How do the three Miller indices (hkl) denote planes orthogonal to the reciprocal lattice vector?

There are two equivalent ways to define the meaning of the Miller indices: via a point in the reciprocal lattice, or as the inverse intercepts along the lattice vectors.

The reflecting plane are parallel to a plane that includes the following three points:

$$\frac{\mathbf{a}_1}{h}; \frac{\mathbf{a}_2}{k}; \frac{\mathbf{a}_3}{l}$$

To show that the plane defined by these three points is perpendicular to $h \mathbf{b}_1 + k \mathbf{b}_2 + l\mathbf{b}_3$, we can calculate the dot product between vectors in the plane and the supposedly perpendicular reciprocal lattice vector. To get vectors in the plane, we simply subtract pairs of points in the plane.

$$\frac{\mathbf{a}_1}{h} - \frac{\mathbf{a}_2}{k}$$
$$\frac{\mathbf{a}_1}{h} - \frac{\mathbf{a}_3}{l}$$

Now we multiply it with the reciprocal lattice vector:

$$(\frac{\mathbf{a}_1}{h} - \frac{\mathbf{a}_2}{k}) (h \mathbf{b}_1 + k \mathbf{b}_2 + l\mathbf{b}_3) = \mathbf{a}_1 \mathbf{b}_1 - \mathbf{a}_2 \mathbf{b}_2 = 1 - 1 = 0$$
$$(\frac{\mathbf{a}_1}{h} - \frac{\mathbf{a}_3}{l}) (h \mathbf{b}_1 + k \mathbf{b}_2 + l\mathbf{b}_3) = \mathbf{a}_1 \mathbf{b}_1 - \mathbf{a}_3 \mathbf{b}_3 = 1 - 1 = 0$$

The mixed terms vanish because they are orthogonal. Because the dot products are equal to zero, we have shown that the vector is perpendicular to the reflecting plane.

  [OP] My understanding is that the real (real space) lattice vector is described by the Miller indices as $h{\mathbf  {a_{1}}}+k{\mathbf  {a_{2}}}+\ell {\mathbf  {a_{3}}}$, where the $\mathbf{a}_i$ are linearly independent basis vectors.

No, the significance of $h, k, l$ in defining reflecting planes in terms of the unit cell vectors is as described in the textbook (inverse intercepts along the lattice vectors).

  [OP] The last point that I'm having a bit of trouble with is how a plane can be defined in such a way that it is orthogonal to three basis vectors

That is not how it is defined. It is defined by the intersections of a plane with the three axes, i.e. through three points that are in the plane.


How to determine the greatest d orbital splitting?

The colour at which the complex absorbs reflects the wavelength of the d–d* electronic transitions. However, this colour is not the same as the transmitted colour (which you see), but is instead complementary to the transmitted colour. Therefore, a complex that appears purple is actually absorbing lower-energy light than a complex that appears red.


Why don't molecules of ionic compounds exist?

Ions are electrostatically bound.  One need only separate them with inegrated inverse squared distance work.  If you mix multiple ionic compounds in solvent, a melt, gas phase, the original paired associations are spontaneously scrambled.  Benzene plus xylene does not give you toluene.  "Separating" benzene is not electrostatic attraction.

Quaternized glycine, $\ce{^{+}N(CH3)3-CH2-C(=O)O^{-}}$, is betaine.  It is ionic, but the ions are covalently bound into a molecule.  Where are the discrete formula units in alumina?  You can draw lines to connect closest atoms, but those are not bonds. (Note fractional atom counts for atoms embedded in planes, edges, and corners.)


The effect of temperature on equilibrium

In order to understand an answer to this you need to be able to differentiate between system and surroundings
The system is the reactant mixture , now for an endothermic reaction the temperature of the reactant mixture will increase , however the temperature of the surroundings will decrease since heat is lost by the surroundings to the system , let's assume that an endothermic reaction took place in a test tube , if we put our hand around the test tube our palm will feel cooler since heat was transferred from our palm , however that heat was transferred to the reaction mixture hence the temperature of the mixture was raised . Hence this increase in temperature means we have altered the equilibrium conditions and thus excess heat needs to be lost therefore the eq will shift towards the exothermic reaction . However a decrease in temperature means the eq needs to gain heat which is in the form of an endothermic reaction


Do aromatic dienes undergo the Diels-Alder reaction?

Yes, you are correct.  Aromatic compounds are stabilized by their resonance energy. This means that the activation energy for reaction of an aromatic compound will typically be greater than that for a similar reaction of a non-aromatic compound.

That said, aromatics will react as the diene component in a Diels-Alder reaction if

the dienophile is extremely reactive (o-benzyne for example)

if the aromatic system is destabilized (for example by steric interactions or structural constraints) so that the resonance stabilization is decreased

or if the aromatic system is large enough that loss of one aromatic ring does not significantly destabilize the entire system.  Anthracene is such an example where the middle ring will react via a Diels-Alder reaction.

(image source)


Why is the pKa of the thiol group in N-acetylcysteine higher then the pKa of the thiol group in cysteine?

Shown below are the relevant ionization events of cysteine and N-acetylcysteine. The question is asking about the second ionization of each, the deprotonation of the thiol groups, but we see a similar difference in pKas for the carboxylic acid groups as well. The difference in each case is an ammonium group vs. an acylated nitrogen. It's true that the acetyl group is an electron withdrawing group, but it turns out that the cationic ammonium group is even more inductively electron withdrawing. For both the first and second ionizations, the difference in electron withdrawing power is enough to lower the pKa by >1 pKa unit.

iad22agp's explanation is a good way of rationalizing this. Looking at the first ionization, cysteine is cationic while N-acetyl cysteine is neutral. Since the rest of the structures are similar, we would assume that it's easier to take a positive charge away from a positively charged compound than a neutral compound. Similarly looking at the second ionization, it's easier to take a positive charge away from a neutral compound than a compound that's already negatively charged.

pKa refs: cysteine, N-acetylcysteine


Genarally, sulfur doesn't disproportionate like that: in fact, it typically behaves very similarly to chlorine (see Balancing the redox disproportionation of chlorine by half-reactions), which upon heating with alkali disproportionates to chloride $\ce{Cl-}$ and chlorate $\ce{ClO3-}$:

$$\ce{3$\overset{0}{\ce{Cl2}}$ + 6 OH- ->[$\Delta$] 5 $\overset{\mathrm{-I}}{\ce{Cl}}${}^- + $\overset{\mathrm{+V}}{\ce{Cl}}$O3- + 3 H2O}$$

Without heating instead of chlorate $\ce{ClO3-}$ hypochlorite $\ce{ClO-}$ is formed, so this is an irrelevant process in this case. Back to sulfur: when reacting with alkali, sulphide $\ce{S^2-}$ and sulfite $\ce{SO3^2-}$ are formed as the primary products according to the scheme:

$$\ce{$\overset{0}{\ce{S}}$ ->[OH-] $\overset{\mathrm{-II}}{\ce{S}}${}^2- + $\overset{\mathrm{+IV}}{\ce{S}}$O3^2-} \tag{1}$$

Half-reactions and net equation for (1):

\begin{align}
\ce{2 S + 4 e- &-> 2 S^2-} \tag{red1}\\
\ce{S + 6OH- &-> SO3^2- + 4e- + 3H2O} \tag{ox1}\\
\hline
\ce{3S + 6 OH- &-> 2 S^2- + SO3^2- + 3 H2O} \tag{redox1}
\end{align}

Note that (red1) has been multiplied by 2 to even the number the electrons in both (red1) and (ox1). The net equation (redox1) is obtained as a sum of both half-reactions.

Now, lets assume that the reaction you've written takes place, where where thiosulfate is formed instead of sulfite. In fact, it might as well happen when boiling solutions of sulfites with sulfur for a long time:

$$\ce{$\overset{0}{\ce{S}}$ ->[OH-] $\overset{\mathrm{-II}}{\ce{S}}${}^2- + $\overset{\mathrm{+II}}{\ce{S}}${}_2O3^2-} \tag{2}$$

Half-reactions and net equation for (2):

\begin{align}
\ce{2 S + 4 e- &-> 2 S^2-} \tag{red2}\\
\ce{2S + 6OH- &-> S2O3^2- + 4e- + 3H2O} \tag{ox2}\\
\hline
\ce{4S + 6 OH- &-> 2 S^2- + S2O3^2- + 3 H2O} \tag{redox2}
\end{align}

Reaction (redox2) would be the net reaction for the process you asked about, even though I'd say it's more likely a hypothetical one. Keep in mind that for as long as the reaction takes place in aqueous solution and the redox process involves acid-base transformations (like formation of oxoanions in this case), you need to take water into account.


Synthesis of 2-methyl-4-nitrophenol from benzene?

There is a lot of good thinking in your proposed synthetic strategy.  The one major shortcoming is, as @Mithoron points out, $\ce{OH^{-}}$ is not an electrophile.

We need to introduce a methyl, hydroxy and nitro group onto a benzene ring.  If we look into our synthetic "toolbox" here are some thoughts that might come to mind:

A methyl group can be readily introduced using the Friedel-Crafts alkylation reaction with$\ce{CH3Cl}$. However, be aware that there are some general issues with the Friedel-Crafts alkylation reaction.  For example, if the alkylating agent can rearrange to a more stable carbocation it will likely do so ($\ce{CH3Cl}$ can't rearrange).  Also, the mono-alkylated product is more reactive than the starting material, so polyalkylation can occur (but in your case, we can use an excess of benzene - it's cheap - and get a good yield of the desired product toluene).  Finally, the Friedel-Crafts alkylation does not work well with deactivated aromatic rings. Therefore, the comment suggesting we run the alkylation after introducing the nitro group is incorrect; we must use the Friedel-Crafts alkylation before we introduce the nitro group.
There aren't any easy ways to electrophilically introduce a hydroxy group into an aromatic ring.  When I need an aromatic hydroxyl substituent, I think of the Sandmeyer reaction.
The nitro group can be introduced via straightforward electrophilic nitration.

There are a lot of ways to pull this synthesis together, but here is a relatively direct proposal.

The first step is our Friedel-Crafts alkylation to prepare toluene.
Next, we'll introduce our ortho hydroxy substituent.  To do this let's first sulfonate using fuming sulfuric acid (oleum) and block the para position. Blocking the para position will simplify our next step - we won't have to separate ortho and para isomers and we will get a higher yield of only the desired ortho (to the methyl group) product. Now the sulfonic acid group (meta-director) and the methyl group reinforce one another and will direct our next electrophile to the position ortho to the methyl group.
Now we nitrate using nitric and sulfuric acid and get only the product where the nitro group is ortho to the methyl group.
Next, we heat the reaction. The sulfonation reaction is readily reversible, by heating our reaction we can readily remove the sulfonic acid substituent since it has served it's purpose and is no longer needed.
Now it's time for the Sandmeyer reaction. We first reduce the nitro group with tin and hydrochloric acid to an amino group.  Then we diazotize it and decompose the diazonium salt in the presence of acidic water to introduce our hydroxyl group.
Finally we nitrate one more time to introduce the nitro group.  The directing effect of the hydroxyl group is much stronger than that of the methyl group and will primarily direct the nitro group into the position para to the hydroxyl.


Molecular orbital diagram of a complex including an oxido ligand

Although chemists and chemistry students alike often indiscriminatorily use the $\mathrm{t_{2g}}$ and $\mathrm{e_g}$ descriptors for the d-orbitals of a d-block metal in coordination complexes, those designations are only truly correct in an octahedral environment. The vast majority of coordination complexes, however, are not octahedral but distorted octahedra, octrahedroids with non-symmetric ligands and others. This is also the case for oxido-ligand coordinated complexes that you are asking about.

To restrict the discussion to a single species (although it can easily be expanded to others) I will craft my answer explaining the situation in the vanadyl cation $\ce{VO^2+}$. I will discuss the pentaaquaoxidovanadium(IV) cation $\ce{[VO(H2O)5]^2+}$ as Ballhausen and Gray analysed.[1] I will label the direction of the oxido ligand $+z$ (in line with their scheme).

Constructing the molecular orbital scheme

The first important thing to note is that the complex indeed does not display octahedral $O_\mathrm{h}$ symmetry but $C_\mathrm{4v}$. That is great for us because it facilitates practically every step in the construction of molecular orbitals. A character table for the $C_\mathrm{4v}$ point group can be found in orthocresol’s collection (Note: MathJax-heavy link. Give it time to load). In the tables below I have given the different metal and ligand orbitals and their irreducible representations.

$$\textbf{Table 1:}\text{ irreducible representations of the metal orbitals in }\ce{[VO(H2O)5]^2+}\\
\begin{array}{cccc}\hline \text{irrep} & \text{s orbital} & \text{p orbital} & \text {d orbital} \\ \hline
\mathrm{a_1} & \mathrm{4s} & \mathrm{4p}_z & \mathrm{3d}_{z^2}\\
\mathrm{a_2} & & & \\
\mathrm{b_1} & & & \mathrm{3d}_{x^2 - y^2}\\
\mathrm{b_2} & & & \mathrm{3d}_{xy}\\
\mathrm{e} & & \mathrm{4p}_x, \mathrm{4p}_y & \mathrm{3d}_{xz}, \mathrm{3d}_{yz} \\ \hline \end{array}$$

$$\textbf{Table 2:}\text{ irreducible representations of the ligand (group) orbitals in }\ce{[VO(H2O)5]^2+}\\
\begin{array}{cccc}\hline \text{irrep} & \ce{O^2-} & \ce{$eq${-}H2O} & \ce{$ax${-}H2O} \\ \hline
\mathrm{a_1} & \mathrm{p}_z & {\sigma_1} \atop {\pi_1} & \mathrm{p}_z\\
\mathrm{a_2} & & \pi_2 & & \\
\mathrm{b_1} & & {\sigma_2} \atop {\pi_3} & \\
\mathrm{b_2} & & \pi_4 & \\
\mathrm{e} & \mathrm{p}_x, \mathrm{p}_y & {\sigma_3, \sigma_4}\atop{\pi_5, \pi_6, \pi_7, \pi_8} & \mathrm{p}_x, \mathrm{p}_y\\ \hline\end{array}$$

If required, I can supply images of the π and σ symmetric orbitals of the equatorial ligands once I am back at a PC that has ChemDraw (not before the 8th of January). It is not always fully obvious which orbitals interact with which ones; however, it is very obvious that the π2 ligand group orbital is nonbonding with respect to the $\ce{L\bond{->}M}$ bond.

In a second round, we need to consider the strength of interaction. The oxido ligand is much closer to the vanadium centre ($d(\ce{V=O}) \approx170~\mathrm{pm}; d(\ce{V\bond{<-}OH2}) \approx 230~\mathrm{pm}$) meaning that any interactions with that will be considerably stronger. (There are hints pointing towards an even more elongated axial aqua ligand, but that would be secondary.)

The resulting scheme you construct should show a strong interaction of the oxido ligands $\mathrm{e}$ orbitals ($\mathrm{p}_x$ and $\mathrm{p}_y$) and the metal’s $\mathrm{e}$ orbitals ($\mathrm{d}_{xz}$ and $\mathrm{d}_{yz}$). This can be interpreted as the oxido ligand donating both p-type lone pairs to vanadium; while the resulting bond is typically depicted as $\ce{V=O}$ a better depiction may be $\ce{{V^-}#O+}$. The other interactions of $\mathrm{e}$ orbitals are considerably weaker.

The orbital scheme as published

Ballhausen and Gray, who I mentioned previously, also included a quantitative scheme in their paper, albeit having omitted the ligand-metal π interactions with the exception of those of the oxido ligand.[1] Thus, their scheme is a little bit simpler than the one I would have drawn. The low-lying σ orbital is the oxido ligand’s; the orbitals labelled π are obviously the oxido ligand’s $\mathrm{p}_x$ and $\mathrm{p}_y$ orbitals. Due to the lack of aqua-π orbitals, the $\mathrm{d}_{xy}$ orbital appears unmodified. The $\mathrm{e}$ orbitals are raised notably, as is the $\mathrm{b_1}$ orbital ($\mathrm{d}_{x^2-y^2}$) due to the ligand’s σ interactions. Finally, the most destabilised metal d orbital is $\mathrm{a_1}$ corresponding to the $\mathrm{d}_{z^2}$ orbital.

The resulting scheme can be seen as an atypical Jahn-Teller distortion of an octahedral complex. In typical Jahn-Teller distortions, the $z$-ligands would be moved away resulting in a stabilisation of orbitals with $z$ contribution. Here, one $z$-ligand is moved closer resulting in a destabilisation of anything including a $z$ contribution.

Reference

[1]: C. J. Ballhausen, H. B. Gray, Inorg. Chem. 1962, 1, 111. DOI: 10.1021/ic50001a022.


Does the carbocation rearrangement in a SN1 reaction specify a change in degree?

The most common carbocation rearrangements involve [1,2] hydride or alkyl shifts. As a typical result a secondary carbocation is converted to a more stabilized tertiary carbocation.

In the course of these reactions, the migrating group and the positive charge change place, they move to opposite directions.

This is not the case in your example.

I fail to see how your transformation can be described as a cationic rearrangement!


I suggest taking the experimental route and exposing another test piece of the plastic you are working with to the hardener. Keep an eye out for any softening, swelling, change of color and any other indicators of a reaction occuring.


Which reagent can I use to remove carbon from oven?

As a disclaimer, this is for information only.  Anyone using this information assumes all associated risks.  The manufacturer's label should be followed for any cleaning products.

The exact chemical composition of the goo is not precisely known. It is likely either tar or creosote existing as some form of hydrocarbons.  These hydrocarbons are tangled up to the point it is extremely difficult to dissolve.  The temperature required to burn this material is likely too high for anything short of a ultra high temperature industrial furnace.

A good way to clean it is to use oven cleaners which take advantage of the saponification reaction. Saponification will loosen the material; however, some scrubbing and/or scraping will still be involved.

For safety when oven cleaning be sure to have good ventilation.  Wear gloves, eye and face protection, as well as not having skin exposed. Caustic cleaning compounds and vapors from the cleaning are hazardous.

Most oven cleaners have multiple components.

A caustic solution causes a saponification reaction.  The saponification reaction is commonly used to make soap from saponifiable oils.  Because the oils in the material to be cleaned are likely saponifiable,  the saponification reaction makes the material much more likely to dissolve in water.
A foaming component creates a foam. When foam bubbles pop, the popping action works to break up the tangled bonds.
A compound may be included to ensure the entire solution will stick to a non-horizontal surface.
Another compound may be included to seal the solution so as to prevent oxidation.  Oxidation may interfere with the saponification.

The trick to getting oven cleaners to be effective, in my experience, is in the timing.  Timing is important because the contents of tars are not exactly known, so the softening reaction is brief.

A warm surface is easier to clean; however, a hot surface will dry the cleaner before it has a chance to work.
The oven should not be powered when attempting any cleaning.
When any oven cleaner is first applied, it is important to leave the cleaner alone and give it time to work. Ten minutes is usually sufficient.
Once dissolution occurs there is a brief window when the tar or creosote can be easily removed with a nylon brush or sometimes even a paper towel.
When the materials dissolved, vapors which were previously trapped in the tars may be released.  These released vapors are not good to breathe.
Sometimes it is necessary to get deeper layers of tar.  The cleaner can be reapplied up to three times before the solution dries if necessary.  Each time allow it to go undisturbed for ten minutes.  (In my experience, by the fourth application, cleaning the tar is simply not worth the effort.)
If too much time passes and the solution dries, then the tar is more difficult to clean.  The caustic solution can also damage metal components.
Rinse the oven of all cleaner when done.  Caustic residues can damage the oven if left on the surface.


Why must NaOH be kept dry during storage?

Without a proper closed container, sodium hydroxide readily attracts water and become a viscous solution (concentrated sodium hydroxide). This solution readily attracts carbon dioxide in air and reacts with it.

$$\ce{CO2(g) + H2O(l)<=>H2CO3(aq)}$$
$$\ce{2OH^{-}(aq) + H2CO3(aq) -> 2H2O(l) + CO3^{2-}(aq)}$$

(The $\ce{OH^{-}}$ in the second equation comes from $\ce{NaOH}$.)

This reaction decrease the amount of $\ce{NaOH}$ and thus the titration result will contains error.

Actually, the first equation can happen even without sodium hydroxide. Carbon dioxide in air is known to spontaneously dissolved into water and become carbonic acid. This is why the distilled/deionized water which in contact with air will have pH of around 6 after some time. With the presence of sodium hydroxide in water, the second equation greatly drives the equilibrium of the first solution to the right. Concentrated sodium hydroxide is used in some experiment as a carbon dioxide absorber.

Another possible problem that might occur is that concentrated sodium hydroxide can react with glass. Althought this is a very slow reaction, it might causes some unexpected error or ruin the glassware.


Clausius-Clapeyron - Slope of the best-fit-line

Using your formula, $\mathrm{slope}=\frac{y_2-y_1}{x_2-x_1}$, and eyeballing the most extreme points on the chart, $\mathrm{slope} = \frac{0-6.2}{0.005 - 0.0033}$, I get a slope of 3647 K, pretty close to the "official" answer.

If you want to get more fancy, you could use a software program to fit this data.

my_data <- data.frame(vapor_pressure = c(1.1, 6, 28, 95, 293, 521),
                      temperature = c(200, 220, 240, 260, 280, 300))

# linear regression on transformed variables
my_data$ln_vp <- log(my_data$vapor_pressure)
my_data$inverse_T <- 1/(my_data$temperature)
linear_fit <- lm(data = my_data, formula = ln_vp ~ inverse_T)
linear_fit$coefficients['inverse_T']

inverse_T: -3787.46829114867

This answer is even closer to the "official" answer.  The remaining difference is probably due to rounding errors or differences in fitting algorithms used.  It only affects the answer by 0.3%.


AAS background correction with Zeeman effect

The very short answer is that properly applied magnetic fields, via the Zeeman effect, can cause atomic absorption bands to split into multiple bands, and these bands have different polarization properties. This is the simulation model I use to demonstrate the spliting and polarization behavior:

This model demonstrates the transverse and longitudinal Zeeman effects.  The theory of Zeeman splitting is covered elsewhere, but the basic idea is this: in a magnetic field, atomic transitions may be “split”, because the levels associated with the atomic transitions are no longer degenerate (due to those pesky magnetic quantum numbers and the associated selection rules for allowed changes in them).  Hence, the transition splits into several transitions.  The simplest case, that of ordinary Zeeman splitting, involves a splitting into three bands: 1 $\pi$ band and 2 $\sigma$ bands.  In the transverse field case, where the magnetic field is transverse to the light propagation direction, the $\pi$ band is linearly polarized parallel to the magnetic field direction.  It does not spectrally shift as a consequence of the application of the magnetic field.  The two $\sigma$ bands are spectrally shifted, by an amount proportional to the strength of the applied magnetic field, to each side of the $\pi$ band.  The $\sigma$ bands are linearly polarized orthogonally to both the light propagation direction and to the magnetic field direction.  The light propagation is the “z” direction.  In the transverse magnetic field orientation, the magnetic field axis is along the “x” axis.  Therefore, the $\pi$ component absorbs light that is plane polarized along the x axis.  The two $\sigma$ components absorb light that is plane polarized along the “y” axis. The next figure shows the unshifted $\pi$ band in the center and the 2 $\sigma$ bands shifted out of harms way:

So the light source is linearly polarized and the $\pi$ band cannot absorb it: it could only absorb the orthogonal polarization, which is not present. The 2 $\sigma$ bands would be able to absorb the light source's polarized light, except that the magnetic field has spectrally shifted them out of harm's way. Thus, when the magnetic field is on, only the broad background absorbance absorbs light. When the magnetic field is off, the background still absorbs light and the atoms do as well.

In the anomalous splitting situation, it is possible to have rather complex splittings, even neglecting hyperfine splitting complications, isotope effects, etc.  In the simplest anomalous splitting, there are two $\sigma$ bands and two $\pi$ bands.  The $\pi$ bands are spectrally shifted, but are still polarized parallel with the magnetic field.  Their spectral shift is less than that of the $\sigma$ components.  In the longitudinal Zeeman configuration, the magnetic field is along the light propagation axis, so there is no $\pi$ component to the absorbance because light cannot be polarized along its propagation direction.  The two $\sigma$ components are circularly polarized, with opposite handednesses, and are spectrally shifted by an amount proportional to the strength of the applied magnetic field. The next figure shows the splitting for calcium, in the longitudinal Zeeman configuration:

Both of these Zeeman configurations are used for background correction in atomic absorption spectroscopy.  The basic idea behind all background correction schemes, in atomic absorption spectroscopy, is this: perform two absorbance measurements and take their difference.  Both measurements should be designed to do an excellent job of measuring the interfering background absorbance, but only one of the measurements should do an excellent job of measuring the desired atomic absorbance.  The other measurement should do a poor job of measuring the desired atomic absorbance.  Differencing then discriminates against the interfering background absorbance, because it is common mode, and it yields the desired atomic absorbance, because it is differential mode.  Just how well the scheme works depends on a variety of factors, so, as it happens, there are several viable background correction schemes in use.

The Zeeman effects are useful for background correction because, by suitable use of polarizers and magnetic fields, the desired atomic absorbance can be effectively “turned off”.  This makes it possible to perform a measurement where the absorbance measured is almost exclusively due to just the interfering background absorbance.  As noted above, this measurement is one of the two needed for the differencing involved in background correction. Hope this answers your question.


As with all NMR analysis problems like this, it is good to have a strategy to interpret data. Perhaps rather than looking at your spectrum and trying to rationalise your structure, look at your structure and predict what you might expect to see in your spectrum. You'll find it much easier to rationalise the appearance of your peaks, especially in the case of overlapping peaks as you have here.
So, in your molecule, what splitting would you expect?

You have 11 proton environments, so your integrals look fine. Your chloroform doublet is  a bit of a worry though. Perhaps your sample needs to be mixed a little better? You've got 3 distinct spin systems that will give characteristic splittings:

trans double bond. These two peaks will appear as distinct doublets, with couplings of around 12-17Hz. I can see one very clearly. The other, which we know to be upfield of the one at 7.77 because of the tenting is probably under the mess of stuff between 7.5-7.6. Being next to the carbonyl will have strong influence on the chemical shifts of these, and through a bit of electron-arrow pushing we can show that the proton closest to the carbonyl is shielded, and the other deshielded. Hence, the signal at 7.77 is likely to be your proton β to the carbonyl group (and closest to the para-substituted benzene ring)
para-substituted benzene ring. Again a distinct splitting pattern; these will look like a pair of doublets (actually a second order pattern) with coupling about 7.5Hz. I can see one probably at 7.4. Its partner will also be under the mess between 7.5 and 7.6
a mono-substituted benzene ring, which will have three sets of signal, with ratio 2:2:1. A carbonyl group is strongly electron withdrawing, deshielding the ortho protons, which is probably the signal up at about 8. The others.... under the mess between 7.5-7.6.

This is clearly a homework/lab problem, so this should be enough to keep you going until the next question.


Will water allow salts to be dissolved in methanol (that are otherwise poorly soluble)?

This is a highly complicated question, and the answer can't be derived by logic alone.

Maybe the liquid phase will remain mixed, and just dissolve some salt (more than would dissolve in methanol, but less than would in pure water). Or maybe it will separate into water-rich and alcohol-rich layers, with nearly all dissolved salt staying in the former. This may be considered an extreme case of salting out.

I know for sure that separation does occur in water/ethanol/$\ce{CaCl2}$ system, even though ethanol by itself is miscible with water in any ratio, as is methanol. Then again, $\ce{CaCl2}$ is much better soluble in water than $\ce{NaCl}$, so the latter might not be strong enough to produce the same effect.

The bottom line is that you can't tell for sure until you try it. Chemistry is an experimental science, after all.


Why does lithium forms no ethynide on reacting with ethyne?

Sodium alkynide was formed by reacting sodium and acetylene but in case of lithium alkynide, it can be formed by reacting acetylene with organolithium reagents like BuLi, Lithium amide, LiHMDS. From wikipedia:

  Lithium amide, LiHMDS, or organolithium reagents, such
  as butyllithium, are frequently used form lithium acetylides:

  $\ce{H-C#C-H + BuLi ->[THF][-78℃] Li -#-H + BuH}$

The above process is well described in this paper which @Tyberius linked in the comment section.


Radiationless energy transfer: how to arrive at the equation that give R0 between two chromophores?

I think I found the answer to my own question through some help from some classmates. What the professor did, was to solve the equation mathematically by converting $R_0^6$ to $1$.

So first, the efficiency of the transfer $E_\mathrm{eff}$ is actually a ratio of the actual transfer efficiency divided by the maximum efficient (which we should assume is "one", unless we are told otherwise.

The next step is a bit hard to explain, as I am not that good in math, but I will try my best. What I understand is that you can "gather" the two $R_0^6$ by multiplying both the numerator and denominator by what we should think of as the number one, but we actually divide by $R_0^6$ and in this way we don’t have to include $R_0^6$ that we divide by on both sides of the equal sign - as it is technically the number one (mathematically allowed), as following:
$$\frac{E_\mathrm{eff}}{E_\mathrm{eff,max}} = \frac{\frac{R_0^6}{R_0^6}}{\frac{R_0^6}{R_0^6}+ \frac{r^6}{R_0^6}}$$

We then get as following:
$$\frac{E_\mathrm{eff}}{E_\mathrm{eff,max}} = \frac{1}{1 + \frac{r^6}{R_0^6}}$$

So by applying some math and assumptions we can solve the equation like this, and what I have explained in my question should therefore be correct.


What is the purpose of the electrolyte in the half-cell where oxidation is taking place?

The primary role of the anode chamber electrolyte (the anolyte) in this case is to provide sufficient conductivity in that chamber.

If deionized water were used in the anode chamber, the conductivity of that liquid would be extremely low. Thus, it's important for there to be something dissolved there in order for any appreciable amount of current to pass through the overall electrochemical cell. The overall electrochemical circuit involves both electronic conduction (in the wires and electrodes) and ionic conduction (in the two electrolytes), and it will not function efficiently/effectively if any element of that circuit is too poorly conductive.

That said: It's not crucial that it be specifically $\ce{ZnSO4}$, however.  $\ce{NaCl}$, $\ce{H2SO4}$, $\ce{NH4NO3}$, or a wide variety of other chemicals might work just as well (or even better!).  However, a sulfate salt makes sense here because (at least):

It provides relatively high conductivity
Its salts with metal cations, including $\ce{Zn^{2+}}$, are relatively water-soluble.
In this specific cell configuration, using sulfate makes sense because it's the same anion as in the $\ce{CuSO4}$ used for the catholyte, and thus you don't have to worry about counter-diffusion of different anions across the salt bridge.

It further makes sense to use $\ce{ZnSO4}$ specifically as the anolyte salt because it provides the same cation as you'll be forming from your anode reaction. This will result in a more stable electrochemical cell, because having a different, potentially electrochemically-active cation in the anolyte would most likely change both the thermodynamic and the practical operating characteristics of the cell.


If you imagine particles to be like billiard balls (the way you're taught in school) then this of course makes no sense. The only way you can understand it is to accept the quantum mechanical description of particles: that they are probability distributions that exhibit both wave-like behaviour and particle-like behaviour depending on how you measure them.

Once you accept that a particle is best imagined as a probability wave smeared out through space then the Heisenberg uncertainty principle is a rather simple mathematical result relating to Fourier transforms. The more concentrated some function $f(x)$ is, the more spread out its Fourier transform $f(k)$ is. And since position and momentum space are Fourier transforms of each other (see here) then this gives rise to the uncertainty principle.

In other words, if you measure the position very precisely (highly peaked $\psi(x)$) then the momentum representation $\psi(k)$ will become very broad (imprecise), and vice versa.


It's subliming (aided in this case by decomposition) and re-depositing.

Substances that turn directly from solid to vapor (e.g. naphthalene, water ice) are said to sublime, rather than evaporate. The inverse, changing from vapor to solid, is termed deposition.

When water vapor deposits on very cold surfaces, such as "dry ice", it grows into amazing dendritic crystals:


How to find out the energy gained from hydrolysis of Dimethylsulfoniopropionate?

When the literature refers to "hydrolysis" of DMSP, it is referring to the following cleavage reaction, even though no water is involved in the overall reaction:

DMSP --> DMS + acrylate + H+

See for example at page 156 of Biological and Environmental Chemistry of DMSP and Related Sulfonium Compounds: "hydrolysis of DMSP to DMS and acrylic acid".

For the above reaction:

$\Delta G = -11.8$ kcal/mol

Source: MetaCyc Reaction: 4.4.1.3

This means 11.8 kilocalories (49.4 kilojoules) of energy are released for each mole of DMSP that reacts.


What is viscosity in terms of chemistry?

More or less similar…I will try to put brief concepts of viscosity in terms of pharmaceutics which is somewhat another branch of chemistry, where it is very much applicable in drug design.

Viscosity is an expression of the resistance to flow of a system under an applied stress. The more viscous a liquid is, the greater is the applied force required to make it flow at a particular rate.

  Viscosity is defined in terms of the force required to move one plane
  surface past another under specified conditions when the space between
  is filled by the liquid in question. More simply, it can be considered
  as a relative property, with water as the reference material.

Viscosity implies that the liquid flows even under the smallest stress and does not return to its original shape or form once the stress is removed.

Viscous deformation, i.e. viscous flow, occurs in accordance with Newton’s law,

$\ce{σ=ηỳ}$

  where the applied stress σ results in flow with a velocity gradient ỳ
  or rate of shear. The proportionality constant η is termed viscosity,
  while its reciprocal is called fluidity.

Viscosity has also been
 described as the internal friction in the fluid as it corresponds to the resistance of the fluid to the relative motion of adjacent layers
 of liquid.

The viscosity of simple liquids (i.e, pure liquids consisting of small molecules and solutions where solute and solvent are small molecules) depends only on composition, temperature, and pressure. It increases moderately with increasing pressure and markedly with decreasing temperature.

There is quite a lot to discuss on viscosity in depth but these are just shallow concepts to understanding the term.

References

Martin’s Physical Pharmacy and Pharmaceutical Sciences : Rheology
Remington: Essentials of Pharmaceutics: Rheology
Ansel’s Pharmaceutical Dosage Forms and Drug Delivery Systems : Special Solutions and Suspensions
Aulton ’s Pharmaceutics The Design and Manufacture of Medicines 4th edViscosity Rheology, and flow of fluids


Expanding @Mithoron's comment into an answer, $\ce{^{106}Ru}$ is used in radioisotope thermoelectric generators that power satellites. Plutonium-238, curium-244 and strontium-90 are the most often cited candidate isotopes for this purpose but other isotopes such as polonium-210, promethium-147, caesium-137, cerium-144, cobalt-60, curium-242, americium-241 and thulium isotopes have also been studied.
Apart from being an energy source, it also finds its application in medicine. $\ce{^{106}Ru}$ is used in cancer radiation therapy, especially to treat eye and skin tumors.


What would be the mechanism of the given reaction?

Bottom line: $\ce{HSO4-}$ is a really bad nucleophile, and it doesn't enter in the picture.

(I realize after scanning that I'm missing the "+ H+" alongside the final product.)

Question for extra credit: after the second step, why is the third step (carbocation migration) favorable rather than directly reforming the double bond on the right-hand side ring?


Relation between number of equivalents of reactants

Step 1: Balance the equation. Your equation as written $\ce{ Sb2S3 + F2 -> SbF5 + SF6}$ is not balanced. You do not have the same numbers of sulfur, antimony, or fluorine atoms on each side of the equation. You cannot say anything about equivalents or stoichiometry with an unbalanced equation. Hint: Start with $\ce{Sb}$ and $\ce{S}$ and balance $\ce{F}$ last.

Step 2: Determine the relationship between $x$ and $y$. Since $x$ is antimony and $y$ is sulfur and they are in the same compound initially, they have a relationship (but they are not equal). In $\ce{Sb2S3}$, there are 2 sulfur atoms for every 3 antimony atoms. What does that tell you that the ratio of $\frac{x}{y}$ must be?

Step 3: Use what you now know about $x$ and $y$ and the correct stoichiometric equivalents (which should give you $z$) to show that $x+y=z$.


If we prepare a liquid-liquid solution of equimolar concentration, we know that the more volatile liquid component will have a greater composition in the vapour phase.

Let us keep increasing the concentration of the less volatile component in the solution. Initially, even though the solution is richer in the less volatile component, the vapour phase will be richer in the more volatile component but there will come a time when the vapour phase will be richer in the less volatile component. It may so happen that the composition of the vapour phase becomes equal to the composition in the solution phase. At this point, azeotrope is formed.

Note that if azeotrope is formed at all, both the solution and the vapour phase will be richer in the less volatile component.

Take the example of ethanol-water solution.
Ethanol boils at 78.4 °C, water boils at 100 °C. Azeotropic mixture is obtained at about 95% percent of ethanol. Note that the less volatile component is present in greater amounts when the azeotropic mixture is formed.

I will keep adding more to it as it comes to me.


What are the effects of Lithium Dioxide on the body?

Is it safe to inject Lithium Dioxide into our body?

No. It will do bad things.

  Does it have any side effects?

Yes. It will do bad things.

As lithium is an alkali metal, it does not usually form covalent bonds, so we should treat whatever it makes as ionic. The common form of lithium oxide is $\ce{Li2O}$, which is not that scary. "Lithium dioxide" would be $\ce{LiO2}$. This implies that that lithium is tetravalent, as $\ce{Li^4+}$. This is not possible, as it requires lithium to give away 4 electrons but it does not have 4, only 3. This means that oxygens are in the superoxide form: is $\ce{O2-}$ instead of the more stable $\ce{O^2-}$. That is definitely not stable in isolation at ambient conditions (unlike other alkali superoxides) and I'm not sure where you will get that stuff.

The result of that is that one molecule (or whatever it is) of $\ce{LiO2}$ will really want to convert itself to $\ce{Li2O}$. The superoxide ion is not going to do fun things. For each $\ce{O2-}$ you're going to need 3 more electrons to make it stable, so that is very strong oxidiser.

As a comparison to what this could do to you, I would guess it would be worse than injecting pure hydrogen peroxide (you have about 5% of that in hair bleach), or pure sodium hypochloride (similar 5% in kitchen bleach). This is definitely not going to treat "palladium poisoning" because it will probably kill you beforehand.

That said, using an oxidiser (such as lithium dioxide) to treat noble metal poisoning (such as palladium) seems dubious to me. I'm not a doctor, but I reckon that there are better method to treat that. So why lithium and palladium? I guess is sounds more credible than "using titanium to treat potassium poisoning" but less credible than "using dysprosium to treat beryllium poisoning", both of which make absolutely no sense as "using lithium to treat palladium poisoning".


Solubility of Gold in low concentrations

Whenever you prepare very dilute working standards, it's good practice to do a stability study to see how long you can store them. There are many factors that can influence the stability of these solutions. The choice of container can make a big difference. Dilute metal solutions should be stored in polyethylene or polypropylene containers, not glass. Another well-known problem is photoreduction. Gold solutions should be stored in opaque containers and be exposed to light as little as possible. Some solutions may also be sensitive to atmospheric exposure. (though I don't think it's an issue for gold)

I don't have experience with working with gold in hydrofluoric acid, but such a large change in such a short period of time in the absence of the above factors may indicate some actual reactivity with your matrix, for which matrix-matching might help, though you may be better off reformulating your solutions if that's the case. It may also be an option to just make fresh working standards very frequently or increase storage life through refrigeration.


What is the product after treating ethene successively with chlorine and water, sulfuric acid, and potassium hydroxide?

The first reaction is a hydrochlorination reaction:

The second reaction is an acid catalyzed ether synthesis. The the solution is full of other 2-chloroethanol, which is capable of acting as a nucleophile in an $\mathrm{S_N2}$ reaction to eject $\ce{H2O}$. This is reaction will have a much lower energy barrier than the dehydration you proposed, which requires the extraction of an aliphatic hydrogen. The ether synthesis mechanism can be seen here:

The final step is a series of two $\mathrm{E2}$ reactions (shown condensed here):

$\hspace{4cm}$


There are a number of different effects:

Rayleigh scattering is the predominant effect for the tiniest particles, much smaller than the wavelength of light, i.e. before they've coalesced. It is dependent on the fourth (!) power of the frequency of the radiation, so short wavelength light is scattered toward the eye. Looking through the colloidal mixture, it should look redder.
Tyndall scattering applies to larger colloidal particles, and it too is dependent on the fourth power of the frequency of the radiation and preferentially scatter bluer light, so the yellow of the sulfur would be less apparent.
Finally, particles coalesce and scattering becomes less important than reflection from the larger particles.
Perhaps sulfur nano-particles behave a bit differently from bulk material due either to quantum effects or to surface effects. See Elemental Sulfur Nanoparticle Coarsening... If someone has experience with nanoscale sulfur, please feel free to comment or edit this.


Number of resonating structures (hypothetical)

Resonances structures are a model of the quantum mechanical orbital mixing.  To my mind they are a complete farce.  They are also quite useful

A model is a simplification of a problem, usually based on ignoring negligible factors.  So this question really distills down to what is negligible and what isn't?

That depends on roughly two factors:

The conditions of reaction.

Are the conditions strongly ionic or acidic?  Then charge separation may become relevant.

The rigor of the problem.

Is this for a high school course? A Ph.D. thesis? A bar-room chat with your mates?

Unfortunately, like most of chemistry, there is not an absolute answer here.


Maximum required order for the Boys function

I think your logic seems fine.

When you make your look up tables, you want to make a table for each $n$ up to some value of $n_\text{max}$. Consider you want to calculate an integral that only requires $n=2$, then it would be very inefficient to calculate $F_{12}$ and then down recur.

In your equation (9.8.12) we can notice that in the sum we have the Boys function as $F_{n+k}$. According to "Molecular Electronic Structure Theory", we might need the first $6$ terms of the sum. I.e. when doing your tabulation you will need to calculate all $n$ up to $n_\text{max}+6$.

Also when you do your tabulation you need a maximum range of your tabulated Boys function, i.e. a maximum value for where you can start using the long range approximation afterwards (else your tables will be HUGE). The long range approximation is given as:

$$F_n(x)\leq\frac{(2n-1)!!}{2^{n+1}}\sqrt{\frac{\pi}{x^{2n+1}}}\tag{9.8.10}$$

For which values of $x$ to start using this approximation for different $n$ you can take a look at Table II in Efficient recursive computation of molecular integrals over Cartesian
Gaussian functions. I am sure newer litterature on this topic exist, but I haven't encountered it.

As a side note, you can also calculate the Boys function with out tabulation (this might be less effective). By using the following formula:

$$F_n(x)=\exp(-x)\sum_{i=0}^\infty\frac{(2n-1)!!(2x)^i}{(2n+2i+1)!!}$$

This formula makes an implementation of the Boys functions much easier. If you are making an integral code from scratch, I would suggest to implement the above formula. You can always speed up the code later. Note with the above formula you can also just calculate the Boys function for the maximum $n$ you need and then use downwards recursion. If your goal is to implement the Boys function via. tabulation, then you can also use the above formula to generate the values for your look-up tables.


what keeps an identical atoms of an element together?

well atoms of the same element will typically either, forma metallic lattice, a giant covalent structure ( a macro-molecule), or from a simple molecular structure. but all the bonds will either be metallic or covalent bonds. if it forms a simple covalent structure, there will be van der Waal's forces acting between the molecules to keep it as a solid or liquid.


What decides the order of priority of functional groups in organic compounds?

It’s an arbitrary agreement by IUPAC, although note
that there is some correlation between the
oxidation state of the carbon and the priority
(more oxidized groups tend to be higher
priority). Generally, the functional group which shows its characteristics more dominantly even when present with other group gets a higher priority.


Solubility of oxygen in liquid hydrogen; is anything near 1:2 (molar ratio) possible?

It would be the other way round.
Since the oxygen is the first to liquify, it would be hydrogen dissolved into liquid oxygen.

In this article they show the following calculated vapour liquid equilibrium diagram.

The 2:1 mixture, $x_{\ce{H2}}=0.67$, would exist at around 85 K and 50 bar.


How to assign overlapping multiplets in 1H NMR spectra?

Referring to your comment on Buttonwood's answer:

Unfortunately, in this case I only have 1D HNMR and IR available to analyse the product

If this is really the case, then you are essentially out of luck. It is not possible to extract much information from a series of overlapping multiplets, which may well be coupled to each other (these strong coupling effects are likely the cause of the "complex splitting" you mention). It may be the case that by going to a higher field you can alleviate some of this, as multiplets will be more dispersed along the frequency axis. However, you will quickly run into a hard limit this way, since our maximum field (as of the time of writing) is ~1.2 GHz and even that is likely not enough to fully resolve your overlapped multiplets.

With that out of the way, let's talk about some other NMR techniques which will be useful in this situation. I want to preface this with a few points:

These are general suggestions which are broadly applicable to a variety of different problems; they are not tailored to the specific molecule which you asked about, and indeed you may find that some of them are less suitable for your specific compound. For the specific case at hand I would suggest talking to a NMR specialist at your institution, if this is at all possible.

There is a caveat in that many of these techniques have slightly poorer performance in situations where there is strong coupling, although it by no means makes them useless.

You may find that you need more than one of these to get a complete understanding of the $\ce{^1H}$ spectrum (especially in light of the previous caveat).

I assume here that the overlapping multiplets belong to the same species, as is the case in the question. In the situation where you are studying a mixture of different species which may have peaks that overlap with each other, the following techniques will still remain very useful, but there are more options which will help to separate the different subspectra based on molecular properties (e.g. relaxation rates or diffusion coefficients). These are outside the scope of the current answer.

Finally, or further reading about all of these techniques I recommend Claridge, T. D. W. High-Resolution NMR Techniques in Organic Chemistry, 3rd ed.; Elsevier: Amsterdam, 2016..

(1) 2D correlation NMR
This has already been mentioned by Buttonwood, and is the standard answer to issues of resolution in 1D spectra. The downside is that 2D spectra are very, very rarely run with high resolution, partly because of time constraints, but especially so if heteronuclear decoupling is used (e.g. in a HSQC). Thus, instead of measuring well-resolved multiplet structure (which would be transferable to the 1D spectrum, thus allowing you to assign the 1D), you are more likely to get broad blobs that are centred on a certain frequency. This tells you nothing about the multiplets actual proton spectrum, it only tells you that at (say) 1.46 ppm you have some kind of multiplet that corresponds to a certain proton in your structure.
Despite this, one should not neglect the amount of information that is provided by standard 2D spectra (COSY, HSQC, HMBC, NOESY). Often even having a blob is enough to figure out which proton is which, even if it may not tell you the exact shape of the underlying multiplet.
(2) Pure shift NMR
This is a series of techniques which remove all coupling information, thereby collapsing multiplets into singlets and giving a 1D spectrum with exactly one singlet per (chemically inequivalent) proton. I would consider the TSE-PSYCHE method[1,2] to be the "state of the art" in this regard, especially with regard to strong coupling. This is very powerful but suffers from the same drawback as 2D NMR does: it removes all multiplet structure, so on its own does not lend any insight into the original $\ce{^1H}$ spectrum. Nevertheless, I would say that it is really worth having a pure shift spectrum in addition to some of the other methods, as it can greatly aid in assigning peaks.
Here's an example of a pure shift spectrum, versus the original proton spectrum of the same compound (personal data).

(3) 2D J NMR
A 2D J spectrum shows proton chemical shifts along one axis and proton multiplet structure along another axis. What this means is that every multiplet in a $\ce{^1H}$ spectrum is separated along a secondary axis according to its chemical shift. This allows you to not only figure out the position of each multiplet (which both of the above already provide you) but also the shape of each individual multiplet.
In particular, pure shift J spectra[3] are really the best versions in this regard as these provide much higher resolution and do not take a lot of time. To be technical, these don't provide "pure shift" characteristics because the 2D J $f_2$ axis (the horizontal axis) is already "pure shift", with the coupling information already separated out onto the $f_1$ axis. What they do is to convert the lineshapes in the spectrum from a "phase twist", which has long "tails" in both dimensions, to a pure absorption Lorentzian which is a single sharp peak. This is an example of how it looks (personal data again):

You can extract vertical traces from this spectrum to view each individual multiplet, with its chemical shift given simply by the value along the $f_2$ axis (horizontal axis).
(4) Selective excitation experiments
This refers to a family of 1D experiments where you selectively excite one multiplet and then perform some kind of mixing in order to look at spins that are coupled / near to the spin that you excited. A selective 1D NOE experiment (for example) would tell you which spins are in close proximity to a given spin that you chose to excite; it is analogous to acquiring only one row of a 2D NOESY. You could also run selective 1D TOCSY experiments, which would tell you which spins belong to the same spin system as the one you excited. By choosing an appropriate TOCSY mixing time you can limit the "spin diffusion" to one or two coupled spins away.
The benefit of this over the original 2D experiment is twofold: first, it takes much less time; and secondly, you can acquire much higher-resolution data. This allows you to disentangle different overlapping multiplets based on their coupling partners or spatial proximity to other spins. For example let's say you have two multiplets, corresponding to protons A and B, that are overlapping, but only proton A is in close proximity to proton C (whose multiplet is well-resolved). Then you can run a selective 1D NOE experiment where you excite proton C, and the resulting spectrum should include only multiplet A and not B. The same can be applied for scalar couplings via 1D TOCSY.
(5) Chemical shift-selective filters
If you find that you need to excite a peak inside an overlapped region, then the idea of a chemical shift selective filter may be useful. This seeks to excite one multiplet out of a series of overlapping ones, and traditionally involves recording multiple spectra and adding them up, allowing the multiplet of interest to add up while averaging all other multiplets to zero by virtue of them having a different phase. However, there is a recent implementation called GEMSTONE[4] which does this very cleanly in just a single experiment without needing to sum anything up. In my limited experience it works quite well although some of the experimental parameters need to be optimised for each peak to obtain best performance. This particular demonstration that I have (personal data, again) isn't so impressive because the overlap isn't as severe, but the paper has slightly more interesting examples.

References

Foroozandeh, M.; Adams, R. W.; Meharry, N. J.; Jeannerat, D.; Nilsson, M.; Morris, G. A. Ultrahigh-Resolution NMR Spectroscopy. Angew. Chem. Int. Ed. 2014, 53 (27), 6990–6992. DOI: 10.1002/anie.201404111.
Foroozandeh, M.; Morris, G. A.; Nilsson, M. PSYCHE Pure Shift NMR Spectroscopy. Chem. Eur. J. 2018, 24 (53), 13988–14000. DOI: 10.1002/chem.201800524
Foroozandeh, M.; Adams, R. W.; Kiraly, P.; Nilsson, M.; Morris, G. A. Measuring couplings in crowded NMR spectra: pure shift NMR with multiplet analysis. Chem. Commun. 2015, 51 (84), 15410–15413. DOI: 10.1039/c5cc06293d.
Kiraly, P.; Kern, N.; Plesniak, M. P.; Nilsson, M.; Procter, D. J.; Morris, G. A.; Adams, R. W. Single‐Scan Selective Excitation of Individual NMR Signals in Overlapping Multiplets. Angew. Chem. Int. Ed. 2021, 60 (2), 666–669. DOI: 10.1002/anie.202011642


Justify the value of the product of the uncertainty in position and momentum

The commutator $[\hat{A}, \hat{B}]$ of two operators $\hat{A}$ and $\hat{B}$ is also an operator, so when using the Robertson uncertainty relation we have to take the expectation value of the commutator first and then take the absolute value of it.

But for $\hat{x}$ and $\hat{p}$ the commutator $[\hat{x}, \hat{p}]$ is simply a constant $\mathrm{i} \hbar$, so that taking the expectation value is trivial: $\langle [\hat{x}, \hat{p}] \rangle = \mathrm{i} \hbar$. Then we take the absolute value of this result which is $\hbar$ and divide it by two to get $\frac{1}{2} \hbar$ as expected.


For $\ce{Pb^2+_{(aq)} + Cu_{(s)}<=> Pb_{(s)} + Cu^2+_{(aq)}}$, your potentials are the wrong way around. The cathode is the electrode at which the reduction occurs so it's lead minus copper.

For $\ce{2 Ag+_{(aq)} + 2 Br^{-}_{(aq)}<=> Ag_{(s)} + Br2_{(l)}}$, I don't know where you got .76 V for bromide, but that's not the right number. Also, consider the $K_{sp}$ of $\ce{AgBr}$ and think about whether a non-redox reaction is occurring.

For $\ce{3Li^+_{(aq)} + Cr_{(s)}<=> 3Li_{(s)} + Cr^3+_{(aq)}}$, your potentials are backwards again and the potential for $\ce{Cr}$ is wrong. Consider how easily elemental lithium is oxidized; just putting it in water will evolve hydrogen and oxidize the lithium, so anything that reduces the lithium cation would have to be a phenomenal reducing agent.

For $\ce{2Fe^3+_{(aq)} + 3Cd_{(s)}<=> 2Fe_{(s)} + 3Cd^2+_{(aq)}}$, you have the right order, but the wrong potential for $\ce{Cd}$.

If you're having trouble keeping the order straight, it may help to think of it this way:
$$E_{\mathrm{cell}}=E_{\mathrm{reduction}} + E_{\mathrm{oxidation}}$$ where $E_{\mathrm{oxidation}}=-E_{\mathrm{reduction}}$. It's the same equation, but instead of keeping track of anode and cathode, the half-cell where reduction is happening will match what the table says and you use the potential as-is (since it's a table of reduction potentials), and the half-cell where oxidation is happening will be backwards from that in the table, so you negate the potential.


Molpro wavefunction symmetry specification

Yes, this is the correct way to find the irreducible representation of the molecule. To obtain the energy ordering of the molecular orbitals (which determines the order in which they are filled for the ground state molecule), you need to run a molecular orbital theory computation (e.g. Hartree-Fock or density functional theory).

The spin number in Molpro is not particularly intuitive. For a molecular term symbol such as $^1B_2$, spin information is conveyed as $2S+1$, where $S$ is the total spin quantum number. In the specific case of a $^1B_2$ state, the molecule has a total spin quantum number of $0$. However, Molpro users must specify spin as $2S$.

As noted in the comments, however, specification of the irreducible representation or spin is generally unnecessary for Molpro computations. For Z-matrix inputs, Molpro determines the symmetry automatically, and in most cases this will be the symmetry the user desires. Molpro also automatically sets the spin if the user doesn't specify it. By default, Molpro uses $S=0$ for molecules with an even number of electrons and $S=1$ for those with an odd number of electrons. Since a large number of molecules that are kinetically stable under typical terrestrial conditions have an even number of electrons, the default spin should work well for many users. Molpro users who work on excited electronic states or molecules with unpaired electrons are much more likely to need to specify the spin.


Your understanding is correct. The usual explanation given in introductory classes is VESPR theory. The oxygen atom has four bond-like "things" coming out of it. Two of those are the covalent bonds with the hydrogen bonds. The other two are the two lone pairs. VESPR theory calls this a steric number of four.

The slightly more advanced explanation comes from molecular orbital theory. Oxygen's valence electrons are all $2\text{p}$ electrons; hydrogen's electron is $1\text{s}$. The only way to mix those orbitals to get enough bonding orbitals is by using two $\text{sp}^3$ hybrid orbitals to generate the two bonds. The $\text{sp}^3$ orbitals are arranged in a tetragonal geometry.


In the Favorskii rearrangement, why doesn't the base grab the alpha hydrogen of the carbon bearing the halogen?

Since it is much easier to discuss based on actual schemes, see scheme 1 for a general Favorskii rearrangement.

Scheme 1: General Favorskii rearrangement reaction conditions. Image taken from Wikipedia where a full list of authors is available.

And because it is also helpful, scheme 2 shows one of the most accepted mechanisms of the Favorskii rearrangement — the one used by both Wikipedia and Kürti-Czakó.[1] One should note, however, that a second mechanistic proposal[2] explains the reaction outcome well and is favoured by some calculations.[3] Because it does not require any deprotonation, it is outside of the scope of this answer.

Scheme 2: ‘Cyclopropanone’-mechanism of the Favorskii rearrangement. Image taken from Wikipedia where a full list of authors is available.

Your question asks why carbon 6 (from the usual numbering of 2-chlorocyclohexanone) is deprotonated to form the corresponding enolate and not carbon 2.

The most important notice here is the base used $\ce{NaOH}$ and the conditions employed (not mentioned in the image, but an example on organic-chemistry.org mentions room temperature). For comparison, table 1 shows a few $\mathrm{p}K_\mathrm{a}$ values from the Bordwell data — unfortunately, an α-chloroketone is not included in this data.

$$\textbf{Table 1: }\mathrm{p}K_\mathrm{a}\text{ values of different relevant acids in }\ce{H2O}\\
\begin{array}{ccc}\hline
\text{acid} & \mathrm{p}K_\mathrm{a}\left(\ce{H2O}\right) & \mathrm{p}K_\mathrm{a} \left(\ce{DMSO}\right)\\ \hline
\ce{Me2CO} & 19.3 & 26.5\\
\ce{PhCOMe} & 18.3 & 24.7\\
\ce{PhCO-CH2F} & & 21.7\\
\ce{Et2CO} & & 27.1\\
\text{3,3-dimethylcyclohexanone} & & 24.8\\
\ce{H2O} & 15.7^* & 31.4\\ \hline \end{array}\\
^*\text{ Other sources say 14.}$$

Obviously from this data, the choice of solvent is import when considering the acidity of a given substance — note the marked decrease in acidity of water in $\ce{DMSO}$. However generally, water is a weaker acid than carbonyl compounds; Professor Mayr taught us to estimate $\mathrm{p}K_\mathrm{a}$ values: alcohols $\approx 15$, ketones $\approx 20$. That means, the deprotonation is in no way complete; it is in equilibrium. This equilibrium will lead to a thermodynamic product distribution: carbon 2 will be deprotonated slightly more than carbon 6, but the most significant species will be the unenolised ketone.

Now as soon as carbon 6 has enolised, it is able to attack carbon 2 to form the bicyclo[3.1.0]hexan-6-one. This is an intramolecular attack and thus very fast. Once this has happened, the formation of the hydrate (6-hydroxybicyclo[3.1.0]hexanolate) is favoured due to the strained three-membered ring. And the subsequent intramolecular migration is also very fast. This leads to a stable product with little or no means for any back reaction; as soon as the initial attack has taken place, things are pretty much settled.

On the other hand, an aldol addition would not only require the enolate (doesn’t matter which one) to be around for some time; it is also inherently prone to retro-aldol dissociation. Usually in organic chemistry, aldol reactions are performed selectively by using kinetic deprotonation (e.g. $\ce{LDA}$; $\mathrm{p}K_\mathrm{a}(\ce{HN(iPr)2}) \approx 35$) ensuring that the forward reaction is strongly favoured. Hydroxide bases may be used in aldol reactions, but then generally the reaction is heated to reflux and the final product is the condensation rather than the addition.

Thus, to sum it up:

Both enolisations occur
The conditions do not favour aldol addition; they will revert readily
Once the Favorskii path has been entered, the rearrangement is basically irreversible.

Notes and References:

[1]: L. Kürti and B. Czako: Strategic Applications of Named Reactions in Organic Synthesis. Background and Detailed Mechanisms, Elsevier Academic Press, Burlington, MA, USA, 2005, page 164.

[2]: The alternative, semi-benzilic mechanism is initiated by the nucleophilic attack of a hydroxide ion on the carbonyl group. The thus formed 2-chloro-1-hydroxycyclohexanolate anion undergoes 1,2-migration with chloride displacement to form the product cyclopentyl carboxylic acid in a single step.

[3]: V. Moliner, R. Castillo, V. S. Safont, M. Oliva, S. Bohn, I. Tuñón, J. Andrés, J. Am. Chem. Soc. 1997, 119, 1941. DOI: 10.1021/ja962571q.


Balancing redox equations with oxygens in multiple oxidation states

This is a rather unusual case of what is discussed in answers like this one, where we circumvent problems with multiple atoms being oxidized or reduced by considering whole compounds as oxidizing or reducing agents.

Here, the whole-compound redox-active material is $\ce{CrO5}$, and as in peroxide disproportionations generally this is both an oxidizing agent and a reducing agent.  We thereby render

$\ce{CrO5 -> Cr^{3+} + (5/2) O2 + 3 e^-}$

$\ce{CrO5 + 10H^+ + 7 e^- -> Cr^{3+} + 5 H2O}$

We then apply the usual method of multiplying the first reaction by $7$ and the second one by $3$ to balance the electrons leading to

$\ce{10 CrO5 + 30 H^+ -> 10 Cr^{3+} + (35/2) O2 + 15 H2O}$

and reducing to "lowest (whole number) terms"

$\ce{4 CrO5 + 12 H^+ -> 4 Cr^{3+} + 7 O2 + 6 H2O}$

All that remains now is to add the spectator ions if desired, and we're done.


Okay, let's tackle this. First of all you worked out correctly that there are four different combinations, resulting in four different peaks. You have also given the probability for finding a $\ce{{}^{35}_{17}Cl}$ is $a=33.3\%\approx\frac13$ and the probability for finding a $\ce{{}^{37}_{17}Cl}$ is $b=66.7\%\approx\frac23$. Since you have three positions to fill, this will result in a binomial formula of third order, $(a+b)^3$. You can solve this easily with Pascal's triangle, $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$. You spectrum should therefore look like this table:
\begin{array}{lrr}\hline
   & \text{m/z} &\text{abundance}/\%\\\hline
\ce{[{}^{35}_{17}Cl_3]^{+}}                & 105   &  3.7  \\
\ce{[{}^{35}_{17}Cl_2{}^{37}_{17}Cl]^{+}}  & 107   & 22.2  \\
\ce{[{}^{35}_{17}Cl{}^{37}_{17}Cl_2]^{+}}  & 109   & 44.4  \\
\ce{[{}^{37}_{17}Cl_3]^{+}}                & 111   & 29.7  \\\hline
\end{array}

The following snowball diagram should further demonstrate this:


Sugar solutions have a neutral pH in themselves, but it makes your body acidic. Why?

It is not proven that "sugar makes your body acidic"!

Your body's pH is very tightly regulated by the body's internal systems; it is also different in different parts of the body - the stomach is acidic (1.0-2.5), the intestine are mildly basic (jejunem 7-9) terminal ileum 7.5 reference here. Blood pH is 7.35, and any deviation from this is indicative of serious illness.


Why dont we use potassium hypochlorite for bleach

Potassium hypochlorite can be used as a bleach as an alternative to sodium or calcium hypochlorite. It is not that it is not used but its uses as bleach is not common as compared to sodium or calcium hypochlorite. Enviro Klor™ is a potassium based bleach (12.5% Potassium Hypochlorite Bleach alternative). As written here:

  Enviro Klor™ potassium hypochlorite is a very versatile EPA registered
  (#63838-10, #63838-CA-01) biocide with all the power of a hypochlorite
  bleach and is approved for use wherever typical sodium hypochlorite
  bleach is, including:

  Recirculating cooling water systems
  Sewage and wastewater treatment
  Disinfection of drinking water
  Public water systems
  Pulp and paper mill process water systems
  Swimming pool water disinfection
  Sanitization of food contact services
  Laundry sanitizers
  Agricultural applications

  ADVANTAGES

  Bleaching and Laundry

  Allows claims for a more environmentally friendly household bleach
  Provides a sodium-free solution to laundry discharges

  Industrial & Institutional (I&I)

  Chemical Compounding – allows creation of sodium free formulas
  Replacement chemical in CIP cleaners
  More stable in certain formulations

  Agriculture

  Useful as a disinfectant for wastewater applied to land for    irrigation, conserving water
  Since potassium is one of the three required soil nutrients, there is    little objection to the introduction of potassium ions onto
  land.

Here's a table of comparison:

You can see that available $\ce{Cl2}$ (wt %) is more in case of sodium hypochlorite for which sodium hypochlorite is preferred as bleaching agent. Also, byproduct of potassium hypochlorite is potassium chloride for which potassium hypochlorite is used in agricultural purposes where soil needs potassium. Another thing is that potassium hypochlorite may disproportionates to potassium chlorate which is a strong oxidant and may react vigourously with water and ignite or explode.


How would a temperature profile in wall change if the rate of heat transfer increases

If the heat transfer in fluid 2 is fast, the thickness of the boundary layer wall - fluid 2 is very small (the 'not so linear' part of the temperature profile on that side)

One can then approximate that the temperature of fluid 2 directly at the wall is equivalent to the bulk temperature.

So the temperature profile through the wall would be steeper, indicating a larger overall heatflux - see it as a self fullfilling prophecy: fast heat transfer will allow you to transfer heat faster


How to identify the parent chain if two chains have the same number of skeletal atoms and the same number of double bonds?

The most important simplified criteria for the choice of a principal chain in such compounds are:

greater number of suffixes
longest chain
greater number of multiple bonds
lower locants for suffixes
lower locants for multiple bonds

The corresponding wording of the nomenclature rules (taken from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)) reads as follows:

  P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES

  When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.

  P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).

  (…)

  P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].

  (…)

  P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. These criteria are illustrated in P-44.4.1.1 through P-44.4.1.12.

  The senior ring, ring system, or principal chain:

  (a) has the greater number of multiple bonds (P-44.4.1.1);

  (b) has the greater number of double bonds (P-44.4.1.2);

  (…)

  (h) has the lower locant for an attached group expressed as a suffix (P-44.4.1.8);

  (…)

  (j) has the lower locant(s) for endings or prefixes that express changes in the level of hydrogenation, i.e., for ‘ene’ and ‘yne’ endings and ‘hydro/dehydro’ prefixes (P-44.4.1.10);

  (…)

Since the compound that is given in the question does not have any principal characteristic group (a characteristic group chosen for citation at the end of a name by means of a suffix or a class name, or implied by a trivial name), the first criterion (greater number of suffixes) given in P-44.1.1 is not relevant in this case.

The second criterion given in P-44.3.2 is the longest chain. However, the  chains A–C and B–C both have eight skeletal atoms.

Also, both chains have three double bonds. Therefore, the criteria given in P-44.4.1 (a) (greater number of multiple bonds) and (b) (greater number of double bonds) cannot be used to reach a decision.

The criterion given in P-44.4.1 (h) (lower locant for the suffix) is not relevant since the compound that is given in the question does not have any suffix.

Finally, the criterion given in P-44.4.1 (j) stipulates that the principal chain shall have the lower locants for the double bonds. Therefore, the principal chain for the compound that is given in the question is the octa-1,4,6-triene from A to C rather than the octa-2,4,6-triene from B to C since the locant set ‘1,4,6’ is lower than ‘2,4,6’.

Thus, the complete name for the compound that is given in the question is (3E,4E,6E)-3-ethylideneocta-1,4,6-triene.


All reaction sources: Chemiday

As I said in the comment section that aluminium hydroxide is amphoteric i.e. It dissolves in acid, forming $\ce{[Al(H2O)6]^3+}$ (hexaaquaaluminium) or its hydrolysis products and dissolves in strong alkaline solutions, forming $\ce{[Al(OH)4]−}$ (tetrahydroxidoaluminate). So, when it is dissolved in base, it forms a sodium aluminohydroxo complex salt. Why? Because as @Pritt mentioned, aluminium has vacant orbital for which free electron can reside there forming a complex salt.

$$\ce{Al(OH)3 + NaOH(conc.) → Na[Al(OH)4]}$$

$$\ce{Al(OH)3 + 3NaOH → Na3[Al(OH)6]}$$

No worries. There is also an acid-base reaction where the two combines to form a salt and water but this occurs at very high temperature and you won't find this type of reaction occurring in laboratory. This reaction can only be find in industries.

$$\ce{Al(OH)3 + NaOH ->[1000 C] NaAlO2 + 2H2O}$$

If only aluminium was used, there would be evolution of hydrogen gas from a very similar exothermic reaction. Note for simplicity, the species is written  $\ce{NaAlO2}$ but it actually contains $\ce{[Al(OH)4]−}$ ion or perhaps the $\ce{[Al(H2O)2(OH)4]−}$ all due to presence of vacant orbital(see sodium aluminate).


Phosphate buffer from phosphoric acid and K2HPO4?

I suggest you first consider the following reaction:
$$\ce{H3PO4 + 3K2HPO4 -> 2HPO4^{2-} + 2H2PO4- + 6K+}$$

It appears, that transforming all $\ce{H3PO4}$ to equal amounts of $\ce{HPO2-}$ and $\ce{H2PO4-}$
requires 3 mole equivalents of $\ce{K2HPO4}$. If you add 3 mole equivalents of $\ce{K2HPO4}$ you will end up in a situation where the concentration of $\ce{[HPO2^{-}] = [H2PO4^{-}]}$, i.e. at the $\ce{pH} = pK_{a2} = 7.21$. Then, I suppose you use the $\ce{HH}$-equation to figure out the rest.

At pH = pka2 = 7.21 the concentration of [H2PO4(-)] = [HPO4(2-)] = 0.40 M. This is because we have added 3 mole equivalents of K2HPO4 to 50*0.2 = 10 mmole of phosphoric acid, i.e. we have reached a total concentration of phosphoric acid protolytes of (3*50*0.2 + 50*0.2)/50 = 0.80 M.

Now, since we wanted to reach pH = 7.0, we have theoretically added too much of K2HPO4. We suppose the excess amount is equal to x.

At pH = 7.0: [HPO4(2-)] < [H2PO4(-)].   At  this pH, only HPO4(2-) and H2PO4(-) are present in significant amounts in the solution. We can then calculate the following:
7.00  = 7.21 + log ([HPO4(2-)] - x/[H2PO4(-)]) = 7.21 + log (0.4 - x)/0.4) => x = 0,1533.

Now, initially we had 50*0.2 mmole of phosphoric acid. To reach pH = 7.0 we should then add 3*50*0.2 - 0.1533*50 mmole =  30 - 7,66(5) = 22,34 mmole of K2HPO4 or 3.8(9) gram.

Please check the calculations.


A quick Google search shows the density to be $11.5~\mathrm{g~cm^{-3}}$.

You might have had trouble finding it since it is actually called plutonium (IV) oxide, as plutonium is an f-block metal (specifically an actinide) displaying multiple  oxidation states (commonly III, IV, V, and VI, while VII is rare).


How to determine the empirical formula for terephthalic acid from its combustion products?

You are on the right track, but if you google terephthalic acid, you will know that your formula is not correct.

One mistake you are making is using the final mass of the $\ce{CaCl2}$ and $\ce{NaOH}$ and not the change in mass. For example, let's take the $\ce{CaCl2}$:

First, we don't know what the initial mass of $\ce{CaCl2}$ is, so we cannot calculate the final mass the way you are trying to do. We are only given the mass of the terephthalic acid (0.4953 g). The calcium chloride and sodium hydroxide are not part of this mass.

However, we know that the calcium chloride increased in mass by 0.1613 g, presumably by absorbing water (you are correct that the calcium chloride is absorbing water). So, the mass of water produced by the combustion reaction is 0.1613 g.

Similarly, we only know the change in mass of the sodium hydroxide (1.0495 g), which occurred by absorbing carbon dioxide. So the mass of carbon dioxide is 1.4095 g.

The second mistake is to assume that all of the mass of the carbon dioxide and water produced must come from the original mass of the terephthalic acid. Not so! We have excess oxygen added contributing mass to the reactant side, and this oxygen is also in the products.
$$\ce{C}_x\ce{ H}_y\ce{ O}_z\ce{ +}\left(\frac{4x+2y-z}{4}\right)\ce{ O2 -> }x\ce{CO2 +}\frac{y}{2}\ce{ H2O}$$

We need to do more maths to figure out how much of the leftover mass in the original sample is oxygen. Here is the beginning of the setup, now that you have some sense of the mass of water and carbon dioxide produced:

Mass of water produced $\ce{->}$ moles of water produced $\ce{->}$ moles of hydrogen in the water produced (and in the original sample!) $\ce{->}$ grams of hydrogen (in sample)
Mass of carbon dioxide produced $\ce{->}$ mass of carbon (in sample)
Mass of oxygen in sample (by subtraction) $\ce{->}$ moles of oxygen in sample

You are doing everything correctly after this.


Prove that a 10-Degree Temperature Increase Doubles the Rate Constant (k), when the Activation Energy is Approximately 50 kJ/mol

Your method and your mathematics seem perfectly fine, and your calculated result is correct. You are also quite correct that the result will change depending on the value of $E_\mathrm a$ that you choose.

The problem you're running into is the assumption that the stated rule of thumb holds exactly, regardless of temperature. It doesn't.

I would set up the problem as follows. Consider the ratio $\rho$ between the reaction occurring at some Kelvin temperature $T$, and at $T+10$:

$$
\rho = {e^{-E_\mathrm a\over R\,\left(T+10\right)}\over
        e^{-E_\mathrm a\over RT}}
     = e^{{E_\mathrm a\over R}\left({1\over T} - {1\over T+10}\right)}
     = e^{{E_\mathrm a\over R}\cdot {10\over{T^2+10T}}}
$$

Plotting $\rho$ as a function of $T$ for $E_\mathrm a = 50~\mathrm{kJ\over mol}$, it becomes immediately clear that the rule of thumb does not hold over a very wide temperature range:

In fact, it only holds strictly for one specific temperature, which is why your math led you to a single temperature as your answer. By my calculations, using a value of $8.3144598~\mathrm{J\over mol\, K}$ for $R$, that temperature is $289.56~K$, which is equal to your $T_1=290~K$ to within three significant figures.

The gray box in the figure marks where the rule holds to within $20\%$ – that is, where $\rho$ falls between $1.6$ and $2.4$. This corresponds to a temperature range of $257~\mathrm K$ to $353~\mathrm K$.


There's something called Hill System. This is the system of writing chemical formulas.
In this system the carbon atoms are first, then hydrogen atoms and then other in alphabetical order. When the formula contains no carbon or hydrogen atoms, then all the elements are sorted alphabetically.

source: http://en.wikipedia.org/wiki/Hill_system

So it should be: $\ce{CH4}$

However, the Hill system is not the only method of ordering elements in a chemical formula. We can instead choose to list the elements based on their electronegativity. ManishEarth explains this in his answer to this question.


Why does increasing temperature for an endothermic reaction always increase equilibrium constant?

The first equation, the corresponding change in G,  and the equilibrium constant is supposed to be related to the following two initial and final states:

State 1:  Pure reactants in separate containers at temperature T and pressure 1 bar

State 2:  Pure products in separate containers at temperature T and pressure 1 bar

The $\Delta G$ you are talking about refers to the change between these two thermodynamic equilibrium states.  But please note that these  thermodynamic equilibrium states that are not in chemical reaction equilibrium with each other.  So, to get the equilibrium constant for the reaction, you are using the $\Delta G$ for these two states which are not in chemical equilibrium.  To understand how this plays out, you need to review the derivation of the relationship between this $\Delta G$ (actually denoted $\Delta G^0$ to distinguish the specific states 1 and 2) and the equilibrium constant K.

Now, when you change the temperature, all three quantities in the equation for $\Delta G^0$ change:  T, $\Delta H^0$, and $\Delta S^0$, not just T.  This is why the Van't Hopf equation gives the correct result for the effect of temperature on the equilibrium constant, and the other approach you described does not.


Can negative (abnormally) lattice enthalpies during the formation of ionic compounds exist?

Short answer: Not for "normal" compounds.

First off, let's consider what the lattice enthalpy means. As you've written it, these lattice enthalpy is part of the Born-Haber cycle of formation of an ionic solid from a metal and halogen.

As you see from your table, the energies decrease with:

Small charges (i.e., these are higher for $\ce{Al2O3}$ or $\ce{CaO}$ because of higher ionic formal charges)
Large inter-atomic distances (i.e., increased ion distance between $\ce{Cs}$ and $\ce{I}$ yield lower electrostatic attraction).

The Born-Lande and Kaputinskii equations give some numeric estimates, e.g. (Born-Landé)

  $E = \frac{N_AMz^+z^- e^2 }{4 \pi \epsilon_0 r_0}\left(1-\frac{1}{n}\right)$

  where:

  $N_A$ = Avogadro constant;
  M  = Madelung constant
  $z^+$ = charge number of cation
  $z^−$ = charge number of anion
  e = charge on an electron (elementary charge)
  $\epsilon_0$ = permittivity of free space
  $r_0$ = distance to closest ion
  n  = Born exponent, typically a number between 5 and 12, determined experimentally or derived theoretically.

(Note that I've inverted the formula from Wikipedia's version to match your reaction.)

None of these can give you a negative lattice enthalpy.

How could it happen? Well, you'd need something where the solid is less stable than the gas-phase ions. Consider that in an ionic solid, the electrostatic attraction between anion and cation is always favorable. Since gas-phase ions aren't stable by themselves, I don't see how this is possible.


Seperating out iron from ferrous sulfate

Decomposing ferrous sulfate to get iron(III) oxide should work:
$$\ce{2FeSO4 ->[\Delta] Fe2O3 + SO2 ^ + SO3 ^}$$
Iron content can be measured from iron(III) oxide. Commercially, iron is recovered from waste ferrous sulfate in the form of magnetite ($\ce{Fe2O3}$) by co-precipitation and magnetic separation$\ce{^{[1,2]}}$. Iron content is measured from magnetite. It was observed that after processing the magnetite, the grade of iron in magnetite concentrate increased from 62.05% to 65.58% and the recovery rate of iron decreased from 85.35% to 80.35%.
References

Wang YU, Ying-lin PENG, Ya-jie ZHENG Recovery of iron from waste
ferrous sulphate by co-precipitation and magnetic separation,
Transactions of Nonferrous Metals Society of China, Volume 27, Issue
1, January 2017, Pages 211-219, DOI: https://doi.org/10.1016/S1003-6326(17)60024-4

Yu, W., Peng, Y. & Zheng, Y. Effect of purification pretreatment on the recovery of magnetite from waste ferrous sulfate. Int J Miner Metall Mater 23, 891–897 (2016), DOI: https://doi.org/10.1007/s12613-016-1304-2


Can Hartree-Fock mix more orbitals than there are electrons?

First some general considerations

Using more orbitals than electrons is not just possible, but almost essential for Hartree-Fock. Although in practice one can do calculations in a so called minimal basis set, this approach is lacking a lot of accuracy. Usually one adds additional basis functions, which leads to basis sets classified as double zeta (triple, quadruple and so on are also possible).

For open shell systems even a minimal basis set includes more atomic orbitals than electrons are available. For example if you want to calculate a single carbon atom, you would have one basis functions for each of the $1s$, $2s$ and all three $2p$ orbitals. This gives a total of 5 atomic spatial orbitals (=10 spin orbitals), but only 6 electrons. (This is not really a good example, as occupying just one of the three $2p$ orbitals breaks the spherical symmetry of the atom, but I am leaving the scope of your question ...)

Virtual orbitals

Since all molecular orbitals (MOs) are expressed in this basis of atomic orbitals (AOs), it means we will get as many MOs as we put AOs in. If ordered by their energy, we first get a set of occupied MOs until all electrons are distributed. After that comes a set of virtual (or unoccupied) MOs. Those are still soluations of the Roothaan-Hall equations, i.e. they are eigenfunctions of the Fock-Operator, but they do not contribute Hartree-Fock energy.

However, virtual orbitals are important in correlation methods going beyond Hartree-Fock, e.g. Coupled Cluster or Pertubation Theory. Here the virtual orbitals become partially occupied and thus influence the ground state energy.

Equations

The ground state energy in case of closed-shell HF reads

\begin{equation}
  E_0 = 2\sum_a h_{aa} + \sum_a\sum_b 2J_{ab} - K_{ab}
\end{equation}

where the sum only goes over the occupied spatial orbitals, which equals half the number of electrons.
And for the Fock operator we have
\begin{equation}
  f(1) = h(1) + \sum_a 2J_a(1)-K_a(1)
\end{equation}
with the same restrictions to the summation. Since the Fock operator is applied to all MOs, this means the effect (coulomb repulsion) of the occupied orbitals is considered in each virtual orbital, but the virtual orbitals do not affect the occupied ones.

(Equations are taken from the book by Szabo and Ostlund, p. 134)


Since $pH = -log(a_{[H^+]})$, $[H^+]_{initial}$ is irreverent when discussing the equivalence point. You can always show a full titration curve over all pH for a given acid-base titration regardless of initial proton concentrations. Therefore, equivalence point is independent of initial concentration. That being said, the fact that the equivalence point occurs when $\frac{pH}{dV}$ is at a maximum  has to do with the shape of the pH vs Volume graph. Since titration curves look similar to $y = tan^{-1}(x)$, the maximum slope will occur at the equivalence point (that would be (0,0) for inverse tangent). This can be further confirmed by taking the second derivative and comparing where the graph crosses 0 and comparing to the maximum of $\frac{pH}{dV}$.

As for the derivative, using a derivative calculator to ensure the correct answer, I got:

$2[H^+]-(\frac{C_{A}V_{A}}{V_{A}+V_{B}})+(\frac{C_{B}V_{B}}{V_{A}+V_{B}})$ which simplifies to $2[H^+]+\frac{C_{B}V_{B}-C_{A}V_{A}}{V_{A}+V_{B}}$. A simple way to determine if these are the same is to use a CAS (computer algebra system) and divide the answer provided by one of these answers. If the result is 1, the difference is only in simplification.

I hope this answers the question adequately.


Are alkylidene cycloalkanes planar molecules?

The alkylidene cycloalkane contains only two C($sp^2$) hybridised atoms, ajacent to each other, as in the instance of a "normal" double bond.  Hence I expect that in the first sphere / in $\alpha$ position to this double bond, all four substitutents adjacent to this double bond may be found almost in the same plane including the axis along the $\ce{C=C}$ bond, too.  The difference to a "normal" case of a single / isolated double bond more frequently seen in textbooks is that two of the substitutents of one side of the double bond are tethered together.  On the level of the double bond, they remain planar.

In the allene example, there are three consecutive C($sp^2$) hybridised atoms, with two adjacent $\ce{C=C}$ bonds. This differs from the instance above, and here the allene protons of one side and their carbon atom they are bond to describe -- as you correctly referred to -- a plane  almost perpendicular to the plane defined by the other two allene protons and carbon atom of the other side of this molecule.  At the level of the double bonds, they are non-planar.

Regarding the methyl group across the cyclohexane, I expect this (again) is preferentially oriented equatorial, rather than in axial alignment towards the main plane of the cyclohexane ring.


I believe that ethylene glycol in this reaction is used as a solvent and Aniline is used as a reducing agent. Ethylene glycol solution heated with chloroauric acid produces $Au^{3+}$ ions (first 20 min) which is then reduced to $Au^{0}$  (3hr reaction) in the presence of aniline. Because there is no stirring during those 3 hrs (so less agitation of the reaction mixture)there is less formation of gold nanoparticle and more gold nano flakes are formed.

May be this paper will help you, it has a similar mechanism


The textbook statement you gave ("solvent is always the one present in larger quantity") is simply a convention that makes it easier for us to talk to each other about mixtures of substances.  We have other words for other types of mixtures.  For example, in your ethanol-water example, you are correct in stating that it doesn't make much sense to talk about solute and solvent.  So, we call these two liquids "miscible".  They mix together in all proportions.  A good demonstration that captures the essence of different types of mixtures is the phase diagram of phenol-water:

Here we have a feature that is called a "miscibility gap".  There is a region, defined by the rainbow shape in the diagram above in which phenol and water separate into two distinct phases.  If we had pure water and added just a little bit of phenol to it at room temperature, we would find that it would dissolve into the water up to about 7 or 8 wt%. So, phenol would be the solute and water the solvent.  Adding more than this amount pushes the mixture into the miscibility gap so that now the phenol and water would mostly separate into two separate phases.  Let's say we add enough phenol to get us to 40 wt%.  If we heat the two phase mixture up to 67 C then it will suddenly become miscible.  Now only one phase exists and we can add as much phenol as we want to it without having to worry about two phases forming again.  There is no solute or solvent here.  Just a mixture of two liquids that exist in a single phase.


How do surfactants spread in the case of multiple materials?

A surfactant is a material that has two regions, one that tends to be attracted to its environment, and another that is repelled by the environment. The amount to which they will be attracted to one interface or another depends upon whatever would minimize the total free energy of the system. Thus, unless the interactions for water/oil, water/air, and water/surface were equivalent, you should see that the surfactant preferentially moves to the interface where it will have the biggest impact on the free energy.


You are going about naming them incorrectly, try breaking it into two parts: the conjugate base of a carboxylic acid and an alkyl group.

$\hspace{5.9cm}$

Combining the two, we get:

$\hspace{6.9cm}$

Both the carboxyl and alkyl carbon chains can be numbered, with the carboxyl chain starting at the carboxyl carbon, and the alkyl chain starting at the carbon that places its bond to the ester oxygen at the lowest position. See below:

$\hspace{5.5cm}$

To refer to a specific carbon, simply say "at position $x$ on the carboxyl/alkyl chain."


Canonical structures of (1E)-prop-1-en-1-yl acetate

May it look like this? In those two possibles ways?


Why ionisation enthalpy of early lanthanides are expected to be greater than early actinides?

This data is incorrct, and I am saying so because the Ionisation energy of the actinoid is a decent bit higher than the corresponding lanthanide

In each case IE 1 is significantly higher for lanthanoids, IE 2 is still higher for lanthanoids and IE 3 is comparable.

$\ce{Ce}$ $534.4$, $1050$, $1949$

$\ce{Th}$ $587$, $1110$, $1930$

These were two examples where i showed all 3 IEs

For, the sake of further clarification, here is a the IE1 of the first few

$\ce{Pr}$ $527$, $\ce{Pa}$ $568$

$\ce{Nd}$ $533.1$, $\ce{U}$ $597.1$

$\ce{Pm}$ $540$, $\ce{Np}$ $604.5$

$\ce{Sm}$ $544.5$, $\ce{Pu}$ $584.7$

(units used are $kJ/mol$)

The reason you stated in your doubt is absolutely correct, as even if we were to conclude at these aren't accurately determined, we see this phenomenon elsewhere in the periodic table also, where the values are accurately determined.

$\ce{Sn}$ is the lowest IE element of GP 14 and not $\ce{Pb}$.
$\ce{Fr}$ has a higher IE than $\ce{Cs}$.
$\ce{Ra}$ has a higher IE than $\ce{Ba}$.

All due to the same reason poor shieling, which increases the effective nuclear charge, which increases nuclear attraction, which in turn leads to a higher IE.

References

Wikipedia data page


When a wfn-file is loaded into Multiwfn and one tries to convert it into a molden file via 100 > 2 > 6, it says Error: This function works only when input file contains basis function information.

So I guess, that as the wfn file lacks important information, it’s not easily possible to convert it into a molden file.


Why are the bond angles in sulfur dichloride and oxygen dichloride not the same?

Why should the bond angles be the same? You have different atoms every time leading to different molecules. Different molecules have different properties; that goes both for macroscopic properties (they will have different melting and boiling points, their liquid phases will have different densities and refraction indices) and microscopic ones (i.e. their molecular structure will be different). Indeed, there are three-atom molecules with a $90^\circ$ bond angle and those with $180^\circ$. I’m not sure whether it ever gets any smaller than $90^\circ$ but I suspect ‘no.’


Osmotic pressure of a solution of a protein?

Ok let's start with $\text{OP} = \text{T}\times\text{R}\times\text{M}$

$\text{T} = (24.0 + 273) \text{K} = 297 \text{K}$

$\text{R} = 0.08206 \dfrac{\text{atm}\cdot\text{L}}{\text{mol}\cdot\text{K}}$

$\text{M} = \dfrac{\text{moles}}{\text{L}} = \dfrac{\frac{37.5 \text{ g}}{1.50 \times 10^4 \text{ g/mol}}}{0.025 \text {L}} = \dfrac{2.50\times10^{-3}\text{ mol}}{0.025\text{ L}} = 0.100 \dfrac{\text{mol}}{\text{L}} $

$\text{OP} = 297\text{ K}\times0.08206 \dfrac{\text{atm}\cdot\text{L}}{\text{mol}\cdot\text{K}}\times0.100 \dfrac{\text{mol}}{\text{L}} = 2.44 \text{ atm} $

So I get your answer too...