How would I determine the relative acidity of substances?

In my homework one of the question was which of the following salt will produce a neutral solution(If any)? A) KI (Potassium Iodine) or B) SrS (Strontium Sulfur)

For (A) both K+ and I- will essentially stay as the ions. So assuming the solution is neutral to start, it will remain neutral.

But for (B), in a neutral solution, $\ce{Sr^{2+}}$ will stay as the ion,  $\ce{S^{2-}}$ will react as:

$\ce{S^{2-} + H2O <=> HS^- + OH-}$

so the solution will become slightly more basic.

"Order the following from most acidic to least. $\ce{SO3}$, $\ce{CO3}$ and $\ce{CO2}$".

I can't really make sense of this question since $\ce{CO3}$ is highly unstable, and there are three isomers.

Bubbling  $\ce{SO3}$ and $\ce{CO2}$ gases into water would produce acidic solutioons via reactions

$\ce{SO3 + H2O -> H2SO4}$

$\ce{CO2 + H2O -> H2CO3}$


Are there terms for ions of chlorine other than chloride?

Assuming that "term" in your question refers to chemical nomenclature, the authoritative source of information would be the current edition of IUPAC Red Book [1]. From [1, p. 70]:

IR-5.3.2.2 Monoatomic cations
The name of a monoatomic cation is that of the element with an appropriate charge number
appended in parentheses.
[…]

$\ce{I+}$ iodine(1+)

Analogously, the name of $\ce{Cl+}$ is chlorine(1+).
Further from [1, pp. 72–73]:

IR-5.3.3.2 Monoatomic anions
The name of a monoatomic anion is the element name (Table I) modiﬁed so as to carry the
anion designator ‘ide’, either formed by replacing the ending of the element name (‘en’,
‘ese’, ‘ic’, ‘ine’, ‘ium’, ‘ogen’, ‘on’, ‘orus’, ‘um’, ‘ur’, ‘y’ or ‘ygen’) by ‘ide’ or by directly adding ‘ide’ as an ending to the element name.
Examples:

chlorine, chloride

[…]
Charge numbers and radical dots may be added as appropriate to specify anions fully.
Examples:
14. $\ce{O^2-}$ oxide(2−), or oxide

So, the name for $\ce{Cl^2-}$ is chloride(2−).
Reference

IUPAC. Nomenclature of Inorganic Chemistry, IUPAC Recommendations 2005 (the “Red Book”), 1st ed.; Connelly, N. G., Damhus, T., Hartshorn, R. M., Hutton, A. T., Eds.; RSC Publishing: Cambridge, UK, 2005. ISBN 978-0-85404-438-2.


What is the stochiometric equation of the sucrose and potassium chlorate reaction with sulfuric acid?

In this paper [Sugar Dehydration without Sulfuric Acid: No More Choking in the Classroom! T.P. Silverstein and Y. Zhang. J. Chem. Ed. 75 (6) 748-749 (1998)], the authors outline two reactions involving sucrose and sulfuric acid. The first one does not include $\ce{KClO3}$ and utilizes a significant amount of $\ce{H2SO4}$ in order to fuel the dehydration of sucrose: this is not the reaction you're interested in.

The second reaction detailed herein makes use of a very small amount of sulfuric acid to start the reaction (a drop is the commonly-cited amount): the sucrose is oxidized by $\ce{KClO3}$. This is the reaction you want to examine.

The authors give the reaction stoichiometry as
$$\ce{C12H22O11(s) + 8KClO3(s) \xrightarrow{H2SO4} 12CO2(g) + 11H2O(l) + 8KCl(s) + heat}$$

They note that the fumes from the reaction contain $\ce{Cl2, ClO}$, and $\ce{ClO2}$, but the amounts are not quantified. I suspect (but don't know for sure) that the amounts are finite but negligible. And even with the scant amount of sulfuric acid used here, I think you're safe going with the reaction as presented above.


Which has a greater dominance: hydration enthalpy or lattice energy?

The dominant factor depends on situation to situation.Here is an explanation i found helpful while predicting solubilities

The hydration energy is of Form A÷r1 + B÷r2 .

The lattice energy is of form C÷(r1+r2).
Here r1 and r2 are radius of cation and anion while A,B,C are constants.

When Size of anion is large example Iodine,Lattice energy does not decrease with increase in radius but hydration energy decreases drastically since r2>>r1(See formula)
Therefore in such cases,Solubility of Salts decreases with radius of cation.
Order- LiI is more than NaI is more than KI(in terms of solubility)

Now when anion size is comparable(eg Fluorine),The lattice energy decreases much more rapidly than hydration energy(see formula).Hence Solubility increases with increase in cationic size.Thefore order is LiF is less than NaF is less than  KF


What determines the elastic limit of a substance?

First let me clarify that there is no way to calculate the elastic limit of a substance - it has to be measured. Viscoelastic theory is pretty well-developed for only being about 150 years old, but it's not that well developed.

To answer your question, though, lots of factors affect elasticity in general and the elastic limit in particular.

It would probably be the most helpful to think about what is happening on the molecular scale. In any viscoelastic material, there is a range of deformations (strain, or the response to stress) and timescales over which molecules will go back to pretty much the same relative position that they were in before the deformation occurred. That's elasticity. What makes this happen? Each atom or molecule sits in a little potential energy well that exists because of its position relative to other atoms in the substance. Atoms like to be close to each other, but not too close. As a result, when given the opportunity they will arrange themselves so that as many atoms as possible are the optimal distance apart(resulting in the lowest energy).

When a material is deformed, we are moving atoms relative to each other. When one atom moves, its energy increases. This also changes the potential energy surface seen by its neighbors. If it moves a little bit, then the change is not usually great enough to cause the neighbors to move by much. On the other hand, for big movements, the effect can be great enough that the other nearby atoms rearrange themselves in order to find the next lowest energy level.

When the relative motion of atoms due to mechanical deformation is small enough that most of them go back to their original positions once the applied stress is removed, then we say the stress/strain was within the elastic limit.

When the atoms are permanently rearranged, the shape of the material is changed and we say that we have exceeded the elastic limit.

The ease with which atoms can move out of position (you can think of this like the depth of the wells), and their tendency to move back to the lowest energy (which can be thought of as the "steepness" of the wells) are the two main factors that affect how much stress/strain is needed to permanently deform a material. There are lots of things that can affect each of these, which is why there is no comprehensive theoretical model that can predict the elastic limit of materials based on molecular structure alone.

I can summarize a few important factors here, though:

degree of crystallinity in polymeric materials
type and number of dislocations in crystalline materials
length of time that the stress/strain is applied, in conjunction with the temperature of the material (this related to a phenomenon called time-temperature superposition)
structure of the molecules making up the material (e.g. linear vs. crosslinked or branched polymers, fcc vs. bcc crystal structure, or type of atoms in an alloy)
mesoscopic structure (e.g. crystalline grain size and shape, aggregation of nanoparticles, or long-range order in polymers)
macroscopic structure (e.g. fiber orientation in composite materials)

There are many more! And each of the ones I listed is practically an entire field of its own in materials science.

To answer the second part of your question - the cloudiness you are referring to is called "crazing" in polymer science. It is basically caused by molecules pulling apart from each other, without covalent bonds breaking. It can be reversed when it happens before the elastic limit is reached because the molecules can still go back to their original positions. If it doesn't happen until after the elastic limit, then the material is permanently deformed, and the molecules cannot return to their original position.


How does water flows through pipes in cold countries?

From the Wikipedia page on Frost Line

  The frost line—also known as frost depth or freezing depth—is most commonly the depth to which the groundwater in soil is expected to freeze. The frost depth depends on the climatic conditions of an area, the heat transfer properties of the soil and adjacent materials, and on nearby heat sources. For example, snow cover and asphalt insulate the ground and homes can heat the ground (see also heat island). The line varies by latitude, it is deeper closer to the poles.

So they just dig the pipes in deep.

  Canada: Can use table of estimates based on freezing index degree days.
  Ontario: Map of frost penetration depths for Southern

  Ottawa: 1.8 metres (5.9 ft) [7][8]
  Windsor: 1.0 metre (3.3 ft) [7][8]

Here's a  soil temperature vs depth graph.

Source


What is the driving force for imine formation?

In practice, a drying agent is sometimes added to the reaction: I've used molecular sieves before, but a cursory database search brings up a few other options, e.g. magnesium sulfate, silica gel, or Montmorillonite (this list is probably not exhaustive). These remove water from the system and pull the equilibrium position over to the right-hand side.

That's one possibility, but I don't think that's all there is to it - I think a lot of imine formations proceed perfectly ok even without dehydrating agents being added to them. However, you'll have to hear about these from somebody else, as I don't know what would drive those reactions. It could well be that simply adding an excess of either amine or carbonyl drives the reaction to completion.

Interestingly, a 2009 paper by Saggiomo and Lüning1 describes their investigations into supposed imine formation in water. Apparently, both the carbonyl compound (benzaldehyde or salicylaldehyde) and amine (aniline) do not dissolve in water - not particularly surprising. The imine did not form appreciably in water and only started to form when water was removed during workup and purification.

P/S This is just the top hit I found on Google, it's hardly a complete literature search.

Saggiomo, V.; Lüning, U. On the formation of imines in water—a comparison. Tetrahedron Lett. 2009, 50 (32), 4663–4665. DOI: 10.1016/j.tetlet.2009.05.117.


What attributes cause a substance to be nonconductive or conductive

Conductivity requires mobile electrons. Metals tend to have loosely held valence electrons so metals are generally conductive. Metals have loosely held valence electrons for two reasons: metals tend not to be very electronegative (able to stabilize electron density) and metals tend to be somewhat big atoms by nature of many metals having a d-group of electrons. D electrons are further from the nucleus on a time average basis than s electrons or p electrons.

However, conductors do not have to be metals. Consider graphite. Graphite is $\ce{sp^2}$ hybridized. Graphite is made solely of carbon atoms - and carbon isn't a metal! The mobile pi electrons in graphite due to resonance allows graphite to be a good conductor.

I should also note that conductivity does not always imply mobile electrons; solutions can be conductive too and this is due to mobile electrolytes (ions). For example, you will find that deionized water is completely unconductive. However, regular tap water is generally conductive due to the presence of ions (e.g. chloride ions from chlorination). Gatorade is also very conductive because it is especially formulated to contain a lot of electrolytes.


How do d-orbitals interact with molecular orbitals?

You are on the right track with your answer.

Notice that the LUMO is a $\pi^*$ orbital, meaning that filling it with electrons would weaken the bonding. When nitrogen and iron interact, electrons from the d-orbitals on iron are well aligned, in terms of symmetry and energy, to fill into $\ce{N_2}$'s $\pi^*$ orbital. This forms a bond between $\ce{Fe}$ and $\ce{N_2}$, but at the cost of weakening the bond between the nitrogens. This is a very important concept in inorganic chemistry referred to as Pi Backbonding.

The classic example given of a ligand that exhibits $\pi$ backbonding in metal complexes is $\ce{CO}$, which is isoelectronic to $\ce{N_2}$ and has the same molecular orbital diagram (though the orbitals themselves look different due to the asymmetry of $\ce{CO}$).


Hydrogen atom transfer between enantiomers of rac-1-phenylethanol

Did you continue reading the paper? :) The source is important and is what was needed to answer the question!

I'm just going to quote from it.

  In DKR [dynamic kinetic resolution] a chemical catalyst is added in order to racemize the enantiomers, ensuring that all the substrate can be utilized. [...] Highly successful catalysts are often Ru(II) complexes [...] The metal complex catalyses racemization by a hydrogen transfer mechanism (Scheme 2).

So, the supposed hydrogen transfer occurs via coordination to a metal, followed by β-hydride elimination. Readdition of the metal hydride across the opposite face of the resultant ketone (acetophenone in this case) produces the other enantiomer.


Meaning of 1,1,1,2 in 1,1,1,2-Tetrafluoroethane & similar

The numbers indicate which carbons the substituents - the fluorine atoms - are on. The root of the molecular name is "tetrafluoro," indicating the presence of four fluorine substituents. So now where are they? That's what the numbers tell us; the "1,1,1" part indicates that there are three fluorine substituents on the first carbon, and "2" tells us there is one more fluorine substituent on the second carbon.

Which is the first carbon and which is the second carbon? The numbering scheme should be selected such that it gives the lowest number at the first point of difference.

This way, we can write "1,1,1,2" to indicate the presence of the four fluorine substituents rather than "1,2,2,2."


In Chemistry, is it possible to use the old representation of the atom as a nucleus surrounded by orbiting particles?

When the fixed-orbit particle model is taught in general chemistry classes, it is usually presented as Bohr's fixed circular orbit model.

However, this was not the culmination of the deterministic-orbit particle model.  In 1916 Sommerfeld expanded the model to include precessing elliptical orbits, and made the theory relativistic. The varying of the eccenticity of the ellipses permits multiple angular momentum values for a given energy.

The Sommerfeld relativistic model correctly predicts the fine structure of the hydrogen atom.

In this one aspect, the deterministic-orbit particle model is superior to the non-relativistic Schrodinger model.  The Sommerfeld model correctly predicts the fine structure, whereas the Schrodinger model does not.


Why does gaseous HCl not change dry blue litmus paper to red?

This is because gaseous HCl  does not dissociate to give up protons and chloride ions.
But, in aqueous solutions HCl does dissociate and exist in ionic form:

$$\ce{H2O(l) + HCl(aq) -> H3O+ (aq) + Cl−(aq)}$$

Thus it acts as a typical Brønsted acid (proton donor) in an aqueous environment. In this case, water is gaining a proton $\ce{(H+)}$, so it is a base, while HCl is giving one away, so it is an acid.

Therefore it is able to turn the blue litmus paper red (a behaviour typical of acids).

For related topic please see: How does the litmus pH indicator work? as suggested above in comments.


If you have a molecule or ion with multiple sites that can be protonated (as you do here), and the sites can interact with each other (as they can here), the affinity for protons drops rapidly as more sites are protonated.

The proton affinity of the ion in your second reaction is a couple of orders of magnitude less than in your first reaction because of this effect.

Piling a third proton on there would be far less favorable than that---oxalic acid is a weak acid, and you're forcing it to act as a base by making it accept a proton1. You should stop with the two reactions you have.

It's really the same thing with hypochlorite ion; putting one proton on is fine (you get HClO, which is a weak acid). Putting a second proton on is tough, because you'd be making an acid act as a base.

1. This kind of thing does happen, though such protonated acids will quickly lose their extra proton to the surrounding water molecules. If you're using glacial acetic acid as a solvent, your acid ion is $\rm CH_3COOH_2^+$ instead of $\rm H_3O^+$,
and $\rm CH_3COO^-$ instead of $\rm OH^-$.


IUPAC Recommendations for Concentration Units (ppb, ppm)

There are not this kind of problems with the IUPAC recommendation. The only point in it is to specify the parts of  what we refer to, nothing more.
In such a sense it seems clearer than expression as molality or molarity.

Take an amount of atmosphere at given conditions. The concentration of Ar or whatever component can surely be expressed in terms of mole number in a million or whatever number of moles that amount contains.

As for the rock analysis case in the question, note that IUPAC umol/mol is an example, it doesn't mean it should be preferred to mg/kg (or ul/l etc.). The measured amount remains the most convenient both practically and mathematically.

I add that following recommendations would - perhaps - also help clear another practical con of unspecified ppm and percentage concentration values.  Often people forget that 1 l --> 1 kg of solution is acceptable only with very dilute solutions and when the solvent is water and go around telling and even publishing % concentration in various solvent so that the reader must assume a wrong procedure in order to follow up. Explicitly saying mg/kg instead of ppm should evidence that when mg starts to be hundred, than one has to switch concentration units or being more careful when reporting.


Try drawing the structure of the conjugate base $\ce{CF_3COO^-}$. In general, the more stabilized the negative charge of the conjugate base is, the more the equilibrium favors that form, thus the more the acid dissociates, thus the "stronger" the acid is.

So for a stronger acid (lower $\ce{pK_a}$), the negative charge must be more stabilized. When comparing $\ce{CF_3COOH}$ (TFA) to $\ce{CH_3COOH}$ (we'll use acetic acid to represent a "typical" carboxylic acid), TFA is stronger than acetic acid because its conjugate base can better stabilize the negative charge. If we draw the structures of the conjugate bases (I encourage you to do so), we will see that they both can stabilize the negative charge via resonance in the carboxylate functional group. However, TFA also has three highly electronegative fluorine atoms which withdraw electron density "through the single bonds" via induction.

As stated above, the $\ce{-CF_3}$ moiety is an electron withdrawing group, by the inductive effect. So our negative charge (rather, the electron density it represents) will be slightly drawn through the bonds toward the fluorines, thus it is further delocalized and thus more stable.

Note that we do not have convenient structural diagrams to show induction like we do resonance, but you should be able to imagine the electrons "sloshing" more towards one side of a bond/functional group. It's the same concept as a good old polar bond like $\ce{H-Cl}$; we say Cl "hogs" the electron density and pulls it away from H. In the same way $\ce{-CF_3}$ will draw electron density toward itself.

So to summarise, in general:

Stronger acid ~ more stable conjugate base. Whenever you want to qualitatively rationalize the relative strengths of two acids, draw their conjugate bases and determine which is more stable. (If your acid was neutral, its conjugate will be negative, and it will be easier to use this method than for a positive acid/neutral conjugate.) Stabilizing factors are size and electronegativity of the atom bearing the charge, resonance delocalization of the charge, hybridization of the orbital holding the charge, induction delocalization of the charge, and hyperconjugation which donates electron density and may destabilized a negative charge. SERHIH, roughly in that order of significance.

In this example, TFA and acetic acid are tied at size & electronegativity because both place the charge on an oxygen atom, tied at resonance because both may use the carboxylate functionality, tied at hybridization because the atoms in a carboxylate are necessarily $\mathrm{sp^2}$, but now induction is present in TFA and not acetic acid, so we expect TFA will be a stronger acid.


What is the mass fraction of trichloroethene in the liquid phase using vapour pressure and mole fraction in gas phase?

As far as the question is concerned, you do not need the vapor pressure. In order to calculate the mass fraction of $\ce{C2HCl3}$, we first have to calculate the molar mass of $\ce{C2HCl3}$ and $\ce{CHCl3}$.
$$M(\ce{C2HCl3}) = \pu{131.38 gmol-1}$$
$$M(\ce{CHCl3}) = \pu{119.37 gmol-1}$$
With these two calculated we can calculate the absolute mass relative to $\pu{1 mol}$, due to the fact of $\frac{\pu{0.73 mol}}{\pu{1 mol}}$ being relative to a total of $\pu{1 mol}$ of molecules. Now if we calculate the absolute masses, we get:
$$M(\ce{C2HCl3}) = \pu{131.38 gmol-1} * \pu{0.27\frac{mol}{mol}} = \pu{35.47 gmol-1}$$
$$M(\ce{CHCl3}) = \pu{119.37 gmol-1} * \pu{0.73\frac{mol}{mol}} = \pu{87.14 gmol-1}$$
And the total mass in the system results in:
$$m_\text{total}=\pu{35.47 gmol-1}+\pu{87.14 gmol-1}=\pu{122.61 gmol-1}$$
And with the total relative mass given, we can calculate the mass fraction of $\ce{C2HCl3}$, as follows:
$$w(\ce{C2HCl3})=\frac{\pu{35.47 gmol-1}}{\pu{122.61 gmol-1}}=0.29$$
$$w(\ce{CHCl3})=\frac{\pu{87.14 gmol-1}}{\pu{122.61 gmol-1}}=0.71$$


Possible carbocation rearrangement in substitution reaction to give two different products

Is compound I formed by $\mathrm{S_N}$2 and compound II formed via $\mathrm{S_N}$1 after a carbocation rearrangement? Or is there some rearrangement I'm not aware of?

For the first question, the answer is probably not because the mechanism of a given reaction is mainly based on the condition used in the reaction (at least at the undergraduate level of teaching). The the answer for the second question is yes, you are unaware of the mechanisms called $\mathrm{S_N}$1' and $\mathrm{S_N}$2' (as Shoubhik R Maiti mentioned in the comments section). This type of reactions are common when the allylic compounds are the substrate, which is appropriately called allylic rearrangement or allylic shift. Allylic shift is possible under either $\mathrm{S_N}$1 or $\mathrm{S_N}$2 conditions, and commonly known with $\mathrm{S_N}$1' and $\mathrm{S_N}$2' notations, respectively.

Strictly speaking, if you have used condition favoring $\mathrm{S_N}$1 reaction, the intermediate is a resonance stabilized allylic carbocation. Thus nucleophile would attack at two positions (either carbon 1 or 3) to give the resulting products, I and II. For instance, see your reaction in a protic solvent ($\mathrm{S_N}$1), which would given the same type of product mixture (for convenience, I have used the labelling to show the difference):

Molecular modeling calculations would shows you the allylic secondary carbon position is more stable than the allylic primary carbon (recall your ab initio calculations for kinetic versus thermodynamic control in Org.Chem.II lab course; a simple and effective lecture demonstration of thermodynamic versus kinetic control, see: Ref.1) in intermediate carbocation. Accordingly, compound II would be the major product.

Similarly, if you are using the condition favoring $\mathrm{S_N}$2 reaction (strong nucleophile in a polar aprotic solvent) in this type of reaction, it is also possible to get two product due to $\mathrm{S_N}$2' mechanism. Usually, a bulky leaving group in $\mathrm{S_N}$2 conditions or bulky non-leaving substituent on the framework which give rise to significant steric hindrance would thereby increase the conjugate substitution ($\mathrm{S_N}$2'). Following scheme show the mechanism for two products:

Late Edit:
Since it seems OP is still cannot see how compound II is formed, I included the arrow-pushing mechanism for $\mathrm{S_N}$2 versus $\mathrm{S_N}$2' products:

References:

I. J. McNaught, "Thermodynamic versus kinetic control: A lecture demonstration," J. Chem. Educ. 1978, 55(11), 722 (https://doi.org/10.1021/ed055p722.1).


How Did Wittig Come to Discover Lithium-Halogen-Exchange?

Georg Wittig, discoverer of the Wittig reaction, was studying the effect of ring strain on the strength of two neighboring carbons; by adding enough strain he expected to coax C-C bonds into forming diradicals:

Unfortunately tetraphenylbenzocyclobutane rearranges to triphenyldihydroanthrancene instead of forming a diyl. This and other “failures” did not deter Wittig; they elected to continue by replacing phenyl groups with anisyls. To their surprise, reactions with anisyls gave unexpected products. They focused their efforts on a simple derivatization with benzil to characterize the major product. They were trying to synthesize $p$-anisyldiphenylcarbinol, but instead what happened is:

Continuing along they discovered phenyllithium would perform the same reaction. As part of their investigation they varied other substituents on the aryl halide and discovered an even more peculiar product which violated their chemical intuition. Wittig was the first to speculate a lithium-bromide exchange reaction in 1938, below is the reaction he and coworkers carried out in diethyl ether giving a high yield (95%), the preferred product led them to infer it went through an aryllithium intermediate:

This same reaction occured with iodine, bromine, and chlorine. Yet, fluorine gave products via an entirely different mechanism. This led them to postulate an aryne intermediate. They then set off in a new direction, by trying to weaken carbon-hydrogen bonds with a neighboring quaternary ammonium, this led them to discover N-ylides. It was found this more readily occurred with phosphonium, where they discovered P-ylides. Now you can see the rough outline of the path required to discover the Wittig reaction.

How Does Lithium-Halogen Exchange Work?

This initial discovery by Wittig and then Gilman encouraged further efforts because it allowed the formation of a number of previously inaccessible organometallics. In the above reaction, the Grignard reagent cannot simply replace phenyllithiun (some Mg cases have been reported for other halides, but Li is the stalwart metal).

Today this reversible lithium-halogen exchange reaction can be formulated as,
$$\ce{RLi + R’X <=>[{K}]  R’Li + RX}.$$

As an equilibrium process, we know that the more stable lithium species is preferred. Interestingly when $\ce{R’}$ = aryl (particularly bromides and iodides) this is best written as, $$\ce{RLi + R’X <=>> R’Li + RX}.$$

When $\ce{R’} = $ allylic or benzylic, Wurtz coupling is favored. $\ce{R’} = $ alkyl tend to give unsatisfactory results. In general, the equilibrium favors those species which most stabilize the carbonanion intermediate. Depending on the specific halide, side-product formation is a problem: $\beta$-elimination, $\alpha$-metalation, or Wurtz-type reactions can occur.

Gilman discovered that alkyllithiums, for example t-butyllithium, were more adaptable in their reactivity than aryllithiums, which explains their widespread success in the modern chemical literature.

How the reaction proceeds remains an unresolved problem, however, there is evidence to suggest the following mechanisms are at play:

(1) intermediates proceeding through an “ate” complex

Perhaps unsurprising to you by now, Wittig also hypothesized the existence of an "ate" complex intermediate:

This is in other words, a nucleophilic attack of the halide by the alkyllithium with a carbonanion leaving group.

(2) radical mechanisms

A one electron transfer from the nucleophile to halide thereby producing a radical intermediate that then can give rise to a number of products. There are data from EPR and radical caging experiments to suggest radicals at least form.

(3) mixtures of (1) and (2)

As with many systems, the truth is both mechanisms compete with each other; of course which pathway, if any, dominates will depend on the solvent, temperature, substrate, and certainly the lithium species used.

tert-Butyllithium requires 2 equivalents, resulting in a non-equilibrium process where: $$\ce{RX + (Me)3CLi -> RLi + (Me)3CX}$$ $$\ce{(Me)3CX + (Me)3CLi -> (Me)3CH ^ + (Me)2C=CH2 ^ + LiX}.$$


Mistake in Wikipedia article on fugacity?

I think you have misinterpreted the passage on Wikipedia a little. Of course, you are right that the chemical potential is given by the formula

\begin{align}
 \mu_{i} (p_{i}, T) &= \underbrace{\mu_{i}^{*}(p^{0}, T)}_{= \, \mu_{i}^{0}} + RT \log a_{i} \\
 &= \mu_{i}^{0} + RT \ln \left(\phi_{i} \frac{p_{i}}{p^0} \right)
\end{align}

where $R$ is the universal gas constant, $T$ is the temperature, and $\phi_{i}$, $p_{i}$, $a_{i}$, and $\mu_{i}$ are the fugacity coefficient, the partial pressure, the activity and the chemical potential of the $i$th component in the system, respectively.
But the passage you are quoting says that the contribution of nonideality to the chemical potential of a real gas is equal to $RT \ln \phi$ and in the equation

\begin{align}
 \mu = \mu_{\text{id}} + RT \ln \left(\phi \right)
\end{align}

$\mu_{\text{id}}$ is not the same as $\mu^{0}$. Rather it is meant in the following way: For an ideal gas the fugacity coefficient is $1$ by definition and so the chemical potential of an ideal gas, $\mu_{\text{id}}$, is given by

\begin{align}
 \mu_{\text{id}} = \mu^{\text{id}}_{i} (p_{i}, T) = \mu_{i}^{0} + RT \ln \left(\frac{p_{i}}{p^0} \right)
\end{align}

If you use the mathematical property of logarithms that $\log(a \cdot b) = \log(a) + \log(b)$ you can rewrite the equation for the chemical potential of real gases to

\begin{align}
 \mu_{i} (p_{i}, T) &= \underbrace{\mu_{i}^{0} + RT \ln \left(\frac{p_{i}}{p^0} \right)}_{= \, \mu_{\text{id}}} + RT \ln \left(\phi_{i} \right) \\
 &= \mu_{\text{id}} + RT \ln \left(\phi_{i} \right)
\end{align}

and so you see, that the contribution of nonideality to the chemical potential of a real gas is indeed equal to $RT \ln \phi$ and you get to the equation from the Wikipedia article.


Do non-ergodic chemical reactions exist?

The time point of the radioactive decay of a single atom cannot be predicted, even given all experimentally available information, and is not even predictable theoretically by any possible means.

Just isolate a single atom and observe its radioactive decay, and you have the purely random system you strive for. Could also take more than one atom and observe fluctuations with Geiger counter.


Well, actually when we evaporate aqueous chromium(III) chloride, the hydrate of the crystal will be gone, so we get anhydrous chromium(III) chloride ($\ce {CrCl3}$). This type of anhydrous produce violet color.

But if we crystallizes aqueous of the chromium(III) chloride, the hydrate won't go, so we get hydrous chromium(III) chloride (usually hexahydrate, like $\ce {CrCl3.6H2O}$). This crystal produce dark green color.


How do I remove tetrabutylammonium perchlorate from an organic synthesis?

It's tempting to use phase separation with an aqueous phase, but you're clearly having issues.  One option is to evaporate the volatile solvents off, add toluene, cyclohexane or heptane and try water or another additive to boost the ionic strength. The approach using NaOH suggests this may not be successful.

A less conventional approach may be to partition between two organic solvents e.g. acetonitrile or methanol with heptane or methylcyclohexane. The solubility of TBAP appears good enough in alcohols to dissolve well in that phase, while your amide seems more likely to have less polar solvent solubility. Your methanol-MeCN mix may convolute things,  so maybe try and strip off the methanol (and likely some MeCN will be lost, but you can top up) or if all else fails,  evaporate both solvents and redissolve from scratch.


How can I define alkenes using graph theory?

Perhaps you could just represent them as simple weighted graphs? Edges of weight 1 represent single bonds, while edges of weight 2 represent double-bonds.

*Edit: For condition #3 simply pick a canonical ordering for the vertices/atoms. See

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

http://depth-first.com/articles/2006/08/12/inchi-canonicalization-algorithm/

for examples


In the alpha helix structure as well as the beta pleated structure, the atoms are arranged in such a way that the number of hydrogen bonds is maximum. H bonding is a stabilizing factor, and consequently, any molecule (or polymer) would like to maximize their number. This answers your question as to why the protein would fold into one of these shapes.
Coming to why alpha and beta structures are not easily interconvertible, it is once again because of hydrogen bonding. The H bonds between H and electronegative atoms like oxygen and nitrogen are difficult to break, and hence we cannot simply smash an alpha helix into a beta pleated structure, since in doing so, we would inevitably have to break some hydrogen bonds. Thus, whether a protein exists in alpha or beta form depends on how it was formed.


Why is the equilibrium constant being calculated differently here?

It's a matter of definition, really.  Equilibrium implies a reversible reaction, thus the same reaction could he rendered as $\ce{A<=>B}$ or$\ce{B<=>A}$.  So we can render the equilibrium constant as $[\ce{B}]/[\ce{A}]$ or as $[\ce{A}]/[\ce{B}]$.  We need to define which is used, of course, when we give a numerical value.


Near-IR spectroscopy measures the broad overtone and combination bands of some of the fundamental vibrations (only the higher frequency modes)

Indeed that sentence should be taken as merely anticipating what the book will treat in following chapters.

Nevertheless, to satisfy your impelling curiosity the statement trace a combination of facts that leads to the actual spectrum of a given molecule.

Overtones IR absorption bands are those resulting from breaking of selection rules. Instead of having just transitions between adjacent vibrational levels, basically the 1<-- 0 one, transitions with a delta nu of 2 are observed as well. In other words the vibrations aren't those of an harmonic oscillator but those described by a Morse's potential.

Combination modes arise from the combination of normal modes, ie they happen when different vibrations are simultaneously excited. Normal modes which are unseen can contribute to these combination modes, as the coupling alter the symmetry.

High frequency: we are speaking about NIR, which is the high frequency range for molecular motions, and it is where vibrational overtones typically occur. The lower frequency part of the spectrum is dominated by the fundamental modes, and except for its fingerprints part and fine details, it has a strong bond related character rather than a molecular one.

If this does not help but instead brings out new terms or doubts, I would suggest that you treat the introductory part of the book for what it is, namely an introduction.

Alternatively you can refer to this link which provides a condensed explanation and includes schemes, further links, and definitions:

http://bit.ly/31Mc6lT


Can Ca and K exist as an ionic compound?

As a creator of this memetic illustration, I feel obliged to post a brief answer.

Potassium is known for forming several metal alloys, such as a widely used Sodium-potassium alloy NaK.
However, CaK doesn't exist [1]:

  The Ca-K system was found by Klemm and Kunze [67Kle] to exhibit virtually complete immiscibility in both solid and liquid states. No compounds were found by X-ray diffraction.

  Cited Reference
  *67Kle: W. Klemm and D. Kunze, "Systems of Alkali and Alkaline Earth Metals," Proc. Int'l. Symp. on Alkali Metals, London Chem. Soc., Special Publ. No. 22, 3-22 (1967). (Equi Diagram; Experimental; #)

References

Pelton, A. D. The Ca−K (Calcium-Potassium) System. Bulletin of Alloy Phase Diagrams 1985, 6 (1), 34–34. DOI: 10.1007/BF02871178.


Why second ionisation energy greater than the first?

The ionisation energy basically measures how tightly an electron is held. That in turn is determined by a couple of factors, namely:

Nucleus-electron attraction. If this increases, the ionisation energy increases.
Electron-electron repulsion. If this increases, the ionisation energy decreases.

More formally these can be quantified by $Z$, the nuclear charge, and $\sigma$, the shielding parameter. The effective nuclear charge, which is hopefully a term you have come across, is defined as

$$Z_\mathrm{eff} = Z - \sigma$$

and the larger $Z_\mathrm{eff}$ is, the larger the ionisation energy. You don't necessarily need a formula to prove this, as it should be pretty intuitive. Larger effective nuclear charge $\implies$ electron is held more tightly $\implies$ more energy needed to ionise. But here is the mathematical relationship:

$$\text{IE} \propto \left(\frac{Z_\mathrm{eff}}{n}\right)^2$$

As long as you are staying within the same subshell, i.e. the principal quantum number $n$ is not being changed. So, the ionisation energy is determined by $Z_\mathrm{eff}$.

When you remove an electron, you decrease the electron-electron repulsions; this means that $\sigma$ decreases. As a result $Z_\mathrm{eff}$ decreases and the ionisation energy decreases. (Technically, there would be an increased attractive force ($\propto kq_1q_2/r^2$) from the nucleus, because the electrons on average will now be closer to the nucleus. However, within this inorganic chemistry framework, the nuclear charge $Z$ is simply the number of protons in the nucleus and therefore remains constant.)


Terminal alkynes, including phenyl acetylene, react with iodine (in MeOH) to give 1,2-diiodoalkenes according to this JOC paper  here. Presence or absence of base will not alter the outcome. Presence of AgNO3 gives some di-iodo ketone which can then give rise to the iodoform reaction.


Is it possible to make micelles to pick up Microplastics from water?

Lyophobic means solvent-hating, and here the solvent is water, so the lyophobic ends (presumably the tails, if we are discussing a normal detergent) will associate to form the hydrophobic core of the micelles. The headgroups meanwhile will preferentially interact with water, forming an interface between solvent and core.

If you add detergent to a solution containing charged plastic particles, the detergent tail is unlikely to interact with the particle unless the particle also displays amphiphilic character. More importantly, headgroups with the same charge as the particle surface will discourage association with the particle. So your proposal lacks a viable aggregation mechanism for at least one reason (charged and hydrophobic regions are not likely to associate) and possibly for a second (oppositely charged regions of detergent and particle will repel). So you need to research carefully:

the expected charge of the particles
the choice of detergent, ideally with a headgroup with charge opposite that of the particle, and which resists precipitation due to the hard water but will encourage flocculation of the particles
scalability, which is a question regarding the volume of water you intend to treat

There is always the possibility that the solution contains ions resulting in charge inversion of the charged particle, another factor to consider.

You'd be well served to look further into the steps used to treat wastewater.


Why does lactose-free milk last so much longer than regular milk?

Apparently the process used by manufacturers (ultra-pasteurization) to produce lactose-free milk is responsible for the longer shelf life.

One source reports:

  Lactose-free milk is pasteurized at a higher temperature than regular milk. The process, known as ultra-pasteurization, is designed to remove the bacteria content entirely, giving lactose-free milk a refrigerated shelf-life of 60-90 days, compared with regular pasteurized milk, which retains some bacteria. It has a shelf life of 1-3 weeks.

A producer (in the United States) of organic dairy products (Organic Valley) describes ultra-pasteurization and the extended shelf-life here as:

  Ultra Pasteurization (UP), or Ultra High Temperature (UHT) pasteurization, is the process of heating milk to approximately  280 °F for just 2 seconds and then chilling it back down rapidly.  The result is milk that's 99.9% free from bacteria.  This pasteurization process creates an extended shelf-life for milk products of up to three times the length as HTST pasteurization; while providing the same wholesome, quality dairy product.  This allows us to distribute UP milk products regionally as well as to other areas of the country that might not have access to our dairy products. This is referred to as 'ultra pasteurized' on the milk package.  Organic Valley offers its quarts and some of its half gallons of milk with Ultra Pasteurization (UP).


How does "breathable" siloxane water sealant work?

According to How silicone water repellents work,
"Because silicones have a very open molecular structure, water vapor trapped in the substrate can easily slip through. Water molecules, however, are too large to enter."

If I remember correctly, siloxanes' Si-O bonds are not intrinsically hydrophobic, but the long molecular chains with outward-facing hydrophobic methyl groups, and thermal molecular motion, prevent liquid water from seeping through, which presents less of a barrier to water vapor and other low-molecular-weight gases.


Why does an increase in reduced mass lead to a decrease in frequency?

Consider a simple molecule $\ce{A-B}$ composed of atoms $\ce{A}$ and $\ce{B}$ and let the bond connecting them be described as a spring.  Then the stretching and contracting of the $\ce{A-B}$ bond (spring) can be considered as a simple harmonic oscillation and is described by Hooke's law.
${\nu = \frac{1}{2 \pi} \sqrt{\frac{k}{m}}}$
where ${\nu}$ is the frequency of the stretch,  ${k}$ is the force constant describing the specific spring and ${m}$ is the mass of the system.  In our $\ce{A-B}$ example, the mass of the system is described by the reduced mass ${\mu}$, where
${\mu = \frac{m_A m_B}{m_A + m_B}}$
So our equation becomes
${\nu = \frac{1}{2 \pi} \sqrt{\frac{k}{\mu}}}$
From this equation we can see that as the reduced mass increase's the stretching frequency will become smaller.


Why do chiral biological molecules only exist as one enantiomer? Does it have any advantage?

Are all the biochemicals that our body uses enantiomerically pure or are racemic mixtures too?

Many molecules exist in both forms in nature. One fun example are the enantiomeric terpenoids R-(–)-carvone and S-(+)-carvone. The R-form smells like spearmint while the S-form smells like caraway. The difference in smell shows that properties other than the optical activity are different for two enantiomers.

  Why, then, is one enantiomer preferred over the other?

Nucleic acids

The_Vinz stated correctly in the comments that for replicating structures like RNA, choice of the enatiomer was random; once established, one chiral form prevailed. DNA building blocks have the same chirality as those of RNA because they are made by the same biochemical pathway.

Amino acids

One amino acid, glycine, is not chiral. Many amino acids are found in both forms. L-amino acids are used to make proteins, but D-amino acids are made in bacteria and used in the context of cell walls and natural antibiotics.

Proteins

Proteins made by ribosomes (from amino acids attached to tRNA by tRNA-synthetases) use L-amino acids exclusively. The tRNA-synthetases are highly specific (including stereo specific), and they don't link tRNA to D-amino acids. It helps that there are very little D-amino acids made in a typical cell. Why one form was chosen over the other is probably luck again. Why all amino acids have the same chirality at the alpha carbon is more intriguing. Some are made from the same precursor, so that will contribute. Right-handed alpha helices require that the amino acids in them be L-amino acids. If proteins had a mixture of L- and D-amino acids (e.g. all alanines are D-alanines but all aspartates are L-aspartate), alpha helices would be more constrained in the possible sequences, and some of them would be left-handed.

All other molecules

Most steps in the synthesis of biomolecules are catalyzed by enzymes. Enzymes, as chiral catalysts that have lots of interactions with reactants, are often highly stereospecific. So the presence or absence of enzymes catalyzing certain reactions largely determines which products are made, and there is no additional cost of making a chiral product from non-chiral precursors (very different from a typical lab synthesis).


This type of consecutive reaction does not really come to steady state for very long, as seen in the figure, i.e there is only a small time over which $d[mTC]/dt \approx 0$.  The decay rate of mTC is faster than that of Mo and so mTc appears quickly and decays slowly. (The amount of mTC is multiplied by 10 in the figure is the amount formed is small].

The figure is drawn assuming  $k_1 = \ln(2)/65, \, k_2 =\ln(2)/6 $ and $\ce{[Mo]_0} = 1$ and that the other species are zero at $t=0$ The equations are
$\displaystyle \ce{[Mo]=[Mo]}_0e^{-k_1t}$ and
$$\displaystyle \ce{[mTc] = [Mo]}_0\frac{k_1}{k_1-k_2}\left( e^{-k_1t}-e^{-k_2t}  \right)$$ The amount of Tc is found as $\ce {[Tc] = [Mo]_0-[Mo]-[mTc]}$

The maximum amount of mTC, which is assumed to be 'steady state'  is found by differentiating.

The time at maximum is $\displaystyle t_m= \ln(k_2/k_1)/(k_2-k_1)$ which is 22.7 hours so the 'steady state' is established just before this and lasts a little past this time. How much in either way is somewhat subjective unless you use a definition of say $\pm 5$ % of the maximum etc.


Can ethyl acetate isolate cinnamaldehyde from cinnamon oil?

Yes, ethyl acetate can be used to isolate cinnamaldehyde from cinnamon oil, though there are better non-haloalkane options. In a study that aimed to optimize the extraction of cinnamaldehyde from cinnamon powder, it was found that from methanol, ethanol, ethyl acetate, and water, methanol was the most efficient. The exact method used in the study was:

Collection of plant material: Bark of cinnamon was collected from a local market at Visakhapatnam, AP. The bark was cleaned and dried under sunlight for $24\ \mathrm{h}$. The dried bark was powdered and used as a raw material and stored in an air tight container. Cinnamon powder was sieved by using different particle sizes ranging from 354 to 125 microns.
Preparation of the extract: Cinnamon powder ($2\ \mathrm{g}$) was added with ethanol ($25\%$) and methanol ($25\%$) in different flasks and the volume was made $50\ \mathrm{mL}$. The solution was soaked for $1\ \mathrm{d}$ and $3\ \mathrm{d}$ respectively. After the soaking time, the solution was filtered using Whatman No.1 filter paper and the filtrate solution was heated to $78^\circ \mathrm{C}$ and $65^\circ \mathrm{C}$ and made up to $50\ \mathrm{mL}$ with distilled water and hexane and incubated for $2\ \mathrm{h}$.


Determination of timescales using characterization methods

The absorption of a photon en route to an electronically excited Franck-Condon state typically typically proceeds in the femtosecond range.

The lifetime of an electronically excited singlet state mostly is in the range of several nanoseconds. In some cases, however, fluorescence lifetimes in the range of several tenth of nanoseconds have been observed.

In cases where the initial excited singlet state undergoes intersystem crossing to a triplet state, much longer phosphorescence lifetimes (seconds range) are observed.

In order to detect a transient, its lifetime must be longer than the laser pulse needed to generate it.

Once you've populated a state, you may either use

time-resolved absorption spectroscopy to record it's spectrum and the decay of the absorption at a particular wavelength over time or
use time-resolved emission spectroscopy to measure the decrease of fluorescence (or phosphorescence emission) intensity at a particular wavelength over time.


A lone pair on the ring oxygen forces Chlorine leave. The anomeric carbon in the resulting oxocarbenium ion is sp 2 hybridized, so that part of the molecule is planar. (An oxocarbenium ion has a positive charge that is shared by a carbon
and an oxygen.)

When the alcohol approaches from the top of the plane, equitorial isomer is forrmed. Similarly if alcohol apporoches from bottom of sp2 hybridised oxo carbenium ion , then ,axial isomer is formed .So the given chloro compound should give two isomers (as shown in figure below 2& 3)..

Which is the major product ?

Substituents on a cyclohexane ring prefer the equatorial position to relieve steric strain. cyclohexane ring in chair form with substituent in equitorial position ring-flipps to chair in which the substituent is axial.

When an endocyclic heteroatom ( oxygen, nitrogen, or sulfur) is  at a position adjacent to the methoxyoxy group,the hetroatom substituent now prefers the axial orientation (despite of the associated steric strain).The equilibrium preference of most such compounds for the alpha-isomer is referred to as the anomeric effect.

Such behavior is explained as

In α-methyl glucoside, the non bonding HOMO with a pair of electrons on the ring oxygen is antiperiplanar to the antibonding LUMO of C-O bond in methoxy group. This allows hyperconjugation between them and thus by stabilizing the α-form.
In axial position,such interactions are possible.Whereas, in equitorial the methoxy group is at equatorial position and cannot involve in hyperconjugation since it is not antiperiplanar to the lone pair on ring oxygen. Therefore  equitorial isomer is less stable than the axial isomer.

Other Explanations:

The  dipole interactions of the two ether functions are  in the equatorial conformer are nearly parallel;while, in the axial conformer these dipoles are oriented in opposite directions.

image source :https://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/carbhyd2.htm#anomr

Electrostatic Repulsion: Decreased electrostatic interactions are favored

All the above combination of factors result in equitorial isomer being less stable than the axial isomer.

Referance

https://www.scripps.edu/baran/images/grpmtgpdf/Krawczuk_Nov_05.pdf

Organic Chemistry ,SEVENTH EDITION , Paula Yurkanis Bruice


The concentration of your acid is given as a mass concentration.

Without knowing the molar mass, the molar concentration and the amount of substance in a volume of $15~\mathrm{cm}^{-3}$ are not accessible.

However, knowing that

$\mathrm{1~L = 1~dm^{-3} = 1000~cm^{-3}}$

and therefore

$15~\mathrm{cm}^{-3} = 0.015~\mathrm{dm}^{-3}$

the mass of the acid (in $15~\mathrm{cm}^{-3}$) is given as

$m = 10.95~\mathrm{g\,dm^{−3}\cdot0.015~\mathrm{dm}^{-3} = 0.16425~g}$


What is the significance of heating value?

Note that the mentioned values are molar heats of combustion. Propane is a larger molecule than methane and it is therefore obvious that more energy per mole is released.
If we replace the molar amount by molar mass, the resulting specific heat of combustion has the opposite relation:
$$\begin{align}\text{Methane:}&&\frac{\pu{890kJ/mol}}{\pu{16g/mol}} &=  \pu{55.6kJ/g}\\[0.5em]
\text{Propane:}&&\frac{\pu{2200kJ/mol}}{\pu{44g/mol}}&=\pu{50kJ/g}\end{align}$$
For the enviromental aspect of $\ce{CO2}$ emissions, the relevant value is the heat of combustion per mole of carbon atoms. It is $\pu{890 kJ/mol}$ for methane, but $\pu{2200 kJ/mol}$ for propane, which is roughly $\pu{730 kJ/mol}$ for carbon. For methane, both numbers are the same, as a propane molecule contains 3 carbon atoms, while a methane molecule just 1.
So methane has a an advantage as energy source, as it has higher energy production per the same $\ce{CO2}$ emission.
PLA is partially already oxidized, therefore it has significantly lower energy content per 1 kg of produced $\ce{CO2}$, compared to  PP.
PLA has an energy content of $\pu{16.3MJ/kg}$
PLA has structure [-O-CH(CH3)-CO-]n, with basic unit containing 3 C + 4 H + 2 O
$\pu{1 kg}$ of PLA contains 36/72 kg of carbon
PLA has an energy content of $\pu{16.3 * 72 / 36 MJ/kg} \simeq \pu{32.6 MJ/kg of C}$ .
PP has an energy content of $\pu{46.4MJ/kg}$
PP has structure [-CH2-CH(CH3)- ]n, with basic unit containing 3 C + 6 H
$\pu{1 kg}$ of PP contains 36/42 kg of carbon
PP has an energy content of $\pu{46.4 * 42 / 36 MJ/kg} \simeq \pu{54.1 MJ/kg of C}$ .
To get the same energy by the burning of the stuff,
PP releases $54.1/32.6 \simeq 1.66$ times less $\ce{CO2}$ than PLA.
Or, with the same emission of $\ce{CO2}$, PP releases 1.66 times more energy than PLA.


Bolt head corrosion in soup. Toxic? Zinc? Cadmium?

You can test for cadmium at home, but the ~US$30 price for the test kit (with 1,5-diphenylcarbazone) is probably more than the soup is worth, and the test might not be sensitive enough to detect sub-acute toxic amounts of Cd.

Considering the highly toxic nature of the metal, you might check with your family physician if you and anyone else sampling the soup should have a blood or urine test.

As has been mentioned, the small amount of zinc that might have been absorbed had the bolt been galvanized (unlikely, anyway) is not significant. The US NIH recommends ~10 mg of Zn in the diet per day as a nutrient.


Do weak acid/weak base neutralisation reactions go to completion?

Weak acids and bases do indeed remain in an equilibrium state with a certain concentration of the free acid and the free base alongside a certain concentration of the deprotonated acid anion and the protonated base cation.

In general, you can consider any acid-base reaction to be essentially the following:

$$\ce{HA <=> H+ + A-}\tag{1}$$

Which in turn means that the $\mathrm pK_\mathrm a$ value is defined as in $(2)$:

$$K_\mathrm a = \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}\tag{2}$$

Reactions $(1)$ and $(2)$ are usually written out in water with hydronium in place of a naked proton but the same principle applies. Of course, a base reaction can be considered essentially the reverse, meaning that the acid constant typically given is actually $\mathrm pK_\mathrm a (\ce{HB+})$ rather than anything directly corresponding to the free base. If we now take the reaction of a (weak) acid and a
(weak) base, we get equation $(3)$ and the equilibrium constant as given in $(4)$:

$$\begin{align}\ce{HA + B &<=> A- + HB+}\tag{3}\\[0.7em]
K &= \frac{[\ce{A-}][\ce{HB+}]}{[\ce{HA}][\ce{B}]}\tag{4}\end{align}$$

We can now perform simple mathematics with $(4)$ to arrive at the modified equation $(5)$ as below:

$$\begin{align}K &= \frac{[\ce{A-}][\ce{HB+}]}{[\ce{HA}][\ce{B}]}\tag{4}\\[0.7em]
&= \frac{[\ce{A-}][\ce{HB+}][\ce{H+}]}{[\ce{HA}][\ce{B}][\ce{H+}]}\\[0.7em]
&= \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]} \times \frac{[\ce{HB+}]}{[\ce{B}][\ce{H+}]}\\[0.7em]
K &= \frac{K_\mathrm a(\ce{A})}{K_\mathrm a (\ce{HB+})}\tag{5}\end{align}$$

We have thus arrived at a way to determine the equilibrium constant of the proton transfer reaction just from the acidity constants of both participants. If both acids weak and have similar $\mathrm pK_\mathrm a$ values — as ammonium and acetic acid do — the equilibrium constant will be close to $1$ and thus both sides of the equation will have similar concentrations.

Using $(5)$, you can also predict a ‘degree of completion’ for any acid-base reaction.


What is the meaning of the "dot" notation in chemical formulas?

By writing $\ce{AB.xCD}$ chemists mean that there are CDs are found in the crystalline framework of AB. The most common example of this is water trapped inside the crystal structure of ionic compounds. (See water of crystallization in wikipedia)
An example that's often taught is $\ce{CuSO4.5H2O}$. See that 5 that's a representative of $x$? It means that water can be bound only with intermolecular interactions, but it exists in the structure in a stoichiometric ratio.

Source
Also, according to this article

In summary, when a dot is used to break a formula into subunits, it may signify ignorance of how the subunits are structurally related, as in our inorganic example; or it may correspond to actual significant structural subunits, as in our organic example; or it may represent the combining ratios of the binary starting materials required for the synthesis of the compound, as in our phase diagram example.

As a result, for example in direct addition reactions, where there is ignorance about the structural relation of the compounds, we use a dot.


Classification of the Inside of a Grapefruit

According to Wikipedia:

  In chemistry, a heterogeneous mixture consists of either or both of a)
  multiple states of matter or b) hydrophilic and hydrophobic substances
  in one mixture

The inside of a grapefruit is clearly made up of multiple states of matter (solid and liquid), and is thus heterogeneous. Text books sometimes contain errors, and that appears to be the case here.


How does base size affect the ratio of kinetic/thermodynamic enolate?

One problem with comparing product ratios using $\ce{LiN($i$Pr)2}$ vs $\ce{Ph3CLi}$ is that it's not really an apples-to-apples comparison - one is a lithium amide base, whereas the other is a alkyllithium that is highly stabilised by conjugation.

I think this illustrates two important points:

Things are more complicated than they first appear, ie. are the species we draw really representative of the species undergoing reaction?
Analysis of substrate and reagent structures will only get you so far, eg. for kinetically controlled processes, analysis of competing transition states is necessary.

Regarding the first point, there have been dozens of papers looking into the structure and reactivity of lithium amide and alkyllithium bases, particularly by the research group of David Collum. LDA is probably the most heavily studied. It turns out that these species are much more complex than what is implied by the structural formula. For instance, LDA in a coordinating solvent, such as THF, exists as a disolvated dimer:

How does the steric bulk of this species compare to the LDA structure we normally draw? Likewise, what is the actual structure of $\ce{Ph3CLi}$ in solution?

Even with a particular solvent and conditions, transition states have been put forward for the deprotonation event that vary markedly in structural features between substrates. For example, in a review of LDA (Angew. Chem. Int. Ed. 2007, 46, 3002-3017), Collum and coworkers show the following two different transition state structures for deprotonation of an ester using LDA in $\ce{tBuOMe}$ vs THF:

So, if we can get a good idea of what the transition state for LDA vs $\ce{Ph3CLi}$ deprotonation looks like, the complexity is such that high level calculations, eg. DFT, would be needed to explore the difference in selectivity. Even if we could qualitatively say which transition state involves the more bulky species, there may be features of the transition state, eg. conformational, stereoelectronic, that only become apparent through the modelling. Sometimes, it's not just which reagent is bigger, but details of how the two reactants come together at the transition state that determines the favoured outcome.


Why is carbonic acid a weaker acid than acetic acid?

Before you ask "why", ask "if".

Carbonic acid is not weaker than acetic. In fact, it is stronger, just as one might expect by looking at that extra electron-withdrawing substituent. The trouble is, carbonic acid is also pretty unstable. At any given moment, most of it exists as $\ce{CO2}$. This leads to much lower apparent acidity.

Think of it this way: you add some acetic acid to a carbonate salt. An equilibrium sets in, much closer to the starting compounds than to the products. In other words, you'll have just a very tiny amount of $\ce{H2CO3}$. But as soon as you have it, it starts decomposing, thus shifting the equilibrium more and more to the right.

Look again at that Wikipedia page, it says:

  3.6 ($\rm pK_{a1}$ for $\ce{H2CO3}$ only), 6.3 ($\rm pK_{a1}$ including $\ce{CO2(aq)}$)...


Residual hydroxyl groups in silicone resins

Silicones are produced by hydrolyzing chlorosilanes( in which chlorine atoms are replaced the hydroxyl groups). These hydroxyl groups undergo condensation polymerization with removal of water molecules to form polysiloxanes or silicones.

Now, residual hydroxyl groups in your question means that some hydroxyl groups has not undergone condensation yet. Solvent is evaporated and resin is heating resulting condensation of residual hydroxyl groups to form cured resinous film.


Effect on the partition function upon the mixing of two ideal gases

The fact that $Q_{\mathrm{initial}} = Q_{A}Q_{B}$ implies that the two subsystems, respectively consisting of gas A and gas B, are independent. Because $\log Q_{\mathrm{initial}} = \log Q_{A} + \log Q_{B}$, it follows that various extensive thermodynamic quantities (e.g., internal energy, free energies, entropy) of the total system are simply given by the sum of the corresponding quantities of the subsystems.

The partition functions are doubled because each gas occupies double the initial volume after mixing, i.e., $V \rightarrow 2V$.


Is greater relative AA battery capacity at high currents indicative of greater capacity at low currents?

This is a great question and I'm sorry it wasn't answered sooner. I'll give the caveat that I am not a card-carrying electrochemist, and I'm definitely not a battery researcher, even though I have worked with them in the past. I'll give my best effort.

As you note, in principal, the capacity of a battery is tracked with units of mA$\cdot$hr. In theory, this should be constant no matter the current draw.

That doesn't mean the capacity is uniform across a particular size/shape designation. The "AA" type specifies a particular shape/volume dimension, but different chemistry will give different electrochemical capacities. For example, I have plastic shells that you put a AA battery to fit "C" or "D" types - the downside is less electrochemical capacity (~2500 mA$\cdot$hr for a typical AA vs. ~8000 mA$\cdot$hr for a C cell).

One part of the electrochemical capacity is how much of the volume is filled with electro-active materials, rather than electrolyte, fillers designed to dissipate heat, or for other safety margins.

In practice, as seen in the showdown, nothing ever matches the theoretical capacity, and different batteries and different chemistries yield different practical capacities depending on the current draw.

Results @ 200 mA
Results @ 1000 mA

We see that the lithium cells get ~3000 mA$\cdot$hr under either current. The alkaline cells get ~2500 mA$\cdot$hr at low current, and much less under high current ~1500 mA$\cdot$hr.

So what's the difference?

First, as we mentioned, not every battery has the same capacity, even of the same battery type (size/shape/volume). Assume for the sake of argument that we have two 3000 mA$\cdot$hr batteries: one lithium (of some sort) and one alkaline.

Above a certain current threshold, lower currents will let batteries get closer to their theoretical capacities.

The problem is actually simple. Besides electrons, in every battery ions move through an electrolyte.

Alkaline cells use mobile $\ce{OH-}$ ions.
Lithium cells use mobile $\ce{Li+}$ ions.

At the end of the day, discharge requires ionic conductivity to match the electrical current. So which is likely to have a higher ionic conductivity? Well, $\ce{Li+}$ is smaller and usually has higher ionic conductivity.

In principal, at low current, an alkaline cell with the same capacity rating as a lithium cell, say 3000 mA$\cdot$hr should have about the same capacity. I'm not sure I've ever seen an alkaline AA battery with such high capacity -- they usually come around 2500-2700 mA$\cdot$hr theoretical capacity.

Finally, you ask if manufacturers "tune" the batteries. Put simply, yes. Besides safety considerations, different cells add compounds to minimize self-discharge, corrosion, shelf-life, etc. These can affect the amount of electrochemical capacity.

UPDATE

I realized I should clarify the very low current comment. If the current is low enough, then the rate of ionic mobility will be high enough for the chemical part of a battery to keep up with the electrical draw. So decreasing current draw will improve the capacity, up to this point - and then the real capacity should remain the same no matter how low the current draw.


Why does an OH bond absorb a higher frequency than a CH bond?

First, let me just put down the approximate $\ce{X-H}$ stretching frequencies (where, $\ce{X = O, C}$) just as a reference.

$$\begin{array}{c}
 & \ce{X-H}\ \text{stretch}\ \mathrm{(cm^{-1})}  \\ \hline
\ce{X = O} & 3600  \\ \hline
\ce{X = C} & 3000 \\ \hline
\end{array}$$

Now, as a simplification I will first discuss a diatomic molecule, thought of as a two balls of some mass $\mathrm{m}$ linked together with a spring. Solving, Schrodinger's equation for this system (called a harmonic oscillator), yields the following relation:

$\mathrm{\bar{\nu}} = \frac{1}{2\pi c}\sqrt{\frac{k}{m}}$

For, dissimilar masses, (say, $\mathrm{m_1}$ and $\mathrm{m_2}$), we can switch to the centre of mass coordinates, and the $\mathrm{m}$ in the preceding equation is replaced with the reduced mass ($\mu$) defined as,
$\frac{1}{\mu} \equiv \frac{1}{m_1}+\frac{1}{m_2} $

Consequently, $\ce{C-H,O-H}$ bonds in general have much higher stretching frequencies than do corresponding bonds to heavier atoms other than $\ce{H}$.The mass effect becomes evident when deuterated isotopes are examined. The stretching frequency of a free $\ce{O-H}$ bond is ca. $3600\ \mathrm{cm^{-1}}$, but the $\ce{O-D}$ equivalent is lowered to ca. $2600\ \mathrm{cm^{-1}}$. Since deuterium has a mass twice that of hydrogen, the mass term in the equation changes from 1 to 1/2, and the frequency is reduced by the square root of 2.

Naively, one would expect the frequency of the $\ce{C-H}$ bond to be higher, based on the preceding discussion, but there is another factor at play, the force constant $k$. This, is different for $\ce{O-H}$ and $\ce{C-H}$ bonds, and depends on the bond strength (stiffness of the spring)

Bond strengths can be estimated from bond disassociation energies (BDEs), which is defined as the standard enthalpy change when a bond is cleaved by homolysis, with reactants and products of the homolytic cleavage at $\mathrm{0\ K}$. This values are frequently tabulated, and can be easily found online (here for example; caveat the values here are at $\mathrm{298\ K}$)

$$\begin{array}{c}
 & \ce{X-H}\ \text{BDE}\ \mathrm{(kJ\ mol^{-1})}  \\ \hline
\ce{X = O} & 428.0\ (21)  \\ \hline
\ce{X = H} & 337.2\ (8) \\ \hline
\end{array}$$

Anyway, this tells you that the $\ce{O-H}$ bond is stronger; Why? Well, simply look at the approximate atomic radii of Oxygen and Carbon. The Oxygen is about $\mathrm{48\ pm}$ while carbon is about $\mathrm{67\ pm}$. The longer atomic radii implies a longer $\ce{C-H}$ bond length, ergo a weaker bond.

Stronger bond, implies larger $k$, which in turn means greater frequency of vibration.

So, the $\ce{O-H}$ bond is NOT weaker. Remember, when we define bond disassociation energy, we are looking at homolytic cleavage. I really don't understand the point you try to make about electronegativities.


Why does Methyl Carbocation have p orbital

You might want to read up on hyperconjugation.

Basically, the R groups (if NOT a hydrogen) can donate some electron density into the carbocation’s empty p orbital, stabilising the positive charge and increasing the ion’s stability. This is the reason for Markovnikov’s Rule when it comes to Sn1 mechanisms; the tertiary carbocation is most stable, followed by a secondary one, while a primary/methyl carbocation hardly exists at all.


Reduction of 2-cyclohexenone to cyclohexanol

According to Johnson, M. R.; Rickborn B. Sodium borohydride reduction of conjugated aldehydes and ketones J. Org. Chem., 1970, 35, 1041.
DOI: 10.1021/jo00829a039:

2-Cyclohexenone can certainly be reduced fully. It has very similar reactivity to cyclopentenone. In this report, mixtures of products were observed. The experimental indicates the researchers used 0.5 eq of $\ce{NaBH4}$ per carbonyl.

Furthermore, the use of Luche conditions can again help you favor only 1,2 reduction:


Yes, due to electrostatic coupling. The conductors and the liquid would form a capacitor, causing back-and forth motion of the ions (and in polar molecules). This is an effective way to heat something without contact, as in dielectric heating.

This capacitance enables liquid crystal displays and E-Ink to display images using little or no current when not actively switching.


Aromatic rings that complex "face on" to a transition element are acting as pi acceptors all the time, including cyclopentadienide complexes.  And they act simultaneously as pi donors too.

Let's look at the familiar cyclopentadiende pi orbitals from a different perspective.  We know that there are three bonding orbitals that should be occupied and two antibonding orbitals that shouldn't be -- until we introduce the d orbitals of a transition element.  As illustrated by https://www.chemtube3d.com/orbitalsm-cyclopentadienyl/, now overlap occurs with the possibility of all the ring orbitals bonding to the metal:

So, in a complex such as ferrocene, the filled cyclopentadienyl pi orbitals share their electrons with three of the transition element d orbitals and the latter element, in turn, donates electrons back to the two unoccupied pi orbitals of the ring.

Of course these interactions sacrifice aromaticity in the cyclopentadienyl rings, but it's a bargain with the bonds forming between carbon and the transition element.  Aromaticity is not so much broken as it is superseded by a further bond deolcalization.  In the especially well-known case of ferrocene, the energy bargain is especially good because when all is said and done with the iron/cyclopentadienyl electron exchange, you've exactly filled the bonding orbitals and left the antibonding orbitals empty.  The diagram from this source shows the molecular orbital structure in the case of a $3d$ metal.  For ferrocene where the iron(II) center provides six additional electron, they fill the orbitals in the box right up to the metal $3d$ orbital level (the HOMO is the $a_{1g}$ orbital inside the box) but not beyond.  Note that the filled $e_{2g}$ orbital has a component from the antibonding orbitals of the cyclopentadienide, showing the pi acceptor action of the latter.

Now compare this outcome with magnesocene.  Magnesium has no valence d orbitals or d electrons.  So the pi electron exchange between pi and d orbitals is not happening.  In this case the cyclopentadienyl ion can still act as a donor to the metal s abd p orbitals, but this does not make for as good bonding because (1) the overlap is not as good and (2) magnesium, like most Group 1 and 2 metals, has relatively diffuse valence orbitals that make for relatively poor overlap anyway.  Unable to develop the full covalent bonding network seen in transition element cyclopentadienide complexes such as ferrocene, magnesocene has greater ionic character -- so much as to form ions in some solutions -- but much less stability.


How to determine polarity in large molecules?

Once you move up to "larger" molecules like this, the definition of a "polar" molecule becomes a bit fuzzy. Rather, you will find that molecules contain polar and non-polar groups which all contribute to the overall characteristics of the molecule. It really depends on the context that you are interested in.

If you are curious about more macroscale properties, such as boiling point or solubility, a variety of group contribution methods have been developed to provide ballpark estimates based purely on the contributions of individual components of the molecule.

While group contribution methods will primarily give you an idea of bulk properties, you may also be interested in lower level behavior such as micellization or specific protein binding which will depend on the location and arrangements of all of the polar and non-polar groups on a molecular level.


offset in measuring decomposition potentials

Ok I will answer your  last question first. If you have two identical electrodes(i.e. same electrode material), the cell potential will be zero and there won't be any issues there.

You have to use inert electrodes so that you are not measuring the dissolution of the electrode itself.

The one part I am not sure you understand properly is the current growing. You want to vary the potential until the current raises exponentially when continuing to sweep. This will be a very sharp rise that will be hard to miss.


Take a look at Kelly, P. M., Francis Rose, L.R., The martensitic transformation in ceramics — its role in transformation toughening, Progress in Materials Science, 2002, Vol.47(5):463–557, doi:10.1016/S0079-6425(00)00005-0.

Basically, a transformation is martensitic if it occurs from a collective shear motion of the atoms where each atomic displacement is less than the distance between atoms.  There is no diffusion (atoms moving to other lattice sites). The definition is the same as in a metal, but in a ceramic you have multiple atom types in the crystal, that's all.


Why is the ester carbonyl peak in benzocaine IR spectrum so low?

Let's first have a look at the structure of benzocaine, to see how it is different from an usual aliphatic ester:

One thing that should be immediately obvious is that the ester is attached to an aromatic ring (which has an amine group). The aromatic ring would participate in conjugation with the ester oxygen, particularly because the amine group is strongly electron-donating (via conjugation).
This is the IR spectrum of benzocaine (from Chemical book):

The peak at $\mathrm{1682\;cm^{-1}}$ can be assigned to the ester C=O stretch. Usually aliphatic esters have IR stretches around $\mathrm{1730-1750\;cm^{-1}}$. Esters attached to unsaturated systems or aromatic groups have lower stretches, as the conjugation weakens the C=O bond.
Compare the $\ce{C=O}$ IR stretches of some esters:

Ethyl acetate  : $\mathrm{1752 cm^{-1}}$
Ethyl benzoate  : $\mathrm{1726 cm^{-1}}$

The presence of the benzene group has reduced the stretch wavenumber by $\mathrm{\sim30 cm^{-1}}$.

methylparaben  : $\mathrm{1682 cm^{-1}}$

Here the $\ce{-OH}$ group is quite strongly electron-donating, and it brings down the ester C=O stretch wavenumber even further, and now the IR stretch resembles that of benzocaine. $\ce{-NH2}$ is also a stronlgy electron-donating group like $\ce{-OH}$.


What could be causing my bath salts containing baking soda to explode out of their containers?

I suspect the seasonal introduction of humidity is the root of the problem, which is readily taken up by the magnesium chloride. Then, the following possible reaction:
$\ce{ MgCl2 (aq)  + 2 NaHCO3 (s) -> 2 NaCl (aq)  + Mg(HCO3)2 (aq)}$
Now, aqueous Magnesium bicarbonate exists only in solution (so decomposes on drying), is not stable on standing (per my experience), or warming (whereupon it also decomposes):
$\ce{ Mg(HCO3)2 (aq) ->  MgO (s) + H2O (l) + 2 CO2 (g)}$
where the pressurized buildup of carbon dioxide gas is likely causing a problem.
Solution: Try adding a drying agent or removing the very hygroscopic Magnesium chloride from your formulation.


How to calculate the mass of air that fits into a balloon of a certain size?

You're right in using the ideal gas law, but the numbers are wrong. Assuming atmospheric conditions, the pressure is 101,325 Pascal and the temperature is likely to be higher than 0°C (273 K). So the mass of the gas (in grams) would be:

$$m = \frac{\pu{101325 Pa} \times \pu{500 m3} \times \pu{28.8 g mol-1}}{\pu{8.31 J K-1 mol-1} \times \pu{290 K}^{[1]}}$$

where you can choose an appropriate value for [1] for your location. All values (except for the molar mass) are in SI base units, as described here.


What are the product(s) of red phosphorous when burned?

The major product formed is -  Tetraphosphorus decaoxide ($\ce{P4O10}$)

Infact you could have googled it. There are many interesting videos on youtube.

Check this out -  Burning of Red Phosphorus


Why is the stoichiometric coefficient in the denominator in the rate expression?

From the stoichiometric coefficients $\nu_{i}$ in your reaction equation

$$\ce{2A + B → C}$$

you can see (as you already correctly stated) that for each molecule of $\ce{B}$ consumed, $\nu_{B} = -1$, two molecules of $\ce{A}$ must be consumed as well, $\nu_{A} = -2$. Similarly for each molecule of $\ce{B}$ consumed one molecule of $\ce{C}$ must be formed, $\nu_{C} = +1$ (and analoguously for $\ce{A}$ and $\ce{C}$).
This will also be reflected in the changes of the amounts of those species, $\Delta n_{i}$, during the reaction, e.g. if the amount of $\ce{B}$ changes by $2 \, \mathrm{mol}$ you can expect a change in the amount of $\ce{A}$ by $4 \, \mathrm{mol}$. So, the ratios of the $\{\Delta n_{i}\}$ and the $\{\nu_{i}\}$ will be equal, i.e.

\begin{align}
  \frac{\Delta n_{\ce{A}}}{\Delta n_{\ce{B}}} = \frac{\nu_{\ce{A}}}{\nu_{\ce{B}}} \ ; && \frac{\Delta n_{\ce{A}}}{\Delta n_{\ce{C}}} = \frac{\nu_{\ce{A}}}{\nu_{\ce{C}}} \ ; && \frac{\Delta n_{\ce{B}}}{\Delta n_{\ce{C}}} = \frac{\nu_{\ce{B}}}{\nu_{\ce{C}}}
\end{align}

Rearranging these equations a bit leads you to:

\begin{align}
  \frac{\Delta n_{\ce{A}}}{\nu_{\ce{A}}} = \frac{\Delta n_{\ce{B}}}{\nu_{\ce{B}}} = \frac{\Delta n_{\ce{C}}}{\nu_{\ce{C}}} \ .
\end{align}

Now, divide these equations by the reaction volume $V$ and use $\frac{\Delta n_{i}}{V} = c_{i} \equiv [i]$ to introduce concentration changes

\begin{align}
  \frac{\Delta [\ce{A}]}{\nu_{\ce{A}}} = \frac{\Delta [\ce{B}]}{\nu_{\ce{B}}} = \frac{\Delta [\ce{C}]}{\nu_{\ce{C}}} \ .
\end{align}

Finally, if you divide now by the reaction time $\Delta t$ and use the values for the stoichiometric coefficients, you get the equations from your question:

\begin{align}
  \frac{1}{\nu_{\ce{A}}} \frac{\Delta [\ce{A}]}{\Delta t} &= \frac{1}{\nu_{\ce{B}}} \frac{\Delta [\ce{B}]}{\Delta t} = \frac{1}{\nu_{\ce{C}}} \frac{\Delta [\ce{C}]}{\Delta t} \\
  \frac{1}{-2} \frac{\Delta [\ce{A}]}{\Delta t} &= \frac{1}{-1} \frac{\Delta [\ce{B}]}{\Delta t} = \frac{1}{1} \frac{\Delta [\ce{C}]}{\Delta t} \\
-\frac{1}{2} \frac{\Delta [\ce{A}]}{\Delta t} &= -1 \frac{\Delta [\ce{B}]}{\Delta t} = 1 \frac{\Delta [\ce{C}]}{\Delta t} \ .
\end{align}


Do enzymes that digest D-glucose react with L-glucose?

It is like the problem of the key and the lock. If you put the good key in a given lock, it will go all right.It will open the lock. But if you make up a new key which is the image of the original key, it will not open the lock.


Your second (B) way of thinking is correct.

In the solution there are twice as many OH- ions as H+ ions.

All the H+ from HCL will react with half the OH- from calcium hydroxide.

Half the OH- will remain ($10^{-4} OH-).

Then use $K_w$ to determine the final concentration of H+.


Are the best buffers those that are in a 1:1 ratio with its conjugate acid/base?

Yes, the best buffers are those that are in a 1:1 ratio with its conjugate acid/base, but in a somewhat indirect way; it's because buffer capacity is maximum when $\mathrm{pH}=\mathrm{p}K_\mathrm{a}$, and for that value of $\mathrm{pH}$, the proportion of the acid/base conjugates is 1:1.

The contribution of an acid/base conjugate pair to buffer capacity is

$\beta_{\ce{HA/A-}} = C_\mathrm{A} \cfrac{10^{-\mathrm{pH}-\mathrm{p}K_\mathrm{a}}}{\left( 10^{-\mathrm{pH}} + 10^{-\mathrm{p}K_\mathrm{a}} \right)^2} \ln{10}$

(For derivation, check this response to another question)

The maximum of buffer capacity is achieved when $\mathrm{pH} = \mathrm{p}K_\mathrm{a}$, and is

$\beta_\mathrm{max}= \cfrac{1}{4} C_\mathrm{A} \ln{10}$

As you deviate from $\mathrm{pH}=\mathrm{p}K_\mathrm{a}$ in either direction, concentrations also deviate from 1:1 (as per Henderson-Hasselbalch) and buffer capacity will fall.

Note that the maximum buffer capacity does not depend on the strength of the acid or base, only on the total concentration of the conjugate system $C_\mathrm{A}=\ce{[HA]}+\ce{[A-]}$.

What $\mathrm{p}K_\mathrm{a}$ will determine is at which $\mathrm{pH}$ value will this maximum buffer capacity be achieved, so you will choose an acid/base pair with a $\mathrm{p}K_\mathrm{a}$ that will allow the system to act as a buffer around the $\mathrm{pH}$ that you want to preserve; ideally by choosing a pair with a $\mathrm{p}K_\mathrm{a}$ equal to the target $\mathrm{pH}$, which is when you will obtain the maximum buffer capacity.


as one moves along the period, the atomic radii decrease. You see, this decrease leads to an increased net Z-effective and hence, the electrons come very close. In the case of Nitrogen, which exists as a diatomic molecule - this shrinking in size is less than that for oxygen and fluorine.
    the melting point is determined by the intermolecular interactions as well as the geometry of molecules. greater the number of bonds to break, higher is the melting point. For oxygen and especially Fluorine, there can be a forced compression, but yet again - these small atoms repel each other because of their high electron density. This repulsion lowers the melting and boiling points.

As for noble gases like Argon, they do not need to form any sorts of bond. The only kind of forces acing between two atoms are very weak London forces. These London forces are responsible for the freezing of Helium, at temperature as low as -273K, very close to absolute zero!!

I hope that helps!!


The ideal gas law states
$$ pV = nRT \Leftrightarrow \frac{n}{V} = \frac{p}{RT}\ .$$

For the density $\rho = m/V$ and the mass $m = M\cdot n$ with the molar mass $M$, we get a new formula for the density:
$$ \rho = M\frac{n}{V} = M\frac{p}{RT}$$

With the definition of the specific volume we get
$$ v = \frac{RT}{Mp}\ . $$

So unless you know the molar mass of your parcel of air, you will not get out a numerical value for the specific volume.


What is the nature of the bonding in cyclotriphosphazene?

The bonds between the phosphorus and nitrogen in phosphazenes are best described as having a significant ionic component with each of the N atoms having a negative charge and each of the phosphorus atoms having a positive charge. There is also some negative hyperconjugation resulting from the donation of electrons from the nitrogen atoms' lone pairs to antibonding orbitals  between the P atoms and the R groups. The degree of negative hyperconjugation increases with the electronegativity of the R groups and it imparts a partial double bond character to the bonds between the P and N atoms.
For more information see http://pubs.acs.org/doi/abs/10.1021/ic0511266


Do chemical garden coral formations leach toxins?

Experiments to test if toxins leach from silicates do exist; however, these experiments involve dangerous chemicals.  For that reason these tests should be left to qualified lab personnel and not described here.  Performing these experiments are expensive and would need to be performed on every combination of methodologies to make a "chemical garden".  Because so many combinations exists, quality control on such an experiment would be difficult.

Experiment aside, a "chemical garden" in a fish tank is a bad idea.  While silica is chemically insoluble,  it can be converted biologically into orthosilicic acid.  This form of silica is soluble. This process in nature is called biological weathering.

Chromium, lead, and perhaps some of the other compounds are carcinogenic to humans.  These compounds, even in an immobilized form, are not worth the risk and should be nowhere near a fish tank.


Why do cotton clothes take a longer time to dry as compared to synthetic clothes?

Cotton fiber is primarily cellulose:

The red atoms are oxygen, and the white atoms are hydrogen. Whenever you have hydrogen attached to something extremely electronegative (such as oxygen), there is a potential for hydrogen bonding to occur:

As you can see, the electronegative oxygen atom draws electron density away from the more electropositive hydrogen atom, which results in a very polar bond. The positive charge on one molecule attracts the negative charge on another, and a very strong physical bond forms.

Now compare cellulose with something like nylon:

Since water can hydrogen bond to cellulose, it "sticks" much better - we say that it is hydrophilic. Since it can't form hydrogen bonds to nylon, we say nylon is more hydrophobic.

The same is true for most synthetic polymer fibers, and as a result, cotton absorbs water much more readily than other fibers, and tends to "hang on" to the molecules for longer - which means cotton fibers take longer to dry.


Nomenclature priority of prefixes consisting of identical letters but containing different locants

The compound that is given in the question consists of a cyclohexane ring with two substituent groups: 1-chloroethyl and 2-chloroethyl.

The correct numbering of locants is stipulated by Rule 14.4 in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book):

  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:
  (…)

However, the given case is not explicitly mentioned in this rule.

The correct numbering is actually indirectly given by Rule 14.4 (g):

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

Since the correct order of citation as a prefix in the name is given by Rule P-14.5.4:

  P-14.5.4 When two or more prefixes consist of identical Roman letters, priority for order of citation is given to the group that contains the lowest locant(s) at the first point of difference.

Thus, ‘1-chloroethyl’ is cited before ‘2-chloroethyl’ in the name, and thus a low locant is assigned first to the 1-chloroethyl group.

Therefore, the complete name for the compound that is given in the question is 1-(1-chloroethyl)-4-(2-chloroethyl)cyclohexane.


Why are some materials more flammable than others?

Firstly, for a material to be flammable it needs to contain something that is oxidisable by air oxygen, typically carbon. This explain why hydrocarbons and other carbon bearing materials typically burn well but minerals, rock, glass, water etc do not. In the latter materials the elements are already in a high Oxidation State and cannot be further oxidised.

For a generic hydrocarbon the oxidation reaction by air oxygen is given by the following reaction equation:

$$\ce{C_nH_{n+2}(g,s,l) +\frac{3n +1}{2}O_2(g)-> nCO_2(g) +\frac{n +2}{2}H_2O(l)}$$

The second major factor that affects flammability is volatility. The more volatile a material is, the more flammable it will be. The hydrocarbons illustrate that well. Methane is the most volatile and flammable of all of them and by the time we get to the solid ones like paraffin wax, flammability has been greatly reduced because these aren't volatile at all.


Conductance and Conductivity of acid base titration?

As the concentrations of $\ce{HCl}$ and $\ce{NaCl}$ are equal, that concentration can be regarded as the "concentration" of the solution. If you divide both sides of your first equation (which includes $κ$) with that concentration you would get your second equation (which includes $λ$). So both equations are valid. The problem would arise if the concentrations of $\ce{HCl}$ and $\ce{NaCl}$ are different, you couldn't define the concentration of the solution, and you couldn't calculate it's molar conductivity, therefore only the first equation would be usable. Note that the Kolrauch's Law can also be expressed through $κ$.


Which one of the following is optically active?

How to determine the optical activity when more than one ring are present? Is it sufficient to check those two mentioned condition?

As Ivan Neretin accurately noted above:

the rules for determining optical activity of a molecule are the same for all molecules, whether it has one ring, several rings, or no rings at all.
axes of symmetry are irrelevant in determining the optical activity of a molecule. We only have to consider center and plane of symmetry.

That said, the given molecules have several chiral centers, which is a great sign that the molecule may be optically active.

If you will try to find the plane of symmetry, you won't find any. While the rings may appear planar on the paper, they (even the fused ones) actually exist as conformations in 3D.

Moreover, the center of symmetry is a point that requires identical groups equidistant from itself; a quick glance over both the molecules shows there are no such centers.


Nomenclature Precedence Order for Unsaturation and Hydrocarbon Chain

Unsaturated bonds take precedence over "branches". Consider these bonds as functional groups and name the compound, it will make you number carbons correctly.

Also the compound you mentioned above can't possibly exist as it has 5 bonds on $\ce{C_3}$.


Proton attached to N or O in HNMR spectroscopy

It depends on the condition of your experiment if a coupling of a proton of an amino or hydroxyl group with a proton attached to the $\alpha{}$-carbon is observable, or not.  As pointed out by the indicator set by @Mithoron, intermolecular exchange processes need to be taken into account.

If your alcohol (or amine) to be analyzed is very pure, and the deuterated solvent you deploy is both very pure, and dry, and free of acids or bases catalysing proton exchange, than you may observe $^3J$-couplings you expect to see.  For example a freshly opened sample of deuterated DMSO (used as solvent) may contribute to this observation.  See the peak data of the 1H NMR spectrum of ethyl alcohol in DMSO here.
If you aim to analyze an alcohol, not well dried, analyzed in deuterated chloroform or methanol that was frequently exposed to ambient (humid) air, then indeed instead of a triplet of an hydroxyl group  ($\ce{-OH}$)  attached to a methylene group ($\ce{-CH2-{}}$) all what may be spot is a broad singlet signal.  The same reference, 90 MHz 1H NMR in deutero chloroform provides one example.  (Note the chemical displacement of the hydroxyl protons altered, too.)


Does organic chemistry somehow defy the periodic table?

While all chemistry is unified because all chemical compounds are made from things in the periodic table, that isn't like mathematics and much of physics. In these sciences many important complexities are derivable from a simple set of axioms.

Some chemistry is simple to derive or deduce from the periodic table but not all of it. The electronic properties of iron atoms tell you a lot about many iron compounds; the location of alkali metals in the table tells us a lot about their common properties.

But carbon is odd. The properties of simple compounds follow simple periodic patterns. But there are more organic compounds (consisting of carbon skeletons, usually with C-C bonds) than all other chemical put together. The complexity and variety of carbon chemistry is almost unfathomably large. None of this is obvious or predictable from the periodic table. You can't deduce the enormous variety from the simple properties of carbon atoms or anything given in the periodic table.

That is probably the sort of distinction your lecturer was aiming at.


Why would decreasing ionic strength of buffer disrupt hydrogen bonds

At high ionic strength, the poly-A tails of mRNA bind oligo-dT that is covalently attached to beads in a column (source). The binding interaction is based on base stacking and hydrogen bonding. When you decrease the ionic strength, the repulsion between the (negatively charged) phosphate groups becomes stronger (is shielded less by counter ions), lowering the binding strength to the point that the mRNA elutes.

The length of the oligo-dT on the beads is chosen such that the mRNA binds sufficiently strong at high ionic strength, but is released at low ionic strength. The appropriate length of oligo-dT is probably on the order of 8-10 nucleotides.

  decreasing the ionic strength of the buffer so that the hydrogen bonds will be disrupted

This statement does not capture the situation accurately. What is really being disrupted is the overall binding of mRNA to the beads. The hydrogen bonds, as short range interactions, should not be directly affected by ionic strength. The stacking interactions might weaken with decreasing ionic strength. Long range ionic interactions, however, are significantly weakened by increasing ionic strength because counter-ions bound to the phosphate backbone will strongly shield the ionic interactions (water also shields it compared to vacuum or non-polar solvents). In this case, the ionic interactions are repulsive, so binding will get weaker at low ionic strength because that is when the repulsive interactions are strongest.


Five-pointed crown alkane C25H40: does it exist?

These polyspiro compounds such as the one you gave are known as [m,n]rotanes, where m is the size of the central ring and n is the size of the decorating rings. So, your compound would be called [5,5]-rotane.
A Reaxys search unfortunately doesn't show anything for [5,5]-rotane, but in general, many such compounds have been made. A lot of the articles are in German or French (there are quite a few such articles). However, I did find one in English: J. Am. Chem. Soc. 1998, 120 (2), 317–328.
The authors synthesised and obtained X-ray structures of a number of rotanes, among them [5,4]rotane, which is the closest to the [5,5] that you are looking for. I quote:

In [5.4]rotane (7), the cyclopentane ring adopts a slightly distorted envelope conformation: C(1), C(5), C(9), and C(17) form an almost perfect plane (mean deviation 0.021 Å), while the flap atom C(13) is located 0.684 Å above this plane. The cyclobutane rings at C(13) and C(9) are nearly planar (φ = 9.7°) and planar (φ = 0.2°), respectively. The remaining three cyclobutane rings are moderately puckered (φ = 23.1, 26.5, and 28.2°).

The torsion angle φ of cyclobutane refers to the angle between the two planes that define the molecule. A completely planar four-membered ring would have φ = 0°, and an increasing value of φ indicates greater deviation from planarity:

As you can see in the X-ray structure, the five attached cyclobutane rings do eclipse each other to a certain extent: the obvious exception is the ring fused to C(13). This is not a very surprising conclusion at all. The envelope conformation of unsubstituted cyclopentane is known to suffer from torsional strain (i.e. unfavourable eclipsing interactions between substituents). Carey and Sundberg's Advanced Organic Chemistry, Part A writes (pp 162–3):

There is minimal angle strain in cyclopentane, but considerable torsional strain is present. Cyclopentane is nonplanar and the two minimum energy geometries are the envelope and the half-chair. [...] The planar portions of both conformations have torsional strain owing to C−H and C−C bond eclipsing.

There's a ton more information in the paper, and I'd recommend giving it a read if you are interested - I just tried to pick out the parts directly relevant to your question here, since you indicated that you were interested in the conformation of such a compound.


How to extract pure potassium carbonate from ash?

The composition of the wood ashes varies with the nature and the origin of the wood. In the average, it is a mixture of $40$ % - $70$ % $\ce{CaCO3}$, $5$% - $10$% $\ce{MgCO3}$, $5$% - $10$% $\ce{K2CO3}$, $5$% - $10$% $\ce{Na2CO3}$, $2$% - $5$% $\ce{SiO2}$, $2$% - $5$% $\ce{Ca3(PO4)2}$, $0.5$% -$2$% $\ce{Al2O3}$, <$1$% $\ce{NaCl}$, <$1$% $\ce{KCl}$, plus traces of manganese and iron. So it is difficult to purify, and/or to extract any of these compounds, as a lot of these compounds are insoluble in water: $\ce{CaCO3, MgCO3}$, $\ce{SiO2, Ca3(PO4)2, Al2O3}$. The only soluble compounds are $\ce{K2CO3}$ and $\ce{Na2CO3}$, plus traces of $\ce{NaCl}$ and $\ce{KCl}$, and they form less than $20$% of the total mass of the ashes. The solution obtained from washing the wood ashes is a mixture of sodium and potassium carbonate. The problem is how to separate those two carbonates. Fractional crystallization does not seem easy. I would rather transform them into nitrates by adding nitric acid : at $0°$C, $\ce{KNO3}$ is  not very soluble ($5$ g/$100$ mL), which is much less than $\ce{NaNO3}$ ($73$ g/$100$ mL). So obtaining $\ce{KNO3}$ from the mixture by cooling a hot solution should be relatively easy.
Ref.: J. Hartwig, Ann. der Chem. u. Pharm. XLVI, p. 97. A. Wurtz, Dictionnaire de chimie pure et appliquée, Vol. 1, p. 641 (1891).


why does adding a non volatile solute not decrease the boiling point?

Decrease of the solvent activity due its decreased molar fraction is bigger than the solvent activity increase due eventual solvent solute interaction.

So the pressure versus solute molar fraction curve may not be linear as by the Raoult law, but is monotonous.


How can an aromatic ketone be reduced in the presence of a nitro group?

You will most likely be successful if you convert the ketone to the corresponding tosylhydrazone and reduce that with $\ce{NaCNBH3}$ (DOI).

A previous study (Bull. Chem. Soc. Jpn., 1974, 47, 2323-2324) has shown that the tosylhydrazone of cyclohexanone can be reduced to cyclohexane with $\ce{NaBH4}$ in polar aprotic solvents.

$\ce{NaCNBH3}$ is an even milder hydride transfer agent, that has been widely used in the reduction of imines and iminium salts. A review by C.F.Lane, Sodium Cyanoborohydride - A Highly Selective Reducing Agent for Organic Functional Groups in Synthesis, 1975, 3, 135-146 gives some background information on the versatility of this reagent.

The nitro group is definitely safe with $\ce{NaCNBH3}$. However, considering the high toxicity and the price for this reagent, you might want to run a small scale reduction of your tosylhydrazone with the much cheaper $\ce{NaBH4}$ first.


Determination of carbon content of steel by liquid chemistry

I think cuppellate is a foreign word meaning "burn" essentially.

The basic technique would be to have the iron sample in a small ceramic boat which is put into a hot stream of $\ce{O2}$ so the iron and carbon both oxidize. The iron oxide stays in the boat, and the $\ce{CO2}$ is sweep away and trapped/reacted somehow. We talking small samples 10 grams perhaps.

If you try to dissolve the steel in an acid solution then the $\ce{CO2}$ gas is formed in solution. It is really hard to get a small amount of $\ce{CO2}$ out of solution given that you are starting with a small amount to begin with. Also if you try to scale up such a reaction it gets to be a problem. A $2~\mathrm{cm^3}$ block of iron takes a long time to dissolve in acid.

So choosing the analytical method depends on a lot of factors. Time per sample, sensitivity, and overall cost being the main ones.