Acidity of aldehydes and ketones and enolate formation

I would like to first comment on the reasoning given as to why aldehydes are more acidic than ketones. The usually provided reasoning is, as stated here:

H atoms are regarded as having no electronic effect : they don't withdraw or donate electrons.
Alkyl groups are weakly electron donating, they tend to destabilise anions (you should recall that they stabilise carbocations). This is because they will be "pushing" electrons towards a negative system which is unfavourable electrostatically. Hence, the anion of a ketone, where there are extra alkyl groups is less stable than that of an aldehyde, and so, a ketone is less acidic.

We need to be clear of what we are measuring when we measure the acidity of an aqueous solution of an aldehyde. It is important that we recall what is mentioned in this post, on the acidity of benzaldehyde, as it relates to the acidity of aldehydes in aqueous solution. Most aldehydes exist in aqueous solution with a significant proportion of their hydrated forms, which are geminal diols. Hence, it is very likely that the $\mathrm pK_\mathrm{a}$ values we have obtained are for the $\ce {-OH}$ protons of the hydrate and not for the $\ce {C-H}$ protons at the $\alpha$ position of the aldehyde.
Let's take a look at acetaldehyde. Ethanol has $\mathrm pK_\mathrm{a}$ value of $15.9$ and the hydration equilibrium constant for acetaldehyde is $\ce {1}$, as mentioned here. Based on the guideline that geminal diols have $\mathrm pK_\mathrm{a}$ $2.5$ units less than that of the monohydric alcohol, the predicted $\mathrm pK_\mathrm{a}$ value for acetaldehyde is $13.4$. The experimental measurement gives $13.57$, as seen from here, which is very close to the value determined via calculation with the guidelines. This closeness of the values suggests that the hydrate is indeed formed and the acidity that we are measuring for the aqueous aldehyde is indeed that of the hydrate.
If this is what we are measuring, then clearly the reasoning highlighted in the box may not even be correct. In fact, we can't even be sure that the $\ce {\alpha}$ $\ce {C-H}$ proton of the aldehyde is more acidic than that of a ketone since we won't be able to measure their acidities, at least not in water.
With the clarification I have made, I hope that there is no longer any inconsistency in the issues you have mentioned. In particular, it is correct to say that under thermodynamic control, the intermediate with greater stability (e.g. the enolate with the more substituted alkene) is formed. Regarding your 2nd question, I am unsure of what you mean but I will just say that kinetic and thermodynamic control refer to different reaction conditions resulting in the reaction undergoing different pathways to thus form different products.
As an extension, we can also consider if the $\mathrm pK_\mathrm{a}$ values given for the ketones are representative of the acidity of the $\alpha$ protons or that of the $\ce {-OH}$ protons of the hydrate. Let us consider acetone, with $\mathrm pK_\mathrm{a}$ given as $\ce {19.16}$ here. The hydration constant for acetone is $1.4 \times 10^{-3}$ and the $\mathrm pK_\mathrm{a}$ of 2-propanol is $16.5$ (as seen here). Based on the same sequence of calculations and the same guideline, the predicted $\mathrm pK_\mathrm{a}$ value is $16.9$. The significant difference between the predicted value and the experimentally measured value suggests that the acidity of the acetone is not just due to that of the hydrate $\ce {-OH}$ protons and that there may be significant deprotonation of the $\alpha$ protons taking place.
An important question that we would like to know the answer to is what then is the $\mathrm pK_\mathrm{a}$ value of an $\alpha$ proton of an aldehyde. Well... It seems like due to hydration of the aldehyde, we would not be able to directly measure the acidity of this proton. However, one would reasonably estimate that it is close to that of the ketone $\alpha$ hydrogens. Again, that is something we may not be able to measure directly for the same reason. An educated guess would be in the range of $18-22$ as it needs to be more lower than the $\mathrm pK_\mathrm{a}$ of esters but higher than that of alcohols.


How to determine rotational states of a molecule?

Molecules have only 3 (or 2 if linear) rotational modes.  A molecule with n atoms has 3n-6 vibrational states, 3n-5 if it is linear.  The "6" or "5" that is subtracted represents the 3 translational modes that all molecules can move in (x, y, and z directions) plus the 3 rotational modes (rotation about the x, y and z axes).  The 3 rotational modes reduce to 2 if the molecule is linear since you can't observe rotation about its molecular axis (the axis that contains the $C_{\infty}$ symmetry element).


How can a thermometer ever show the actual temperature of an object if the object loses heat to the thermometer?

I'll start by mentioning that there's no such thing as an exact measurement—there is always some measurement error.  The only observations that can be numerically exact are counted numbers of discrete objects (e.g., the number of electrons in a neutral carbon atom is exactly 6).  And I say "can be", because if the numbers are sufficient large, even with counted numbers there can be errors (we see this with vote counting).
Having said that, your idea is correct—if the heat capacity of the temperature probe is significant relative to that of the object being measured, then the measurement can significantly change the temperature of your object.
This is a particular concern when measuring small objects.  For such applications, researchers can employ (as Jon Custer mentioned in his comment) non-contact thermometry.  See, for instance:
https://www.omega.co.uk/temperature/z/noncontacttm.html


How does Carbon Dioxide have a degree of freedom of 6?

There are always 3N total independent degrees of freedom for a molecule, where N is the number of atoms. These come about because when each atom moves, it has three independent degrees of freedom: its position in each of the x, y, and z directions.

Now, having independent degrees of freedom for each atom isn't all that useful. Instead, we can make combinations of different degrees of freedom. The important thing when doing so is that the number of independent degrees of freedom are preserved: it's just that what a particular degree of freedom does to the atoms changes.

The standard breakdown of degrees of freedom subtracts out global movement in each of the three directions. So you have 3N total degrees of freedom, but you can set aside 3 of them as translation of the whole molecule in each of the x, y and z directions, leaving (3N-3) degrees of freedom.

Likewise, it's standard to subtract out the whole molecule rotation. For most larger molecules, there's three different degrees of rotational freedom: rotation around each of the x, y, and z directions. But for linear molecules like $\ce{CO2}$, one of those rotations (around the axis of the molecule) doesn't actually change the position of the atoms. Therefore it's not a "degree of freedom" which counts against the 3N total. So while for non-linear molecules there are (3N-3-3) = (3N-6) degrees of freedom which are independent from the global rotational and translational ones, for linear molecules there are (3N-3-2) = (3N-5) degrees of freedom which are independent from the global rotational and translational ones. -- A quick clarification. The reason why we ignore this rotation is not because the center of mass doesn't move. The center of mass doesn't move for any of the global rotations: in the typical assignment of degrees of freedom the axis of rotation goes through the center of mass. Instead, the reason the rotation is ignored is that none of the atoms move due to the "rotation".

So since $\ce{CO2}$ has three atoms and is linear, it has 3*3 - 5 = 4 degrees of freedom which are independent of the global rotation and translation. We call these the vibrational modes. There's different ways you can decompose them, but the most useful one is 1) the symmetric stretch (both oxygens going out and in at the same rate while the carbon sits still), 2) the asymmetric stretch (the carbon going back and forth while the oxygens sit more-or-less still), and 3&4) two different out-of plane bending modes. (If the molecule is aligned along the x-axis, one each where the carbon moves in the y and z directions.)

That's it for $\ce{CO2}$: any movement you make - any position you put the atoms in - can be thought of as a combination of those nine movements (three global translations, two global rotations, and four internal vibrations/bending).


Can a saturated hydrocarbon have side chains?

Standard organic chemists’ understanding is that saturated/unsaturated is a word pair entirely independent of branched/linear/cyclic. Thus, all six combinations are possible:

branched/saturated (e.g. neopentane or 2,2-dimethylpropane)
branched/unsaturated (e.g. 2,3-dimethylbut-2-ene)
linear/saturated (e.g. n-pentane)
linear/unsaturated (e.g. hex-3-ene)
cyclic/saturated (e.g. cyclohexane)
cyclic/unsaturated (e.g. cyclohexene)

I chose these examples for a reason. The central carbon atoms in both neopentane and 2,3-dimethylbut-2-ene are only bonded to other carbon atoms while all other carbons in the branched examples are methyl groups ($\ce{CH3}$). Yet, no organic chemist will classify neopentane as ‘unsaturated’ — so clearly there must be an additional criterion other than ‘maximise the number of bonds to hydrogen atoms’.

On the other hand, hex-3-ene has almost maxed out its number of hydrogen atoms: two carbons with three, two carbons with two and only two carbons which are lacking a hydrogen each. But that does not make it any less unsaturated. Note that cyclohexane and hex-3-ene have the same formula $\ce{C6H12}$ and thus the same carbon/hydrogen ratio. Yet, the cyclic compound is considered saturated and the linear isn’t.

Thus, forget about the maximum hydrogen definition. A saturated molecule is one that is made up of single bonds only.

Note: Of course, cyclic compounds can also be branched or ‘substituted’. But that doesn’t add too much.


Some higher-molecular weight alkanes definitely have an odor, but they aren't as intense as most of the compound types you'll find stinking up an organic lab.

"The odor threshold for hexane is 130 parts per million (ppm), with a faint peculiar odor reported." ---http://www.epa.gov/ttnatw01/hlthef/hexane.html
"[Heptane is a] colorless liquid with a mild gasoline-like odor." ---http://www.cdc.gov/niosh/docs/81-123/pdfs/0312.pdf
"Alkanes have relatively high odor detection threshold (37 to 50 mg/m$^3$ for dodecane) as compared to other VOC groups such as sulphides (0.0003 to 0.16 mg/m$^3$  for
dimethylsulphide), phenolics (0.022 to 4 mg/m$^3$  for phenol), and nitrogen heterocycles (0.0004 to 0.0008 mg/m$^3$  for skatole) (O'Neill and Phillips, 1992)" ---http://www.bae.uky.edu/jbicudo/odorVOC_02a.pdf

According to the following paper, there are apparently at least two olfactory receptors  in rats that respond to alkanes; one accepts alkanes with areas around 3.5 nm$^2$, and another  responds to hydrocarbons with areas of about 5 nm$^2$. The authors note that response to cycloalkanes is higher than for normal alkanes, so the shape matters too.

"Concanavalin A reveals olfactory receptors which discriminate between alkane odorants on the basis of size." ---http://www.ncbi.nlm.nih.gov/pubmed/2803264


How exactly do sulfites prevent oxidation of wine?

Sulphites work as anti-oxidants in wine because they are very easy to oxidise to sulphates. Schematically:

$$\ce{SO_3^{2-} + \text{oxidant}\to SO_4^{2-}}$$

This ease with which the sulphite ion is oxidised to the sulphate ion means that sulphite added to wine tends to 'mop up' any oxidisers (like air oxygen) that might get into the wine, rather than the alcohol (and maybe other substances in the wine) becoming oxidised.

The sulphite method has the added advantage that it, as well as its sulphate reaction product, is quite harmless to humans.


Maximum observed optical rotation through a polarimeter

This is something I discovered after an attempt at a more mathematically rigorous investigation of how to interpret polarimetry results. It's not that hard, but I don't think you'll ever find a similar analysis in a book because in practice the details are pretty much unnecessary for experimental purposes (though fun mathematically!).

Before we begin, it's very important to distinguish two types of optical rotation. The first is the real optical rotation, $\alpha_{real}$. This is the twist which a substance actually causes on the plane of polarized light as it travels through the optically active solution, and therefore is the rotation which is part of the specific rotation equation, shown further below. The numerical value of this angle can be any finite real number, in degrees or radians. There is no constraint on how much an optically active solution is "allowed" to twist a plane of polarized light; the fact that you can't directly see how many complete twists the plane of polarized light made is entirely an experimental limitation.

This experimental limitation is what gives rise to the other type of rotation: the observed or apparent rotation, $\alpha_{obs}$. This is an angle which is also real number, but which is limited between $0$ and $180$ degrees or $0$ and $\pi$ radians (or you could say between $-90$ and $90$ degrees or $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ radians, it makes no conceptual difference). The observed rotation is what you measure in a polarimeter, and in principle it cannot be directly put into the specific rotation equation.

Why is $\alpha_{obs}$ allowed only in the range $[0, 180]$ degrees? Because you can't observe a difference between an angle $\alpha_{obs}$ and $\alpha_{obs} + 180\times n$ (where $n$ is a positive or a negative integer). Look at a figure such as this (source) and try to convince yourself of this. If it helps, remember that two identical polarizers on top of eachother with an angle of $0^\circ$ let light through completely, but rotating one will block more and more light until they are perpendicular ($90^\circ$) and no light at all comes through, at which point increasing the angle will brighten the light until it reaches a maximum again at $180^\circ$ before once again starting to dim. This means that, in polarimetry the observed brightening or dimming by rotations of plane-polarized light are cyclic with period $180^\circ$, not $90^\circ$ as you suggest.

Turning the previous sentence into a simple equation, we obtain:

$$\alpha_{real} = \alpha_{obs} + 180\times n \ \ \ , \ \ n \in \mathbb{Z} $$

(The angles in this answer will always be treated in degrees, though for simplicity, in the equations I shall forget about the $^\circ$ unit and treat angles as pure numbers*).

Now we use the equation for optical rotation to find the specific rotation $[\alpha]^T_\lambda$, which characterizes a substance when measured at a temperature $T$ with a monochromatic light source of wavelength $\lambda$:

$$\alpha_{real} = [\alpha]^T_\lambda \times l \times c$$

, where $l$ is the optical path and $c$ is the concentration of the (enantiomerically pure) optically active substance. Substituting that into the previous equation:

$$[\alpha]^T_\lambda = \frac{\alpha_{obs} + 180 n}{lc}$$

So for each individual measurement of $\alpha_{obs}$ made, there are infinite solutions for $[\alpha]^T_\lambda$. This comes as no surprise.

But if we take several measurements, then we can find a value of $[\alpha]^T_\lambda$ as a unique solution to the equation, right? Speaking in strictly mathematical terms, no. Any finite amount of measurements, even with perfect accuracy and precision, will still yield infinite possible values of $[\alpha]^T_\lambda$. In fact, there are infinite levorotatory and dextrorotatory solutions, so in principle you can't even determine the direction of rotation by this method! See the graph below:

EDIT: In my previous graph, I incorrectly equated the values of $\theta$ with $\alpha_{real}$. The actual identity comes from $\mathrm{tan}\ \theta = \frac{d\alpha_{obs}}{dc} = \frac{d}{dc}(\alpha_{real}+180n)=\frac{d\alpha_{real}}{dc}=\frac{d}{dc}(lc[\alpha]^T_\lambda)=l[\alpha]^T_\lambda$, from which $\theta = \mathrm{arctan}\ (l[\alpha]^T_\lambda)$. This however does not change the theoretical development of the answer.

Note that since the measurements are cyclic with period $180^\circ$, straight lines wrap around. The two black points are two measurements of $\alpha_{obs}$ at two different concentrations. There is an "obvious" red line connecting them, but it need not be the correct one. There are infinite lines which cut both points; I have represented two more perfectly valid solutions with the green and blue lines. Each line which cuts all points represents a possible solution, with an inclination given by the angle $\theta = \mathrm{arctan}\ (l[\alpha]^T_\lambda)$ (see derivation above). From this, $[\alpha]^T_\lambda = \frac{\mathrm{tan}\ \theta}{l}$. Interestingly, the red and blue lines are represent dextrorotatory solutions to the problem, while the green line indicates a levorotatory solution for the same compound. There are infinite more solutions, characterized by steeper lines which wrap around more times. See the comments for information about how they come into play.

However, in practice, materials only have a finite and usually relatively small ability to perform optical rotation. Most of the more commonly encountered substances have a value of $[\alpha]^T_\lambda$ which tops out at a few hundreds. Some extreme examples are helicenes, for which specific optical rotations can reach as high as $\approx 10000^\circ$ in [13]helicene. After performing enough measurements (especially if you do them intelligently, by not measuring at different concentrations which are related to eachother by simple fractions), then you basically reach a situation where you can find an infinite family of solutions such as $[\alpha]^T_\lambda = -534 + 462785\times n$, in which case the true specific rotation is probably levorotatory with a value of just $-534^\circ$.

As a small additional comment, usually the filters in an optical polarimeter are rotated until the least amount of light comes through as the human eye is better at detecting small differences in light levels in darker conditions, which will make for a more accurate measurement of the observed rotation.

Originally I had answered this question by bringing up modular arithmetic, but while it's interesting it can possibly make the answer harder to understand at first. I have relocated my previous argument here.

The first equation shown is equivalent to:

$$\alpha_{obs} = \alpha_{real}\ \ (\mathrm{mod\  180})$$

The $\mathrm{mod\  180}$ part comes from modular arithmetic, which may sound scary and new, but it's just normal mathematics, with the small quirk that every time a number appears out of the range $[0, 180]$, you just add or subtract $180$ as many times as necessary for the answer to appear between $0$ and $180$.

* In reality some of the $\alpha_{real}$ and $\alpha_{obs}$ variables in certain equations above would have to be divided by $^\circ$, so that all equalities are dimensionally valid. Equivalently, the modulus could be expressed with the unit as $\mathrm{mod\  180^\circ}$, but I'm not sure if the modulus is allowed to have physical dimensions, even if I don't see any issue with it.


What does it mean to take reading up to 0.05 mL?

I believe what your teacher means by this is that the error for your measurement is within $0.05~\mathrm{mL}$ of the estimate, since $0.05~\mathrm{mL}$ is half the value of the smallest measurable volume in a typical $50~\mathrm{mL}$ burette. This is typically how the error of an instrument reading is determined.

The readings you gave would then be: $23.56~\mathrm{mL} \pm 0.05~\mathrm{mL}$ and $23.44~\mathrm{mL} \pm 0.05~\mathrm{mL}$.


Can we force a reversible reaction in equilibrium to achieve 100% completion?

The two questions are sort of related, so i'll answer to the best of my ability.

A reaction in dynamic equilibrium can get close to completion depending on its equilibrium constant, which is a ratio between products and reactants at a certain temperature. No, the equilibrium constant won't change if you add more reactants in an attempt to push the reaction to completion. We refer to Le Chatelier's principle which states that if one were to add another parameter, say a reactant, the reaction will act to stabilize the stress that has been placed on it.

Take this reaction $\ce{PCl3 + Cl2 <=> PCl5}$. If I were to add more $\ce{PCl3}$, the change we would witness is a shift in equilibrium, that is to say, the system will react to consume more reactants and produce more products. What we see essentially is the system reacting to stabilize the changes we made.

This principle isn't just limited to adding reactants or products, however. We see the system tries to stabilize changes made to pressure, temperature, and volume. If we increase pressure, how will the system act to stabilize it? Pressure is a result of the gas particles hitting the walls of the container. Thus we would see the system reduced the number of gas particles in order to reach equilibrium again. Essentially, it will shift to the side with the least amount of gas particles. Volume and pressure are obviously related in this principle. An increase in volume is a decrease in pressure and a decrease in volume is an increase in pressure.

Take this reaction for example $\ce{CaCO3(s)<=>CaO(s) + CO2(g)}$. If we increase the volume (decrease the pressure), we would see the system shift to the right because it has the most moles of gas and we need more gas particles to fill the volume and stabilize the pressure.

Lastly, when changing heat we view it as a product or a reactant for exothermic and endothermic reactions respectively. An exothermic reaction has heat written as a product $\ce{A <=> B} + \text{heat}$ and an endothermic reaction has heat written as reactant. Thus, when we change the heat of an endothermic or exothermic reaction, we are changing its reactants or products and we treat it in the same way we treat a change in concentration.

The point is that the equilibrium constant remains constant, otherwise it would be called the equilibrium variable. We do, however, see a change in equilibrium position.


Does citric acid react with carbonated water?

The chemistry you're describing the the one you expect: CO2 is formed, and released from the liquid. However, your question is more about kinetics than about the final product. We can assume that slow release of gas leads to larger bubbles, because nucleation rate for the bubble formation is lower. Now, in fizzy drink, there is a large quantity of carbon dioxide accommodated by the liquid due to pressurization. In your case, it will depend on the concentration of your lemon juice (natural lemon juice has pH ~ 2), and more importantly on that of the carbonated water: I just tried with a few grams of sodium bicarbonate dissolved in a glass of water in my kitchen, and the kinetics and bubble size really depends on initial carbonate concentration. (I haven't been able to make pictures or video, because I had to handle my kids who were very fast drawn to this impromptu experiment.)

Also, I second Richard's comment: there are many additional chemicals in lemon juice, some of which surely act as surfactants… which changes bubble formation and kinetics.


Why does potassium have the electronic configuration of 2.8.8.1?

You could actually have the 2.8.9 configuration, putting the last electron in a 3d rather than 4s orbital.  But the 2.8.8.1 one is lower in energy and thus more stable.  How electrons fill subshells in neutral atoms to make the lowest energy configuration can be seen here.


Chemical processes and chemical reactions difference

According to IUPAC:

Chemical reaction
A process that results in the interconversion of chemical species.
Chemical reactions may be elementary reactions or stepwise reactions
(It should be noted that this definition includes experimentally
observable interconversions of conformers.) Detectable chemical
reactions normally involve sets of molecular entities as indicated by
this definition, but it is often conceptually convenient to use the
term also for changes involving single molecular entities (i.e.
'microscopic chemical events').

A chemical reaction is a kind of process in which interconversion of chemical species takes place. Because of this, we can say that a chemical reaction is a chemical process.
Chemical process can be a broad term, which describe any process that is commonly studied by chemists because they are closely related to chemical reactions. It is somewhat diffuse, as boundaries of the field of chemistry.
Industries use a variety of procedures to obtain substances. Chemical processes/reactions take place. If your interest is realising what is done in industries, do not get stuck in these subtleties. Most of the times they are not intentionally distinguished, and their usage is customary in the field.


Why electronegativity instead of electropositivity

The first thing that should be highlighted here is that electropositivity is simply the opposite of electronegativity, any of the two can be used interchangeably with the necessary modifications to the sentence.

It is true though that electronegativity is more commonly used then electropositivity. For example, the Wikipedia article on the subject is electronegativity, and mentions electropositivity as its opposite. I believe there are two reasons for that, related to each other.

First: Consistency

It's very important in the scientific community, from reasons that go from facilitating students understanding to the writing of papers, that standards are used, and that everyone is talking about the same thing. It's easier if everyone is talking about the same thing, even if they are the complete opposite. It just makes us think faster. Why choose electronegativity then?

Second: How we explain chemistry

It's unecessary to explain why and how the study of the electrons "behaviour" in atoms and molecules are important to chemistry. Usually, when explaining a phenomenon, we talk about where the electrons are "going to". We say a reaction occurs because an electron is taken by an atom, or maybe because it's donated by one. We explain the polarity of the H-O bond by saying that the oxygen will attract the electrons of the covalent bond a lot more than hydrogen will, and not (usually) that the hydrogen doesn't attract the electrons as much as oxygen does.

Again, it's obvious that you can explain everything by means of electropositivity, but, in my experience, we usually explain things by saying which atom/group has the property that makes electrons go to them with higher value, rather than saying which atom/group has the property that makes electrons leave them with higher value.


Why does a tetrahedral molecule like methane have a dipole moment of zero?

Karsten's answer is excellent, but here is a figure that shows the mathematics involved:

The central atom (green) is at the center of the cube, the four other atoms (purple) are at alternating vertices and the geometry should be clear.

Alternatively, if you orient the molecule so one peripheral (purple) atom is directly "above" the central (green) atom, then each of the other three atoms is just $1 - \theta$ ($\approx 70.52877940 ^\circ$) away from being directly "under" the central atom, so each contributes $\cos(1 - \theta)$ times the bond dipole. This is the downward component of the bond dipole from one of the lower atoms.

But $\cos(1 - \theta) = 1/3$, so this is simply (bond dipole)/3, and there are three of these lower atoms, so the three downward components exactly balance the one upward component.


Final temperature of a adiabatic process

Your error stems from a miscalculation of $\mathrm{V_f}$ in the first calculation. As mentioned in the comments, it appears that you actually exponentiated by $\gamma$ rather than $\frac 1 \gamma$. When you differentiate by $\frac 1 \gamma$ , you should obtain $\mathrm{V_f}=0.0094 \mathrm{m^3}$.

Plugging this into your expression for $\mathrm{T_f}$, you should obtain:$$T_f = T_i\left(V_i\over V_f \right)^{R\over C_v} = 310\left(0.00782\over 0.0094 \right)^{8.314\over 20.8} = 288\mathrm{K}$$

Similarly, inserting this into the ideal gas equation, you should obtain: $$\mathrm{T_f= {P_fV_f \over R}}={(2.5)(0.0094) \over(8.206\times10^{-5})}=288\mathrm{K}$$ (The ideal equation will be slightly off as written due to rounding).

In regards to the textbook solution, the final temperature could not possibly be higher than the initial. An adiabatic expansion always reduces the temperature. Intuitively, we can understand this by noting that an expansion with no heat exchange should decrease the internal energy and thus decrease the temperature of the system.


False positive in lucas test for primary alcohol due to formation of tertiary carbocation by rearrangement?

In order to understand this, you need to have this basic knowledge of the kinetics of multistep reactions

In chemical kinetics, there are two ways to deal with multiple-step mechanisms

Rate determining step method- Here one specific step is the slowest. So we consider all steps after this step to be equally fast. The rate-determining step is the slowest step of a chemical reaction that determines the speed (rate) at which the overall reaction proceeds. The rate-determining step can be compared to the neck of a funnel. (1)
Steady-state approximation method- This is used when no step is the clear slowest. The steady-state approximation is a method used to estimate the overall reaction rate of a multi-step reaction. It assumes that the rate of change of intermediate concentration in a multi-step reaction is constant. (2)

So, in the Lucas test, the formation of the carbocation is the clearly slowest first step. This governs the reaction rate and the rest of the steps don't really matter. So, although it might seem deceptive at times if the R.D.S. is slow then the whole net reaction in slow.

Hence, the rate of formation of the one-degree carbocation is so less, that the entire rection does not proceed. So what you can theoretically say will be the only difference in say a Neopentyl alcohol and a 1-alkanol would simply be that after the super slow first step, the rest of the reaction will be much faster in the case of Neopentyl alcohol. But, due to the slow first step your given example will give an absolutely accurate negative Lucas test


How to convert an oxirane to an alkanol with one additional carbon atom?

You're right: The reaction of the alkene with perbenzoic acid forms an oxirane.

My suggestion for the next step is:

Draw the epoxide.
Draw the final product.
Map the atoms: Count and assign the carbon atoms. Do you spot any difference?

Supposed that the oxygen atom of $\ce{OH}$ group in the final alkanol is the same atom introduced by the action of perbenzoic acid, the oxirane was apparently opened.

Which general ways to open an oxirane exist?

Try to think in general synthons first.

One of the two steps in the sequence might just be a neutralization, quenching or workup step.

UPDATE

With a bit of drawing, atom counting and mapping, you will have realized  (by now) that:

The final alkanol has one carbon atom more than the oxirane intermediate.
The additional methyl group was added to the less substituted side of the oxirane.

Now, try to think in synthons, such as "$\ce{CH3+}$" and "$\ce{CH3-}$", rather than in reagents.

Where would a "$\ce{CH3+}$", thrown at the oxirane, add? Probabaly at the centre of high(est) electron density. This is the oxygen atom, and thus the wrong position. Apparently, "$\ce{CH3+}$" is not your synthon.

What about "$\ce{CH3-}$"? It's a strong nucleophile that would preferably attack centres of low(er) electron density, such as carbon atoms with electron withdrawing substituents. You have two of these, namely the two carbon atoms that constituted the $\ce{C=C}$ double bond and which are now in your oxirane. If "$\ce{CH3-}$" attacks one of these carbon atoms, the oxirane is cleaved in such a way that one $\ce{C-O}$ bond is broken and the oxygen atom gets the negative charge. Note that sterical hendrance comes into play here too.

In summary,"$\ce{CH3-}$" seems to be the right synthon. Now, you have to translate the synthon, "$\ce{CH3-}$", into a reagent that you can either find in the lab or make in situ. In order to get the negative charge on the carbon atom, it might be a good idea to attach an electropositive to it.

Consequently, a Grignard reagent, $\ce{CH3MgBr}$ is a good choice.

As I mentioned initially, while trying not to spoil the fun of solving, the last step is just an aqueous, acidic workup to release the alkanol ;-)


What reaction will ensue when sulfur trioxide is added to excess barium hydroxide solution?

Basically yes.
Sulfur trioxide reacts with water to form sulfuric acid (this is extremely dangerous!  do not try it at home) in a strongly exothermic reaction which, unless done under very precise control, produces a fine mist of concentrated sulfuric acid vapor (not nice).
$$\ce{SO3 + H2O -> H2SO4}$$
If you are doing this with an excess of barium hydroxide then the sulfuric acid will be deprotonated in two steps as you suggest.
$$\ce{H2SO4 + OH- <=> HSO4- + H2O}$$ $$\ce{HSO4- + OH- <=> SO4^2- + H2O}$$
Barium sulfate will precipitate from the solution, driving the reaction to completion.
$$\ce{Ba^2+ (aq) + SO4^2- (aq) -> BaSO4 (s)}$$
I couldn't find any mention of barium hydrogen sulfate as a compound anywhere so I take this to mean that it is not stable and therefore will not precipitate.


SMART string for alkyl group excluding alcohols

How to find carbon

with four connections
not -CH3
and only connected to carbon

As in your SMILES there are only implicit hydrogen, you do not need to count Hs.
Count the explicit bonds and exclude the carbon with one explicit bond to avoid -CH3.
[CX4;D2,D3,D4]
To find the carbons that are connected to others than carbon
C[!C]
Now you can use recursive SMARTS, and set the second pattern to NOT.
[CX4;D2,D3,D4;!$(C[!C])]
Now only carbons would be found that match both patterns.
CX4   C with total connections of 4
D<n>  number of explicit bonds
$()   recursive SMARTS
!     NOT
;     AND
,     OR
In cyclohexanol it will find

and here is a dummy molecule for testing it.
CCC(F)C=CCC1CCC(O)C(C(C)(C)C)C1


Long-term storage of Ascorbic Acid mixed with Sodium Bicarbonate

The reason that your solid mixture is clumping and hardening inside your airtight container is that the reaction between the two compounds is producing water as follows:

$$\ce{C6H8O6 + NaHCO3 <=> NaC6H7O6 + CO2 + H2O}$$

where $\ce{C6H8O6}$ is ascorbic acid.

This reaction proceeds rapidly in the presence of excess water, which allows the reacting species to dissolve and thus have ready access to each other. This results in the observed fizzing from the release of carbon dioxide.

In the solid phase, the initial reaction is limited to the surfaces of the particles coming into contact with each other. As the reaction proceeds, the once purely solid material becomes wet via the acid-base reaction producing water, and the remaining materials can somewhat dissolve, facilitating migration of the reacting species so that on a timescale of hours all of your material has reacted.

Regarding the stability of storing the solid mixture vs. storing the solids separately, you are probably better off storing them separately until use. Both ascorbic acid and sodium ascorbate are susceptible to being oxidized by long-term exposure to air. Although water itself shouldn't act as an oxidant here, it could facilitate reaction with oxygen in the air in a similar manner in which it facilitates the acid-base reaction, as discussed above.


Choosing LC-MS modifier - why only acids?

Stanislav, that is a Nobel Prize winning question. In short, nobody knows nitty-gritty details of how the electrospray ionization imparts a charge to the analyte. When Fenn (reinvented) the electrospray phenomenon and used for ionizing large biomolecules he was shocked and gave an analogy that it is like making the elephants fly. He got a Nobel Prize as well and a lot of patent issues in the university. There are three electrospray ionization models, you can read on Wikipedia, https://en.wikipedia.org/wiki/Electrospray_ionization#Ionization_mechanism

The key point in electrospray is that you need an electrolyte (forget about acids or bases). All you need is a conducting liquid between the electrode to generate the electrospray. In that process, the ion is created. There are studies, I am sure, on the effect of pH on the efficiency of ionization. I can tell what to avoid in electrospray: TFA (because it forms ion pairs), hexanes, all other pure non-conducting organic solvents. So you you can get the point that ionization in the solution has little to do with ionization in the electrospray.

Now coming to your chromatography part, which is purely empirical. The reason for avoiding bases in mobile phases is that it they will eat (=dissolve) the silica columns. Silica is not pH stable above 8. Silica based HPLC columns can tolerate low pH like 2.5, and below that the bonded starts to peel off, e.g. if you use a C-18 column and use pH 1, it will become a bare silica in a few days.


Is glass an amorphous solid or supercooled liquid?

You are correct: it's not both.

Why? What you've heard is nonsense. Very old window panes are thicker towards one side because of how they were made at the time. It became common practise to put the thicker edge at the bottom, because it was less likely to exceed the acceptable stress on the thinner area.

Contemporary windows have been found with the thicker edge to one side, or even the top. Gravity on Earth has been consistently towards the core since at least the Middle Ages, so the "flowing glass" hypothesis is false.

Glass is an amorphous solid by definition (per Karl's comment) and is not a super-cooled liquid (per Prabhdeep Singh's comment).


How can I predict if a reaction will occur between any two (or more) substances?

Can I predict the products of any chemical reaction?

In theory, yes!

Every substance has characteristic reactivity behavior. Likewise pairs and sets of substances have characteristic behavior. For example, the following combinations of substances only have one likely outcome each:

$$
\ce{HCl + NaOH -> NaCl + H2O} \\[2ex]
\ce{CH3CH2CH2OH->[$1.$ (COCl)2, (CH3)2SO][$2.$ Et3N] CH3CH2CHO}
$$

However, it is not a problem suited to brute force or exhaustive approaches

There are millions or perhaps billions of known or possible substances. Let's take the lower estimate of 1 million substances. There are $999\,999\,000\,000$ possible pairwise combinations. Any brute force method (in other words a database that has an answer for all possible combinations) would be large and potentially resource prohibitive. Likewise you would not want to memorize the nearly 1 trillion combinations.

If more substances are given, the combination space gets bigger. In the second example reaction above, there are four substances combined: $\ce{CH3CH2CH2OH}$, $\ce{(COCl)2}$, $\ce{(CH3)2SO}$, and $\ce{Et3N}$. Pulling four substances at random from the substance space generates a reaction space on the order of $1\times 10^{24}$ possible combinations. And that does not factor in order of addition. In the second reaction above, there is an implied order of addition:

$\ce{CH3CH2CH2OH}$
$\ce{(COCl)2}$, $\ce{(CH3)2SO}$
$\ce{Et3N}$

However, there are $4!=24$ different orders of addition for four substances, some of which might not generate the same result. Our reaction space is up to $24\times 10^{24}$, a bewildering number of combinations. And this space does not include other variables, like time, temperature, irradiation, agitation, concentration, pressure, control of environment, etc. If each reaction in the space could somehow be stored for as little as 100 kB of memory, then the whole space of combinations up to 4 substances would require $2.4 \times 10^{27}$ bytes of data, or $2.4\times 10^7$ ZB (zettabytes) or $2.4\times 10^4$ trillion terabytes. The total digital data generated by the human species was estimated recently (Nov. 2015) to be 4.4 ZB. We need $5.5\times 10^5$ times more data in the world to hold such a database. And that does not even count the program written to search it or the humans needed to populate it, the bandwidth required to access it, or the time investment of any of these steps.

In practice, it can be manageable!

Even though the reaction space is bewilderingly huge, chemistry is an orderly predictable business. Folks in the natural product total synthesis world do not resort to random combinations and alchemical mumbo jumbo. They can predict with some certainty what type of reactions do what to which substances and then act on that prediction.

When we learn chemistry, we are taught to recognize if a molecule belongs to a certain class with characteristic behavior. In the first example above, we can identify $\ce{HCl}$ as an acid and $\ce{NaOH}$ as a base, and then predict an outcome that is common to all acid-base reactions. In the second example above, we are taught to recognize $\ce{CH3CH2CH2OH}$ as a primary alcohol and the reagents given as an oxidant. The outcome is an aldehyde.

These examples are simple ones in which the molecules easily fit into one class predominantly. More complex molecules may belong to many categories. Organic chemistry calls these categories “Functional Groups”. The ability to predict synthetic outcomes then begins and ends with identifying functional groups within a compound's structure. For example, even though the following compound has a more complex structure, it contains a primary alcohol, which will be oxidized to an aldehyde using the same reagents presented above. We can also be reasonably confident that no unpleasant side reactions will occur.

If the reagents in the previous reaction had been $\ce{LiAlH4}$ followed by $\ce{H3O+}$, then more than one outcome is possible since more than one functional group in the starting compound will react. Controlling the reaction to give one of the possible outcomes is possible, but requires further careful thought.

There are rules, but they are not few in number. There are too many classes of compounds to list here. Likewise even one class, like primary alcohols (an hydroxyl group at the end of a hydrocarbon chain) has too many characteristic reactions to list here. If there are 30 classes of compounds (an underestimate) and 30 types of reactions (an underestimate), then there are 900 reaction types (an underestimate). The number of viable reaction types is more manageable than the total reaction space, but would still be difficult to commit to memory quickly. And new reaction types are being discovered all the time.

Folks who learn how to analyze combinations of compounds spend years taking courses and reading books and research articles to accumulate the knowledge and wisdom necessary. It can be done. Computer programs can be (and have been) designed to do the same analysis, but they were designed by people who learned all of the characteristic combinations. There is no shortcut.


Polymers that aren't brittle at the cryogenic temperatures found on Titan?

According to NASA1 the main factor inhibiting the widespread use of polymer materials in space environments is oxygen erosion, the materials tested so far to be proven effective in space are  Teflon, Mylar, Kapton and Tedlar.
Further experimentation is being conducted on board ISS according to their website2.
I do not believe there are polymers currently capable of handing the cold, pressure, and oxygen erosion of Titan, as such, the probe3 Huygens sent to Titan was constructed of aluminum coated Kapton, which seems to be the industry standard for space environments.
References:

https://www.nasa.gov/centers/glenn/shuttlestation/station/misse.html
http://www.spacestationresearch.com/tag/polymer-erosion-and-contamination-experiment-polymers/
https://en.wikipedia.org/wiki/Huygens_(spacecraft)


Hybridisation of the negatively charged carbon atom in a carbanion

Hybridization is a concept, which is often misconceived. One has to understand, that it is a mathematical concept, which explains a certain bonding situation (in an intuitive fashion). In a molecule the equilibrium geometry will be formed respecting steric and electronic interactions and possibly interactions with a medium or external field. The geometric arrangement will not be formed because a molecule is hybridized in a certain way.

Hybridisation is therefore only a result of a certain geometric arrangement of atoms in a molecule. A linear combination of all available (atomic) orbitals will form molecular orbitals (MO). These can then be rearranged (recombined) in a certain way to form localised molecular orbitals (LMO). These can then be used to interpret a bonding situation in a simpler theory.

Each LMO is expressed as a linear combination of the atomic orbitals, hence it is possible to calculate contributions of the atomic orbitals and describe these also as hybrid orbitals.

In a very simple case methane, the $\ce{C-H}$ bonds are formed from contributions of carbon's $\ce{2s, 2p}$ and hydrogen's $\ce{1s}$. For the carbon this may be interpreted as a $\ce{2sp^3}$ orbital.

In the simple case of $\ce{{}^{-}CH3}$, as Uncle Al already pointed out is isoelectronic to ammonia, the bonding situation may be also described in terms of $\ce{2sp^3}$ orbitals. As well as ammonia (or amines in general) it may also undergo (nitrogen) inversion.
(source)

In this inversion scenario the hybridisation of the carbon (nitrogen) changes from $\ce{2sp^3}$ to $\ce{2sp^2}$ to $\ce{2sp^3}$.

With larger carbanions this becomes a little more complicated. As long as $\ce{R=Alkyl}$, hyper conjugation is not strong enough to prevent pyramidalisation. However, the inversion berrier is furthermore reduced.

If any of the $\ce{R}$ groups are different from alkyl substituents, then I would expect that the environment around the carbon becomes (more) planar, i.e. the contribution of carbon's $\ce{2s}$ orbital increases. That may have various reasons, but some kind of conjugation or hyperconjugation will be involved. When these $\ce{R}$ groups become pretty bulky, I would expect the same, due to steric interactions.

Summary

Please understand hybridisation as a tool to understand bonding situations. Therefore it is a result of a geometric alignment of atoms in a molecule. It is not the driving force to form a certain geometric constellation.


How many Au atoms can be extracted from 0.250 L of seawater?

You also need to convert the grams of gold into moles in order to use Avogadro's number.

Atomic weight of gold = $\pu{196.967 g/mol}$
$$
\begin{align}
\frac{\pu{1.5E-04 g/ton}}{\pu{196.967 g/mol}}&= \pu{7.615E-07 mol/ton}\\
\pu{7.615E-07 mol/ton}\times \pu{6.022E+23 atoms/mol} &= \pu{4.586E+17 atoms/ton}\\
\frac{\pu{4.586E{+17} atoms/ton}}{\pu{2000 lb/ton}}&=\pu{2.293E{+14} atoms/lb}\\
\frac{\pu{2.293E{+14} atoms/lb}}{\pu{0.454 kg/lb}}=\pu{5.056E{+14} atoms/kg}&=\pu{5.056E{+11} atoms/g}\\
\pu{5.056E{+11} atoms/g} \times \pu{1.03 g/ml}&=\pu{5.208E{+11} atoms/ml}\\
\pu{5.208E{+11} atoms/ml}\times \pu{250 ml}&= \pu{1.302 E{+14} atoms}/\pu{250 ml}\\
\end{align}
$$
There would be $1.302 \times10^{14}$ atoms of gold in $\pu{0.250 L}$ of seawater.


Yes, the calcium ion could lead to precipitation. The solubility of $\ce{CaCO3}$ in distilled water is about 15 mg/L, which is about 0.15 mM calcium ion if there is no other source of carbonate.

The solubility constant for $\ce{CaF2}$ is about $4\times 10^{-11}$, which means that we can only have 0.5 mM fluoride ions before precipitation will start. That's well below the solubility of NaF.

However, the presence of additional carbonate could be used to reduce the calcium concentration. Similarly, any addition of acid will convert carbonate to bicarbonate and increase the maximum calcium concentration.

UPDATE:
In the above answer, I assumed a basic familiarity with solubility products. For those not familiar with those, here's more detail.

The key quantitative measure of solubility of ionic compounds is the solubility product, usually indicated as $K_{sp}$, which is the product of the concentrations of the separate ions.

For $\ce{CaCO3}$, we have $K_{sp}=[\ce{Ca^2+}][\ce{CO3^2-}]$. Reported values vary somewhat, but are typically around $2\times 10^{-8}$.

Likewise, for $\ce{CaF2}$, we have $K_{sp}=[\ce{Ca^2+}][\ce{F-}][\ce{F-}]$, and the reported values are around $4\times 10^{-11}$. Thus, the condition for keeping fluoride in solution is

$[\ce{Ca^2+}][\ce{F-}][\ce{F-}]<4\times 10^{-11}$.

If $\ce{CaCO3}$ is dissolved in distilled water to maximum solubility, the calcium ion concentration is $\sqrt{K_{sp}}\approx 0.15$ mM, which is the basis of the calculation above.


Subtracting standard electrode potentials

Let's take a look at what you're doing specifically for computing the potential difference. The potentials at the cathode and the anode are all relative to some standard. Importantly, it must be the same standard. Let's refer to this as ground.

$$E_{\mathrm{cathode}} = V_{\mathrm{cathode}} - V_{\mathrm{ground}}$$
$$E_{\mathrm{anode}} = V_{\mathrm{anode}} - V_{\mathrm{ground}}$$

The $V$'s here represent absolute potentials on some arbitrary scale.

What's the change in potential when I move a test charge from the anode to the cathode? Since potential differences are path independent, we can look at this via states: moving the charge from anode to cathode is the same as moving the charge from anode to ground and then from ground to cathode.

The potential difference is expressed as:

$$E = (V_{\mathrm{cathode}} - V_{\mathrm{ground}}) - (V_{\mathrm{anode}} - V_{\mathrm{ground}}) = E_{\mathrm{cathode}} - E_{\mathrm{anode}}$$

The key is the cancellation of the ground potential, so that we have a potential difference.

This is exactly what are doing when working with half-reactions. And it works because both the reduction and oxidation half reactions are referenced to the standard hydrogen electrode.

Let's look at two half-reactions:
$$\ce{A -> A+ + e-}\ ,\ \ E^{⦵} = -E_{A}$$
$$\ce{B+ + e- -> B}\ ,\ \ E^{⦵} = E_{B}$$

As a general rule, we tabulate reduction potentials, so let's treat the $\ce{A}$ reaction with a reduction potential:

$$\ce{A+ + e- -> A}\ ,\ \ E^{⦵} = E_{A}$$

Now, both of these half reactions are relative to the standard hydrogen electrode.

In the reaction
$$\ce{A + B+ -> A+ + B}$$
we can look at the equivalent process of moving an electron from $\ce{A}$ to a proton in a standard hydrogen electrode, and then taking a electron from dihydrogen in the standard hydrogen electrode and moving it to $\ce{B}$.

The potential difference is:
$$E^{⦵}_{\mathrm{cell}} = E_{B} + (-E_{A}) = E_{B}-E_{A}$$

This is exactly the same process because the anode potential relative to ground is the reduction potential of the oxidation half-reaction. The potential difference is computed by subtracting the potential from the other one.

Alternatively, if you already expressed the anode potential as a oxidation potential (inverse of the reduction potential), then the minus sign is implicitly included, and you can just add the two potentials.


The product you were using is very likely a "rust converter". Used to convert rust to a stable products before painting . I have used it several times , usually I just rinse the  surface , let it dry and paint . It is also used for industrial maintenance coating of poorly prepared surfaces ( those not sand blasted). That is ;it is commonly used and when I worked with coatings I never heard of a problem. Baking soda will neutralize it . I once inspected super phos storage tanks : it starts as 107 % phosphoric acid used to blend liquid fertilizers. When I stepped out of the tank and walked across the limestone gravel ,each step of the rubber boots left some jelly-like acid which foamed on the gravel. The people that worked at the facility thought nothing of it; from that I conclude neutralized phos acid it not a problem . Sorry this is all anecdotal ,but I have never needed to research the situation.


Is the absorbent substance in nappies biodegradable?

To keep it simple: there are as many different polymers as suppliers of the superabsorbent polymer used in diapers.https://en.wikipedia.org/wiki/Superabsorbent_polymer#Current_synthesis.

So, answer to #1. Some are, some aren't. As far as #2, define what you mean by "biodegradable". Take a diaper, put it into a composting bin, and turn it over every week for a million years, and I'd expect little to remain. It is likely based on your question that you are unaware that some diapers are marketed to be "biodegradable" and some aren't. The reality is that the term has no standard meaning - it's marketing hype. However, it's also true that some polymers will be inherently much more prone to decomposition than others and that some of these are incorporated into some diapers.

Your best bet is to call (or visit) the customer service of the specific diaper manufacturer you are interested about. Note that the nih link provided in a different answer doesn't mention any use of the sodium salt for any major brand.  Pampers (P&G), I think, dominates the market and sells 9 different disposable diaper brands. I didn't see any mention on their site about biodegradability, which was an important controversy some generations back. Since even "highly biodegradable" materials can take centuries to decompose in landfills, the question is a bit meaningless. (If you want biodegradable, use toilet paper (not seriously, since TP is designed to fall apart rapidly, too rapidly for practical use in a wearable product).)


Why does alkyl rearrangement happen in the Hofmann rearrangement?

The ‘motivation’ is that migration is the only way in which the reaction proceeds in a productive manner towards the product. From seeing the mechanism alone, you have no idea if the reaction of a particular substrate is rapid or slow, if it requires higher or lower temperatures etc. Only if you start taking these into account, you can tell whether the compound in question reacts rapidly or not.

I would not be surprised to see vastly different reaction times between compounds, depending on whether the $\ce{R}$ group easily migrates or not.


What will be the product when 3,4-dibromocyclohexene is reacted with sodium followed by heating with selenium?

Assuming your Product A is correct then product B should be benzene. Selenium act as a mild oxidizing agent at temperature 220 °C to 330 °C. In the above reaction selenium oxidizes cyclohexadiene to form benzene.

Below is a reaction in which 1-methyl-1,3-cyclohexadiene is converted into toluene.

you can also go to this link for more info


How to measure/know the moles of carbon dioxide (as dry ice) when in KOH solution

In this particular case weighing would work with limited accuracy (I am going to explain the limitations later in the answer).

Since the only compound leaving the reaction is the unreacted $\ce{CO2}$, it is valid to use weighing. Everything else is either sticking around or is already included in the initial weighing. So by weighing everything prior to reaction and after "all unreacted $\ce{CO2}$ is vaporized", you can calculate the amount of "reacted $\ce{CO2}$".

(weight of reacted $\ce{CO2}$) = (Weight of initial $\ce{CO2}$) - (weight of unreacted $\ce{CO2}$)

(Weight of initial $\ce{CO2}$) = (50 g) - (condensed water from atmosphere [estimate])

(Weight of unreacted $\ce{CO2}$) = (deficit in pre vs post reaction weighing)

pre reaction weight: 50 g $\ce{CO2}$ + 25 g $\ce{KOH}$ +100 g water = 175 g

post reaction weight: weight of the whole solution after all unreacted $\ce{CO2}$ is escaped from the solution (make sure the whole solution is at room temperature and make sure there is no bubbling occurring).

I understand there are some practical limitations, mainly :

1- There might be unreacted/frozen $\ce{CO2}$ still hanging around, Try to shake well. Also when weighing post reaction, make sure your solution is slightly warm to assist unreacted $\ce{CO2}$ to evaporate fully from the reaction.

2-Since you have dry ice, some water from atmosphere will condense into your reaction, this will lead to underestimation of the "leaving compound". The way to overcome, is to keep your reaction vessel as small as possible and control its inlet by a cold trap to make sure no water from atmosphere gets in. Also account for amount of water usually present in 50 g dry ice, by examining the water content on a similar sample beforehand.

pH:
Your initial pH is clearly not right. concentration of your $\ce{OH-}$ is way higher than just 0.1 M. Let's assume that it is due to high pH value being out of the calibration range of the pH-meter and let's assume that your pH=10 after addition of $\ce{CO2}$ is correct.

One will be able to calculate the amount of unreacted $\ce{CO2}$ from reaction balance. You are starting with 25 g (0.446 moles) of $\ce{KOH}$ and 50 g (1.14 moles) of $\ce{CO2}$. For every $\ce{CO2}$ reacting, 2 $\ce{KOH}$ are used up. So IF all(or most) of $\ce{KOH}$ is used up, you should end up with 0.446/2~ 0.22 moles of $\ce{K2CO3}$, which means the reacted CO2 is about 0.22 moles(~9.7 g), and unreacted is 0.92 moles(~40.3 g).
Why do I make the assumption that all the $\ce{KOH}$ is reacted, notice your final pH is 10 , meaning that your [OH-] concentration is about 0.0001 M, your starting concentration was 4.4 M.

What stops the reaction from using all $\ce{KOH}$ to be used up and dropping the pH to neutral zone? You are using excess $\ce{CO2}$, so we expect all the $\ce{KOH}$ to be used up. This has to do with part of $\ce{CO2}$ escaping the reaction as gas, also since the temperature is freezing the water (by dry ice) you may not have complete agitation which hinders the reaction procession to some extent.

I still think that you should take the weighing method seriously. Using weighing methods will provide confirmation for your calculations.


Is the answer key correct?

Yes, they have a very good approximation of the pH, and their way of solving it is straightforward and conventional.

  After doing a BCA-table, I got that the remaining moles of HCOOH = 0.003moles

The sum of formic acid and formate has to remain 0.027 mol. You could start out with 0.027 mol of formic acid and turn some of it into formate by adding NaOH. Or you could just mix them as described in the problem. However, you are claiming that there is a total of 0.015 mol of formic acid plus formate, so you lost some atoms there. As the comments state, there is no reason to think about where they got the sodium formate (stock room?).

  If the answer key is correct, why does the pH decrease?

Decrease or increase depends on your frame of reference. If you compare it with a 1:1 buffer where the pH = pKa, it decreased because there is a higher concentration of acid than conjugate base.

If you compare it to the pH of 0.3 M formic acid (roughly pH = 2.1), then it increased because you added conjugate base (neutralized and diluted the acid).


Is the ring or the chain chosen as the parent structure in octan-2-ylbenzene?

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

  P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)

  Two methods are recognized to name systems composed of rings and chains
  (exclusive of linear phanes).

  (1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.

  (2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.

  (…) For selection of a preferred IUPAC name, see P-52.2.8.

  P-52.2.8 Selection between a ring and a chain as parent hydride

  Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).

Method (1): “octan-2-ylbenzene” (preferred IUPAC name; ring preferred to chain)

Method (2): “2-phenyloctane” (chain has greater number of skeletal atoms)

The function Convert Structure to Name of ChemDraw correctly yields “octan-2-ylbenzene”.


What will form in the hydrolysis of urea in urine?

The hydrolysis is the first of two reactions that occurs in aqueous urea. This reaction forms one equivalent of ammonia and one equivalent of carbamic acid. Carbamic acid rapidly decomposes to form gaseous carbon dioxide and another equivalent of ammonia:

The ammonia then simply participates in an equilibrium where it reacts with water to form ammonium hydroxide, or so it is commonly described. There is actually a fair amount of debate on this topic, but it typically written like this: $$\ce{NH3 + H2O <<=> NH4OH}$$


Why are the dispersion forces in CS2 stronger than the dipole-dipole forces in COS?

Although individual dispersion forces are weak, they are cumulative, and increase with molar mass.  As a general rule, boiling point increases with molar mass.

Polar molecules will have higher boiling points when compared to molecules with similar molar masses.  For example, ethanol($\ce{CH3CH2OH}$) has a higher boiling point than dimethyl ether ($\ce{CH3OCH3}$).

$\ce{CS2}$ is ~16 g/mol heavier than COS.


The answer you seek can be found here uwyo.edu/roddick/site/alkane_dehydrogenation.html , the catalytic cycle of dehydrogenation of alkanes by Iridium complexes does not follow a simple syn addition such as in the hydrogenation of alkenes. In the hydrogenation of alkenes, the $\ce {H2}$ dissociates and is added syn to the substrate in a well understood mechanism.

In the dehydrogenation though, the coordination of the alkane to the Ir metal center occur first, then a migratory insertion forming a 5-coordinated trigonal bipyramidal complex, then a β-elimination occur forming an unstable 6-coordinated complex which readily loses the alkene (final product). The catalyst is then regenerated by a hydrogen acceptor which can be a bulky alkene (so its product can't participate in the cycle due to steric hindrance). If no regio inducing group is present in the catalyst, the formation of the Saytzeff product (most substituted alkene) is favored.


Thermal decomposition products of polyurea?

Here's one abstract:

  The primary fragmentation processes in the thermal decomposition of polymers were studied in detail on a series of structurally related polyureas by direct pyrolysis with a mass spectrometer. Our results indicate that polyureas I–III undergo a quantitative depolycondensation process analogous to that observed for N-monosubstituted polyurethanes. The thermal decomposition of polyureas IV–VI proceeds by intramolecular hydrogen transfer processes that occur at higher temperatures with respect to depolycondensation. Polycarboxypiperazine VI is decomposed by a single-stage decomposition mechanism that leads to fragments with amino end groups and carbon oxide.

I took a look at the article.  Depolycondensation is accurate: most of the polyureas examined decomposed to the "parent" isocyanates and amines.  The temperature range examined was 150°C to 350°C.

If you're interested in what happens at higher temperatures, I would guess a decomposition to mixtures of CO, ammonia, water, and possibly methane and/or hydrogen.  But I didn't find any references that would support me after a quick look


The difference between the carbonyl oxygen (red in colour) and alkoxy oxygen (blue in colour) is that the highly distorted pi-electron density (green shaded) towards to the carbonyl oxygen, due to more electro-negativity of oxygen that is the feature which is more on this oxygen than alkoxy oxygen . That feature provide the cavity for the H+ ion from acid, and the distortion of sigma- electron density at alkoxy oxygen is comparatively less than the distortion of  pi-electron density at carbonyl oxygen.


What are the differences and similarities between an aliphatic and an alicyclic compound?

Aromatic compounds (wiki) (about.com) (study.com) (chemguide) (khan)

Aromatic compounds are compounds that contain one or more aromatic rings.

Each aromatic ring, in its unsubstituted form, has the chemical formula $\ce{C_nH_n}$ (e.g. benzene), or with one or more of the $\ce{CH}$ groups replaced with an atom from Group V or Group VI, usually $\ce{N}$ (e.g. pyridine) or $\ce{S}$ (thiophene) or $\ce{O}$ (furan). They are usually stabilized by resonance.

Aliphatic compounds (wiki) (about.com) (study.com)

Aliphatic compounds are hydrocarbons (wiki, compounds containing only $\ce{C}$ and $\ce{H}$) which are not aromatic.

An example would be hexane (wiki).

Alicyclic compounds (wiki)

Alicyclic compounds are aliphatic compounds that contain one or more rings.

An example would be cyclohexene (wiki).

Gallery

Benzene, aromatic (courtesy Introduction to Organic Chemistry):

Furan, aromatic (courtesy Wikipedia):

Cyclohexane, alicyclic (courtesy Jarad Chemicals):

Cyclohexene, alicyclic (courtesy Exporters India):


Does Genistein remain in resonance stabilised form in physiological pH or acidic pH?

The first acid dissociation constant for genistein ($\mathrm{p}K_\mathrm{a1}$) is $6.51 \pm 0.20$ (Ref.1), mainly because of hydrogen bonding between $\ce{C_{(4)}}$ carbonyl oxygen and $\ce{C_{(5)}-OH}$, similar to that in acetylacetone (see the diagram):

Therefore, in physiological $\mathrm{pH}$ ($\mathrm{pH} \approx 7.3$) it should be partially ionized. If it is in acedic solution, $\ce{C_{(4)}}$ carbonyl oxygen would be partially or fully protonated based on the acid strength of the solution. Furthermore, the energy minimized three-dimensional structure of genistein shows the hydrogen bonding as depicted in following image (See Figure A; Ref.1):

References:

Yoshiyuki Mizushina, Kazuaki Shiomi, Isoko Kuriyama, Yoshihiro Takahashi, Hiromi Yoshida, “Inhibitory effects of a major soy isoflavone, genistein, on human DNA topoisomerase II activity and cancer cell proliferation,” International Journal of Oncology 2013, 43(6), 1117-1124 (https://doi.org/10.3892/ijo.2013.2032).


Relation between ΔH and ΔE for the reaction in gaseous phase at constant temperature and pressure

By definition, $$\Delta H=\Delta E+\Delta(PV)$$For a reacting ideal gas mixture at constant temperature, $$\Delta(PV)=(\Delta n)RT$$where $\Delta n$ is the change in the number of moles between reactants and products.  Therefore, $$\Delta H=\Delta E+(\Delta n)RT$$For the reaction under consideration, $$\Delta n=-\frac{1}{2}$$


How to do Lowdin symmetric orthonormalisation?

You obtain the diagonalization of $S$ as defined
$$
S = U \cdot s  \cdot U^{\dagger}
$$
(which holds for the case of symmetric $S$, which it is) and apply the power needed
$$
S^{-\frac{1}{2}} = U \cdot s^{-\frac{1}{2}}  \cdot U^{\dagger}
$$

For more background, try the application section of Wikipedia's article on diagonalizable matrices.
As far as understanding it goes, I'm afraid not much beyond doing some model calculations on 2D or 3D vectors can be done.

I do not have a strong background in parallel computing, but all operations involved are well-known and your linear algebra framework (BLAS, anyone?) will document how to run any of this in parallel and what storage may be needed for it. Note that, typically, diagonalization is not the most expensive step of a self-consistent field calculation (Hartree-Fock or DFT) anyway, so parallelization may not be critical for this step.


Diagonalization of Hessian H= PkP$^{-1}$: what do the elements of P matrix physically mean?

Are the $q$s in $\mathbf{P}$ coordinates of the atoms in molecules?

Close, but not quite. The $q_{i,j}$ represent the displacements from the equilibrium geometry of the appropriate atom/coordinate $i$, for each mode j. Thus, for example, $q_{5,9}$ corresponds to the displacement from equilibrium of atom $2$ along axis $2$ (typically the $y$-axis) when the molecule is undergoing vibration in mode $9$.

  If so, aren't they always changing since they are vibrating, rotating...?

There are no changes due to translation or rotation. It is customary to "project out" the translational and rotational motion from the Hessian and the geometry.  Qualitatively, this means that we simply ignore the translation and rotation and consider the molecule as "fixed" in its equilibrium configuration (allowing for vibrational 'wiggling'), with that configuration always oriented in the same way relative to the coordinate system.

As for the vibrational motion, though: yes, the $q_{i,j}$ values do technically also 'wiggle' as the molecule vibrates. However, the magnitudes of the perturbations $\delta q_{i,j}$ are in many cases very small, and can be neglected. In those cases where they cannot, the harmonic oscillator approximation to the vibrational motion breaks down and is no longer a useful model for vibration along that mode.

See my answers here and here for some links to literature on situations where this occurs (emphasis on the 'internal rotation' topics, though some of the 'anharmonic' and 'Coriolis' references might be somewhat relevant). Here is a Chem.SE question about nitrogen inversion that might be of interest.  More generally, a Google search for "low-frequency vibrational modes" should turn up lots of interesting reading.


Is it possible to hold and rotate molecules to relativistic speeds around axis in observable way?

Rotation speed cannot be made arbitrarily close to the speed of light. Molecular rotation is quantized, and every molecule has a finite number of rotational levels, just like it has a finite number of vibrational levels based on the shape of its potential energy surface (the same can be said of vibrations: the molecule cannot vibrate arbitrarily quickly without dissociating).

However forces like the van der Waals force do get noticeably affected at the highest supported molecular energy levels. Specifically the $C_6/r^6$ van der Waals force becomes $C_7/r^7$ at certain distances, due to the Casimir-Polder effect in quantum electrodynamics (an intrinsically relativistic theory of the quantum world). The reason for this is that the derivation of the $C_6/r^6$ van der Waals force does not take into account the fact that information cannot travel at infinite speed, so it actually takes a finite amount of time (based on the finite speed of light) for one atom to "feel" the force of a distant atom. This "slowing down" of the energy transfer, is called "relativistic retardation" because "retardation" means slowing down. This causes the force to become weaker for larger distances ($1/r^7$ is much smaller than $1/r^6$ for larger $r$).

This is usually not taught at the undergraduate level, or even graduate level courses, so there is no easy "textbook" example to demonstrate it, so unfortunately we have to resort to publications in scientific journals, where the audience is far more advanced: But in the image below, I hope you can at least see that "with retardation" the experimental values here are much closer to the theoretical values of Tang (2009), versus the experimental values derived from a model "without retardation":

This shows evidence that the theoretical idea of "relativistic retardation" is in fact a much better description of the experiment than the non-relativistic theory, so it is indeed fair to say that the effect of light (electromagnetic radiation) having finite (rather than infinite) speed is observed in a molecule. The molecule here was Li$_2$ and the above image is from Figure 6 of this paper which deals with deriving information from experimental spectra of a molecule:

http://scienide2.uwaterloo.ca/~rleroy/Pubn/09JCP-Li2AX.pdf


London dispersion forces and distance relationship

London dispersion forces (or van der Waals) can be rationalized (classically) as a relationship between induced dipole moments.

Some electrostatic (Coulombic) relationships:

Two Point Charges: $V = \frac{q_1 q_2}{4\pi\epsilon_0 r}$ (i.e., $r^{-1}$)
Point Charge and Permanent Dipole: $V = \frac{\mu_1 q_2}{4\pi\epsilon_0 r^2}$ (i.e., $r^{-2}$)
Two Permanent Dipoles: $V = \frac{\mu_1 \mu_2}{4\pi\epsilon_0 r^3}(1 - 3 \cos^2 \theta)$ (i.e., $r^{-3}$)

The last assumes that the polar molecules are stationary with angle $\theta$ between the dipole moment vectors. If the molecules rotate then it falls off to $1/r^6$.

In general, the interaction of a $2^n$-pole (dipole, quadrupole, octupole, etc.) with $2^m$-pole is $V \propto \frac{1}{r^{n+m-1}}$.

So why the $1/r^6$ for dispersion? Here's a quick justification. Let's take the system of a permanent dipole that's creating a induced dipole moment in another molecule:

$\mu_2^* = \alpha_2 E$

where $\alpha_2$ is the polarizability in the second molecule and E is the electric field generated by the polar molecule:

$E = \mu \sqrt{1 + 3\cos^2 \theta} / 4\pi\epsilon_0r^3$

The potential energy is given by:

$V = \alpha_2 E^2 / 2$

$V \propto r^{-6}$

Since there are multiple dispersion interactions (two dipole moments averaged over all rotations, permanent dipole-induced dipole, hydrogen bonding, and induced dipole-induced dipole moment) the general potential energy is classically approximated as:

$V = \frac{C}{r^6}$


Why is it necessary to avoid even traces of moisture from a Grignard reagent?

Grignard reagents are highly reactive and react with any source of proton to give hydrocarbons. Even water, alcohols, amines are sufficiently acidic to convert them to corresponding hydrocarbons.

$\ce{RMgX + H2O -> RH + Mg(OH)X}$

Where, $\ce{R}$ is a hydrocarbon.


What is the effect of temperature during cryotrapping?

It depends on the temperature regime you of your cryo-trap.  If you are trapping in the temperature range from dry ice to water ice then you must be sure any significant water vapor in you sample has been removed.  If you are using liquid nitrogen, there is some danger of condensing oxygen.  This can be mitigated by using the enough-warmer-that-this-doesn't-happen-with-air liquid argon.


This is not the answer OP would expect because I couldn't find metal complexes of type $\ce{[M(H3O)_x]^{y+}}$ (the reason for non-existence(?) is mentioned in the comments) but I did find a complex of an organic compound where hydronium ion act as a ligand i.e a complex of 18-crown-6 and hydronium ion(emphasis mine):

The observation that cyclic crown ethers can bind cations with high
affinity and selectivity has had a tremendous impact in chemistry. The
complex between 18-crown-6 (18c6) and the hydronium ion, $\ce{H3O+}$,
is a prototypical representative of this important host−guest paradigm
and has been the subject of numerous experimental and theoretical
studies. The computational evidence presented so far has favored a
structure with three linear $\ce{O−H···O}$ hydrogen bridges as the
lowest, stable minimum, whereas a bifurcated arrangement has turned
out to be a transition state for $\ce{H3O+}$ rotation 4−5 kcal/mol
higher in energy.

The closest metal complex I could find where hydronium ion is present as an entity is hydronium zinc(II) pyrazine-2,3-dicarboxylate(again not the type of complex OP is looking) where 2,3-PZDC (μ-pyrazine-2,3-dicarboxylato-N,O,O′) act as ligand(emphasis mine):

The structure of dihydronium
[catena-bis(μ-pyrazine-2,3-dicarboxylato-N,O,O′)zinc(II)],
$\ce{(H3O)2[Zn(2,3PZDC)2]}$, is composed of polyanionic ribbons of
zinc(II) ions linked by double bridging 2,3-PZDC ligand molecules.
Each ligand uses an N,O bonding moiety formed by one carboxylic group
[$\ce{Zn–O}$ 2.071(2) Å; $\ce{Zn–N}$ 2.184(2) Å] and a monodentate oxygen atom of
the other carboxylate group [$\ce{Zn–O}$ 2.092(2) Å]. Coordination around the
zinc(II) ion is strongly distorted octahedral. Hydronium cations
$\ce{H3O+}$ link the ribbons by hydrogen bonds.

References

Hydronium Ion Complex of 18-Crown-6:  Theory Confirms Three “Normal” Linear Hydrogen Bonds
Michael Bühl, Ralf Ludwig, Rachel Schurhammer, and, and Georges Wipff,
The Journal of Physical Chemistry A 2004 108 (51), 11463-11468
DOI: 10.1021/jp045879+
Michal Gryz, Wojciech Starosta & Janusz Leciejewicz (2005) "Doubly bridged molecular ribbons in the structure of an ionic complex, hydronium zinc(II) pyrazine-2,3-dicarboxylate", Journal of Coordination Chemistry, 58:11, 931-935, DOI: 10.1080/00958970500055492


First of all, there's no pinacole in the first place, so pinacole/pinacolone rearrangement isn't possible. Your rearrangement reaction/hydride shift wouldn't be called pinacole/pinacolone rearrangement.

Second, we have to think about why this would happen. Such rearrangements happen because we produce a more stable carbocation. In case of pinacole/pinacolone, if we do the rearrangement we end up with:

Which is well stabilized and after deprotonation a stable compound.

In case of sugars it's a little bit different, since it's not an alcohol in 1-position, it's a hemiacetale. This is extremly important since it has quite different reactivities.

We can cleave of water under much weaker acidic and mild conditions than in case of alcohols. The produced carbocation is quite stable since we got an oxygen stabilizing the cation, therefore no rearrangement is happening.


A very simple way is using molecular mechanics: it only applies to very simple molecules, but it should be enough for your purpose. Avogadro is a free software which allows you to build structures and to minimize them via molecular mechanics, through a variety of "methods" (force fields).
The software outputs an estimate of the energy of your geometry, which you can compare to that if other geometries (for instance, boat vs chair cyclohexane).

Note that these results are very approximate, the methods work properly only in the case of neutral and covalent molecules, but anyway they provide good results for a qualitative analysis.

More accurate methods, capable of working with more complex molecules (in order of accuracy: semiempirical methods, like PM3, PM6, AM1... << Ab initio methods like HF and DFT methods like B3LYP, M06, PBE...) exist in other software packages, but this is a good starting point for learning.

Molecular mechanics works by treating angles, bonds, etc. as subject to forces (Imagine them as springs) which constrain them to an "optimal" conformation, as seen in sets of typical and simple molecules.

The other methods work by solving quantum-mechanical equations, with or without some approximations, and thus can treat "atypical" geometries which don't work with molecular mechanics.

To make a example, molecular mechanics works well for whole sets of typical molecules like our amino acids, but fails to predict the geometry of strange molecules, like tetrahedrane, which shows very atypical angles.

Within these limits, you can use molecular mechanics for free and with not much computational and software hassle


The mass of a particular isotope is determined by mass spectrometry. Hopefully you know this technique. In order to obtain a high degree of precision, the beam must go through several mass spectrometers coupled in series, some of which are quadrupoles or octupoles. It would take hours to explain the developments of this technique.


How much energy does it cost to have electron configurations that are not in accordance with Hund's rules?

Each energy level of a given electron configuration is described by a  Russell-Saunders term symbol, assuming LS coupling. The ground state term symbol is predicted by Hund's rules. Tables of atomic energy levels identified by their term symbols have been compiled by NIST. So, the only thing you have to do, is to compare the energies of the terms describing the electron configurations you are looking for with their respective ground states.

In the following pictures the notation $m_{S} = + \frac{1}{2} \overset{\scriptsize{\text{def}}}{=} \, \uparrow$ and $m_{S} = - \frac{1}{2} \overset{\scriptsize{\text{def}}}{=} \, \downarrow$ will be used for the spin quantum number $m_{S}$.

The data for carbon can be found here. The ground state for carbon is ${}^{3}\mathrm{P}_{0}$ - it is assigned the energy $0.00 \, \mathrm{cm}^{-1}$. From your question it's not 100 % clear what exact electron configuration you want. I assume it is either ${}^{3}\mathrm{P}_{2}$ or ${}^{1}\mathrm{S}_{0}$ therefore I will give the values for both of them (for comparison: the thermal energy at room temperature is about $200 \, \mathrm{cm}^{-1}$).

The data for nitrogen can be found here. The ground state for nitrogen is ${}^{4}\mathrm{S}_{3/2}$ - it is assigned the energy $0.00 \, \mathrm{cm}^{-1}$. From your question it's not 100 % clear what exact electron configuration you want. I assume it is either ${}^{2}\mathrm{D}_{5/2}$ or ${}^{2}\mathrm{P}_{3/2}$ therefore I will give the values for both of them.


Why does Cu(OH)2, which is generally insoluble due to low Ksp become soluble in excess of a base, such as NaOH?

Also, can the equilibrium $$\ce{Cu^{2+} + 6H2O <=> [Cu(H2O)6]^{2+}}$$ participate in the solubility process in any case?

This equilibrium does contribute to the solubility of copper(II), but probably not in the way it appears you are envisioning it to.  When copper(II) is fully dissolved in moderately acidic solution, as long as the anion of the acid is a weak ligand for copper (e.g., $\ce{ClO4-}$ or $\ce{NO3-}$), the copper ions are dissolved as the aquo complex, $\ce{[Cu(H2O)6]^{2+}}$.

As the $\mathrm{pH}$ is adjusted upward with, e.g., $\ce{NaOH}$, though, other reactions come into play (see below) and the hexaaquo species contributes negligibly to the overall solubility behavior.

  Where does the "excess" criteria come into the picture?

To answer this question, one really must first understand what's actually going on when solvated metal ions undergo acid/base reactions.  As I indicated above, in acidic solution copper(II) exists as an aquo complex, typically written as the hexaaquo species, $\ce{[Cu(H2O)6]^{2+}}$.  As the $\mathrm{pH}$ rises, the following series of acid-base ("hydrolysis") reactions occurs$^\ddagger$:

$$
\ce{[Cu(H2O)6]^2+ + OH- <=> [Cu(H2O)5(OH)]+ + H2O}\tag{1}
$$

$$
\ce{[Cu(H2O)5(OH)]+ + OH- <=> [Cu(H2O)4(OH)2]^0 + H2O}\tag{2}
$$

$$
\ce{[Cu(H2O)4(OH)2]^0 + OH- <=> [Cu(H2O)3(OH)3]- + H2O}\tag{3}
$$

$$
\ce{[Cu(H2O)3(OH)3]- + OH- <=> [Cu(H2O)2(OH)4]^2- + H2O}\tag{4}
$$

(For simplicity, I have omitted the $\ce{M2L2}$ hydrolysis complex reported in Vuceta$^\ddagger$, and have written the above in terms of addition of alkali so as to highlight the similarities among the series.)

So, to some extent, the copper can be treated much like a polyprotic acid: as hydroxide ion is added, the copper is able to "absorb" it, forming water/hydroxide complexes with increasing numbers of hydroxides attached.

Per the low $K_\mathrm{sp}$ of $\ce{Cu(OH)2}$, however, the following (fictitious) solubility relation must also be considered:

$$
\ce{ [Cu(H2O)4(OH)2]^0 (aq) <=> Cu(OH)2 (s) + 4H2O} \tag{5}
$$

Eq. $(5)$ is a re-interpretation of the typical solubility expression:

$$
\ce{[Cu^2+][OH^{-}]^2} \leq K_\mathrm{sp} \tag{6}
$$

Qualitatively, Eq. $(5)$/Eq. $(6)$ indicates that the copper existing as $\ce{[Cu(H2O)4(OH)2]^0}$ is limited to a certain concentration in solution, depending on the conditions of the system.

Thus, as base is added to an acidic solution of copper and the equilibria are swung to a range that favors $\ce{[Cu(H2O)4(OH)2]^0}$/$\ce{Cu(OH)2}$, much of the copper precipitates out.  But, as more base is added, eventually there's enough hydroxide present that the equilibria can start to shift toward $\ce{[Cu(H2O)3(OH)3]-}$ and $\ce{[Cu(H2O)2(OH)4]^2-}$, and the copper will start to redissolve.

This, finally, leads back to question: what does "excess" mean in this context?

To help answer this question, I input the stability constants from Vuceta$^\ddagger$ into HySS, a program that is designed to simulate these sorts of equilibria and that is able to handle precipitates like $\ce{Cu(OH)2 (s)}$.  The HySS file used to generate the below figures can be downloaded from Google Drive here.  The below figure shows the simulated speciation diagram for $10~\mu\mathrm M$ total copper over a $\mathrm{pH}$ range of $4$ to $14$ (all ligating water molecules are omitted from the labels; click the image for a larger version):

As expected, the copper is fully dissolved at an acidic $\mathrm{pH}$ of $4$, but starts to precipitate at around $\mathrm{pH}~7$.  Coincidentally, the concentration of the dissolved $\ce{[Cu(H2O)4(OH)2]^0}$ is predicted to be equal to the "concentration" of the precipitated $\ce{Cu(OH)2 (s)}$ (mass of solid divided by the liquid volume) over much of the insoluble range. Starting at about $\mathrm{pH}~11.5$, the base is in sufficient excess (~$316$-fold) over the copper to begin to re-dissolve it.  Only around $\mathrm{pH}~13$ or so does all of the solid finally re-dissolve, at a ~$10,000$-fold excess of base.

At the higher copper concentration of $10~\mathrm{mM}$, re-dissolution has hardly begun to happen at all by $\mathrm{pH~14}$ (click image for larger version):

This is perfectly sensible, as the base is only in excess of the copper by ~$100$-fold at that $\mathrm{pH}$.  In order to observe re-dissolution at this copper concentration, one would presumably have to use extraordinarily concentrated alkali in order to achieve a sufficient excess of base.

$^\ddagger$ Vuceta & Morgan, Limnology and Oceanography 22(4): 742, 1977, doi:10.4319/lo.1977.22.4.0742


Osmotic pressure

is consequence of non zero net water diffusion,
which is consequence of non equal water activities on both sides of semipermeable membrane,
which is consequence of the fact dissolved solutes decrease activity of water.

Osmotic pressure of a free solution is formally an external pressure needed to be acting on this solution to keep diffusion equilibrium with pure solvent on the other side of a semipermeable membrane.
Applying external pressure then leads to the well known reverse osmosing, widely used in production of deionized water ( replacing expensive distillation ) from sweet water, or in desalination of sea water.
Osmosis continues until fully counteracted by extenal, e.g. atmospheric pressure. E.g. if there is osmotic pressure $\pu{1 atm}$, osmosis leads up to $\pu{10 m}$ high solution column ( eventually corrected by solution density ). Similarly, you would need to use external pressure $\pu{p > 1 atm}$ to start reverse osmosis.
If a book says the primary reason is hydrostatic pressure difference, it is wrong. We can have 2 solutions of the same density, but different osmotic pressures. Combining them on the opposite membrane sides would lead to disbalancing of solution levels, even if they would be initially the same and even if the hydrostatic pressure difference would be zero.


Why does the US not manufacture sodium thiopental themselves?

I […] can't help wondering why USA […] can not manufacture this chemical themselves.

They can and they did!

Back in the nineteen thirties, Ernest H. Volwiler and Donalee L. Tabern filed a couple of patents on sedative thiobarbituric acid derivatives for the Illinois-based Abbott Laboratories. Among these was US 2,153,731 for sodium thiopental.

Abbott Laboratories produced it for a long time and when the company was restructured, the still Illinois-based split-off Hospira continued the production.

Note that it is not a "bad compound" as such and actually, it is still listed as an alternative to propofol on the WHO Model List of Essential Medicines.

According to a 2010 article in the German ÄrzteZeitung (online, 29.09.2010), the FDA was aware of the thiopental shortage and a speaker of Hospira told the NYT that his company is rather unlucky about the fact that this anaestetic found it's way into death row.

Around this time, the production of thiopental at the facilities in Illinois was seemingly stopped.

In a 2011 article in the ÄrzteZeitung (online, 24.01.2011) it is stated that Hospira had initially announced to continue the production at a facility in North Carolina and in Liscate, Italy.

However, it seems that the production in North Carolina never started. Moreover, the Italian authorities demanded a declaration that the thiopental produced in Liscate would not be used for lethal injections. Obviously, Hospira could not comply.

Personally, I think that Hospira wasn't overly unlucky about the legal situation in Europe. They are a health care company and I suppose that they prefer to make profit by improving the medical situation of their customers, rather than by killing them!


Why is 3,6-dimethylpiperazine-2,5-dione optically active in spite of having a centre of symmetry?

There is no evidence (e.g. aromaticity) to support your claim that the compound is planar. In fact, crystalographic studies [1] revealed existence of both cis- and trans-3,6-dimethylpiperazine-2,5-dione. Corresponding CCDC numbers are LCDMPP01 and TRDMPP01, respectively — feel free to open the links and play around with the 3D structures in JSmol to understand the difference.

According to Sletten, in cis (ʟʟ) modification the ring is puckered resembling a skewed boat conformation, whereas the ring in trans (ᴅʟ) molecule is nearly planar:

Figure 1a. cis-3,6-dimethylpiperazine-2,5-dione (drawn from LCDMPP01)

Figure 1b. trans-3,6-dimethylpiperazine-2,5-dione (drawn from TRDMPP01)

References

Sletten, E. Conformation of Cyclic Dipeptides. The Crystal and Molecular Structures of Cyclo-D-Alanyl-L-Alanyl and Cyclo-L-Alanyl-L-Alanyl (3,6-Dimethylpiperazine-2,5-Dione). J. Am. Chem. Soc. 1970, 92 (1), 172–177. DOI: 10.1021/ja00704a028. (Open Access)


What is the IUPAC name of fullerene like molecules?

I think I've found relevent sources here:

https://old.iupac.org/publications/pac/1997/pdf/6907x1411.pdf
http://www.sbcs.qmul.ac.uk/iupac/fullerene/

However, I still wonder if we have another name for all those hydrocarbons with 3-dimensional structures. Should they be regarded as a subclass of "aromatic" or do we have another name for them?


It depends. Bread has moth small molecules (like 2-Acetyl-1-pyrroline). These molecules will just pushed aside by the blade. You have hydrogen bond network between water molecules. You break those. Bread has large polymers like starch. Knife breaks those.

A cleaner system to study is a polyethylene. You can freeze it in N2, break under inert atmosphere and study by EPR spectroscopy. You will find free radicals at the breaking point. This is because it costs less energy to break a bond then to move one molecule relative to others.


For hydrogen atom, L^2 and Lz can be obtained as eigenvalues for a particular wave function. But that does not completely specify the angular momentum vector. How to get about this problem?

Well, you can't. The different components of the angular momentum operator $\hat{\mathbf{L}}$, i.e. $\hat{L}_{x}$, $\hat{L}_{y}$, and $\hat{L}_{z}$, do not commute, so you cannot simultaneously determine all of them. But the squared angular momentum operator, which gives you the squared length of the angular momentum vector commutes with the components of $\hat{\mathbf{L}}$, so you can simultaneously determine the length of the angular momentum and one of its components. Which component you choose is immaterial, usually $\hat{L}_{z}$ is chosen. What you get out of those two operators can be pictured like this: you know the length of the angular momentum vector and its z-component; the other two components are undetermined but because you know the length they can only take on values that lie on a circle centered around the ${L}_{z}$ axis, whose offset is determined by the value of ${L}_{z}$ and whose radius is determined by $| \sqrt{\mathbf{L}^{2}}|$. For more details have a look at this Wikipedia article or any general physical chemistry or quantum mechanics textbook.

  Also, in the orbital approximation (ie the solutions obtained after
  ignoring electronic repulsions in the Hamiltonian) of the many
  electron atom( say helium) how does one go about finding total orbital
  angular momentum?

Ignoring relativistic effects the Hamilton operator of a system consisting of $M$ nuclei and $N$ electrons, which are considered to be point charges takes on the form

\begin{align}
  \hat{H} &= \underbrace{ \sum_{j=1}^{M} - \frac{ \hbar^{2} }{ 2 M_{j} } \mathbf{\nabla}_{ \mathbf{R}_{j} }^{2} }_{ = \, \hat{T}_{\mathrm{K}} } + \underbrace{ \sum_{i=1}^{N} - \frac{ \hbar^{2} }{ 2 m_{\mathrm{e}} } \mathbf{\nabla}_{ \mathbf{r}_{i} }^{2} }_{ = \, \hat{T}_{\mathrm{e}} } + \underbrace{ \frac{1}{2} \sum_{i \neq j}^{M} \frac{ e^{2} Z_{i} Z_{j} }{ 4 \pi \epsilon_{0} | \mathbf{R}_{i} - \mathbf{R}_{j} | } }_{ = \, \hat{V}_{\text{K--K}} }  \\
  & \quad \underbrace{- \sum_{i=1}^{N} \sum_{j=1}^{M} \frac{ e^{2} Z_{j} }{ 4 \pi \epsilon_{0} | \mathbf{r}_{i} - \mathbf{R}_{j} | } }_{ = \, \hat{V}_{\text{e--K}} } + \underbrace{ \frac{1}{2} \sum_{i \neq j}^{N} \frac{ e^{2} }{ 4 \pi \epsilon_{0} | \mathbf{r}_{i} - \mathbf{r}_{j} | } }_{ = \, \hat{V}_{\text{e--e}} }
\end{align}

where $ \mathbf{R}_{j} $, $Z_{j}$, and $M_{j}$ are the coordinate, the charge number, and the mass of the $j^{\mathrm{th}}$ nucleus, respectively and $ \mathbf{r}_{i} $ is the coordinate of the $i^{\mathrm{th}}$ electron and $m_{\mathrm{e}}$ is the electron mass.
Furthermore $e$ is the elementary charge und $\epsilon_{0}$ is the vacuum permittivity.
The different terms of this Hamilton operator can be explained as follows: $\hat{T}_{\mathrm{e}}$ and $\hat{T}_{\mathrm{K}}$ are the terms representing the kinetic energy of the electrons and the nuclei, respectively, $\hat{V}_{\text{K--K}}$ describes the coulomb interaction between the nuclei, $\hat{V}_{\text{e--e}}$ is the interelectronic repulsion, and $\hat{V}_{\text{e--K}}$ represents the coulomb interaction between the electrons and the nuclei.
Using the Born-Oppenheimer approximation you can seperate the electronic degrees of freedom from the ones of the nuclei and treat the electronic problem independently and for the many-electron atom you also only have one nucleus to consider so the terms $\hat{T}_{\mathrm{K}}$ and $\hat{V}_{\text{K--K}}$ drop out of the Hamiltonian. If you now also ignore the electron-electron interaction you are basically left with a Hamilton operator describing two independent electrons moving in the electrostatic field of the nucleus:

\begin{align}
  \hat{H} &= \underbrace{ \sum_{i=1}^{N} - \frac{ \hbar^{2} }{ 2 m_{\mathrm{e}} } \mathbf{\nabla}_{ \mathbf{r}_{i} }^{2} }_{ = \, \hat{T}_{\mathrm{e}} } + \underbrace{ \sum_{i=1}^{N}  \frac{- e^{2} Z }{ 4 \pi \epsilon_{0} | \mathbf{r}_{i} - \mathbf{R} | } }_{ = \, \hat{V}_{\text{e--K}} }
\end{align}

For this you can easily seperate the $N$-electron problem into $N$ one-electron problems

\begin{align}
  \hat{H} &= \sum_{i=1}^{N} \hat{h}_{i} \qquad \Rightarrow \qquad \Psi(\mathbf{r}_{1}, \ldots, \mathbf{r}_{N}) = \psi(\mathbf{r}_{1}) \cdots \psi(\mathbf{r}_{N})
\end{align}

where each one-electron Hamiltonian $\hat{h}_{i} = - \frac{ \hbar^{2} }{ 2 m_{\mathrm{e}} } \mathbf{\nabla}_{ \mathbf{r}_{i} }^{2} - \frac{e^{2} Z }{ 4 \pi \epsilon_{0} | \mathbf{r}_{i} - \mathbf{R} |}$ and one-electron eigenfunction $\psi_{n}(\mathbf{r}_{i})$ basically satisfy a Schroedinger equation for the hydrogen-like atom

\begin{align}
  \hat{h}_{i} \psi_{n}(\mathbf{r}_{i}) = \varepsilon_{n} \psi_{n}(\mathbf{r}_{i}) \qquad \text{with} \qquad \varepsilon_{n} = -\frac{m_{\mathrm{e}} e^{4} Z^{2} }{ 32 \pi^{2} \epsilon_{0}^{2} \hbar^{2}} \frac{1}{n^{2}} \qquad n = 1, 2, ...
\end{align}

Using spherical coordinates, $x, y, z \to r, \theta, \phi$, you can rewrite the kinetic energy operator $- \frac{ \hbar^{2} }{ 2 m_{\mathrm{e}} } \mathbf{\nabla}_{ \mathbf{r}_{i} }^{2}$ in terms of $\hat{\mathbf{L}}^{2}$ so that the Hamilton operator becomes

\begin{align}
  \hat{h}_{i} = \frac{ - \hbar^{2} }{ 2 m_{\mathrm{e}} r^{2}} \frac{ \partial }{ \partial r } \biggl( r^{2} \frac{ \partial }{ \partial r } \biggr) + \frac{ 1 }{ 2 m_{\mathrm{e}} r^{2} } \hat{\mathbf{L}}^{2} - \frac{ Z e^{2} }{ 2 m_{\mathrm{e}} r }
\end{align}

where $r$ is the distance between nucleus and electron. The operator $\hat{\mathbf{L}}^{2}$ commutes with the Hamilton operator $\hat{h}_{i}$ because $\hat{\mathbf{L}}^{2}$ commutes with itself and does not involve the variable
$r$. Likewise, the operator $\hat{L}_{z}$ commutes with $\hat{h}_{i}$ because it
commutes with $\hat{\mathbf{L}}^{2}$ and also does not involve the variable $r$.
Thus the operators $\hat{\mathbf{L}}^{2}$, $\hat{L}_{z}$, and $\hat{h}_{i}$ have simultaneous eigenfunctions:

\begin{align}
  \hat{h}_{i} \psi_{n}(r_{i}, \theta_{i}, \phi_{i}) &= \varepsilon_{n} \psi_{n}(r_{i}, \theta_{i}, \phi_{i}) \\
  \hat{\mathbf{L}}^{2} \psi_{n}(r_{i}, \theta_{i}, \phi_{i}) &= \hbar^{2} \ell_{i} (\ell_{i} + 1) \psi_{n}(r_{i}, \theta_{i}, \phi_{i}) \qquad \qquad \ell = 0, 1, 2,..., n-1 \\
  \hat{L}_{z} \psi_{n}(r_{i}, \theta_{i}, \phi_{i}) &= \hbar m_{i} \psi_{n}(r_{i}, \theta_{i}, \phi_{i}) \qquad \qquad m_{i} = -\ell_{i}, -\ell_{i} + 1, \ldots , \ell_{i} - 1, \ell_{i}
\end{align}

Thus, you have a well defined angular momentum for each electron in the system.