Molecular orbital diagram and irreducible representations for dinitrogen

Because this is a diatomic molecule, there are no group orbitals. Put another way, the group orbitals are the molecular orbitals. Knowing the nitrogen atomic orbitals (AOs) and their irreducible representation (irrep) labels is enough.

Since we'll work in the $D_{\mathrm{2h}}$ point group, we need its character table:

$$\begin{array}{c|cccccccc|cc} \hline
D_\mathrm{2h} & E & C_2(z) & C_2(y) & C_2(x) & i & \sigma(xy) & \sigma(xz) & \sigma(yz) & & \\ \hline
\mathrm{A_g} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & & x^2,y^2,z^2 \\
\mathrm{B_{1g}} & 1 & 1 & -1 & -1 & 1 & 1 & -1 & -1 & R_z & xy \\
\mathrm{B_{2g}} & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & R_y & xz \\
\mathrm{B_{3g}} & 1 & -1 & -1 & 1 & 1 & -1 & -1 & 1 & R_x & yz \\
\mathrm{A_u} & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & & \\
\mathrm{B_{1u}} & 1 & 1 & -1 & -1 & -1 & -1 & 1 & 1 & z & \\
\mathrm{B_{2u}} & 1 & -1 & 1 & -1 & -1 & 1 & -1 & 1 & y & \\
\mathrm{B_{3u}} & 1 & -1 & -1 & 1 & -1 & 1 & 1 & -1 & x & \\ \hline
\end{array}$$

Using the nitrogen atom's electron configuration $\ce{1s^{2} 2s^{2} 2p^{3}}$ as our minimal basis, no d-orbitals will be present, so we can assign irreps to each AO right away, in part because we assume the principal rotation axis (the one of highest order) is aligned along the z-axis:

$$\begin{array}{cccc}
\hline
\mathrm{s}     & \mathrm{p}_x    & \mathrm{p}_y    & \mathrm{p}_z    \\
\hline
\mathrm{A_{g}} & \mathrm{B_{3u}} & \mathrm{B_{2u}} & \mathrm{B_{1u}} \\
\hline
\end{array}$$

Then, proceed by starting to form the standard MO diagram for a diatomic, but add the irrep labels to each AO:

Note that I haven't spaced the energy levels properly; the $\ce{1s}$ should be much lower than it is relative to the $\ce{2s}$. More on this later. Regardless, because this is a homodiatomic, all AOs will mix at each energy level, even the core $\ce{1s}$ AOs. Then, attempt to form the MOs:

This is probably wrong, but it's a starting point. Again, energy levels will be discussed later. Finally, add symmetry labels to each MO, remembering that

MO symmetry is derived from AO symmetry,
numbering is consecutive within each irrep, not with respect to the set of all MOs, and
lowercase is used to signify that these are MOs; uppercase is for the irreps themselves.

This is our final MO diagram for $\ce{N2}$. To form the diagram for $\ce{N2^+}$, remove an electron from $\mathrm{4a_g}$. This ignores orbital relaxation effects, but for the purposes of working this out on paper, it should be fine.

Now for the matter of relative and absolute energy levels. It is probably possible to get the correct relative ordering of the MO energy levels. Here, I assume that since AO mixes with an identical partner on the other atom, the splitting for $\ce{1s}$ would be the same as $\ce{2s},~\ce{2p_z}$, etc. Since I drew the $\ce{1s}$ too high, the $\mathrm{3a_g}$ is almost certainly too high, and perhaps should even go below what is labeled as $\mathrm{2a_g}$. The way to confirm this, and the only way to get absolute energy levels, is to perform a quantum chemical calculation. Since we've used a minimal basis for the drawing, we'll stick a minimal basis in the calculation. Here is a Psi4 input file:

molecule {
    N 0.0 0.0 0.0
    N 0.0 0.0 1.0975
}

set {
    basis sto-3g
    scf_type direct
    df_scf_guess false
    cubeprop_orbitals [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
}

e, wfn = energy('hf', return_wfn=True)
cubeprop(wfn)

and from its output:

Orbital Energies (a.u.)
-----------------------

Doubly Occupied:

   1Ag   -15.518067     1B1u  -15.516124     2Ag    -1.442840
   2B1u   -0.722491     1B2u   -0.573123     1B3u   -0.573123
   3Ag    -0.539495

Virtual:

   1B2g    0.281319     1B3g    0.281319     3B1u    1.123476

It looks like I might have messed up the diagram, because the ordering isn't what's expected (why isn't $\mathrm{B_{1u}}$ degenerate with the other two p-orbitals?), and the virtual degenerate p-orbitals have gerade symmetry. Time to plot!

In my haste, I forgot an important point. When you make the MOs from AOs, you're making two linear combinations:

\begin{align}
\psi_{s} &= \frac{1}{\sqrt{2}} (\chi_{l} + \chi_{r}) \\
\psi_{a} &= \frac{1}{\sqrt{2}} (\chi_{l} - \chi_{r}),
\end{align}

which means that an antisymmetric combination of two s orbitals will look like a $\mathrm{p}_z$ orbital. This doesn't fully explain everything, but is a good starting point. The lesson up to now is that while drawing the diagrams by hand is a useful exercise, it is not enough to do only that when the goal is to perform a correlated ab initio electronic structure calculation.


Molar heat of combustion with different fuels

This question is asking which substance has the greatest heat density, or energy released per mass combusted (something like J/g).

$\ce{C2H2}$ has a mass of $26.04\ \mathrm{g/mol}$. Since the $\Delta H$ in its combustion is $1\,630\ \mathrm{kJ/mol}$, you will be looking at a heat density of $\frac{1\,630\ \mathrm{kJ/mol}}{26.04\ \mathrm{g/mol}} = 62.6\ \mathrm{\frac{kJ}{g}}.$

Similarly, for $\ce{C6H14}$, you get $\frac{1\,300\ \mathrm{kJ/mol}}{86.178\ \mathrm{g/mol}} = 15.1\ \mathrm{\frac{kJ}{g}}.$

For $\ce{C2H6}$, you get $\frac{1\,560\ \mathrm{kJ/mol}}{30.07\ \mathrm{g/mol}} = 51.9\ \mathrm{\frac{kJ}{g}}.$

It looks like you were on the right track.


I don't have a full recipe, but if you're fine with black ink, try to boil some crushed or grounded organic material containing large amounts of gallic acid (or other hydrolyzable tannins) with rusty iron and some vinegar (acetic acid).

Suitable organic materials that come to my mind are oak bark or green tea.

You should end up with an ink that darkens over time. Note that this ink will probably clog fountain pens but should be fine for calligraphic pens.


How can I identify whether a substance is KOH or NaOH?

A flame test should tell you pretty quickly. Make a small wire loop (preferably steel or nichrome), heat in a bunsen burner or propane torch flame until red hot and all the contaminants are burned off, then dip it in your solution and put it in the flame. If sodium is present, you should get a bright yellow flame, and if it is potassium, the flame will be a dim purple (lilac) color. But a small amount of sodium present can mask that color.


How to draw the 2D representation of ammonia molecule

According to the rules, yes, Ammonia can be drawn with three of it's atoms in one plane and the other outside the plane. Here's an example, along with a 3D projection:

However, what do you comprehend from this structure? This structure does not convey the trigonal pyramidal geometry of the molecule very convincingly. It might be mistaken for trigonal planar at a glance, and this defeats the purpose of using the wedge/dash notation. Ammonia (and several other tetrahedral molecules) are represented in that manner since it is easiest to interpret their 3d structure from a 2d image.


The halide ion is involved, indirectly.  The hydrogen halide has to protonate the alcohol first, and this requires dissociation of the proton from  the halide ion.  Chloride ion is more strongly basic than iodide ion, an effect we do not see in water (leveling effect) but can see it in the reaction with the alcohol.  So we can protonate more of the alcohol at any time, with hydrogen iodide, making more opportunities to then form the carbocation, than with hydrogen chloride.

Hydrogen fluoride would be way off the scale here, in the direction of less reactivity.  It is only a weak acid in water, let alone in this reaction mechanism.  It's one manifestation of fluorination being more difficult than other halogenations.


How to find the vitamin C content of orange juice?

Iodine and iodide will form triiodide in aqeous solution. This is a common way to increase the solubility of $\ce{I2}$ in water. For a titration, one will use excess $\ce{KI}$ and control the formation of iodide by adding $\ce{KIO3}$ using a burette.

In response to the comment:

$\ce{KI}$ contains iodide anions. I strongly suspect that an excess of $\ce{KI}$ was used (more than fivefold compared to $\ce{KIO3}$), leaving enough iodide anions to form the triiodide. Also note that effectively a $\ce{I-}$ can be subtracted from either side of the second equation of the question.


Why hot water is more effective than cold water in cleaning

Many physical and chemical reactions are affected by temperature. Usually higher temperature causes the reaction to occur faster, and lower temperature will slow it down.
Water solubility (the ability of substances to go into solution in water, i.e. bond with water molecules) is a reaction that is affected by temperature for many substances. Changes in temperature may increase or decrease the solubility of a certain substance in water, depending on the substance properties.

Since high temperature will cause water molecules to move around relatively faster, it will also cause the hydrogen bonds between them to form and break at a higher rate (that is why hot water vaporizes more quickly than cold water). In addition, the high temperature increases the water's energy, and this energy may be used to break bonds between dirt and curtains for which colder water will not have enough energy.
This means that the detergent/water solution will be able to dissolve (form bonds with) the dirt faster and better, because the bonds between the detergent and water and the bonds between the detergent and dirt will form faster and will be strong enough to prevent the dirt from reforming the bonds with the curtains. As mentioned earlier, these bonds will also break faster, but the presence of a detergent helps to stabilize the solution and keep the dirt dissolved in the water. Once the dirt is dissolved in the water, the curtains are clean, and the dirty water can be easily discarded.

In summary, hot water helps the cleaning process (i.e. moving the dirt from the curtains to the water) because it can dissolve the dirt faster and break strong bonds (up to a limit, of course) of dirt with the curtains.

It is worth mentioning that between the detergent and the hot water, the detergent contributes more to the dissolution of dirt in the water than does hot water alone (that is the same reason it is better to wash your hands with soap and water than with water alone). However, for optimal results, using both detergent and hot water will clean the curtains better (and faster).

Here are some images I found online that may help clarify this point:


Is a Racemic Mixture always an equi-molar solution of two enantiomers, both in the same state?

Yes and yes. A racemic mixture is equimolar in both enantiomers. Otherwise you would call the reaction enantioselective with a so-and-so selectivity.
A racemic mixture contains both enantiomers in the same state, as energy differences between enantiomers are so minuscule that you would never get one solid and one liquid at the same time.

  Is it necessary for both the mixture to be liquid in nature, or does racemisation also occur in the gaseous or solid state?

Racemisation is the process where you start out with a pure enantiomer and in the end get a mixture. For complete racemisation you end up with an equimolar mixture as discussed above.

It is thus possible that racemisation occurs in gaseous and solid phases, not only in solution or liquid. However, the reaction times will differ significantly. I could imagine that the solid racemisation reaction will take a long time.

  Can the two enantiomers still form a Racemic mixture if they are initially at different temperatures?

Higher temperatures just mean that the racemisation process proceeds faster. In the end, the warmer enantiomer will have racemised further on than the colder one, but for $t\to\infty$ the result will be the same.


Yes, the only crystal system that is guaranteed not to be birefringent is cubic, because all of the other crystal systems have at least one axis distinct from the others. What matters is the symmetry not the centring, so this is equally true for face-centred or body-centred as for primitive crystals.

Sadly, however, there is no way to determine the crystal structure, or even the crystal system, simply from consideration of the formula - indeed, crystal structure prediction and prediction of polymorphism are active research fields. Counterexamples to your first generalisation include ammonium chloride, bromide, and iodide; to your second include ammonium fluoride and silver iodide.


Calculate Gibbs energy of dinitrogen pentoxide using its vapor pressure and sublimation

First, here is a way to solve it that avoids calculating entropy:

Calculate the enthalpy from the given data.
Calculate the vapor pressure at 25 $^\circ$C. This is equal to the equilibrium constant for the process.
Calculate $\Delta G^\circ$ as -RT ln(K).

This way is efficient because you already set up the Clapeyron equation in the first step and you just use it again in the second. You could even solve 1. and 2. graphically, and then go from there.

  I got within a few significant figures when I used Tsublimation.

You need to calculate $\Delta S^\circ$, i.e. the change in entropy when all species are at standard state. At the normal temperature of sublimation, $\ce{N2O5(g)}$ is at the standard state, so that is the temperature you need to use. The enthalpy is not concentration dependent, but the entropy and the Gibbs free energy are.

  I am not sure if I even solved the problem correctly[...]

You solved it correctly in that you got the correct answer. You did not solve it as the problem author intended, I suspect, because you had to look up the normal sublimation point (unless you used the Clapeyron equation to estimate it).


How to convert the D'Arcy and Watt model from activity to concentration in mol/m^3

The general relation between activity (effective concentration) and concentration is as follows (see Wikipedia for more detail):
$$a_i = \gamma_i \frac{c_i}{c^\circ}$$
The standard concentration is defined as $c^\circ=1~\pu{mol dm^-3}$. For extremely dilute systems it is possible to just use the concentration instead of the activity (often used as a first order approximation, too):
$$\lim_{c_i\to0}\gamma_i\approx1 \implies a_i\to c_i$$
Depending on the accuracy necessary for your model, you might have to look up the specific values.

In any case you should again look up the definition of activity (or relative humidity) in D'Arcy and Watt's paper, because the activity defined for the above relation could be larger than 1.


Why are nitrogenous hydrocarbons sometimes colorful?

Introduction from Color Introduction

"What causes a compound to appear as being colored? The answer to this question lies in the way that electromagnetic energy interacts with matter. The portion of the electromagnetic spectrum that is of interest to this discussion is the visible portion, or what we call light. This visible light includes all the colors that we perceive: violet, blue, green, yellow, orange, red, and the various combinations of these. Although light can interact with matter in several different ways, this discussion will limit the modes of interaction to absorption and reflection."
In most cases color comes from the excitation of electrons due to an absorption of energy performed by the chemical. But - there is still ... :)
"(...) The vast majority of simple inorganic (e.g. sodium chloride) and organic compounds (e.g. ethanol) are colorless. Transition metal compounds are often colored because of transitions of electrons between d-orbitals of different energy.
Organic compounds tend to be colored when there is extensive conjugation, causing the energy gap between the HOMO and LUMO to decrease, bringing the absorption band from the UV to the visible region. Similarly, color is due to the energy absorbed by the compound, when an electron transitions from the HOMO to the LUMO.(...)" Wiki - color of chemicals

You can also assume that if a compound has double or triple conjugated (2 or more) bonds it can be colorful.
To answer your question: in alanine there is almost no chance to make double bond in amine group (e.g. $\ce{=NH}$), but nitrogen group ($\ce{O-N=O}$) has a capability to conjunct with other double bonds (can be also delocalized electrons or pi electrons - e.g., nitrobenzene)


Multi-step organic synthesis using benzene

@Waylander has provided a concise route to the nitro compound in question. I am also in accord with his comments regarding the OP's effort and his assessment that benzene, as a starting material, is not a choice by a practicing chemist. Now that he has opened the door to a solution, I offer a more traditional approach.
Classical nitration of ketoacid 7 should be reasonably selective in that the introduction of the nitro group in 8 is both para and meta directed. Keto acid 7 arises from indene 6 via exhaustive oxidation. To access the indene, I would start with benzaldehyde (2), whose preparations from benzene (1) are numerous. The Perkin condensation 2 $\rightarrow$ 3 is known. Hydrogenation of 3 forms carboxylic acid 4, which can be cyclized to indanone 5 with polyphosphoric acid. Friedel-Crafts techniques may also be employed. Reduction of this ketone followed by dehydration leads to the critical indene 6.


What are the factors behind making ice faster?

The simplest explanation would be that it isn't (pure) water at all. It is most likely a supersaturated solution of sodium acetate - $\ce{CH3COONa}$ (or sometimes sodium thiosulfate - $\ce{Na2S2O3}$) in water.
A supersaturated solution is a solution that contains more solute than normally soluble at those conditions. So as soon as a rough surface is provided for nucleation (formation of crystals), the solute starts crystallizing and comes out of solution as a big crystal.
The rough surface used over here is most probably the rough end of the burnt straw. This produces enough disturbance to cause crystallization. Usually an already prepared crystal of pure sodium acetate is used for nucleation but only if you want high purity crystals.
This is a pretty common trick (also the original video is from trick life -- as mentioned on the youtube description --) and most people who don't know about this phenomenon are usually easily tricked (then again, the "ice" formed would actually be hot since it is an exothermic reaction but you can't feel it through a video right :) ).
See this page and this page if you want more info.


Why apart from when building atoms in the first place, the 3d is the lower energy orbital?

In P element, the energy of 3d is lower than 4s. Look at the link provided below. Only in K and Ca, 4s has lower energy than 3d.

http://www.cdeep.iitb.ac.in/nptel/Core%20Science/Engineering%20Chemistry%201/Slide/lect6/6_6.htm

Another reason could be because the symmetry of hybridized orbital with d orbital will be more preferred in bonding with that of s orbital, and the molecule obtain more energy compensation with sp3d over sp3s.


Can I anodize an aluminum coating on steel to have better corrosion inhibition for the steel?

The aluminum oxide layer formed by anodizing will have substantially higher corrosion and abrasion resistance than non-anodized aluminum.  For most applications, simple Type II sulfuric acid anodizing is sufficient.  For maximum protection, Type III hard anodizing will yield a much thicker coating with substantially higher corrosion and abrasion resistance. If you require a thinner film, the older Type I chromic acid anodizing process may be used.

If you plan to paint or dye the part, you will most likely need to stick to Type II sulfuric acid anodizing or Type I chromic acid anodizing.  Type III hard anodizing, depending upon the alloy, will result in a dark grey to an almost black finish, so dyes will not show. Paint adhesion is also an issue with Type III.

See MIL-A-8625 for more complete information about the various types of aluminum anodizing.


Why are there two structures of L-Glucose in the Fischer projection?

It looks to me like the left one is just wrong.
It looks like who made the image took D-glucose and just flipped the 5-OH.

The result? The molecule at left is not glucose, but L-Idose!


Is it possible to react chlorine with hydrogen under a UV lamp during the electrolysis of salty water?

I advise you not to attempt a reaction of hydrogen and chlorine gas by light!

Do not attempt it unless you know exactly what you are doing and have seen the experiment be performed by others!

You’re actually rather lucky that one of the following happened:

the black light gave the wrong wavelength
there were too small concentrations of hydrogen gas and chlorine gas to react with each other.

The reaction of hydrogen and chlorine is a radical chain reaction and rather exothermic. There is no change in the amount of substance involved, so entropic effects do not play a role. However, it is an explosive reaction and was often used in the introductory inorganic chemistry course at my university to give off a loud bang (and send the tin can in which the gases were mixed flying).

The reaction is triggered by light energy interacting with chlorine; the first (initiation) step is the generation of chlorine radicals by the following equation:

$$\ce{Cl2 ->[$h\nu$] 2 Cl^.}\tag{1}$$

This chain then propagates according to reactions $(2)$ and $(3)$.

$$\ce{Cl^. + H2 -> HCl + H^.}\tag{2}$$

$$\ce{H^. + Cl2 -> HCl + Cl^.}\tag{3}$$

The light needed for the first step must supply chlorine’s bond dissociation energy; blue (visible) light, not UV light is actually required. But just for your safety please let me restate:

The reaction is explosive! Do not attempt it at home under any circumstance! Do not attempt it unless you know exactly what you are doing!


Why does CO2 lowers the pH of water below 7?

Carbon Dioxide (CO2) readily dissolve in water and form Carbonic Acid (i.e H2CO3 (aq) )

This is the formation of bonds.

Then Carbonic Acid (i.e H2CO3  (aq)  ) dissociate in water as follows.

So water gets H+ ions, so that cause water acidic.

The following shows dissociation of Carbonic Acid (i.e H2CO3 (aq) ) more clearly.

Carbon Monoxide (CO) do not readily react with water in room temperature in standard conditions. So it do not dissolve in water.


However, in my opinion, there should be a proportionality between the concentration and the peak intensity, not the peak area.

There is a proportionality between both peak area vs. concentration and peak height vs. concentration.

Peak height is proportional to the instantaneous amount of analyte that is transiting the detector.
Peak area is proportional to the sum of all of analyte moleucles that have transited the detector.
From 1 and 2 you might be able to infer the relationship betweeen peak height $h$ and area $A$: $$A = \int{h(t)\;dt}$$
People are usually interested in the total amount of substance injected into the column.  (If they know the injection volume, they can calculate the concentration from this value.)  The total amount of substance would be calculated from the peak area.
The maximal peak height is proportional to the peak area, but only if the peak "shape" is constant.  Here are some scenarios where peak shape will not be constant:

You want to compare injection #2 you made on your column two years ago to injection #2000 that you made recently.  Due to column degradation, the recent injections have much more noticeable peak tailing.  Because the longer tail leads to wider, asymmetric peaks, there will be less total height at the peak maximum relative to "better", symmetric peaks.
You change the flow rate of the HPLC.  All peaks are narrower, and thus, higher.
The scan rate of your detector changes, and the detector reports "counts" or "intensity" rather than counts per time.  This is actually very common for mass spectrometers, but less common for absorbance detectors.  If you scan twice as fast in MS, your peak heights go down by approximately twofold, but you have data points twice as often, so the peak area is relatively unchanged.
Your detector undersamples the peak.  Say your chromatography is nice and the "real" peak shape is nicely gaussian, but is ~ten seconds wide and your detector only gives you a data point every two seconds.  Say that due to very small random drifts in retention time, on some injections one of the data points coincides exactly with the maximum of the "real" peak, but in others, it is a little bit off.  The peak heights in this scenario will vary considerably more than the peak areas. (If the recorded maximum is off-center from the true maximum, there will be two data points that are higher, i.e. "closer" to the maximum than if the data maximum coincides with the true maximum, so integration will partially correct this error.)  Essentially, the peak maximum is a single-point sample from the gaussian distribution of the peak, while the area is a several-point sample from the gaussian distribution, so it has better sampling properties.

  Is there any mathematical relation between the peak area and the concentration?

Yes, there is, but it depends on the peak shape.  For perfectly gaussian peaks, $$h_{max} = \frac{A}{\sigma \sqrt{\tau}}$$

where $\tau$ is $2 \pi$ and $\sigma$ is the width of the peak, which is related to the full-width at half-maximum of the peak by $\mathrm{FWHM} = 2 \sigma \sqrt{2 \ln 2}$.

However, for non-gaussian peaks, this relationship does not hold.


You can attempt to use a copper item as the anode in a vinegar solution; the $\ce{Cu}$ is oxidised to $\ce{Cu^2+}$ and the acidity of the vinegar prevents the precipitation of $\ce{Cu(OH)2}$. The copper acetate formed in solution will migrate to the cathode, where you can place your item to be plated.

However, as my own experiment has shown, too high a current will result in a black deposit, not pink as the copper crystals are plated to quickly to form large crystals and thus scatter light. So use a low current!


Equilibrium constant for esterification process

The key point is pure liquids. In context of the esterification equilibrium, it considers the reactant concentration in the common solvent - which may or may not be  one of the reactants.

Note that if the solvent is at the same time a reactant or a product, which concentration (activity) can be considered constant, then there can be defined the equilibrium constant for this solvent, with the solvent excluded from the equilibrium.

$$K_{aqua} = \frac{[\ce{CH3COOCH2CH3}]}{[\ce{CH3COOH}][\ce{CH3CH2OH}]}$$

$$K_{ethanol} = \frac{[\ce{CH3COOCH2CH3}][\ce{H2O}]}{[\ce{CH3COOH}]}$$

$$K_{ethylacetate} = \frac{[\ce{H2O}]}{[\ce{CH3COOH}][\ce{CH3CH2OH}]}$$

$$K_{acetic acid} = \frac{[\ce{CH3COOCH2CH3}][\ce{H2O}]}{[\ce{CH3CH2OH}]}$$


How to determine the packing geometry of compounds with contradicting coordination number and radius-ratio values?

You've calculated your radius ratio value and you know you're between 0.414 and 0.732, so cubic is off the table, based on your chart.

You're looking at either square planar or octahedral. Your chart also says that ionic compounds do not use square planar packing and the coordination number would have to be 4. LiBr is an ionic compound (like NaCl) with a coordination number of 6, therefore you're looking at octahedral.

From there you can determine your "specific" answer.


How do you figure out the positional and chain isomer possibilities of a compound?

Is the parent hydrocarbon unbranched? That would be n-butane, $\ce{C4H10}$. Replacing one hydrogen atom by chlorine would lead to $\ce{C4H9Cl}$. Apparently, you're not having that!
Move a methyl group in n-butane, yielding 2-methylpropane, $\ce{C4H9}$. Here, replacing one hydrogen atom by chlorine gives $\ce{C4H8Cl}$. You're not having that either!
Does your target molecule have $\ce{C=C} $double bonds? Calculate the number of double bond equivalents in $\ce{C4H7Cl}$! I count one.

Work your way down from here.


Does 1-bromo-4-chlorobenzene have a permanent dipole moment?

As others have stated before in the comments, the molecule itself has a dipole moment, but it is indeed very, very small, to the point where you would say it is negligible. To illustrate this, I have performed a few quick calculations, rounded to two decimal places.
Remember that the dipole is defined from negative to positive pole, i.e. $\displaystyle \ominus \xrightarrow{~\mathbf{q}~}\oplus$.
\begin{array}{lr}\hline
\text{Method}  & \mathbf{q}(\ce{Br-C4H4-Cl})~/~\mathrm{D}\\\hline
\text{BP86/def2-SVP}           &     0.07\\
\text{BP86/def2-TZVPP}         &     0.11\\
\text{B3LYP/def2-TZVPP}        &     0.09\\
\text{B3LYP+D3(BJ)/def2-TZVPP} &     0.09\\
\text{PBE0/def2-TZVPP}         &     0.10\\
\text{M11/def2-TZVPP}          &     0.10\\
\text{B2PLYP/def2-TZVPP}       &     0.09\\
\text{MP2/def2-TZVPP}          &     0.06\\
%\text{CCD//MP2/def2-TZVPP} &     ----\\
\hline\end{array}

In this case the dipole is directed towards the chlorine. Here are the NBO (natural bond orbital) charges for the CCSD(T)//MP2/def-TZVPP level of theory:

We can see that the chlorine (light green) is essentially neutral, while the bromine (dark green) is slightly positive polarised. From this we would assume the dipole moment to be directed towards the bromine. This is also what you would expect from the electronegativity differences.
Slightly different values are obtained from an QTAIM (quantum theory of atoms in molecules) analysis on the MP2/def2-TZVPP level of theory, where the volume is cut off at an electron density smaller than 0.001.

 Normalization factor of the integral of electron density is    1.000001
 The atomic charges after normalization and atomic volumes:
      1 (C )    Charge:    0.071729     Volume:    68.316 Bohr^3
      2 (C )    Charge:    0.078474     Volume:    79.386 Bohr^3
      3 (C )    Charge:    0.088508     Volume:    78.816 Bohr^3
      4 (C )    Charge:   -0.122884     Volume:    73.483 Bohr^3
      5 (C )    Charge:    0.088513     Volume:    78.816 Bohr^3
      6 (C )    Charge:    0.078474     Volume:    79.386 Bohr^3
      7 (H )    Charge:    0.026701     Volume:    46.501 Bohr^3
      8 (H )    Charge:    0.025890     Volume:    46.475 Bohr^3
      9 (Br)    Charge:   -0.097472     Volume:   260.461 Bohr^3
     10 (H )    Charge:    0.025886     Volume:    46.475 Bohr^3
     11 (H )    Charge:    0.026702     Volume:    46.501 Bohr^3
     12 (Cl)    Charge:   -0.290521     Volume:   214.364 Bohr^3

However, the calculated dipole moment points in the opposite direction. Reasons for this are similar to the case of carbon monoxide (see here). In short: it is basically because bromine is larger and more diffuse than chlorine. The σ-lone pairs, which are mainly responsible for the direction of the dipole moment, are further away from the atomic centres of the molecule for bromine compared to chlorine.
This causes a larger charge difference than the polarisation of the bonding structure.

Above graphic corresponds to the QTAIM analysis, the bold blue line represents the electron density contour of 0.001.


Why are these two methods of converting from ppm to g/m3 equivalent?

Using the ideal gas law
$$
pV = nRT = \frac{m}{M}RT
$$
you can express the density $\rho = \frac m V$ in terms of molar mass, pressure, ideal gas constant, and temperature:
$$
\rho = \frac{pM}{RT}
$$
With the units you use in your second example, this yields a density in $\mathrm{g\,m^{-3}}$, whereas you insert a density in $\mathrm{kg\,m^{-3}}$ into your first equation. Thus, I will write your first equation as
$$
C[\mathrm{g\,m^{-3}}] = \frac{C[\mathrm{ppm}] \cdot \rho}{10^6}
$$
where the density is in $\mathrm{g\,m^{-3}}$.
Inserting the expression for $\rho$ derived above into this equation, we get
$$
C[\mathrm{g\,m^{-3}}] = \frac{C[\mathrm{ppm}] \cdot \rho}{10^6} = \frac{C[\mathrm{ppm}]}{10^6}\frac{pM}{RT}
$$
which is identical to your second equation.


Accuracy of FP-LAPW vs Plane Wave Pseudopotential: examples

In all cases FP-LAPW will be better than Pseudopotential + Plane Waves calculations. However, FP-LAPW calculations are better at calculating properties which involve core electrons (e.g. core level shifts, hyperfine structures). Such calculations are often not possible with Pseudopotentials + Plane Waves, or are more complicated.This is because PP+ PW calculations do not treat core electrons explicitly.  Furthermore,  FP-LAPW calculations do not depend on pseupotentials. Although one can test pseudopotentials to check for transferability, there is no guarantee that the pseudopotential works for the system that you are simulating. As a result, "artifacts" due to pseudopotentials can be avoided with FP LAPW.


Yes, the picture should be sufficient to decide between viscose and cotton; see photo of cotton at http://whrhs-forensic-chem-spring-2010.wikispaces.com/Fiber+Analysis?responseToken=906bf6a76a5e268a00e54c8bce6927ba.

Recall that cotton was the cell wall of a plant cell, so it retains some of the shape of the cell. (Silk, on the other hand, is excreted from spinnerets, as are man-made fibers, so it looks more like viscose).


Can there be any substance whose number of electrons and protons are different at neutral state

It's certainly possible, you just need other charged particles that aren't electrons or protons.

For example, a positron is positively charged, so something that's made out of an electron plus a positron would work. See e.g. https://en.wikipedia.org/wiki/Onium, or this answer on SE, for a few examples.


You are not wrong in your understanding of a water softener, nor are you wrong in your understanding of a reverse osmosis filtration system.

  Does the RO filter care whether the supply water is salty or hard?

I would like to make the point that your RO filter is incapable of caring about the hardness of the water. This is more about carefully choosing your words more than anything else. I was also tripped up by this when I first began chemistry, but it was something I learned to do and I challenge you to do the same.

  Is it more effective at filtering one or the other?

Yes, but only slightly. Here you can find a list of the typical rejection characteristics of RO filters. The lower bound for the percent amount of multivalent ions ($\ce{Ca^{2+}}$ and $\ce{Mg^{2+}}$) filtered is higher than the lower bound for the percent amount of monovalent ions ($\ce{Na+}$ and $\ce{K+}$) filtered, but the upper bounds are very close in value.

Overall you could say that RO systems are more efficient at filtering hard water, but the difference is not very large. Most systems will likely be performing considerably above the lower bounds of the typical rejection characteristics anyway, so this should not be a deciding factor in deciding to feed your RO hard or soft water.

  Is one going to influence the life of the RO filter more than the other?

Yes! In an ideal RO filtration system, all of the filtered materials would simply drain away, and pure water would flow to the storage tank, but nothing is ideal. If you feed hard water into your RO system, there will be scaling on the semi-permeable membrane that will eventually reduce its filtered water output. This can cause lack of flow will create a potentially damaging buildup of pressure elsewhere in the system. I would therefore recommend that you keep your setup the way it is!


What is the reason for including excited states in the wave function in post-Hartree Fock methods?

I think that here it is not accurate to say "starting from the Hartree-Fock picture", instead of it, I find better to say "starting from the Hartree-Fock method". When this method is carried out we get a lot of orbitals and that allow us to construct a lot of Slater determinants, even if, we can be only interested in one of them.

Given a complete set of functions $\{f_i(x)\}$, any function $g(x)$ can be thought as a linear combination
$$g(x) = \sum c_i f_i(x).$$

For a two variable function $g(x_1, x_2)$, fixing $x_2$ we have
$$g(x_1, x_2) = \sum c_i(x_2)f_i(x_1).$$
But as $\{f\}$ is a complete set:
\begin{align}
c_i(x_2) &= \sum d_{ij} f_j(x_2) &\text{and}&&
g(x_1,x_2) &= \sum d_{ij}f_i(x_1)f_j(x_2).
\end{align}

This reasoning is trivially extended to a function of $n$ variables.

For an $n$-fermions system, the wave function depends of $n$ spin-orbitals ($n$ variables). To get a mathematic valid form, we only need to antisymmetrize the generalization of the expressions above.

Take in mind that none of these determinants are the true state (they are not eigenfunctions of the true hamiltonian). The same happens with the orbitals (that are eigenvectors of the Fock operator).

So, the use of excited determinants arises naturally from a math view point. Just because we already have functions for the determinantal expansion, and we know that they are pretty good in some sense, because the are the best we can get at least for one determinant (the Hartree Fock solution).

It is not needed to think configuration interaction as a mixture of stationary quantum states (ground and excited states), as said above they truly aren't.

Direct to your question: We can think the problem just as a decomposition of the wave function, somewhat like Fourier series, where the functions involved are the Slater determinants, just because it is convenient. But, this doesn't imply a physical meaning.

Response to comment:

  Also this discussion is quite old, I would like to add that the term "picture change" is frequently used in relativistic quantum chemistry, when (effective) Hamiltonians are compared to each other. E.g., when approximating the four-component (4c) Dirac equation by an effective two-component (2c) formalism, one usually introduces further approximations. In this respect, "starting from the Hartree-Fock picture" is a correct terminology, as it implies certain approximations on the Hamiltonian. However, the better way is usually to end up with a certain picture, compared to starting from it... (TheFox)

I agree: By choosing an effective Hamiltonian one is working on a class of approximations that build up a picture for a set of analogous systems. When considering correlation we have in mind a "picture" of the system. Being that H-F is a mean field method, it belong to some picture, but I feel that when dealing with Coupled Cluster the H-F method is just a part of a bigger procedure consisting in finding some functions in order to obtain a solution for another picture. In such sense, while the H-F method itself can be understood as the method for obtaining a solution of the H-F picture, in this context is just part of method for solving a problem in another picture. I added the word here for strictness.


Computational program not using full cpu power

Most applications in theoretical chemistry are memory and or memory bandwidth bound. Especially due to shared memory bandwidth it can be bad for your performance if you use too many processes.

In addition you benefit from additional processes only if your algorithm is more or less parallel. If there are too many serial parts/communication overhead you will have idle cores, if your process number is too high. (Amdahl's law)

Take home message:

Time the execution of your code with a non trivial example.
More is not always better.


Why are transition metals said to have different valencies?

I think it's kind of a poor use of terms here. When talking about transition metals, the valency generally refers to the oxidation state of the metal atom. E.g: hexavalent chromium is $\ce{Cr^6+}$, not necessarily a chromium bonded to six other things.


First question

From P.248, Rudiments of Chemistry:

  'p' block elements contain both metal and non-metal and we observe a gradation of properties (such as electropositivity, electronegativity, reducing or oxidising property etc.) as we move from left to right in a period.

Second question

From the Wikipedia article on atomic orbital:

  They are derived from the description by early spectroscopists of certain series of alkali metal spectroscopic lines as sharp, principal, diffuse, and fundamental. Orbitals for ℓ > 3 continue alphabetically, omitting j (g, h, i, k, …)


How many electrons are around the platinum in an ethenetriplatinate complex?

Looking up your proposed sum formula, I arrived at a quantum mechanical paper that used said complex (triangular $\ce{Pt3}$ moiety, whereonto an ethylene molecule coordinates onto the longer edge of the triangle) as a modelling study for the interaction of ethylene with platinum metal surfaces.[1]

As far as I can tell,[2] the authors in now way account for a surface of platinum; their three platinum atoms in vacuum are supposed to model an entire platinum surface. I cannot say whether that type of approach is typically used or whether it is considered good by quantum chemists.

Searching further, I also found a 1981 paper describing $\ce{[Pt3(C2H4)(cod)2(cot)2]}$, a neutral complex that includes two additional cycloocta-1,5-diene ($\ce{cod}$) and two additional cycloocta-1,3,5,7-tetrene ($\ce{cot}$) ligands.[3] In this complex, the bonding situation and electron count can be well answered, but that wasn’t your question.[4]

Furthermore, I find a complex $\ce{[Pt3(C2H4)]-}$ to be highly questionable: Platinum is in the 10th group, ethylene is a singlet molecule so adding a single additional electron will create a radical anion. Another reason why your question and answer cannot be serious, since in no way can a radical end up at $16$ electrons. I would even go a step further and say that it is likely an unstable complex that would decompose depending on which species are close to it. Note that the authors of the first paper calculate an uncharged complex $\ce{[Pt3(C2H4)]}$.

Notes and references

A. Cruz, V. Bertin, M. Castro, Int. J. Quantum Chem. 2000, 80, 298. DOI: 10.1002/1097-461X(2000)80:3<298::AID-QUA3>3.0.CO;2-2.
I am not a quantum chemist, and my last calculation efforts date back to a research project during my master’s studies. Those acquainted with methods will have to comment on how good the paper is.
N. M. Boag, J. A. K. Howard, J. L. Spencer, F. G. A. Stone, J. Chem. Soc., Dalton Trans. 1981, 1051. DOI: 10.1039/DT9810001051.
Two platinum atoms with 18 electrons in a square-planar environment, one with 16 electrons in a trigonal environment. The former coordinated by $\ce{cod}$ and $\ce{cot}$ (each contributing two double bonds), the latter coordinated by two $\ce{cot}$ molecules (each contributing one double bond) and an ethylene. The three platinum centres are too far apart to be considered bonded, according to the authors.
The bonding of the two 18-electron platinums to their respective $\ce{cot}$ unit is described as $\ce{cot^2-}$ and $\ce{Pt^2+}$, resulting in oxidation states of $\mathrm{+II}$ and $\mathrm{\pm 0}$ for platinum.


Let's do it step by step.

$C = 2$. The reason is that you have two components: $\ce{HBr}$ and $\ce{C4H6}$. The third compound, $\ce{BrC4H7}$, is simply a combination of both of them. Remember what is $C$: It's the minimal number of components needed to describe all phases in the system. Since $\ce{BrC4H7}$ can de described as $1\times\ce{HBr} + 1\times\ce{C4H6}$, it is not an independent component.
$P=1$ or $P=2$. It depends on how much the reactions has progressed. Basically, you have one phase on each side of the reaction. On the left, you have one gas phase that has a mixture of both $\ce{HBr}$ and $\ce{C4H6}$. On the right, you one one liquid (?) phase: $\ce{BrC4H7}$. So here are the possibilities:

The reaction has not started yet. All you have in the one gas phase. In that case, $P=1$ and $F=2-1+2=3$.
The reaction is going as written, but not completed yet. In this case not all gas is consumed yet, but liquid exists. Then, $P=2$ and $F=2-2+2=2$.
The reaction is finished. All gas is now liquid. Then we're back to $P=1$ and $F=2-1+2=3$.

So what happened in stage 2? Why did you lose a degree of freedom? Well, it depends on what what the driving force for the reaction. For example: Why did gaseous $\ce{HBr}$ and $\ce{C4H6}$ combine to form liquid $\ce{BrC4H7}$? Was it because of cooling or heating? In this case, the lost degree of freedom is $t$. Was it pressure? Then the lost degree of freedom is $P$.

$X$ is a bit problematic. There is not enough information given in the question. In my opinion, $X$ was fixed to begin with: I assume that once the reaction is complete, none of the reactants remain. Thus their ratio is fixed to 1:1 and $X$ is constrained. Had it not been the case, you would also get differing $P$ in stage three because you would have two phases, not one.


What would the Electronic Configuration of Ru²⁺ be?

You've already figured out that the electron configuration of $\ce{Ru}$ is $\ce{[Kr]\,4d^7\,5s^1}$.

All you need to know know is the order of energies for the $\mathrm{4d}$ and the $\mathrm{5s}$ orbitals. You might want to have a look at the Aufbau principle (and possible pitfalls/exceptions) and another thread on this site.


Is there a "Comprehensive Solid State" book?

I think the closest comprehensive analogical literature source for solid state materials to the aforementioned series like Elsevier's Comprehensive Coordination Chemistry II, Elsevier's Comprehensive Organic Synthesis or Wiliey's Inorganic Syntheses would be Springer's Springer Series in Solid-State Sciences currently consisting of 223 titles:

  The Springer Series in Solid-State Sciences consists of fundamental scientific books prepared by leading researchers in the field. They strive to communicate, in a systematic and comprehensive way, the basic principles as well as new developments in theoretical and experimental solid-sate physics.

As for the universally recommended textbooks on solid-state chemistry in general, there are the following titles issued for the past 15 years:

West, A. R. Solid State Chemistry and Its Applications, 2nd ed.; John Wiley & Sons, Inc: Chichester, West Sussex, 2014.
Simon, S. H. The Oxford Solid State Basics, 1st ed.; Oxford University Press: Oxford, 2013.
Smart, L.; Moore, E. Solid State Chemistry: An Introduction, 4th ed.; CRC Press: Boca Raton, FL, 2012.
Tilley, R. J. D. Understanding Solids: The Science of Materials; J. Wiley: Chichester, West Sussex, England; Hoboken, NJ, USA, 2004.
Müller, U. Anorganische Strukturchemie, 6., aktualisierte Aufl., unveränd. Nachdr.; Studium; Vieweg + Teubner: Wiesbaden, 2009. (in German)
Dronskowski, R. Computational Chemistry of Solid State Materials: A Guide for Materials Scientists, Chemists, Physicists and Others; Wiley-VCH: Weinheim, 2005.


Why are NMR typically taken for a power of 2 scans?

If you mean why do we process raw NMR data using $n^2$ data points, then @Klaus's answer is correct. Although, it is actually usually possible to record an arbitrary number of time domain data points; it is just the processing algorithm is limited to how many data points it can handle. You will either lose the extra time domain data (if using less frequency domain data points) or add zero filling (if using more frequency domain points).

If you mean the number of scans as the number of transients added together to give the final raw FID, then this comes about from the requirement for phase cycling a pulse sequence in order to remove unwanted NMR signals that may arise from

imperfections in hardware phase (giving rise to quad images)
coherent noise
generation of undesired magnetization arising in multi-pulse experiments

Here, though, the number of scans is not limited to $n^2$. Historically, the number of scans for a 1D spectrum would typically have been confined to a multiple of 8, as a phase cycle requirement. As you start to need to run greater numbers of scans due to sensitivity issues, then you would need to consider the accumulation of noise as well as signal, as @ron has alluded to (doubling the number of scans only give a $\sqrt{2}$ improvement in S/N; you need to run 4 times as many scans to double S/N). It is through this combined relationship that the number of scans, then, adheres to a $n^2$ increase. If 8 scans (required by phase cycling) didn't give sufficient S/N, then we need to double or quadruple the number of scans to see a real improvement in S/N. And so of for ns=16 or 32 or 64.
For more complicated experiments, phase cycling may have required multiples of 16 or more. I remember running a dqf-COSY experiment which had a phase cycle of 24, so the number of scans had to be a multiple of 24. Yes, these experiments took many, many hours to run.

However, (there's always a however) in modern spectrometers (anything newer than about 20 years old), the phase cycling restrictions are largely removed through improvements in hardware design and implementation of gradient pulses. So now we see for most basic 1D experiments, such as a 1H, there is a minimum number of scans of 1. We run many high quality, high resolution, quantitative analyses of samples using long acquisition, single scan experiments.
For multi-pulse experiments, there is still a requirement to phase cycle, but even for many 2D and 3D experiments, this is limited to multiples of 1, 2 or 4. Very few experiments these days require large numbers of scans to account for phase cycling.


Calculating rate constant and order of a multi-step reaction

The method you've indicated in your solution is correct. Regarding:

  Then will the rate constant of the whole reaction be equal to the product of the equilibrium constant of the fast reaction and the rate constant of the slower reaction?

In your question, the final $K'=K_1\cdot K_{eq}$, but this is not always true.

Remember that you've to adjust the final rate law expression in such a way such that it is free from all intermediates. In doing so, you might need to multiply the $K$ of another reaction, as you did above. But you may also need to divide or multiply the square/cube/etc. of another reaction.

Hence, $K_{final}$ is not always equal to $K_1\cdot K_2\cdot...\cdot K_n$. So, your assumption is incorrect.


What are the magnetic properties of [Ni(S2C2Ph2)2]^z's 3 oxidation states?

$\ce{Ni^{2+}}$ is a $d^8$ metal. In a square planar orbital splitting, all $8$ electrons are paired, hence it is diamagnetic. Even when the complex is neutral, the electron loss is at the ligands (by definition, non-innocent), and they are antiferromagnetically coupled, which means that their spins "cancel" each other. Since there are two, the net moment is zero.

All complexes are $\ce{Ni(II)}$, so assuming all complexes are square planar, all $d$ electrons are paired up and the magnetic moment from these is 0. Both the $z=-2$ (no radicals) and $z=0$ (antiferromagnetic coupling) area diamagnetic. The $z=-1$ complex is paramagnetic. There is a single unpaired electron in one of the ligands that can't couple to the other one, where all of them are paired up.


Fluorine Preparation:How can this be explained on basis of Lewis Acid-Base Reaction?

$\ce{2 K2MnF6 + 4 SbF5 → 4 KSbF6 + 2 MnF3 + F2↑}$

Occurs in two stages. First there is a (Lewis) acid-base reaction

$\ce{2 K2MnF6 + 4 SbF5 → 4 KSbF6 + 2 MnF4}$

where the Lewis acid $\ce{SbF5}$ accepts a lone pair from a fluorine in the manganese complex, and the adduct then falls apart to leave $\ce{MnF4}$. This is thermally unstable at ambient temperatures (Housecroft and Sharpe) and falls apart in the redox reaction

$\ce{2 MnF4 → 2MnF3 +F2}$


Rotational-vibrational spectroscopy of a molecule

I'm not quite sure what you mean by 'vibrational levels composed of lower rotational energy levels' so I try to explain the whole thing, briefly.
In analysing the electronic, vibrational &  rotational spectra of molecules the Born-Oppenheimer approximation has to be used. This is applicable because the vibrational period (a few femtoseconds) is far faster than rotation period (a few picoseconds) thus these motions do to interact appreciably with one another to a good approximation. This allows the energy levels of, rotation and vibrational motions to be added together.
A normal mode vibration is the motion of all atoms in the molecule in a fixed phase relationship with one another. For N atoms there are $3N-6$ normal modes ($3N-5$ for a linear molecule). How how far and in what direction each atom moves has to be determined by calculation, but the overall symmetry species ($A_1, B_g, E_u$ etc) of the total motion of all atoms  is given by the Point Group of the molecule, $C_{3v},D_{2h}, O_h$ etc. The symmetry species determines the relative phase of each atoms motion with respect to all the others. By 'phase' is meant that as some bonds contract  others extend but they always do this together.
As energy levels add each vibrational normal mode has its own stack of rotational levels. (Because the B.O. approximation is not exact a more detailed analysis in needed to add the interaction between vibration and rotation.)
The figure shows a typical spectrum (calculated)  of a (perpendicular band) of a symmetric top molecule such as $\ce{PCl3 , CHCl3} \mathrm { ~or~ } \ce{CH3Cl}$. The total spectrum is shown on the bottom line and its separation into individual PQR rotational bands above this.  There will be similar complicated features for each vibrational normal mode. Clearly the analysis of spectra such as this is non-trivial :)
(In symmetric top molecules two of its three moments of inertia are the same but different from the third. Two quantum numbers are needed J for total angular momentum and K to fix the angular momentum about the symmetry axis and this is why there are so many rotational bands for a single vibration.)

(Spectrum is taken from G. Herzberg, ' Infra-red and Raman Spectra of Polyatomic Molecules')


What is the driving force for Claisen rearrangements?

The formation of the carbonyl group is a gain in thermodynamic stability over the vinyl ether. In your example, the monosubstituted alkene is a "wash". In any case, the alkyl substitution pattern of the non-enol ether double bond is not a problem.
The heat of this Claisen rearrangement, first conducted by Hurd and Pollack1 in 1938, can be calculated by determining the heat of formation (ΔHfo) of allyl vinyl ether (1) and 5-hexenal (2). Heats of formation of allyl vinyl ether (1), hexanal (3), 1-hexene (4) and hexane (5) are known (NIST and Chemeo). These heats of formation are shown in black. However, the ΔHfo of 5-hexenal (2) is not accessible in tables but it can be calculated. The heat of hydrogenation of monoalkyl substituted alkenes is ~125 kJ/mol (~30 kcal/mol). The reduction of 1-hexene (4) to hexane (5) is -124.6 kJ/mol, which is computed from the respective heats of formation of 4 and 5. Applying this heat of hydrogenation to the reaction 2 ---> 3 gives a ΔHfo of -128.2 kJ/mol for 5-hexenal (2). Accordingly, the Claisen rearrangement of allyl vinyl ether (1) is exothermic by -100.6 kJ/mol.

1. C. D. Hurd and M. A. Pollack, J. Am. Chem. Soc., 1938, 60, 1905.


What do locants with prime symbols (e.g. 4’) mean?

4 means the fourth atom and the prime symbol specifies the part of the molecule (one ' means the secondary structure), so 4'-hydroxyacetanilide means you have acetanilide with a hydroxide group on the fourth atom of the secondary structure.
Naming organic compounds works like this:

  Find the longest continueous chain of carbon atoms. Called the parent chain this chain establishes the base alkane name for the compound.
  Number the carbon atoms in the parent chain beginning with the end closest to any substituents.
  Using the correct substituent name (methyl, ethyl, chloro,etc.), and the carbon atom number on the parent chain.
  If a particular substituent occurs more than once, use a prefix (di-, tri-, tetra-) to indicate the number of those substituents.
  List the alkyl substituents in alphabetical order.

Source: http://intro.chem.okstate.edu/1014/lecture/organic.html

Here are the IUPAC guidelines about the nomenclature of organic compounds:
http://old.iupac.org/reports/provisional/abstract04/BB-prs310305/Chapter1.pdf


Why isn't the side chain of arginine totally protonated at low pH?

It may be hard to see from a 2D-drawing, but there are simply no more available lone pairs on arginine than the ones protonated in form $\ce{A}$.

The guanidine group ($\ce{H2N-C(=NH)-NHR}$) is isolobal to a urea group, and indeed guanidine behaves much like urea except for the common differences between $\ce{C=O}$ and $\ce{C=NH}$. The lone pairs of $\ce{-NH2}$ and of $\ce{-NHR}$ are actively taking part in resonance with the $\ce{=NH}$ group’s double bond much like an amide structure. One can equivalently say that those lone pairs are delocalised into the $\ce{C=N}$ $\pi^*$ MO. Hence, only the $\mathrm{sp^2}$ hybridised lone pair on the $\ce{=NH}$ nitrogen is basic in any way and can be protonated.

Of course, you can further protonate arginine theoretically. However, it would take superacidic media.


Determine stable oxidation state from Ionization energy

You don't necessarily need to have your element binding with oxygen to be oxidized, per se. Unless it is, of course, specified that we're talking about a reaction with oxygen. So, don't misguide yourself on focusing on the oxygen here :) After all, if we, for instance, think about burning of potassium, it is being oxidized with oxygen, however, it still only makes $\ce{K+}$ and not $\ce{K^2+}$!
The trick in this question is to look for the values, or a magnitude, of those ionization energies in question. To remove the first electron, you need to provide $\pu{7 eV}$, to remove the second one, you need to provide $\pu{12.5 eV}$. So the difference between 7 and 12.5 is not that big if we compare it with the difference between $\pu{12.5 eV}$ and $\pu{42.5 eV}$ that is required to remove the third electron. If explaining it in simple terms, the first to energy "bumps" are easy to "climb", while the last one is way high and you really need to provide a sufficient amount of energy to get that high. Thus, removing the first two electrons is relatively easy, while removing the third one is not.
This gives you the most stable oxidation state for this element to be +2. If we perform a reaction of this element (which looks like $\ce{Sr}$) with any other electronegative element, say chlorine, it would still have a charge of +2.


What chemical properties of ethanol make it usable for drinks as compared to that of methanol?

The problem arises from the metabolized products of methanol. Methanol oxidizes in the liver by an enzyme called alcohol dehydrogenase to formaldehyde which is further metabolized to formic acid by another enzyme called aldehyde dehydrogenase. This formic acid is the source for acute toxicity associated with methanol poisoning. Accumulation of this chemical in the blood deprives cells of oxygen by inhibiting the enzyme cytochrome c oxidase in their mitochondria, a key element of the respiratory electron transport chain.
Formic acid, together with formaldehyde, are responsible for nerve damage, blindness, and other unpleasant effects associated with methanol poisoning.

Note that ethanol is also metabolized in the same way by the same pair of enzymes to ultimately form acetic acid but human can tolerate acetic acid to an extent because it is less toxic compared to formic acid and thus can be consumed like for example, vinegar. In fact, ethanol is used as a remedy for methanol poisoning as it acts as a competitive inhibitor by more effectively binding and saturating the alcohol dehydrogenase enzyme in the liver, thus blocking the binding of methanol with the enzyme rendering it useless.
References

https://theskepticalchemist.com/methanol-toxic-ethanol/ (above reaction source)
https://metode.org/metodes-whys-and-wherefores/why-can-we-drink-ethanol-but-not-methanol.html
https://en.wikipedia.org/wiki/Methanol_toxicity


Why does 1,2-dimethylcyclohexane only possess three stereoisomers?

Quoting , ORGANIC CHEMISTRY second edition by JOSEPH M. HORNBACK, UNIVERSITY OF DENVER

1,2-dimethylcyclohexane has two chirality centers.

The cis-diastereomer has a plane of symmetry bisecting the ring bond between
the two methyl groups  and therefore a meso compound.It is not chiral and does not rotate plane-polarized light.The trans-diastereomer exists as a pair of enantiomers.

The chair conformation of cis-1,2-dimethylcyclohexane  shows that it is chiral.
But due to  ring-flip , another  conformation ,its  enantiomer of the original conformation is also formed.These conformers interconvert rapidly at room temperature, therefore cis-1,2-dimethylcyclohexane is not chiral.

Further I quote

"stereochemical analyses such as the preceding one can be done with drawings using planar rings" pg.235.

A molecule cannot be optically active if its chiral conformations are in
equilibrium with their mirror images. We consider such a molecule to
be achiral. PG- 193 - ORGANIC CHEMISTRY  .EIGHTH EDITION
L.G. WADE , JR . WHITMAN COLLEGE


Difference between Sildenafil and Sildenafil citrate

In many pharmaceuticals (and many recreational drugs) the choice of form is determined by the convenience of the manufacturing and delivery

Many drugs (not just sildenafil) are manufactured as salts or simple derivatives of the active molecule. For example, the anti-inflammatory steroid fluticasone is usually delivered as the propionate or furoate esters and the recreational drug cocaine is usually delivered as the hydrochloride salt. In all this cases the active ingredient is exactly the same in the body.

But there are two factors that matter as well as the active ingredient.

One is how easy it is to obtain a convenient form that is pure and easy to make. Amines, for example, are much less prone to produce nice crystals than their salts (sildenafil and cocaine are amines). Since the industry needs to have a good route to a verifiably pure compound and since crystallisation is a good way to obtain pure compounds, they often experiment with different salts to find one that gives a satisfactory end result. This is convenient for the manufacturer and makes no difference to the active ingredient which is determined by the pH of the biological environment.

The second reason is the ease of delivery of the drug. The physical form and chemistry of the substance or salt do matter for the bioavailability. If the drug is taken as a pill, it has to be well absorbed in the digestive system (which means it should, usually, dissolve quickly in the stomach). Salts are more likely to be rapidly soluble than simple amines (though this depends on the form as much as the fact they are a salt though the form is more controllable if you have nice crystals to start with in manufacturing).

The illegal drug cocaine is an interesting case. It is easy to maufacture it as a salt and it is commonly available as the hydrochloride. The hydrochloride dissolves in nasal linings giving a high. But some prefer a faster, more intense (and also, unfortunately, more addictive) high obtained by using the vaporised freebased amine. This is easy to make by reacting the hydrochloride with household baking powder. But it can't be sniffed in that form (as a powder), it has to be heated and vaporised for effective absorption. The vapour is absorbed more rapidly directly into mucous membranes, making a faster hit. But this isn't as convenient as sniffing a finely ground powder and would not be a convenient way to distribute legal pharmaceuticals (imagine the police attention if you had to heat a spoon full of fluticasone to relieve your blocked nose).

So, in short, convenience of manufacture and delivery frequently lead to the ideal form of a drug being a salt and not the pure active ingredient.


"Estrontium" is not used as an element name in any language. It appears that the error traces to a single user by the name of Alejo Miranda (listed as from Ecuador) who has posted a large collection of periodic tables and individual element clip art to shuttersock, iphoto, etc. in which "estrontium" is incorrectly used as the English name for strontium.  The Spanish periodic table correctly uses estroncio.  The link to the gllery at shutterstock is

https://www.shutterstock.com/g/Alejo+Miranda?searchterm=estrontium&search_source=base_gallery&language=en&sort=popular&measurement=px&safe=true


What happens when mixing elemental mercury with sodium hypochlorite solution

There are several redox reactions can be taken place when you add household beach on elemental mercury metal. However, this question is more like a homework question. Thus, I give you insight to the answer and you may read a bit and find out what's happening. This reaction is studies and results have been published (Reference 1), the abstract of which states that:

  The absorption of elemental $\ce{Hg}$ vapor into aqueous hypochlorite was measured in a stirred tank reactor at $25$ and $\pu{55 ^{\circ}C}$. $\ce{NaOCl}$ strongly absorbs $\ce{Hg}$ even at high pH. Low $\mathrm{pH}$, high $\ce{Cl-}$ and high-temperature favor mercury absorption.  Aqueous free $\ce{Cl2}$ was the active species that reacted with mercury. However, chlorine desorption was evident at high $\ce{Cl-}$ and $\mathrm{pH}<9$. Overall second-order reaction was observed between mercury and chlorine with an apparent second-order rate constant of $\pu{1.7\times 10^{15} M^{-1} s^{-1}}$ at $\pu{25 ^{\circ}C}$ and $\pu{1.4\times 10^{17} M^{-1} s^{-1}}$ at $\pu{55 ^{\circ}C}$. Gas-phase reaction was observed between $\ce{Hg}$  and $\ce{Cl2}$ on apparatus surfaces. Strong mercury absorption in water was also detected with $\ce{Cl2}$ present. Results indicate that the chlorine concentration, moisture, and surface area contribute positively to mercury removal.

I hope you will figured it out when you consider the standard reduction potentials of $\ce{Hg}$ and $\ce{OCl-}$ in any electrochemical series. For example, see Reference 2.

References:

L. L. Zhao, G. T. Rochelle, “Mercury absorption in aqueous hypochlorite,” Chemical Engineering Science 1999, 54(5), 655-662 (https://doi.org/10.1016/S0009-2509(98)00263-2).
S. G. Bratsch, “Standard Electrode Potentials and Temperature Coefficients in Water at $\pu{298.15 K}$,” Journal of Physical and Chemical Reference Data 1989, 18(1), 1-21 (https://doi.org/10.1063/1.555839).


Why won't removing the salt bridge cause the battery to recharge?

The ions in the salt bridge never leave the salt bridge.

That's not true.  The ions literally have to move to balance the charge of the gain / loss of the electrons at the electrodes.  (You even allude to this in your question.)  One of the ways that a battery can run out is that the bridge is used up.  You could views this by using a deeply colored ions (e.g. MnO4-) to see the migration.

While the salt bridge is in place, there is no build up of charge.  Electrons moving through the wires are offset by the ions moving through the salt bridge and into the respective half-cell.  Thus, if you remove the salt bridge, there is no net charge that will cause anything to reverse; there is no 'force' that will cause ions to revert.

If you want to reverse the reaction (and charge the battery), you have to force electricity back in the other direction, using more voltage than what you got out of the battery.  This doesn't work for all batteries.  (I.e. this is why only some batteries are rechargeable).  As you force electricity backwards through a battery, any movement of ions from the salt bridge means that the half-cell might now be contaminated with chemicals other than the ones you want in your half-reaction (unless you designed your salt bridge just right).  Thus, the "right" ions may not go back to being uncharged (or vice versa).


Mechanism of sulfur trioxide reaction with water to make sulfuric acid

I've drawn a more correct mechanism for the reaction of dilute $\ce{SO3}$ with water in the liquid phase:

$$\ce{SO3 (aq) + 3H2O (l) -> SO4^2- (aq) + 2H3O+ (aq)}$$

$\ce{SO3}$ is a strong electrophile, enough to react quickly with water, which is a relatively weak nucleophile. A water molecule is added to the structure, facilitated by the dislocation of a $\ce{S=O}$ bonding electron pair in $\ce{SO3}$ to the oxygen atom, forming the intermediate structure shown. Additional water molecules then quickly abstract hydrogen atoms from the intermediate (actually, the first abstraction is far easier than the second, so it might have been more proper to show each proton abstraction separately in sequence). The final product of the reaction is sulfate anions and hydronium cations ($\ce{H2SO4}$ is a strong acid in water, so it doesn't really exist as a molecule. The second proton isn't nearly as acidic as the first, but in dilute solution both are ionized). The formal global reaction is shown below the mechanism. Notice that at every point the formal charge on the central sulfur atom is exactly the same, zero.

Now, with regards to our little discussion about the actual structure of $\ce{SO3}$. Classical valence bond theory, enhanced with the concept of hypervalency and VSEPR theory, attributes the structure shown, with three double bonds. The Wikipedia article for $\ce{SO3}$ mentions another structure, based on modern valence bond theory computational calculations, with one double and two single bonds. This structure is quite interesting as it does not invoke hypervalency. Indeed, hypervalency is thought to be an old concept, and not really necessary in more modern theoretical treatments of chemical bonding.

However, if one opts to display the structure according to modern valence theory, they would do well to make it explicit. Almost all structural formulae today are drawn in the context of classical valence bond theory. Certain assumptions from the classical theory do not carry over to the modern theory. For example, the authors of the article "Modern valence bond theory", Chem. Soc. Rev., 1997,26, 87-100 (free access mirror) find that the bonding in $\ce{N2O}$, described classically as a mixture of the canonical forms $\ce{^{-}N= N^{+}=O}$ and $\ce{N≡N^{+}-O^{-}}$, is best described as $\ce{N≡N=O}$. Even with five bonds, the central nitrogen atom has a calculated occupancy close to 7, i.e. it is not hypervalent. Since the structural construction rules of classical and modern valence bond theory aren't the same, the structures generated from both do not mesh and cannot be compared directly. The simplest thing to do, then, is to draw $\ce{SO3}$ according to classical valence bond theory, with three double bonds.


Show the vibrational frequency of fluorine molecule anion is much lower than that of fluorine molecule

You should definitely post the MO diagram of $\ce{F2}$ that you worked out. When trying to solve such a problem you should try to analyse the question as thoroughly as possible to make sure that you collect all the information available to you beforehand. Especially important here is that the question already tells you that the vibrational frequency of $\ce{F2^{-}}$ is lower than that of $\ce{F2}$. So the first question you should ask yourself is: Under what circumstances is a vibrational frequency lowered and which relation do I know that links the vibrational frequency to some variable that is connected to the MO diagram of $\ce{F2}$? I give you a hint: You will need the equation of a harmonic oscillator

\begin{equation}
 \omega = \sqrt{\frac{k}{m}}
\end{equation}

where $\omega = 2 \pi f$ is the angular frequency, $f$ is the frequency, $k$ is the force constant (in your case of the bond and thus its a measure of the bond strength) and $m$ is the reduced mass of the oscillating system. (The harmonic oscillator is one of those simple model systems that can help you quite often when you want a starting point for the investigation of much more complicated ones.) Now ask yourself: Which of the variables $k$ or $m$ might have a relation to the MO diagram and how will they be related?


What is the purpose of Gaussian, itself?

There are several interrelated reasons (disclaimer: I am a GAMESS developer).

Gaussian has a lot of features and interoperability.  To take your PM3 example. Yes, the results of a PM3 calculation are identical in GAMESS and Gaussian. But until recently you couldn't do a PM3 calculation in solution using the PCM method in GAMESS.  Also, what if you want to do a PM6 calculation instead. In GAMESS this is currently only possible only for elements up to F.

Because of this Gaussian has been the program of choice for many people, which also means they have trained a lot of people to use Gaussian.  Learning to use a new (and perhaps free) program well can be very time consuming, so it might be more cost effective to pay for a program you know how to use well.  And chances are you probably have to learn how to use several new programs to get all the functionality of Gaussian.


How to find volume of substance from molar concentration

Nothing has been said about the actual volume of the solution, only concentration of the solution has been given.

Assume that $c$ moles of solute is present in 1m$^3$ of solution.

Then mass of solute=c*molar mass

volume of solute * density of solute = c* molar mass

From this equation, volume of solute can be known.

Clearly, if the volume of the solution increases, the volume of the solute will also increase. Since the volume of the solution has not been given, it must give you a hint to take the above assumption.


Can you experimentally determine activation energy if the rate expression and rate constant are unknown?

This is valid only provided that the reaction is strictly first-order and provided that the “specific end-point” is a fixed concentration ratio. Note that the slope of $\ln(t_\text{end})$ vs. $\frac{1}{T}$ is $\frac{E_a}{R}$ but the intercept is not $\ln(A)$.

If the rection is first order ( $X {\xrightarrow{k}} Y$) then the product concentration at time t is:

$[Y]_t = [X]_0 .e^{k. t}$

ie:

$\ln\left( \frac{[Y]_t}{[X]_0}\right) = k. t $

If the end point is some fixed concentration ratio ($\frac{[Y]_{end}}{[X]_0}$), the time to reach the endpoint is:

$ t_{end} = \frac{\ln\left( \frac{[Y]_{end}}{[X]_0}\right)}{k}$

If we substitute this into the Arrhenius equation ($\ln(k) = \ln(A) - \frac{E_{a}}{R.T}$), we get:

$\ln(t_{end}) = \ln\left( \ln\left( \frac{[Y]_{end}}{[X]_0}\right)\right) - \ln A + \frac{E_{a}}{R.T} $

so a plot of $ln(t_{end})$ vs. $\frac{1}{T}$ will have a slope of $\frac{E_{a}}{R}$


Why is the residual proton resonance in deuterated benzene a singlet?

To a very good approximation, $J_\ce{HD}$ in a deuterated molecule is related to $J_\ce{HH}$ in the corresponding protonated molecule by the gyromagnetic ratios:

$$\frac{J_\ce{HD}}{J_\ce{HH}} = \frac{\gamma_\ce{D}}{\gamma_\ce{H}} = \frac{\pu{41.066 * 10^6 rad s-1 T-1}}{\pu{267.522 * 10^6 rad s-1 T-1}} = 0.15$$

In benzene where $^3J_\ce{HH} = \pu{7.6 Hz}$ and $^4J_\ce{HH} = \pu{1.4 Hz}$ (Reich NMR tables), we would thus have $^3J_\ce{HD} \approx \pu{1.2 Hz}$ and $^4J_\ce{HD} \approx \pu{0.2 Hz}$. The four-bond meta coupling would be all but invisible, and the three-bond ortho coupling if resolved would give rise to a $1:2:3:2:1$ quintet.

On a lower-field spectrometer this relatively small coupling is usually not properly resolved. A suitable window multiplication, though, will reveal all. I don't have a d6-benzene spectrum on hand, but I do have a $\pu{500 MHz}$ d8-toluene spectrum here. The two spectra on the left are the aromatic residual peaks, but at the bottom-left a Gaussian window function has been used to improve resolution (at the cost of signal:noise ratio). This allows us to resolve the three-bond ortho couplings of $\pu{1.1 Hz}$.

At the bottom-right is the signal corresponding to residual $\ce{C6D5CD2H}$. Since the two-bond H–D coupling has a larger magnitude ($^2J_\ce{HD} = \pu{2.2 Hz}$), it is much easier to resolve, and can be seen even with the normal exponential window function (which sacrifices resolution for signal:noise ratio improvement).

Similarly, in d6-DMSO (shown below at $\pu{500 MHz}$), the two-bond coupling is more easily resolved.


Clarification about the difference of internal energy formula

Your book has driven you crazy with its stupid sign convention.  If the gas is expanding against constant pressure, then the equation is still $$\Delta U=Q+W$$ but, in this case, the work done on the system by the surroundings is negative (i.e., W is a negative number, because the system is doing work on the surroundings).  So not all the heat added to the system is being used to increase its internal energy.  The increase in internal energy is less than if the gas was not pushing back its surroundings.


Which of the following octahedral complexes absorbs the photon with the longest wavelength, λ?

Since CN- is a strong field /pi-acceptor ligand, therefore, it will create larger splitting. So for transition higher energy will be needed. It means that in case of Complexes with CN- higher energied photons i.e. photons with shorter wave length will be absorbed.


Are refractive or density based Brix meters better for use with alcoholic drinks?

Once I got going, this turned out to be a lengthy answer. So here's the summary: the alcohol will introduce a significant error in both methods. The density measurement will probably be closer to the actual sugar content. Th density measurement will underestimate the sugar, and the refractive index will overestimate the sugar.

Both the refractive index and the density-based methods work for measuring sugar content because adding sugar to water raises the refractive index and the density in amounts proportional to the concentration of the added sugar. As you and others noted in that thread, ethanol interferes with that measurement because it also affects the density and refractive index of the resulting solution. So how large is that effect?

To estimate this effect, I am going to restrict myself to the 15-35% alcohol (by volume) range you suggested. The charts I found for the density and refractive index of water/ethanol mixtures gives the ratio by weight; your range converts to 12-28% by weight. Pure water has a density of 0.99823 g/mL, and a refractive index of 1.3330. 12% ethanol has a density of 0.97910 g/mL, and a refractive index of 1.3410. 28% ethanol has a density of 0.95710 and a refractive index of 1.3524. Note that adding ethanol increases the refractive index, but decreases the density. I'll come back to that at the end.

This page gives a table of the index of refraction and density of sucrose solutions in water, as a function of the sucrose weight fraction. The 20 g sucrose per 100 mL solution is approximately 20% by weight, so that will have a refractive index of 1.3639, and a density of 1.0810 g/mL.

The density and refractive index changes for both sucrose and ethanol are pretty linear over the range you're interested in, if you take them independently. Assuming that the sucrose effects and the ethanol effects are independent (NB: this may not be a good assumption), the refractive index measurement will be higher by 0.01-0.02, which corresponds to an overestimate of 7-14% sucrose. The density measurement will be lower by an amount 0.02-0.04 g/mL, which corresponds to an underestimate of 5-10% sucrose. So it looks like the hydrometer measurement will be a bit better, but the two methods will be off in different directions. It's a noticeable effect: depending on the method and the alcohol content, you might be off by somewhere between a quarter and half of your intended sugar content.

There is one neat thing you could do to take advantage of the fact that the alcohol moves the density in the opposite direction to the sugar. If you had a bottle of liquor with unknown alcohol content and unknown sugar content, you could take both the density and the refractive index measurement. Again, assuming the four effects are linear, and the sugar effects are independent of the alcohol, you could write two equations for the refractive index $RI$ and density $\rho$ as functions of the alcohol and sugar: $RI = Ax + By, \rho = Cx + Dy$. $x$ and $y$ are your alcohol and sugar content in whatever units you prefer, and $A, B, C,$ and $D$ are the empirical parameters describing each of the four effects. If you take a measurement for $RI$ and for $\rho$, you can solve for $x$ and $y$.

PS: to answer one question suggested by the comments of the [cooking.se] question you linked, density-based brix meters (I'm used to seeing them called hydrometers) are typically used for home brewing beer and wine because the only source of alcohol is the fermenting sugar. You take a density measurement before fermentation begins to get a baseline. Then, at any point after fermentation begins, the difference between the current density and the baseline density is directly proportional to the alcohol content. I'm sure the same reasoning applies to the refractive index method, but I've never used that for beer-brewing (I've only used it for my real research).  See Cutaia, A.J., Reid, A-J. and R.A. Speers,  2009. Examination of the relationships between Original, Real and Apparent Extracts, and Alcohol in Pilot Plant and Commercially Produced Beers. J. Inst. Brew. 115:318-327.


Mismatch of plane of symmetry vs Gold Book's definition for a pseudochiral carbon

The Gold Book definition for a meso compound is

  the achiral member(s) of a set of diastereoisomers which also includes one or more chiral members.

So, a meso compound has chiral subsets. Here we would have three such subsets:

Leftmost carbon (2R)
The centermost (pseudochiral 3S)
The rightmost (4S) carbon

Even though  the centermost carbon is pseudo chiral, the plane of symmetry still exists. The compound remains meso, achiral.

Nomenclature is just for naming a compound.

The following two molecules are diastereomers, not enantiomers.


The color of the most sulfides of p- and d-block elements

The band gap between conduction band and the valence band plays a role in color of these compounds. The good example is $\ce{SnS2}$, which is bronze color (Image $\bf{A}$; $\ce{SnS2}$ is used in decorative coating where it is known as mosaic gold) and have a band gap of $\pu{2.18-2.44 eV}$ (Image $\bf{B}$; Ref.1). When an electron in the conduction band falls down to the valence band, it loses energy and the lost energy is emitted as a photon ($h\nu$). This photon has an energy equal to the band gap (Image $\bf{C}$). In mathematics, it is: $E_\mathrm{gap} = h\nu$. The crystals of $\ce{SnS2}$ and its calculated band gap is shown below:

The compounds is colored when the band gap is between $1.8$ and $\pu{3.0 eV}$ (visible range). The compounds with larger band gaps $(\gt \pu{3.0 eV})$ are either white or transparent crystals.  Those with smaller band gaps $(\lt \pu{1.8 eV})$ give  black color crystals. For example, the band gap of $\ce{ZnS}$ is $\pu{3.54 eV}$, the color of which is white. Meantime, the color of $\ce{CdS}$ (Pigment name: Cadmium yellow) is yellow because its band gap is $\pu{2.45 eV}$. On the other hand, the colors of $\ce{CdSe}$ and $\ce{CdTe}$, which have the same structure of $\ce{CdS}$, are black because their band gaps are $\pu{1.74 eV}$ and $\pu{1.49 eV}$, respectively. Also, the color of $\ce{PbS}$ is black because its band gap is $\pu{0.37 eV}$.
For some crystals, the band gap change with temperature (Ref.1). Hence, for those, the color depend on temperature.
References:

Lee A. Burton, Thomas J. Whittles, David Hesp, Wojciech M. Linhart, Jonathan M. Skelton, Bo Hou, Richard F. Webster, Graeme O'Dowd, Christian Reece, David Cherns, David J. Fermin, Tim D. Veal, Vin R. Dhanak, Aron Walsh, “Electronic and optical properties of single crystal SnS2: an earth-abundant disulfide photocatalyst,” J. Mater. Chem. A 2016, 4, 1312-1318 (https://doi.org/10.1039/C5TA08214E).


Your mechanism is missing an important step: between the oxidative addition of palladium into the $\ce{C-Br}$ bond and the alkene insertion there should be an alkene annealing forming a palladium-alkene complex. This is displayed in the mechanism scheme of a generic Heck reaction below, which I sheepishly borrowed from Wikipedia.

Scheme 1: Catalytic cycle of a Heck reaction. Image taken from Wikipedia, where a full list of authors is available. Note that the β hydrode elimination in step C should be syn and not anti or gauche as displayed.

The annealing step and the π complex are important because they influence the Heck reaction’s selectivity. Due to the avoidance of steric stress, both residues will adopt an anti-type conformation as depicted in structure 3. On the (syn-selective) insertion of the alkene into the $\ce{C-Pd}$ bond, this geometry is kept resulting in the palladium catalyst ending up on the higher-substituted alkene carbon. For the inverse insertion, both organic residues — in your case two phenyl groups — would have to occupy the limited space on one side of the catalyst.

The subsequent β hydride elimination can theoreticly occur in both directions in your molecule. However, eliminating towards the hydroxy group allows keto-enol tautomerism with further stabilisation. Since β hydride elimination is a reversible process, the reaction will select the ketone over the allylic alcohol that would be formed upon opposite β hydride elimination.

The alkene insertion into the $\ce{Pd-C}$ bond, is not reversible. Two $\ce{C-C}$ single bonds are formed at the expense of a palladium-π interaction resulting in a notably downhill process.


Why are there more fluoride compounds formed with Xenon?

The reason that we discovered more fluorides of xeon than helium, argon, and krypton is quite obvious and you might already know it. As lighter noble gases have more stable shell configuration, it is more difficult to make any kind of compound, including fluorides.
Also, the higher amount of fluorine atoms in a molecule of the compounds means drawing more electron from the noble gas, which is thermodynamically not preferred. Thus, $\ce{KrF4}$ and $\ce{KrF6}$ are far less stable than $\ce{KrF2}$. The quantum chemical calculation confirmed this (Dixon 2007). The same reference said that they are stable, but have not been synthesized yet, and the synthesis would be very difficult.

For radon it is a different problem. Briefly, there might be more fluoride compounds of radon than xenon, but the synthesis is difficult to be confirmed.
Even though radon is less inert than xenon and its compounds were predicted and calculated to be more chemically stable than xenon compounds (Chernick 1962, Thayer 2005), it is difficult to study. Radon is is rare and radioactive (half-life of 3.8 days). Studying chemical compounds in trace amount is a difficult task. There are claims for discoveries of $\ce{RnF4}$ and $\ce{RnF6}$ (also in the above references), but confirming that you have synthesized new compounds in trace amount is extremely difficult because the evidence is not obvious. $\ce{RnF8}$ was calculated to be unstable, but should be more possible than $\ce{XeF8}$.

References:
1 David A. Dixon, Tsang-Hsiu Wang, Daniel J. Grant, Kirk A. Peterson, Karl O. Christe, Gary J. Schrobilgen, "Heats of formation of krypton fluorides and stability predictions for $\ce{KrF4}$ and KrF6$\ce{KrF46}$ from high level electronic structure calculations," Inorganic Chemistry 2007, 46(23), 10016-10021 (https://doi.org/10.1021/ic701313h).
2 C. L. Chernick, H. H. Claassen, P. R. Fields, H. H. Hyman, J. G. Malm, W. M. Manning, M. S. Matheson, L. A. Quarterman, F. Schreiner, H. H. Selig, I. Sheft, S. Siegel, E. N. Sloth, L. Stein, M. H. Studier, J. L. Weeks, M. H. Zirin, "Fluorine Compounds of Xenon and Radon," Science 1962, 138(3537), 136-138 (DOI: 10.1126/science.138.3537.136).
3 John S. Thayer, "Relativistic Effects and the Chemistry of the Heaviest Main-Group Elements," Journal of Chemical Education 2005, 82(11), 1721-1727 (https://doi.org/10.1021/ed082p1721).


Why does the degree of dissociation change when we dilute a weak acid even though the equilibrium constant is constant?

See my comment above.

As a numerical example, take acetic acid ($\ce{AcOH}$), which has $K_\mathrm{a} = 1.8 \cdot 10^{-5}$.

This means that:

$$K_\mathrm{a} = \frac{[\ce{AcO-}][\ce{H+}]}{[\ce{AcOH}]}$$

And the total nominal concentration of acid is:

$$C_\mathrm{a} = [\ce{AcOH}] + [\ce{AcO-}]$$

Combining these two equations, you can see that the % of dissociated acid is:

$$\alpha = \frac{[\ce{AcO-}]}{C_\mathrm{a}} = \frac{K_\mathrm{a}}{K_\mathrm{a} + [\ce{H+}]}$$

i.e. not a constant, but a value that depends on the $\mathrm{pH}$. So no, there is no need for the 'degree of completion' of the reaction to be a constant.

You can even calculate the explicit concentrations of all the species.

Given that:

$$C_\mathrm{a} = \left(\frac{1}{[\ce{H+}]} + \frac{1}{K_\mathrm{a}}\right) \cdot ( [\ce{H+}]^2 - K_\mathrm{w}^2)$$

you can see that, for $\mathrm{pH} = 5$ you need:

$$C_\mathrm{a} = \left(\frac{1}{10^{-5}} + \frac{1}{1.8 \cdot 10^{-5}}\right) \cdot \left((10^{-5})^2 - (10^{-14})^2\right) = 1.5554 \cdot 10^{-5}$$

Then:

$$[\ce{AcOH}] = \frac{C_\mathrm{a} \cdot [\ce{H+}]}{K_\mathrm{a} + [\ce{H+}]} = 5.555 \cdot 10^{-6}$$

$$[\ce{AcO-}] = C_\mathrm{a} - [\ce{AcOH}] = 9.999 \cdot 10^{-6}$$

So the % of dissociated $\ce{AcOH}$ is:

$$\alpha = \frac{[\ce{AcO-}]}{C_\mathrm{a}} \approx 64 \%$$

Now repeat for $\mathrm{pH} = 3$:

$$C_\mathrm{a} = \left(\frac{1}{10^{-3}} + \frac{1}{1.8 \cdot 10^{-5}}\right) \cdot \left((10^{-3})^2 - (10^{-14})^2\right) \approx 0.05656$$

$$[\ce{AcOH}] = \frac{C_\mathrm{a} \cdot [\ce{H+}]}{K_\mathrm{a} + [\ce{H+}]} \approx 0.05556$$

$$[\ce{AcO-}] = C_\mathrm{a} - [\ce{AcOH}] \approx 0.001$$

$$\alpha = \frac{[\ce{AcO-}]}{C_\mathrm{a}} \approx 1.8 \%$$

So you see, in a more dilute mixture the $\mathrm{pH}$ is higher, and the dissociation reaction of acetic acid is 'more complete', whereas at higher concentration the pH is lower, and there is proportionally less acetate anion.

All of these systems are at equilibrium.


Equilibrium and Completeness of Reaction

The basic stoichiometry is done to understand the basic concepts of moles and ratios in reactions. Later on, as the reactions become more complex, alot of them don't go to completion and so for a lot of them, you need equilibrium constants. This being said, not all reactions end partway, some of them do go to completion. The most basic example is disassociation of salt ions in solutions. From basic stoichiometry, you learn that the ratio is one to one (like NaCl -> Na+ + Cl-). So Moles of NaCl added = the moles of Na and Cl ions in water. So basic stoichiometry is sometimes (not always) applicable but is essential for the learning process.


What is the difference between saturated and unsaturated fatty acids?

Saturated fatty acids are fatty acids with no double bonds and unsaturated fatty acids are fatty acid with one or more double bonds. This is the biggest difference between a saturated and unsaturated fatty acid.

There is two types of structural configuration for unsaturated fatty acids due to the presence of double bond; cis (kink/bend in fatty acid struture) and trans(No kink/bend in fatty acid struture).

.

  cis fatty acid (wikipedia)

  A cis configuration means that adjacent hydrogen atoms are on the same side of the double bond. The rigidity of the double bond
  freezes its conformation and, in the case of the cis isomer, causes
  the chain to bend and restricts the conformational freedom of the
  fatty acid. The more double bonds the chain has in the cis
  configuration, the less flexibility it has. When a chain has many cis
  bonds, it becomes quite curved in its most accessible conformations.
  For example, oleic acid, with one double bond, has a "kink" in it,
  whereas linoleic acid, with two double bonds, has a more pronounced
  bend. Alpha-linolenic acid, with three double bonds, favors a hooked
  shape. The effect of this is that, in restricted environments, such as
  when fatty acids are part of a phospholipid in a lipid bilayer, or
  triglycerides in lipid droplets, cis bonds limit the ability of fatty
  acids to be closely packed, and therefore could affect the melting
  temperature of the membrane or of the fat.

Iodine value test is used to determine degree of unsaturation in a unsaturated fatty acids; that is to measure number of double bonds present in a unsaturated fatty acids.

This link will explain the health benifits of saturated and unsaturated fatty acids


Does magnesium carbonate exist? what makes magnesium bicarbonate different from calcium bicarbonate

Both of these reactions are thermal decomposition reactions, but the difference basically occurs in the polarizing power of the cations $\ce{Ca^2+}$ and $\ce{Mg^2+}$ .

The thermal decomposition of $\ce{Ca(HCO3)2}$ is just the reverse reaction of the lime water test for the detection of $\ce{CO2}$ , where the carbonate combines with $\ce{CO2}$ and $\ce{H2O}$ to give the bicarbonate, so it stands to reason that the bicarbonate will break down to give the reactants we started with.

We can therefore, imagine that a similar decomposition will occur with $\ce{Mg(HCO3)2}$ to yield $\ce{CO2}$, $\ce{H2O}$ and $\ce{MgCO3}$. And in fact, a similar reaction does occur on drying the bicarbonate solution.

However, a small amount of $\ce{Mg(OH)2}$ is also obtained as a side product.
The reason is that, $\ce{CO3^2-}$ is relatively quite a large-sized anion.
Also, if you recall the diagonal relationship between $\ce{Li}$ and $\ce{Mg}$, then sizes of both $\ce{Li+}$ and $\ce{Mg^2+}$ will end up being similarly quite small relative to other members of their respective groups. Thereby,Fajan's rule will become important to consider here, and the carbonates of both these cations will experience similar internal polarization, and tendency to decompose relatively.

Hence, since we know that $\ce{Li2CO3}$ is the only Group 1 carbonate to decompose at laboratory temperatures to give $\ce{Li2O}$ and $\ce{CO2}$ ,(see this) we can expect $\ce{MgCO3}$ to also  break down into $\ce{MgO}$ and $\ce{CO2}$ on prolonged heating.

$\ce{CO2}$ being a gas will escape, and since Group 2 bicarbonates exist only in aqueous form, so the magnesium oxide produced will hydrolyse within the solution to produce the hydroxide, that is, $\ce{Mg(OH)2}$.

Note: Decomposition of $\ce{Li2CO3}$ is only "easy" relative to other members of Group 1. It still has a pretty high decomposition temperature of 1300 °C approximately

This value of 1300 °C might still not be accurate, see the green Note under 'Heating the carbonates' for more insight.


Lead-Acid Battery: Standard Electrode Potential

Your reactions are wrong.

$$\ce{Pb(s) + HSO^−_4(aq) -> PbSO4(s) + H+(aq) + 2e−}$$
$$\ce{PbO2(s) + HSO^−4(aq) + 3H+(aq) + 2e− -> PbSO_4(s) + 2H2O(l)}$$

Lead acid batteries are fairly reversible, hence their use as rechargable batteries. Anything that produces a gas, like in your reduction reaction, is unlikely to be reversible.


Dependence of steric hinderance on enol content

How can hydrogen form two bonds?

In the case of an enol, hydrogen does not form two full bonds.  What the drawing below is trying to show with the enol hydrogen is that one bond (the O-H with the solid line) is elongated, partially broken, while the other bond (the O..H with the dotted bond) is also elongated and only partially formed.  It's as if the partially broken OH bond and the partially made OH bond are each "half a bond", so all together the electron sharing is such that there is still just one full bond to hydrogen

  How does the steric repulsion destabilize the enol form?

Look at the bottom part of the drawing where we have inserted an alpha-methyl group into the 1,3-diketone.  Notice the steric interactions indicated by the arrows.  The situation looks similar to that seen with cis-2-butene.  The methyl groups are large enough (larger than a hydrogen) and close enough to one another that their electron clouds "bump" into one another causing what we often call "steric destabilization."  Remember that trans-2-butene is more stable than cis-2-butene because the trans isomer does not have the adverse steric interaction present in the cis isomer.  It is that same type of steric interaction that occurs here with the enol when an alpha-methyl group is present, and just like in cis-2-butene, it destabilizes the enol.  Therefore less of the enol will be present at equilibrium when an alpha methyl group is added to the enol.