't comment on the accuracy or otherwise of you calculation but that does not seem to a problem just now.

In the spectrum you show and in similar ones, the electronic transition energy is at approximately the mid-point of the first absorption and emission peaks, at approx 22800 cm$^{-1}$ in your figure. The different bands you see are due to vibrational transitions from various normal modes 'on top', as it were, of the electronic 0-0 transition. The energy gaps between peaks in the absorption spectrum measure the vibrational frequency in the excited state and vice versa for the fluorescence spectrum. Look up a Jablonski diagram to see generic sketches vibrational levels and transitions.

The number and magnitude of the vibrational transitions depends on the shape of the potential energy surface and is governed by the Franck-Condon factors which are $|\int \psi_e\psi_g|^2$ between the ground (g) and excited state(e).

The width of the lines in solution spectra (such as the one you show) is partly also due to rotational lines  but mainly due to collisions with solvent shifting and broadening energy levels. Consequently if you look up gas/vapour phase spectra at high resolution, such as for benzene as there a many published spectra, you will see many narrow lines corresponding to different vibrational normal modes and if resolution is high enough, even rotational lines.


Performing an SN1 reaction in a polar aprotic solvent

The rate of an $S_N1$ reaction may be increased by increasing the polarity of the solvent of reaction.  Because, the more polar a solvent (as measured by the dielectric constant / relative permitivity, the easier to separate ions from each other; here: carbocation and its counter anion.
Contrary to polar-aprotic solvents (think DMF, DMSO) eventually leading to «naked ions», polar-protic solvents (like ethanol, methanol, water) may weakly interact with these ions; similar to a Lewis acid/base reaction, they stabilize the ions to stay in proximity to each other as an ion pair.  Eventually, it is by experiment to figure out which effect is dominating at the reaction conditions (structure of the substrate and reagents, concentration of starting material and reagents, temperature, etc.).


Why do amides require much harsher conditions for hydrolysis than esters?

Dimethyl acetamide (DMA) is less susceptible to hydrolysis than ethyl acetate (EA). This statement is true for aqueous acid or base. The infrared (IR) frequency for the carbonyl group of DMA is 1662 cm-1 while EA is 1742 cm-1. The lower the frequency, the lower the energy and the more, relatively speaking, single bond character in the carbonyl group. Thus, the carbonyl group of DMA has less double bond character than that of EA. Because nitrogen can sustain a positive charge better than oxygen, resonance is more pronounced in DMA than in EA. Because resonance is stronger in DMA, its carbonyl group is less susceptible to nucleophilic attack. The carbonyl bond length of EA would be expected to be shorter than that for DMA because the more double bond character, the shorter the bond. It is no different than the carbon-carbon bond length in the series ethane, ethylene and acetylene.


The following order is experimentally found:

$$\ce{BF3}<\ce{BCl3}<\ce{BBr3}<\ce{BI3}$$

This stands in contrast to what is expected when the electronegativity of the halides is considered. A more electronegative halide should be able to stabilize the negative charge at $\ce{B}$ in the Lewis acid-base complex better, and this would suggest that $\ce{BF3}$ would be the strongest Lewis acid.

However, the nature of the $\ce{B-X}$ bond needs to be taken into account. The change of hybridization from $\ce{sp^2}$ in $\ce{BX3}$ to $\ce{sp^3}$ in $\ce{[BX4]-}$ involves the loss of partial $\pi$ bonding which results from the overlap of the $\ce{p_z}$ orbitals of $\ce{B}$ and $\ce{X}$. The $\ce{p_z}$ orbitals of $\ce{B}$ and $\ce{F}$ are of similar size, because both elements are members of the same period. Therefore, they have the best overlap and form the strongest $\pi$ bond. Because the reaction with a Lewis base would require some energy to break this bond, $\ce{BF3}$ is a comparably weak Lewis acid. The larger $\ce{p_z}$ orbitals of the heavier halides have increasingly less overlap with the smaller boron $\ce{p_z}$ orbital. Therefore, the $\pi$ bonding is weaker and formation of the tetrahedral Lewis acid-base adduct via rehybridization becomes energetically more favorable.


Enantiomeric excess (ee) is defined as:

$$\text{ee} = \frac{\alpha(\text{sample})}{\alpha(\text{pure enantiomer})}\tag{1}$$

Here, we are given the specific rotation of our sample ($-45^\circ$) and that of a pure enantiomer ($-53^\circ$). To calculate the ee, all you need to do is to plug these two values into equation $(1)$ and work out how to turn a fractional number into a value in $\%$.

   The answer is $85~\%~\text{ee}$.


Reaction of benzene diazonium salts with different compounds

As to your first reaction that does not utilize copper, a reference to the source would be useful. Nonetheless, the Meerwein Arylation Reaction has been reviewed in Organic Reactions. The reaction is conducted in the presence of catalytic cuprous ion. Meerwein proposed an aryl cation but a radical mechanism appears plausible. Reduction of the aryldiazonium salt 1 produces aryl radical 2 and cupric ion. Addition of the aryl radical to cinnamic acid 3 to the double bond to give the more stable of two possible radicals, ---namely--- the benzylic radical 4. This radical is oxidized by cupric ion to produce benzylic cation 5 and the catalyst cuprous ion. Loss of CO2 forms the stilbene 6.


What are the products of the Grignard reaction with methyl-magnesium iodide?

Your reaction scheme doesn't make much sense. $\ce {R'=O}$ is not a ketone. Is $\ce X$ supposed to be $\ce Me$? This is confusing since $\ce X$ is most commonly used as the symbol of a halogen.
I would rewrite as:
$$\ce {R'RC=O + MeMgI +H_2O-> R'RC(Me)OH + Mg(OH)I}$$
$\ce {R'(R)C=O}$ is implied.

Make sure you use very dilute aqueous sulphuric acid since your product is a tertiary alcohol. I would rather use saturated aq. $\ce {NH_4Cl}$. Even water will do but I suspect that the work-up might not be as smooth (a less water soluble magnesium salt will be formed). There will be no iodine formed and in any case I don't find the idea of recycling iodine cost effective. It will cost you more to isolate iodine that can be pure enough to work in the notoriously capricious Grignard reagent formation. In fact I would suggest you to buy a commercially available solution of $\ce {MeMgX \ (X=Cl, Br, I)}$ since preparing a Grignard can be very challenging and you will also have to handle toxic and volatile $\ce {MeI}$, have access to dry ether/THF etc. I assume you follow a detailed experimental procedure from the literature.


Total number of octahedral void(s) per atom in a cubic closed pack structure

Finally got it, it was a misunderstanding. I ignored to read the word ‘per atom’ in a hurry. So clearly the answer is 1.


How to optimize graphene geometry with Gaussian?

The decision for a bond to be visually represented as a single, double, or triple bond in a molecular visualization program (e.g. GaussView) is based on rather arbitrary distance rules that often do not represent the actual bond order. Unless you are investigating the bond order in other ways (e.g. Wiberg bond index), what you're seeing for the bond order is nothing more than a guess. If you switch visualization programs, you'll likely find the bond orders to change.

If, indeed, the actual geometry of the graphene is completely unphysical after the geometry optimization, here are some questions to ask yourself:

How good is the initial structure you gave the optimization algorithm? Are you using a template graphene structure that should be close to the final geometry? If you made it "from scratch", it would be worthwhile to apply symmetry to your structure prior to running the calculation in Gaussian.
How have you terminated your graphene? You say you're using Gaussian, so I doubt you are using periodic methods. Have you terminated all dangling bonds with hydrogens, for instance?
In line with Point #2, what is the size of your graphene "cluster" model? Perhaps the cluster you chose is too small.
Ensure you don't have various input file errors (e.g charge/multiplicity).


It appears to be one of degrees of bonding, without a clear distinction. As stated in this Wikipedia article:

  Physisorption, also called physical adsorption, is a process in which
  the electronic structure of the atom or molecule is barely perturbed
  upon adsorption.$\mathrm{^{[1][2][3]}}$

A more thorough description is given later in the article:

  "The fundamental interacting force of physisorption is caused by van
  der Waals force.   [...]   In comparison with chemisorption, in which
  the electronic structure of bonding atoms or molecules is changed and
  covalent or ionic bonds form, physisorption, generally speaking, can
  only be observed in the environment of low temperature (thermal energy
  at room temperature ~26 meV) and the absence of the relatively strong
  chemisorptions. In practice, the categorisation of a particular
  adsorption as physisorption or chemisorption depends principally on
  the binding energy of the adsorbate to the substrate."

1) K. Oura; et al. (2003), Surface Science, An Introduction, Berlin: Springer, ISBN 978-3-540-00545-2
2) M. C. Desjonqueres; et al. (1996. Corrected printing 1998), Concepts in surface physics (2nd ed.), New York: Springer-Verlag, ISBN 3-540-58622-9
3) Hans Luth; et al. (1993), Surfaces and interfaces of solids, Springer-Verlag, ISBN 978-3-540-56840-7


As according to Molecular Orbital theory , when anti- aromatic compound is formed, the electron goes to anti bonding, and is unpaired even, which reduces it's stability.,
Or we can say that in anti- aromatic,the plane of electron cloud twists there bonds between carbons, making it unstable, thus resonance is not possible.
Antiaromaticity is a characteristic of a cyclic molecule with a π electron system that has higher energy due to the presence of 4n electrons in it. Unlike aromatic compounds, which follow Hückel's rule ([4n+2] π electrons) and are highly stable, antiaromatic compounds are highly unstable and highly reactive. To avoid the instability of antiaromaticity, molecules may change shape, becoming non-planar and therefore breaking some of the π interactions.
Greater the delocalisation energy(resonance energy) of a compound, more stable it is. The resonance energy can be calculated using Huckel Molecular Orbital Theory(HMOT). Antiaromatic compounds have zero resonance energy hence are unstable.
According to HMOT,
Delocalisation energy=( Pi electron binding energy) - (total energy of isolated double bonds in a classical structure)


Pressure exerted by a particle of ideal gas

Here's your confusion:
You need to consider two different things:

The momentum transfer per particle per collision.  There, since we assume an instantaneous collision, it doesn't make sense to try to figure out force from acceleration. [I suppose you could do this using limits, and maybe there are applications in which that does make sense, but adding that complication is completely unnecessary for introductory kinetic theory, given that there is a simple and direct way to calculate the effect of each collision.]  And the simple and direct way is to use the change in momentum, which is $2mv_x$.

Note: Your statement that "The particle is assumed to travel perpendicularly towards one of the sides before bouncing back off it" is incorrect.  Rather, in determining the momentum transfer per collision, we use the component of the velocity perpendicular to the wall, namely $v_x$.

The number of collisions per unit time.  Here is where the average time it takes for the particle to travel from collision to collision is relevant. This is where ${\Delta t}$ is being used -- not to calculate the momentum transfer/collision, but to determine the frequency of collisions per particle.  And you need the average momentum transfer/collision, and the frequency of collisions per particle, to determine the average force per unit area contributed per particle.

So, in summary: For momentum transfer per collision, we don't consider time, since we can get this directly from $2mv_x$.  For number of collisions per unit time, we need to account for the time between collisions.


Based on Recovery of Iodine with Activated Carbon from Dilute Aqueous Solutions,Petrochemical Center of Excellence, Department of Chemical Engineering, Amirkabir University of Technology, No.424, Hafez Ave., Tehran, Iran , you should get good recovery of your iodine by heating in a solution of NaOH. According to the paper:

  The optimal desorption conditions obtained by the software are as
  follows: temperature of NaOH solution: $\mathrm{50^oC}$, concentration of NaOH
  solution: 0.5 % and agitation speed: 300rpm. Desorption experiments
  under these optimal conditions showed 98.9% desorption. The Taguchi
  design software results also showed that in the desorption section,
  the contributions of each factor on desorption percentage are as
  follows: concentration of NaOH solution (55%), temperature of NaOH
  solution (39.4%) and agitation speed (4.6%).
  [...]
  the necessary time for equilibrium desorption is obtained as 6h.

The paper does not discuss the algorithms used to calculate the relative importance of the different parameters, but obviously temperature and NaOH concentration are the important factors to getting such good recovery.


Time to effervescent an effervescent tablet

Dissolution in general

Dissolution is normally a function of several aspects of the solid and liquid and one of them is the interfacial area between the solid and the liquid. A common relation for the dissolution rate which is not always exact, but shows all trends/dependencies very well is the following:

$$\frac{\mathrm dm}{\mathrm dt}=A \frac{D}{d}\left(C_\mathrm s-C_\mathrm b\right) $$
where $A$ is the interfacial area, $D$ is the diffusion coefficient of your solid molecules in the liquid, $d$ is the thickness of the solute boundary layer and $C_\mathrm s$ and $C_\mathrm b$ are the concentrations of the solid molecules at surface and in the bulk (this term is the driving force).

Dissolution of a whole or broken tablet

To make a simple calculation for the whole vs. the broken tablet let us do the math.  Assuming that we have a disc-like tablet the surface area is 2 times the frontal area $+$ the area of the sides.

$$A=2 A_\mathrm f+A_\mathrm s $$
For the whole tablet
$A_\mathrm f=\frac{\pi}{4}D_\mathrm t^2$ and $A_\mathrm s=\pi h D_\mathrm t$

For the broken tablet
$A_\mathrm f=\frac{\pi}{4}D_\mathrm t^2$ and $A_\mathrm s=\pi h D_\mathrm t  + 2 h D_\mathrm t$

Here I assume that the tablet is cleanly broken in the exact middle of the disc. So you can already see that the surface area is bigger for the broken tablet, but how much will depend on the specifics of the tablet. Rewriting the area we can see how the different aspects of the broken tablet will influence the area:
$$A=\pi D_\mathrm t \left[\frac{D_\mathrm t}{2}+h\left(1+\frac{2}{\pi}\right)\right] $$

From this equation it is clear that if $D_\mathrm t \gg h$ the height term doesn't do much and therefore it will not matter much that the tablet is broken.

For a final check: let's assume some typical numbers now. A typical effervescent tablet is about $D_\mathrm t=2\ \mathrm{cm}$ in diameter and $h=3\ \mathrm{mm}$ thick. This will result in the following surface areas for the broken and the whole tablet: $A_\text{whole}=8.2\ \mathrm{cm^2}$ and $A_\text{broken}=9.4\ \mathrm{cm^2}$.

This means that the difference in dissolution rate will only be about $20\ \%$ which is probably hard to notice due to other effects that disturb the comparison. One such effect is that $d$ depends on the amount of stirring which will be different already when you drop the tablets in the glass in a slightly different way (which is unavoidable, because you have 1 broken and 1 whole tablet).


Does hydrogen bonding contribute in solubility of a substance

Ethoxyethane, better known as "diethyl ether" or even just "ether", can form hydrogen bonds with water.  But fitting the ether molecules into the water solvent means you have to break up some water-water hydrogen bonds, and the ether molecules cannot replace enough of those hydrogen bonds to dissolve freely.

Of course, ether does form some hydrogen bonds, which makes is more soluble than it would be without the hydrogen bonds (compare ether with, for example, diethyl sulfide).  But most (or at least the most common) freely water-soluble organic compounds have either ions to generate ion-dipole interactions with the water, or hydroxyl groups that readily fit with the hydrogen bonding structure of the water (think of ethanol or sugars).


The image below (Taken from the Evans' lectures, which you might like to consult as there is a whole lot more detail in them) sums up the situation.
Essentially you're interested in how good a leaving group is attached to the boron. The triflate is excellent and as such fully dissociates, whilst the chloride remains (at least partially) in association.
Depending on whether the group on boron dissociates, the 'boron' is significantly more/less bulky, changing the conformation from which enolisation is preferred, which in turn leads to E/Z enolates.
The choice of base is also important. With boron triflates, DIPEA is often used as it is just that bit bulkier and so biases even more towards the Z enolate. In the boron chloride aldols, the bulky DIPEA is a bit too much, and so Et3N is commonly used.
If you study the TS's below, it might be apparent why the boron chloride aldols are typically a little less selective than the boron trifoliate aldols.

Source: Dave Evans' Harvard CHEM207 Lecture notes.


Safety concerns for radical halogenations of methane in sunlight?

Let's have a look at some fundamental data and the chemistry involved.

Chlorine is an inflammable gas with a boiling point of -35 °C and a nice yellowish-green colour. It is heavier than air.

It is highly irritant and toxic. It is a strong oxidant. With water, it partly decomposes to hydrogen chloride and hypochlorous acid. Your mucous membranes are wet. Inhalation of chlorine results in severe corrosive injuries. Inhale a tad longer to face dyspnoea, coughing of blood and asphyxia. Did I mention the pulmonary oedema and fatality?

Methane is a flammable gas with a boiling point around -160 °C and a flash point around -180 °C. The lower explosion limit of methane in air is around 5%.

Open flames and sparks are not your friend here.

For the mixture of methane and chlorine, red light districts are safe.
The bond dissociation energy of chlorine is $243\, \mathrm{kJ \cdot mol^{-1}}$, you will need blue light ($\lambda$ < 490 nm) to initiate the self-propagating reaction:

$$\ce{Cl2 ->[h\nu] 2 Cl*}$$

Then the chain reaction starts:

$$\ce{Cl* +\  CH4 -> HCl + CH3*}$$
$$\ce{CH3* +\  Cl2 -> CH3Cl + Cl*}$$

All that goes on and on; termination steps are radical recombinations:

$$\ce{2 Cl*  -> Cl2}$$

$$\ce{2 CH3*  ->  C2H6}$$
$$\ce{ CH3* +Cl* ->  CH3Cl}$$

Apparently hydrogen chloride gas is is produced in the reaction, together with methyl chloride.

Methyl chloride is a flammable gas (boiling point around -24 °C, flash point around -20 °C, lower explosion limit in air around 8%).
GHS hazard statements file it with H-351: suspected of causing cancer.

If you perform the reaction with a large excess of chlorine, $\ce{Cl*}$ radicals will attack methyl chloride, generate dichloromethane via the chloromethyl radical $\ce{*CH2Cl}$, etc. Tetrachloromethane ($\ce{CCl4}$) would be the final product then.

That was the healthy and boring part!

When fluorine talks to chlorine, it goes like in the musical "Annie, Get Your Gun":

Anything you can do I can do better!

Particularly when it comes to running havoc on your tissue!

As far as the photolysis of fluorine in the presence of methane is concerned: People have done that!

Typical conditions were:

Deposit methane in a fluorine matrix at 15 K
Co-deposit methane/argon and fluorine/argon samples at 12 K
Trap fluorine in a methane matrix at 15 K

Yes, these guys had no sense for adventure - but they weren't blown away either!


I think the question aims at the composition of carbonic acid solutions. The concentration of carbonate ions ($\ce{CO3^{2-}}$) depends on the pH of the solution. At low pH values the concentration is low, at high pH values the concentration is high. So the question is if the concentration of $\ce{CO3^{2-}}$ at pH 8 is high enough to precipitate calcium carbonate.

$\ce{H2CO3}$: black, $\ce{HCO3^{-}}$: violet, $\ce{CO3^{2-}}$: turquoise, $\ce{H+}$ and $\ce{OH-}$: grey


What are the functional group names of halogenoalkanes?

They don't have a functional group name similar to that of hydroxyl/phenyl/etc. Instead, you've to name them as if they were an alkyl branch - "haloalkyl" - and that too within parentheses.

Have a look at this simple comparison:

Notice that both 2,3-dimethylbutyl and chloromethyl are both within parentheses, whereas phenyl isn't. Therefore don't mistake "haloalkyl" to be a short name as "phenyl" is. There simply is no such short name, and you've to name it as an alkyl chain instead.


Stability of alkenes by hyper conjugation

You are confusing the mixing of orbitals with the promotion of electrons from one orbital to another. You're thinking of this:

$\hspace{50 mm}$

I don't blame you, because the phrasing "delocalisation into an antibonding orbital" implies that, but that is not actually what happens.

In what actually happens, the two original orbitals are said to mix. They cease to exist. They overlap to give two new MOs, one which is even lower in energy and one which is higher in energy, like in this diagram:

$\hspace{45 mm}$

The "new MO 1" has characteristics of both the original $\ce{C-H}$ $\sigma$ as well as the original $\ce{C=C}$ $\pi^*$, which is why we loosely say "donation into the $\pi^*$ orbital". However, don't take it too literally. The phrase means that two orbitals are mixed. It does not mean that electrons are promoted from one orbital to another.


Should rates and rate constants be expressed in terms of pressure or concentration?

The ambiguity doesn't lie so much in the definition of the rate, because as long as the constants $a$ and $b$ in $ap_\ce{X}$ and $b[\ce{X}]$ are chosen appropriately, the two expressions are entirely equivalent. So you can define the rate either way, and there is no physical difference, not even a mathematical difference. It is somewhat akin to asking whether you should factorise the number $12$ as $2 \times 6$ or $4 \times 3$. It doesn't matter how you factorise $12$, it's still the same number.
The ambiguity is actually in the definition of the rate coefficient, $k$. The constants $a$ and $b$ above, though directly proportional to each other, are obviously different: and only one of them can be the "true" rate coefficient $k$.
Before launching into the answer proper I should mention that there is no physical difference between using the pressure and the concentration. At the end of the day, it doesn't change the actual rate at which your system is reacting: it's just an arbitrary choice as to what physical constants you want to include in the rate coefficient $k$.
That said, the IUPAC Gold Book seems to come down quite strongly on the side of the concentration, meaning that $k$ is the same as the second constant $b$. They do not make any reference to the phase of the reaction, so it should probably be assumed that they intended it to hold for any general phase. (If anyone knows of a more specialised definition, please point it out!)

order of reaction, $n$ (DOI: 10.1351/goldbook.O04322)
If the macroscopic (observed, empirical or phenomenological) rate of reaction ($nu$) for any reaction can be expressed by an empirical differential rate equation (or rate law) which contains a factor of the form $k[\ce{A}]^\alpha[\ce{B}]^\beta\cdots$ (expressing in full the dependence of the rate of reaction on the concentrations $[\ce{A}]$, $[\ce{B}], \cdots$)
[...]
The proportionality factor $k$ above is called the ($n$th order) 'rate coefficient'. Rate coefficients referring to (or believed to refer to) elementary reactions are called 'rate constants' or, more appropriately 'microscopic' (hypothetical, mechanistic) rate constants.

This approach is quite consistent throughout. For example, the rate of reaction is also expressed in terms of concentrations, at least under constant-volume conditions:

For the general chemical reaction:
$$\ce{aA + bB -> pP + qQ + \cdots}$$
occurring under constant-volume conditions, without an appreciable build-up of reaction intermediates, the rate of reaction $\nu$ is defined as:
$$\nu = -\frac{1}{a}\frac{\mathrm{d}[\ce{A}]}{\mathrm{dt}} = -\frac{1}{b}\frac{\mathrm{d}[\ce{B}]}{\mathrm{dt}} = \frac{1}{p}\frac{\mathrm{d}[\ce{P}]}{\mathrm{dt}} = \frac{1}{q}\frac{\mathrm{d}[\ce{Q}]}{\mathrm{dt}}$$
The quantity $$\dot{\xi} = \frac{\mathrm{d}\xi}{\mathrm{d}t}$$ defined by the equation:
$$\dot{\xi} = -\frac{1}{a}\frac{\mathrm{d}n_\ce{A}}{\mathrm{dt}} = -\frac{1}{b}\frac{\mathrm{d}n_\ce{B}}{\mathrm{dt}} = \frac{1}{p}\frac{\mathrm{d}n_\ce{P}}{\mathrm{dt}} = \frac{1}{q}\frac{\mathrm{d}n_\ce{Q}}{\mathrm{dt}}$$
(where $n_\ce{A}$ designates the amount of substance $\ce{A}$, conventionally expressed in units of mole) may be called the 'rate of conversion' and is appropriate when the use of concentrations is inconvenient, e.g. under conditions of varying volume.

One tangential question I can anticipate is: what's the difference between the expressions $r = -(1/a)(\mathrm{d}[\ce{A}]/\mathrm{d}t)$ and $r = k[\ce{A}]$? The answer is: the former is a definition. The rate is literally defined by how fast the reactants disappear, which is a universal truth, applicable to all reactions. The latter is an empirically observed expression, which only holds true for a small subset of reactions (i.e. first-order ones), and it says that the rate at which the reactant disappears is directly proportional to its concentration.


Why should the scalar product of the components along two axes remain unchanged after transformation?

Your question
The scalar product depends on the length of the vectors and the angle between them. Rotations, translations and mirror operations do not change distances or angles (i.e. don't distort objects), so the scalar product is unchanged as well. There are other operations that do (stretching, warping, etc).
The proof
Without knowing the context, the proof seems wrong. Equation [2.18] is not general and only works for orthogonal coordinate systems. To give a simple example: if the b axis is not perpendicular to the c axis, and we rotate on a twofold along c, the b axis will not be transformed into -b, but rather into a linear combination of b and c. What this proof shows is that if axes are orthogonal, they are orthogonal (tautology).
What is true is that if there is a two-fold axis along c, you can always choose the two other axes such that they are orthogonal to c. This is a choice, however, and other choices are possible.

In the figure, b and c are not perpendicular. Points at the origin, at the tip of b and the tip of c are equivalent by translation. If there is a crystallographic twofold rotation along c, there is another equivalent point at the tip of the rotated b. The rotated b is not the same as negative b, as the figure illustrates by counter-example. However, if you want perpendicular axes (and that is the choice recommended by the international tables), you can switch from the axis b to the thin line (shown as dot-dash-dot-dash), giving you a new axis perpendicular to c. In the case shown, the lattice is now face centered.


Creating a chemical solution with water and chlorine

Is your measuring instrument made for measuring chloride or chlorine ? It is not the same. In your first sentence, you speak of chloride. Later on you speak of chlorine. Anyway, the best thing to do is to buy a bleach solution. The amount of chlorine is printed on the label. It is often $14$% chlorine $\ce{Cl_2}$. By diluting a sample (maybe $1$ mL) of such a solution in 1 liter water, you obtain a solution $0.14$ gram chlorine $\ce{Cl_2}$ per liter, and you can dilute it still more.


What does the "h" "k" and "(5.2.2.02,6)" mean in Dibenzo(h,k)-4-oxatricyclo(5.2.2.02,6)undec-8,10- diene-3, 5-dione

The systematic (but not necessarily preferred) name tricyclo[5.2.2.02,6]undecane describes the structure of octahydro-1H-4,7-ethanoindene (see this answer on how to name tricyclo compounds):

Expanding the name from tricyclo[5.2.2.02,6]undecane to 4-oxatricyclo[5.2.2.02,6]undec-8,10-diene-3,5-dione adds the functional groups, which leads to the structure of 3a,4,7,7a-tetrahydro-4,7-etheno-2-benzofuran-1,3-dione:

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), isomers of fused polycyclic ring systems are distinguished by lettering each peripheral side of the parent component using the italic letters a, b, c, etc., beginning with a for the side numbered ‘1,2’, b for ‘2,3’, etc.
For example:

Accordingly, fusion of two benzene rings to the above-mentioned parent structure 4-oxatricyclo[5.2.2.02,6]undec-8,10-diene-3,5-dione at the relevant sides yields the name dibenzo[h,k]-4-oxatricyclo[5.2.2.02,6]undec-8,10-diene-3,5-dione.
(Note that this name does not necessarily correspond to the preferred IUPAC nomenclature and notation.)


Combination of polydisperse polymer mixtures

It should not be difficult to derive either of the number-averaged molecular weight or weight-averaged molecular weight by their definitions. To avoid ambiguity, I will use $i$ as the index of each actual length of chains and $b$ as the index of batches.

First, the definitions:

$\bar{M}_\mathrm{n,final}=\frac{\Sigma_in_iw_i}{\Sigma_in_i}$

$\bar{M}_\mathrm{w,final}=\frac{\Sigma_in_iw_i^2}{\Sigma_in_iw_i}$ where $n$ and $w$ are number and molecular weight of each chain length.

The numerator of $\bar{M}_\mathrm{n,final}$ is, in fact, the total mass of the mixture. Hence, $\Sigma_in_iw_i=\Sigma_bw_b$. The denominator is the total number of chains. Hence, $\Sigma_in_i=\Sigma_bn_b=\Sigma_b\frac{w_b}{\bar{M}_\mathrm{n,b}}$.

It is slightly more complicated in the case of $\bar{M}_\mathrm{w,final}$ but you got it correct. Feel free to ignore my derivation. The numerator does not have a straightforward definition. It has to be derived from the $\bar{M}_\mathrm{w,b}$ instead. $\Sigma_in_iw_i^2=\Sigma_b\Sigma_{i,b}n_{i,b}w_{i,b}^2=\Sigma_b\bar{M}_\mathrm{w,b}w_b$. Since we have already defined the $\Sigma_in_iw_i$ above, we can have the final equation.

$\bar{M}_\mathrm{n,final}=\frac{\Sigma_bw_b}{\Sigma_b\frac{w_b}{\bar{M}_\mathrm{n,b}}}$

$\bar{M}_\mathrm{w,final}=\frac{\Sigma_b\bar{M}_\mathrm{w,b}w_b}{\Sigma_bw_b}$

$\bar{M}_\mathrm{w,final}/\bar{M}_\mathrm{n,final}=\frac{\Sigma_b\frac{w_b}{\bar{M}_\mathrm{n,b}}\Sigma_b\bar{M}_\mathrm{w,b}w_b}{(\Sigma_bw_b)^2}$


Software for visualizing molecular force vectors

As long as the output from your harmonic frequency calculation can be read directly using Avogadro, it will do exactly what you want.

Notice that:

The normal mode of interest is selected on the right-hand side.
"Force" is selected on the left-hand side; this box is present when "Display Settings..." is highlighted in the top toolbar.


Why, in an applied magnetic field, is electron with spin parallel to the magnetic field higher?

In addition to the other answer, the energy  of a magnetic moment in a magnetic field is $E=-\bar{\bf{\mu}}.\bar{\bf{B}}$ where the dot indicates the dot product between the two vectors $\bar{\bf{\mu}}$ and $\bar{\bf{B}}$, thus the energy is lowest when the magnetic dipole is parallel to the field $B_0$; $E=-\mu B_0$. The sign of $\mu$ is opposite for nuclear and  electron spin angular momentum.The magnetic moment of a nuclear spin is defined as $\mu=+\gamma I$, where $I$ is the nuclear angular momentum and if the field is along only the $z$ direction  $\mu_z=\gamma I_z=m_z\gamma \hbar$ (Joule/Tesla) where $\gamma$ is the magnetogyric ratio, (unit radians/Tesla/sec) the components are $m_z$, the projection, magnetic or azimuthal spin quantum number on the $z$ axis, the axis parallel to the magnetic field $B_0$. $I_z$ is the $z$ component of the nuclear spin angular momentum. (The magnetogyric ratio is $\gamma = g\mu _0/\hbar$ where $\mu_0$ is the nuclear magnetron and $g$ the nuclear spin factor, different for each type of nucleus). The energy is then $E=-m_z\gamma\hbar B_0 $. For a proton $m_z=\pm 1/2$.

The spin magnetic moment of the electron is defined in a similar way to that for nuclear spin but with a negative sign; $\mu_e=-\gamma _e I_e$ where $I_e$ is the electron spin angular momentum and $\gamma _e$ the electron magnetogyric ratio. (  $\gamma _e=g_e\mu_B/\hbar$,  $g_e$ is the electron $g$ factor, $g_e =2.0023..$, and $\mu _B$ the Bohr magnetron. Some authors give a negative $g_e$, in this case the negative sign is removed from $\mu _e$).
(The nuclear magnetogyric ratio can be positive or negative, e.g. $\ce{26.75.10^7 rad T^{-1} s^{-1}}$ for protons. For a few types of nuclei e.g. $\ce{^{15}N, ^{29}Si}$ $\gamma$ is negative. The magnetogyric ratio for the electron is much $\ce{larger 1.76 10^11 rad T^{-1} s^{-1}}$.)


Calculating the molar mass of a gas given state function values

The mistake you made was the value of $R$. If pressure and volume are taken in $\mathrm{atm}$ and $\mathrm{L}$, then the value of $R$ is $0.0821$. (Not $0.82..$).

Now, $$n=\frac{0.9645\times 0.255}{0.0821\times 821.15}=0.0036$$

$$M=\frac{1.22}{0.0036}=334.412\mathrm{g}$$


Why do higher-mass isotopes have higher melting and boiling points than lower-mass isotopes?

As you already mention in your question, melting and boiling of a substance is related to the inter-particle forces. You also correctly assumed that the same number of electrons results in identical (electronic) bonding properties, which basically is a different formulation of the Born-Oppenheimer Approximation (as in the BOA the nuclear mass is assumed to be infinite, different isotopes should have the same electronic structure).

So what is the difference then?

Well, it's in the vibration between the different particles. Consider a number of (neutral) particles in a certain phase (it can be solid, liquid or gas) at a certain distance with respect to each other. Because of dispersion forces between the particles, they interact and to a first approximation, every particle experiences a harmonic force $F(x)=-k(x-x_e)$. The magnitude of this force depends on the distance between the particles and on the electronic structure of the particles and in principle not on the mass. However, as you might know, a harmonic force in quantum mechanics results in quantized energy levels given by

$$
E_v=h\nu_e(v+\frac{1}{2}),
$$
where $E_v$ is the vibrational energy of the $v$th level and $h$ is Planck's constant. What is important to realize is that the energy of the lowest state is not equal to zero but has a value of $h\nu_e/2$, which is referred to as the zero point energy. Of course the potential is not really harmonic but rather anharmonic so that at a certain distance between the particles the force becomes negligible and the particles dissociate (or melt or evaporate). The lower the zero point energy, the larger the binding energy. The constant $\nu_e$ is given by $\nu_e=\frac{1}{2\pi}\sqrt\frac{k}{\mu}$, where $k$ is the harmonic force constant and $\mu$ is the reduced mass. The constant $k$ is the same for the different isotopes as it depends on the electronic potential between the particles but the reduced mass is of course different for the different isotopes. As you can see from the definition of $\nu_e$, the heavier species have a smaller zero point energy and thus a larger binding energy and will require higher temperatures to melt and boil.


Decreasing magnitude of peaks in cyclic voltammogram?

Perhaps your observation is an indication of irreversible reaction conditions.  More specifically, a tight deposit on the electrodes hampering the advancement of the electrochemical reactions. This happens sometimes with some less soluble organic dyes used organic electronics, too.  Take a look if the electrodes are still shiny, maybe they need to be cleaned again.  If this is the case, try out a different solvent/electrolyte system.

One aiding companion to trouble shoot is this.


Most periodic tables quote the molar mass of an element as the average mass of the element given the natural abundance of the isotopes. Chlorine, for example, has a standard atomic weight of 35.45 which is a weighted average of ~75% 35Cl and ~25%  37Cl, the two common isotopes.

But isotopes of hydrogen and oxygen are rare. So the vast majority of water molecules will not contain them. So, at least to one decimal place, most water will have a molar mass of 18.0.

At it's natural abundance of about 0.25% 18O won't noticeably affect the molecular weight of water until you can measure it to at least 3 decimal places of precision. Slightly simplifying (assuming pure isotopic masses are exact integers, which they are not), water containing that isotope at its natural abundance will have an an average mass of 18.005 rather than 18.000.

When working with normal (unenriched) substances the best assumption is that they will contain the normal isotopic abundance of the component elements (which are well known and, for most elements, don't very much by source). But the effect on molecular mass will often be small and you will need a fair amount of precision to notice it. If you are not working to very precise masses, it is fair to assume natural isotopic abundance.


As mentioned by @Mockingbird, nickel oxide is somewhat soluble in potassium cyanide which will dissolve the oxide to form nickel cyanide, $\ce{Ni(CN)2}$ and potassium tetracyanonickelate, $\ce{K2Ni(CN)4}$  but potassium cyanide is very, very toxic.

Do not use it until and unless you are a professional chemist.


I don't see planes of symmetry through H atoms or OH groups. Actually, these planes are not present. It is easy to check this by looking at the position of the CH3 group after reflection in one of these planes: you find CH3 at a completely different position (none of the original positions is mapped and hence you can distinguish the starting configuration from the final one).


Does DCl have a higher rotational partition function (qrot) than HCl? And if so why?

The partition function is the sum over states, which means over all the energy levels, thus
$$Z=\sum _i g_i \exp(-E_i/(k_BT))$$
where $g_i$ is the degeneracy of each level with energy $E_i$. The rigid rotor rotational energy levels of a diatomic molecules are $$E_J= \hbar^2J(J+1)/(2I)$$ where I is the moment of inertia $I=\mu r^2$ for reduced mass $\mu=m_1m_2/(m_1+m_2)u $ (in kg) and bond length r. The degeneracy of rotational levels is $g_J=2J+1$.

As the moment of inertial of DCl is greater than that of HCl (due to reduced mass effect) the energy levels are lower in energy in DCl at the same J than in HCl and so at the same temperature more levels are populated and the partition function is greater. This means that the exponential term in the partition function is greater in DCl at a given J and so this increases the partition function. (recall that $e^0 = 1$ and $e^{-\text{big number}} \rightarrow 0$). The energy levels of DCl and HCl are shown in the left figure and the terms in the partition function on the right as $g_J \exp(-E_J/(k_BT))$. (The curves are plotted for clarity assuming that J is  continuous not discrete). The respective partition function is the sum of the values at each J in the right-hand figure.

(To plot the curves I used a mass of Cl = $35$ and H and D, $1$ & $2$ respectively. The unified atomic mass unit $u = 1.660\cdot 10^{-27}$ kg, and assume a temperature of $300$ K. The bond length is $74.15$ pm.)

(Note that the often used integral form of the partition function has a systematic error that does not imporove as J increases.)


Can I convert elemental nitrogen to nitrous oxide?

There are several ways to fairly safely make $\ce{N2O}$ in the laboratory that don't involve the difficulties associated with trying to react molecular nitrogen.  The following reactions all are mentioned in this Wikipedia article which I suggest reading, along with the references.  Depending on the quantity you are making, I'd be very cautious about using the thermal decomposition of ammonium nitrate as this can kind of blow up on you (literally). Of course every reaction has it's own risks and it's up to you to decide what you are most comfortable with.

I think the direct oxidation of ammonia over a manganese dioxide-bismuth oxide catalystis an intriguing method$\mathrm{^{(1)}}$

$$\ce{2 NH3 + 2O2 -> N2O + 3 H2O}$$

This also seems like a fairly straight-forward route:

$$\ce{2 HNO3 + 8 HCl + 4 SnCl2 → 5 H2O + 4 SnCl4 + N2O}$$

And this seems to me a bit of a more complex synthyses:

$$\ce{2 (NH2)2CO + 2 HNO3 + H2SO4 → 2 N2O + 2 CO2 + (NH4)2SO4 + 2H2O}$$

Again, I'll say molecular nitrogen makes a lousy starting material for most syntheses, including $\ce{N2O}$.

1) Suwa T, Matsushima A, Suziki Y, Namina Y (1961). "Synthesis of Nitrous Oxide by Oxidation of Ammonia". Kohyo Kagaku Zasshi, Showa Denka Ltd. 64: 1879–1888.


When you fill a bath, Is it different that you fill it directly or non-directly?

If the beaker is at or below the outlet of the faucet, the two values should be nearly identical.  There are very small effects, like water sticking to the outside of the beaker as it overflows or the surface of the water changing shape constantly (splashing out more at some times and less at others as a result.)  Those are insignificant for practical purposes.

If the top of the beaker is above the outlet of the faucet, that is, the faucet is submerged in the beaker, then there will be a small addition to the height to which the water must be pumped.  It's only centimeters, but the pressure increase and flow decrease is entirely measurable and would have a small but significant and predictable effect over the course of running a bath.


Question on assumption in Michaelis-Menten kinetics

I think it is more about expressing[S]. If we assume the substrate concentration is much higher than the enzyme concentration, then we can ignore the substrate that is bound as ES in the derivation of the MM equation.  If the enzyme concentration is in the same range as the substrate (or if the enzyme affinity is very high), then the amount of substrate bound as ES cannot be ignored, and the derivation is not simplified. In this latter case, [S] in the equation can't be considered as the total, but must be corrected by the amount bound.


Instead of a typical oil/dye ink,  erasable pens use a compound closer to rubber cement, which acts like a liquid when a bit of pressure is exerted. 1
A bit of history and a link to patents can be found here
EDIT: Be sure to check out the other answer here about thermochromic ink too.


Do most chemical reactions happen at constant pressure?

What happens during a reaction is important insofar as reversibility is concerned (a reversible path being a special way of performing a reaction in which equilibrium is sustained throughout), but not when computing $\Delta H$. The condition $\Delta H = q$ is derived from the basic law of conservation of energy and the definition of enthalpy. Because H is a state function, to compute the change in its value you need to know only the initial and final states of the system (and of course the values of H corresponding to those states). For $\Delta H = q$ to hold requires that (1) only expansion (pV) work be performed and that (2) the initial and final pressures of the system equal each other.
You can dream up scenarios in which you start at mechanical equilibrium (balanced pressure) between surroundings and system but end up at a constrained system volume (rigid). For instance, if an explosion moves a piston to a constrained setting within a cylinder. In that case it is not true that the enthalpy change and heat are equal. On the other hand, if the explosion is constrained such that the volume can expand until pressures balance (and external constraining pressure is constant), then enthalpy change and heat are equal.


Why there is a color fading ( to colorless) when Cerium Ammonium Nitrate (CAN) is mixed with ethanol or butanol?

Recall that Ce(IV) is a pretty strong oxidizing agent. So oxidation of alcohols and fading of the color is expected. The mystery lies in the immediate intense red color. It has been relatively recently studied in great detail and well known long time ago.
See these references:

V. Briois, D. Lützenkirchen-Hecht, F. Villain, E. Fonda, S. Belin, B. Griesebock, and R. Frahm, "Time-Resolved Study of the Oxidation of Ethanol by Cerium(IV) Using Combined Quick-XANES, UV−Vis, and Raman Spectroscopies," J. Phys. Chem. A 2005, 109(2), 320–329 (DOI: https://doi.org/10.1021/jp046691t).

Lewis Brewster Young and Walter S. Trahanovsky, "Oxidation of organic compounds with cerium(IV). VII. Formation constants for 1:1 cerium(IV)-alcohol complexes," J. Am. Chem. Soc. 1969, 91(18), 5060–5068 (DOI: https://doi.org/10.1021/ja01046a021).

Well, you can see Ce(IV), alcohols form transient 1:1 complexes with alcohols, but such complexes don't last long because of oxidation. In ref. 1, one can deduce that alcohol replaces one of monodentate ligand (nitrate), and thus absorption spectrum of this complex changes for a few seconds.


How can the iron in the reaction of iron (ii) chloride and zinc not go to zero oxidation state?

Generally, to predict redox reactions, we consider the Standard Reduction Potentials at $25\ ^\circ\mathrm{C}$: The bigger the difference between the potentials of two redox couples, more favorable is the reaction.
$$E^0(\ce{Fe^3+ /Fe^2+})=0.77\ \mathrm{V}$$
$$E^0(\ce{Fe^3+ /Fe})=-0.04\ \mathrm{V}$$
$$E^0(\ce{Fe^2+ /Fe})=-0.44\ \mathrm{V}$$
$$E^0(\ce{Zn^2+ /Zn})=-0.76\ \mathrm{V}$$
So, when $\ce{Fe^3+}$ reacts with $\ce{Zn}$, we get $\ce{Fe^2+}$ (not $\ce{Fe}$) and $\ce{Zn^2+}$.

As for your reaction: when $\ce{Fe^2+}$ reacts with $\ce{Zn}$, we get  $\ce{Fe}$ and $\ce{Zn^2+}$.


What determines the boiling point of esters of the same number of carbon atoms but different structural formula?

In this case you have consider the degree of branching of the isomer, this influences the strength of intermolecular forces. Think of how tree logs with lots of branches sticking out cannot stack together with each other as easily as a pile of logs with no branches. Branched chain isomers have less contact and therefore weaker intermolecular forces and lower boiling point and straight chained isomers have better contact and thus stronger molecular forces and higher boiling point.


Molality of the solution prepared from a hydrated salt

Let's denote a hydrated salt as $\ce{M_pA_q*xH2O}$ ($\ce{M}$ – metal, $\ce{A}$ – anion). Then molality $b$ of the aqueous solution of $\ce{M_pA_q}$ is defined as

$$b=\frac{n(\ce{M_pA_q})}{m_\mathrm{tot}(\ce{H2O})}\, ,\label{eq:1}\tag{1}$$

where $n(\ce{M_pA_q})$ – amount of water-free salt; $m_\mathrm{tot}(\ce{H2O})$ – total mass of water.

Note that the amounts of the hydrated and the water-free salt are equal:

$$n(\ce{M_pA_q})=n(\ce{M_pA_q*xH2O})=\frac{m(\ce{M_pA_q*xH2O})}{M(\ce{M_pA_q*xH2O})}\, ,\label{eq:2}\tag{2}$$

where $M$ – molar mass.

Total amount of water is found as the sum of water added to dissolve the salt and the water "stored" within the hydrated salt itself:

$$
\begin{align}
m_\mathrm{tot}(\ce{H2O}) &= m(\ce{H2O})+m_\mathrm{hydr}(\ce{H2O})\\
&= m(\ce{H2O})+n_\mathrm{hydr}(\ce{H2O})\times M(\ce{H2O})\\
&=m(\ce{H2O})+x\times n(\ce{M_pA_q*xH2O})\times M(\ce{H2O})\\
\eqref{eq:2}\quad\Rightarrow\quad&=m(\ce{H2O})+x\times m(\ce{M_pA_q*xH2O})\times \frac{M(\ce{H2O})}{M(\ce{M_pA_q*xH2O})}\label{eq:3}\tag{3}
\end{align}
$$

Now, putting \eqref{eq:2} and \eqref{eq:3} back in \eqref{eq:1}, one can obtain uniform expression for molality of a hydrated salt:

$$b=\frac{m(\ce{M_pA_q*xH2O})}{m(\ce{H2O})\times M(\ce{M_pA_q*xH2O})+x\times m(\ce{M_pA_q*xH2O})\times M(\ce{H2O})}\label{eq:4}\tag{4}$$

   Plugging in the numbers for sodium carbonate decahydrate:
 $$b=\frac{\pu{13.8 g}}{\pu{103.5 g}\times\pu{286.14 g mol-1}+10\times\pu{13.8 g}\times\pu{18.02 g mol-1}}=\pu{4.3E-4mol g-1}$$

A side note regarding the notations and units [1, p. 28]:

  The term molal and the symbol $\pu{m}$ should no longer be used because they are obsolete. One should use instead the term molality of solute $\ce{B}$ and the unit $\pu{mol/kg}$ or an appropriate decimal multiple or submultiple of this unit. (A solution having, for example, a molality of $\pu{1 mol/kg}$ was often called a $1$ molal solution, written $\pu{1 m}$ solution.)

Reference

Thompson, A.; Taylor, B. N. Guide for the Use of the International System of Units (SI). NIST Special Publication 811, 2008.


Why is water considered ionic in fuel cells but otherwise covalent?

Water is a covalent compound. But as we know no compound is 100% ionic or 100% covalent. Since the electronegative difference (END) between hydrogen and oxygen is very high it is a strongly polarised covalent compound therefore the electron cloud is shifted towards the oxygen atom and therefore water has a very high % ionic character.


Does knowing the ratio of masses of the elements in a compound lead to the unique chemical identity of the compound?

No, knowing the mass ratios is not sufficient by itself. In the absence of additional information (for example, molar mass) that would only be enough information to determine the empirical formula, which is the formula that contains the smallest integral ratios of atoms. (E.g., the empirical formula of $\ce{C2H6}$ would be $\ce{CH3}$.) There are (theoretically) infinitely many molecules with the same atomic and mass ratios, yet differing in the actual molecular formula.

Further, even given the additional information of molar mass (so that the molecular formula can be determined), the existence of isomers precludes identification of a singular, specific compound without more sophisticated analysis. For example, $\ce{C3H6}$ could be either cyclopropane or propene. $\ce{C3H6O}$ could be as innocuous as acetone, or it could be highly carcinogenic epoxypropane. With more atoms, the number of possible isomers grows very rapidly. There's a nice table giving the number of isomers of alkanes of different lengths in this article. By the time you get to $\ce{C30H62}$, the number is $4111846763$. (Note, this is only considering structural isomers; accounting for stereoisomerism, the number would be even more astronomical.)


How does Combinatorial Chemistry differ from Rational Drug Design?

From a medical point of view the problem in drug discovery is to find a compound that causes a desired outcome, not one that interacts with a specific target. For instance, if you want to reduce mortality due to infection with Covid, you have many strategies that may be pursued at the molecular level, from vaccination to interference with viral propagation to reduction of symptoms or boosting of the immune response.
The difference between a combinatorial strategy and a rational strategy is to a first approximation simple: in a combinatorial strategy compounds are tested for activity without immediate concern for how they bring about an outcome. There is no "a priori" information to guide compound selection other than the common properties of known drugs (as encoded eg by Lipinski's rules) and the ability to procure or synthesize compounds. Therefore the "rational" approach in this case is to create as large a library and to screen as many compounds as is reasonable given cost and time.
Rational drug design takes into consideration a molecular target (such as a protein) and perhaps the mechanism of action of a known lead compound and designs future potential drug compounds around that target and/or lead. That narrows the set of compounds under consideration reducing time and cost.
In general, the strategies of combinatorial and rational design are not mutually exclusive. It is typical to start a rational strategy after early leads have been identified some other way, say through combinatorial chemistry. It is also common to design a set of compounds during a stage in rational drug design using a combinatorial strategy applied to develop next generation leads from a starting compound.


The two distribution plots shown here are for a lower temperature $T_{1}$ and a higher temperature $T_{2}$. The area under each curve represents the total number of molecules whose energies fall within particular range. The shaded regions indicate the number of molecules which are sufficiently energetic to meet the requirements dictated by the two values of $E_{a}$ that are shown.

It is clear from these plots that the fraction of molecules whose kinetic energy exceeds the activation energy increases quite rapidly as the temperature is raised. This the reason that virtually all chemical reactions (and all elementary reactions) are more rapid at higher temperatures.

Activation energy seems to be more of a material or reaction pertinent . Say a reaction has $10J$ as $E_{a}$ to proceed but has $20 J$ kinetic energy - I would say the the additional energy is transferred/carried on by the products after reaction is complete. So having more K.E shouldn't affect the character of a reaction but only its kinetics .


A carbocation is electron deficient. In an electron-neutral state, carbon would have four valence electrons that form four bonds resulting in an overall octet structure. To form a carbocation, one of these bonds needs to be removed, resulting in a sextet on the carbon atom; and one could also consider carbon now only having three valence electrons that form three bonds. Whichever picture we choose, carbon is lacking electrons.

(A more true picture would, of course, involve orbitals and their interactions, but still leave us with the conclusion that there are too few electons around the carbon atom.)

The three possible neighbours differ, amoung others, in their electronegativity — their ability to draw electrons towards them, away from what is next to them. Oxygen is the strongest, nitrogen second strongest and sulphur is the weakest.

In the case of the carbocation, I have already shown that it is electron-deficient. If we place a more electronegative element next to it, that element will draw the electrons away from the cation even more; this makes the carbon even more electron-deficient and thus destabilises the system. On the other hand, if a more electropositive element such as silicon were neighbouring the carbon, it would push electrons towards the carbon — exactly what the cation wants since it’s lacking electrons.

By this reasoning, sulphur, being the least electronegative element on the list should best stabilise the carbocation in question. But the answer gives nitrogen; so there must be a second effect. This would be mesomeric stabilisation.

For all three elements, one can draw a resonance structure where instead of a single $\ce{C-X}$ bond and a lone pair on $\ce{X}$ one has a $\ce{C=X}$ double bond shifting the positive charge to $\ce{X}$. This resonance structure is considerably less prevalent with sulphur (a third period element) than with oxygen or nitrogen; the reason being the spacial distribution of sulphur’s p-orbitals not fitting too well with carbon’s. Oxygen and nitrogen, on the other hand, have p-orbitals of a similar size and distribution so mixing works well. (Yes, this is also more complex on a more realistic level, but this visualisation works well enough for the simple case here.)

The question is now, whether the resonance stability provided by oxygen is stronger or weaker than that of nitrogen and if the difference is larger or smaller than the energy difference caused by the different electronegativities discussed earlier.

Again arguing that oxygen is more electronegative and thus draws electrons towards it, it would equally be less happy to share its unbonded electron pair with carbon to relieve carbon’s electron deficiency. Thus, one may expect nitrogen to provide a greater stabilisation to a neighbouring carbocation.


How can some enzymes work faster than the diffusion rates of the reactants allow?

The conceptually easiest case is that of a positively charged active site with a negatively charged substrate. The substrate (i.e. reactant) enters the active site with kinetics that are faster than diffusion because there is a long-range electrostatic interaction.
Catalase works near the diffusion limit. The document the OP cites is a hypothesis, without experimental data. It suggests that catalase acts at a distance to catalyze reactions outside of its active site. I would not worry about this hypothesis until there is some experimental evidence supporting it.
Wikipedia has a list of 9 examples of diffusion limited enzymes:

Acetylcholinesterase
β-lactamase
Catalase
Carbonic anhydrase
Carbon monoxide dehydrogenase
Cytochrome c peroxidase
Fumarase
Superoxide dismutase
Triosephosphate isomerase

Some of them have neutral substrates, so the electrostatic attraction would not apply to them. There are other hypothetical mechanisms, though, like weak binding sites on the surface of the enzyme that increase the local concentration of substrate, for example.


What is the difference between transaldolase and transketolase in the pentose phosphate pathway?

An aldol is a $\beta$-hydroxy ketone or aldehyde, and is the product of an aldol addition.

The enzyme that cleaves the bond formed (red) is known as an aldolase. The transaldolase reaction begins by aldol cleavage that is, cleavage of the bond in red.

Transketolase on the other hand simply cleaves off the ketone group from the substrate. It does not perform an aldol cleavage.

Apart form this, the mechanisms of action of the two enzymes vary considerably. For instance transketolase requires thiamine pyrophosphate (TPP) as a cofactor but transaldolase does not.


Relationship between the symmetry number of a molecule as used in rotational spectroscopy and point group

This is not in general true
Consider molecules a point group not containing inversion symmetry, e.g. $C_2$ hydrogen peroxide

The $C_2$ group has only two elements, $E$ and $C_2$, and the $C_2$ rotation operation maps between two identical arrangements of atoms.  Both the symmetry number and order of the group are 2.
Matrix representations
All rotations in 3D space can be represented by a orthogonal 3x3 matrix with determinant 1, and the composition of rotations about the origin can be represented by matrix multiplication of the transformation.
More formally: the group SO$(3)$ (3D rotations) is isomorphic to the group of orthogonal real 3x3 matrices with determinant 1 under multiplication.
The number of symmetry elements in a point group formed from rotations (and which hence preserve chirality) will be the same as the number which can be represented by such matrices and the same as the number of equivalent orientations of the molecule (the symmetry number) as they are all representation of the same thing - 3D rotations.
Inversions
The group of all transformations that keep the origin fixed, including reflections as well as rotations is O$(3)$ - or the product of SO$(3)$ with the set $\lbrace I,-I\rbrace$ or the inversion operation.  It maps rotations onto improper rotations (which include reflections - improper rotation by 0 degrees) and hence all symmetries in 3D space.
All point groups with inversion symmetry, such as the 3 listed in the question, contains this $-I$ element.  $-I$ will multiple all the elements of the rotation matrix by $-1$ and create an orthogonal 3x3 matrix with determinant -1.  By composition with each of the rotations elements of the point group it creates another element - an inversion, improper rotation or reflection - doubling the size of the group without increasing the symmetry number.
Conclusion: chirality matters
This rule is true for non-chiral groups.  You can invert the molecule to get another symmetric copy for every rotationally equivalent copy, so the order of the group is twice the symmetry number.
For chiral groups that will create the enantiomer which isn't symmetrically equivalent so the order of the group and symmetry number are the same.


Determining number of unique hydrogens in lactupicrinal 1H NMR

Your question doesn't sound like you care about exact splitting, so I'm going to ignore magnetic equivalence/inequivalence. If you do care, you need to take that into account.

For the purposes of this discussion, equivalent means chemically equivalent.

You also need to keep in mind that NMR is achiral. In other words, opposite enantiomers will produce the same spectrum.

The standard procedure is to replace the atom of interest with something that is distinguishable from the regular atom. I usually replace an $\ce{H}$ atom with an $\ce{H}*$, but anything will do, provided it's still a hydrogen but different from the others.

With this replacement you've created a new molecule.

For two specific hydrogen atoms, you can perform the replacement operation to generate two molecules.

Now, compare these two molecules.

If they are different molecules (e.g., geometric isomers, (E)-/(Z)- isomers, cis-/trans- isomers), then the original hydrogen atoms are heterotopic and should have different chemical shifts. Their environments are different.
If the molecules are diastereomers, then the two hydrogens are diastereotopic. Since we can distinguish diastereomers by achiral methods, so can NMR. That means that the two original hydrogen atoms will have different shifts.
If the molecules are enantiomers, then the two hydrogens are enantiotopic. They are different, but because NMR can't distinguish between enantiomers, it can't tell that these two hydrogen atoms are different. They will have the same shift.
If the two molecules are identical, then the two hydrogens are homotopic. Nothing can distinguish them. They are chemically equivalent, and by NMR, they will have the same chemical shift.

Try applying this test to the hydrogen atoms on the ethyl group and the hydrogen atoms on the ring.


You obviously cannot ‘filter water out’ as water is a liquid at STP and the solvent in which your reaction occurs in. However, you can evaporate your solution whereby the solvent gets removed and the lower vapour pressure contents (mainly solids or oils) will remain. While water can be more difficult to evaporate at a standard rotavap, it is by no means impossible.


I would guess that benzoic acid forms dimers or trimers connected by hydrogen bonds in nonpolar solvents. To the outside, these only show the phenyl rings.

The colligative properties of those solutions would clearly show how many individual particles they contain. I´ll leave the literature research for you. ;)


Is the flame temperature of a burning fuel affected by pre-heating the fuel?

Yes and no: preheating a mix a little bit may increase the final temperature somewhat, but adding even more initial heat has a decreasing effect on the final temperature.

There are two competing factors that limit the final temperature in a closed system:

The total amount of energy as the sum of the initial amount of thermal energy and the chemical energy, divided by the thermal mass [mass times heat capacity] of the system, and
The tendency for a reaction to be forced backwards by applying additional energy.

For example, $\ce{2H2 + O2 -> 2H2O}$ + 286 kJ/mole, but water dissociates more as the temperature rises. At 2,200 °C water is about 3% dissociated. At 3,000 °C  it is about 50% split, so half the chemical enegy is unavailable.

BTW, thermodynamically, a heat engine produces power based on the difference in temperature, so preheating a fuel-air mixture actually decreases efficiency, though it might be needed to vaporize the fuel (e.g. in Otto-cycle engine in cold weather).


What does radiofrequency really do in NMR & MRI?

The radiofrequency pulses and the gradients have completely different purposes in NMR. Radiofrequency pulses are the basic building block of NMR experiments, you use them to manipulate your spin populations and coherences. While each proton has a different frequency depending on its environment, you generally can't (or don't) target individual protons with your pulses. In most experiments you hit either all protons with your pulses, or a specific group of protons.

Gradients have different purposes in NMR and MRI, what they allow you to do is to change the magnetic field depending on the position inside your sample. In NMR this allows you to remove unwanted signals and select specific coherences.

In MRI, the gradients are what makes imaging possible. They allow you to encode the spatial position of your protons into their frequency. In NMR you look at all the molecules in your sample at once, you don't care where your molecules are floating around. In MRI the position is important, as you want to know what the protons at a particular point in the body are doing. Gradients also allow you to only look at a slice of the body and not all of it at once.

I'll skip phase encoding in gradients as it is MRI-specific and I don't know much about it. Each part of your question has whole books written about it, so this is a drastically simplified answer.


Ammonia and nitric acid can be considered corrosive materials (though the real question would be "corrosive towards which material?"). The thing here is that ammonia and nitric acid are corrosive for strictly opposite reasons: ammonia is a base and nitric acid is an acid. So in a mixture of 1 equivalent(1) of ammonia and 1 equivalent(1) of nitric acid, both effects will cancel and the resulting mixture is no longer corrosive.

You do not create nitrogen (with such a mixture) for your lucerne, but you provide it with a form of nitrogen(2) which can be transformed by bacteria before being absorbed by your plant. Though, you do not want to mix ammonia and nitric acid because it is dangerous. As Mithoron said, you would better buy the product.

The chemistry and biology of soils are extensively (?) studied nowadays because it appears that monoculture exhausts soils and an agroecological approach can drastically outperform the monocultures we employ in western world.

(1) This requires a bit of calculations.

(2) Use of the term "nitrogen" can be very misleading as it can describe the "nitrogen gas" which composes 80% of our atmosphere (but is quite inert) or any source of nitrogen atoms for a plant.


Yes, the substance is a reducing agent which will reduce $\ce{KMnO4}$ into $\ce{MnO2 }$ in neutral or weakly basic medium.


Specific latent heat of fusion of a noble gas

When you heat a substance (not at its melting point or boiling point), the energy supplied goes to increase the kinetic energy of the atoms/molecules in the substance.

When the same heat is applied to the substance at its melting point (or boiling point), the heat goes to increase the potential energy of the atoms/molecules in the substance. You could refer to the potential energy as the energy used to break the inter-particle bond. In ice, there are relatively strong forces of intermolecular attraction, which makes the ice hard.

Now, temperature is directly related to the kinetic energy of the atoms/molecules of a substance, and since the kinetic energy of the atoms/molecules does not increase during a change of state, the temperature remains constant.

At least, this is what happens at a macroscopic level and in the long term. But if you heat ice at one particular point (at particle level and for a short period of time), that point will melt faster and the temperature will tend to rise there, but this heat energy will then be distributed to the other parts of the ice still cold for melting.

It is only when all the ice has molten than the heat supplied will contribute once again to the increase in kinetic energy, that is, temperature.

And for the last part, every substance that undergoes a change of state has latent heat.


Why ZnS can be seen as the solution is made alkaline? Explain with reactions

The final solution contains $\ce{Zn^{2+}}$ ions, which reacts with $\ce{H_2S}$ to produce white $\ce{ZnS}$. But you may not avoid that maybe some of the initial $\ce{Cu^{2+}}$ ions still remains in the solution. And these rare $\ce{Cu^{2+}}$ ions do react with $\ce{H_2S}$ to produce Copper sulphide $\ce{CuS}$ which is dark black.
$$\ce{Cu^{2+}  +  H_2S  ->  CuS  +  2 H^+}$$
So the obtained final precipitate is probably a mixture $\ce{ZnS}$ and some $\ce{CuS}$ ; it is probably not perfectly white, but grey or even black. To be able to see the white color of $\ce{ZnS}$, you add some drops of ammonia $\ce{NH3}$ in the dark mixture containing $\ce{ZnS}$. This will dissolve the copper sulphide, producing a solution containing $\ce{[Cu(NH3)4]^{2+}}$.
$$\ce{CuS  +  4 NH3 -> [Cu(NH_3)_4]^{2+}  + S^{2-}}$$
So the black component of the precipitate will disappear, and you will be able to see the white color of insoluble $\ce{ZnS}$.


Crystallization of solutions and liquids

Crystallization is rightly called by you a process.  Simplified, there are two steps to be considered: Nucleation and crystal growth.  For the sake of simplification, let me consider the crystallization of a molecular compound in absence of ionic charges.

Nucleation refers to the initial stage, where once isolated molecules of the dissolved substance assemble to a first crystal germ. This is energetically hindered because these particles of especially small size have unfavourably high ratio of particle surface energy per molecule in these just formed particle (or perhaps better: aggregates). This energetic drawback has to be overcompensated. In addition, molecules loose degrees of freedom accessible in solution, at first: dynamic freedom, and secondly, depending how strictly rules are followed to build the crystal, a number of orientations in space otherwise available and accessible--In a perfect crystal, the structural close environment of one molecule is going to be repeated for all molecules, the very next ones and for their next ones, over ranges of Angstroms to Centimetres.

Once such a germ is formed,

Crystal growth more or less refers to simply attachment of additional molecules according to these "(energetically justified) previously determined rules" to yield a bigger entity.  The need of oversaturation of the solute is less prominent than during nucleation, yet still existent to energetically favour this process. In this description, crystal growth is again simplified, too.  It may happen, in the (sub)process of crystal growth, conditions are met that allow crystallisations in realisations even more energetically favourable than at early stages of crystal growth so that a compound preferentially forms a polymorph A initially, then ones in polymorph B. And often a crystallization does not yield only one (perhaps even single)crystal, rather than multiple ones.  So nucleation may occur in one spot of the solution while simultaneously crystal growth already takes place on germs already formed in other parts of the solution.

Complementary, think this way: it is every time easy to dissolve a perfect crystal (already considering entropy), but easily difficult to bring molecules back into a perfectly ordered crystal.

The science and art (like craftsmanship) of crystallization is of high importance in our daily life, spanning areas of electronic devices (semiconductors) till pharmaceutics. Journals like Crystal Growth & Design or CrystEngComm are examples concentrating on this subject.


Froth floatation process - why CuSO4 is activating floating character of ZnS?

I'm going to consult the literature later, but for me it seems to be as follows.

In the compounds discussed, there are following cations:

$\ce{Fe^2+}$. It is a transition metal with a moderate affinity to oxygen, a slightly larger affinity to sulfur and a high affinity to cyanide.
$\ce{Zn^2+}$. It is a small cation with a high affinity to oxygen, a slightly larger affinity to sulfur and a moderate affinity to cyanide.
$\ce{Pb^2+}$. It is a large cation with a moderate affinity to oxygen and a slightly larger affinity to sulfur.

Xanthate bind to the surface via the sulfur atoms. So, in a basic cyanide solution the binding sites on the surfaces of pyrite and sphalerite are both completely filled with cyanide and hydroxide respectively (please, note that sulfur in xanthate is a lot less nucleophilic than in sulfide form). Copper has a high affinity to sulfur, a lot larger than zinc or iron. So it binds to xanthate and to the sulfur in either pyrite or sphalerite. However, the surface of pyrite is still occupied with cyanide, which binds really strongly there. And in an acidic environment, cyanide is transformed into $\ce{HCN}$ and thus the pyrite surface is open to binding with xanthate.


Concentration of dissolved lead(II) bromide

The given value $K=7.9\times10^{-5}$ actually is the value of the solubility product constant $K_\mathrm{sp}$ of $\ce{PbBr2}$

$$K_\mathrm{sp}=a_{\ce{Pb^2+}}\cdot a_{\ce{Br-}}^2=7.9\times10^{-5}$$

which approximately is (when incorrectly neglecting activity coefficients $\gamma$, the difference of molality $b$ and concentration $c$, and the correct units)

$$K_\mathrm{sp}\approx \left[\ce{Pb^2+}\right] \cdot \left[\ce{Br-}\right]^2$$

When $\ce{PbBr2}$ dissolves in water according to the given chemical equation, the resulting concentrations of $\ce{Pb^2+}$ and $\ce{Br-}$ correspond to

$$\left[\ce{Br-}\right]=2\cdot\left[\ce{Pb^2+}\right]$$

These equations can be used for calculating a value for $\left[\ce{Pb^2+}\right]$.


Is this mechanism for the formation of a tosylate correct?

No, $\ce{Cl-}$ will not remove the $\ce{H+}$ because it is not basic enough. The $\mathrm{p}K_\mathrm{a}$ of $\ce{Cl-}$'s conjugate acid ($\ce{HCl}$) is $-7$. The $\mathrm{p}K_\mathrm{a}$ of $\ce{NEt3}$'s conjugate acid ($\ce{HNEt3+}$) is $10.75$ (source). This means that $\ce{HCl}$ almost completely dissociates in solution, and is very unlikely to form $\ce{HCl}$ molecules. $\ce{NEt3}$, on the other hand, is much more basic and will easily pick up the $\ce{H+}$.

Also the mechanism is incorrect. A lot of the arrows are wrong. See here (pyridine serves the same function as $\ce{NEt3}$):


What is the order of nomenclature priority among an alkyl side group, a double bond, and a triple bond?

Your so-called priority table (the “table of functional group priorities for nomenclature” shown here) is relevant for the choice of a characteristic group for citation at the end of a name by means of a suffix or a class name. It does not apply to the numbering of locants. This particular table could also be misleading since it mingles the concepts of classes, suffixes, endings, and prefixes.

The most important simplified criteria for the numbering of simple acyclic compounds are:

lower locants for suffixes
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

  P-14.4 NUMBERING

  When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:

  (…)

  (c) principal characteristic groups and free valences (suffixes);

  (…)

  (e) saturation/unsaturation:

    (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;

    (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);

  (f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;

  (g) lowest locants for the substituent cited first as a prefix in the name;

  (…)

Since both compounds that are given in the question do not have any principal characteristic group (a characteristic group chosen for citation at the end of a name by means of a suffix or a class name, or implied by a trivial name), the first criterion (greater number of suffixes) given in P-14.4 (c) is not relevant in this case.

The systematic name of the first compound is 6-methylhept-2-yne rather than 2-methylhept-5-yne since low locants are given first to the ‘yne’ ending and the locant ‘2’ is lower than ‘5’.

The systematic name of the second example is (4E)-hex-4-en-1-yne rather than (2E)-hex-2-en-5-yne since low locants are given to multiple bonds as a set (even though this may give ‘yne’ endings lower locants than ‘ene’ endings), and the locant set ‘1,4’ is lower than ‘2,5’. Only if a choice remains, preference for low locants is given to the double bonds.


How do Lewis acids interact with 1,4-diketones and amines to form an alkaloid?

This is a variant of the Mannich reaction. The methoxymethyl groups on the amine are a form of captured Mannich intermediates or masked formaldehyde equivalents. The Lewis acid unmasks them by binding to the oxygen of the methoxy group, creating the functional equivalent of H2C=N+. These are reactive electrophiles which react with the active methylenes of the B-ketoester to give the alkaloid.

More on the Mannich reaction here


Deciding which α-Η gets abstracted in aldol condensation

There have been tons of articles, review articles and even books written on the subject. The tl;dr version is that it depends on the conditions you use.

In general, the less-hindered enolisation (in your case: enolising towards the methyl) will be more rapid. This leads to the possibility of kinetic control to selectively generate the terminal enolate by very strong bases and low temperatures such as LDA or even BuLi if you’re daring (both at $\pu{-78^\circ C}$).

On the other hand, the more substituted enolate is more stable thermodynamically as you noted. Thus, if you use a weak base that binds and deprotonates reversibly and if you use relatively high temperatures or longer reaction times, the more stable enolate will be formed. An example for these conditions would be an alcoholate base at room temperature.

Unfortunately, I can give you no simple rules. You will have to check the literature or your lecture notes to see which combinations of reagents give which selectivity.


Why is cytosine far more vulnerable to hydrolytic deamination than both adenosine and guanosine?

Cytosine deaminates to form uracil at a finite rate in vivo

Pyrimidine bases e.g cytosine in DNA are more susceptible to spontaneous deamination than are the purine bases because there are more thermolabile.

  This was demonstrated by the previous observation of significant
  cytosine deamination but no guanine deamination upon heating of the
  synthetic copolymer poly(dG-dC):poly(dG-dC) and the homopolymer
  poly(dG):poly(dC)

Reference: Journal of Biological Chemistry -  Nitric Oxide-induced Deamination of Cytosine and Guanine in Deoxynucleosides and Oligonucleotides

Another research article: Cytosine Deamination Plays a Primary Role in the Evolution of Mammalian Isochores agrees with the theory of temperature as the major cause of sponteneous demination in cytosine as it states:

  Because a 10°C increase in temperature in vitro increases the rate of
  cytosine deamination 5.7-fold, cytosine deamination must be highly
  dependent on body temperature.

Cytosine residues in DNA (2) can undergo hydrolytic deamination to yield uracil residues (3)

The reaction may proceed either via attack of hydroxide on the neutral nucleobase or attack of water on the N3-protonated base:

  In single-stranded DNA, and at base mismatches in duplex DNA,
  deamination proceeds much faster (t1/2 ~ 200 years) presumably due to
  increased solvent accessibility of the base.   Deamination of guanine
  and adenine residues in DNA is much slower, occurring at only 2–3% the
  rate of cytosine deamination.

Alkylation of the N3-position of cytosine and reactions that lead to saturation of the 5,6-double bond in cytosine and 5-methylcytosine accelerate deamination.

Activation-induced cytidine deaminases catalyse the conversion of cytosine residues to uracil residues in single-stranded regions of DNA.

Reference:  An Overview of Chemical Processes That Damage Cellular DNA: Spontaneous Hydrolysis, Alkylation, and Reactions with Radicals

Another factor which apparently "contributes" to susceptibilty of cytosine to hydroysis is its presence in both DNA and RNA.

DNA is not susceptible to alkaline hydrolysis while RNA is alkali labile and is readily hydrolysed by hydroxide ions. On the other hand RNA is relatively resistant to the effects of dilute acid, but gentle treatment of DNA
with acid leads to hydrolysis of purine glycosidic bonds and the loss of
purine bases from the DNA.

These different conditions  (i.e acidic and basic conditions) to which the bases are exposed to appears to have a "double effect" on cytosine. (and additionally , as a pyrimidine it is more thermolabile)

Other reference

Biochemisty (Grisham)


Thermal stability of alkali metal hydrides and carbonates

Down the group, atoms of the alkali metals increase in both atomic and ionic radii, due to the addition of electron shells. This results in the charge density of their corresponding cations decreasing down the group. Thus, as we go down the group, the cations become "softer".

Now, note that the hydride ion is "hard", having high negative charge density. While the carbonate ion is "soft", having a lower negative charge density.

Applying the "hard-soft acid base" (HSAB) principle, we would expect the "hard" ions to form more stable compounds with each other and the "soft" ions to form more stable compounds with each other.

Since the heavier alkali metal ions are softer, we would expect them to form more stable compounds with carbonate. As for the harder lighter metal ions, we would expect them to form more stable compounds with hydrides.


Welcome to parallel worlds of trivial and systematic names. We also use formic acid, acetic acid, acetyl chloride.

You can name them all systematically as methanoyl chloride, methanoic acid,  ethanoic acid and  ethanoyl chloride. But in the real lab life, nobody call them like that.

Formyl chloride has  its name from
formyl-  as acyl derived from formic acid, the same as acetyl- from acetic acid.

Technically, formyl chloride is an aldehyde as well, as formyl is de facto an aldehyde group, being formally a member of the serie :

$$\ce{Cl-CH2OH <> Cl-CHO <> Cl-COOH}$$

But by naming conventions, both trivial and systematic, being a derivate of an acid replacing $\ce{OH}$ by $\ce{Cl}$ takes precedence to naming it as an aldehyde.

As a sidenote, it is rather funny aldehyde names are derived from the respective acids and not alcohols.  Because the aldehyde etymology says aldehyde="alcohol dehydrogenatus", so it could be as well methylaldehyde and ethylaldehyde, instead of formaldehyde and acetaldehyde.


what happens to hydrogen released in addition polymerization?

Consider the reaction stoichiometry.

You begin with ethylene (that is, CH2CH2).
The polyethylene chain is a chain of single-bonded carbons. Hence, it is a  -[...]-CH2-CH2-CH2-CH2-[...]- chain).

As you can se, no hydrogens are removed by the monomer.
So, no hydrogens are released during polymerization.

What is happening in that reaction is an electron transfer, not an atomic transfer: the double bond of ethylene, full of electrons, reacts with the radical and forms a more stable structure with a radical carbon (a radical carbon has three bonds, in this case with C, H and H, and an unpaired electron).
The carbon radical reacts with another ethylene monomer: the "old" radical carbon now forms four bonds (C, H, H, plus the new C from the ethylene monomer), the new ethylene monomer loses its double bond, and forms a carbon radical (again, the radical carbon has three bonds and an unpaired electron).

This is the logic of that tipe of polymerization: no hydrogen is involved.


Functional difference of Benedict's solution and Fehling's solution

Fehling's solution consists of two parts, Fehling's A and Fehling's B. Fehling's A consists of aqueous copper(II) sulfate, while Fehling's B consists of aqueous sodium potassium tartrate ($\ce{NaK(C4H4O6)}$). Benedict's reagent consists of sodium carbonate, sodium citrate ($\ce{Na2(C6H6O7)}$), and copper(II) sulfate dissolved in water. It's pretty clear now, that the main difference between the two reagents is the complexing agent (citrate in Benedict's and tartrate in Fehling's).

Fehling's reagent was invented by Hermann von Fehling in 1849. You notice that this solution is given as two separate solutions, and is only mixed together while performing a test. This is because the complex formed when mixed ($\ce{[Cu(C4O6)2]}$) is kinetically favored, and that $\ce{Cu(OH)2}$ will eventually precipitate out due to being favored energetically. That's why the two solutions are stored, and only mixed during testing.

On the other hand, Benedict's reagent was discovered by Stanley Rossiter Benedict somewhere in between 1904 to 1936. Thus Benedict's reagent was developed after Fehling's was. The citrate complex in Benedict's reagent is much more stable that $\ce{Cu(OH)2}$, so Benedict's reagent doesn't precipitate on standing for long, unlike Fehling's. This is quite advantageous since you wouldn't have to mix two solutions together to perform the test. This is where Benedict's reagent is quite handy.

Both the reagents have a similar way of oxidizing aldehydes. The copper complex oxidized the aldehyde to a carboxylic acid, and while reducing itself from $\ce{Cu^2+}$ to $\ce{Cu+}$
 ($E^\circ = \pu{0.154 V}$). This reduction potential is sufficient to oxidize aldehydes to carboxylic acids. There is not much difference here though.

Neither of the reagents can oxidize aromatic aldehydes, due to the fact that they lack α-hydrogens, and thus oxidizing them requires stronger oxidizers (like Tollens’ reagent).

In short, Benedict's reagent is kind of like an "improved" version of Fehling's reagent, which is much more stable due to the difference in the complexing agent.