At what composition is Maximum deviation from Raoult's law observed?

There's no definitive answer to your question. You might want to look up azeotropic mixtures.
Essentially, to make an azeotrope, you need a set of liquids with really high deviation from "ideal" behaviour, and at the composition where the deviation is maximum, the slope is zero. At that composition, there isn't much variation in vapor pressure as you vary the mole fractions, and so both components of the liquid boil equally, as stated here:

An azeotrope or a constant boiling point mixture is a mixture of two or more liquids whose proportions cannot be altered or changed by simple distillation.Because their composition is unchanged by distillation, azeotropes are also called (especially in older texts) constant boiling point mixtures.

Now, if you look at a few examples of azeotropes, it becomes clear that the maximum deviation is not achieved at the same composition ratio:

Sulfuric acid (98.3%) / water, boils at 338 °C
||Nitric acid (68%) / water, boils at 120.2 °C at 1 atm||
Perchloric acid (71.6%) / water, boils at 203 °C||
Hydrofluoric acid (35.6%) / water, boils at 111.35 °C

References: Azeotropes, wikipedia


Energetic Placement of Atomic Orbitals in the HCl Molecular Orbital Diagram

I'd say there are two important things to consider here: Firstly, $\ce{Cl}$ is more electronegative than $\ce{H}$. And secondly, as you proceed from the left to the right in any row of the periodic table, the s and p orbitals go down in energy. This is a consequence of the fact that the valence electrons do not screen each other effectively. Thus, the addition of one proton to the nucleus and one electron does not cancel; instead, the valence electrons "feel" an increased nuclear charge.
The electronegativity difference and the reduced screening among the valence electrons seems to be enough to shift the $3\ce{s}$ and even the $3\ce{p}$ orbital of $\ce{Cl}$ below the energy of the $1\ce{s}$ orbital of $\ce{H}$. The experimental state averaged ionization potentials (IP) could serve as a good guide for the placement of starting orbitals in an orbital interaction diagram. For the $\ce{H}$ $1\ce{s}$ orbital you have an IP of $-13.6 \, \mathrm{eV}$. For the $\ce{Cl}$ $3\ce{s}$ and $3\ce{p}$ orbitals you have IP's of $-25.2 \, \mathrm{eV}$ and $-13.7 \, \mathrm{eV}$, respectively.$^{[1]}$ So, this indicates that the $3\ce{p}$ orbital of $\ce{Cl}$ does indeed lie slightly below the $1\ce{s}$ orbital of $\ce{H}$.

Update:

Here is a table of the experimental state averaged ionization potentials, in electron volts, for the main group atoms. However, those $\ce{p}$ atomic orbitals for groups 1 and 2 and for the $\ce{s}$ and $\ce{p}$ orbitals of the sixth row are not experimentally known and hence, calculated values (indicated by an asterisk) have been used.$^{[2,3]}$ The latter include relativistic corrections. The table is taken from the 2nd Edition of the book Orbital Interactions in Chemistry by T. A. Albright, J. K. Burdett, and M.-H. Whangbo.

[1] L. C. Allen, J. Am. Chem. Soc. 1989, 111, 9003 ; L. C. Allen, J. Am. Chem. Soc. 1992, 114, 1510; L. C. Allen, Acc. Chem. Res. 1990, 23, 175.

[2] A. Vela, J. L. Gazquez, J. Phys. Chem. 1988, 92, 5688.

[3] J. P. Desclaux, At. Nucl. Data Tables 1973, 12, 311.


Octasulfur is just one of few different allotropic forms of sulfur. It happens to be the most stable form at room temperature, but different conditions favor different forms. Thus, for instance, sulfur vapor usually contains $\ce{S, S2, S4, S6, S8}$, and the higher the temperature, the lower is the amount of $\ce{S8}$ molecules since they break up into smaller allotropes of sulfur.

So, while two sulfur atoms can share a pair of electrons by forming a diatomic disulfur molecule with a double bond, so can also eight atoms, but by forming a puckered ring structure with few single bonds. Then it is basically the question of thermodynamic and kinetic stability, which form would dominate others at some particular conditions.


Unsaturation of hydrocarbons and derivatives

In your question, you are talking about two different things (although they are related). First one is the definition of unsaturation (in organic chemistry). Well, Dictionary.com gives fairly okay definition:

  (An organic compound) having a double or triple bond and capable of taking on elements or groups by direct chemical combination without the liberation of other elements or compounds, as ethylene, $\ce{CH2=CH2}$; undersaturated.

However, this definition would mostly apply to non-aromatic compounds. Since theyaromatic compounds are well stabilized by resonance, they would seldomely undergo addition reaction with reactive reagents like $\ce{Br2}$ to give saturated compounds (even though, theoretically, aromatics are unsaturated compounds). Thus, I'd call the definition of unsaturation of organic compound is the one with one or more $\pi$-bonds (e.g., acetone has one $\pi$-bond between $\ce{C-O}$). The number of $\pi$-bonds is the measure of unsaturation (remember, a triple bond is a combination of two $\pi$-bonds).

The second thing in your question is degree of unsaturation. That is the most frequent one used in structure elucidation. The degree of unsaturation of given molecule (or Index of $\ce{H2}$ Deficiency) would only give you the sum of double bonds, triple bonds, and/or rings present within the molecule. For example, if the degree of unsaturation of given compound is 1 (e.g., $\ce{C6H12}$), that means the compound is an alkene (with one double bond; e.g., hexene) or a monocyclic alkane (e.g., cyclohexane).

If you look at the saturated linear alkanes, their molecular forrmula is $\ce{C_nH_{2n+2}}$. Take a $\ce{H2}$ away from the molecule, which gives you $\ce{C_nH_{2n}}$ (like $\ce{C6H12}$, one degree of unsaturation), and so on. Thus, benzene ($\ce{C6H6}$) lacks four $\ce{H2}$ molecules (eight hydrogen atoms; $\ce{C6H14-C6H6}$), and its degree of unsaturation is 4 (3 $\pi$-bonds and a ring).


How to know it when I see a covalent network?

A good source of information on this topic is the website Tetrahedra of Structure, Bonding & Material Type, from which the above image is taken.

An extreme example of the metal corner should be cesium, the ionic corner CsF, and the molecular corner $\ce{F2}$.  At the top is diamond carbon.

The network covalent apex is truncated on the basis that a network covalent material requires an element with the electronegativity of carbon or less (less than or equal to 2.55).

Then the challenge is to distinguish a boundary between network covalent and the other three types

network covalent vs. metal

network covalent vs. ionic

network covalent vs. molecular

Examples of network covalent compounds other than diamond and silicon dioxide include silicon carbide, silicon nitride, boron nitride, aluminum phosphide, gallium arsenide, aluminum oxide.  However, anytime there is an electronegativity difference there is some ionic component.

Tin can exist in metallic and network covalent allotropes.

Selenium can exist in a network covalent allotrope or as molecular $\ce{Se8}$

Going from network covalent to molecular covalent, think of how there is diamond, graphite, nanotubes, fullerenes and even small molecules like $\ce{C4}$


Is the Fischer Esterification mechanism very simplified?

That is the wrong question to ask. Mechanisms are putative pathways. Accepted mechanisms are mechanisms that we have not found evidence to refute. There is no way to determine whether a mechanism is the true mechanism.


Units for dissociation constant and relationship to Gibbs Free Energy

free energy change of -32 kcal/mol, which yields something on the order of 10^-24 at 298K.

Seems the main mistake is a sign error.  Also "log" must mean "natural logarthim" the way the equation in the question is written to be correct.

If the free energy change is negative, the equillbrium constant is $>1$

Should be $10^{\frac{32\frac{kcal}{mol}}{2.303\times0.00199\frac{kcal}{Kmol}\times{298K}}} = 10^{23}$


I'm not sure what you mean by "cementing" a wall. However, cement (typically Portland cement) is the paste that binds together the aggregate (gravel and sand) in concrete. The better the bond between the cement and the aggregate, the higher the resultant compressive strength of the concrete.

When cement is mixed with water in concrete it binds to the aggregate through a chemical reaction known as "hydration."

http://en.wikipedia.org/wiki/Hydration_reaction

A hydration reaction consumes water as a reactant, which is why you must keep concrete wet for several days after it is placed. If the concrete dries out then it has no water to continue the hydration reaction. The end result then is that you're left with cement that is not fully adhered to the aggregate (less compressive strength).

Careful though - you only need to keep the concrete wet after it has set up and is curing. Adding too much water to the mix (giving it a high water to cement ratio, w/c) actually drastically reduces the compressive strength of the concrete.

Curing compounds are especially helpful for properly curing concrete. (speaking from a heavy civil general contractor's point of view) We use this product all all the time:

http://www.wrmeadows.com/1100-clear-resin-water-based-concrete-curing-compound/


What is the best hydrogen bond acceptor in 2-amino-N-(3-hydroxyphenyl)acetamide?

As a general rule of thumb, a nitrogen is usually a better hydrogen bond acceptor than an oxygen. There is a brief overview on Wikipedia with a couple of values and where they were derived from.

This makes sense intuitively, too. Ammonia is a stronger base than water as it is a better proton acceptor. An amine is a stronger base than an alcohol.

In your example, 2-amino-N-(3-hydroxyphenyl)acetamide, there are three functional groups (or four if you want to count the aromatic ring, too): a primary amine, a secondary amide, and a phenolic alcohol.

The phenolic alcohol is rather acidic and not such a good hydrogen bond acceptor, as some of the lone pairs are in conjugation with the aromatic ring.
The amide is also a delocalised system, which makes the nitrogen pretty unavailable for hydrogen bonding. For this functional group, if it were separate, the π-orbital would be the HOMO, and this is where the protonation/ proton-accepting will happen; due to the delocalisation this is less so the case. An amide usually is not a good hydrogen-bond acceptor.
The amine is primary, which means there is much space around it and it can directly interact with the solvent. This is a good hydrogen-bond acceptor.

Therefore neither oxygen will be the best hydrogen-bond acceptor.


Why does the solubility of LiCl increase with temperature if the dissolution is exothermic?

The key to the answer lies in this comment in the answer to the linked OP:

  the dissolution of NaOH is exothermic, yet it's solubility increases with temperature.

So in short it's the opposite of #2: dissolving the first couple of molecules of $\ce{LiCl}$ is exothermic, but then if you dissolve more it become endothermic. The heat of dissolution to form an infinitely dilute solution is not the same as the heat associated with formation of a saturated solution.

Alternately you might say that the heat of dilution of $\ce{LiCl}$ is also exothermic (see e.g. Reference 1).

Reference

YC Wu, TF Young, Journal of Research of the National Bureau of Standards Vol. 85, No.1, January- February 1980, p. 11-17.


Why is a 7-endo radical cyclisation favoured over a 6-exo in this synthesis?

I'm the co-first author of this paper. I want to thank Mathew Mahindaratne for the helpful and thoughtful discussion. I just want to add one more result when I was doing the synthesis, please see the attached. The beauty of total synthesis is that we never 100% sure what the product we get, but we can always find a way to explain the results.


Getting the pressure of a system by series expansion

Since this is clearly an exercise, I don't think it's appropriate to give the full solution, but I see no harm in pointing you in the right direction.

You are aiming to convert an expansion in $z$ into an expansion in the number density $\rho=N/A$. These are often called virial expansions. Your first equation can be written
$$
\rho = q\left[z+\frac{z^2}{4}\right] \qquad\text{where}\qquad
q = \frac{2\pi}{(h\beta c)^2} .
$$
I have defined $q$ just to cut down on needless clutter.
The brute force approach to this would be to solve this equation, to give $z$ in terms of $\rho$,
and then substitute this expression into your series formula for $P/K_BT$ as a function of $z$. You can do this; it's a bit fiddly, since it is a quadratic equation in $z$, but it could be done.

For your purposes, a slightly more straightforward approach is to assume that you will get a result of the form
$$
z = c_1 \rho + c_2 \rho^2 + \ldots
$$
where you need to determine the coefficients $c_1$, $c_2$ etc.
Hopefully you can see why there is no need for a $\rho$-independent term $c_0$, when you look at the equation for $\rho$.
Also, it should be obvious that you don't need to consider more terms like $c_3$ etc, if you are only interested in the final result having the first couple of powers of $\rho$.

So you determine the coefficients $c_1$, $c_2$ by substituting that expression for $z$ into the right hand side of the equation for $\rho$,
and equating the coefficients of corresponding powers of $\rho$ on each side.
The term linear in $\rho$ will give the coefficient $c_1$ immediately:
$$
[\rho^1]:\qquad 1 = q \, c_1 .
$$
The quadratic term in $\rho$ will give an equation for $c_2$, which
also involves $c_1$,
$$
[\rho^2]:\qquad 0 = \text{I'll leave you to work this out.}
$$
Since you just determined $c_1$, this will be easy to solve to get $c_2$.

So now you have $z\approx c_1 \rho + c_2 \rho^2$ and you can insert this expression into the given series expansion for $P/K_BT$,
only keeping the terms in $\rho$ and $\rho^2$.
If you do all this carefully, you will get the answer provided.


What are examples of group 14 elements forming dπ - pπ bonding?

For $d\pi-p\pi$ bonding we need to consider other elements of group 14 except $\ce{C}$. (i.e, $\ce{Si, Ge ,Sn}$ etc.) which have got a vacant $d$ orbital of proper symmetry and energy to come into play bonding with attached group  having lone pair of electron in p orbital of comparable energy.
    For example $\ce{SiF4}$. $\ce{GeCl4}$ etc.
In tetrahedral geometry the orientation of vacant  d orbital of the central atom (dxy. dyz,dxz) may overlap with filled p orbital of the ligating atom  and hence $d\pi -p\pi$ bonding is possible.


What is the acid dissociation constant of cacodylic acid?

I found this article (TW. Shin, K. Kim, IJ. Lee, J. Solution Chem. 1997, 26, 379. doi: 10.1007/BF02767677) with precise analysis of the pKa of cacodylic acid - it says that in 25 °C it's estimated to be 6.27. Values lower than 6 are wrong, but found often. Sometimes it's said to be a diprotic (which it isn't) with first pKa 1.57 and second 6.27 - pubchem says 1.57 citing Lide, D.R. (ed.). CRC Handbook of Chemistry and Physics. 76th ed. Boca Raton, FL: CRC Press Inc., 1995-1996., p. 8-45 - it might be one of the sources of problem.


$$\ce{Cu(NO3)2}+2\ce{NH4OH} =>\ce{Cu(OH)2}(s)+2\ce{NH4NO3} $$
Essentially, we're trying to find the Molarity of $\ce{NH4NO3}$ since the copper filters out(as you stated)

First solve for limiting reagent. $$10mL\ce{Cu(NO3)2}*\frac{1L}{1000mL}*\frac{1.15mol}{1L}=0.0155mol\ce{Cu(NO3)2}$$
$$20mL\ce{NH4OH}*\frac{1L}{1000mL}*\frac{0.85mol}{1L}=0.0170mol\ce{NH4OH}*\frac{1}{2}=0.0085$$

Why do we divide mols of $\ce{NH4OH}$ by 2? Remember that for every molecule of $\ce{Cu(NO3)2}$, there must be 2 molecules of $\ce{NH4OH}$. $$\frac{2\ce{NH4OH}}{\ce{Cu(NO3)2}}$$ for this reason we need to divide the amount of $\ce{NH4OH}$ by 2 to compensate for stoicheometry.
So actually, $\ce{NH4OH}$ our limiting reagent.
Since you didn't give my any solubility constants, I'm going to assume 100% of the Copper became  $\ce{Cu(OH)2}(s)$ and all of the ammonium became $\ce{NH4NO3}$. In the real world, perciptates never form at 100%.

0.0085 mols of $\ce{Cu(OH)2}$ is produced. You want $\ce{NH4NO3}$  Can you use stoichometric ratios to proceed from there? Molarity - (mol/L). Make sure you add up the two volumes to find the Total volume.


Comprehensive resource on diazo coupling?

If you prefer books as secondary reference, the The chemistry of diazonium and diazo groups in the Patai series (back 1978) may be one source.  More recently, the two books about diazochemistry by Zollinger, 1994 here and here.

Maybe not as comprehensive as the two sources above, but offering an overview are Merino's critical review about diazocoupling reactions in Chem. Soc. Rev. (doi 10.1039/C0CS00183J) in 2011, and Oger's overview about their application in flow chemistry in Organic Chemistry Frontiers (doi 10.1039/C5QO00037H) in 2015.


Why does the "gap" between atomic radii get smaller down the periodic table?

The first thing to remember is that the "shell" model is an over-simplification of quantum chemistry. A better picture is that the electrons represent "clouds" (i.e., electron density).

Your question basically boils down to "why don't 5th and 6th row atoms get much larger?"

Quantum mechanics can give a relatively easy formula for the "average" radius of a given hydrogen orbital

$$\langle r \rangle = a_0 n^2 \left( 1 + \frac{1}{2} \left[ 1- \frac{l(l+1)}{n^2} \right] \right)$$

Here, $a_0$ is the Bohr radius, the expected radius of the hydrogen atom, and $n$ is the principal quantum number and $l$ is the angular quantum number (e.g., l=0 for "s" orbitals and n=5 and 6 for Rb and Cs).

It's pretty obvious that as n gets large for s orbitals (l=0), that formula approximates:

$$\langle r \rangle \approx \frac{3 a_0}{2} n^2$$

(The left side becomes smaller and smaller, so the coefficient gets closer and closer to 3/2.)

So for a hydrogen atom (one electron), the 5s and 6s orbitals in fact get very large.

Your question is about atoms. So we have to consider many electrons. Basically, this is another representation of the lanthanide contraction caused by relativistic effects.

Indeed, in that last link, there's a formula for the atomic radius after relativistic corrections.

For atomic numbers 11 and 19, the relativistic correction is 0.99 (i.e., no real change to the radius). For atomic numbers 37 and 55, it's 0.963 and 0.916, respectively. So ~9% of the "ideal" radius is lost with $\ce{Cs}$.


Nucleophilicty order of various nucleophiles

Like most concepts that are not captured by a single number, nucleophilicity is not a simple (or at least, relatively simple) concept like acidity.

Nucleophilicity depends a lot on orbital alignment and interaction, in addition to thinking about charges. It's also confusing because nucleophility is not always married to the stability of the adduct. For example, iodide is a great nucleophile, but alkyl iodides are also highly reactive.

In your specific case, since the two nucleophiles have the same charge, you should look at other factors. Bromide is significantly larger than hydroxide, and it is a much better nucleophile, at least when we're considering carbon electrophiles. Indeed, we see bromide used in nucleophilic catalysis but not hydroxide.


Can a copper(II) chloride be added to liquified butane?

If we had butane under pressure in a vessel as liquid and this was to be in contact with a solution of copper chloride in water then I think that nothing will happen. The butane and the water will form two seporate layers.

However if we were to add a copper extraction agent to the butane such as LIX-860 (a phenolic oxime reagent) then it would be possible to convert some of the copper from aqueous copper chloride into an organic soluble copper complex. Then the copper would be extracted into the butane layer.

Aother option would be to add copper chloride powder to the butane solution of LIX-860, I think that this could react but the reaction would be slower. Also I think that the hydrogen chloride formed as side product from the reaction would have a greater inhibiting effect on the reaction.

Another option would be to use a lipophilic chloride ionic liquid such as aliquat 336 to dissolve the copper chloride. This could react (slowly) with powdered copper(II) chloride to form 2(R4N+).[CuCl4]2- which would have some solubility in liquid butane. From my own experience of doing solvent extraction I know that aliquat 336 in alkanes is not a great system. When the metal loading becomes moderate the organic phase can separate into the organic solvent (diluent) and a liquid metal bearing layer.

In short I hold the view that pure butane will not be able to dissolve copper chloride, but with a suitable additive it will be possible to get a solution of copper in butane. The copper will be there as a copper complex.

I think that if we were to place water in a sealed container and put pure butane gas in the head space that the concentration of the butane in the aqueous layer will be very low. The copper chloride is likely to stay in the aqueous layer as hydrated copper ions and chloride anions if the concentration is low. If the concentration is high then it will be able to form chlorocomplexes of copper.


What should be the name of this compound containg both double and triple bond?

According to R-3.1.1 of IUPAC, Commission on Nomenclature of Organic Chemistry:

Numbers as low as possible are given to double and triple bonds as a set, even though this may at times give "-yne" a lower number than "-ene". If a choice remains, preference for low locants is given to the double bonds. Only the lower locant for a multiple bond is cited except when the numerical difference between the two locants is greater than one, in which case the higher-numbered locant is cited in parentheses (see R-0.1.4)

Numbering from the right side of the molecule as seen in the question figure will assign the locants (2,4) to the double and triple bonds respectively.
On the other hand, following a similar numbering process from the left side of the molecule as seen in the diagram will yield (3,1) as the locants for the double and triple bonds respectively.
Now, as per IUPAC Recommendations 1:

P-14.3.5 Lowest set of locants
The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference; for example, the locant set ‘2,3,5,8’ is lower than ‘3,4,6,8’ and ‘2,4,5,7’.

Therefore, since (1,3) is a lower set of locants as compared to (2,4) as per the above definition, the IUPAC name of the compound should be pent-3-en-1-yne
Note that this molecule can also exhibit stereoisomerism by the virtue of it's double bond, so depending on whether the hydrogens are cis or trans to each other, the complete name of the molecule could be (3 Z)-pent-3-en-1-yne or (3E)-pent-3-en-1-yne, respectively.


What does gases at constant pressure mean?

Without a clearer setup it's difficult to say whether someone misspoke, whether you misunderstood precisely what your teacher was saying, or whether your teacher was incorrect.

If the setup is that there is a single piston that contains two gases, and a weight is placed upon it, then:

The volume of each of the gases in that piston is equal to the entire volume of the piston.  The volume of any gas is the entire volume of its container.  If the gases are ideal, then the volume of the piston will be the sum of the volumes of the piston if each of the gases were kept in it individually.
The pressures of the two gases sum to the pressure exerted on the piston by the weight (and the outside atmosphere.)  The pressures will be constant as long as the amount of gas present, the applied pressure, and the temperature are kept constant.

If the setup is that there are two pistons that each contain a gas, and an identical weight is placed upon each (or a single piston filled with two gases, one at a time) then:

The volume of the two pistons will be proportional to the number of moles of ideal gas that it contains, as long as both are also at the same temperature (we're already assuming the same pressure due to the identical weight.)
The pressures of the gases in the two pistons will be equal, because they are opposing the same weight + outside atmospheric pressure.

As to your other questions:
Gases at different pressures do indeed exert constant pressure, assuming things like constant temperature, constant volume, and constant number of moles of gas.  The pressure of a gas is the result of huge numbers of gas particles striking the walls of their container every second.  The rate at which those particles strike and rebound and the force with which they do so determines the pressure.  Because there are sooooo many gas particles in any reasonable quantity of gas, the fluctuation in the rate and intensity of those collisions is nearly constant (to many, many decimal places.)

Gases at constant pressure do not necessarily exert equal pressures.  If I pump up the front tire of a bicycle to twice the pressure as the back tire, each has a constant pressure, but they are different.  If I put two moles of gas A and one mole of gas B into a sealed container, the pressure due to A will be twice that of the pressure due to B, but both will be constant, as will their sum, the total pressure of the container.


Is the application of Le Chatelier's principle wrong in this case?

Try doing the calculation using the fact that $K_x$ actually is not constant at a given temperature.  It depends on total pressure as well as temperature.  Assuming you maintain ideal gas behavior $K_p$, and only $K_p$, is constant when temperature is fixed.  In terms of that equilibrium constant you should find the reaction following Le Chatelier's Principle both of your cases.


What is the relation between volume strength and normality of H2O2?

The n- factor must be obviously wrong because two electrons are involved in the reduction of peroxide to oxygen. Write the half cell yourself. Indian chemistry textbooks should have pity on the world and stop teaching these equivalents to young students. This is a thing of the 18th century. Those who invented this concept moved on.


Traditional understanding of trajectory is of no use for electrons. Granted, on macroscale you may detect something close to movement and trajectory, (electron beams and electron cannons take their names from this) but it still far from what we have in classical mechanics. Ideally you should abandon any attempts to relate quantum and classical mechanics and view quantum mechanics as an abstract mathematical construct.
Incorrect question. If you have a system of two or more particles, you should consider the wavefunction of combined system, not separate wavefunctions for separate particles. Still, for electrons being fermions, there is a strict rule that wavefunction must be zero for positions in phase-space where two electrons of same spin are at same location and since electrons are of same charge, probability of having two electrons of different sping close to each other is usually lower than to have them at some distance.
Yes, only positive probabilities. Though, this does no apply to wavefunction itself, it may be negative at some point (actually, its value can be a complex number).


What's the simplest way to etch Nickel and Chromium sputtered layers?

Nickel can be etched using diluted (1:1 V) $HCl$. 50 nm will be dissolved within seconds.

The etching solution for the chromium layer contains $\ce{Ce(NH4)2(NO3)6}$+$\ce{CH3COOH}$+$\ce{H2O}$
in ratio 165g : 45 ml (con.) :100 ml DI water. The chromium layer is dissolved within 5 minutes of etching.


Is the effectiveness of chemical catalysts affected by temperature the same way enzymes are?

This is a somewhat complex question to answer because even your ‘simple’ enzyme example isn’t as simple as it looks. Let me discuss enzyme optimum temperatures first.

For any reaction that is exergonic — and that explicitly includes enzyme catalysis — an increase in temperature raises the reaction rate. This is due to at least two microscopic features:

higher temperature means more and faster movement thus a higher probability of collisions
each collision has a greater chance of being energetic enough to supply the required activation energy

This is condensed in van ’t Hoff’s postulate that an increase of the reaction temperature by $10~\mathrm{^\circ C}$ leads to an increase of the reaction rate by a factor of $2$ to $4$. Therefore, we would initially expect all enzymatic processes to be faster with increased temperatures.

However, the enzyme itself is a highly fragile system that is held together a good deal by noncovalent interactions. These can break apart under elevated temperatures causing the well-balanced enzymatic structure to break down (‘denature’)  into a seemingly random protein chain. This also breaks the catalytic centre meaning that the enzyme can no longer catalyse efficiently. Taking these two effects together, we arrive at the optimum curve of enzyme catalysis.

Chemical catalysts come in a number of shapes and varieties. Consider for example:

the catalyst (mainly platinum as I heard at school) in your car’s exhaust
the complex catalytic oxide system in the Haber-Bosch process
transition metal catalysts such as $\ce{[Pd(PPh3)4]}$
acid catalysis, i.e. dissolved $\ce{H+}$ is the catalyst
organic catalysis
…

Most of these will have some equivalent of a denaturation temperature; e.g. once platinum metal (or palladium, if on charcoal for typical organic hydrogenations) melts, it can no longer act as a catalyst for the reaction in the way it could in solid state. Thus, there should be a ‘denaturation equivalent temperature’ for most chemical catalyst systems at some point. Some compounds — the Haber-Bosch catalysts jump to mind — will, however, probably melt at such high temperatures that you would need to live on the sun’s surface for that to happen. Therefore, we could say that no limit exists for all intents and purposes.

Catalysts such as tetrakis(triphenylphosphane)palladium(0) need to dissociate at least one ligand to be catalytically active. An increase in temperature should enhance that dissociation. Therefore, many homogenously palladium-catalysed reactions are indeed sped up by a moderate increase in temperature. However, side reactions also speed up.

Organocatalysts and simple proton catalysis should be among the most temperature-insensitive — the former because all bonds should be covalent, the latter because it is just a reactive, activating species that cannot be poisoned. However, these two require a medium in which they act; the boiling point of the solvent is the highest possible temperature for their reactions.

Thus basically: yes, all chemical catalysts have some maximum useable temperature somewhere, but for most it does not matter for practical intents and purposes. Also, there are no simple models to explain this like in the enzyme case, hence it isn’t typically taught.


What actually happens when 'energy is released'?

Thermodynamics works perfectly well by assuming that matter does not need to be made of discrete particles, i.e atoms and molecule are not necessary. It was developed before the structure of atoms and molecules  were known so it cannot inform us about anything at a molecular level. It is concerned primarily about energy and entropy.

If you break a bond in a diatomic molecule then all that can happen is that the two atoms move away from one another with the excess energy used to break the bond  shared between the atoms in the form of kinetic energy. The 'excess energy' is the energy above the dissociation energy. The atoms will be 'hot' and if in a gas (or solvent) of inert molecules will collide with these and their energy will be spread out among all the molecules raising the temperature by just a little bit.

In a polyatomic molecule if the products are lower in energy than reactants the excess energy  will raise the temperature of the surrounding solvent or gas. To break a bond the activation energy has to be overcome, once this happens the product has more potential energy than the reactant, i.e. more vibrational and rotational energy levels are populated than in the reactant, see figure. This is because the potential well of the product is lower than that of the reactant. At the top of the activation barrier is the transition state.

Once in the product potential the extra energy is lost quickly, a few picoseconds, to any solvent surrounding the molecule. Energy is also lost to other vibrational /rotational energy levels in the product molecule. These are vibrations not involved in the bond breaking.  Eventually all this energy is spread out in the surrounding solvent (or gas) raising its temperature. The kinetic energy change in a reaction may not be that large if the energy is taken up by the vibrations and rotations. This is unlike the case when an atom is formed as a product.

The figure shows how the product has more vibrational energy than  the reactant after surmounting the barrier and transition state (blue dot) which is the energy and coordinates at which the molecule changes from reactant to product. The energy difference, leading to heating, is that between the two lowest energy levels in each potential.


Why does chlorination of toluene stop at two chlorine atoms?

It doesn't stop, it just gets a lot slower the more chlorine atoms you add. From the abstract of an article titled Photochemical side-chain chlorination of toluene[1]:

  Where $k_1$, $k_2$, and $k_3$ are the rate consts. for the successive reactions: $\ce{PhMe}$ → $\ce{PhCH2Cl}$ → $\ce{PhCHCl2}$ → $\ce{PhCCl3}$, resp.; $r = \frac{k_1}{k_2}$ and $s = \frac{k_2}{k_3}$, values of $r$ and $s$, resp., were 6.0 and 5.7 at 100° and 8.4 and 8.5 at 40°, for batch operation, with light varying from ultraviolet to yellow.

[1] Haring, I. H. G.; Knol, H. W. Chemical and Process Engineering (London) 1964, 45 (10), 560–567.


One of the authors has recently established a research group at my university, so I spoke to a member of their lab:

The corona phase is not exactly a phase in the thermodynamic sense, but rather the portion of the polymer in contact with the nanotube. In the below image (taken from the Nature Nanotechnology paper in question), the blue portion and the parts of the red portion in contact with the tube would be the corona phase.


Which invariant reactions are congruent?

You are correct, none of the reactions are congruent reactions.

Per IUPAC GoldBook:

  Congruent Transition: A transition in which the two-phase equilibrium of melting, vaporization or allotropism of a compound involves phases of the same composition.

The key term here is "phases of the same composition." Since all of the reactions have two phases transforming into one phase (or vise-versa) the reaction is not congruent.

See the phase diagram for silicon and titanium. Congruent melting occurs for $\ce{Ti5Si3}$ and $\ce{TiSi2}$ where you observe a $\ce{\color{red}{comp A (s)} -> \color{red}{comp A (l)}}$ transition, whereas at the eutectics you have $\ce{\color{blue}{comp B (s)} + \color{green}{comp C (s)} -> \color{red}{comp A (l)}} $ transition where $\mathrm{B \ne C \ne A \ne B}$. Thus the eutectics are not congruent transitions.


Why isn't viscosity considered a property of solids?

The concept of deformations of solids under loading has a long history. It just isn't called 'viscosity' - instead this is just part of the mechanical properties of solids.

Relevant places to start include:

Wikipedia on Plasticity

Wikipedia on Deformation

The Non-Destructive Testing org on Elastic/Plastic Deformation

In the purely elastic regime, if you pull on a steel bar it will extend under the load, and will return to its original dimensions when the load is removed. If you apply too large of a load, it will plastically deform, so that the dimensions will not be the same when the load is removed. Under some circumstances even fairly small loads over long times can result in permanent deformation.

Physically, these can be a result of:

Dislocation formation and movement in crystals (e.g. slip bands)
Grain boundary motion
Creep (point defect formation and diffusion)
Crazing in plastics (formation of open volumes)
And a number of other, less common, mechanisms.

These topics are typically covered in metallurgy and materials science courses/textbooks. Again, even in amorphous solids, one would not tend to call it viscosity - that is reserved for liquids.


Quoting from Weller, M.; Overton, T.; Rourke, J.; Armstrong, F. Inorganic Chemistry, 6th ed:-

  $\ce{KMnF3}$ is prepared by the addition of $\ce{KF}$ to $\ce{Mn^2+}$ solution. [...] Fluorination of $\ce{MnF2}$ in presence of an alkali metal fluoride leads to the complex Mn(IV) fluorides of composition $\ce{M2MnF6 (M = K, Rb, Cs) }$.

So, the reaction is all right.


Jahn-Teller Metal : The new state of matter

The Jahn-Teller metal is not one of the 3 (or 4 including plasma) states that occur naturally on Earth; this state allows a metal to have the properties of an insulator and a superconductor.

It is an intermediate state of matter confirming the Jahn-Teller distortion; by interacting with certain electrons (a more detailed description can be found in this Science Advances magazine article, a distorted crystalline structure allows a solid depicting characteristics of an insulator to also becomes a conductor.

The main breakthrough this brings is having a superconductors. Superconductors are extremely useful because they possess zero resistance, except the conditions require them to be cooled to extreme temperatures.
This state nonetheless suggests reproducing superconductors at a less extreme temperature.

Making it an economic way to make efficient electronics.


How do we calculate N factor and why is it useful in calulations?

There are some different applications.
For one, it tells you how many times an acid can lose a proton. For instance, you can have a diprotic ("with $\ce{2 H+}$") molecule, with $n=2$. For instance, $\ce{H2SO4}$, deprotonates to $\ce{HSO4-}$, which can then go to $\ce{SO4^{-2}}$.
Why is that important? Take an acidic reaction with $\ce{H2SO4}$. If the extent of reaction is strongly forward, you would treat $\ce{H2SO4}$ as if it had twice its concentration to take into account that it has two protons it can lose. If you only accounted for the first proton you would be underestimating this effect.
It's also known as equivalent concentration, which might make it sound less foreign to you.
In redox reactions, it shows how many electrons an agent can accept/donate, and in precipitation reactions it measures the # of ions that'll precipitate out of a solution. For all these situations the n-factor or equivalent concentration corrects for multiplicity, since now we know not all molecules are made equal. Some acids can deprotonate twice, some reagents will lose 4 electrons in redox while others may lose just 2. To consider these differences we use this concept.
Sources: Wikipedia


Determine Cu+2 and SO4-2 concentration using Potentiometry?

What is the range of the components? This is very important because this will determine if titration or photometry is used (or a combination of both).

Also interfering components should be known. For eaxample. Barium is known to react with a lot of other components (solubility table). Also EDTA reacts with oter components (Also depending on the pH (EDTA complexation constants)

Titration:

Copper and sulfate can be determined in one analysis using and copper ISE. This is done by adding an NH3 buffer (pH of solution = 10) and adding an exxess of Barium nitrate. Barium will react with SO4 and the free barium is titrated. Titrate the solution with EDTA. This first inflection point will be copper, the second one the free barium.

Keep in mind that EDTA reacts with a lot of other components. Also if the concentration difference (molar) between copper and barium is large, high titrant usages are to be expected

Photometry:

Copper can be determined using an variaty of chemical substances (e.g. cuprizone, Bathocuproine). The choise depends on range, interfering components ect..

Sulfate is possible. However, titration is preferred.


Can an ethyl ester of L-proline be synthesised from the amino acid and ethyl alcohol in toluene with azeotropic distillation of water?

I think this is unlikely to work under the conditions described as the catalytic PTSA is tied up as its proline salt. If you reflux under D-S conditions you will just remove the EtOH due to the azeotrope with Toluene.


Dipole moment of (1s,2s,3s)-1,2,3-trichlorocyclopropane

The issue stems from the fact that you may not have understood what each bond notation means.
Here, the solid wedge indicates that the bond is toward us and so a 3-D view of the same molecule would look as follows:

So as you can see the dipole moments – instead of cancelling, add up. So simply by elimination option (d) is the right answer
For trans-1,3-dichlorocyclobutane, the reason for the zero dipole moment is explained in ron's answer to the question "Does trans 1,3-dichlorocyclobutane have zero dipole moment?".

[...] Conformers A and B both have $C_$ symmetry (the only symmetry element is a plane that bisects the ring and contains the two cyclobutane carbons bearing the substituents) and therefore do have a dipole moment. However, the dipole moments of conformers A and B are equal and opposite, so when flipping is rapid the dipole moment averages out to zero.

Therefore (d) is the correct answer and not (c).


Balancing the chemical equation using Oxidation Number Method

It's hard to pinpoint what the problem here is since I don't have the mentioned book to compare an answer with.
The simplest way is write down half-reactions for reduction (red) and oxidation (ox) processes once you've assigned oxidation numbers (denoted above the symbols of the elements which are participating in a redox reaction), then balance the number of the transferred electrons and the algebraic sum of both half-reactions will yield in a complete balanced redox reaction:

$$
\begin{align}
\ce{\overset{+7}{Mn}O4- + 2 H2O + 3 e- &→ \overset{+4}{Mn}O2 + 4 OH-} &\quad |\cdot 2 \tag{red}\\
\ce{\overset{-1}{Br}^- + 6 OH- &→ \overset{+5}{Br}O3- + 3 H2O + 6 e-} &\quad \tag{ox}\\
\hline
\ce{2 MnO4- + Br- + H2O &→ 2 MnO2 + BrO3- + 2 OH-} \tag{redox}
\end{align}
$$

As a sanity check, you can see that there are no protons $\ce{H+}$ participating in either of the reactions, and the resulting medium is indeed basic (slight excess of $\ce{OH-}$), which is in a good agreement with the conditions mentioned in the problem.


How do we know the number of s and p orbitals in a shell?

The number of orbitals in a shell is restricted by the possible values for the quantum numbers. Those come directly from the solution of the Schrödinger Equation for the Hydrogen atom (except for the spin, but let's keep it simple here), and are applied to all other atoms as well.

The quantum numbers and their possible values are:

the principal quantum number $n=1,2,3,4,5 \dots$, which relates to the period
the angular momentum quantum number $l=0,1,\dots, (n-1)$ which defines the shape of the orbitals $s$, $p$, $d$ and so on
the magnetic quantum number $m_l=-l,-l+1,\dots,+l$ further define the shape, e.g. $p_x$ or $d_{z^2}$
the total spin for electrons is always $s=\frac{1}{2}$ (this may be different for other elementary particles)
the relative spin of an electron is $m_s=-\frac{1}{2},+\frac{1}{2}$, and represents $\alpha$ and $\beta$ or up and down spin

Each electron needs to have its own unique set of quantum numbers (within one atom). This is know as Pauli Principle.

Thus for $s$ orbitals ($l=0$) there is only $m_l=0$ possible, hence there can only be one for each shell ($n$). Having four $2s$ orbitals would violate the Pauli Principle.
Additionally having a $2d$ orbital would also not be allowed as $l=2$ would be larger than $(n-1)=1$. So the $n=2$ shell is restricted to one $2s$ orbital ($m_l=0$) and three $2p$ orbitals ($m_l=-1,0,+1$).


Measuring the % of pectin in a fibre after enzymatic treatment with pectinase

Note that I do not have professional experience in the textile industry!

Supposed that the stems have been (mechanically) treated, and the "bark" is removed, the ruthenium red method should work.

In Crystal-Structure of Ruthenium Red and Stereochemistry of its Pectic Stain by Clarence Sterling, Am. J. Bot. 1970, 57, 172-175, the specific binding of Ruthenium Red (RR, CAS-RN: 11103-72-3) along a pectin chain, that is, between a $\ce{COOH}$ of one galacturonic acid unit and the $\ce{OH}$ of the next, has been described.

In a more recent publication, Qualitative distinction of carboxyl group distributions in pectins with ruthenium red, published in Bot. Bull. Acad. Sin., 1999, 40, 115-119 (PDF), the method was applied to orange peel pectines of different origin. Here, the dye-pectin adducts were precipitated by addition of $\ce{CaCl2}$ and centrifugated. Then, the absorbtion of the supernatant solution and a stock solution of the dye at $\lambda = 534\,\mathrm{nm}$ were compared.

Consequently, it should be possible to take fibre samples, suspend them and detemine whether they decrease the absorption of RR (= still contain pectin).

Are you planning to bleach these fibres in a way recently discussed in another question? It might be interesting to figure out whether these conditions (alkaline hydroperoxide) are sufficent to (partly) degrade the pectin.


What is the amount of limiting reagent in this chemical reaction?

$4.59$ is not "something" (you haven't specified the unit) of an $\ce{Al}$ mole, but your amount of $124\mathrm{~g}$ of $\ce{Al}$ is exactly $4.59$ moles of $\ce{Al}$, it's just a unit change. In reaction you wrote, every two moles of $\ce{Al}$ reacts with one mole of $\ce{Fe2O3}$. This may help you to find the solution.


White organic solid that forms CO2 with NaHCO3, forms a diazo dye withe beta-naphthol, and reacts with bromine water

(i) indicates presence of 'strong' acidic group: -(SO3H) and -(CO2H) since both are strongly acidic to protonate NaHCO3 and yield CO2.

(ii) indicates presence of Primary Amine which can fully undergo diazotization reaction and then with beta-naphthol form azo dye (coloured dye).

(iii) excess of bromine water forming s-tribromoaniline directly indicates that Option 2(b) is correct.

An article on point (iii): journal_iisc


What are the effects of radio waves on matter?

This is a bit of a stretch, but the Zeeman and Stark effects are linked to radio frequencies, in that line splitting of a visible spectral frequency (~10^15 Hz) can give rise to a beat-frequency in the RF. Your answer of NMR, though, is more applicable. The most general term for precession in a magnetic field is Larmor precession.


In the reaction that you have mentioned above, the reaction essentially stops after the reaction has taken place in a 1:1 ratio. That is because the electrophilicity of the polar bond of the acid derivative has reduced sharply so it cannot undergo further reaction.
However, had the nucleophile been $\ce{RLi}$ instead of $\ce{RMgX}$ ($\ce{RLi}$ is more ionic and hence a better nucleophile), further reaction would've been possible, although it provides the product in low yield. R attacks C=O resulting in the formation of $\ce{R-C-O- Li+}$ (The other groups remain as they were- the only difference being that it is $\ce{NHLi}$ instead of $\ce{NHMgX}$)


Conc. vs Time Graph for Second Order Reaction

Let's take a single-component second-order reaction:

$$\ce{2A -> B}$$

If it is a 2nd-order reaction, the rate will be proportional to the concentration of $A$, squared:

$$\frac{dA}{dt} = -kA^2$$

That equation can be integrated:

$$\frac{dA}{A^2} == -k dt$$

$$-(A^{-1}-A_0^{-1}) = -k(t-t_0)$$

And solving for $A$:

$$A = \frac{1}{k(t-t_0) + A_0^{-1}}$$

That is the equation of a hyberbola.  It goes to zero more slowly than a the expression for a first-order reaction, i.e. an exponential decay.  This makes sense because as $A$ gets more and more diluted as the reaction occurs, collisions between $A$ molecules become rarer and rarer.

See this web page for more information.


What does the "s" mean in s-cis and s-trans conformations

The s stands for single, as in single bond. It means the double bonds are cis or trans to each other based around the single bond.

It is a common enough expression to search for, if more information is required.


Why is only one part of buffer in the reactants side?

A buffer is a solution that is meant to prevent the pH from drifting away from its intended value in both directions (i.e. the solution should neither become more acidic nor more basic than desired). For the buffer to accomplish this, two components are added one of which is intended to counteract acidic additives the other to counteract basic additives.

Whether or not these two components are conjugate Brønsted acids and bases does not matter; a solution of e.g. ammonium acetate will also buffer. Every liquid inside the human body is buffered by multiple components to a very narrow pH range, so many different compounds together can also produce a buffer effect. However, the combination of an acid and its conjugate base is typically the most effective and narrow-range buffer system available in a typical lab context — and also the easiest to set up and use.

While in the buffered solution the two components of a buffer can indeed be said to react; this can be proven with isotope labelling. For example, if you had $\ce{^13C}$-labelled acetic acid and $\ce{^12C}$-labelled acetate, these two would react in the following way:

$$\ce{CH3^12COO- + CH3^13COOH <=> CH3^12COOH + CH3^13COO-}\tag{1}$$

This reaction does not, however, describe the buffering action.

Buffering itself can only take place if some acidic or basic compound is added to the solution as hinted above. Depending on what is added either equation $(2)$ or equation $(3)$ will take place.

$$\begin{align}\ce{CH3COO- + H3O+ &<=>> CH3COOH + H2O}\tag{2}\\
\ce{CH3COOH + OH- &<=>> CH3COO- + H2O}\tag{3}\end{align}$$

The first equation $(2)$ is the one if something acidic is added, equation $(3)$ takes place if a basic compound is added.

As you can see, one component of the buffer basically exists to quench basic additives (the acidic part) while the other only exists to quench acidic ones (the basic part). Therefore, in a buffering equation you will only see one of the two active at once.


Why does the reaction between 2‐methylbut‐3‐en‐2‐ol and hydrogen bromide show allylic shift?

Even thought the immediate carbocation formed is tertiary, the stability of the double bond should also be taken into account. A more substituted double bond is also favored/stable. In this instance it seems that it's a competition between the double bond's stability and the carbocation's stability. There could be other situations where this is not the case, but you would need experimental data to back up stability (there's no universal ranking system that I know of to say when a double bond outweighs a positive charge).

In short, I would say that the relative stability of a tertiary double bond (compared to a tertiary carbocation) is what would justify the major product being labeled as it is.


Platinum Orbitals in Cisplatin Forming Bonds with LCAOs of Chlorine

I figured it out!

We have to start by looking at d orbital shapes:

To form bonds, the d orbitals of platinum must have the same symmetry (i.e. shape) as the molecular orbital that's approximated by the LCAO.

We start with the bonding orbital, which looks like this: (sorry about the pencil-and-paper phone picture).

We can see we need a d orbital with exactly one lobe in the zx plane (x is pointing to the right in the above). The only d orbital that fits the requirement is the $\textrm{dz}^{2}$.

Moving on  to the anti-bonding orbital, which I can't post because I don't have at least 10 reputation, we see that it has four lobes with two nodal planes. It looks like a $\textrm{d}_{xz}$ orbital, in fact.

We can see we need a d orbital in the zx plane that has two lobes. Clearly, this is the $\textrm{d}_{xz}$ orbital.

Easier than I thought once I got the concepts down!


Comparison of COSY and TOCSY for detection of small couplings between protons

The TOCSY transfer also depends on the size of the J coupling, you don't get around that.

Generally speaking it is found that cross peaks due to direct coupling build up to their maximum intensity in a time which is of the order of $1/(2J)$. Peaks which arise from transfer through two successive couplings take longer to build up, and if three successive couplings are involved, even longer mixing times are needed
Understanding NMR spectroscopy, James Keeler, page 230

The TOCSY has the advantage of creating in-phase signals instead of the anti-phase signals of the COSY, which increases sensitivity. But that doesn't mean that it is generally more sensitive.

Whether or not the overall intensity of the COSY or TOCSY cross peaks is greater depends intimately on the choice of the experimental parameters and on the molecule being studied. However, the general experience is that for small to medium-sized molecules, COSY is the preferred experiment for detecting small couplings.
Understanding NMR spectroscopy, James Keeler, page 229

I'm not sure if you can get a general answer for your case, you might have to directly compare the experiments for the kind of molecule you want to look at. You could also try longer mixing times for the TOCSY and see if they help with your small couplings (which they should, as long as relaxation doesn't eat up that advantage).


Most activated position on para-terphenyl for EAS

So, how do I find the most activated position in this compound?

Start by drawing resonance structures of the various possible intermediates (sigma complexes) formed when $\ce{Br^{+}}$ attacks the different ring positions in p-terphenyl. Whichever intermediate has the most resonance structures is likely to be the most stable (lowest energy) intermediate. The lowest energy intermediate will have the smallest activation energy and consequently be the kinetically favored product (barring steric effects).

I've drawn just a few of these resonance structures in the following diagram.

In the top row I've drawn a few of the possible resonance structures for electrophilic attack at the para position on a terminal phenyl ring.  You can draw many more resonance structures delocalizing the charge around the various rings, but notice that it is possible to delocalize the positive charge over all 3 aromatic rings. In the second row I've drawn a few resonance structures showing charge delocalization when electrophilic attack occurs in the center ring.  Importantly, in this case charge can only be delocalized over 2 of the aromatic rings.

This suggests that based on resonance effects, electrophilic attack on a terminal benzene ring would be preferred over attack at the central benzene ring.

Further, we would expect ortho and para attack on the terminal rings to be preferred since the phenyl substituent acts as an activating o-p director in electrophilic aromatic substitution. The rate of attack at the ortho position might also be decreased somewhat due to the steric effect of the adjacent phenyl (actually biphenyl) substituent.

In a paper by Shafig and Taylor (J. Chem. Soc., Perkin Trans. 2, 1978, 0, 1263-1267. DOI: 10.1039/P29780001263), they explore your question by studying the electrophlic protonation of tritiated p-terphenyl.  After correcting the rates for statistical effects (e.g. there are twice as many ortho positions on the terminal rings as there are para positions) they find that:

the para position (your positions $10, 16$) reacts at a relative rate of $273$
the meta position (your positions $9, 11, 15, 17$) reacts at a relative rate of $1.54$
the ortho position (your positions $8, 12, 14, 18$) reacts at a relative rate of $176$, and finally,
the $4$ identical positions on the middle ring (your positions $2, 3, 5, 6$) react with a relative rate of $59.1$

These results support our predictions that

attack on the terminal rings will be preferred over attack on the central ring and
that attack at the ortho and para positions is preferred over attack at the meta position on the terminal rings.


Relationship between Gibbs free energy, enthalpy, entropy and rate of reactions

I'm having a little trouble constructing the bigger picture of chemical reactions.

Let's go step by step.

  From the collision theory we know that molecules react if: 1) they collide 2) they collide bearing a sufficient amount of energy to surmount the activation energy and collide at a specific geometric orientation.

Yes, in order to react, molecules taking part in a (bimolecular) reaction need to:

be in contact;
be correctly aligned to each other (a random alignment may not react);
have enough energy to overcome the activation barrier ($\Delta G^\neq$).

  From the relationship between Gibbs free energy, enthalpy, entropy and temperature we know: $\Delta G = \Delta H - T\,\Delta S$,
  and that a reaction is only thermodynamically favored when $\Delta G < 0$. And even when $\Delta G < 0$ the reaction may still not occur at a measurable rate because it has too high an activation energy.

What you're describing is the Gibbs reaction energy: the Gibbs energy between products and reactants.
This indeed tells us how thermodynamically favourably the reaction is ($\Delta G < 0$).

But the phrase in bold above is wrong.
The reason is that the Gibbs activation energy ($\Delta G^\neq$) is the Gibbs energy difference between transition state and reactants.

As Mithoron mentioned, Gibbs reaction energy would give you the equilibrium constant for the reaction:

$$K = e^{-\tfrac{\Delta G}{R T}}$$

  From the Arrhenius equation we know: $k = A ~\mathrm e^{-E_\mathrm a / (RT)}$

Generally, from Eyring theory,

$$k = \frac{k_B T}{h c_0} e^{-\tfrac{\Delta G^\neq}{R T}}$$

Comparing to Arrhenius equation, see:

$$k = \frac{k_B T}{h c_0} e^{-\tfrac{\Delta H^\neq - T \Delta S^\neq}{R T}}
= \frac{k_B T}{h c_0} e^{\tfrac{\Delta S^\neq}{R}} e^{-\tfrac{\Delta H^\neq}{R T}}$$

Thus,

$$A = \frac{k_B T}{h c_0} e^{\tfrac{\Delta S^\neq}{R}} \\
E_a = \Delta H^\neq$$

That's what you would find from an Arrhenius's plot.

  Linking all these together:
  1. Reactions only occur when particles can collide with each other.
  2. Colliding particles will only react if they bear sufficient energy (can be increased through temperature and/or adding other external energy sources) and have correct orientation (chances can't be increased?).
  3. Only particles in thermodynamically favored reactions have enough energy to overcome $E_\mathrm a$ and so will spontaneously occur without external energy applied. But they may occur extremely slowly if they have a very high $E_\mathrm a$ (according to the Arrhenius equation).

  Is this reasoning correct?

The reasoning is correct, except for point three above (in bold), for which the same error is found as before: reactions may have a big $K$ (small $\Delta G$) but small $k$ (big $\Delta G^\neq$) or vice-versa.


Reaction of solid metals with a solution of metal ions

You can explain these experiments with the Standard Electrode Potential. With these experimental values you know which element/compound releases more energy upon receiving a comparable amount of electrons. As a rule of thumb, alkalimetals and alkaline earth metals release less energy than, for example, elements from the halogen-group.

With this table at hand, you can predict that $\ce{Mg}$ will be oxidized and the $\ce{Cu^2+}$ reduced, because of the potential for the two following reactions:
$$\ce{Mg^2+ + 2e^- -> Mg(s)} \quad E^\circ=−2.372\,\text{V}$$
$$\ce{Cu^2+ + 2e^- -> Cu(s)} \quad E^\circ=0.337\,\text{V}$$

Reducing copper-ions will release more energy, so the following chemical reaction will apply:
$$\ce{Mg +Cu^2+-> Cu + Mg^2+}$$

In an analogous fashion, for the second question:
$$\ce{Ag^2+ + 2e^- -> Ag(s)} \quad E^\circ=+1.98\,\text{V}$$
$$\ce{Ca^2+ + 2e^- -> Ca(s)} \quad E^\circ=−2.868\,\text{V}$$

Reducing $\ce{Ag^2+}$ will release more energy - since silver is already in its solid state, the following reaction will not take place:

$$\ce{Ag +Ca^2+-> Ag^2+ + Ca}$$

Note: To see how those potentials or potential difference relate to energy see, for example, the Nernst Equation / Gibbs free energy


Confusion in solubility expressions in Henry's Law

Sander (Ref. 1) has compiled a useful review of Henry's law constants in water that includes an introduction showing notation and conversions.
There are two types of Henry's law constants:

Solubility constants convert from pressure to concentration in solution (solubility):

$$c=Hp$$

Volatility constants convert from concentration in solution (solubility) to pressure:

$$p=Kc$$
Usually it is possible to determine which constant is reported by inspecting the units.
References

R. Sander. Compilation of Henry’s law constants (version 4.0) for water as solvent. Atmos. Chem. Phys., 15, 4399–4981, 2015.
www.atmos-chem-phys.net/15/4399/2015/doi:10.5194/acp-15-4399-2015


It's actually not the hydroxymethelene group that determines the α/β status of a sugar, instead it's the relative orientation of the anomeric hydroxyl to the stereocenter which determines the absolute stereochemistry (D/L) of the sugar. This just so happens to match up with the same side/opposite side heuristic for the common six carbon pyranoses, but that's a quick rule of thumb, not the official definition:

  α (alpha), β (beta): Relative stereodescriptors used in carbohydrate nomenclature to
  describe the configuration at the anomeric carbon by relating it to
  the anomeric reference atom. For simple cases the anomeric reference
  atom is the same as the configurational reference atom. Thus in
  α-d-glucopyranose the reference atom is C-5 and the OH at C-1 is on
  the same side as the OH at C-5 in the Fischer projection.

IUPAC Gold Book

  The configuration at the anomeric centre (that derived from the
  carbonyl carbon) is denoted alpha- (α-) or beta- (β-) by reference to
  the stereocentre that determines the absolute configuration. In a
  Fischer projection, if the substituent off the anomeric centre is on
  the same side as the oxygen of the configurational (D- or L-) carbon,
  then it is the α--anomer. If it is directed in the opposite direction
  it is the β-anomer.

CAZypedia - includes diagrams and examples, including for five carbon pyranoses.


In the cold some paraformaldehyde (formaldehyde polymer) or trioxymethane (a cyclic condensation product of three formaldehydes) may precipitate.
It will depend on how much methanol has been added to avoid the polymerization products.
Normally it will contain 10 to 15% methanol.

In reality, it is not necessary to store formaldehyde solution in the cold.


Which will be oxidised between water and Copper?

The question is not fully complete (not your mistake but someone who wrote this exercise). I am not surprised that you are confused. The question is indeed silent about the electrodes. You should have asked the question: What is the cathode made of? What is the anode made of before starting the question. The second trick in your question in your question is, which I liked, is that it is an example of exhaustive electrolyte. You passed a huge current which consumed all free metal ions and the solvent can also be reduced or oxidized.

So let us look at your problem:

At the cathode you have two explicit choices: (i) Reduction of Cu ions which are floating in the solution, (ii) Reduction of the solvent, which is water in your case to hydrogen! There is nothing else which can be reduced here as per your question.

At the anode you still have two implicit (hidden) choices: (i) Oxidation of water and (ii) Oxidation of the anode material.

Your solved questions bluntly assumes that the electrode is inert, say it is made of carbon or Pt or gold, whatever but inert. So are left with only one choice. Oxidation of water. However, let us say they actually told you that the anode is made of Copper:

Oxidation potential of Cu

Cu(s) ---------> Cu$^{2+}$(aq) + 2e$^{-1}$    -0.34 V

Oxidation potential of water

2 H$_2$O(l)---------> O$_2$(g) +  4 H$^+$(aq) + 4e$^{-1}$    -1.23 V

Which one do you think will get oxidized? The Cu electrode instead of water!

This is the process and principle behind copper plating.

When your examiners are quiet about electrode materials, assume the solvent is going to oxidize, i.e., water.


A septum sealed chloroform bottle is likely to have been packed under nitrogen to increase shelf life. Unless you are doing this under a dry nitrogen atmosphere, I'm not sure that decanting into smaller containers is likely to improve shelf life.

Chloroform needs to be stored in a cool, dry, well-ventilated place, out of direct light. Your choice of storage container is likely to be influenced by how long you intend to store it for, but an amber glass bottle with a ptfe-lined cap is probably best suited for longer storage. Rigid polyethylene and polypropylene do not have the best chemical resistance for long-term storage, but if they are air-tight containers will be probably be ok only for short term storage.

Be sure to read and understand the MSDS for chloroform fully before doing this. You will find storage recommendations and incompatibilities listed. DO NOT use containers or lids made from or containing aluminium, magnesium and other metals.


Why are these ions classified as isoelectronic?

Core Electrons don't count. It's the valence electrons that are counted (outermost shell). The valence shell of all 3 is empty. So they all have 0 electrons in their valence shell.


Carbonate + acid reactions - How to recognize these type of reactions?

The reaction you showed is actually a salt "exchange" reaction (also called salt metathesis), in which a salt is transformed into another one during the course of the reaction. In your example when calcium carbonate is treated with hydrochloric acid, calcium chloride is formed. At the same time, carbonic acid is generated and it spontaneously decomposes into carbon dioxide and water. The equilibrium is shifted towards the products due to the release of carbon dioxide gas from the reaction mixture, thus driving the reaction to completion.


Differences between Liquid and Solid Infrared Spectrometry

Some basic comments can be found in Spectrometric Identification of Organic Compounds. Although it is primarily concerned with interpretation, it has some notes on the types of instrumentation and samples that would typically be encountered in an (in)organic chemistry lab. I'll only reproduce parts that mention the quality of the obtained spectrum.

  Vapor-phase spectra resemble those obtained at high dilution in a nonpolar solvent: Concentration-dependent peaks are shifted to higher frequency compared with those obtained from concentrated solutions, thin films, or the solid state (see Aldrich, 1989).

  Thick samples of neat liquids usually absorb too strongly to produce a satisfactory spectrum. (...) For example, thick samples of carbon tetrachloride absorb strongly near 800 cm-1; compensation for this band (using a reference beam with pure solvent) is ineffective since strong absorption prevents any radiation from reaching the detector.

  Solids are usually examined as a mull, as a pressed disk, or as a deposited glassy film. The suspended particles in a mull must be less than 2 microns (in diameter) to avoid excessive scattering of radiation. The quality of the spectrum (when using a pellet) depends on the intimacy of mixing and the reduction of the suspended particles to 2 microns or less.

  In general, a dilute solution in a nonpolar solvent furnishes the best (i.e., least distorted) spectrum. Nonpolar compounds give essentially the same spectra in the condensed phase (i.e., neat liquid, a mull, a KBr disk, or a thin film) as they give in nonpolar solvents. Polar compounds, however, often show hydrogen-bonding effects in the condensed phase. Unfortunately, polar compounds are frequently insoluble in nonpolar solvents, and the spectrum must be obtained either in a condensed phase or in a polar solvent; the latter introduces the possibility of solute–solvent hydrogen bonding.

One of the reasons you don't want very concentrated samples (for structural determination) is that you increase the amount of intermolecular interaction present in the sample, which can change peak positions and widths. A prototypical example is water:

For the vapor (gas phase), the water molecules are almost non-interacting, leading to well-defined and narrow peaks. The structure here may still be from some small number of environments where a molecule is interacting with a handful of others, but it may also be rotational fine structure (see rovibrational spectroscopy). In the (condensed) liquid phase, these individually-identifiable peaks coalesce into one broad peak due to the many hydrogen bonding environments induced by collsion with other molecules. The ~400 cm-1 redshift of the peak center originates from the nature of a single hydrogen bond:

where the molecule on the right (the h-bond acceptor) donates electron density into the $\ce{O-H}$ bond of the molecule on the left (the h-bond donor), specifically an antibonding orbital, which weakens the bond. Weaker bonds are lower in energy, and wavenumbers (cm-1) are units of energy. Finally, for the solid phase, most common polymorphs of ice have regular structure with a finite number of hydrogen bonding environments, which leads to slightly more distinct peak structure in the solid over the liquid. These peaks are further redshifted due to the decreased $\ce{O-O}$ distance in ice over liquid water (2.76 versus 2.82 Å), leading to even greater charge transfer between water molecules. Broadening and shifting can be also seen for glycine peptides of different lengths prepared as both a solid mull and in continuous films due to the terminal $\ce{-NH3 / -NH2+}$ and backbone $\ce{-NH}$ groups.

I don't have any references for it right now, but you can also obtain peak broadening due to solute–solvent chemical exchange; I have only heard of this for hydrogen.

To summarize,

increasing the number of distinct chemical environments for a particular vibrational oscillator is going to broaden the peak associated with that oscillator, and
intermolecular interactions that directly alter the electron density of the vibrational oscillator are going to shift its peak position; whether it will redshift or blueshift roughly corresponds to a decrease or increase in bond order.

References

Silverstein, Robert M.; Webster, Francis X.; Kiemle, David J. Infrared Spectrometry. Spectrometric Identification of Organic Compounds, 7th ed.; John Wiley & Sons, Inc.: New Jersey, 2005; pp. 79-80.
http://www1.lsbu.ac.uk/water/water_hydrogen_bonding.html
Blout, Elkan R.; Linsley, Seymour G. Infrared Spectra and the Structure of Glycine and Leucine Peptides. J. Am. Chem. Soc. 1952, 74, 1946-1951.


Theoretical Calculation of Lattice Parameter from Space Group

No - the spacegroup only gives information about symmetry, not the lattice parameters.

The lattice parameters depend on what is in your unit cell, ie your compound - but even if you know your compound, this is not straightforward, as you don't know its arrangement. And if you do, you should already know your lattice parameters ;)

That being said: if you have additional information - e.g. the radius of a protein you crystallise - you might be able to get a rough estimate of what you are expecting as your minimal unit cell parameters. But this does not take into account any stoichiometry in the crystal, which is impossible to know in advance, so multiples thereof are just as likely.


Are all PP(Polypropylene), i.e. resin code 5 plastics, safe for microwave and steaming?

No.

Just because the polymer is of one or the other kind does not let you say what industrial process actually synthesized it. Subsequently it is impossible to know what trace chemicals are left in the polymer packaging.

If if isn't labelled as microwave-safe, it isn't!

They test the package for it's microwave compatibility, not one type of polymer each! It seems that resin code 5 polymers are often used for microwave-proof packaging, but again: You don't know how it was made, you have no idea what's in there, don't microwave it (together with food or something else that you then ingest).


Why does steric hindrance cause a d8 complex to have a tetrahedral geometry rather than a square planar geometry?

For a given number of ligands surrounding a central atom there is always an arrangement in which the distances between the ligands is maximised and which typically also results in the maximum possible angles between then. You can deduce them logically if you start at a single ligand (which has no steric clashes whatsoever).

A second ligand should approach the same centre from exactly the opposite side leading to a linear arrangement and a bond angle of $180^\circ$.

A third ligand should approach the same centre perpendicular to the existing bond axis. Upon its approach, the two ligands already present will ease away from the approaching one. The final arrangement is trigonal planar with bond angles of $120^\circ$.

A fourth ligand should approach the same centre in a perpendicular manner again; this time relative to the plane of the three existing ligands. Again, they will ease away to make more room for the incoming. The final arrangement is a tetrahedron with bond angles of $109.5^\circ$.

Going from a tetrahedron to a square-planar arrangement means squashing the ligands together i.e. reducing the space between them. Thus, if there are no electronic arguments favouring a different arrangement any tetracoordinated entity should assume a tetrahedral environment to minimise steric strain.


Why exactly does molten NaCl explode, when it is poured into water?

Understand that sodium chloride is not made up of sodium metal and chlorine gas but of sodium ions and chloride ions, held together by ionic interactions. Under these thermal conditions (liquefication), the compound will not decompose into its elements and therefore all reactions you suggested which include elemental chlorine or sodium cannot occur.

Dissolution of sodium chloride in water is neither strongly exothermic nor strongly endothermic, so any effects stemming from the dissolution are neglegible.

Also note that you neglected an important variable in your calculations. As far as I can tell, you only calculate how to arrive at melting-point hot solid sodium chloride. To liquefy, additional melting enthalpy has to be applied, a further reservoir of energy to draw from. This melting enthalpy is of course released upon rapid crystallisation, but you should really subsume it into the heat energy difference altogether.

Finally, thermal decomposition of water is not exactly a process that will happen quickly, as exemplified by (water) steam temperatures that can be huge. Check out this unrelated answer of mine about how much energy is required to heat water to $100~\mathrm{^\circ C}$ and boil it.

So all things considered: The principle contribution to the explosion is the rapid heating of water including boiling, leading to tightly compressed gaseous $\ce{H2O}$ which expands explosively. This concurs with the recrystallisation of $\ce{NaCl}$ which accounts for a non-neglegible portion of the energy added, but which is technically already present in molten $\ce{NaCl}$. All other processes are minor or impossible.


Boiling and Melting point of Lithium (or any other element that does not occure in free state)

Your premise here is wrong, pure metal Lithium exists and can be handled without much difficulty.

However, materials which are extremely reactive can always be kept in an inert gas when you want to measure their physical properties. Of course, there are compounds which can don't need an excuse to explode (and  will explode in a vacuum/inert gas as well), in which case it is harder to find out BP/MPs (one can put them at low pressure and then extrapolate, but that's not as accurate).


How can I estimate how much humidity is present in a NaOH granulate?

For a rough estimate, you could do the following:

Weight a sample of it and heat it above 100°C for a while (say, one hour?).
Weight the sample afterwards.

The difference should be the amount of water given away and is an estimate (from below) of how much water there was in the sample.

But be aware that you may have some sodium carbonate depending on storage conditions.


Ordinary differential equation for molecule collision

There are two types of molecular simulation (in a broad sense), namely
1) Monte Carlo simulation
2) Molecular dynamics simulation
And they can be mixed together to give rise to new algorithm. According to your problem statement I am assuming you are trying to use molecular dynamics simulation. There are lots of things you should consider for your simulation system. According to your problem statement your atoms have position and velocity, but the problem is when they will collide each other they will loose energy and will come to a halt after few time steps. Either you can use thermostat to keep the temperature constant (Anderson thermostat and Nose-Hoover thermostat), or you can use constant pressure molecular dynamics simulation. For a beginner I would like Anderson thermostat as this is very simple technique. Then you have to use Newton's second law to determine the force and velocity on each atoms. After that you can use numerical integration technique to find new position based on F=ma formulation. Now you will repeat this task to get the trajectory of individual atoms.

If you want to find the diffusion coefficient of atoms, you can use lattice gas models or Ising models based on random Monte Carlo movement. I wrote a python code for finding diffusion coefficient. Also you can use Markov Chain Monte Carlo (MCMC) method to determice the stochastic evolution of the system.

For your better understanding I would suggest to use Understanding Molecular Simulation by Daan Frenkel & Berend Smit


Does humidity increase oxygen levels in the air we breathe?

As to the answer in the title, no, increased humidity does not increase oxygen levels in the air.  In fact it does the exact opposite, since the air in each breath will have some fraction displaced by water molecules instead of the normal nitrogen and oxygen mix we breathe.  It's a relatively small effect, with water making up at most a few percent of the air under high humidity/high temperature conditions that are safe for people to inhabit for reasonable periods of time.  That would mean that oxygen content would be decreased by about 1% in your steam room (again, this assumes you're not doing something dangerous with the temperatures and humidity.)

As to your other questions, you've got serious methodological issues to deal with.  Your first goal in finding the reason for the significant differences you've noticed should be verifying whether those significant differences actually exist, or whether they are based on things like confirmation bias or the simple effects of being in a boiling hot room vs outdoors in Colorado.  How you do that is really more of a psychology experiment, as has been suggested by the comments above, and would be better suited for the Cognitive Sciences Stack Exchange.


Can I predict the preferred resonance structure of carbon monoxide from its molecular orbital scheme?

On the favoured Lewis structures

Well yes, you have correctly drawn a minus sign next to the oxygen in the rightmost structure and similarly a plus sign next to the oxygen in the leftmost structure. However, these are not true charges! These are formal charges. They can be representative of real charges (cf. $\ce{SO4^2-}$ but they do not have to be. They derive from the way Lewis structures are drawn and not from any physical significance.

Choosing a good Lewis structure requires balancing a lot of effects. Main group atoms should get electron octets (hydrogen dublets) is one of the main tipping points to take care of. In the leftmost structure, all atoms have electron octets. The central structure has an octet (on oxygen) and a sextet (on carbon) — less favoured. The rightmost structure only gives cabon a quartet, plus a positive charge. This combination is so strongly disfavoured that not even the negative charge on oxygen can save anything.

Minimising the total formal charges is a second balancing effect which would prefer the central structure over the other two; however, it is less strong than the octet idea. And finally, structures that have formal charges distributed according to electronegativity are slightly favoured over those that don’t, but as I said, the former rules are more important.

Remember that Lewis structures are just a representation; and not a particularly good one at that! You can sort-of derive things like bond orders or free electrons from them, but for most uses, they are inferior to the molecular orbital model.

On the MO diagram

  Because the HOMO of CO is closer to the atomic orbital of O

Stop right there. That’s not how MO diagrams work.

The general gist of these diagrams is that orbitals with a lower energy will be closer to the electronegative atom(s) while those with higher energy will have stronger contributions by the electropositive partner. This is the HOMO we are talking about in a diatomic molecule. It has to have a larger contribution by carbon.

Reasoning: Carbons atomic orbitals are higher per se. The higher the energy of a hybrid orbital, the higher the contribution of higher energy atomic orbitals must be. Remember that you need to distribute one entire atomic orbital across all the molecular orbitals that have the same (correct) symmetry. The contributions from each atom (squared) need to add up to one. It’s a lot easier for the atom to contribute to orbitals closer to the original energy.

Of course, the negative pole of the dipolar molecule is closer to oxygen. Why? All populated low-energy molecular orbitals have a stronger contribution by oxygen while the high-energy unpopulated ones have a stronger carbon contribution.

A side note, that was mentioned in the question’s comments: You have 5 $\sigma$ orbitals and 2 $\pi$ orbitals populated in the MO drawing. 1$\sigma$ and 2$\sigma$ can be thought of the 1s orbitals of carbon and oxygen, leaving five orbitals for bonding. One of them happens to have an antibonding structure (nodal plane through the bond axis) so you get $4 - 1 = 3$ a triple bond between carbon and oxygen. (This only works this easily because there are only two atoms.) Please also check out the pictures in the other thread

I’m not sure if I touched all the points you wanted answered concerning the molecular orbitals in here. If not, comment and I’ll be back to try to explain what I missed. There’s a lot more that you can read out of it if you want to give it a try.


Which of the following show hyperconjugation

I guess you've only considered one resonance form of compound C. If you consider the other major resonances, there are alpha C-H bonds that can participate in hyperconjugative effects:


Photochemical reduction of benzophenone: why inverted flask?

I think the reason for inverting the the round flask is to capture sunlight very efficiently. A round bottomed flask filled with a fluid is like a ball lens. Sun rays can be considered parallel and see how they focus. If light is coming from all directions, every part will be illuminated well as the Sun moves. See ball lenses here ball lenses

The clue comes from the instruction in the procedure given in your link "Then the flask is exposed to the best available sunlight. "


why bond energy of hydrogen molecule is twice the chlorine molecule?

Because there are shielding electrons, meaning that the atoms can't come as close to each other. The outermost electron is the one being shared to fill a p orbital in chlorine where as it fills the s orbital in the hydrogen


There are several interesting aspects to your question.

  How are neutral amines effective bases in organic chemistry?

First of all, recall the definition of a base. Anything which accepts a proton (or donates a pair of electrons) is a base. Organic bases are very strong and corrosive! These bases should not be undermined, even though they may sound weak (and their stench is worse than inorganic bases). Your lye-free oven cleaner has an organic base, an neutral organic amine such as ethanolamine, which dissolves dirty grease stuck in the oven. This molecule has no charge yet it is a strong base, acting just like sodium hydroxide based oven cleaner in terms of performance.

  Thus, hydroxides and alkoxides are superior alternatives to these
  amines.

The second reason to avoid hydroxides and alkoxides is their solubility in organic solvents. Neutal organic amines easily dissolve in toluene and a variety of organic solvents. NaOH or NaH remain as a suspension.

  MOMCl  uses large bulky amines to deprotonate the alcohols, rather
  than using hydroxides

Again you have to think about the reaction solvent. Is an inorganic hydroxide soluble in your solvent? Then comes an important point. Why use bulky amines? Their sole purpose is to act as a proton sponge in the reaction mixture and prevent the formation of free acid. Bulky amines are preferred so that they don't react easily with the alkyl chlorides or acyl chlorides, thus prevent the formation of their own undesired side products.


Your first set of compounds are identical.
In the Fischer projection of a chiral centre, you can rotate three of the connected groups in a clockwise/anti-clockwise direction without changing the configuration of the chiral centre.
Upon performing this operation (keeping the top fixed, rotating bottom three clockwise) on the first compound in the first set, we see that the methyl group now occupies the position of the $\ce{-CH3}$ group in the second compound and the $\ce{-CH3}$ in the first occupies the position of the methyl group of the second.
Now, $\ce{-CH3} \equiv -\text{Me}$. Therefore, the first two compounds are now identical. As yukelid stated, identical compounds are not diastereomers.
However, repeating the same for the second case does yield two different compounds that are non-mirror images yet non-superimposable. Hence they are diastereomers.
Enantiomers and diastereomers are only seen for chiral compounds since achiral compounds would have symmetry that makes them identical. So, chirality is needed for enantiomers and diastereomers.


What is the exothermic reaction of superglue with cardboard?

I believe this is due to the fact that cyanoacrylates polymerize by anionic (or sufficiently strong neutral nucleophile) addition. In particular, they react with hydroxyl ($\ce{-OH}$) groups; paper, being primarily cellulose, is abundant in hydroxyls. The bond formation involved in the polymerization reaction is inherently exothermic (generating a $\ce{C-O}$ bond on initiation and many subsequent $\ce{C-C}$ bonds as the polymerization proceeds), and with a large surface area and many reactive groups, I could certainly imagine the reaction being surprisingly violent. The Wikipedia page on cyanoacrylates notes its reactivity with cotton fibers (again, being chiefly composed of cellulose), and the MSDSs for various similar compounds warn of exothermic reaction with paper and similar materials (q.v., 1, 2, 3).


I have chanced upon the answer in an inorganic chemistry textbook (Essential Trends in Inorganic Chemistry by D. M. P. Mingos) I was reading. It turns out that this trend only holds for the boron trihalides and does not apply to the aluminium trihalides. However, the book did not provide the explanation for the exception of boron.

My thinking is that perhaps, backbonding is not present in the aluminium trihalides, possibly due to incompatibility of the p atomic orbitals for pi bonding between aluminium and the halogen atoms. Does anyone else have any insights to share?


$
\begin{align}
(n_{\ce{NaOH}})_i= \pu{0.2 mmol}\\
(n_{\ce{HCl}})_i= \pu{0.1 mmol}\\
(n_{\ce{NaH2PO4}})_i= \pu{0.1 mmol}\\
(n_{\ce{Na3PO4}})_i= \pu{0.05 mmol}
\end{align}
$
At first the $\ce{NaOH}$ will react with $\ce{HCl}$ as per the following reaction:
$$
\begin{align}
\begin{array}{ccccc}
\ce{NaOH} & + & \ce{HCl} & \ce{->} & \ce{NaCl} & + & \ce{H2O} \\
\pu{0.2 mmol} & & \pu{0.1 mmol} & & \pu{0 mmol} & & \\
\pu{0.1 mmol} & & \pu{0 mmol} & & \pu{0.1 mmol} & & \\
\end{array}
\end{align}
$$
Now, the remaining $\ce{NaOH}$ will react with $\ce{NaH2PO4}$ as per following:
$$
\begin{align}
\begin{array}{ccccc}
\ce{NaOH} & + & \ce{NaH2PO4} & \ce{->} & \ce{Na2HPO4} & + & \ce{H2O} \\
\pu{0.1 mmol} & & \pu{0.1 mmol} & & \pu{0 mmol} & & \\
\pu{0 mmol} & & \pu{0 mmol} & & \pu{0.1 mmol} & & \\
\end{array}
\end{align}
$$
Now the resulting solution is  a buffer of $\ce{HPO4^2-}$ and $\ce{PO4^3-}$. So the Henderson equation can be used.
$
[\ce{PO4^3-}]=\frac{0.05}{45} \pu{M}\\
[\ce{HPO4^2-}]=\frac{0.1}{45} \pu{M}\\
$
$$
\begin{align}
\begin{array}{ccc}
\mathrm{pH} & = & \mathrm{pK_{a_3}} + \log(\frac{[\ce{PO4^3-}]}{[\ce{HPO4^2-}]}) \\
\mathrm{pH} & = & 12.35 + \log(\frac{0.05}{0.1})\\
\mathrm{pH} & = & 12.35 + \log(\frac{1}{2})\\
\mathrm{pH} & = & 12.04 \\
\end{array}
\end{align}
$$


It is strange that you are concerned about diethyl ether (which has been used as an anesthetic), but you are not concerned about THF. From the US Pharmacopoeia:

  Permitted Daily Exposure
  Tetrahydrofuran 7.2 mg/day
  Diethyl Ether 50 mg/day
  Ethyl Acetate 50 mg/day

If you dry the product thoroughly (above 120°C), you should not have any problems with residual solvents.


What lab equipment did Marie Curie use to isolate radium?

After searching Google Books, I found autobiographical notes from Marie Curie. Apparently, it was radium chloride that was isolated first, not elemental radium. Salts can often be precipitated from solution at much lower temperatures than those needed to melt many inorganic materials, and you typically wouldn't need to shield a solution from air, so ampoules wouldn't be needed. This would be the case for fractional crystallization, which she mentions in her autobiographical notes. (She also mentions spending whole days "mixing a boiling mass with a heavy iron rod". I'm not sure if she's referring to mixing molten material or a solution in water or acid.) She also says they eventually used a factory; as time passed and their work became "more and more known", she was able to do the "crude processes" in a factory, allowing herself "more time to the delicate finishing treatment".
source: the book Pierre Curie: with Autobiographical Notes by Marie Curie


What is the compound called when it is in solution?

I believe the term you are looking for is solute; a solute is a substance dissolved in another substance, known as a solvent.


In your first reaction, you added concentrated sulfuric acid to anhydrous potassium chloride. In your equation, you wrote $\ce{KCl (aq)}$, which is incorrect. Under dilute conditions, that reaction would not take place because all reagents and products would be in aqueous ionic state, meaning there is no reaction.
In your second reaction, the equation is completely wrong. It is true that elemental $\ce{Cl2}$ a stronger oxidizing agent than elemental $\ce{Br2}$ or elemental $\ce{I2}$. But in current situation is ionic $\ce{Cl-}$ versus ionic $\ce{I-}$ where both are aqueous. If all reagents are in ionic form and expected products are also ionic, there is no reaction. See your reaction below:
$$\ce{H+(aq) + Cl-(aq) +K+(aq) + I-(aq) <=> H+(aq) + I-(aq) +K+(aq) + Cl-(aq)}$$
There is no gas, liquid or solid formed, but all ions. Therefore, there is no reaction.


Why aren't resorcinol and catechol classified as dihydric alcohols?

As @Loong pointed out in his comment, resorcinol and catechol are phenols, not alcohols. Although both classes of compounds contain an O-H group, they are different in that the oxygen in phenols is bonded to an arene ring such as benzene or napthalene.
Importantly, a lone pair on the phenol oxygen is able to interact with the aromatic $\pi$ system via resonance, which gives phenols characteristically different properties to normal alcohols.

The lone pair delocalisation increases the electron density in the ring dramatically at the ortho and para positions, as indicated by the negative charge at these positions in the resonance structures. This makes these positions much more reactive towards electrophiles than they are in unsubstituted benzene.
The lone pair delocalisation also occurs in the phenoxide ion, the conjugate base of phenol, which makes it much more stable than a regular alkoxide ion, and therefore makes phenols much more acidic than alcohols.


is there a lead compound that crosses the blood-brain barrier when ingested?

Tetraethyl lead (bp 180 °C) and tetramethyl lead (bp 110 °C) are easily resorbed through the lungs and the intact (!) skin. They pass the blood-brain barrier without any problems. The half-life for the elimination from the brain is about 500 days.

(Source: Günter Fred Fuhrmann, Toxikologie Für Naturwissenschaftler: Einführung in Die Theoretische und Spezielle Toxikologie, Springer-Verlag, 2006, p 209)

These compounds are neurotoxic and the literature is full of reported cases of intoxications and - sometimes - successful treatment, such as here.


Why do these two calculations give me different answers for the same acid-base titration?

Part of the reason why there is such a large discrepancy is because in your second method of calculating $[\ce{HA}]$, you failed to take into account the fact that your vinegar was diluted from $3.1\mathrm{~cm^3}$ to $50\mathrm{~cm^3}$.

Effectively, if the original concentration of $\ce{HA}$ was $x$, then your dilution would have brought it to a concentration of $\frac{3.1}{50}x$. And you plugged the $\mathrm{pH}$ of this diluted solution into the the $K_\mathrm{a}$ equation. Let's adjust for that dilution:

$$\begin{align}
\frac{3.1}{50}x &= 0.041\mathrm{~mol~dm^{-3}} \\
x &= 0.661\mathrm{~mol~dm^{-3}}
\end{align}$$

As you can see, there is still some difference between the two values. However, you can calculate the theoretical $\mathrm{pH}$ of a $1.067\mathrm{~mol~dm^{-3}}$ solution of $\ce{CH3COOH}$, using the fact that the $\mathrm{p}K_\mathrm{a}$ is $4.6$. Here, I will make the simplifying assumptions that the final and initial concentrations of $\ce{HA}$ are the same, i.e. $[\ce{HA}] = c_{\ce{HA}}$, and also that $[\ce{H+}] = [\ce{A-}]$:

$$\begin{align}
\frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]} &= 10^{-4.6} \\
&= 2.511\times 10^{-5}\mathrm{~mol~dm^{-3}} \\
[\ce{H+}] &= \sqrt{(2.511\times 10^{-5})(1.067)}\mathrm{~mol~dm^{-3}} \\
&= 0.00518\mathrm{~mol~dm^{-3}} \\
\mathrm{pH} &= 2.28
\end{align}$$

You could go all the way and do mass-balance and charge-balance equations in order to avoid making that assumption, but I doubt it alters the result significantly. So, if you really had a concentration of $1.067\mathrm{~mol~dm^{-3}}$, your observed initial $\mathrm{pH}$ would not have been $3.0$. There is likely some experimental error present, or perhaps the $\mathrm{pH}$ meter that you used to measure the initial $\mathrm{pH}$ is not calibrated well.


Differences in Gibbs activation energies for rotation around a bond

In both molecules, there is π delocalization across the $\ce{C=C}$ bond which extends across the two carbonyl groups on the alkene. In order to maximize the overlap of the p orbitals involved in this conjugated system, it means that this portion of each molecule has to be flat. It is not a problem for the conjugated system to be flat in molecule B, but in molecule C you have three bulky methyl groups near the conjugated system. Steric clashes in molecule C between the t-butyl group and the carbonyl group on the same side of the $\ce{C=C}$ bond mean the carbonyl will rotate out of the plane of the conjugated system to avoid "bumping into" the t-butyl group, lowering the overlap between the p orbitals in the conjugated π system and thus destabilizing the molecule.

Because molecule C does not experience as much favorable pi electron delocalization as  molecule B, molecule C sits at a relatively higher energy than molecule B. The transition states of the two molecules, however, are not so different in energy. The Gibbs energy of activation is defined as the difference in Gibbs energy between the transition state of a reaction and the reactants (where our "reaction" is a rotation about the $\ce{C=C}$ bond). As molecule C sits at a higher energy relative to molecule B, but the transition states are relatively similar, it makes sense that there is a lower barrier to rotation about the $\ce{C=C}$ bond in molecule C.