A contradiction, probably from confusion

Processes in nature that lead to an increase in the entropy of the system are in general favourable, but entropy alone is not the deciding factor!

If we were to follow a full proof we would have to consider Carnot engines and such but for now the result that we are interested in is called the Clausius statement. It is the Clausius statement that really defines the second law and the direction of spontaneity. The Clausius statement leads us to define two parameters depending on the experimental conditions: either the Helmholtz free energy $A$ (when working at constant volume) or the Gibbs free energy $G$ (under conditions of constant pressure). In practice it is much easier to work with $G$ since we can expose our reaction mixture to the atmosphere and call it constant pressure.

It is these thermodynamic potentials $G$ and $A$ that determine the overall thermodynamic picture of spontaneity of a given process.

If we work with the Gibbs potential, after some mathematics we can easily express by the following equation:

\begin{equation}
G=H-TS
\end{equation}
Where $H$ is the enthalpy of the system, $T$ is the temperature and $S$ is the entropy. For a given process it is the Gibbs energy that determines the spontaneity of the reaction. Crudely put, the starting material will have a certain $G$ value and the products will have a certain $G$ value, if the change in $G$ over the course of the reaction is negative then we say it is spontaneous. If it is positive it is not spontaneous.

We can clearly see that the condition for spontaneity is not solely dependant on the entropy alone rather it is also a function of the enthalpy too. Further the entropic contribution to the $G$ is dependent on the temperature indicating that there will be a certain threshold temperature for an endergonic process.

An endergonic process is characterised by having a positive enthalpy change, requiring energy to occur. If the "amount" of energy the process requires is very large then the change in enthalpy will also be large and as we can see this makes the change in $G$ for the process become less favourable.

Moving on, we can now see that a given process will have a certain energy barrier... an activation energy if you like. Whether a process is thermodynamically spontaneous or not one has to consider the Kinetics too! A simple yet ubiquitous model for Kinetics is the Arrhenius equation which in general says the rate of reaction decreases for processes with a higher energy of activation.

Turning the later part of your question. The most stable phase (state of matter) is dependent on the conditions of the reaction. The most stable phase for the given combination of temperature and pressure will prevail in the reaction mixture. In practice this leads to the state with the lowest value of $G$.

Solids actually liberate energy when they come together, especially ionic solids! This energy is called the Lattice energy and it is relative to all the reactants in their gaseous state (as ions). (look at Born-Lande equation for more info)! So although gaseous species will have a higher entropy due to all the rotational vibrational freedom they have ... the state of matter is dictated by $G$.

Without the full mathematical argument behind what I have said I feel my answer will be incomplete, but I hope it helps you :)


How do I explain the existence of these 2 peaks in this mass spectrometry?

In electrospray ionization, a molecule can end up with an extra proton or an extra cation. Either one of these +1 charged species is accelerated and detected in the mass spectrometer (negative charged particles or neutral species are not; if there are multiple cations, they would supposedly repel each other once the solvent is gone, and not fly together).

So a molecular formula that fits the data is $\ce{C43H82NO8P}$. The lighter species detected in the spectrometer would be $\ce{C43H82NO8PH+}$, and the heavier species would be $\ce{C43H82NO8PNa+}$. The masses given by https://www.sisweb.com/referenc/tools/exactmass.htm for these species match the given values pretty well. The difference in mass is not due to fragmentation, but to a cation piggy-backing with the molecule of interest (instead of the usual proton).

The data does not distinguish between phosphatidylcholines and phosphatidylethanolamines. You can, however, infer the number of C=C double bonds in the fatty acids attached to the glycerol.

I had to look up the unit "Th" on wikipedia:

  The thomson (symbol: Th) is a unit that has appeared infrequently in scientific literature relating to the field of mass spectrometry as a unit of mass-to-charge ratio. The unit was proposed by Cooks and Rockwood naming it in honour of J. J. Thomson who measured the mass-to-charge ratio of electrons and ions.

HRMS stands for high resolution mass spectroscopy, I think.


It seems that this compound has not yet been characterized, although traces of it could be found in mixtures of $\ce{H2SO4}$ and $\ce{SO3}$. The most recent paper I could find is this one from 2012, showing that salts of the respective anion exist and theoretical calculations imply that the acid is stable with respect to decomposition into $\ce{H2S2O7}$ and $\ce{SO3}$. The conditions under which it may be stable are however unknown.


What does it mean when it is said that an sp³ orbital has 25% s character?

Hybridisation is a purely mathematical concept, which makes it possible to explain experimentally found structures. The most prominent example for this is methane, where you can consider the central carbon atom to be $\mathrm{sp^3}$ hybridised. Formally, the $\mathrm{s}$ orbital and the three $\mathrm{p}$ orbitals can be linearly combined to form four equivalent orbitals. Hence $\mathrm{sp^3}$ is a contraction of $\mathrm{s^{\frac{1}{4}}p^{\frac{3}{4}}}$. Therefore you can see, that the orbital has $25\,\%$ $\mathrm{s}$ character and $75\,\%$ $\mathrm{p}$ character.
Generally, the concept of hybridisation has to be treated as a concept only, it is nothing which can be physically observed. It is a mathematical model to understand geometries better. Please also note, that hybridisation is a function of the geometry, not the other way around. Having said that, it is fairly obvious, that I am not a big fan of the concept. I would not go as far as Alexander Grushow* and preach to abandon the whole thing, but I advise caution in the use of this concept to not obtain wrong conclusions.

  Despite significant experimental evidence and theoretical advances to indicate that hybrid atomic orbitals do not exist and do not appropriately describe molecular bonding, their description still permeates chemical education at many levels, and the model still finds its way into modern chemical literature.

* “Is It Time To Retire the Hybrid Atomic Orbital?” Alexander Grushow, J. Chem. Educ., 2011, 88 (7), 860–862.


Titration of sodium carbonate and nitric acid

To solve this problem, you need to know the pKas of carbonic acid.

The pKas are 6.35 and 10.33.

If you add 1 equivalent of acid to $\ce{Na_2CO_{3}}$, the product will be $\ce{NaHCO_{3}}$ (sodium bicarbonate).

In this case, you are going 0.4-.377 = 0.023 moles past the point of 1 equivalent.

I would model the final solution as 0.354 moles of $\ce{HCO_{3}^-}$ and 0.023 moles $\ce{H_2CO_{3}}$ , and use the pKa of 6.35 and the Henderson-Hasselbalch equation to calculate pH.

The proposed answer 10.77 is incorrect.


Determine the quantitative composition of a solid sample containing Co, Mn, Al, C, and O

It very much depends on element concentrations and the accuracy required for the total assay, but, I would do the metals by ICP-AES or AA and C by a classical combustion technique. ICP-MS would be unnecessarily expensive and since you are looking at compositional levels rather than low trace levels, it just wouldn't buy you anything.

If you know oxygen is the only other element present at compositional (i.e. percent) levels then don't bother with the finickier oxygen analysis and just determine it by difference. If carbon is present at at least a few percent then you only need a few mg, leaving enough to do multiple digestions for the metals analysis if desired.

If you want to do direct oxygen analysis, it can be done by classical pyrolysis techniques, though this presumably high-metals matrix would require a special technique (Mertz) and this analysis could have relative uncertainties of greater than 10%.


Why does water attack the more substituted carbon in oxymercuration?

Using 1-methyl-1-cyclohexene as an example, the mercurinium ion is not symmetrical. If it were, the positive charge would all be on mercury. The ion is unsymmetrical with part of the positive charge on the more substituted carbon. Why this carbon? Follow Markovnikov's Rule: tertiary carbocations are more stable than secondary carbocations. This bias is manifested in the transition state permitting water to attack at the more substituted carbon. The addition is anti. I chose 1-methyl-1-cyclohexene as an example because adding a tertiary butyl group to the 4-position produces more subtle aspects of the reaction mechanism.

While structures 1 and 2 above are both tertiary alcohols, only alcohol 1 is formed. This selectivity is controlled not only by the anti-addition but also by the more favorable chair-like transition states (red arrows) over the boat-like transition states (blue arrows). These issues are provided in greater detail at
http://ursula.chem.yale.edu/~chem220/chem220js/StudyAids.html#Mechanisms as PowerPoint and pdf files.


Given an external electromagnetic field, the orbitals do split and even an otherwise isolated atom is no longer spherically symmetrical. For a mostly magnetic field, this is called the Zeeman effect, and for an essentially electrostatic field it is the Stark effect.

Of course, this begs the definition of an "isolated" atom. As @Karl mentions, how do you measure the shape without interacting with the atom? Any incident photon, for example, would be "letting the cat out of the box."


Why do we use the external pressure to calculate the work done by gas

If the piston is frictonless and massless, then, if you do a force balance on the piston, you must have that the force per unit area that the gas exerts on the inside face of the piston will always be equal to the external force per unit area that one imposes on the outside face of the piston.  The sudden drop in pressure on the outside face of the piston causes the gas to undergo an irreversible expansion.  During an irreversible expansion, the local pressure within the cylinder becomes non-uniform, so that the average pressure of the gas differs from the force per unit area at the piston face.  As a result, the ideal gas law (or other equation of state) cannot be applied globally to the gas in the cylinder.  In addition, during an irreversible expansion, there are viscous stresses present in the gas that allow the force per unit area at the piston face to drop to the new lower value while requiring that force to match the external force on the outer face.  So the work done by the gas on the piston is equal to the external force per unit area times the change in volume:  $$W = \int{P_{ext}dV}$$  This equation is always satisfied, irrespective of whether the expansion is reversible or irreversible.


How to write the redox reaction of chlorate to chlorine and oxygen using half rections?

This is tricky and one way to think about it that I can offer is to treat $\ce{ClO3-}$ as a combination of $\ce{Cl^5+}$ and $\ce{O^2-}$. Since the problem is that the starting material is undergoing disproportionation with different elements being reduced/oxidised, if you could split the starting material into the two different elements, then you could circumvent the issue. Obviously the bonding in the chlorate ion is not ionic but this procedure allows you to construct "half-equations" and then an overall equation.

I would not suggest that you write this in an exam.

The two half-reactions are then:

$$\begin{align}
\ce{Cl^5+ + 5e-} &\longrightarrow \ce{1/2Cl2} \tag{1} \\
\ce{O^2-} &\longrightarrow \ce{1/2 O2 + 2e-} \tag{2}
\end{align}$$

$[(1) \times 2] + [(2) \times 5]$ gives

$$\ce{2 Cl^5+ + 5O^2- -> Cl2 + 5/2 O2}$$

In order to "re-form" your chlorate, you need three oxide ions per Cl ion, so you need to add one more oxide to both sides:

$$\ce{2 Cl^5+ + 6O^2- -> Cl2 + 5/2 O2 + O^2-}$$

Now you can get back your chlorate:

$$\ce{2 ClO3- -> Cl2 + 5/2 O2 + O^2-}$$

Now it's highly unlikely that there are going to be free oxide ions hanging around, so we just protonate them by adding $\ce{2 H+}$ on both sides. The oxide ion simply becomes water:

$$\ce{2 ClO3- + 2H+ -> Cl2 + 5/2 O2 + H2O}$$

  I understand that 12 electrons would be lost from oxygen if it all went to $\ce{O2}$ but only 10 electrons are gained by the two chlorine (V)'s so you have to use a water to make the charge balance work out.

Yeah, if all six oxygens became $\ce{O2}$ then they would lose 12 electrons. Therefore, one of the oxygens does not become $\ce{O2}$; it becomes $\ce{H2O}$ and remains as oxygen(-2). It is not about charge balance.


Is there a phase diagram available for hydrogen fluoride?

Linde Gas contains a vapor pressure curve.  Note that this does not extend to Venusian surface temperatures which are above the critical point.

Hydrogen fluoride may not lift as well as you think.  Its gas phase may associate through hydrogen bonding so molecules could be heavier than just HF.  See the note near the end of the safety information here and note that the association could persist to higher temperature if you apply higher HF pressure than one atmosphere.


short answer: The record is based on the old record by Bannister 1934 as first description of this phase . The fine intensity scaling 1:1000 on the record card is based on a more recent re-investigation.

recording the data:

Beside the point detector / Geiger counter technique mentioned in Jon Custer's comment -- still highly valuable in powder diffraction analysis -- a common method was to record the diffraction pattern on a photographic film.

In the case of powder samples, you used for example Guinier cameras:

(source)

with a stripe of a film placed in the inner of the housing.  After the experiment, the film was developed; and the more diffraction intensity, the darker the band on the developed film.  The following image depicts a simultaneous analysis of four samples:

(source).

In case of single crystals, one method was to place the oriented sample in front of a film, for example with a Buerger precession camera

(source)

to yield clichés like

(source)

where the location of the spots provides information about symmetry distance relationships of the lattice.  (Relative intensity of the spots equally has a meaning, but beyond the scope of this question here.)

In both cases, the intensity of the signals were attributed by eye (e.g., strong, medium, weak), or later by densitometry and densitometers in numbers. While these film methods today are superseeded in convenience by positional detectors and CCD arrays (no wet chemistry to develop the films, quick numeric read-out), you still find these noted in volume C of the International Tables of Crystallography, section 7.1 «Detectors for X-rays».

(source)

your phase:

You find the first entry about $\ce{BiOCl}$ as entry #1011175 published 1934 in Nature, and #9011782 published 1935 in Mineralogical Magazine by the same author as indexed by the COD with their .cif files. You may use these to predict how a powder diffraction pattern of this phase should look like:

And indeed, there are some signals expected to be much weaker, than others, e.g., reflection $(103)$ (blue rectangle) would be weak and maybe not spot.

Interestingly, the AMS database equally offers access to the Bismoclite data, including a copy of Bannister's publication in Mineralogical Magazine, too (here).  And table 2 (loc. cit.), based on the photographs and their visual readout lists only 16 entries.  You find all of them in the subsequent re-investigation in 1983 (here, view text file next to download diffraction data), now with a total of observations equal to 37.

Thus, it is plausible the record card you accessed refers either to this set (1983) of data (or an  even more recent study) to list $(h,k,l)$ and their diffraction intensities, but credits Bannister for the first description of the very same phase in the 1930s.


Chemical formula of ethanol and other carbon compounds

$\ce{CH3CH2OH}$, and $\ce{C2H5OH}$ or $\ce{EtOH}$ (Ethyl OH) tells you specifically what are the bonds in the molecule. $\ce{CH3H6O}$, which would be the equivalent of writing ethane like $\ce{C2H6}$, on the other hand, tells you nothing, since it can be both ethanol or dimethyl ether. Bigger formulas can represent thousends of different compounds if written like that.

The reason why $\ce{C2H6}$ is used instead of $\ce{CH3CH3}$ is because it can only represent ethane - it's a very simple compound with no isomers, so you can write it both ways.

You can write ethanol both ways you suggested, although I've never seen $\ce{C2H5}$ instead of $\ce{Et}$ anywhere but in examples. What you shouldn't do is write it like $\ce{C2H6O}$.

If you were asking what is the logic of $\ce{CH3CH2OH}$:

What it's telling you is that the compound it is representing [ethanol] consists of a carbon atom bonded to 3 hydrogen atoms and to another carbon atom, which is bonded to 2 hydrogen atoms and 1 oxygen atom, which is finally bonded to 1 hydrogen atom, as follows:


Do you include the "10 H2O" in the formula for borax?

You are talking about the waters of hydration in crystal structure $\ce{Na2B4O7 \cdot 10H2O}$, and the answer is invariably—yes.

These water molecules account for 47.24% of the crystals weight! If you wanted to make $1\: \mathrm{L}$ of $1\: \mathrm{M}$ of $\ce{Na2B4O7}$, and you did not account for these waters of hydration, you would end up making a $0.4818\: \mathrm{M}$ solution. Less than half the desired concentration!

This is the result of a double effect. Not only are you not adding as much salt as you thought you were, but because the salt contains water it is diluting the solution even further. In the case above $95.05\: \mathrm{mL}$ of water are actually added to the solution just in the salt!


When should we not draw a Lewis dot structure?

Short answer: no, there's nothing preventing you from drawing arbitrarily complex molecules according to the conventions of Lewis dot structures. They are not special, and the rules for drawing them are settled by convention and have evolved since they were first developed in Lewis's original paper. They're just one type of diagrammatic rendering of molecular structure which (probably due to their simplicity and explicitness) has caught on in pedagogy. They don't see much use beyond that. Read on for further details.

It's worthwhile to precisely define our terms here. The IUPAC Gold Book provides the following definition for a central atom:

  The atom in a coordination entity which binds other atoms or groups of atoms (ligands) to itself, thereby occupying a central position in the coordination entity.

By that definition, the term is reserved almost exclusively for coordination complexes with a distinct metal atom at the center and ligands bound to it (with some exceptions for purely molecular Lewis adducts, e.g., amine oxides), where the electrons of the bond(s) originate on the ligands, rather than each atom donating an equal number of electrons in the formation of the bond(s). For non-coordinate covalent bonds, if the bonds were to be split homolytically (i.e., such that an equal number of electrons are distributed to each atom), the result would be separate, neutral atoms. Were coordinate covalent bonds to be split this way, that would not be the result.

Obviously, that definition is highly limited, and it's not the prevailing definition in my experience. The other usage of the term is essentially to refer to any non-terminal atom (i.e., one that has two or more bonds to separate atoms). There's nothing inherent to Lewis structures that prevents them being used to represent molecules containing multiple such central atoms. They're simply one representation of bonding. In practice, they don't see much use, chiefly because they're substantially less economical than condensed structural formulas or the skeletal bond-line structures (introduced in their prototypical form by Kekule), and because pure Lewis dot structures, as they were originally conceived, are exceedingly poor at representing stereochemistry.


Is there any chemical preservative better than sodium benzoate?

I just want a second opinion about allowed food preservatives.

Allowed is crucial here and it might strongly depend of where the product is to be sold. You might want to have a look at national legislations.

For Germany, the Zusatzstoff-Zulassungsverordnung lists allowed preservatives and their maximum concentrations in different products.
For Canada, Health Canada provides a List of Permitted Preservatives.


Why are fluorides more reactive in nucleophilic aromatic substitutions than bromides?

The key point to understanding why fluorides are so reactive in the nucleophilic aromatic substitution (I will call it SNAr in the following) is knowing the rate determining step of the reaction mechanism. The mechanism is as shown in the following picture (Nu = Nucleophile, X = leaving group):

Now, the first step (= addition) is very slow as aromaticity is lost and thus the energy barrier is very high. The second step (= elimination of the leaving group) is quite fast as aromaticity is restored. So, since the elimination step is fast compared to the addition step, the actual quality of the leaving group is not very important, because even if you use a very good leaving group (e.g. iodine), which speeds up the elimination step, the overall reaction rate will not increase as the addition step is the bottleneck of the reaction.

Now, what about fluorine? Fluorine is not a good leaving group, but that doesn't matter as I said before. It is not the leaving group ability of $\ce{F-}$ which makes the reaction go faster than with, say, bromine or chlorine, but its very high negative inductive effect (due to its large electronegativity). This negative inductive effect helps to stabilize the negative charge in the Meisenheimer complex and thus lowers the activation barrier of the (slow) addition step.
Since this step is the bottleneck of the overall reaction, a speedup here, speeds up the whole reaction.

Leaving groups with a negative mesomeric effect (but little negative inductive effect) are not good at stabilizing the Meisenheimer complex, because the negative charge can't be delocalized to them.


With no knowledge of chemistry, is it realistic to undertake a biochemistry class?

I would say that taking a biochemistry course without any background in chemistry or biology would be very difficult, but it would be possible.

  The course description lists the following prerequisites: "Basic knowledge in general and organic chemistry". How important are these subjects in biochemistry?

The info page link that you gave would not work for me, so I can't tell how in-depth this course would be, but I know that the undergraduate biochemistry courses I have experience with expect you to be familiar with (at a minimum):

Chemistry

Bonding theory
Molecular structure
Nomenclature of organic molecules
Polarity and its relation to hydrophobicity/hydrophilicity
Basic organic reaction chemistry (particularly Lewis acid/base and functional group chemistry)
Basic reaction kinetics
Basic thermodynamics

Biology

Macromolecules (proteins, carbohydrates, fats)
Cell structures
Basic anatomy/physiology

All of these are fundamental to just about every concept in biochemistry, and they will probably expect you to already be familiar with if not good at them.

  What are some good references (books) for self-learning these subjects? Is Stryer's Biochemistry a good book for self-learning?

That I can't answer. I can tell you that Voet & Voet's "Biochemistry" is a good biochem book, but I am not sure if it would be suitable for self-learning. I don't have experience with Stryer's book. Voet & Voet also have a "Fundamentals of Biochemistry" book that might be better - but again, I don't have any experience with that one and can't compare to Stryer. In my personal experience, I have found that reading as many different books as possible is the best way to self-study, but your results may vary.

I can give you some suggestions for general chem and organic books:

General Chemistry

Brown, Lemay, and Bursten's "Chemistry: The Central Science"
Tro's "Chemistry: A Molecular Approach"

These are both very easy to read and clearly explain all of the fundamental concepts, with lots of example problems worked out in detail.

Organic Chemistry

Wade's "Organic Chemistry"

This is the only organic chem book I have much experience with, but I remember it being very easy to understand.

For biology textbooks, I'm afraid I can't be much help.

  Are biochemistry and biostatistics related at all?

Only in the sense that biostatistics is statistical methods applied to problems found in biology, and many problems found in biology are related to biochemistry. You would certainly want to learn some basic biochemistry if you were going to work on bioinformatics problems, for example.

  Are there many applications of statistics in biochemistry?

There are - particularly in areas like design of experiments, genomics, proteomics, and drug screening, and other various assays.


The only right choice is D, because there is an ester, even though this ester is intramolecular, being made of the bridge -CO-O- in the lower part of the molecule. The aldehyde group is -CHO in the R.H.S. The carboxylic acid is in the upper part of the middle. And there are two alkene groups in the cycle at L.H.S.


Calculate Rate constant, k, for a consecutive reaction where k1<<k2

In a consecutive reaction A-B-C, A decays as $A=A_0\exp(-k_1t)$ with $A_0$ the initial amount (initial B and C are assumed to be zero) and B rate expression $\displaystyle \frac{dB}{dt}=k_1A-k_2B$  which when substituting for $A$ and integrating gives $\displaystyle B=A_0\frac{k_1}{k_2-k_1}\left(\exp(-k_1t) -\exp(-k_2t  \right)$. C is found as $A_0-A-B$

then $\displaystyle C= A_0\left(1+\frac{k_2\exp(-k_1t)-k_1\exp(-k_2t)}{k_1-k_2}  \right) $

In your case $k_1<<k_2$ then $\displaystyle C= A_0\left( 1 - \exp(-k_1t)  \right) $ which is almost your equation.

Taking logs gives $\displaystyle\ln\left(\frac{C}{A_0}\right)=\ln\left(1-\exp(-k_1t)\right)$ and this is what you should use.

[To go further you will have to expand the series for log and exponential but this will lead to a polynomial and will only be an approximation anyway.  The first few terms are $\log(k_1t)  - k_1t/2 + (k_1t)^2/24\cdots$; very messy.]


How much UV lamp power is needed to initiate radical polymerization?

Sounds interesting! These hand lamps typically have two tubes with emission maxima around $\lambda$ = 254nm and 360 nm, respectively. The latter should be just right to excite AIBN and isn't cut off by Pyrex or comparable glas.

More power is obtained from a high pressure mercury lamp, such as the classical HPK 125W, but these lamps are normally used in cylindrical irradiation vessels with immersion cooling.

If you want a complete system with top- or side irradiation, replacable tubes, metal housing and ventilation have a look at the portfolio of companies like LuzChem.


How much salt (or any other substance one can find in a kitchen) do I need to add to make water boil at 104 °C?

According to [1, pp. 281–282], solution of sodium chloride $\ce{NaCl}$ prepared by dissolving 25 g of salt in 100 g of water has boiling point of 104.6 °C.
Additional data is available in the following table for the aqueous solutions of common salts and bases.
English transcription; column 1: Compound; columns 2–6: Concentration, g/100 g water — boiling points $(t_\mathrm{b.p.},~\pu{°C})$.

Note, however, that there is a significant drawback of using boiling salt solution due to the shift of boiling point upwards as the water boils off.
You either have to add water to the mark from time to time, or use a bath with another heat carrier and a thermal sensor.
References

V.A. Rabinovich and Z.Y. Havin. (Eds.) "Kratkii khimicheskii spravochnik" (Brief chemical handbook), 2nd ed., Khimiya, Leningrad, 1978. (in Russian)


Is circularly polarized light ever used to isolate particular chiralities?

Yes. For some chiral compounds, each enantiomer in a racemic mixture absorbs certain frequencies of light differently depending on the circular polarization type of the light.

That means that if you can pump in sufficiently intense and energetic (e.g. ultraviolet) circular polarized light pulses into the racemic mix of such compounds, you may in some cases increase the rate of breakdown of the enantiomer that absorbs that particular circular polarization (left or right), resulting in an increase in the relative concentration of the other enantiomer.

You can view the first page of a paper on asymmetric photolysis using circularly polarized ultraviolet light here.


Average or individual molar heat capacity?

They both are same.

If you plug $C_{\mathrm{m},p,\mathrm{avg}}$ in your second equation, you will get your first equation. The average $C$ is calculated as a mole fraction weighted average.


Calculating the molar iodine concentration of a CT contrast agent

I highly recommend not looking for literature from non-authoritative sources such as drugs.com: that's a paid-advertising site and is full of partial/misinformation.

For your compound, see the paperwork from the people who make it - Bracco Imaging.

Each $\mathrm{mL}$ of Isovue-300 contains more than what you list (namely, water, sodium hydroxide, and hydrochloric acid). That means your approach is close.

The wording about the 300 $\mathrm{mg}$ of organically-bound $\ce{I}$ per $\mathrm{mL}$ of solution makes sense when you consider the atomic weight of $\ce{I}$ (about 127 $\mathrm{mg}\,\cdot\mathrm{mmol}^{-1}$) and that 3 atoms of iodine comes out to 381 $\mathrm{mg}$. Thus, for that slightly dilute solution, the 300 $\mathrm{mg}$ makes sense. It's not an additional 300 - it's what's in the $\mathrm{mL}$ as-is.

So, the molarity of $\ce{I}$ is (I abbreviate iopamidol as "iop" below to save space):

$$\left({612\,\mathrm{mg\,iop}\over 1\,\mathrm{mL}}\right)\cdot
\left({3\,\mathrm{mmol}\,\ce{I}\over 777.09\,\mathrm{mg\,iop}}\right)\cdot
\left({1\,\mathrm{mol}\,\ce{I}\over 1000\,\mathrm{mmol}\,\ce{I}}\right) \cdot
\left({1000\,\mathrm{mL}\over 1\,\mathrm{L}}\right)
= 2.36\,\mathrm{M}$$


I highly doubt that it could exist. The co-linear arrangement, $C_{\infty\mathrm{v}}$, of $\ce{Cl-O-S-Cl}$ is a second order saddle point on the DF-BP86+D3(BJ)/def2-SVP level of theory on the $S=0$ potential energy surface. It has two imaginary bending modes, $\nu_1=371.39\mathrm{i}$, $\nu_2=108.23\mathrm{i}$. (Please note that they are actually perpendicular to each other.)

That basically means, that distortion along these modes will lead to an arrangement of the nuclei which is lower in energy. Since this is a second order saddle point, this arrangement can itself only be a transition state (first order saddle point), which also cannot be considered stable. (Compare also here.)
Therefore distortion among $\nu_1$ would probably lead to $\ce{Cl-O}$ and $\ce{S-Cl}$ fragmentation. While distortion along $\nu_2$ might preserve the $\ce{S-O}$ bond, fragmenting the chlorine atoms/ chloride ions - but that is pure speculation of a hypothetical case.

Let's try a very tempting ad-hoc explanation of why this is the case. In that case let us assume all atoms are $\mathrm{sp}$ hybridised and the bonding axis is $z$. The Lewis structure that seems to be valid has therefore three lone pairs ($\mathrm{sp}$, $\mathrm{p}_x$, $\mathrm{p}_y$) at each chlorine and two lone pairs ($\mathrm{p}_x$, $\mathrm{p}_y$) at the oxygen and sulfur. The three σ bonds are formed by the $\mathrm{sp}$ orbitals.
That would result in four lone pairs that are co-aligned in the same plane. Overlap is inevitable, and in the most simplest terms of VSEPR theory, we would call that lone pair repulsion. (Please note that for strictly VSEPR theory we would have to assume tetrahedral hybridisation.)
In other words, same spin electrons have to avoid occupying the same space and since the orbitals are so close together this will rise the energy a lot. You can probably call it Pauli repulsion.

The MO picture is too messy, so that I won't show it here. Basically most of the occupied orbitals are anti-bonding with respect to one of the bond. The HOMO is of π symmetry and anti-bonding with respect to every bond.


Which factor should I take into priority to determine reactivity of of Halides?

If you need to consider the stability of the carbocations obtained after chloride leaves, we can, as Klaus commented already, use Hückels rule, which more or less states that fully $\ce{sp^2}$ (i.e. usually $\ce{sp^2}$, as long as there is atleast one full p-orbital) hyberdized  compounds with $4n+2$ electrons are particularly stable or aromatic. The same type of compounds that have $4n$ are particularly unstable, anti-aromatic. Of course a carbocation has an empty p-orbital, so compunds I and III will be aromatic and compound II anti-aromatic. Compound I is fully benzenoid and also less strained than compound III and will hence be more stable, so I > III. Then follows the allylic carbocation IV which is obviously more stable than its non-allylic counterpart V. And last but not least the anti-aromatic compound II.

In conclusion I > III > IV > V > II, as the answer indicated.


Different statements of second law of thermodynamics

Please use simple layman terms, I am simply trying to grasp the idea.

Thermodynamics is a rigorous field, which is one of the things that makes it so powerful. Explaining something in simple layman terms will make it less powerful and more mysterious. One way around this is to pick and choose some parts that are necessary to answer the question and ask you to suspend your disbelief while following the argument. In this case, we need to know that in a ideal heat engine, the change in entropy of the reservoirs is:
$$ \Delta S_\mathrm{reservoir} = \frac{Q_\mathrm{rev}}{T}$$
Also, the overall change in entropy will never be negative (and for an ideal heat engine, it will be zero).

What I get from second law is that heat cannot be completely converted into work and hence efficiency of heat engine is always less than one. Please explain what it has to do with entropy.

You can figure that out because if you extract heat from a reservoir, the entropy of it increases. Your process has to increase the entropy of something else by at least the same amount, so you can't just turn thermal energy into work.

Also from second law, entropy of any isolated system never decreases. How are the two statements similar?

If it never decreases, it is zero at best. You can apply this general statement to the specific case of an ideal heat engine. In this process, you have to heat up the cold reservoir to match the entropy decrease of taking heat out of the hot reservoir. Because we are dividing by $T$, and it is different for the two reservoirs, we can extract some of the energy in form of work without breaking the second law. The higher the temperature difference, the more work we can extract for a given amount of heat transfer.


According to the general methodology described in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), a double bond is expressed by changing the ending ‘ane’ of the name of the corresponding saturated parent structure to ‘ene’.

  P-31.1.1.1 The presence of one or more double or triple bonds in an otherwise saturated parent hydride (…) is denoted by changing the ending ‘ane’ of the name of a saturated parent hydride to ‘ene’ or ‘yne’. (…)

For example, the name ‘hexane’ is changed to ‘hex-1-ene’.

If the compound contains multiple double bonds, a multiplicative prefix is placed before the ending ‘ene’ and the letter ‘a’ is inserted between the root of the name of the parent structure and the multiplicative prefix.

  P-31.1.1.2 The multiplying prefixes ‘di’, ‘tri’, etc., are placed before endings denoting unsaturation to indicate the number of multiple bonds of each kind, as required, for example, ‘diene’ and ‘triyne’. For euphonic reasons, when the endings ‘ene’ and ‘yne’ are preceded by a multiplying prefix and a locant the letter ‘a’ is inserted. (…)

See also:

  P-16.8.2 For euphonic reasons, the letter ‘a’ is inserted between the root of the name for polyenes, polyynes, and polyenynes, and the numerical multiplicative prefix ‘di’, ‘tri’, etc., preceding the ending ‘ene’ or ‘yne’.

Therefore, the correct name for the compound given in the question is hexa-1,3-diene (not hex-1,3-diene).


Do Ligands electrons fill the d-orbitals of the metal ion?

You are mistaken. The electron pairs donated in the coordinate bonds by the ligands do not fill the d orbitals of the metal ion.

The basis of crystal field theory is the electrostatic repulsion between the d orbitals of the central metal ion and the electron cloud of the ligands. This repulsion then leads to the breaking of the degeneracy of the orbitals, giving rise to disparity in energy between different sets of d orbitals. This arises due to the different spatial orientations of the 5 d orbitals (i.e. $xy, yz, xz, x^2-y^2, z^2$). Thus, they experience different amounts of repulsion from the ligands when the ligands are oriented in different geometric configurations (i.e. octahedral, square planar, tetrahedral etc.) around the central metal ion.

For more details on crystal field theory and an alternative explanation known as ligand field theory (based on molecular orbital theory), you can refer to the following discussion:

In an octahedral complex, what happens to the electrons donated by the ligand?

Since the d orbitals are not filled up by any ligand electrons, the electronic transitions between the orbitals of different energy levels (such as in the case of $t2g$ and $eg*$) are certainly possible.

Now, you may wonder: Where do the donated electrons go?

Well... based on valence bond theory, they are used in the formation of covalent bonds between the central metal ion and the ligands. These covalent bonding interactions are described as the overlap of the ligands' atomic orbitals with the electron pairs with the hybrid atomic orbitals of the central metal ion.

For example, in the case of a hexaaquairon (III) complex, the iron (III) ion has 5 electrons, each singly occupying one d orbital. As the d orbitals are occupied, the vacant 4s, 4p and 4d orbitals are hybridised to give 6 $sp^3d^2$ orbitals to be used to overlap with the atomic orbitals of the ligands.

Thus, the donated electrons actually are "shared" between the valence atomic orbitals of the ligand and the valence hybrid orbitals of the metal ion.

Hope I have clarified your doubts.


How Fittig reaction gives good yield if phenyl cations are unstable?

The most probable mechanism for Wurtz reaction or Fittig reaction involves formation of organic free radicals and not cations. Like in Wurtz reaction, phenyl free radicals are formed in Fittig too.


Regarding the probability density of finding an electron in between a bond

Bonds are in general not maxima of the electron probability distribution. Local maxima are typically centered at the nuclei. This is due to the Coloumb potential. The electron can lower its potential energy by getting closer to the nuclei so they will always try to maximize their probability around the nuclei, where all the positively charged protons are bundled. This is a bit oversimplified but in essence the reason why you see the maxima at the H position or in general at the position of nuclei.
To see bonds, you should look at the difference of the probability density of two lonely  hydrogens and a H2 molecule. I.e. calculate the electronic wavefunction for a single Hydrogen Atom and then add the plots of two such single hydrogen calculations placed at the positions as in H2. Compare this with the plot coming from the H2 calculation and you should see that the H2 distribution has a higher density in between than the two unbonded hydrogens simply taken together.
Bonds as we draw them in simple chemistry are not rigorously defined mathematically and more or less a useful approximation to rationalize reactions without doing difficult calculations. There are ways to obtain a mathematical definition of bonds but these definitions are not neccessarily agreeing with the way that we draw bonds in general chemistry. One such model is Atoms in Molecules which evaluates the electron density and its topology to define bonds. But there are other models, for example Electron localization functions and i am sure that there are more, unknown to me.
Overall it is important to realize that bonds are a useful concept but they have no unambiguous clear cut underlying physical definition. Physics/Quantum mechanics yields no equation that unambiguously defines a chemical bond.


How is using Jones reagent different than Potassium dichromate in Sulfuric acid for the oxidation of alcohols?

Sodium or potassium dichromate or chromium trioxide in dilute aq. sulfuric acid all give rise to Jones reagent. This is used, though less than it used to be due to toxicity and cost issues, to oxidize secondary alcohols to ketones and primary alcohols to carboxylic acids via the aldehyde hydrate. In the case where the aldehyde does NOT form a hydrate with water (e.g. benzylic and allylic aldehydes), then that aldehyde will be stable to the conditions and be the major reaction product. Aliphatic aldehydes will generally give carboxylic acids.

Read more here.


Indeed yes, and it's been used in all kinds of airless situations, including torpedos and rockets and I think even submarines.  It's hydrogen peroxide, which when exposed to a catalyst will oxidize and reduce itself:

2 H2O2(l) -> 2 H2O(g) + O2(g)

You'll notice that the oxygens on the left are in oxidation state -1.  Half of them become reduced to the O in water (oxidation state -2) and half become oxidized to the O in O2 (oxidation state 0).


How to rationalise the coordination number of CsCl versus NaCl?

Since both compounds are of the form $\ce{AB}$ (and not $\ce{AB2}$ or $\ce{A2B3}$/etc), the coordination numbers will be equal for the ions. Similarly, for $\ce{AB2}$, the coordination number for $\ce{A}$ will be twice that of $\ce{B}$.

What happens is that each ion is connected to more ions of the opposite type in $\ce{CsCl}$. Each $\ce{Cs+}$ ion is connected directly to 8 $\ce{Cl-}$ ions, and vice versa. The net effect is that their ratio stays the same, they just have more connections.

Look at it this way: There are four men and four women in a room. First, each man holds hands with only one woman and vice versa. Now, we tell each man to hold the hands of two women, and we tell each women to do the opposite. Now, we shift it up to 3 hands (with the reasonable approximation that humans have an infinite supply of hands), and finally 4. Try this on a piece of paper -- you'll see that the man:woman ratio doesn't need to change as long as the "handholding number" (coordination number) for each person is the same.

Taking a look at the actual lattices:

For $\ce{CsCl}$ (body-centered cubic):

Note that each $\ce{Cs+}$ ion touches 8 $\ce{Cl-}$ ions and vice versa. Also note that the $\ce{Cs:Cl}$ ratio is indeed $1:1$. For each $\ce{Cs+}$ ion, there is a corresponding $\ce{Cl-}$ ion on its NE diagonal tilting into the plane of paper, and for each $\ce{Cl-}$ ion there is a corresponding $\ce{Cs+}$ ion on the SW diagonal tilting out of the plane of the paper.

Here's $\ce{NaCl}$ (face centered cubic, close packing):

Here each $\ce{Na+}$ ion touches 6 $\ce{Cl-}$ ions and vice versa. We can define a similar pairing as before -- each $\ce{Na+}$ ion has a corresponding $\ce{Cl-}$ ion to its east, and each $\ce{Cl-}$ ion has a corresponding $\ce{Na+}$ ion to its west.


Does there exist a dye that will primarily bind with water in a water/oil emulsion?

You are asking for water-soluble dyes. Typical applications are dyeing methods in microscopy and food coloring.

Water-soluble dyes for microscopy: e.g. Eosin Y, Methylene Blue, Ethyl Green.

See also this table: Wikipedia in German: Lebensmittelfarbstoff - Deutschland.


What piezoelectric materials can be made at home?

The top of my list for exhibiting piezo activity and can be easily purchased and is also transparent..

Quartz

That's about it. It's used everywhere - as an oscillator for quartz crystal watches, quartz crystal microbalances, etc.

Most of the other piezoelectric options:

Lead Zirconium Titanate (PZT)
Barium Titanate ($\ce{BaTiO3}$)
Zinc Oxide ($\ce{ZnO}$)

Aren't transparent. They're white ceramics. You can easily purchase them online from most scientific supply companies, but they don't meet your transparency criteria.

Polyvinylidene difluoride (PVDF) is claimed to be transparent (e.g., here) but you would need to apply a high electric field to pole it, and if the formulation isn't crystalline enough, you may not get a high piezo response.

My group is designing some materials that would be good, but they're still in the research stage. Stay tuned.


NMR spectrum of plasma treated glucopyranoside

The usual approach to determining change by NMR following a reaction or procedure is to compare your final spectrum to your starting sample's spectrum, run with identical parameters. That is, look at the spectrum you ran before your plasma treatment and look for changes in peak position and intensity. Within Topspin (which your spectrum here has been processed with) this is very easy to do; you can run a dual display and overlay your spectra on the screen, and even compute a difference spectrum.

Of course, if you are having problems assigning the final spectrum, you should again go back to the spectrum of your starting material and compare to the assignments you made for that. That will make it easy to identify which parts of the molecule have undergone some change. Of course, you might expect that you haven't had complete clean conversion, and there may be a number of products constituting only partial conversion. Otherwise, you are potentially having to characterise a completely new molecule.

You do have the spectrum for your starting material, don't you?


Your suggestion of resulting corresponding hydrazone by the reaction between acrolein and hydrazine is correct. It is normal hydrazine-hydrozone reaction you perform in idetifocation of aldehyde and ketone carbonyl groups. Usually, E-isomer is preferred (Ref.1). E- to Z-isomerization can be occurred in presence of trace amount of acid (Ref.1). I expect that may be the case during constant heating. As a result, Z-isomer can undergo intramolecular Michael addition as depicted in following mechanism to give dihydropyrazole:

References:

Shigehisa Uchiyama, Masanori Ando, Shohei Aoyagagi, "Isomerization of aldehyde-2,4-dinitrophenylhydrazone derivativesand validation of high-performance liquid chromatographic analysis," Journal of Chromatography A 2003, 996(1-2), 95–102 (https://doi.org/10.1016/S0021-9673(03)00542-9).


Polyethylene is generally considered suitable for food handling and even for warming food. Unlike some plastics such as polycarbonate, it does not need an additional plasticizer that could be toxic.

That said, the melting point of polyethylene is ~1150C, and I've found that it tends to shrink-wrap around microwaved food. For that reason, polypropylene, m.p. ~1300C, is often used for food containers intended for microwave oven use and for automatic dish-washing.

BTW, both polyethylene and polypropylene absorb some of the natural lycopene coloring of tomatoes or other carotenoid pigments and slowly develop a pale brown discoloration that cannot be removed, but that is of no significance.


The thing is that $\ce{HCO3-}$ can act as an acid and a base. These substances are  called amphiprotic (other examples include $\ce{HSO4-}$ and $\ce{H2PO4-}$). So it can donate a proton to water to form $\ce{H3O+}$ but it can also accept a proton from water to form $\ce{OH-}$. So really, the  question is which process occurs to a greater extent?.

You can't answer that question by simply looking at the molecule. Instead we need to know the $\mathrm{K_a}$ values for $\ce{H2CO3-}$ and $\ce{HCO3-}$. By comparing these values and doing a few calculations we can determine whether $\ce{HCO3-}$ is basic or acidic. It turns out $\ce{HCO3-}$ is actually basic.

However I am assume that equilibrium constants are probably above the level that you are learning at. So this is just one of these things which you should just know. For example you should know that despite $\ce{HCO3-}$ being basic, $\ce{HSO4-}$ and $\ce{H2PO4-}$ are actually acidic.


Could I get a hint on the structure of this molecule based on HNMR spectra?

That is not a quartet, but probably two overlapping signals. You always have to pay attention to the signal ratios and not just the number of split signals. The ratio here is 2:2:2:1, that is not a quartet ratio. A quartet would be 1:3:3:1.

There is a certain distortion in strongly coupled systems, the "roof effect" you can also see in this spectrum. This distorts the ratios a bit from the ideal ones, but not so much as to explain the ratio in the pseudo-quartet here.


Inductive vs resonance effects and the acidity of phenol

The acidity of $\ce{A-H}$ is a measure of where the following equilibrium lies.

$$\ce{A-H <=> A^- + H+}$$

This means that we are comparing the relative stabilities of the products and reactants ($\ce{\Delta G=-RTlnK}$).

In the present example (assuming your $\mathrm{p}K_\mathrm{a}$ for propen-2-ol is really being measured in water, note too that in water the relative concentration of this enol of acetone is ~ $10^{-8}$, how can you measure the $\mathrm{p}K_\mathrm{a}$ of something so dilute?) we are comparing the relative stability between [phenol and phenoxide] to the relative stability between [propen-2-ol and the corresponding anion].  Just because the two systems have the same difference in relative stabilities does not mean that phenol and propen-2-ol have the same stability.  In other words, just because the two systems have similar $\mathrm{p}K_\mathrm{a}$'s this cannot be interpreted to mean that the additional resonance structures in phenol and phenoxide are not significant contributors to both their stability and description.

EDIT: Jerepierre's comment as worked up by Dissenter (Addressing jerepierre's point)

in his edit Dissenter wrote,

  Given this, we can conclude that the reactants are less stable in the acid-base sense than the products

I agree

Dissenter further posits,

  acetone enolate is less stable than phenoxide. Phenol is also less stable than acetone enol.

That's where I disagree.  We know that the reactants are less stable than the products, but I don't see how we know which reactant(s) is less stable than which of the products.

Let's say the reactants are $10\,\mathrm{kcal/mol}$ (made up number) less stable than the products.  Is phenol $5\,\mathrm{kcal/mol}$ less stable than phenoxide and acetone enolate $5\,\mathrm{kcal/mol}$ less stable than acetone enol?  Or is it $10$ and $0$ or $0$ and $10$, or $15$ and $-5$?  How can you break down the overall reactant/product difference of $10\,\mathrm{kcal/mol}$ further?


Ozone has a bent structure with a partial positive charge on the central oxygen atom and partial negative charges on the two terminal oxygens. It has a permanent dipole moment of 0.53 D, according to Wikipedia.


The Masterton Hurley (6th ed.) textbook has a bond dissociation energy of +499 kJ/mol for breaking water into a hydrogen atom and the hydroxyl radical OH. The bond dissociation energy for breaking the hydroxyl radical into another hydrogen atom and an oxygen atom is given as +428 kJ/mol.

You could make the argument that you are starting out with a situation where all electrons are paired and end up with two radicals in the first step, whereas in the second step you already start with unpaired electrons (the hydroxyl radical has seven valence electrons).

In the title of your question, you mention the hydroxide ion. This is not a relevant species because bond energies usually refer to homolytic bond cleavage, with one electron each remaining with the two fragments. It would require heterolytic bond cleavage to go from water to the hydroxide ion (water reacting as a Bronsted acid). This reaction merely requires +55 kJ/mol in aqueous solution.

Some sources contribute to the confusion between hydroxide ion and hydroxyl radical in this context. For example, Boundless states:

  For instance, the HO-H bond in a water molecule requires 493 kJ/mol to break and generate the hydroxide ion (OH–). Breaking the O-H bond in the hydroxide ion requires an additional 424 kJ/mol. Therefore, the bond energy of the covalent O-H bonds in water is reported to be the average of the two values, or 458.9 kJ/mol. These energy values (493 and 424 kJ/mol) required to break successive O-H bonds in the water molecule are called ‘ bond dissociation energies,’ and they are different from the bond energy. The bond energy is the average of the bond dissociation energies in a molecule.

They incorrectly refer to the hydroxide ion (OH-) and should be referring to the hydroxyl radical (OH).


Leaving Group Tendencies in Amide Hydrolysis

You can imagine following situation in this hydrolysis:
$$\ce{R-C(=O)NH2 + OH- <=>[$k_1$][$k_2$] R-C(NH2)(OH)O- \\->[$k_3$] R-C(=O)OH + H2N- ->[$k_4$] R-C(=O)O- + NH3}$$
Keep in mind that:
$$\ce{R-C(=O)NH2 + OH- ->[$k_1$] R-C(NH2)(OH)O- }$$
$$\ce{R-C(NH2)(OH)O-  ->[$k_2$] R-C(=O)NH2 + OH-}$$

In this reaction, $k_1 \gt k_2$ because of LeChateliers principle since $[\ce{OH-}]$ is in excess thus forward reaction is in favor.
Also, $k_2 \gt k_3$ because of the reason given by OP ($\ce{OH-}$ is better leaving group than $\ce{H2N-}$).
Since acid-base reaction is much faster than any other reaction, we can conclude that $k_4 \gt \gt k_3$.

Therefore, as soon as $\ce{R-C(=O)OH + H2N- }$ formed they would convert to $\ce{R-C(=O)O- + NH3}$ quickly so that $\ce{R-C(=O)OH + H2N- }$ can not able to go back to $\ce{ R-C(NH2)(OH)O-}$ intermediate. That means, when boiling with extra basic solutions, amide will be hydrolyze slowly to its corresponding carboxylate.


Need of refining aluminium and Hoope vs Hall–Héroult processes

I think the difference is due to the way eddy currents form Magnetohydrodynamics.

As the current flows through the conductive liquid, a magnetic field is produced. This in turn causes the liquid to be pushed (at right angles to the current and field).

In one version of the cell, this current causes stirring of the aluminum, alumina, flux mix. This is more efficient from a time and electrical point of view. The trade-off however is that some of the impurities which would otherwise get captured in the slag (waste), end up mixed into the final product.

Edit: Thinking about it some more, the reason for the mixing may be due to where the $\ce{CO2}$ gas is formed. ($\ce{Al2O3 + C_{electrode} -> Al +CO2}$) Bubbles formed a the bottom of the cell would rise up and cause more stirring than if they formed near the top of the cell.


Why is the most probable KE for a monoatomic ideal gas exactly 1/3 of its average KE?

There is no physical reason. It's bare mathematics. The probability distribution in terms of energy is asymmetric, for it is bounded by 0 from below and not bounded at all from above. Asymmetric distributions are known to have (generally speaking) different values for mean, mode, and median.
Think of the exponential distribution: its average value is some positive number, but the mode (the most probable value) is $\bf 0$.

Moreover, 3 is not even some intrinsic characteristic of the system. If you rewrite the distribution in terms of velocity (the way it is commonly done), you'll have an altogether different situation. All of a sudden, the ratio of mean to mode is... $2\over\sqrt\pi$, how would you like that?


What is responsible for black tarnish on silver?

There is no such thing as "being reactive". Say, you have A and B; some compounds tend to react with A and not with B, and some would react with B and not with A. Which is more reactive? You can't tell; you can't even ask this question, for it's meaningless.

Black tarnish on silver is mostly silver sulphide, as confirmed experimentally. I'd even say that silver does not form a black film on being exposed to air, unless there is some source of hydrogen sulphide nearby (raw eggs, sewage pools, or something else).

It is just that $\ce{Ag}$ tends to prefer $\ce{S}$ over $\ce{O}$. Many other heavier metals are like that, and many lighter metals have different preferences. This is explained (kind of) by the hard/soft acids/bases concept.

So it goes.


A 1.00 L gas at 100 °C and 500 Torr contains 70.0 % He and 30.0 % Ne by mass. What is the partial pressure of He?

You're getting a different answer because partial pressure is proportional to
the fraction of each gas in moles, rather than by weight. Why is that?
Looking at $pV=nRT$, you can see that the ratio between partial
pressures is the same as the ratio between number of moles (since $V,
R, T$ are the same for both gases).

So, if you respect proportions,
your answer will come out the same whether you consider $1.00\ \mathrm{L}$ of the mix
or some unspecified number of liters. This lets you avoid a messy
calculation with $pV/RT$.
You may as well assume that there are 70 g of helium and 30 g of
neon. Find how many moles of each gas that represents. The partial
pressures of each gas are in the same proportion as the number of
moles. With this and the total pressure, you'll get the answer.


It is not a chemical problem, but a trivial mathematical problem.

If you have an equation $x \cdot y = c$, where $x$, $y$ are variables and $c$ is constant, then if $x$ increases, $y$ must decrease, otherwise $c$ is not a constant.

The key part is to understand that  chemical equilibrium means existence of 2 ongoing opposite chemical reactions of the same rate. Like

$$\ce{H2O <=> H+ + OH-}$$

where the rate of ion creation equals the rate of ion recombination.

If there is an excess or deficit  of either of  $\ce{H+}$   or $\ce{OH-}$ ions, the rate of their recombination changes, while the rate of their creation remains the same. As consequence, the product of their concentrations converges quickly towards $K_\mathrm{w}$ to be in the equilibrium again.

The ammonia reacts :

$$\ce{NH3 + H+ <=>> NH4+}$$

That creates deficit of $\ce{H+}$. As consequence, dissociation

$$\ce{H2O -> H+ + OH-}$$

Is faster then recombination

$$\ce{ H+ + OH- -> H2O}$$

An alternative reaction mechanism is
ammonia reacting with water:

$$\ce{NH3 + H2O <=> NH4+ + OH-}$$

what directly produces the excess of $\ce{OH-}$ that recombines with the most of the present $\ce{H+}$.

The production of $\ce{OH-}$ and elimination of $\ce{H+}$ continues, until their concentrations satisfy both the basicity constant of $\ce{NH3}$ ( or equivalently acidity constant of $\ce{NH4+}$ ) and the ion product of water  $K_\mathrm{w}$

$$K_\mathrm{w}=[\ce{H+}][\ce{OH-}]$$

$$K_\mathrm{b, \ce{NH3}}=\frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]}=\frac{[\ce{NH4+}]K_\mathrm{w}}{[\ce{NH3}] [\ce{H+}] }=\frac{K_\mathrm{w}}{K_\mathrm{a, \ce{NH4+}}}$$


Where does the pH come from in the equation of Standard Hydrogen Electrode?

There are a few logarithm rules that you need to apply here.
\begin{align}
E &= \mathrm{RT \over F}\ln {a_{H^+} \over (p_{H_2}/p^0)^{1/2}}\\
  &= \mathrm{RT \over F}\ln a_{H^+} - \mathrm{RT \over F} \ln \left({p_{H_2} \over p^0}\right)^{1/2}\\
  &= \mathrm{\ln(10) RT \over F} \lg a_{H^+} - {1 \over 2}\mathrm{RT \over F}\ln \left({p_{H_2} \over p^0}\right)\\
  &= -\mathrm{2.303 RT \over F} \mathrm{pH} - {1 \over 2}\mathrm{RT \over F}\ln \left({p_{H_2} \over p^0}\right)
\end{align}

You need a few logarithm rules:

$\log\left({a \over b}\right) = \log(a) - \log(b)$
$\log_b(x) = \frac{\log_a(x)}{\log_a(b)}$
$\log_b(x^r) = r \log_b(x)$


How is the chirality of sidechains written in the IUPAC name?

You have correctly ideentified and numbered the parent structure (the longest chain), which is undec-3-yne.
Including the substituent groups, the complete name is 10-bromo-5-(butan-2-yl)-2-methylundec-3-yne.
Note that names for simple substituent groups derived from acyclic hydrocarbons with the free valence not at the end of the chain are generically called ‘alkanyl’. A low locant is assigned to the free valence (i.e. the suffix ‘yl’) in accordance with the numbering of the chain. In this case, butan-2-yl is such a substituent group.
Also note that the locant ‘2’ in butan-2-yl refers to a different structural element compared to the other locants in 10-bromo-5-(butan-2-yl)-2-methylundec-3-yne. Therefore, parentheses are used around the substituent prefix to separate the locants.
If you need to indicate the stereochemical configuration, the same numbering and separation of locants is used, for example: (5R,10R)-10-bromo-5-[(2S)-butan-2-yl]-2-methylundec-3-yne.


Interpretation of Figure - biofunctionalization

The amount of adsorbed protein is proportional to the area of the 100 nm islands. Figure 4E proves it. The authors misinterpreted their own figure.


Neither of your options is correct.

Ideally an acid base indicator would be colorless at one pH and colored after the transition. Such an indicator is phenolphthalein. So when using phenolphthalein it is indeed impossible to detect when all the phenolphthalein is colored. So you titrate to a point where there is the faintest pink color.

So for bromcresol green you should either stop when the solution starts to turn from blue to blue green (ph about 4.6), or as the solution turns from green-yellow to yellow (ph about 5). It would very difficult to judge the deepest green (at about 4.8) accurately.

So which transition you would choose would depend on balancing between which side of pH 4.8 that you wanted to end on, and which transition was easier to detect visually.

You can look at a list of some acid/base indicators at Wikipedia. There are few that go between colorless to colored. Most are color-1 to color-2.

I also point out that you question contains a fallacy.

    Should titration stop at the intermediate color or a drop past the intermediate

The burette column is marked "to contain" an amount of fluid along its length. So you can let a partial drop form, then touch the tip of the burette to the flask and get a partial drop of titrant. You don't have to titrate in "quantized" drops.


Compounds that adopt the zincblende structure

Generally when you get a preference between two structures due to directional bonding, the first order environment of the cation is different. In this case however both are tetrahedrally coordinated. The Wurtzite structure has a marginally greater Madelung constant (1.641 versus 1.638).

Of the top of my head, I think the dominant factor in favour of the blende structure is stronger directional cation-cation and anion-anion interactions. In the case of Cu(I), the d-orbitals are going to be fairly diffuse due to the low charge, potentially allowing some covalent interaction between adjacent sites. A much better example would be GaAs, where both cation and anion are realtively soft/polarizable, which adopts blende, in contrast to AlN, with hard ionic interactions, which adopts wurzite. I don't think it's really as simple as this because the difference in energy between the two structures is often very small; you can see this in the relative stability of both ZnS polymorphs.


If there are multiple bases in solution and a strong acid is added, will both bases react?

Both based will indeed react with the acid.  But wait, there's more.  When the bases react to form their conjugare acids, the conjugate acid from one base could react with the other base.  When one base is strong and the other weak, this secondary reaction will favor the direction where the strong base gets protonated.  So you see the strong base gets more protons while the weak base tends to be pushed back to the basic form.

What if both bases are similar in strength?  Then we apply equilibrium thermodynamics.  Say hydrochloric acid is added to a solution containing both sodium acetate and sodium nitrite.  The equilibrium between each of the latter two ions is governed by the appropriate acid dissociation constants:

$\ce {HC2H3O2 = H+ + (C2H3O2)-} $  $K_{a1}={\ce {[H+][(C2H3O2)-]}}/{\ce {HC2H3O2}}$

$\ce {HNO2 = H+ + (NO2)-} $  $K_{a2}={\ce {[H+][(NO2)-]}}/{\ce {HNO2}}$

To get the equilibrium that relates the competing weak acids and bases directly, just divide to eliminate the hydrogen ion concentration; in this case

${\ce {[HNO2][(C2H3O2)-]}}/{\ce {[(NO2)-][HC2H3O2]}=K_{a1}/K_{a2}}$

So your acid dissociation constants aren't just about acid dissociation.  They also govern how a limited supply of protons gets distributed between competing bases.


Since $\ce{NaOH}$ is a strong base, the pH can be directly calculated from the analytic concentration (1M):

$\ce{pOH} = -\log\ce{[OH-]} = -\log(1) = 0  $

$\ce{pH} = 14 - \ce{pOH} = 14 - 0 = 14$

For ammonia, you need to consider its reaction in water and the associated $\ce{K_b}$ value,

$\ce{NH3 + H2O -> NH_4+ + OH-}$

The equilibrium concentrations for an unknown $x$ reacted amount (in concentration units) would be

$\ce{[NH3]} = \ce{[NH3]_0} - x$

$\ce{[OH-]} = x$

$\ce{[NH4+]} = x$

$$\ce{K_b = \frac{[NH4+][OH-]}{[NH3]} = \frac{x^2}{[NH3]_0 - x}}$$

By plugging in the proper $\ce{K_b}$ value for ammonia and its analytical concentration ($\ce{[NH3]_0}$), you can calculate $x$, from which

$$\ce{pOH} = -\log(x)$$


You likely formed an emulsion. Many of the flavor compounds in the products you mentioned are negligibly soluble in water but soluble in ethanol. Once you add the ethanol extract to water the much of the ethanol from the extract will mix with into the water but the flavor compounds in the extract will not dissolve in the water, which leaves micro-sized spheres of flavor floating around the water.  these micro-sized spheres have a different refractive index of light and thus disperse the light making it cloudy.  The same thing will happen if you mix water and oil with soap.  If you look at Gatorade or PowerAde sport drinks all of them are cloudy because of the emulsion produced from micro-size pockets of flavor in the drink.


Bicycle tube inflation with 16g CO2 cartridge

Assumptions. The $\ce{CO2}$ will cool down a bit as it expands out of the cartridge (due to the Joule-Thomson effect). But let's assume that you're interested in the pressure in the tire when it warms back up to room temperature, $25~^\circ\mathrm{C}$, so there is essentially a zero temperature change before and after filling. Let's also assume that the full $16~\mathrm{g}$ of $\ce{CO2}$ is contained by the cartridge and tire when they are connected; none of it is vented during filling.

Estimating the volume of the tire and cartridge. A $30 \times 700$ bicycle tire has a $30~\mathrm{mm}$ outside tube diameter and a $70~\mathrm{cm}$ wheel diameter. Let's say the rubber in the tire is $3~\mathrm{mm}$ thick. You then have an inside diameter of about $24~\mathrm{mm}$. If we assume the tube has a circular cross-section, the volume inside the tire when fully inflated is.

$$V = \pi \left(\frac{2.4~\mathrm{cm}}{2}\right)^2\cdot (\pi \cdot 70~\mathrm{cm}) \cdot \frac{1~\mathrm{L}}{1000~\mathrm{cm^3}} = 0.99_5~\mathrm{L}$$

Let's assume the $\ce{CO2}$ cartridge has a volume of $21~\mathrm{cm^3}$. The total volume of the $\ce{CO2}$ in the attached cartridge and tire will then be about $1.0_2~\mathrm{L}$.

If you have $16~\mathrm{g}$ $\ce{CO2}$, and the tire is fully deflated to begin with, the moles of gas is.

$$n = 16~\mathrm{g} \cdot \frac{1~\mathrm{mol}}{44.01~\mathrm{g}} = 0.36_4~\mathrm{mol}$$

Estimating the pressure. Since we're dealing with a gas with significant intermolecular interactions, which is also at pressures above $1~\mathrm{atm}$, let's compare an ideal gas calculation of the pressure
with a van der Waals calculation to see if nonideality makes any difference.

$$\begin{align}
P_{ideal} &= \frac{n R T}{V}\\
          &= \frac{(0.364~\mathrm{mol})(0.082059~\mathrm{L~atm~mol^{-1}~K^{-1}})
              (298~\mathrm{K})}{1.02~\mathrm{L}}\cdot
              \frac{14.696~\mathrm{psi}}{1~\mathrm{atm}}\\
          &\approx 129~\mathrm{psi}\\
P_{vdw}   &= \frac{n R T}{V - n b} - \frac{n^2 a}{V^2}\\
          &= \frac{(0.364~\mathrm{mol})(0.082059~\mathrm{L~atm~mol^{-1}~K^{-1}})
             (298~\mathrm{K})}{1.02~\mathrm{L} - (0.364~\mathrm{mol})
             (0.04267~\mathrm{L~mol^{-1}})}\\
          &~~~~~~ - \frac{(0.364~\mathrm{mol})^2(3.592~\mathrm{L^2~atm~mol^{-2}})}
                    {(1.02~\mathrm{L})^2}\\
          &= 8.43~\mathrm{atm} \cdot \frac{14.696~\mathrm{psi}}{1~\mathrm{atm}}\\
          &\approx 124~\mathrm{psi}\\
\end{align}$$

A tire gauge reads (gas pressure - atmospheric pressure), so you'd read about about $109~\mathrm{psi}$ if you started with a fully deflated tire. If you started with a tire filled with air at $1~\mathrm{atm}$, though, you'd get about $124~\mathrm{psi}$.


The most practical approach is to identify the structure of an organic compound by combination of spectroscopic techniques.  This is why you were provided with UV-Vis, IR, NMR and MS data, because it equally reflects the eventual approach in the lab after product isolation and purification, too.  Hopefully you have sufficient background what theses techniques assess (if not, consider consulting e.g., the spectroscopy section here on ChemSE, or on Chemistry LibreTexts).
Then, start training with simpler molecular structures in first place.  You may use, for example, Field's training booklet from the shelves (either print or/and electronic resource) of your school's library.  The obstacle here: for the majority of the sets, there is no answer in the book (the key is available only to instructors).  You may however list structures likely to be in agreement with the data per set with public references such as the SDBS spectral database.  You equally may train yourself picking an organic compound from Wikipedia and list the features a theoretical UV-Vis IR or MS or NMR spectrum would contain and again compare, e.g. with SDBS.
Yes, there are programs to predict the position and shape of NMR spectra, and you equally may find diazapam with a search via Hill formula on nmrshiftdb2.org, too.  They may help you to internalize how structural features (e.g., CH vs CH2) show in the APT spectrum because you may select either of the type of nuclei to be displayed, or to highlight their attribution of a calculated position of a signal to an atom in a given structure, or their indication of a cross peak e.g. in the HSQC.  Aside from an exam where these programs might not available to you, don't rely too much on them.  E.g., the position of signals in the $\ce{^1H}$-NMR spectrum depends at least on the concentration of your analyte, the choice of solvent (e.g. chloroform or DMSO) may influence if exchangeable protons on alcohols and amines yield a broad signal, or split into a multiplet (if there are coupling nuclei).
The only way to get better in this is training.  If the structure suggested is not in agreement with all of the spectral observations, the assignment is erroneous and one has to start again. It may be demanding, but don't give up.

As a starter:

N-NMR:  there are two types of N with different chemical environment
H-NMR:  there are two sets of alkyl-H; by integration possibly one $\ce{CH3}$ and one $\ce{CH2}$ or integer multiples of both.
APT-NMR:  again indication for a methyl, and a methylene group; count of CH: ... which are in the region expected for aromatic CH, count of C: ... etc.


How can I reverse engineer a commercial cleaning solution's "other ingredients"?

Reverse engineering, or deformulation, is a common industry practice. This is a very common type of request for commercial analytical laboratories, and is generally not cheap. You can discover just how widespread it is, and how many commercial providers there are with a quick Google search.

To get yourself a reasonable starting point, you should request a product MSDS from your suppplier. All suppliers should be able to provide a full MSDS, which will give an approximate ingredient listing and component concentration. Sometimes, you will find ingredients listed as "Trade secret" or "Proprietary", but many times you will get more information. It is very likley that your lens cleaner solution is, in fact, predominantly distilled water; perhaps up to 90-95%. However it is the other minor ingredients that are the active ingredients, and may include alcohols or glycols, anti-fog and anti-static agents, anti-baceterials and preservatives. Examples of lens cleaner MSDS can be found via a google search, and this may give you a useful bit of background information.

Note:
1. all suppliers are legally bound to providing MSDSs upon request in the local idiom (official language).
2. this document is safety-based which means it will (a) provide (useful) information on possibly hazardous materials, (b) assess the hazardous component(s), their type, level and correlate to concentration.
3. websites such as state Chemical authorities (ECHA, FDA, ...) publish ratings, studies and lists (of registered/authorized/forbidden) of chemicals for use in particular applications such as food contact, medical, pharmaceutical devices/products.

The next step is to determine what information you really are after, as this will determine the budget you will need to outlay. If you want to just determine whether the solution is predominantly water, you could try a few things yourself (if you are experiment-savvy). Try smelling it, as a first step; check for evaporation if you run a thin film on a glass slide, test density or melting point. These will all tell you much the same thing, which is, yes, the solution is predominantly water. A simple commercial 1H NMR analysis will show you this also, and confirm water concentration up to about 99.9%. This means minor ingredients will show up, but as long as they make up about 0.1% of the ingredient composition, and they contain hydrogen (sodium bicarbonate will not show up for instance). This method is commonly used when you have a reasonable idea about the identity of the minor ingredients, but can involve extensive analysis if you really are shooting in the dark.

Beyond this, if you actually want to determine the ingredients, and have no idea what they might be, you will need to head down the avenue of GC-MS and/or LC-MS; a combination of chromatography to separate all of the ingredients, and then Mass Spectrometery to try to determine the mass of each component. This type of analysis is generally charged by the hour, and the more variables you are trying to evaluate, the more expensive the procedure. Generally, when analysing a competitor's products, a reasonable knowledge of the types of raw materials exists, and fingerprinting chromatography and NMR traces against raw material libraries saves a lot of time (and money). Sometimes, other techniques such as FTIR, AAS, UV-Vis or ICP may also be used, but these would really only be if the MS methods failed to give you suitable leads.

Commercial analytical labs will be able to give you a very good summary of the best techniques for your sample, but be prepared; the very first question you are likely to be asked is "what is your budget?". Good luck.


You are on the right track when you mention molality.

Because freezing point depression is a colligative property of the solvent, in this case water, all you need to do is to find out which solute has more moles per kg, aka molality.

The equation for freezing point depression for a non-ionic solute is given as:

$$\Delta T_\mathrm{F} = K_\mathrm{F}\times b$$

Where:
$\Delta T_\mathrm{F} =$ freezing point depression
$K_\mathrm{F} =$ cryoscopic constant (dependent on solvent, not solute)
$b =$ molality, the moles of solute per kg of solvent

The key point here is that freezing point depression is linearly proportional to solute concentration (molality).

Since glycerol has a molecular mass of $\pu{91.1 g/mol}$, 1 kg will contain 11.0 moles of solute. Methanol has a molecular mass of $\pu{32.0 g/mol}$, so 1 kg will contain 31.2 moles of solute.

So, for any given price per kg, methanol will give you $31.2/11.0 = 2.9$ times as many moles of solute, resulting in a 2.9 times greater freezing point depression.  Note that you cannot calculate the actual freezing point depression without knowing $K_\mathrm{F}$.


Does the 'X' of Grignard's reagent react?

Lets look at Grignard's reaction route. Key step is nucleophilic addition of R. And MgX is removed with acid on the next step. Nobody cares about X - if you want MgX2 in solution you just add MgX2. The main reason to use Grignard's reagent is nucleophilic addition of R.


Are photolysis reactions considered exothermic or endothermic?

In general it could be either. In your example you will have to work out the heat of formation of ozone and of O atoms (presumably it is an endothermic reaction). In general what the photon is doing is providing a way to overcome an activation barrier in the ground state by opening up a pathway from an excited state to the products. This last step is generally exothermic.

The enthalpy released in the second reaction is less than the photon energy by the difference in energy of the photon and that of products, an oxygen atom in your case as the heat of formation of O$_2$ is zero.

The sketch gives the general idea, I hope, for an endothermic reaction :)

Notes: The barriers may also be called transition states. There may be no barrier in the excited state. The crossing from the excited state potential to the product may be controlled by Landau-Zener behaviour or be a conical intersection. The far right upper line (not labelled) is the potential of the excited state of the product


You need to know how much of each solution you add and the final total volume. The molarity of each one would be

$$\frac{C_1 \cdot V_1}{V_1 + V_2} \quad \text{and} \quad \frac{C_2 \cdot V_2}{V_1 + V_2}$$

where $C_1 = \pu{2 M}$, $C_2 = \pu{1 M}$, $V_1$, $V_2$ - volumes of $\pu{2 M}$ and $\pu{1 M}$ solutions, respectively.


What can be used as substitute to dichloromethane for breaking optical cement bonds?

According to an email i got back from Norland, "It will work if the paint stripper has a chlorinated ketone in it"


Stabilization of multiple bonds through inductive effects

Since electron-withdrawing groups (EWGs) lower orbital energy (we can rationalise this in terms of reduced electron density leading to reduced shielding), they will stabilise multiple bonds - you see this in the Diels-Alder reaction, where EWGs on the dienophile reduce the energy of its LUMO and so accelerate the reaction by reducing the gap between the dienophile LUMO and diene HOMO (conversely, electron-donating groups on the diene destabilise its HOMO by the same logic and so also speed up the reaction).


Acetic acid is present in 'wood vinegar' that you get by thermal destruction with a lack of oxygen.
Steam can be used as a source of heat, but to be of any help in your desired production it would have to be overheated or under very high pressure, about 100 atmospheres. It is still useful as a way to increase the yield a bit.
Likely the best way is to heat the wood to 300*C without oxygen, collect and cool the gaseous products to condense them. You will get wood vinegar.
https://en.wikipedia.org/wiki/Pyroligneous_acid
After this you can separate the products using distillation.
https://en.wikipedia.org/wiki/Distillation
About 70 % of wood mass will be converted during the thermal destruction into wood vinegar.
About 15 % of wood mass will become non-water wood vinegar.
About 5 % of wood mass will become acetic acid.
Small yield, toxic waste, energy intensity of the process, are reasons why this process is not used anymore.


Green and red crystals of [Ni(Ph2EtP)2Br2]

The coordination number of $\ce{[Ni(Ph2EtP)2Br2]}$ is four;
thus, the complex may be tetrahedral

or square planar

(figures taken from here)

The complex $\ce{[Ni(Ph2EtP)2Br2]}$ contains $\ce{Ni^2+}$, which has eight d electrons.

When filled with eight electrons, the tetrahedral structure is paramagnetic, and the energy difference between the highest occupied and the lowest unoccupied orbital is relatively small (4.4 Dq).

The corresponding square-planar structure is diamagnetic, and the energy difference between the highest occupied and the lowest unoccupied orbital is relatively large (10 Dq).

Therefore, the square planar complex absorbs light of higher energy (lower wavelength) than the tetrahedral complex.

The green crystals absorb red light, whereas the red crystals absorb green light. Thus, the red crystals absorb light of higher energy than the green crystals. Therefore, the red crystals correspond to the square-planar complex and the green crystals correspond to the tetrahedral complex.

Furthermore, since the complex $\ce{[Ni(Ph2EtP)2Br2]}$ contains two different ligands, there are cis and trans isomers of the square-planar structure.


Which lone pairs/bonding pairs go into which pi molecular orbitals?

This is a common attempt to visualise what π systems actually mean for the overall Lewis structure. But let me note a couple of things first:

Obviously, there is no $\mathrm{sp^3}$ lone pair on DMF’s nitrogen. Instead, the atom is $\mathrm{sp^2}$ hybridised and the lone pair (if it is there) occupies a p-type orbital.
Remember that the π analysis only deals with the π orbitals (obviously); there is an underlying σ system which is implied but hardly mentioned. Thus, if ‘one bond’ is created, it is actually a double bond, while ‘no bond’ means ‘only the underlying single bond’.

If you want, in simplistic terms, you can add the π orbitals one after the other. The first is fully bonding and has contributions by all three atoms. Since it is fully bonding across two bonds but only contains two electrons, we can say that it contributes $\frac22$ bonding electrons to each bond or a bond order of ½.

The second orbital is not really bonding as it has a perpendicular nodal plane through the central atom (carbon). Thus, we cannot really analyse it in terms of the $\ce{C-O}$ or $\ce{C-N}$ bond order directly. However we can say that the electron density will be stronger at the ends, i.e. on nitrogen and oxygen. Each of these atoms practically experiences $\frac22$ lone pair electrons (do not participate in bonding) or ½ a lone pair.

Consider what would happen if you had filled the third orbital. It would have been antibonding with respect to both the $\ce{C-O}$ and the $\ce{C-N}$ bond. It would have contributed the same bond order of ½ to both — but this time in negative. That would contribute to an overall bond order of zero or the bonding electrons being on either side of the bond as lone pairs.

For your question, this means:

the fully bonding orbital cannot contribute to any lone pairs since it is not compensated for by an antibonding one
the nonbonding orbital contributes the lone pair. Since it only experiences contributions from oxygen and nitrogen, this is where the lone pair will be located.
it would be wrong to look at the Lewis structure, see a lone pair on nitrogen and immediately associate that with the ‘lone pair’ of the question. If anything, consider all relevant resonance forms and all their lone pairs. Also, as soon as it has resonanted to form a bond, it is no longer a lone pair.
finally, as per your side question, note that the $\ce{C-H}$ bond in DMF has a very, very low tendency to deprotonate. That proton is not even mentioned on both the Evans table and the Bordwell data. While it might lead to a carbene, the carbene is not exactly stable. A much more likely deprotonation would be that of a methyl group which would extend the π system. Note that the sequential treatment with butyllithium and DMF is a method to convert electron-rich aromatic rings (e.g. thiophene) into their corresponding carbaldehydes — in this reaction, the proton of DMF is less acidic than that of a thiophene or proton exchange would occur in place of nucleophilic attacks.


Electron configuration of non-hybridized carbon

The answer comes in several layers, as Martin has alluded to. I'll try and give a short summary: each point goes slightly deeper than the previous one.

For an isolated atom, the labelling of the orbitals is arbitrary. That is to say, the $p_x$, $p_y$, and $p_z$ orbitals are all interchangeable. More formally, this reflects the isotropy of the system, which means that all directions are equal. There is nothing that can physically differentiate the $x$-axis from the $y$-axis or the $z$-axis.

You can still arbitrarily assign $x$-, $y$-, and $z$-labels to three perpendicular axes. If you kept these coordinate axes consistent between different atoms, such that every $p_x$ orbital pointed in the same direction, you should not expect to find in all molecules that the $p_x$ and $p_y$ orbitals are always occupied. Instead, on average, you would have one-third of them with the configuration $(p_x)^1(p_y)^1$, one-third with $p_xp_z$, and one-third with $p_yp_z$. Again, this symmetry stems from the fact that there is no physical factor that favours any of the three coordinate axes over each other.

The previous point assumes that you have made a measurement of the electrons, such that you can definitively say that they are in a specific set of orbitals.* But prior to this measurement, there is no reason that they should already be in that state. It's better to posit that an isolated carbon atom is in a superposition of those three states, i.e. $p_xp_y$, $p_xp_z$, and $p_yp_z$. My old question: What constraints are imposed on a wavefunction by the symmetry of the system? is somewhat related, except that it deals with boron (which has the same issue).

Having gone through this, there is still the question of why we write $p_xp_y$ when it's not really accurate. The answer is that it's far easier than dealing with a superposition of states (in (3)), and since space is isotropic, $xy$ is the same as $xz$ and $yz$ as long as you relabel the axes, which means that (2) is more of a technicality rather than a real "gotcha". Also, it's perhaps not worth overthinking this: once you form a bond, all of this ceases to have any meaning. In a molecule, electrons don't remember "where" they come from, so it becomes pointless to quibble over whether it was in the $p_x$ or $p_y$ orbital.

* You can't measure individual electrons in an atom, you can only measure their collective properties. Also, I'm skeptical about whether you can measure a state of $p_xp_z$ given that $p_x$ and $p_z$ are eigenstates of non-commuting operators, but let's leave this aside for now (or assume, perhaps, that we're dealing with $p_{-1}$, $p_0$, and $p_{+1}$, which are all eigenstates of $\hat{L}_z$).


The reasoning behind predominance diagrams

I think you might be making a fundamental mistake when you differentiate between the "current amount of protons" and the "amount at equilibrium." All dissociation constants are based on equilibrium concentrations, and predominance diagrams refer to the concentrations present only after equilibrium has already been established.

Consider the concrete example of the dissociation of ammonium: $\ce{NH4+ <=> NH3 + H+}$. The equilibrium constant expression for this chemical process is:

$$
K_\text{a} = \frac{\ce{[NH3][H+]}}{\ce{[NH4+]}}
$$

The $\mathrm{p}K_\text{a}$ of this process has been experimentally determined to be approximately $9.25$. Therefore, the following equation should hold:

$$
-\log(K_\text{a}) = -\log\frac{\ce{[NH3][H+]}}{\ce{[NH4+]}} = \mathrm{p}K_\text{a} = 9.25
$$

If a given solution has a measured $\mathrm{pH}$ precisely equal to the $\mathrm{p}K_\text{a}$ of the above process, then what we have is the following (keeping in mind that $\ce{[H+]} = 10^{-\mathrm{pH}}$):

$$
\log\frac{[\ce{NH3}]\cdot 10^{-9.25}}{[\ce{NH4+}]} = -9.25
$$

Taking the anti-log of both sides of the equation:

$$
\frac{[\ce{NH3}]\cdot 10^{-9.25}}{[\ce{NH4+}]} = 10^{-9.25}
$$

Then dividing both sides of the equation by $10^{-9.25}$, we find:

$$
\frac{[\ce{NH3}]}{[\ce{NH4+}]} = 1
$$

That is to say, the concentrations of ammonium and ammonia in the solution are perfectly equal when the $\mathrm{pH}$ and $\mathrm{p}K_\text{a}$ are equal. If the $\mathrm{pH}$ is higher (that is to say, more basic), then you'll find that your conjugate base (in this case, $\ce{NH3}$) will predominate by factor of $10^{\mathrm{pH} - \mathrm{p}K_\text{a}}$. Conversely, if the $\mathrm{p}K_\text{a}$ is higher, then the conjugate acid ($\ce{NH4+}$) will predominate by a factor of $10^{\mathrm{p}K_\text{a} - \mathrm{pH}}$.

If you prefer to use an inequality and some generic conjugate acid/base pair $\ce{HA/A- }$, then the logic goes as follows.

If $\mathrm{pH} > \mathrm{p}K_\text{a}$:

$$
\mathrm{pH} > \mathrm{p}K_\text{a}\\
-\log\ce{[H+]} > -\log\frac{\ce{[H+][A- ]}}{\ce{[HA]}}\\
\ce{[H+]} < \frac{\ce{[H+][A- ]}}{\ce{[HA]}}\\
1 < \frac{\ce{[A- ]}}{\ce{[HA]}}\\
\ce{[HA]} < \ce{[A- ]}
$$

If $\mathrm{pH} = \mathrm{p}K_\text{a}$:

$$
\mathrm{pH} = \mathrm{p}K_\text{a}\\
-\log\ce{[H+]} = -\log{\frac{\ce{[H+][A- ]}}{\ce{[HA]}}}\\
\ce{[H+]} = \frac{\ce{[H+][A- ]}}{\ce{[HA]}}\\
1 = \frac{\ce{[A- ]}}{\ce{[HA]}}\\
\ce{[HA]} = \ce{[A- ]}
$$

If $\mathrm{pH} < \mathrm{p}K_\text{a}$:

$$
\mathrm{pH} < \mathrm{p}K_\text{a}\\
-\log\ce{[H+]} < -\log\frac{\ce{[H+][A- ]}}{\ce{[HA]}}\\
\ce{[H+]} > \frac{\ce{[H+][A- ]}}{\ce{[HA]}}\\
1 > \frac{\ce{[A- ]}}{\ce{[HA]}}\\
\ce{[HA]} > \ce{[A- ]}
$$


First, my answer here is relevant: What is chemical formula of Bauxite?, and also follow links inside.
Regarding your question specifically, there is not much difference. However, there are some conventions that come into play here.
The three formulae describe compounds that stoichiometrically have four atoms of oxygen and hydrogen for each atom of silicon.
$\ce{SiO2.2H2O}$
This notation simply says that for each component of "silica" there is a single component of "water". This notation is often used for solids, and never for dissolved species in aqueous solution. In some cases, this can denote that this entire thing has a crystalline structure, in which there are actual molecules of $\ce{H2O}$ held together with "molecules" of $\ce{SiO2}$. This is often called water of crystallisation, but this is not the case here because hydrated silica (which is what this thing is) is not crystalline, but is amorphous.
$\ce{Si(OH)4}$
This is a slightly more simplified way of writing the formula for hydrated silica, withour referring to specific chemical components (such as water or silica). In some cases this indicates that the crystal structure contains hydroxide groups ($\ce{OH-}$), but again, as hydrated silica is amorphous this is not the case here.
$\ce{H4SiO4}$
This is the formula for "silicic acid", which when in solid form, is exactly the same as "hydrated silica". However, this notation is often used when discussing hydrated silica in aqueous solution chemistry, as you correctly inferred. This does not necessarily mean that it is an actual acid, at least in the sense we think of when discussing things like sulfuric or phosphoric acid. We have a bottle of commercially acquired "silicic acid" in our lab. It's not an acid - it's just some hydrated silica, but written in the "acid"-type notation.

Something to remember here, is that some of these types of notation imply that the hydrogen occupies a specific structural site in the crystal lattice. The first implies molecular water, the second implies hydroxide groups, the third implies hydrogen protons.
This is misleading, because in fact hydrated silica can contain all three structural types of hydrogen simultaneously, and figuring out exactly which and their relative abundance is often of little importance in most studies. Adding to the confusion is that the compound is not really crystalline (like opal). And finally to confuse you even more, a ratio of 2-to-1 of water to silica isn't required, and it can vary. In fact, sometimes the formula is written as $\ce{SiO2.nH2O}$ because we simply don't know (and don't care) exactly how much water is in there. Finally, to add even more confusion, some groups use one notation while others use another for exactly the same thing. The same author can even use two different notations in two different papers!


In your series, all of the alkyl benzenes will have roughly the same +I inductive effect.  Where they differ is with regard to the resonance effect.  Hyperconjugated structures such as those drawn below for ethylbenzene are often invoked to explain these differences.

There are two hydrogens in ethylbenzene that are capable of donating electrons into the aromatic ring by hypercojugation.  In isopropyl benzene there is only one and there are none in *tert-*butylbenzene.  Hence ethylbenzene should be the most electron donating of the compounds in your series due to resonance involving hyperconjugation.

Realize that these effects are relatively small, for example the relative rates for the electrophilic nitration of various aromatic compounds is:

benzene=1, toluene=24, tert-butylbenzene=15.7 (reference, page 1060)

Benzene has no substituent (other than hydrogen), so neither resonance or inductive effects play a role, it would be the slowest.  The expected order in your series would therefore be:

ethylbenzene > isopropylbenzene > tert-butylbenzene > benzene


Relation between activation energy and rate constant

This can be answered both conceptually and mathematically.

Conceptually: Let's call the two reactions 1 and 2 with reaction 1 having the larger activation energy. Here is a plot of the arbitrary reactions.

It is clear from this graph that it is "easier" to get over the potential barrier (activation energy) for reaction 2. Thus if we increase temperature, the reaction would get faster for reaction 2 before it gets faster for reaction 1. Therefore, when increasing the temperature by the same amount, the rate constant for the reaction with the lower activation energy will increase more.

Mathematically: You can understand how the rate constant changes with temperature by simply graphing. Here is a very quick and very rough graph on how the rate constant changes with temperature (plotting the equation $k=Ae^{-E/RT}$).

Notice how increasing the temperature causes a quicker increase in the reaction with the lower activation energy $(E_{a2})$. However, this is only an a priori expectation.

You mentioned in a comment above about differentiating. You can use that method as well. Taking the derivative of the equation for $k$ with respect to temperature while assuming $A$ and $E_a$ do not depend on temperature yields

$$\frac{dk}{dT} = \frac{AE_a}{RT^2}e^{-E_a/RT}$$

Edit: As porphyrin pointed out in the comments below, this equation elucidates the true behavior of the temperature dependence. Plotting this function for the two energies, we see that the derivative for the reaction with a smaller activation energy increases faster for lower temperatures. But for higher temperatures, there is a shift in the change of the rate constant. Therefore, the process is temperature dependent. At low temperatures, the rate constant for the reaction with low activation energy changes is more sensitive to temperature change while at high temperatures, the reaction with larger activation energy is more sensitive to temperature change.

As a side note: at very large (infinite) temperatures, both functions approach the same value and there will be an equivalent distribution among the reactions' products.

From comments below: Prove that if we increase temperature in an endothermic reaction, equilibrium proceeds in forward direction.

There are a couple ways to do this. As this is now an equilibrium question, I will keep the regime to equilibrium instead of rate constants. First, let's use fundamental chemistry to get an expected result then use equations to show what the result would be quantitatively.

Fundamentals: Say you have a simple reversible reaction
$$A\rightleftharpoons B$$
Given this reaction is endothermic, energy is required to drive the reaction. Because of this, we can write energy as a "reactant"

$$A + Energy \rightleftharpoons B$$

According to Le'Chatelier, if you increase temperature then the equilibrium must shift towards the products, thereby answering the question.

The Maths: Now, let's apply some math as overkill. The equilibrium constant is dependent on the standard free energy of the reaction given by the following equation

$$\Delta G^o = -RT\ln(K)$$

If we want to understand how the equilibrium constant will change we can rearrange this equation to solve for $K$ and plot as a function of temperature. Doing this and making the substitution for the free energy $\Delta G^o = \Delta H^o - T\Delta S^o$ gives

$$K = e^{-\frac{\Delta H^o}{RT} + \frac{\Delta S^o}{R}}$$

Since the reaction is endothermic, $\Delta H^o > 0$. This gives the same monotonically increasing profile that we see for the Arrhenius equation (plotted above). Therefore $K$ must always increase, and therefore the reaction proceeds in the forward direction, for increasing temperature. Of course, this is making the assumption that $\Delta H^o$ does not change with temperature.


Do buffers only work for species of the opposite variety? (I.e. weak acids buffer strong bases and vice versa)

Think of it in terms of the equilibrium constant:

$$K_{c}\; =\; \frac{[\ce{H3O+}][\ce{HCO3-}]}{[\ce{H2CO3}]}$$
Let's assume for a moment that there are 2 mols of $\ce{H3O+}$, 2 mols of $\ce{HCO3-}$, and 1 mol of $\ce{H2CO3}$. In our pretend scenario, $K_c$ is 4 (however in reality the $K_c$ is much lower and is about $2.5 \cdot 10^{-4}$). When you add an acid, the $\ce{H3O+}$ concentration rises. Lets assume there are now 9 mols of $\ce{H3O+}$. So now the reaction quotient is

$$Q\; =\; \frac{9\cdot 2}{1} = 18$$

Because $Q > K$ the reaction has to shift left in order to compensate and go back to the equilibrium state. If 1 mol of $\ce{H3O+}$ recombines with 1 mol of $\ce{HCO3-}$ to give back 1 mols of $\ce{H2CO3}$ now you have

$$Q\; =\; \frac{8\cdot 1}{2} = 4$$

and because $Q = K$ the reaction is now at equilibrium.

So while the total number of mols of $\ce{H3O+}$ did increase, the effect of it was dampened by the buffer (as it only went up to 8 mols instead of 9 mols).