Non polar molecular and attractive forces

There is more than one factor that matters in the forces between molecules. The tradeoff between these different factors is complex and can't be reduced to a single simple equation or factor. So some of the factors become more important because of molecular or atomic size counterbalancing the factor of distance between the entities.

The collective force that cause molecules and atoms to attract are often referred to as van der Waal's force (though some sources use a more narrow definition that doesn't include all of the components below). The key non-bonding forces involved are:

forces between permanent dipoles
forces between permanent dipoles and induced dipoles
forces between instantaneously induced dipoles caused by quantum fluctuations

In non-polar molecules it is the third of these that dominates. Crudely stated the force arises because of uneven distribution of the electron cloud in a molecule or atom that happens as a result of quantum fluctuations in the cloud. An uneven distribution of electron density is the equivalent of a dipole moment. These dipole moments create induced dipoles in neighbouring atoms and can interact with other quantum fluctuations in the neighbouring atoms. These create small electrostatic forces between the atoms or molecules.

The reason why larger atoms or molecules seem to have larger forces is two fold. Larger molecules, crudely, have larger electron clouds and more ways to interact (shape matters too and large flat molecules "stick" together more than irregular lumpy ones: compare the melting points of benzene and toluene).

In larger atoms the explanation is simpler. Larger electron clouds are less tightly held by there nuclei than those in smaller atoms. Crudely, larger, floppier clouds can have more scope for the quantum fluctuations than tighter-held smaller clouds (also they have a larger area and that maters to the total force). So the size effect increases the strength of the force rather than reducing it as might be expected from simpler arguments.

I'm aware that this is a slightly superficial explanation, but a better one would probably involve enough quantum mechanics to make the average reader's head explode. I think i've captured the key intuitions that matter.


How to separate and analyse a sample of cations of alkaline and earth alkaline cations and ammonium?

In a qualitative analysis, the fast way is always the wrong way. There are already so many issues with the "long" version that I am surprised that this is remotely successful.

The cations of groups 1-3 are always the most difficult to separate and during my time as a teaching assistant were by far the biggest source of error.

The proper course of analysis is

dissolve sample (completely)
precipitate ion
dissolve precipitate
specific reaction
repeat

The reason for separating the cations in the first place is, that they often act in similar ways and the reactions are not very specific. Certain cations should be tested multiple times, too.

All the reactions you are performing are equilibria and very dependent on the overall concentration. None of them form a precipitate quantitatively. Sometimes you might have problems recognising precipitates in the first place.

You need to perform double blind tests for all used chemicals, since already small contaminations may lead to false positives.

The main issue with your proposed strategy is that you will not precipitate aluminium with this strategy, since this is extremely pH dependent.
$$\ce{Al^3+ + 4OH- <=> Al(OH)3 + OH- <=> [Al(OH)4]- }$$
Without a buffered solution at around pH 9 and ageing the solution, there should not form a precipitate. I learned this by adding ammonium sulfide first and raising the pH slowly with ammonia. (This will co-precipitate cobalt, nickel, manganese, zinc, and iron [as sulfides], and chromium [as hydroxide]. This might be useful if you have to analyse for more ions later on.)
If a precipitate forms you can dissolve it with sodium hydroxide and use cobalt(II) nitrate to form Cobalt blue.
You can also test the presence of aluminium with morin as it forms a fluorescent complex with it. Note that $\ce{Mg(OH)2}$ is slightly fluorescent, too.

I cannot remember precipitating magnesium with that strategy in any sufficient amount. I would prefer precipitating it as an ammonium phosphate, which you can try to (re-)crystallise as struvite and look at them with a microscope.
$$\ce{Mg^2+ + NH4+ + PO4^3- <=> MgNH4PO4 v ->C[H2O] MgNH4PO4.6H2O}$$

If you want to precipitate out aluminium and magnesium (and almost every other heavier element) you can use oxine, $\ce{C9H7NO}$ (abbr. $\ce{Hox}$), solution.
$$\ce{ M^{$x+$} + $x$\,Hox <=> [M(ox)_{$x$}] + $x$\,H+ }$$

Then precipitating calcium is straight forward with carbonate solution. Other earth alkaline metals will precipitate, too. You can dissolve this precipitate with acids or you can use the solid to perform flame tests. Since there can only be calcium, strontium, or barium, this should be conclusive. (Of course you could precipitate barium sulfate.)

The main pitfall with your proposed strategy is, that you will not quantitatively precipitate calcium hydroxide and it is as mentioned above quite concentration dependent. The solubility constant is not too small, i.e. $K_\mathrm{sp}=5.5\times10^{-6}$.

If you have precipitated everything so far, then the last thing left over are ammonium, potassium, sodium(, and lithium). The biggest problem is, you cannot mask them properly and you cannot precipitate them. You can use flame tests to determine their presence, but as long as you have sodium in it you might not see anything else. You should use a prism, or a spectrometer to determine the characteristic spectral lines. Apparently phones can do it now, too. Also see publiclab.org... makes you think why separate at all.

You will need to analyse ammonium from the sample. You can add sodium hydroxide, which releases ammonia. You can detect the presence of ammonia in the sample by holding a wetted universal indicator paper over the sample. Additionally, you should be able to smell it.

In other words ...
... be patient with your analysis as there are more pitfalls, false negatives, false positives, contaminations. Don't use short cuts, except atomic emission spectroscopy.


Consider this structure of purine:

The ring on the left contains two nitrogen heteroatoms and is aromatic due to π electron conjugation. Having six π electrons, it obeys Hückel's rule.

For the ring on the right, it contains four π electrons in bonds and the lone pair on the NH group takes part in conjugation. Hence the total number of electrons taking part in conjugation is six, which makes it aromatic by Hückel's rule as well.

Since both the rings of purine are aromatic, the compound is aromatic.

For more information see: Ciesielski, A.; Krygowski, T. M.; Cyrański, M. K.; Balaban, A. T. Defining rules of aromaticity: a unified approach to the hückel, clar and randić concepts. Phys. Chem. Chem. Phys. 2011, 13 (9), 3737–3747. DOI: 10.1039/C0CP01446J.


Major product in reaction of butan-2-one + selenium dioxide

To understand the major factor we have to consider the mechanism of the reaction first. The reaction you mentioned is popularly know as Riley oxidation. The mechanism is as follows:

As you can observe the RDS of the reaction is the tautomerization of the keto form into an enol. Thus the extent of enolization determines the major product in this reaction.
As the reaction takes place in an neutral environment the acidity of the alpha-hydrogen plays the major role in deciding the major enol formed and so the corresponding oxidized product is major.
In short, the major product in Riley oxidation is the product associated with the major enol content which in the given conditions is determined by the acidity of respective alpha-hydrogens.


For what condition do the equations 1 and 2 apply? I think that 2 is for strong electrolytes at low concentration though im not sure. So what about at high concentrations of strong electrolytes? What about weak electrolytes? 2) When would you use equation 1, when would you use equation 2? Are they equivalent?

Equation 1 defines the molar conductivity:
$$[6]\;\;\mathrm{Λ_{m}} = \frac{\kappa}{c}$$
It is generally applicable. You use it to compute the molar conductivity from experimental data (conductivity and molar concentration of electrolyte). Here $\kappa$ is the conductivity. For an ideal measurement cell and electrolyte it is defined as
$$\kappa = \frac{l}{RA}$$
where l and A describe the geometry of the cell. A more general definition is possible for an arbitrary geometry or sample composition.
Equation 2 is called Kohlrausch's law and sure enough was derived by Kohlrausch based on the behaviour of strong electrolytes, to which it commonly applies at low concentrations:
$$[7]\;\;\mathrm{Λ_{m}} = \mathrm{Λ_{M}^\circ} - k\sqrt{c}$$
You might fit this equation to concentration versus molar conductivity data to determine limiting molar conductivities $\mathrm{Λ_{m}^\circ}$, and from that obtain the limiting molar conductivities of the component ions. The parameter k (also obtained from the fit) may be modeled by applying the Debye-Huckel-Onsager theory.
For weak electrolytes one might invoke Ostwald's dilution law.

In what situation should you use equation [8]?

Equation [8] is the law of independent migration of ions:
$$[8]\;\;\mathrm{Λ_{M}}^0 = \lambda{_{M+}}^0+\lambda{_{A-}}^0$$
where the $\lambda$ are limiting ionic conductivities, properties of individual ionic species in a specific solvent at infinite dilution. As written the equation applies to a solution containing a single 1:1 electrolyte. More generally $\Lambda_{M}^0 = \sum_i \nu_i\lambda_i^0$, where the $\nu$ are stoichiometric coefficient of individual ions in the electrolyte solution. The equation allows prediction of the limiting conductivity from the electrolyte composition provided limiting conductivities have been compiled. You can also use the equation to parse limiting conductivities of individual components from data for different simple electrolytes.

One of my questions is to calculate the concentration of impurities in tap water, using the conductivity measurement. How can I do that when I have more than one unknown in the equation?

Conductivity measurements can be used to estimate the concentration of total dissolved solids in tap water. Typically standard conversions factors are used to convert from conductivity to TDS, the conversion factor depending on the water source, as explained in the wikipedia:

The relationship of TDS and specific conductance of groundwater can be approximated by the following equation:

TDS = keEC

where TDS is expressed in mg/L and EC is the electrical conductivity in microsiemens per centimeter at 25 °C. The correlation factor ke varies between 0.55 and 0.8.2.

There are also online calculators (such as from lenntech) that convert between TDS and conductivity. However if you have performed a water composition analysis you can combine this with tabulated conductivities for different ions, as the lenntech website explains:

The EC determination can be done in different ways. One possibility is the usage of an ion specific conductivity coefficient. This coefficient is listed in charts though for their usage it is necessary to have an exact water analysis because every single ion affects the conductivity. An example for such a chart can be found in the „Handbook of Chemistry and physics” 76th edition, S. 5-90“.

See also ISO document ISO 7888:1985 and links from the EPA.


Conditions for Complete hydrolysis of salt

The question is, in fact, a good one, but the concept of weaker/stronger is confused - and confusing!

A way to describe the situation, using the terms in the question is this: strong acids and strong bases undergo complete hydrolysis in water. The energy gained from hydrogen bonding with a water molecule is great enough to effect complete ionization in water. Examples would be HCl and NaOH.

Water lies in the middle here, being able to be protonated by strong acids and deprotonated by strong bases. Hydrolysis is a measure of how much the water is protonated to H3O+ or deprotonated to OH-.

There are also materials which ionize in water, but not so extensively, e.g., acetic acid and amines such as ammonia. They dissolve, but do not react so much and thus hydrolyze to a limited extent. Confusing, perhaps, that we call these materials weak acids and bases but they have strong internal bonds which prevent their total ionization. These strong internal bonds mean that if you did form the the conjugate acid or the conjugate base (in a salt), they would react strongly with a water molecule to form a considerable amount of the corresponding acid or base. Thus the salts would ionize and hydrolyze where the original acid or base would not be said to hydrolyze (much).

The "strength" of the cation may perhaps be better described as a polarity, which, if it is a "strong", favors hydrolysis, i.e., complete ionization, meaning it is strongly attracted to the oxygens of water and less so to its own anion. By this terminology, Na+ is a strong acid because it is strongly attracted to water (oxygens) at the same time OH- would be a strong base.

A "weak" cation, like NH4+, would be one which will not protonate water to a great extent. A solution of 1 M NH4Cl will give complete ionization to NH4+ and Cl-, but we do not call that hydrolysis. The ammonium ion is stable (strong internal bonds) and only weakly protonates the H2O molecules around it, so we call it a weak base, with limited hydrolysis.


Do I need to include calcium chloride in this reaction equation?

From an instructor's point of view, you have accounted for the reaction that has taken place during titration.  The point you are making surrounds the conditions the titration took place under (as you pointed out, solubility is altered, for instance), but not the chemistry of the reaction, in and of itself, which you are focused on.

In essence, you are doing the first titration in aqueous calcium chloride as well, because as soon as the titration begins, what is formed?

That said, I feel I may be missing something more of the other potential rxn conditions or possible expectations of your evaluator.


How do I know with which oxidation numbers I can form a certain kind of compound?

You need to be familiar with some of the compounds earlier, otherwise you can't tell if it exists. Remember facts come first before explanation. It also depends usually on the atom/ion to which it is bonded. for example since Fluorine is most electronegative it can form $\ce{XeF6,XeF4}$,etc. but $\ce{XeH3}$ doesn't form also oxygen forms large oxidation states as it is electronegative too and there is also more energy released as many bonds are formed (ususlly pi-bonds.)


I originally wrote this as a comment, but I will post it as an answer as I'm fairly certain this is most of the answer.

In general, there are two distinct types of electron correlation which can be described. Exchange correlation, which results from the indistinguishability of electrons, and Coulomb correlation which correlates the motions of electrons by considering individual repulsions between electrons.

Hartree-Fock (HF) theory accounts for the exchange interaction, but does not have Coulomb correlation as HF is a mean-field theory. That is, each electron experiences the average potential of the other electrons.

So, there is a sort of imbalance here from the beginning. It is known that the presence of this exchange interaction, which only happens for electrons of the same spin, results in two phenomena called Fermi holes and Fermi heaps. The references in that Wikipedia page are quite good. Essentially, a Fermi heap results when you have electrons of opposite spin. Because they must have a spatially symmetric wavefunction, the ends up being a greater probability of finding both electrons at the same location. This sounds silly, but read the references in the wikipedia page and you'll see what I mean. A fermi hole results when you have two electrons of the same spin. These must have an antisymmetric spatial wavefunction, so there is no probability of finding them in the same location.

If we think about these two effects in terms of electrical repulsion, the Fermi hole is a lower energy configuration than the Fermi heap. Because we almost always have both spin-up and spin-down electrons in multi-electron systems, both of these phenomena will take place, and the overall electron density will be some kind of average of all of these. (This is only really a conceptual framework.) If we take a simple case of a homonuclear diatomic, then these heaps and holes will all average out and we will get a symmetric electron density. If, on the other hand, we take a heteronuclear diatomic, then we see that there will be a tendency for these Fermi heaps to form over the atom with the larger nuclear charge just based on Coulombic attraction.

From here, I believe it is fairly clear that Coulomb correlation will tend to make the effect of these Fermi heaps smaller, and hence the overall separation of charge will decrease rather than increase, which explains why HF tends to overestimate dipole moments. By the same reasoning, the electron density is already polarized due to this heaping effect, so the system is not likely to be as polarizable as if this electron density were allowed to relax as is the case when Coulomb correlation is included.

This is all a very conceptual way of thinking about this, but I don't know that the math describing HF will be all that much more useful.


Implications of orthonormal wavefunctions

If you have two orthonormal states $|\psi\rangle$ and $|\phi\rangle$,
$$
    \langle \psi|\phi\rangle = 0,
$$
there is no guarantee that the term
$$
    \langle\psi|\hat{A}|\phi\rangle
$$
vanishes. All depends on the action of $\hat{A}$ on your state.

As you already noticed, if $|\phi\rangle$ is an eigenstate of $\hat{A}$ with eigenvalue $a_\phi$, the term is obviously zero:
$$
    \langle\psi|\hat{A}|\phi\rangle = \langle\psi|a_\phi|\phi\rangle =a_\phi \langle\psi|\phi\rangle = 0.
$$
But this is only a special case, since $|\phi\rangle$ is not necessarily an eigenstate of $\hat{A}$ in general.

One of the simplest situations where this term does not vanish can occour  when $|\psi\rangle$ and $|\phi\rangle$ are eigenstates of the harmonic oscillator, that we can denote $|n\rangle$ and $|n'\rangle$. As you probably know, these states are orthonormal:
$$
    \langle n|n'\rangle = \delta_{n,n'}.
$$
If you consider the specific case $|\psi\rangle=|n\rangle$ and $|\phi\rangle=|n-1\rangle$, it immediately follows that
$$
    \langle n|n-1\rangle = 0.
$$
Now if you take $\hat{A}= \hat{a}^\dagger$, for which you have $\hat{a}^\dagger|n\rangle=\sqrt{n+1}|n+1\rangle$ ($|n\rangle$ is not an eigenstate of $\hat{a}^\dagger$), you have ($n\geq1$)
$$
    \langle n|\hat{a}^\dagger|n-1\rangle = \langle n|\sqrt{n}|n\rangle = \sqrt{n} \langle n|n\rangle = \sqrt{n}\delta_{n,n} = \sqrt{n} \neq 0.
$$

As I already stated, all depends on the action of $\hat{A}$ on your state.


How derive g and u symmetry labels for orbitals?

g and u labels are only used when the molecule itself possesses the inversion symmetry element. For a diatomic molecule, this means that both atoms have to be the same, i.e. it has to be a homonuclear diatomic molecule $(D_{\infty\mathrm{h}})$.

  How do I carry out the inversion?

$\hspace{25 mm}$

Let's consider the $\ce{O2}$ molecule.
Let's consider the $\pi$ bonding MO. (Or one of them, to be precise, since they occur in degenerate pairs.)
Find the point of inversion of the molecule. Clearly it has to be the midpoint of the O=O bond, i.e. exactly in the middle.
Draw a straight line from one lobe to the point of inversion.
Extend the straight line through the point of inversion. You will see that you have gone from a shaded lobe to an unshaded lobe, indicating that the phase has changed. Therefore, the $\pi$ bonding MO is ungerade, or u.

If you start from the bottom-left lobe, you would draw a line through the point of inversion and end up in the top-right lobe, and you would draw the same conclusion.

  And what when the two overlapping orbitals are not of the same type, e.g. a $\pi$ overlap between d and p orbitals? Will the MO ever have g symmetry?

The parity of such MOs cannot be classified as g or u. However, if we are talking about a homonuclear diatomic, then this wavefunction is not acceptable. You cannot have a wavefunction that has only d orbital character from one atom, but only p orbital character from the other. By symmetry, the contribution from both atom's d orbitals have to be equal.

This arises because the Hamiltonian is invariant to the permutation of identical particles. Essentially, the idea is that both atoms are exactly identical, so there cannot be any way for them to be distinguished from each other. If you had such a wavefunction, and if it was populated, then there would be different amounts of electron density on the two atoms, which would mean that the atoms could then be distinguished. This is quantum-mechanically forbidden.

Such MOs may arise in heteronuclear diatomics. However, heteronuclear diatomics $(C_{\infty\mathrm{v}})$ do not possess the inversion symmetry element, and accordingly their MOs cannot be classified as g or u. Since inversion does not preserve the molecule, there is no requirement that it should preserve the wavefunction.

  What if one atom does not have any d orbitals?

There is no such thing. Every atom has d orbitals.

Whether they have the same energy, or whether they are occupied, is another thing altogether. If the energies are different, or if one atom has occupied d orbitals and the other does not, then it is a heteronuclear diatomic and as I said earlier, the g and u labels are not appropriate in such a case.


Explaining differences in rate of propagation based on monomer structure

The argument presented by your instructor is based on one factor: the stability of the radical terminus of the polymer. But for bimolecular polymerization reactions we need to consider 4 things:

Stability of the radical
Stability of the monomer
Stability of the transition state
The probability of the radical and the monomers being correctly oriented

The first three items affect the activation energy ($E_\mathrm{a}$), while the final item affects the frequency factor ($A$), or pre-exponential factor in the Arrhenius equation:

$$ k = A \exp{\left(\frac{-E_\mathrm{a}}{RT}\right)} $$

For styrene, vinyl acetate (a vinyl ester), and ethylene, I have found the following values for radical polymerization processes.[1,2] These have been calculated to provide an apples-to-apples comparison:

$$\begin{array}{cc}
\hline
\text{Monomer} & E_\mathrm{a}~/~\mathrm{kJ~mol^{-1}} & A~/~\mathrm{L~mol^{-1}~s^{-1}} & T~/~^\circ\mathrm{C} & k_\mathrm{p}~/~\mathrm{L ~mol^{-1}~s^{-1}} \\ \hline
\text{Styrene} & 32.5 & 4.3 \times 10^7 & 60 & 341  \\
\text{Vinyl acetate} & 20.4 & 1.5 \times 10^7 & 60 & 9460 \\
\text{Ethylene} & 27.7\mathrm{-}32.8 & 2.0 \times 10^3 \mathrm{-} 3.0 \times 10^4 & 60 & 0.1\mathrm{-}0.2 \\ \hline
\end{array}$$

Note:

  "the propagation rate coefficient for most monomers is strongly pressure-dependent with large negative activation volumes; thus an increase in pressure leads to an increased $k_\mathrm{p}$ value."[1]

This is particularly relevant for the ethylene example, which is done at $100\ \mathrm{bar}$. At $60\ ^\circ\mathrm{C}$, this should be about $0.15\ \mathrm{g~mL^{-1}}$,[3] which is a factor of 7 less dense than the styrene and vinyl acetate systems.

With those caveats noted, the argument of the vinyl ester having a lower activation energy ($E_\mathrm{a}$) due to the reactants being less stable relative to the transition state seems valid. However, the data for ethylene shows that the pre-exponential factor ($A$) is likely responsible for the lesser rate of propagation displayed by ethylene.

Sources:

Krzysztof Matyjaszewski & Thomas P. Davis, Handbook of Radical Polymerization, 2003, p 199.
Etienne Grau, Jean-Pierre Broyer, Christophe Boisson, Roger Spitz, and Vincent Monteil, Macromolecules 2009, 42, 7279–7281. DOI: 10.1021/ma901622u
Thomas, W. & Zander, M. Int J Thermophys (1980) 1: 383. DOI: 10.1007/BF00516565


Why are pressure or volume not able to change the equilibrium constant?

Just like any sweeping statements made by the General Chemistry textbooks, this statement is also not completely true. When you work at pressure extremes, as in modern day chromatography, such as 1000 or higher bar, large molecules can change their shape, in that case, one can see a change in retention factors which sort of indicates a change in equilibrium constant with pressure under constant volume.

Just for fun, some solvents can become solids at high pressures-but this happens at another high-pressure level.

Coming back to routine reactions, pressure will not affect equilibrium constants because whenever you try to adjust the pressure in a gaseous reaction, the equilibrium concentrations will change in such a way that their ratio remains constant-hence the equilibrium constant does not change.

One way to think about it as that the equilibrium constant is a ratio of forward and backward rate constant. Changes in pressure or volume will not change the kinetic energy of the molecules but only temperature can change the kinetic energy and hence affect the rate constants.

  According to the Ideal Gas Law, temperature is dependent on the
  pressure and volume.

Mathematically, this is incorrect. You would rather say that pressure and volume of a gas are functions of temperature i.e., the converse of what you wrote.


Off top of my head, I would think this is due intramolecular hydrogen bonding (a 5 membered ring), giving rise to a sharp peak at $\pu{3499.14 cm^-1}$, and intermolecular bonding, giving a slightly broader peak at $\pu{3394.45 cm^-1}$. But I don't have any references, this is just a hypothesis.


How the Nernst equation is applied for electrodeposition?

For a metal electrode immersed in a solution of its own salt, the concentration of the reduced form as a solid phase remains constant ($[\ce{M^0}] = \mathrm{const}$), therefore:

$$E = E_0 + \frac{RT}{nF}\ln[\ce{Ox}]$$


Is there a way to create natural Red food coloring from Turmeric in the house kitchen?

Curcumin(the active ingredient in Turmeric) can turn red in two ways: Either by acting as an acid-base indicator or by forming a complex with a phospholipid film, as you stated.

Colour change due to Acid-Base action

The curcumin molecule is a tribasic acid in basic solution, where it exists as its enol tautomer. The loss of one or more protons from the phenolic hydroxyl groups convert the benzene rings into their quinonoid form, which have a larger absorption/emission wavelength than the benzenoid form, giving a red colour.

This indeed works experimentally: a paper highlighting this can be found here. A simple home experiment would also suffice

The left one is a suspension of turmeric in water; the right one is the same suspension, but with a few drops of soap added to it, which make the medium alkaline and cause the turmeric to turn red. Unfortunately, around my kitchen, I could not find an edible compound basic enough to cause turmeric to change colour. Lime is a possible compound that I could test with but I don't have any foodgrade lime powder lying around. Another issue with using this colouring is that any acidic medium will revert the colour back to yellow: this means that sprinkling lemon as garnishing on your dish would change the colour from an appetizing red to a dull yellow.

Colour change due to Phospholipid complex

A lot of papers have been released on this topic, such as this one and this one. The second paper mentions the following synthesis method to prepare curcumin liposomes (curcumin surrounded by a spherical membrane of phospholipid)

  The MFGM phospholipids/soybean lecithin was dispersed in chloroform. Curcumin was dissolved in anhydrous ethanol and thoroughly mixed with the MFGM phospholipids/soybean lecithin dispersion in a 250-mL round-bottomed flask. The mixture solution was evaporated on a rotary evaporator at 35°C, until a thin film was formed on the flask. The film was washed with PBS and completely hydrated for 2 h. After the hydration of thin lipid film with PBS, the lipid dispersion was sonicated for 4 min, and finally the liposomes encapsulated with curcumin were obtained.

For cooking food, this above synthesis method seems unreasonable and overkill. I urge you to experiment with a few methods of forming the red curcumin-phospholipid complex yourself, with soy lecithin/eggs. I would have tried some methods out but I don't have these ingredients at home. see edit.

The advantage of having curcumin in a liposome is that it is much more stable: unlike the acid-base action, curcumin in a liposome remains stable across a wider range of pH's, as shown in this graph. This method does show promise, but as I mentioned, you would have to give this a try yourself.

Other practical solutions

Seasoned Advice would probably give you would get a better answer in this field, but I'll drop in my two cents. Generally, chilli powder is used to provide the red colour to Tandoori. More specifically, Kashmiri chilli powder is specifically used for tandoor preparations, such as tandoori chicken and chicken tikka. This powder is applied as a dressing (in combination with oil, rice flour etc after cooking the food) and during marination (chicken is marinated with curd, red chilli powder and turmeric).

Happy cooking!

References:

Kuntal Maiti, Kakali Mukherjee, Arunava Gantait, Bishnu Pada Saha, Pulok K. Mukherjee; Curcumin–phospholipid complex: Preparation, therapeutic evaluation and pharmacokinetic study in rats, International Journal of Pharmaceutics, Volume 330, Issues 1–2,
Sam-ang Supharoek, Kraingkrai Ponhong, Watsaka Siriangkhawut, Kate Grudpan;
Employing natural reagents from turmeric and lime for acetic acid determination in vinegar sample, Journal of Food and Drug Analysis, Volume 26, Issue 2,
Hong-Hao Jin, Qun Lu, Jian-Guo Jiang; Curcumin liposomes prepared with milk fat globule membrane phospholipids and soybean lecithin, Journal of Dairy Science, Volume 99, Issue 3,
Research gate: Why does curcumin have a red colour in the alkaline condition
Wikipedia: Phospholipids, Curcumin

EDIT: I tried to make a phospholipid complex with turmeric and egg yolks in the kitchen. Around a teaspoon of turmeric was mixed with an equal amount of egg yolk and heated in a water bath: An orangish colour was observed, which was in contrast to the bright yellow of the egg yolk. (I scraped the top layer away which had a larger amount of turmeric to reveal the bottom layer, which had more egg yolk)

A possible conclusion that can be drawn from this experiment is that the complex is indeed forming, but the concentration of phospholipids/curcumin is too small to impart the magnitude of red colour required for food colouring. This pretty much conclusively proves that no easily made derivative of turmeric can be used as red food colouring.


What does a higher half cell potential mean?

That's not quite correct.

Half-reaction potentials are only really meaningful when you complete the redox pair, that is, you must have an oxidation and reduction reaction. The sum of the potentials is the potential for the full redox reaction at standard concentration and state. In that sense, the more positive either of the potentials is, the more favorable the overall reaction will be.

There is no preference for a neutral species in general. If you put sodium in water, the electron is quickly liberated and the end products of hydrogen and sodium ion are much more favored than the reactants of water and sodium.


Why is thallium(III) oxide stable while thallium(III) chloride unstable

The stability of thallium(III) oxide and thallium(III) chloride can be compared by observing two properties:

Size of corresponding ions

Size of $\ce{Tl^3+}$ is 0.95 Å, size of $\ce{O^2-}$ is 1.44  Å and that of $\ce{Cl-}$ is 1.8  Å. So, it pretty seems that difference in size between thallium and oxygen atoms is less than that of difference in size between thallium and chlorine atoms. So, thallium(III) oxide is stable than that of thallium(III) chloride. (data)

Difference in electronegativity

Electronegativity of thallium is 2.04, oxygen is 3.44 and that of chlorine is 3.16. So, difference in electronegativity of thallium-oxygen is more than that of thallium-chlorine. So, thallium(III) oxide is stable than that of thallium(III) chloride. (data)


How can either reactants or products be "favored" at equilibrium?

At equilibrium the rates of the forward and reverse reactions are equal. This means that you can have more product than reactant, or more reactant than product. It just means that the rates of formation of each are equal and so there is no net change in the amount of each. The amount of products and reactants at equilibrium can be favored, their amounts just won't change once they reach equilibrium.
Depending on the temperature and pressure, the composition of the system at equilibrium can change.
Recall that in the reaction $$\ce{aA + bB<=>cC + dD}$$
The equilibrium constant is: $$K=\frac{[\ce{C}]^c[\ce{D}]^d}{[\ce{A}]^a[\ce{B}]^b}$$
These concentrations refer to the concentrations of $\ce{A}$ and $\ce{B}$ at equilibrium.
Therefore in a situation where $\ce{A<=>B}$ and $\Delta G_\mathrm{r}^\circ\lt0$, then $K>1$, meaning that there is more of $\ce{B}$ than $\ce{A}$ at equilibrium (and vice versa).

If you are interested to know why $\Delta G_\mathrm{r}^\circ = -RT \ln K$ gives the standard change in Gibbs energy for a reaction, then you should note that the general form of the equation at any point in the reaction is given by
$$\Delta G_\mathrm{r} = \Delta G_\mathrm{r}^\circ + RT \ln Q$$
Where $Q=\frac{[\ce{C}]^c[\ce{D}]^d}{[\ce{A}]^a[\ce{B}]^b}$ at any point in the reaction. At equilibrium, there is no more Gibbs free energy, as the system no longer tends to the formation of products or reactants. Therefore at this point $\Delta G_\mathrm{r} = 0$, and the original expression given in Atkins is what you get. Also we define this special case of $Q$, saying that at equlibrium $Q=K$, the equilibrium constant.

Source
You might find the above image helpful. The y-axis shows the amount of Gibbs free energy, $G$. The reaction will want to tend towards the state where it has the lowest amount of Gibbs free energy (the local minimum). The slope of the curve can be therefore seen as $\Delta G_\mathrm{r}$. Therefore the point where the system has the least Gibbs free energy will be when $\Delta G_\mathrm{r} = 0$ could occur anywhere in the extent of reaction. This shows that the equilibrium concentrations need not be equal, but shows that the reaction will not progress any further when $\Delta G_\mathrm{r} = 0$.


Constraints on the polar angle (θ) in the particle in a sphere problem

First of all, as orthocresol and porphyrin said, we should keep in mind that $\phi$ and $\theta$ must span different sets, and there are different conventions on them:

Any spherical coordinate triplet (r, θ, φ) specifies a single point of three-dimensional space. On the other hand, every point has infinitely many equivalent spherical coordinates.
[...]
If it is necessary to define a unique set of spherical coordinates for each point, one must restrict their ranges. A common choice is:
$r$ ≥ 0
0° ≤ $\theta$ ≤ 180° ($\pi$ rad)
0° ≤ $\phi$ < 360° ($2 \pi$ rad)

(Source)
In other words (using the convention above), you must have both $\psi(r, \theta, \phi) = \psi(r, \theta, \phi + 2 \pi)$ and $\psi(r, \theta, \phi) = \psi(r, \theta + 2 \pi, \phi)$ for spherical periodicity.
This is the basic symmetry on a sphere.
But the assignment $(r, \theta, \phi) \rightarrow (x, y, z)$ becomes "ambiguous" (not injective) if the function is defined as an application from $[0, \infty) \times [0, 2 \pi] \times [0, 2 \pi)$ to $\mathbb{R}^3$ and hence not invertible.
You can easily see this if you imagine mapping the surface of earth by having a very long rope with its ends fixed at both poles (half diameter, 180°) and an airplane carrying the middle of this rope around the globe (full diameter, 360° movement).
That being said, the mapping $(r, \theta, \phi) \rightarrow (x, y, z)$ must be taken from $[0, \infty) \times [0, \pi] \times [0, 2 \pi)$ to $\mathbb{R}^3$.
But that does not mean that $\psi(r, \theta, \phi) = \psi(r, \theta + \pi, \phi)$, and as higgsss observed, $\psi(r, \theta + \pi, \phi)$ is not necessarily equal to $\psi(r, \theta, \phi)$.
But higgsss also observed that, for some problems, some solutions might have symmetries allowing $\psi(r, \theta, \phi) = \pm \psi(r, \theta + \pi, \phi)$.
Solutions for the hydrogen atom are good examples of this: for the minus sign, p atomic orbitals, and, for the plus sign, d atomic orbitals.
You can see other orbital symmetries by looking at them here.


How to find the valence of transition elements?

For the 2nd and 3rd rows for 4b-7b groups the highest oxidation state equal to the group number is preferable over all else and for the 1st row for 4b-7b groups the highest oxidation state is equal to the group number.
Unfortunately, for all other information you'll need case-by-case approach (i.e. plainly remember it).


Catalysist activation energy - the Brønsted-Evans-Polanyi relation

First, a note on the Bell-Evans-Polanyi (BEP) equation:
The form you've written it in disguises what the constant $\alpha$ tells you. From the wikipedia article, it's said that $\alpha$ is a number between 0 and 1 which tells you how close the transition state is to the reactant state. Thus, this constant actually tells you quite a lot about the shape of potential energy plotted versus the bond coordinate.
To see why this is true, we should look at the derivation given in the article you linked. The BEP equation assumes that the energy of both the reactant and product vary linearly with the bond length of the atom being transferred. This is probably a very bad assumption for finding the actual shape of the well, but the intersection of these two lines seems to give a good approximation for the location of the transition state. Specifically, if one knew the slopes of these lines $m_1$ and $m_2$, we could find $\alpha$ according to $$\alpha=\frac{m_1}{m_1-m_2}.$$
See the wikipedia article you linked for the details on this, but hopefully this motivates the meaning of the constant $\alpha$.
So, what does this have to do with the article you're asking about?
Well, the reaction that is being catalyzed in the figure you have copied is,$$\ce{CO +3H2->CH4 +H2O}$$This seems very unlikely to happen in the gas-phase because the $\ce{CO}$ bond is very strong and we would have to have four molecules collide all at once and magically rearrange their atoms.
Thus, a transition metal surface is used as an adsorption site to get all these atoms close together.

What exactly is on the axes?

Well, now we can see that by the "CO Dissociation Energy" they mean the energy it takes to dissociate the carbon and oxygen atoms once they have been adsorbed onto the metal surface. So, a negative dissociation energy indicates that when $\ce{CO}$ adsorbs, it exists as two separated carbon and oxygen atoms.
Additionally, because they have shown the BEP linear relationship, the slope of that line is $\ce{\alpha}$ and the y-axis is the classical Arrhenius activation energy.
At first, it seems like the best catalyst should be the one which has the smallest CO dissociation energy, but as the article points out,

The reason is that in the limit of weak coupling (Ediss is only a little negative), the BEP relation gives that the barrier for dissociation of the reactants will be high and the rate low. For strong coupling (Ediss very negative) the activation energy of adsorption is small but now the barrier for forming the products will be large.

That is to say, if the reactants adsorb quite strongly to the surface (i.e. Rhenium in the plot) then reaction wil go really slowly because the activation energy (not the on you're seeing on the y-axis here) for forming the intermediates leading to methane will be very high.
On the other hand, if $\ce{CO}$ doesn't adsorb very favorably, the reaction will be slowed down because both $\ce{CO}$ and $\ce{H2}$ need to be adsorbed to react. When they do meet though, because they adsorb weakly, the activation energy to forming the products will be small.
This is why they say,

An optimal interaction strength must exist between these two limits.

The figure below on the right shows this intersection for pure metal surfaces. The closer you are to the peak, the better.

(source: nature.com)
Thus, either Cobalt or Ruthenium would be optimal for his reaction, but as the authors note, these are expensive, so nickel is used in the industrial process. Thus, some kind of alloy which behaves similarly to Ruthenium or Cobalt would be great if it could be found.
Finally, it should be noted that the BEP equation will only work in very specific circumstances where bond stretching is the dominant feature which determines if two things react. Thus, when reactants are adsorbed and thus all they can really do is vibrate, I would expect the BEP equation will do a pretty decent job.
I hope that clarified your questions. Let me know if I missed anything.


Heat of reaction for the decomposition of AgCl means?

The intent of the question is that the reaction:
$$ \ce{Ag+} + \ce{Cl-} \rightarrow \ce{Ag} + \ce{Cl}$$
requires absorbing exactly one photon. That photon must have at least the energy required to perform this reaction. The grey colour produced is caused by the metallic silver ($\ce{Ag^0}$) formed by the reaction.
Because a photon's energy is inversely proportional to its wavelength, $E_{photon}=\frac{h c}{\lambda}$, in order to have the required energy or more to cause the reaction, the photon must have a required wavelength or less.  So there is a maximum wavelength that the photon may have in order to cause this reaction.
You're told that the energy required to decompose 1 mole of $\ce{AgCl}$ is 248 kJ/mol, so you have to figure out the energy per atom/photon and convert that energy to a wavelength.
This is an example of the photoelectric effect - that light comes in quantised particles (set amounts of energy) called photons, and certain processes require absorbing a single photon at a time with enough energy, rather than absorbing more lower energy light over time, as if the light was only a continuous wave.


I think you have it all right. The OH- present must come from the autolysis of the water present, which is far in excess of the HCl. If you figure the pOH of pure water would be 7, then it makes sense to think that adding 1E-06 M (a small concentration) of HCl would lower the basicity (raise the pOH from 7 to 8).

The reason a scale of 0-14 is used for pH is that when you get to either extreme end of the scale (close to 1M in either acid or base), the sheer concentration of ions present, relative to water, is a game changer and the system no longer behaves in a nice linear fashion. So the scale does not work well outside the 0-14 range. Conceptually, you can extend the pH scale down below zero using concentrated mineral acids or strong acids like fluorosulfonic acid, and empirically measure their effectiveness in protonating a very weak base as a proxy for pH.


Why are organolithium and organomagnesium compounds nucleophilic in nature?

When Li reacts with an alkyl halide, a salt (lithium halide) is formed, whose stability is the primary thermodynamic driving force of the reaction: $\mathrm{2Li}+\mathrm{RCl} \rightarrow \mathrm{RLi}+\mathrm{LiCl}$.

Due to the larger difference between the electonegativities of C and Li, the C-Li bond in organolithium compounds has more ionic character and C is more negatively polarized. The higher polarization makes R in RLi a harder nucleophile compared to RMgX, which explains why lithium organyls and Grignard reagents react differently when adding to certain carbonyl compounds.


Rearrangements; Carbocations; E1 Mechanism

1) Will the methyl shift? I'm guessing the answer is an emphatic yes.
  Tertiary carbocations are more stable than secondary carbocations.

Yes, if that is the only way to generate a tertiary carbocation and the C-Methyl bond can more or less line up with the carbocation p orbital, then the methyl would likely shift.  If there are other alkyl groups attached to the carbon initially bearing the methyl group, they could shift too.

  2) Will there still be some secondary carbocation left over? I.e. will
  there be both major and minor products? I'm guessing yes

An equilibrium exists between the various carbocations.  At 300 K, an energy difference of 1.4 kcal/mol results in a power of ten difference in the equilibrium (e.g. from 50:50 to 90:10).  So it really depends on the energy difference between the various carbocations involved, as well as the rate they are captured or proceed to the next step.

Finally, if there is a hydrogen on the carbon adjacent to the carbocation center, then elimination to create a double bond becomes a competing pathway,


Determining the more acidic proton in cyclopenta-1,3-diene

Figure c does not look like a real compound. If it represents the conjugate base of cyclopentadiene, it is lacking a negative sign, and all five protons around the ring are equivalent, so the question of whether the red or blue proton is more acidic makes no sense.

If Figure c represents cyclopentadiene, the acid, it does not clearly show that the red position really has 2 hydrogens. However, the double bond structure shown is consistent with 2 hydrogens in the red position, and the question becomes very simple: in cyclopentadiene, is a methylene proton or a vinylic proton more acidic? Removing a vinylic proton puts a negative charge in the localized sigma system - no aromaticity develops. However, removing one of the methylene protons causes that carbon to go from sp3 to sp2p, and the negative charge goes into the pi system where it can be delocalized over the 5-membered ring. And aromaticity is a characteristic feature of cyclopentadienide ion.


Hydrogen water is a solution of molecular hydrogen $\ce{H2}$ in water, which shouldn't be confused with acidic or basic solutions.  The first scientific reference to hydrogen water appears to be in the journal, Nature Medicine which can be accessed here if you have a personal or institutional subscription to the journal. The claim is that molecular hydrogen can serve as a mild reducing agent and therefore has antioxidant properties; namely, it can reduce hydroxyl radicals:
$$\ce{H2 +2 OH\cdot -> 2H2O}$$
The article has generated a considerable amount of research, with well over 300 journal articles citing this work originally published in 2007.  There seems to be some merit to the claim although in my brief scan of the abstracts, there does not appear to be a large body of evidence suggesting that commercial products are beneficial.  One such article, which is publicly accessible is here.  As mentioned in another answer, the commercial product approach to generating hydrogen water appears to be the oxidation of magnesium in water, which generates $\ce{Mg^2+}$ ions.  As with any other nutritional supplement, I would be very wary of entering this type of treatment without consulting a professional.  Magnesium is an important nutrient, however at high doses it does show signs of toxicity.


Determine the geometry of each nitrogen atom in formamidinium ion

The hybridisation of any atom is determined by the total count of its sigma bonds and localised lone pairs. This is because, while localised lone pairs occupy hybrid orbitals (as in ammonia), delocalised lone pairs occupy pure $\ce{p}$ orbitals instead. Hence, the latter is not counted in determining hybridisation.

The $\ce{N}$ atom labelled "B" has its lone pair in conjugation with a pi bond. This is perfect for delocalisation/resonance to take place. Hence, that lone pair is not counted in determining hybridisation. Thus, it is also a trigonal planar geometry.

Interestingly, notice that the formamidinium ion has two equivalent canonical resonating structures. The lone pair of the atom B on delocalisation, results in a lone pair on A and a formal positive charge on B, which is just the reverse of the structure shown.


Entropy as the driving force for osmosis

(a) The tendency for the solute undergoing diffusion to occupy as large a volume as possible is similar to that of a gas filling the volume available to it and in each case the driving cause is the increase in entropy. In any  distribution of particles there are more ways of distributing them in a larger volume than in a smaller one.

Recall that as the solvent enters the concentrated solution the solution volume increases and this continues until the chemical potential on either side of the semi-permeable membrane is equalised. The pressure is also increased as solution is pushed up into a column, the difference in pressure is the osmotic pressure.

(b) From the statements above you can see that reverse osmosis cannot occur without applying external energy.

Notes: The change in entropy of a gas with volume is $\displaystyle \frac{\partial S}{\partial V}=\frac{R}{V}$. In a dilute solution the mole fraction $x_s \sim 1/V$ where $V$ is the  volume of the solvent (or solution)  with one mole of solute. As $G-G^0=-T S = RT\ln(x_s)$ then $\displaystyle \frac{\partial S}{\partial V}=\frac{R}{V}$. Thus the change in entropy is the same for the dilution of a gas as for the solute in a dilute solution.


How does an energy input actually change a molecule?

It is about transition energy barriers.

Imagine a mountainous landscape with lots of valleys of different depths. Stable molecules are large balls than sit in the bottoms of the valleys. Add some energy and the ball might rise up the valley floor but it will fall back. Add enough energy and the ball can rise high enough to get over the barrier to the next valley and will fall into a different place with a different depth (and the analogy falls down as it also becomes a different ball). So, whatever the relative depths of the valleys, there is a big barrier to getting from one valley to another. Even when one valley is much lower than another (so moving from one to another releases energy) you still have to put energy in to cause the transition.

Chemical reactions are often controlled by those transition barriers between one molecule and another rather than by the absolute energy difference between the molecules. Methane and oxygen will burn to give various compounds and will release energy, but there is a big barrier to starting the burning process. This means that the deoxygenation reactions you describe are possible and won't spontaneously reverse even though they would release energy if they did.


Sample preparation for UV analysis of drug content in capsules/tablets

I'd agree with Zhe that "this is a non trivial problem." Method development is tricky if you don't have a standard analysis procedure.

When you dissolve the capsule, if you do have suspended particles then yes they will interfere with the UV analysis. So do centrifuge.

But think of the removal of the liquid as an extraction. So add more solvent to the centrifuge tube, shake up the particles and centrifuge again. Assuming that you won't over dilute, I'd do the extraction three times, then dilute to the final volume.

The rub here in my mind is that UV is not very selective. I'd doubt that you're going to get a nice flat baseline with one nice peak. So you'll need a "pure" drug sample to compare to your extract samples to see if there is any other gunk causing interference.


The interpretation of the statement depends on exactly what is meant by "overall rate."

The rate-determining step corresponds to the largest free energy difference from a transition state to any preceding stable state (reactant or intermediate).
In the example above, the biggest forward barrier is from the reactant to the second transition state, so the second step is rate-determining. However, the barrier, as identified above, is comprised of both the first and second step, so they both figure into the overall rate of the reaction, as determined by the slow step.
EDIT:
Based on comments, I think a clarification is in order. Note that the above conclusion is not true in the general case. You could make the indicated claim for the depicted reaction specifically, but not, for example, in the case below:

The second step here is still rate-determining because the transition state is involved in the largest barrier. However, the biggest barrier is from the intermediate to the second transition state. In this example, I would not say that the overall rate depends on the first step.


Torquoselectivity: Disrotatory ring opening

No, there is no general orbital symmetry rule for predicting which of the two possible disrotatory motions will be favored.  However, by examining steric effects in the two possible transition states we can often make a good guess as to which product will predominate.  There are always a few "special cases" like the one you referenced in your post, where with certain substituents under certain conditions certain things will happen.  But to my knowledge there is no general rule for this situation.

In the following diagram there are 3 thermal reactions controlled by orbital symmetry (6 pi electrons, therefore disrotatory motion).

The top line shows the symmetry allowed disrotatory closure of trans, trans-1,6-dimethylhexatriene.  There are two possible disrotatory motions - either the tops or the bottoms of the p-orbitals on carbons 1 and 6 can move towards each other in a disrotatory fashion.  One motion produces the dimethylcyclohexadiene with the two methyl groups above the plane of the screen, the other motion produces the mirror image molecule with the methyl groups below the plane of the screen.

Similarly, cis, cis-1,6-dimethylhexatriene can undergo two disrotatory motions to produce the same two products (Note: the cis, cis isomer will react much slower and probably in lower yield; steric interactions between the methyl groups will disfavor the s-trans conformation about the two single bonds that is required for the ring closure to be able to occur).

The bottom line shows the reverse reaction, thermally converting a cyclohexadiene to a hexatriene.  As shown, two possible isomers can be formed.  The ratio of these two isomers will depend on steric interactions that develop during the two possible disrotatory motions.  In this case, we will primarily obtain the trans, trans isomer; the cis, cis isomer has steric problems in its transition state from the two methyl groups bumping into each other (build a model).

In this last reaction, if we replaced the methyl groups with deuterium, we would obtain a roughly equal mixture of the two possible products, cis, cis- and trans, trans-1,6-dideuteriohexatriene.  So again, there is no orbital symmetry "rule" for predicting whether a group would wind up cis or trans on the double bond.  But if we examine the steric effects in the two possible disrotatory transition states, we can often make a good guess about which isomer should predominate.


Removing the hydrazine byproduct from the mitsunobu reaction

Acidic wash is not going to help. As you said you can try a DIAD alternative for example DCAD can give good results (sry no advertising intended here). There is also the option of taking your contaminated product to the next step(s) and hope you can get rid of the impurity there. I would take option A if I have to make a bunch of similar analogues by a Mitsunobu and B otherwise for not wasting too much time


Do halogens bond more commonly with each other or themselves?

There's a lot of circumstantial evidence that heterodiatomics are more stable than homodiatomics. It is usually ascribed to the observation that atoms of different "electronegativity" tend to have stronger bonds than atoms of the same electronegativity.

In your particular example, consider the reaction:

$$\ce{Cl2 + F2 -> 2 ClF}$$

I don't have the tables handy, but I calculate with electronic structure theory that this reaction is exothermic by about 25 kcal/mol. That's quite a bit! (The equilibrium lies far to the right.)

This would suggest that a mixture of Chlorine and Fluorine, heated up to very high temperature and then cooled down, would tend to reform as ClF.


Is this statement for the First Law of Thermodynamics correct?

You can think of the first law of thermodynamics as an energy accounting method. It says that all inputs and outputs of energy should add up to the total change in the energy of a system. The exchanged energy is generally classified as work or heat. By "energy can neither be created nor destroyed," it is meant that nuclear reactions, in which the structure of nuclei changes with accompanying transformation of matter into energy and vice-versa, will not be considered. The reactions considered are ordinary chemical reactions involving making and breaking of chemical bonds resulting from interactions between electrons and nuclei.


How many valid resonance forms can be constructed for aniline?

Your four structures are correct and you could also include your suggested fifth structure (your first and fifth structures are comparable to the resonance structures of benzene). While some of the structures are mirror images, we generally think of particular atoms as being fixed when writing out resonance structures. You can think of the carbons as having a number (which they technically would if you named the resonance structures IUPAC style), which leads to the "mirrored" structures not being identical since the double bonds are between different carbons.


Is “mass-action ratio” identical to “reaction quotient?”

The concepts are extremely closely related.  The difference is that the mass action ratio is a formally a ratio of concentrations, or, for gases, of partial pressures.  The mass action ratio, and the term "mass action" in general, refers to situations where the chemical potential of the reactants and products is directly proportional to the concentration, or pressure, or mass, of the species involved.

The reaction quotient is is a tiny bit more general; the implicit assumption of mass action is not necessarily made.  The reaction quotient is the ratio of chemical potentials of reactants and products.  It works even in systems that don't follow the law of mass action, i.e. where the activity coefficients vary during the equilibration.

For all intents and purposes, the two concepts are the same.  If you want to split hairs, you'd probably do it the way I did here.


They're called "reversible" because the processes can run backwards without additional input of energy. There aren't really any truly reversible processes in the universe, though (some come very close, but the construct of "reversible" is an ideal, not a reality).
Entropy of a system can increase just fine during a reversible process: expanding a gas "slowly enough" is reversible ("slowly enough" means in an infinite number of infinitesimal steps), but the entropy of the gas increases as it expands as it has more microstates to go into. The entropy that remains constant in a reversible process is that of the system plus the surroundings (in other words, the universe).
The process does not need to be reversible to have entropy. However, we measure entropy based on the heat of the reversible process because entropy is a state function (it depends on the beginning and end points of your process, nothing more) while heat is a path function (it depends on the path you take to get there).


What is the difference between dynamic and static electronic correlation

$\newcommand{\el}{_\mathrm{e}}$In quantum chemistry, when a nomenclature in which one distinguishes between "static" and "dynamic" correlation is used, "correlation" referrers to all the deficiencies of the Hartree-Fock (HF) single-determinantal approach. For instance, the the correlation energy is defined as the difference between the exact (non-relativistic) energy and the HF energy (calculated with a complete basis),
$$
  E_{\mathrm{corr}} = E_{\mathrm{exact}} - E_{\mathrm{HF}} \, .
$$

Now, what are the deficiencies of a Hartree-Fock approach?

First, electrons in this model do not instantaneously interact with each other, as they do in reality, but rather each and every electron interacts with the average, or mean, field created by all other electrons. Classically speaking, each electron moves in a way so that it avoids locations in a close proximity to the instantaneous positions of all other electrons. And the failure of the HF model to correctly reproduce such motion of electrons is the first source of $E_{\mathrm{corr}}$. This type of correlation is called dynamic correlation since it is directly related to electron dynamics.

Secondly, the wave function in the HF model is a single Slater determinant, which might be a rather poor representation of a many-electron system's state: in certain cases an electronic state can be well described only by a linear combination of more than one (nearly-)degenerate Slater determinants. This is the second reason why $E_{\mathrm{HF}}$ may differ from $E_{\mathrm{exact}}$ and the corresponding type of correlation is called static, or nondynamic, since it ist not related to electron dynamics.

Interestingly, both static and dynamic correlation effects can be taken into account by "mixing in" more Slater determinants $\Phi_i$ to the Hartree-Fock one $\Phi_0$,
$$
  \Psi\el(\vec{r}\el) = c_0 \Phi_0 + \sum_i c_i \Phi_i \, ,
$$
Here, if $c_0$ is assumed to be close to $1$ and a large number of excited determinants $\Phi_i$ are added each of which is assumed to give only a small contribution, then the method primarily treats dynamic correlation. And if, on the other hand, it is assumed that there are just a few excited determinants $\Phi_i$ with weights close to that of the reference determinant $\Phi_0$, then the method primarily treats static correlation.

An example of a method that recovers primarily dynamic correlation is Møller–Plesset perturbation theory (MP$n$), while multi-configurational self-consistent field (MCSCF) method primarily takes account of static correlation. Note the word "primarily" here and above. It is almost impossible in principle to keep dynamic and static correlation effects separated since they both arise from the very same physical interaction. Thus, methods that typically cover dynamical correlation effects include at high-order also some of the non-dynamical correlation effects and vice versa.


Calculating numbers of moles at equilibrium given the volume of mixture and initial concentration of reactants

Following a discussion that took place in the comments:

If given answers are correct, as Poutnik pointed out, your equation should be :
$$\require{cancel}\ce{A(g) + B(g)<=>2 C(g)}$$ - Mathew Mahindratne
Could you explain how you arrived at that conclusion? - Ollie

This answer solves this question for a general case:
$$\ce{a A(g) + b B(g) <=> c C(g)}$$
Let's assume that initial concentrations of A, B, C are $[A_0]$, $[B_0]$ and $[C_0]$ and the final concentration be $[A]$, $[B]$ and $[C]$.
Now, the given variables are $[A_0]$, $[B_0]$, $[C_0]$ and $[C]$, assuming similar data to the question. Let $\mu$ be change for every mole of C
\begin{align}
\therefore\ &\text{Change}=[C]-[C_0] \tag{1}\\
\implies &\text{Change/mol of C} = \frac{[C]-[C_0]}{c}\tag{2}
\end{align}
Now, building a RICE table we get:
\begin{array}{lc} \hline
 &\ce{a A &+ &b B &<=> &c C} \\ \hline
I & [A_0] && [B_0] && [C_0] \\
C & -a\cdot\frac{[C]-[C_0]}{c} && -b\cdot\frac{[C]-[C_0]}{c} && +c\cdot\frac{[C]-[C_0]}{c} \\
E & [A_0] - a\cdot\frac{[C]-[C_0]}{c} && [B_0] - b\cdot\frac{[C]-[C_0]}{c} && [C_0] + [C]-[C_0] \\ \hline
\end{array}
Therefore, the final concentrations of A and B would be:
\begin{align}
[A] &= [A_0] - a\cdot\frac{[C]-[C_0]}{c} \\
[B] &= [B_0] - b\cdot\frac{[C]-[C_0]}{c}
\end{align}

Now, onto the specific solution, how to figure out $a$,$b$ and $c$ given the above convoluted generalisation?
Point 1: The change in $[A]$ and $[B]$ are both equal. Looking at change in the above RICE table, we deduce:
\begin{align}
a\cdot\cancel{\frac{[C]-[C_0]}{c}}&=b\cdot\cancel{\frac{[C]-[C_0]}{c}} \\
\implies a&=b
\end{align}
Point 2: We notice that for the change of $\pu{0.04 mol}$ of C, we observe a change of $\pu{0.02 mol}$ in A (or B, both work). Similar to point one, we equate the equations of change of A and C.
\begin{align}
2a\cdot\cancel{\frac{[C]-[C_0]}{c}}&=c\cdot\cancel{\frac{[C]-[C_0]}{c}} \\
\implies 2a&=c
\end{align}
Now, we substitute the values of $a$,$b$ and $c$ in terms of $a$, we get:
$$\ce{\cancel{a} A + \cancel{a} B <=> 2\cancel{a} C}$$
which implies:
$$\ce{A + B <=> 2C}$$
Therefore, your textbook is clearly in the wrong, since it states $\ce{A + B<=>C}$.


Computing of strengthening of π-interaction

First off, definitely check out ETS-NOCV, which I believe is a fairly recent addition to ADF which seems to do exactly what you're interested in. Other NBO-esque software may also suit your needs - I seem to recall NBO5 in Gaussian can generate sigma and pi natural bonds.

Regarding doing things the energy decomposition way:

If you have not already seen it, this paper goes about doing (in ADF) what you want to do:

http://pubs.rsc.org/en/Content/ArticleLanding/2012/DT/c2dt31845h

(disclosure - I am a member of the Stranger group)

I did this for a related system a while back but have forgotten the exact details, however fragment bonding analysis is a nifty technique that allows estimation of the energy of single bonds in a molecule as well as decomposition by bond type.

Broadly, you cut your molecule into chunks on either side of the bond you're interested in and generate fragment orbitals from them. (that is, you use the TAPE21 files for these molecular fragments as a basis for the calculation of the full molecule). These fragment orbitals subsume all of the interactions within the fragments so that the total bonding energy only consists of interactions between the fragments.

From here, you can actually remove atomic orbitals from the fragment orbitals. If you prune the orbitals relevant to pi bonding, you can estimate the pi and sigma components by relative energy.

Here's where you should start: ADF manual - Analysis


According to Wikipedia, solubility of aspirin in water is $\pu{3 g/L}$ or $\pu{3 mg/mL}$ at $\pu{25 ^\circ C}$. Thus, it is not hard to make a $0.3\% (w/v)$ aquious aspirin solution (maximum concentration, which equals to $\pu{300 mg}/\pu{100 mL}$ at $\pu{25 ^\circ C}$). Therefore, your target, $\pu{10 mg}/\pu{100 mL}$ at $\pu{25 ^\circ C}$ can be easily achieved without any modifications. You have not mentioned the what different $\mathrm{pH}$ values of buffers you are planning to use. Since $\mathrm{p}K_\mathrm{a}$ of aspirin is about $3.5$ and the amount of it above your target value would dissolve in water (so final $\mathrm{pH}$ would be around $2.12$ for $\pu{300 mg}/\pu{100 mL}$ soution), it is guaranteed to dissolve your desired amount in any buffer with  $\mathrm{pH}$ above $2.12$.
None the less, dissolve aspirin first in small amount of ethanol and diluted it with desired buffer is not bad idea. As a matter of fact, in aspirin preparation lab, we use ethanol to recrystallize the final product, therefore, would not arise any problem on dissolution. Thus, this would be an ideal alternative procedure.


Are there any STP liquids not containing large amounts of hydrogen and oxygen that are safe for human contact?

Perfluorocarbons would be a possibility, $\ce{(-CF2-)_{n}}$, they contain only carbon and fluorine and are relatively low cost and widely available in large quantities.  They are inert (e.g. non-reactive) and depending upon chain length and structure, you could pick one that can be warmed up nicely to a bath temperature.  They are also used industrially in degreasing operations, so they might have some cleaning ability as well.

Some perflurocarbons are actually being investigated as human blood substitutes - they can transport and release oxygen to tissues.  Perfluorodecalin is a prime example. As shown below, it is basically a naphthalene ring in which all of the hydrogens and double bonds have been replaced with fluorine.

Ingesting small amounts would not be a problem. In fact, these same materials have been used in liquid breathing experiments.

There are a number of other scientific aspects of these perfluorocarbons that could play into science fiction.  For example, the small size of the molecules (compared to say, haemoglobin) allows it to travel to smaller places that haemoglobin cannot access due to its larger size. Perhaps the perfluorcarbon could also carry small drugs to relax, rejuvenate, repair body parts, or enhance sensory experiences during the bath. Since these perfluorocarbons act as oxygen transport agents they could also be worked into the story as a bath medium that also scavenges "loose" oxygen.


How do you name this compound using IUPAC system (including steps)?

Identify the parent chain. In this case you have a 6-carbon chain starting at the alkene, wrapping through the chiral center and ending off to the right. 6 makes the prefix 'hex'; the alkene starts at the '1' position; thus, hex-1-ene.
Substituents. In this case they're both at the 3-position so they'll be listed in alphabetical order: 3-bromo-3-methyl
Chiral Center at the 3 position. Numbering the legs of the center according to IUPAC rules gives (1) to the bromine atom, (2) to the carbon atom double bonded to the next carbon atom, (3) the carbon atom single-bound to the next carbon atom, and (4) the methyl group. Rotate this tetrahedral image in our head to where you're looking at the molecule with numbers 1-3 towards you. The numbers rotate clockwise, thus the chiral center is (3R)
String it all together.

(3R)-3-bromo-3-methylhex-1-ene.


None of your methods are correct. If you have a small data set - in this case, 5 values from a titration - then the absolute uncertainty for each individual value is half the range of data. For example, say you get the following results (in ml, from a burette with a precision of 0.05ml): $4.70$, $5.35$, $4.85$, $4.60$ and $5.20$. The range of these results is $5.35-4.60=0.75$. This gives an absolute uncertainty of $\frac{0.75}{2}=0.375$ for each individual piece of data e.g. $4.85\pm0.38$ml.

However, if the absolute uncertainty calculated from half the range is less than the precision of the instrument (in this case $<0.05$ ), then the precision of the instrument should be given as the uncertainty e.g. $5.00\pm0.05$. I think you may have incorrectly calculated the absolute uncertainties.

For small data sets ($<10$ pieces of data), the uncertainty in the mean is then given by $$\Delta {x}_{avg}=\frac{\Delta x}{\sqrt{N}}=\frac{R}{2\sqrt{N}}$$ where $\Delta {x}_{avg}$ is the uncertainty in the mean, ${N}$ is the number of times the measurement was repeated, $R$ is the range and $\Delta{x}$ is the uncertainty in one measurement.

For example, using the example data above, $\Delta {x}_{avg}=\frac{0.375}{\sqrt{5}}=0.168$. The average can then be stated as $4.94\pm0.17$ml


Why elements with high atomic mass (like tungsten) are used in the production of X-rays, in the Cathode-Ray Tube?

If a sample containing a particular element is irradiated (e.g. with photons or, in case of an X-ray tube, with electrons) with an energy high enough to excite inner electron orbitals, X-rays are emitted in the de-excitation. In contrast to the continuous bremsstrahlung energy spectrum, the energy liberated in the transition from the excited to the ground state takes the form of a characteristic X-ray photon whose energy is given by the energy difference between the initial and final states. If the temporary vacancy has been created in the K shell of an atom, then a characteristic K X-ray is liberated when that vacancy is subsequently filled. In particular, if the concerned electron comes from the L shell, than a Kα photon is generated whose energy is equal to the difference in binding energies between the K and L shells.

The K-series X-rays are generally of most practical importance because their energy is greatest. Their energy increases regularly with atomic number $Z$ of the element, for example, only about $1\ \mathrm{keV}$ for sodium ($Z=11$) and about $100\ \mathrm{keV}$ for uranium ($Z=92$). An empirical law concerning the characteristic X-rays that are emitted by atoms is expressed by Moseley’s law.

Hence, in order to produce characteristic X-ray photons with a high energy, K-series X-rays of a heavy element are preferred.

Since only a small fraction of about $1\ \%$ of the energy generated in an X-ray tube is actually emitted as X-rays and the rest of the energy is released as heat, the temperature of the anode target can be very high during the operation of the X-ray tube. Therefore, the anode has to be made of high-temperature materials. You already mentioned the high melting point of tungsten ($Z=74$); whereas for example the heavier elements lead and bismuth cannot be used because of their low melting points.


Removing sodium borate for the test proving thiocyanate ions

Lehrbuch der analytischen und präparativen anorganischen Chemie by Jander et al. [1] suggests to precipitate borate with $\ce{Ba^2+}$ in neutral solution beforehand as barium(II) thiocyanate is one of the few barium(II) salts actually soluble in water, or use an excess of $\ce{Fe^3+}$ to make sure all the interfering anions are precipitated.
From [1, p. 358]:

6 . Nachweis als $\ce{Fe(SCN)3}$
[…]
Störungen: Um den störenden Einfluss von $\ce{F-}$, $\ce{PO4^3-}$, $\ce{AsO4^3-}$, $\ce{H3BO3}$, $\ce{C4H4O6^2-}$, $\ce{C2O4^2-}$ usw., die mit $\ce{Fe3+}$-Ionen Komplexe bilden, auszuschalten, wird $\ce{Fe^3+}$ im Überschuss zugegeben.

From [1, p. 418–419] (emphasis mine):

9. Nachweis von Fe(III) als $\ce{Fe(SCN)3}$
$$\ce{Fe^3+ + 3 SCN- <=> Fe(SCN)3}$$
[…]
Störungen: […] Ferner beeinträchtigen
$\ce{Hg^2+}$-Ionen durch Bildung von wenig dissoziiertem $\ce{Hg(SCN)2}$, $\ce{F-}$ durch $\ce{[FeF6]^3-}$-Komplexbildung, die Anionen organischer Säuren ebenfalls durch Komplexbildung und auch $\ce{PO4^3-}$, $\ce{AsO4^3-}$, Borat-Ionen sowie ein größerer Mineralsäureüberschuss die Reaktion. Es ist
daher ratsam, vor der Prüfung […] die störenden Anionen in neutraler Lösung mit $\ce{Ba^2+}$ abzutrennen.

References

Jander, G.; Blasius, E.; Strähle, J.; Schweda, E.; Rossi, R. Lehrbuch der analytischen und präparativen anorganischen Chemie, 16. Auflage.; Hirzel: Stuttgart, 2006. ISBN 978-3-7776-1388-8. (in German)


The crucial point which explains this is that there are two different forms of conduction happening here.

The first kind of conduction, and the one we are most familiar with is electronic. This is a flow of electrons, and is the conduction which flows in the wire and through the load (a motor, LED etc).

The second kind is the flow which occurs in the electrolyte, and this conduction is ionic. It is a flow of charged particles (ions) within the electrolyte. Even though we often speak of the conductivity of an electrolyte, what is really meant is the electrolyte's ionic conductivity.
This is governed by the mass of the ions, their concentration, temperature and the viscosity of the solution.

Electrolytes do not really conduct electrons at all, which is why there is no 'short circuit' through them. The conductivity arises due to several mechanisms - either the ions themselves move through the solution, or electrons (or protons) hop from one ion to the next, conveying the charge.

This will only happen if there is a means of causing the flow of charge through the electrolyte. When it is open circuit, the electrolyte is overall neutral in charge and no current will flow in it.
If an external circuit is connected, it allows electrons to flow from one electrode to the other, and then the corresponding amount of charge will flow through the electrolyte to balance it out, producing current.


What is "mer", the first ingredient in a type nail polish?

In the organic lab, sometimes we would run reactions that produced a tar-like substance in the reaction flask as an undesired by-product.  It was just a gooey polymer, and it happened often enough that this substance was just referred to as "mer", short for polymer.  "Mer" was therefore a fairly common term used in many (most?) labs.
Since "mer" is listed as the first ingredient in nail polish, that indicates that it is the major component in the formulation.  If you google "nail polish", you'll find a Wikipedia article. Its opening paragraph states:

Nail polish consists of an organic polymer with various additives

I suspect they use the term "mer" to refer to whatever polymer is the main proprietary ingredient in their formulation.  The Wikipedia article also notes that

Nail polish consists of a film-forming polymer dissolved in a volatile organic solvent. Nitrocellulose that is dissolved in butyl acetate or ethyl acetate is common.

Once the volatile solvent evaporates, you are left with a hard polymer coating on your fingernail.


What is the reasoning behind the increasing pKa for organic compounds?

Cyclohexanol is an alcohol. An alcohol contains an $\ce{OH}$ group connected to a tetrahedral carbon.
Phenol is slightly different; it contains an $\ce{OH}$ group connected to a trigonal planar carbon that is part of an aromatic ring.
cyclohexanecarboxylic acid contains a very different functional group, carboxylic acid.

Cyclohexanol has a $\mathrm{p}K_a$ of about 18. It is less acidic than water. It can give up a proton, but the proton is much more likely to be bound to the oxygen than disscociated. Phenol has a $\mathrm{p}K_a$ of about 9 and is considered mildly acidic. cyclohexanecarboxylic acid has a $\mathrm{p}K_a$ of about 4.9; its acidity is similar to hydrofluoric acid.

Let's compare stability of the conjugate bases of cyclohexanol, phenol and cyclohexanecarboxylic acid:

In cyclohexanol, the anion formed by loss of a proton is localized on the oxygen. There is no resonance stabilization.
In phenol, the anion formed by loss of a proton is delocalized. Resonance structures show the negative charge can be shared between the oxygen atom and three of the carbons in the benzene ring.
In cyclohexanecarboxylic  acid, the anion is also delocalized. This time the negative charge is shared between two different oxygen atoms.

Resonance delocalization plays a clear role in stabilizing the conjugate base formed after loss of a proton. Despite the similar bond polarity in cyclohexanol and phenol, the proton is much more tightly bound in cyclohexanol. The greater stability of the phenolate anion compared to the cyclohexanoxide anion makes it easier to remove a proton from phenol than cyclohexanol.

The atoms onto which the charge is delocalized still plays a major role in determining stability. Delocalization onto an additional oxygen atom may be more stabilizing than delocalization onto several carbons.

For more details, please visit: http://employees.csbsju.edu/cschaller/principles%20chem/acidity/acid%20distal.htm


can I mix. trisodium phosphate, potassium hydroxide, and sodium hydroxide together?

I advise against using potassium or sodium hydroxide in homemade cleaning agents. You could easily end up with a mixture that could severely burn eyes (up to blindness), skin or lungs.

Also, grease from fast food restaurants is fundamentally different from gas station grease. The former consists most probably of fat while the latter is a mineral oil product.

Your home-made degreaser, with a hydroxide concentration that could be handled without complete personal protective equipment, might or might not have some effect on fat due to saponification but for removing mineral oil based grease I would not expect it to be effective.


What is the meaning of 'anion water' in coordination chemistry?

I’m pretty sure that ‘anion water’ is a term they use to describe water which coordinates the anions by hydrogen bonding rather than having anything to do with the cations. This is again to be separated from interstitial water, whose job it is to stabilise either of the other water types by hydrogen bonding without interacting with cations or anions at all. (Or maybe, whose job it is to fill voids in the crystal structure).

Ammonia in coordination compounds typically does what water does if it is not interstitial or hydrogen-bonding to anions, thus the distinction and the mention.


What is the boiling point and density of LSD?

Lysergic acid diethylamide doesn't boil. It decomposes. The density is $1.2 \pm 0.1~\mathrm{g~cm^{-3}}$.

Data can be found at  ChemSpider. There's a predicted boiling point of 541°C. However, lysergic acid decomposes at its melting point of about 240°C (CRC Handbook of Chemistry ad Physics, 64th ed.). This suggests that lysergic acid diethylamide does not have a real boiling point - at least at normal pressure.


Why NaBH4 does not reduce carboxylic acid while LiAlH4 does?

The answer is polarity.

Consider the two ions $\ce{AlH4-}$ and $\ce{BH4-}$ (as they exist as ions in a solution).

Now the electro-negativity of $\ce{Al}$ is $1.5$, and that of $\ce{B}$ is $2$, and that of $\ce{H}$ is $2.1$.

So the electro-negativity difference ($\Delta\mathrm{EN}$) between $\ce{Al-H}$ bond is $0.6$ and that of $\ce{B-H}$ bond is $0.1$. It is clearly visible that the $\Delta\mathrm{EN}$ of the $\ce{Al-H}$ bond is greater than that of the $\ce{B-H}$ bond.

Now considering the mechanism, due to the small polarity induced by the carbonyl group of the carboxylic acid, the $\ce{H}$ atom in $\ce{Al-H}$ bond is readily broken due to its high polarity, while the $\ce{H}$ in $\ce{B-H}$ can't do the same because it has a low polarity.

This is the reason why $\ce{NaBH4}$ cannot be used instead of $\ce{LiAlH4}$ for reducing carboxylic acids to alcohols.


How does a catalyst affect the rate equation?

Yes. It is possible for a catalyst to change the order of a reaction since it shows different path to the reaction i.e. reaction occurs through different mechanism. In the new path, rate determining step (slow step) may involve different number of reactant molecules.

For example, in case of reaction catalyzed on metal surfaces, most of the times, the rate does not depend on the concentration of the reactants at higher concentration since now it is surface area that governs the rate.


Silica turning transparent during column chromatography with chloroform

You are correct, sir!  (or madame)  The silicic acid becomes translucent (not transparent) in chloroform because their refractive indexes are similar.  This is a chance to check for air bubbles in the column, which reduces efficiency.  Also, any material still on your column will probably not have the same RI, and will be visible as a white or colored band, probably at the top of the column.


Cyanidine actually looks to me to be a rather highly polar compound. It's essentially a $\ce{C15}$ hydrocarbon having 5 hydroxyl groups and a formal positive charge.

The ratio of hydroxyl groups to the number of carbons is the same as that of water-miscible isopropyl alcohol. (Of course smaller molecules like isopropyl alcohol also tend to be more soluble than larger molecules.)

The most important feature indicative of it's rather high polarity and water solubility is it's formal positive charge. This is analogous to placing a positive charge on a large amine by protination in order to turn the amine into a polar, water soluble compound.


Usually you leave a reaction uncovered to induce crystallization, because this allows the solvent to evaporate. As solvent evaporates, the solubility of the product decreases, causing it to exit solution and begin to nucleate crystals. Once nucleation has occurred, crystals precipitate more readily.

Whether partial or totally uncovered isn't always important depending on how volatile the solvent is or how saturated the solution is. Partially covered will slow the rate, especially if the solvent isn't volatile, but it can block loose particles from falling in. On the other hand, a saturated solution doesn't need much evaporation to go below the solubility limit of your product and instigate nucleation. Whatever your case may be, a watchglass is not a sturdy seal against evaporation so it seems like your product was still able to crystallize.

Also, depending on the heat of the reaction and melting point of the product, sometimes evaporation isn't necessary at all, but it's still convenient to allow as much solvent as possible to evaporate in order to isolate dry crystals. If your solvent dries over night then you don't have to remove it the next day, so as a rule of thumb a reaction product is left uncovered unless there's something air/moisture sensitive present.


Most energy-dense commercial battery available

This has a list of common battery types with energy densities.

Lithium ion primary batteries appear to have the highest specific energy at around 300 Wh/kg. Lead-acid batteries have a specific energy around 10% that of lithium-ion batteries, but are extremely reliable (depth of discharge, self-discharge, etc.), and do not have nearly the same heating/fire issues (though they can explode if you mess with them), which is one of the reasons they have endured so long in automotive applications.

More exotic chemistries have higher specific energies, but I am unsure as to whether they are available to the average consumer.


What are the limitations of a waters of hydration experiment?

You can mention experimental details and how you achieved 1:4 ratio, which may explained if we know the maximum temperature of heating.  CuSO4.5H2O loses water of hydration in steps as a function of temperature showing that those five water molecules are not equivalent. In this figure from German Wiki, follow the green line (ignore blue). The y-axis is % loss in weight and x-axis is the temperature. It clearly shows that water is lost in steps. The labels read Mass change (loss of crystal water). If your heating temperature is not enough you are left with CuSO4.H2O.


How do I calculate how much sodium hydroxide I need to add to get a new pH?

At the beginning, the amount of substance of hydronium ions is:
$$n(\ce{H3O+})=10^{-3.7}  \times 100 \times10^{-3}=10^{-4.7} \mathrm {mol}$$
After adding sodium hydroxide, the amount of substance of hydronium ions is:
$$n(\ce{H3O+})=10^{-4.5}  \times 100 \times10^{-3}=10^{-5.5} \mathrm {mol}$$

The difference represents the amount of substance of hydronium ions that have reacted with hydroxide ions.
$$n(\ce{H3O+})=10^{-4.7}-10^{-5.5}= n(\ce{OH-})$$
The volume of sodium hydroxide to add is:
$$v(\ce{OH-})=\frac {10^{-4.7}-10^{-5.5}}{10}= 1.68\times10^{-6}\mathrm{L}=1.68\,\mu \mathrm{L} $$

Assuming the volume of a droplet to be about 60 micro liter, it is much less than a droplet!


What is the correct value of the Avogadro constant? And how was it derived?

Whenever you're looking for accurate fundamental physical constants, CODATA recommended values are the way to go. As of May 20, 2019, Avogadro's constant is now truly an exact value, with infinite significant digits. Behold!

$$6.022\,140\,76\times10^{23}\ \mathrm{mol^{-1}}$$

It is unlikely this value will change within our lifetimes, which is exciting in its own way!

For historical reasons, I'll leave my previous answer below:

As of 2015, the latest data for the Avogadro constant is from 2014. According to CODATA, the most accurate value is:

$$6.022\,140\,857\times10^{23}\ \mathrm{mol^{-1}}\pm0.000\,000\,074\times10^{23}\ \mathrm{mol^{-1}}\quad\text{(CODATA 2014)}$$

The relative uncertainty in the measurement is thus only 12 parts per billion!

Interestingly, the Avogadro constant may be redefined in the near future to be an exact value, that is, a constant with zero uncertainty by definition, much like the speed of light. This would come as a consequence of redefining the SI kilogram as a function of the number of atoms in an ultrapure monoisotopic $\ce{^{28}Si}$ monocrystalline sphere engineered to extreme precision. A great video on this can be found in the Veritasium YouTube channel.

All that said, I suspect you don't really have to care which constant should be used. All the suggested values differ by one part in a million, which makes essentially no difference for most chemistry.

Edit: As pointed out by Loong in the comments, a few weeks after writing this answer, CODATA released updated values for the physical constants, so I updated this answer for accuracy. The next set of updated values will likely be announced in 2018-2019. For comparison, the previous value was:

$$6.022\,141\,29\times10^{23}\ \mathrm{mol^{-1}}\pm0.000\,000\,27\times10^{23}\ \mathrm{mol^{-1}}\quad\text{(CODATA 2010)}$$

This represents an uncertainty of 44 parts per billion. This means the uncertainty in the measurement has been cut to almost a fourth of its previous value in four years. Go science!


Why do bubbles form when a reactive metal is placed in a less reactive metal's salt solution?

The bubbles are still hydrogen.  Transition metal salts are generally much weaker acids than whatever you put your magnesium ribbon into for your hydrogen control.  So you get less gas for the candle flame to react with when you use the transition metal salt, which I am sure you saw.  That the decreased amount of gas still popped, which is basically a low-intensity bang, should be deemed a positive test for hydrogen.


Succinic anhydride esterification won't go forward

The monoesters of succinic acid can be prepared using Steglich esterification method (Ref.1,2), which is specially suitable for alcohols with the $\ce{OH}$ functional group is in steric hindrance. In this method, dicyclohexylcarbodiimide (DCC) as the coupling reagent and 4-dimethylaminopyridine (DMAP) as a catalyst are used. The reference 1 gave a typical procedure as follows:

To a stirred solution of 10mmol carboxylic acid in $\pu{lO ml}$ anhydrous $\ce{CH2Cl2}$ (in the case of sparingly soluble acids, in DMF) is added $\pu{30-110 mg}$ DMAP and $\pu{20-40 mmol}$ alcohol or thiol ($\pu{10 mmol}$ with alcohols or thiols which are not easily removable without significant loss; $\pu{3.4 mmol}$ with glycerol). DCC is added to the reaction mixture at OT, which is then stirred for $\pu{5 min}$ at $\pu{0 ^\circ C}$ and $\pu{3 h}$ at $\pu{20 ^\circ C}$. Precipitated urea is then filtered off and the filtrate evaporated down in vacuo. The residue is taken up in $\ce{CH2Cl2}$ and, if necessary, filtered free of any further precipitated urea. The $\ce{CH2Cl2}$ solution is washed twice with $\pu{0.5 N}$ $\ce{HCl}$ and with saturated $\ce{NaHCO3}$ solution, and then dried over $\ce{MgSO4}$. The solvent is removed by evaporation and the ester isolated by distillation or recrystallization. Crystalline products can be obtained in pure form by filtration on a short silica gel column (eluent: $\ce{CH2Cl2}$ or $\ce{CHCl3}$).

For example, recently, this method has successfully used to couple Vitamin $\mathrm{D_2}$ to succinic anhydride in order to achieve succinic monoester of Vitamin $\mathrm{D_2}$ (Ref.3). However, there is an US patent publication revealing cheap solvent free method using alkali salt bases such as $\ce{NaOAc, KOAc, Na2CO3}$, and $\ce{K2CO3}$ as catalysts to efficiently prepare mono alkyl esters from cyclic anhydrides (Ref.4).
References:

Bernhard Neises, Wolfgang Steglich, "Simple Method for the Esterification of Carboxylic Acids," Angew. Chem. Int. Ed. 1978, 17(7), 522–524 (https://doi.org/10.1002/anie.197805221)(PDF).
Gerhard Höfle,  Wolfgang Steglich,  Helmut Vorbrüggen, "4‐Dialkylaminopyridines as Highly Active Acylation Catalysts. [New synthetic method (25)]," Angew. Chem. Int. Ed. 1978, 17(8), 569-583 (https://doi.org/10.1002/anie.197805691).
Victar Nyaarugwe, "Synthesis and Characterization of Hybrid Compounds for the Treatment of Bone Related Diseases," Journal of Trauma and Orthopedic Nursing 2018, 2(1:4), 16 pages (PDF).
Mark, B. Erman, Joe W. Snow, Mikhail Y. Lebedev, "Process for making monomenthyl esters," US Patent 2007, US 7,247,743 B1 (PDF).


What is the limiting value of band gap that makes a material semiconductor?

The transition of electron from the valence band to the conduction band is a probabilistic event.  Electrons get this excess energy equivalent to the band gap ($\Delta E$) from the thermal energy ($kT$, where $k$ is the Boltzmann constant and $T$, temperature in Kelvin). Hence, the probability of such an event will be proportional to $e^{-\Delta E/kT}$.

Answering your question, although there is no standard defined value of band gap, you can see since it's an exponentially falling smooth curve that one can define a cutoff $\Delta E$, saying that for band gap $> \Delta E$ there are practically no electrons transiting to the conduction band, qualifying the material as an insulator (i.e., exhibiting insulator-like properties).


You should be careful to distinguish between the magnitude of enthalpy (or some other change in energy) and the sign of the property:

the magnitude tells you how large the change was (yes, self-evidently)
the sign tells you in which direction the change occurred:

a) $\text{system}\rightarrow \text{surroundings}$, $\Delta H<0$, or exothermic
b) $\text{surroundings}\rightarrow \text{system}$, $\Delta H>0$, or endothermic

This is the typical convention used (not the only one).

In the case of the condensation of a gas, the magnitude can vary depending on the strength of intermolecular interactions in the liquid compared to the gas. The sign is negative because condensation at a constant temperature requires dissipation of heat to the surroundings.

  So which one is it, where am I going wrong ??

That the enthalpy of $\ce{H2O(l)}$ is less than the enthalpy of $\ce{H2O(g)}$ means in the scenario you present that
$$\Delta H_2 < \Delta H_1$$
not the other way around. You could also start from the fact that condensation is exothermic (or vaporization endothermic) so that $H_g-H_l=-\Delta_{cond}H =\Delta_{vap}H>0$$^\dagger$, from which $$\begin{align}H_g&>H_l\\ H_g-H_{reagents}&>H_l-H_{reagents}\\ \Delta H_1&>\Delta H_2\end{align}$$

$\dagger$ (Writing "$H_g$" is a bit "dangerous" because it suggests there is an absolute enthalpy scale (rather than relative) but you can assume that here you are using a common reference state for water).


How does Hofmann elimination work in this reaction?

Wow, this is an amazing question. The expected reactivity is strongly dependent on the exact structure.

For starters, trimethylammonium is about the size of a tert-butyl group. So, expect an A value around 4.9, which is very large. So, put that group equatorial on the ring.

Once you do this, you see that there are only 2 hydrogen atoms that are anti-periplanar for an E2 elimination. One is on the axial methyl group (red). The other is on the N-ethyl group (blue). Only these two can easily be deprotonated for elimination. Of these two hydrogen atoms, the one on the axial methyl is very sterically crowded. Specifically, any base has to get around the ring structure and the encroaching flag pole hydrogen atoms. On the other hand, the hydrogen atom on the ethyl group sees no such steric encumberance.

Of the three elimination products, I expect to see:

Almost exclusively ethylene
Maybe a small amount of the methylidenecyclohexane
No methylcyclohexene

EDIT:
@Martin-マーチン was kind enough to compute the structure below. Notice the structural features we've already discussed: the ring conformation and the position of the anti-periplanar hydrogen atoms.


Bicuculline is a chiral molecule.  That means that its mirror images are non-superimposable.  These non-superimposable images are termed "enantiomers."  When you see a "(+)" or "(-)" in front of a chemical name it tells you that only one the enantiomers is present (e.g. it is not a racemic mixture of the enantiomers).  Chiral molecules have the ability to rotate the plane of a polarized light beam that passes through it. Enantiomers will rotate the plane of polarization in exactly equal amounts (same magnitude) but in opposite directions.

Dextrorotary designated as (+), indicates clockwise rotation (to the right)

Levorotary designated as (-), indicates anti-clockwise rotation (to the left)


The equivalent weight of HNO3 in the following reaction

The term equivalent weight is used in conjuction with normality (which is quite obsolete, except in the industry). Anyway, whenever you determine equivalent weight, you have to know beforehand whether you are considering a given "molecule" as an acid or as an reducing/oxidizing agent. This is where the confusion sets in.

Nitric acid as an acid is a monoprotic acid: It's equivalent weight is the same as its formula weight

Calculating the equivalent weight of nitric acid as an oxidizing agent is another story, it depends on how the redox equation is written.

From your equation nitric acid is acting as an oxidizing agent because there is no nitrate ion on the right hand side and the oxidation state of N is different on both sides:

$\ce{HNO3}$ is converting to NO. In the answer above, you already have the equation

$\ce{ NO3−}$ +3e => NO

In this case, the equivalent weight of nitric acid is 1/3 (formula weight).

As you can see equivalent weight is very much dependent on the way you write the equation. This will come with practice.


If you have $n$ functions (e.g. AOs) you can make a maximum of $n$ new linearly independent functions (e.g. MOs).  If you try to make $n+1$ MOs, then any one of them can be expressed as a linear combination of the other $n$ MOs.

The most usual way to enforce linear independence is to enforce orthogonality, i.e. all your MOs have to have zero net overlap with each other. All the "correct" MOs you drew are orthogonal to each other, but the "incorrect" MOs are not orthogonal to any of the "correct" MOs.