Is it wrong to call carbon dioxide "methanedial"?

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

P-66.6.1.1.1 Mono- and dialdehydes derived from alkanes are named substitutively using the suffix ‘al’ added to the name of the parent hydride with elision of the final letter ‘e’ of the parent hydride before ‘a’.

Examples:
pentanal ($\ce{CH3-CH2-CH2-CH2-CHO}$)
pentanedial ($\ce{OHC-CH2-CH2-CH2-CHO}$)
In this context, the term ‘aldehyde’ is unambiguously defined:

P-66.6.0 Introduction.
The class name ‘aldehyde’ traditionally refers to compounds containing the $\ce{-CH=O}$ group attached to a carbon atom. However, nomenclature for aldehydes has been extended to describe a $\ce{-CHO}$ group attached to a heteroatom.

Since carbon dioxide ($\ce{O=C=O}$) does not contain a $\ce{-CH=O}$ group, it is not considered an aldehyde and thus Section P-66.6.1.1.1 does not apply to carbon dioxide.
Therefore, the shortest dialdehyde is ethanedial ($\ce{OHC-CHO}$). (Note that the preferred IUPAC name is ‘oxaldehyde’, which is derived from the retained name of the corresponding carboxylic acid.)


[Co(H2O)6]3+ vs [Co(NH3)6]3+ oxidising water

There are 2 competing principles.

$\ce{[Co(H2O)6]^3+}$ has  strong oxidation effect, but...

$\ce{[Co(NH3)6]^3+}$ is much more stable than $\ce{[Co(NH3)6]^2+}$.

As the consequence, the former from the next 2 equations is significantly shifted to the right.

$$\begin{align}
\ce{  [Co(H2O)6]^3+ + 6 NH3 &<=>> [Co(NH3)]^3+ + 6 H2O} \\ \ce{[Co(H2O)6]^2+ + 6 NH3 &<=> [Co(NH3)]^2+ + 6 H2O}
\end{align}$$

Therefore

$$\frac{c_{\ce{[Co(H2O)]^3+}}}{ c_{\ce{[Co(H2O)]^2+}}}\lt\lt     \frac{c_{\ce{[Co(NH3)]^3+}}}{ c_{\ce{[Co(NH3)]^2+}}}$$

As both complex pairs keep the same redox potential, their standard redox potentials are therefore very different.


Indeed the second carbocation is formed.  It's favored by the chlorine atom being able to delocalize the charge through resonance, draw resonance hybrids for the second cation and you should see it.  In a sense you might say that the carbon bonded to chlorine is more effectively substituted for stabilizing the cation and so the Markovnikov rule is "bent" accordingly.

The usual version of the Markovnikov rule is for hydrocarbon substrates.


How do I chill water without a refrigerator?

I'm really not sure this is a chemistry question.

Having said that, evaporative cooling will do the job for you. Here on the Indian sub-continent a clay/earthenware water container works wonders using the technique of evaporative cooling.

In the absence of earthenware, wrap two pieces of thick cloth of the appropriate size around your containers. Keep both cloth wet/moist, and let nature do the rest.

You may also consider taking a bit of pot luck


Why does solubility of solids sometimes decrease with temperature increase?

In short, for this group 2 hydroxide crystal, the dissolution and solvation process is an exothermic equilibrium. Increasing the temperature drives the reaction in the reverse, favouring crystallisation!

However, it would be nice to rationalise some of these terms, so I will lay out a very brief overview of general solubility factors so you can understand the process in a little more detail.

Solubility is a complex thermodynamic process that takes many factors into account. It can be understood as the process of dissolving a crystal in a solvent, and then forming a solvent cage around the solute (solvation). It is a function of three interactions at heart;

solute-solute
solute-solvent
solvent-solvent

Large solute-solute interactions hinder solubility and are associated with high lattice enthalpies (of crystalline solutes), these must be overcome in order to break up the solid and dissolve it. Solvent-solvent interactions must also be overcome in order to solvate the dissolved solute, therefore for a highly interacting solvent the solubility is reduced.

The formation of ordered solvent structures such as clathrin cages etc orders the solvent, and therefore reduces solvent entropy significantly. Therefore solvents with high degrees of freedom hinder solubility.

Solute entropy favours solvation, this reflects the higher solubility of flexible molecules, this is because in the solid they are all locked up, while in solution their rotational and vibrational modes are populated.

In addition to these if the solute has multiple ionic forms then the solubility will be a function of pH too, so must be held constant in practice. The surface profile of the solid too, as well as the extent of cavitation and porosity are also key factors (since they affect the number of solvent-solute interactions possible).

Importantly the type of solvent into which you are trying to dissolve your crystal in is also of fundamental importance! The rule "like dissolves like" is best understood by the (semi-empirical) general solubility equation,
\begin{equation}
\log S =\frac 12-0.05 (T_m-T)-\log P
\end{equation}
A large partition coefficient (the ratio of how soluble the crystal is in organic solvent to aqueous solvent) reduces the solubility, (and mathematically since $\log x$ follows $x$ in limits).

To further complicate the situation there are several definitions of solubility that we can take. In practice we choose intrinsic aqueous solubility, mainly used for high throughput and speed in drug development.

Solubility is really an experimentalists tool for parametrising the extent of Gibbs energy for the dissolving-solvation process, which in turn is usually modelled by complex thermodynamic cycles via gas phase species or super cooled liquids. Modern research into solubility focuses on finding either molecular descriptors through random forest/QSPR or support vector models etc (which are all just number crunching techniques to find correlations between large sets of data) or through fundamental physics and chemistry of the process.
\begin{equation}
\log S=\frac{(-\Delta G_{sol}/RT)}{\ln 10}
\end{equation}
So we understand solubility as an equilibrium constant for the Gibbs change of the solvation process (whatever particular cycle you use to get that data).

Now that we know something about solubility we can ask ourselves, under what conditions does increasing the temperature reduce solubility?

In practice the temperature can effect the entropic term, but also the enthalpies of the crystal lattices, the heats given out by the solute-solvent interactions etc. It is the balance of these process that determines the solubility of something.

In your case we are observing a calcium hydroxide crystal dissolve in aqueous solvent. As we increase the temperature the solubility decreases. There is an equilibrium between crystallisation and (dissolution + solvation). Increasing the temperature is shifting that equilibrium to favour the crystallisation rather than (as we crudely expect) in the forward. This is simply an enthalpic term, due to the solvation process being an exothermic process, increasing the temperature drives the equilibrium in the reverse by Le Chatlier's principle.

Hope that is a bit clearer, and that you now understand some of the ideas behind solubility. :)


Why a particular enantiomer of a diastereomer shows affinity towards another?

Let's set aside compounds for the moment.

Your hands are chiral objects, that is, they are non-superimposable mirror images of each other [see more here]. If you pick up something that is not chiral, like a cup, there is no difference whether you use the left hand or right hand. The "energy" of those two interactions (right hand + cup or left hand + cup) is the same.

Now, let's say you get a pair of gloves made to fit your hands. If you put your left hand into the right-hand glove, it feels different than if you put your left hand into the left-hand glove. You could think of it as those two interactions (left hand + left glove vs left hand + right glove) having different energies.

The reason for this is that enantiomers of chiral objects interact with non-chiral objects with the same energy, but interact with enantiomers of chiral objects with different energy (ignoring coincidental cases, as @Zhe suggests).

The way one enantiomer of a compound "fits" with enantiomers of other compounds (objects) like proteins or chromatography instruments is difficult to predict. Sometimes it is useful to actually model the interactions, but often, like in chromatography, it's enough to know that there should be a difference in energetics of the enantiomers of a compound with a single enantiomer of another chiral object.


What caused the flashes in the video footage of Chernobyl?

The white flashes seen in the video are the result of radioactive contamination of the original film. Airborne radioactive particles were deposited onto the film and caused a local overexposure of the negative, quite similar to the historic experiment of Antoine Henri Becquerel. That's why the flashes only appear in individual spots in individual frames of the film. (This is also explained in the documentary The Battle of Chernobyl, which also contains these film excerpts.)

There was no corresponding visual phenomenon in the air that could be directly observed by people in Pripyat.

The ambient dose rates measured in 26 locations in Pripyat on 26 April 1986 at noon (i.e. about the time when the film was recorded) were about 0.038–1.9 R/h (i.e. about 0.38–19 mGy/h; with a geometric mean of 2.4 mGy/h). Such dose rates cannot cause any directly visible effects in air.

The flashes on the film are not related to any visual phenomenon that was caused by nuclear reactions, radiation, fire or other chemical reactions at the open reactor during the night of the accident.


What the exchange-correlation hole means in common words?

Assuming that the probability is the common word, one could say that the exchange-correlation hole, is a region of space around an electron in which the probability of finding another electron is close to zero due to electron correlation.

  How do you explain it to a non-literate person?

Long story short is that mathematically electron correlation arises as a consequence of the instantaneous Coulomb repulsive forces between electrons and also as a consequence of the fundamental principle of antisymmetry of an electronic (fermionic in general) wave function. Note that the first kind of electron correlation which is due to Coulomb repulsions is known as the Coulomb correlation while the second one which is due to antisymmetry of a wave function is knows as the exchange correlation. However, since the root of all ab-initio wave function methods, the HF method, already accounts for exchange correlation, the Coulomb correlation is usually also referred to as just the correlation.

With respect to these two different kinds of electron correlation note that exchange-correlation hole does not refer specifically to just the exchange correlation, rather it is the hole due to both types of the electron correlation mentioned. I think that one would better call it, say, exchange-Coulomb correlation hole to avoid an ambiguity, but I'm afraid we're stuck with the exchange-correlation hole term.

Long story

Consider the simplest many-electron system, a two-electron one, and let us examine the following two important events

$\vec{r}_{1}$ - the event of finding electron-one at a point $\vec{r}_{1}$;
$\vec{r}_{2}$ - the event of finding electron-two at a point $\vec{r}_{2}$.

Probability theory reminds us that in the general case the so-called joint probability of two events, say, $\vec{r}_{1}$ and $\vec{r}_{2}$, i.e. the probability of finding electron-one at point $\vec{r}_{1}$ and at the same time electron-two at point $\vec{r}_{2}$, is given by
\begin{equation*}
    \Pr(\vec{r}_{1} \cap \vec{r}_{2})
    =
    \Pr(\vec{r}_{1}\,|\,\vec{r}_{2}) \Pr(\vec{r}_{2})
    =
    \Pr(\vec{r}_{1}) \Pr(\vec{r}_{2}\,|\,\vec{r}_{1}) \, ,
\end{equation*}
where

$\Pr(\vec{r}_{1})$ is the probability of finding electron-one at point $\vec{r}_{1}$ irrespective of the position of electron-two;
$\Pr(\vec{r}_{2})$ is the probability of finding electron-two at point $\vec{r}_{2}$ irrespective of the position of electron-one;
$\Pr(\vec{r}_{1}\,|\,\vec{r}_{2})$ is the probability of finding electron-one at point $\vec{r}_{1}$, given that electron-two is at $\vec{r}_{2}$;
$\Pr(\vec{r}_{2}\,|\,\vec{r}_{1})$ is the probability of finding electron-two at point $\vec{r}_{2}$, given that electron-one is at $\vec{r}_{1}$.

The first two of the above probabilities are referred to as unconditional probabilities, while the last two are referred to as conditional probabilities, and in general $\Pr(A\,|\,B) \neq \Pr(A)$, unless events $A$ and $B$ are independent of each other.

If the above mentioned events $\vec{r}_{1}$ and $\vec{r}_{2}$ were independent, then the conditional probabilities would be equal to their unconditional counterparts
\begin{equation*}
    \Pr(\vec{r}_{1}\,|\,\vec{r}_{2}) = \Pr(\vec{r}_{1}) \, , \quad
    \Pr(\vec{r}_{2}\,|\,\vec{r}_{1}) = \Pr(\vec{r}_{2}) \, ,
\end{equation*}
and the joint probability of $\vec{r}_{1}$ and $\vec{r}_{2}$ would simply be equal to the product of unconditional probabilities,
\begin{equation*}
    \Pr(\vec{r}_{1} \cap \vec{r}_{2}) = \Pr(\vec{r}_{1}) \Pr(\vec{r}_{2}) \, .
\end{equation*}
Note that this is the picture in which electrons are sort of billiard balls, i.e. neutral macroscopic particles.

In reality, electrons are, of course, not like billiard balls; they are charged microscopic particles, and consequently,
\begin{equation*}
    \Pr(\vec{r}_{1}\,|\,\vec{r}_{2}) \neq \Pr(\vec{r}_{1}) \, ,
    \quad
    \Pr(\vec{r}_{2}\,|\,\vec{r}_{1}) \neq \Pr(\vec{r}_{2}) \, ,
\end{equation*}
and
\begin{equation}
    \Pr(\vec{r}_{1} \cap \vec{r}_{2}) \neq \Pr(\vec{r}_{1}) \Pr(\vec{r}_{2}) \, .
\end{equation}
And since the spatial probabilities of finding electrons are inevitably related to their states, the inequality above means that the state of electron-one is not independent of the state of electron-two and vice versa.
This interdependency of the states of electrons is referred to as the electron correlation.

The inequalities above, which are the mathematical essence of electron correlation, hold true for two reasons. First, electrons repel each other by Coulomb forces, and, as a consequence, at small distances between the electrons the conditional probability is less than half of the corresponding unconditional one
$$
    \Pr(\vec{r}_{1}\,|\,\vec{r}_{2}) < \Pr(\vec{r}_{1}) \, ,
    \quad
    \Pr(\vec{r}_{2}\,|\,\vec{r}_{1}) < \Pr(\vec{r}_{2}) \, ,
$$
while at large distances between the electrons the conditional probability is greater than half of the corresponding unconditional one
$$
    \Pr(\vec{r}_{1}\,|\,\vec{r}_{2}) > \Pr(\vec{r}_{1}) \, ,
    \quad
    \Pr(\vec{r}_{2}\,|\,\vec{r}_{1}) > \Pr(\vec{r}_{2}) \, ,
$$
In the extreme case when $\vec{r}_{1} = \vec{r}_{2}$, the Coulomb repulsion between the electrons becomes infinite, and thus, $\Pr(\vec{r}_{1}\,|\,\vec{r}_{1}) = 0$ and consequently $\Pr(\vec{r}_{1} \cap \vec{r}_{1}) = 0$, i.e. the probability of finding two electrons at the same point in space is zero.
The correlation due to Coulomb repulsion is termed as Coulomb correlation and one sometimes speaks about the presence of the so-called Coulomb hole around each electron - a region of space around it in which the probability of finding another electron is close to zero due to Coulomb repulsion.

Secondly, as a consequence of the Pauli exclusion principle, electrons in the same spin state can not be found at the same location in space, so that for electrons in the same spin state there is an additional contribution into the inequalities between conditional and unconditional probabilities above.
This effect is relatively localized as compared to one due to Coulomb repulsion, but bearing in mind the relationship between probabilities and wave functions and taking into account the continuity of the later, it is still noticeable when electrons are close to each other and not just at the same location in space.
This type of electron correlation, which is a consequence of the Pauli exclusion principle, is referred to as Fermi correlation, or exchange correlation, and one sometimes speaks about the presence of the so-called Fermi hole around each electron - a region of space around it in which the probability of finding another electron in the same spin state is close to zero.


What is the difference between the titration of a strong acid with a strong base and that of the titration of a weak acid with a strong base?

The answer is A.

Let's look at a weak acid: acetic acid. It has a $K_a$ of $1.8\cdot  10^{-5}$. From this you can calculate how much it will dissociate in water. Let's say we have a concentration of $1~\mathrm{M}$:

\begin{align}
K_a &= \frac{[\ce{CH3COO^{-}}][\ce{H+}]}{[\ce{CH3COOH}]}\\
1.8 \cdot 10^{-5} &= \frac{x \cdot x}{1~\mathrm{M} - x}\\
x &= 4.233 \cdot 10^{-3}\\
\end{align}

This means, at equilibrium, $[\ce{CH3COOH}] = 1~\mathrm{M}- 4.233 \cdot 10^{-3} = 0.995~\mathrm{M}$. As you can see, by far the largest part is not dissociated!

B is not true, you can still observe the equivalence point. Have a look at the following picture, that shows the titration curve for weak / strong acid titrated with strong base. When choosing an indicator for colorimetric titration select one so that the pH jump at the equivalence point contains the interval $\mathrm{p}K_a\pm1$. Phenolphthalein has a $\mathrm{p}K_a \approx 9$, so to decide if it is a suitable indicator you would check if the pH jumps from 8 to 10 at the equivalence point. It does:

C is not true either. If it was a very slow reaction, you could not titrate accurately, as you would have to wait a long time between every drop of $\ce{NaOH}$.


While one might think that β-sitosterol is an epimer of α-sitosterol, differing by configuration at C-3 (with $\ce{-OH}$ group), it does not seem to be the case.

There are various products and their isomers of phytosterol sitosterol biosynthesis, and their trivial naming seems to be somewhat rudimentary.

All of the sitosterols are 3β alcohols. The Greek letter (beta) designation in ‘β-sitosterol’ seem to be for more general, non-systematic isomers designation. In the field of natural compounds, it has historical reasons – a compound was isolated, described and named, and later was found that it in fact is a mixture of different isomers. (E.g. “α-caryophyllene” and “β-caryophyllene” do even have different number of cycles).

Sitosterols mentioned the Merck Index (14th ed.) are as follows ("natural product" IUPAC name according to last steroid nomenclature shown as well, might slightly differ from those from the M.I.).

(I) α1-sitosterol, (24Z)-4α-methyl-5α-stigmasta-7,24(241)-dien-3β-ol
      (or (24Z)-4α-methyl-5α-poriferasta-7,24(241)-dien-3β-ol)
(II) β-sitosterol, stigmast-5-en-3β-ol
(III) γ-sitosterol, poriferast-5-en-3β-ol

So, β- and γ- sitosterols (II, III) are diastereomers, epimers at C-24 (those differently named parent structure names stigmastane and poriferastane are in fact just (24R)-24-ethylcholestane and (24S)-24-ethylcholestane, respectively; these ethyl carbons are nowadays numbered 241 and 242 instead of older 28, 29). α1-Sitosterol (I) has the double bond not at C-5 but at C-7, additional methyl at C-4, and additional double bond at C-24 (ethylidene instead of ethyl). (And I guess that that ‘1’ subscript might be because of E/Z isomerism if this double bond.)


So as mentioned the Krapcho-Decarboxylation follows a SN2-mechanism and is therefore working best with a methylester.

The solvent choice for DMSO (polar aprotic) is due to low stabilisation of the chloride anion rendering it highly reactive for the nucleophilic attack. The resulting methyl chloride is gaseous at room temperature and leaves the reaction mixture (le Chatelier -> formation of product strongly favoured)
The leaving group of this substitution reaction decays into CO2 and the decarboxylated product. (Again: formation of a gas)

I think that the destabilisation of the cyanide in DMSO and the formation of stable acetonitrile can be considered as the driving force in the second reaction.


In which cases does the 3d subshell take precedence over the 4s subshell?

If we take the definition of "3d subshell takes precedence over the 4s subshell" as being that the configuration has zero 4s electrons, yet at least one 3d electron, then all such cases are ions.

+1

V: 3d^4
Cr: 3d^5
Co: 3d^8
Ni: 3d^9
Cu: 3d^10

2+

All elements from Sc to Zn

3+

All elements from Ti to Ga

Etc.


Is there any easy trick to know if a substance is a Brønsted–Lowry acid or base?

Not perfect, but here are a couple rules of thumb to start with for the most common cases.  The key with both rules is to look at the molecular structure.

A Bronsted acid typically has hydrogen attached to oxygen, sulfur or a halogen, all these being relatively electronegative elements.  The hydrogen could also be attached to another nonmetal, especially nitrogen or phosphorous, if the atom is a positive charge center like the nitrogen in ammonium ion.  If the atom is both a positive charge center and oxygen, sulfur or a halogen, you likely have a strong acid.

A Bronsted-Lowry base typically has a nonbonding electron pair that can combine with a proton, thus resembling a Lewis base.  In stable compounds the nonbonding pair us usually on the same kind of atom as one of the elements mentioned above for acids.  This is one reason Bronsted-Lowry acid-base reactions are typically fast and reversible.


Diffusivity is not dependent on time. It is dependent on temperature (and material). Fick's second law holds because the diffusivity, $D$, is constant with respect to depth, $x$, and time, $t$, so it is treated as a constant in the equation.

$$\frac{\partial c_{x}}{\partial t}=\frac\partial{\partial x}\left(D\frac{\partial c_{x}}{\partial x}\right) = D\left(\frac{\partial^2c_x}{\partial x^2}\right)$$

Fick's second law of diffusion applies to non steady state diffusion, which concentration at a depth $x$, $c_x$ is a function of time.


You have correctly identified that m-nitrophenol should be the weakest acid among the isomers as the resulting anion cannot be stabilised by the $\text{-M}$ effect of nitro group owing to its meta location.
The acidity order of the remaining two nitrophenols can be explained in this way:

Hydrogen bonding is possible in both of them. But note that:

$$\text{H-bonding} \propto \delta^+ \text{charge on $\ce{H}$ atom}$$
Inductive effect (here, $\text{-I}$) is distance dependent. So $\delta^+$ charge on the $\ce{H}$ of ortho-nitrophenol's -$\ce{OH}$ is going to be more than the $\delta^+$ charge the $\ce{H}$ atom of para-nitrophenol's -$\ce{OH}$.

Also,  quoting from this comment:

Usually the $\mathrm{p}K_\mathrm{a}$ of a substance is determined in dilute solution. So just due to dilution, there won't be many/any inter-molecular H bonds, but there can still be plenty of intra-molecular ones.

The answer above the comment relates the strength of these two types of hydrogen bonds to kinetic motion of molecules:

Inter-molecular interaction, such as inter-molecular hydrogen bonding only take place for a short time when two molecules are in each other's vicinity. Shortly after, the molecules separate and the interaction does no longer exist.
Intra-molecular interactions on the other hand are much stronger since the two atoms forming the hydrogen bond remain in each others proximity due to structure of the molecule. In other words, they exhibit strain.

These three factors help us conclude that:

Strength of intra-molecular hydrogen bonding in o-nitrophenol $\gt$ Strength of intermolecular hydrogen bonding in p-nitrophenol

Thus, p-nitrobenzoic acid is more acidic than the o-isomer.

...and since phenol has no ring activating group, it comes out to be least acidic in the given compounds.

This reasoning is incorrect. Phenol has no group that stabilises the negative charge on the resulting phenoxide ion through $\text{-M}$ or $\text{-I}$ effect (a deactivating group). Therefore, it's the weakest acid of all under investigation.


Of course, the answer is resonance.  As described by Benet et al., a basic medium (pH>10) favours the Tollen's reaction. On the other hand, if we want to dissolve phenol in water, we will need OH- to generate phenolate. It means the hydroxyl group of salicylaldehyde is sensitive to the high pH, and this hinders attacking the other hydroxides to the carbonyl group by further stabilizing it.


Why is condensation in dyed beverages colorless?

Given that the majority of beverages such as Gatorade is water, as stated in the comments, the condensate is predominantly water, as can be seen in the Patent Electrolyte blends providing reduced salty taste (which includes Gatorade), that water amounts to well over 99% by composition, of course this will vary. The dyes themselves don't make up much of the volume.

A common food dye used, according to the Indiana University online experiment webpage How Much Red Dye is in My Gatorade®? , is the red dye, also known as 'Allura Red'

its properties according to the Wikipedia page are stated as:

  It usually comes as a sodium salt, but can also be used as both calcium and potassium salts. It is soluble in water; Its melting point is above 300 °C (572 °F).

It is likely that the majority of the food dye based salts would remain in solution.

Any salts (such as the example above), transported with the evaporating/condensing water would be too small of a concentration to cause any visible effects.

A similar question and answer is Does salt vaporise?


I assume "DCM" represents dichloromethane, if so...

There is an old paper that describes the formation of a low molecular weight polymer from the double Friedel-Crafts alkylation of benzene by dichloromethane. That's probably what's happening in your reaction as a minor pathway.  The dissolved polymer and tar is what's producing your black solution.

I actually ran that reaction once, I wanted to prepare that polymer.  It didn't work for me, but at the end, after the aluminum chloride had been removed, I did have a dark solution.


How to read these graphs and what are they called?

This depiction is a ribbon diagram of a protein structure. It shows the protein backbone (alpha carbons and the peptide bonds) smoothed out to a continuous line. In certain areas where the protein adopts alpha-helices or beta-sheets, these are indicated by the elaborate spiral in your picture (alpha-helix) or a wide, flat arrow (beta-sheet; not present in your structure). These depictions are supposed to give you an idea of what the protein looks like.

I would assume the two arrows in the bottom to be two of the coordinate axes of the crystal since ribbon diagrams are typically obtained from crystal structures. However, I cannot know without additional information (e.g. in a footnote of a picture).


How to define the atom numbers in unsaturated hydrocarbons correctly according to IUPAC?

Your first compound should be (3E)-5-ethynylhepta-1,3,6-triene. The 2 conjugated double bonds at the end determine the first carbon.

The second compound should be (4-methylpentyl)cyclobutane. The aliphatic group has less precedence than the cyclobutane group.


Working with your numbers gives me the same result through moles of $\ce{NaF}$.  However, at this point you can multiply by the molar mass of fluoride, 18.998 g/mol, to get grams of fluoride, not moles, specifically, 0.0002835 g $\ce{NaF}$.

Parts per million is a mass-based concentration measure.  It's similar in concept to percent mass, but instead of parts per hundred, it uses parts per million.  Since we know the original mass of this sample, we can calculate the concentration in ppm based on that value.  It will be the same concentration as in the larger sample.

I get the following:
$\frac{0.0002835 \textrm{ g } \ce{F-}}{0.7540 \textrm{ g toothpaste}}\times 1000000 = 376.0\textrm{ ppm }\ce{F-}$


The problem is with the free $\ce{NH}$. Grignard reagents are strong bases ($\mathrm{p}K_\mathrm{a}\sim 50$), strong enough to deprotonate $\ce{R2NH}$. Any Grignard that formed would be immediately quenched by the free $\ce{NH}$. You need to protect, this is typically done with tBOC, more discussion here.


Unusual color from leaking battery. What happened?

It is, as you suspected, likely a copper corrosion product. Copper compounds are usually blue or green, and are used as pigments. Malachite, http://en.wikipedia.org/wiki/Malachite, is one of the likely corrosion products of an alkaline battery and copper terminals.

BTW, you may need to rinse the battery compartment and use a file to remove the corrosion from the contacts to make connection.


How can I find the bond angles for carbamic acid?

A quick google of 'carbamic acid bond angles' reveals many papers in the literature measuring/computing the bond angles in this and similar molecules. See here, for example.

That said, chemists are often not concerned with the exact bond angle, instead, we often look at the structure and make predictions based upon bonding and the substituents around the atoms.

In carbamic acid, the simplest carbamate, we can consider the central carbonyl to be sp2 hybridised, giving it a planar structure with bond angles of 120. Due to conjugation with the nitrogen lone pair, the N can also be considered to be sp2 hybridised, and also have bond angles of around 120.

Although this is quite a crude analysis, it holds up quite well for sp2 centres, with very little variation (except in cases of exceptional steric bulk forcing groups to move closer/further apart, or in strained ring systems). The paper referenced confirms that in this system (carbamic acid), the deviation is as little as 4 degrees.


What are disadvantages of having an impurity?

Impurities are the UNWANTED substances that are present in the substance..
There are Impurities in most of stuff including a lot of chemicals and it's very hard to remove them too
However If impurity exceeds the limit its completely Dangerous..
Like for example :

  Surface water (river or lake)
  Surface water can be exposed to acid rain, pesticide runoff, water runoff, and industrial waste.

  Groundwater (private wells or public water supplies)
  Groundwater can be contaminated by disease producing pathogens, careless disposal of household wastes, and leaching from landfills and septic tanks.

which can lead to Nausea
Lung irritation,Vomiting, Dizziness, Skin rashes and even Death

For the Chemistry and Chemical Industry impurities might slow down the reaction or may lead to formation of unwanted bi products , Might affect the rate of Crystallization - Most common issue I think

There are lots of articles online out there here are some for you : Impurity in medicine
Impurity in Electrical cells
Impurity in water
Effect on Solder machine
you can google and find more


What happens if a polymer's length is equal to its persistence length?

The persistence length is a statistically defined property - in other words, it is defined by looking at many lengths of many polymer molecules over a long period of time. It is the length (L) at which a vector drawn tangent to the polymer is no longer correlated in space with another tangent vector at distance (L). In other words, it tells you how well the position of one section of a polymer is correlated to another section. Beyond the persistence length, you would not expect to find any correlation. Below it, you would expect correlations.

In this case "correlations" are measured by the cosine of the angle between the two vectors. When the angle is 90 degrees on average, the cosine is zero, and when it is zero degrees, the cosine is 1. So, for lengths beyond the persistence length, the amount of correlation is zero, and for lengths much less than the persistence length the amount of correlation is one. For lengths close to the persistence length, the correlation is between 0 and 1.

To give you an idea of how this works, take a look at the following illustrations:

The red lines represent tangent vectors along the backbone of a polymer chain, spaced at regular intervals. We can translate all of these vectors and put their origin at the same point, and then find the angle between them:

If we find another polymer chain, or watch this one for a little while and take another "snapshot" and do the same thing, we might end up with something that looks like this instead:

We can even take the same very long chain and find all of the vectors that are a distance 1 apart, 2 apart, and so on. If we do this for all possible distances over hundreds of polymer molecules, and average over many snapshots for a very long time (nano- to micro- seconds for a melt or something in solution), what we will eventually find is something like this:

The correlations drop off quickly with length, then more slowly, until finally they are completely uncorrelated.

It is important to realize here that at any given time, for any given polymer, the tangent vectors could be almost anything. You wouldn't expect two adjacent bonds to have a very small angle between them, and in general you would expect more flexibility over longer lengths of chain compared to shorter lengths, but you would not be able to predict the shape or angles with any degree of certainty, other than to say that on average, beyond the persistence length the tangent vectors would not be correlated, and that the amount of correlation would decay exponentially with length.

The examples you gave of a stiff rod vs. cooked spaghetti are meant to demonstrate this intuitively, but your question is about what happens when the persistence length and the polymer itself are the same length. The answer is that the polymer will be somewhere between a very flexible piece of cooked spaghetti and a stiff rod. Imagine taking a piece of spaghetti and bending it into a loop, but not enough that it breaks. The circumference of 1/4 of the loop would be something like the persistence length, and the response of the spaghetti in the loop would be similar to a polymer that is about the same length as its persistence length. However, keep in mind that polymers are constantly in motion, and so "flexible rods" and "cooked spaghetti" are not great analogies in the first place. It's more realistic to think of them as constantly wriggling worms, or something like that. Statistical descriptions of correlation lengths and times are the best we can get, unfortunately.

Another concept that might help you visualize this is that of the Kuhn length - this is twice the persistence length, and is the length at which sections of polymer are uncorrelated with each other in space. Practically, it means that the motion of the polymer within the Kuhn length is irrelevant to the motion of the overall chain. On a large scale, the motion of the Kuhn "blobs" is what matters. In your case, the polymer is about as long as the persistence length, which means it is one-half of the Kuhn length. This means motions along the polymer are correlated, and so you can't ignore them (if they are important to your model.)

Practically, what this all means is that your model is probably going to be fairly complicated, because you can't average out or ignore terms that can be ignored in either of the extreme cases.

If you post some more details about your model, I could probably give you more specific information.


What governs the merging of globs of inmiscible liquid?

Emulsion chemistry is actually rather complicated, but I'll try to give a conceptual description. Whenever you have a surface dividing two phases, there is an energy associated with creating a certain amount of surface, expressed as a surface energy density with dimensions of energy divided by area. This represents the energy cost of disrupting the intermolecular interactions of each phase at the interface, e.g. for your water/chloroform system, water can't hydrogen bond across the interface so it takes energy to break all the bonds required to make the surface. The larger the surface area, the more energy it takes. Logically, you can see that the lowest energy state is the one in which the interfacial area is minimized, which is why in the absence of external forces like gravity, liquids assume spherical shapes which give the smallest surface area for a given volume.

When you have many small droplets, there is a large surface area to volume ratio which is a high energy state, so it is energetically favourable for the droplets to form larger droplets. However, the reason it doesn't occur instantaneously is that an energy barrier can exist between the two states, for various reasons (electrostatic repulsion is a big one). In most cases, smaller droplets are lost first because the energy barrier to coalescing is surface area dependant. Larger droplets are more stable. Depending on the liquids, different separation mechanisms can occur with different kinetics. (coalescence, Ostwald ripening) I can't give a definite answer, but I'm guessing in your system you would not see an exponential decay-type half-life as that assumes a stochastic process where the probability is always the same. Since the stability depends on surface area, this isn't true in this case.


Can I use potassium acetate instead of potassium carbonate?

Stay with the carbonate.
In the course of the reaction $\ce{HCl}$ is formed. The aim is to remove this from the reaction mixture both as soon as well as complete as reasonably possible.  Compare

The addition of sodium acetate will increase the complexity of your reaction mixture.  $\ce{HCl}$ is soluble in methanol (reference), and upon dissociation ($\ce{HCl <=> H+ + Cl-}$), the $\ce{H+}$ and the acetate $\ce{OAc-}$ join each other to generate the buffer system of $\ce{HOAc <<=> H+ + OAc-}$.  Think about the $\ce{HOAc}$ as a reservoir, it may release $\ce{H+}$ at any moment you probably do not intend; a prominent example may be an aqueous workup.
The quantity of $\ce{HOAc}$ generated may be small -- in comparison of the methanol used as solvent of reaction.  Still it may complicate an aqueous workup and following extraction, because this organic substance solutes so well in water, but fairly enough in organic solvents such as hexanes, or chloroform, too.  So your extraction of the product may be affected.  A distillation (bp 118 C (pure acetic acid) instead of 65 C (pure methanol) at ambient pressure) ahead of the workup may remove this from your reaction mixture.

with the alternative

addition of $\ce{K2CO3}$.  Darn efficient removal of $\ce{H+}$ for once and forever, because the quench along  $\ce{K2CO3 + 2H+ -> 2K+ + H2O + CO2}\uparrow$ is practically irreversible.  In addition, it allows you to monitor the advancement of the reaction since $\ce{K2CO3}$ is not well soluble in methanol.


How to calculate molar mass using the ideal gas law?

You can first use the ideal gas law to calculate $n$:
\begin{align}
pV &= nRT\\
(\pu{101325 Pa})(\pu{0.250 l}) &= n(\pu{8.314 J K-1 mol-1})(\pu{273 K})\\
\end{align}

Solving for "$n$" gives $n = \pu{0.0111577 mol}$.

Now we use the connection between mass $m$, amount of substance $n$, and molar mass mass $M$:

\begin{align}
  M &=\frac{m}{n}\\
  n &=\frac{m}{M}\\
  \pu{0.0111577 mol} &= \frac{\pu{1.78 g}}{M},
\end{align}

which gives a molar mass of $M = \pu{159.53 g mol-1}$.

Looking at the Periodic Table, we can identify the diatomic gas as being bromine gas ($\ce{Br2}$), which has a molar mass of $\pu{159.8 g mol-1}$.


Degenerate orbitals in the Hydrogen atom

It is true that if you solve the Schroedinger equation the solutions depend only on the principal quantum number $n$ and the energies (often called eigenenergies or eigenstates) have the values predicted by the Bohr formula, $E=-hcR_\infty /n^2$ and energy differences as predicted by the Balmer formula. The calculated energy does not depend on $l$; this degeneracy of wavefunctions with different $l$ is a peculiar and special feature of the symmetry of the Coulomb potential.

However, nature cares little for our equations and experiments show that in fact the energy of the $l$ levels are different in energy to one another. This is called 'fine structure' and the energy shifts are very small as the name suggests. The first cause is explained as being due to the fact that the electron has an intrinsic quantum property called 'spin', which does not mean that it is literally spinning, but crudely speaking that it can be described by the same equations as angular momentum. As the electron is charged its spin produces a magnetic field  which interacts with the magnetic field of its orbital motion, which in turn depends on $l$. This interaction is called spin-orbit coupling, and technically it is a relativistic effect.The second effect is called the Darwin term, for electrons with $l=0$, and the third effect is the Lamb Shift, which needs explanation beyond relativistic effects and requires Quantum Electrodynamics.

The fine structure due to spin-orbit coupling can be observed easily in many atoms so the $l>0$ levels are split for each $n$ and so are no longer degenerate. The common example is the two lines of Sodium 'D' emission. The difference is small, the two transitions are 589.59 nm to 588.99 nm. This is the yellow colour associated with sodium easily seen if you sprinkle salt into a gas cooker's flame and in some types of street lights. The splitting can be measured with a simple spectrometer and is a common teaching lab experiment. The other effects mentions are far,far more difficult to measure and in a Chemistry course you are unlikely ever to meet them.


How many liters of hydrogen gas is produced per gram of aluminium?

Your work is correct.  I have to assume that this is being graded by a computer, so trivial differences between your answer and the answer that it is expecting matter.

Possible source of error 1: Ideal Gas Law

Instead of using $22.4\ \mathrm{L/mol}$ they may have been expecting you to use the ideal gas law.
$$PV = nRT$$
$$101.3\ \mathrm{kPa}\times V = 0.05559\ \mathrm{mol} \times 8.314\ \mathrm{J\ mol^{-1}\ K^{-1}} \times 273.15\ \mathrm K$$
$$V = 1.246\ \mathrm L$$

Possible source of error 2: Units

The computer may be looking for units in liters per gram of aluminum, something like:

$$\frac{\mathrm L}{\mathrm g} \ \text{or}\ \frac{\mathrm L}{\mathrm g\ \ce{Al}}$$

I doubt that this is the case, but it's possible.

Possible source of error 3: Typo

Either you or the person who set up the quiz made a mistake and typed in the answer incorrectly.


The statement is:

  This[collision frequency] must be the same for the two reactions, since both involve collisions of the same particles: a propane molecule and a chlorine atom.

It is clear that n-propyl radical and Isopropyl radical have same carbon structure. In the next step the hydrocarbon radical reacts with a chlorine ($\ce{Cl2}$) and since chlorine is symmetrical the collision frequency cannot depend upon it. It should be easy to judge from the following structures, the very minor difference:

Another reason is: Since the difficult step is the abstraction of a hydrogen and the determination of rates should be done while comparing primary, secondary or tertiary radicals, which have different stability.


Why is sand, sand; and why is glass, glass; While both mainly contain Silicon Dioxide?

Is it crystallization?

You are correct. The main difference is that sand is crystalline and glass is not—it is amorphous.

The main component (> 95%) of common yellow sand is quartz (the mineral whose composition is SiO2). Note that not all sand is quartz. There are white sands containing calcite (CaCO3) and black sand (containing various heavy minerals). But the most common sand is indeed quartz sand: SiO2.

Glass, the type you see in your everyday life, on the other hand, is not composed of pure SiO2. It has a bunch of other additives such as Na, K, B, and others. This is done to modify the properties of the glass and make it more suitable for human use. It doesn't matter much though for our discussion.

So if they are made of the same thing, why the difference? The answer is cooling rate. If you cool molten SiO2 slow enough, the atoms have enough time to organize themselves into a crystalline structure. In the case of pure SiO2, this is a network of SiO4 tetrahedra: One silicon atom surrounded by four oxygens. If it cools too fast, then the crystalline structure does not form. It may be completely amorphous, or form into a sub-microscopic array of SiO2 crystals in various structures (CT-opal for example).

What determines the cooling rate? Well, in the case of glass it is a matter of minutes. You've seen glass making: The glass is molten and very quickly it solidifies to a solid. In contrast, most of the quartz sand you're seeing is actually broken fragments of rocks called granite. This type of rock has abundant quartz in it, and it forms deep underground (as in 10s of kilometers) at very slow cooling rates. While a glass maker can take his glass and let it cool in the atmosphere or in water, molten silicate magma ("glass") deep in the Earth is surrounded by rocks that are in the hundreds of degrees. This slow cooling facilitates crystallization of the SiO2 into quartz rather than glass. How slow is this? At least tens of years, more commonly hundreds or even thousands of years. This is much slower than the seconds and minutes in glass making.

Slow crystallization of SiO2 in the Earth is not the only process. There are actually processes that can take molten silicate magma and cool it very rapidly. This commonly happens in volcanic settings: A volcano throws molten magma (again, made mostly out of SiO2) and throws it into the atmosphere where it can cool very rapidly. It's even faster if the magma erupts into water. The result is volcanic glass: sometimes called obsidian.

The difference in the visual appearance is actually not due to the presence or absence of a crystalline structure. It's mostly related to grain size. If you take window glass and shatter it so it's all small grains, it will look just like sand. On the other hand, if you take quartz, and make it big and smooth (just like window glass), it will look like this:

(By Parent Géry (Own work) [Public domain], via Wikimedia Commons)


How are the boiling points of Tungsten and other metals determined and proved?

According to the 1913 paper THE VAPOR PRESSURE OF METALLIC TUNGSTEN, The Physical Review, vol. II, pp. 329-342:

The vapor pressure (p) of a pure substance is related to its evaporation rate (m) in a vacuum by the following relationship:
$$ m= \sqrt{\frac{M}{2\pi RT}}p$$

Evaporation rates were measured at temperatures in the range from 2400K to 3100K, heating by electric resistance of a filament, and determining the evaporation rate by weighing the filament before and after the evaporation.
The temperature was determined by observing the color and intensity of the light emitted by the filament. (Paper says they will publish a separate paper about the details of the temperature determination technique).
Vapor pressure was extrapolated to a boiling point of 5110 K using the Clausius-Clapeyron relationship.


Inexpensive [about US\$0.30 / liter] marble or limestone scraps, $\ce{CaCO3}$, likely would serve well. You could make an outer bucket with polyethylene netting, so that the pickling bucket would be well below the level of surrounding limestone chips. Coarse netting would not work with powders. Make the neck just large enough to admit the parts.
Better would would be to draw air continuously from above the container through a scrubber stack filled with marble or limestone chips, and possibly vent outside.
Definitely don't use lye, $\ce{NaOH}$, which is perhaps more dangerous than $\ce{HCL}$, as well as hygroscopic! You'd wind up with a puddle of caustic lye solution at the bottom of the outer bucket. And $\ce{NaHCO3}$ would also make a mess, though non-toxic.
BTW, if using large amounts, be careful that a spill into the outer container doesn't release so much $\ce{CO2}$ as to displace air and asphyxiate someone.


According to Jesus de La Fuente, CEO of Graphenea:

  Graphene is, basically, a single atomic layer of graphite; an abundant mineral which is an allotrope of carbon that is made up of very tightly bonded carbon atoms organised into a hexagonal lattice.

He also states that there are three commonly-known naturally occuring allotropes of carbon - Diamond, Graphite, and amorphous carbon. Graphene's structure is similar to that of Graphite, but Graphene itself is much different because it has a $sp^2$ hybridization, compared to $sp^3$ hybridization of carbon in the form of Diamond, and Graphene has a thickness of just 0.345Nm (nanometers). This means that it does not form tetrahedrals like common carbon atoms, making Graphene similar but special compared to other forms of carbon.

Also stating Jesus de La Fuente in Graphite-Graphene Comparison:

  Graphene is fundamentally one single layer of graphite; a layer of $sp^2$ bonded carbon atoms arranged in a honeycomb (hexagonal) lattice. However, graphene offers some impressive properties that exceed those of graphite as it is isolated from its ‘mother material’.

To elaborate, Graphene has special properties that set it apart. There are similarities: stacking layers of Graphene may be similar to creating Graphite; creating a rope of Graphene may be similar to creating Crabon Nano-tubes; both have $sp^2$ hybridizations. However, Graphene has special properties that make it stronger when stacked because of how it is extracted and isolated.

As for the conductivity, Jesus de La Fuente states that it's electronic properties is what sets it apart (See Graphite-Graphene Comparison). Fundamentally, Graphene is a zero-overlap semi-metal, with both holes and electrons as charge carriers, that has very high electrical conductivity by itself. Graphite carbon allotropes have 4 electrons in it's valence shell available for bonding, however, Graphene has 3 electrons that are bonded with other Graphene atoms in the 2D Plane (a.k.a. other Graphene atoms next to it on the same one-atom thick sheet). This makes a single sheet very conductive, because the remaining electron that isn't bonded with other Graphenes in the same sheet are in the 3D Plane (above and below the sheet), causing what is called, pi electrons. However, when Graphene sheets are stacked together, their pi electrons bond with each other, thus reducing the electric conductivity of the Graphene.


One thing you can do before searching for peaks is to smooth the data with a 9 point (or 15 point) moving average. I'm not sure how the ends are handled, but the average of 9 points replaces the data point so noise fluctuations are reduced.


What is the difference between an Aryl Halide and a Vinyl Halide?

No, aryl halides are not a type of vinyl halides.
A compound is called an aryl halide when it contains a halogen group directly attached to an aromatic system. For example,

Whereas a compound is called a vinyl halide when it contains a halogen group directly attached to a doubly bonded carbon atom. For example,

The double bonds in an aromatic system are not the same as the aliphatic double bonds. They differ quite a bit in terms of electron density and thus there derivatives behave very differently is many circumstances.
The term phenylic halide doesn't exist but a very similar term phenyl halide exists which is a subset of aryl halides family.
An aryl halide can be called a phenyl halide when the aromatic system of the compound is specifically the phenyl system($\ce{-C6H5}$). For example,


Stereoisomers of compounds named tribromocyclobutane

The second configuration shown cannot exist simply because of the geometry of the orbitals at the carbon center. The angle between the bromine's (those on the same carbon) is much less than 108° even though there is no forced strain.


Does the enthalpy of solution formula (LE + Hyd) change depending on the question?

Apparently your problem is related to the improper use of negative signs. Let's solve it differently.
Suppose you start from the separated ions $\ce{Ca^{2+}}$ and $\ce{2 F^-}$ in the vacuum. This state has the energy zero by definition.
Now suppose you join the ions to produce $1$ mole of solid $\ce{CaF2}$.  The process is exothermic, and the final state $\ce{CaF2(s)}$ is at the energy level $-2611 kJ/mol.$
Now suppose you are doing another transformation with the same ions. You put them directly into water, so that the ions become hydrated in solution. You obtain a solution of $\ce{CaF2}$ in water, or a solution of $\ce{Ca^{2+}(aq)}$ and $\ce{2 F^-(aq)}$. This transformation is also exothermic, and at the end $\ce{Ca^{2+}(aq)}$ + $\ce{2 F^-(aq)}$ are at the energy level $- 2662 kJ/mol$.
The final levels after these two transformations are $-2611 kJ/mol$ for the final solid, and $-2662 kJ/mol$ for the solution. These energy levels are not very different from one another.
So the passage from the solid state $\ce{CaF2}$ to the solution $\ce{Ca^{2+}(aq)}$ + $\ce{2 F^-(aq)}$ is slightly exothermic, with $\Delta H = -51 kJ/mol$


Why are there only 6 series (Lymen, Balmer, Paschen, Bracket, Pfund and H.Davy) in the hydrogen atomic spectrum? Why can't there be more?

As Ivan commented, there are actually an infinite number of possible series of this type. So your question is really why there are only six named series.  The reason is part of the culture of science.  Typically, a result is named if it is, to use Mithoron's term, sufficiently notable.  Thus we have the Diels-Alder reaction, the Hartree–Fock method, the Wentzel–Kramers–Brillouin (WKB) approximation, the Southern blot, and so on.

Hence the reason why the initial series of hydrogen spectra were named is because they were notable scientific results. [In addition—and perhaps someone more expert in the history of science can comment on this—it is possible that identifying the, say, fifth and sixth series weren't that scientifically notable, but a tradition had become established of naming successive spectra of hydrogen.] But beyond the sixth, finding more series was essentially deemed routine. The last published series I know of is the seventh, identified by Hansen and Strong, but not named after them [Peter Hansen and John Strong, "Seventh Series of Atomic Hydrogen," Appl. Opt. 12, 429-430 (1973)].

That's not to say it wouldn't be experimentally difficult to see higher-numbered series, since they become increasingly faint. Rather, it's that, at this point, finding them isn't interesting science.


As orthocresol mentioned, this is all about relativity, so let's talk about it. I am hardly an expert myself, but I'll try to give an answer to the best of my limited knowledge. For an interesting and accessible overview of the incorporation of relativistic effects into chemistry, I recommend the 2012 review article "Relativistic Effects in Chemistry: More Common Than You Thought" by Pekka Pyykkö (free access!), one of the great names in the field.

The theoretical basis for most of our work as chemists is "old" quantum mechanics, developed up to around 1925, and one of the hallmarks of this era is the Schrödinger equation. The whole concept of orbitals comes as the solution of the Schrödinger equation for the hydrogen atom, (i.e. a two-body system composed of an electron in a spherically symmetric electric field originating from the nucleus). The stationary states in this system are spherical harmonics, described by the quantum numbers $n$, $l$ and $m$ (or $m_l$). The first two numbers determine the shape of the orbital, but we've gotten used to calling the second number by a letter (s, p, d, f, g, ... for 0, 1, 2, 3, 4, ...). So that's what orbital names, such as $\textbf{4s}$, $\textbf{3d}$ and $\textbf{5f}$ mean.

That said, old quantum mechanics is fundamentally incomplete, in a way which very much matters to chemists. In particular, in 1922 the Stern-Gerlach experiment discovered the property of spin. While its existence could be accommodated within the prior framework by "tacking it on" and adding an extra quantum number $s$, there was no theoretical background to explain it. This changed a few years later, when Paul Dirac incorporated special relativity into quantum mechanics. Old quantum mechanics implicitly assumes the speed of light is infinite, and by adequately introducing its finitude, many new experimentally-verified phenomena appear, such as antiparticles and, indeed, spin. In other words, spin is a purely relativistic quantum mechanical property!

The equivalent of the Schrödinger equation in relativistic quantum mechanics is the Dirac equation. Much like the Schrödinger equation, the Dirac equation can be solved for a two-body system composed of an electron in a spherically-symmetric electric field created by the nucleus. The resulting solutions ("Dirac orbitals"), however, are somewhat different. In particular, the quantum numbers $l$ and $s$ are no longer "good" quantum numbers (that is, they no longer describe stationary states, due to spin-orbit coupling), but their sum ($\overrightarrow{j}=\overrightarrow{l}+\overrightarrow{s}$) is. For the electron, $s=\frac{1}{2}$, and thus for each value of $l$, there are now two possible values of $j$, given by $|l+\frac{1}{2}|$ and $|l-\frac{1}{2}|$. This means that the s, p, d, f... subshells from the Schrödinger solutions are now split according to their $j$ values. The s subshells are unique in that they don't split ($|0+\frac{1}{2}| = |0-\frac{1}{2}| = \frac{1}{2}$), but are now termed $\textbf{s}_{1/2}$ subshells. The p subshells, which are triply degenerate via the Schrödinger equation, now split into a single $\textbf{p}_{1/2}$ orbital and two degenerate $\textbf{p}_{3/2}$ orbitals. The d subshells, previously quintuply degenerate, split into two degenerate $\textbf{d}_{3/2}$ orbitals and three degenerate $\textbf{d}_{5/2}$ orbitals. The trend continues similarly for the further subshells. Here is a figure from "Relativistic quantum chemistry: The electrons and the nodes", which will hopefully make it clearer:

So what are the physical consequences of these different mathematical solutions? Firstly, the orbitals change in energy. Orbitals with a lower $j$ for a given $n$ and $l$ have lower energy (e.g., the single $\textbf{4p}_{1/2}$ orbital is lower in energy than the two degenerate $\textbf{4p}_{3/2}$ orbitals). The shape of the orbitals also change; all orbitals with $j=\frac{1}{2}$ are spherical, regardless of $l$.

According to "Contour diagrams for relativistic orbitals":

  Contour plots for $^2P_{1/2}$ and $^2S_{1/2}$ are spherically symmetrical,  while those for $n = 2$ and $3$, $l = 1$, $j= 3/2$, and
  $m = \pm 3/2$ look very similar to those for p orbitals already published in this journal.

Similarly, all orbitals with $j=\frac{3}{2}$ are dumbbell-shaped, including $\textbf{d}_{3/2}$ orbitals. The article brings a helpful schematic:

As additional confirmation, "Pictorial Representations of the Dirac Electron Cloud for Hydrogen-Like Atoms" states:

  As an example of these two theorems the angular charge distributions for the magnetic states $m = \pm 3/2$ for a $^2P_{3/2}$ term are not only the same but are also identical with the two magnetic states $m = \pm 3/2$ of the $^2D_{3/2}$ term. It should be pointed out that the radial charge distributions of the $^2P$ and $^2D$ terms are different. [...]
  In this connection one of the consequences of the Dirac theory is that
  a single p electron or two similar p electrons, depending on whether $j=\frac{1}{2}$ or
  $j=\frac{3}{2}$ respectively, present spherical symmetry. Not only are all S states
  spherically symmetrical, as on the Schrodinger theory, but also all one electron systems having but one valence electron and that electron in a $^2P_{1/2}$ state,
  eg., the normal states of B, Al, Ga, In, Tl.

The article also shows a table of the angular distribution functions for the wavefunction. Note how the function is equal to $1$ for $\textbf{s}_{1/2}$ and $\textbf{p}_{1/2}$ orbitals. That is, there is no angular dependence, implying spherical symmetry.

The article also brings some diagrams and pictures of simulated orbital shapes.

I wouldn't be able to tell you why this is the case any more than I would be able to explain the shapes for the solutions to the Schrödinger equation - that's just how it pans out.

Another difference is that the Dirac orbitals do not possess nodal points or planes. They are now replaced with "pseudo-nodes", where the wavefunction assumes very small values, but never reaches zero anywhere. This is nicely shown in this figure from "Relativistic quantum chemistry: The electrons and the nodes":


How do Chiral centers determine the number of isomers?

That 2^n number of isomers are exactly stereoisomers ( Not "total no. of isomers"). There is no direct formula for calculating the number of structural isomers.Total number of isomers = (number of structural isomers + number of stereoisomers).
Though there are direct formulas for obtaining number of stereoisomers, but there is no direct formula for finding out the total number of structural isomers.
As I said above, we do have formulas for finding out the number of stereoisomers. Here is the exact formulas which you can use to determine number of stereoisomers:
a) When the molecule is unsymmetrical and contains 'n ' chiral carbon atoms:
Total no. of stereoisomers = 2^n
b) When the molecule is unsymmetrical and has even number of stereogenic centres or chiral carbon atoms:
Total no. of stereoisomers = Number of optical isomers + Number of meso forms = 2(n-1) + 2(n/2-1)
c) When the molecule is symmetrical and has odd no. of stereogenic centres
Total no. of stereoisomers = [2(n-1)-2(n/2-1/2)] + 2(n/2-1/2)]


Why Zn has highest ionisation enthalpy in 3d series?

There is band structure formed in the metallic lattice  in the crystals of d and f block elements due to partially filled orbitals. But as you move along the periodic table to 'Zn' , you will find that the orbitals are completely filled hence there is no possibility for inter-orbital mixing. Another thing worth noting is that due to the filled d-orbital, zn and copper behaves more like s-block rather than d-block elements.

"The filled d-shells in these elements contribute little to interatomic interactions, which are dominated by the s-electrons through metallic bonds. Unlike metals with incomplete d-shells, metallic bonds in copper are lacking a covalent character and are relatively weak. "

An alternate explanation can be on the metallic lattice:
Copper has a face centered cubic structure and zinc has a hexagonal close packed structure

reference : wikipedia


How does the pH of fresh glacial meltwater compare to marine seawater?

In general, the pH of seawater can range from 7.5 to 8.4 depending on where you are in the world, this is due to temperature differences, shipping activity, pollution, and several other factors. Reefs are quite sensitive to pH and are usually found in bodies of water with a pH ranging from 8.0 - 8.4. Glacial water should be nearly neutral as most glaciers were formed thousands of years ago when pollution levels were relatively low. It is quite possible that the glacier would actually be very slightly acidic due to the carbon dioxide that would have been trapped in the ice when it formed bubbling through the water creating a very weak carbonic acid. Scientists actually measure the levels of carbon dioxide in the ice to get an idea of what the Earths atmosphere was like thousands of years ago.

Other users please feel free to correct me if I am incorrect in my answer,  hopefully you can find this useful.


Confusion about atomic/molecular orbital terminology

The short answer is that orbitals are defined to within a unitary transformation.

Canonical Hartree-Fock equations

The application of variational principle to the Slater determinant leads to the system of Hartree-Fock equations
$$
    \newcommand{\op}[1]{\hat{#1}}
    \newcommand{\mat}[1]{\mathbf{#1}}
    \op{F} \psi_{i}(1)
    =
    \sum\limits_{j=1}^{n} \lambda_{ij} \psi_{j}(1)
    \quad \text{for} \quad i = 1, 2, \dotsc, n \, ,
$$
which can be simplified by choosing a unitary transformation $\mat{U}$, defined by the way it transforms spin orbitals,
$$
    \psi_{k}' = \sum\limits_{j=1}^{n} u_{kj} \psi_{j} \quad \text{for} \quad k = 1, 2, \dotsc, n \, ,
$$
that makes the matrix of Lagrange multipliers $\lambda_{ij}$ diagonal and leads to the system of the so-called canonical Hartree-Fock equations
$$
    \op{F} \psi_{k}'(1) = \lambda_{kk}' \psi_{k}'(1) \, , \quad k = 1, 2, \dotsc, n \, ,
$$
or, in a much more familiar form using symbol $\varepsilon_k$ for $λ_{kk}'$ and dropping the primes
$$
    \op{F} \psi_{k}(1) = \varepsilon_{k} \psi_{k}(1) \, , \quad k = 1, 2, \dotsc, n \, .
$$
The canonical Hartree-Fock equations can be thought of the eigenvalue ones, for solving which there exist a number of well-developed algorithmic approaches, thus, they are so convenient to work with.

Slater determinant invariance to a linear transformation

It can be shown that Slater determinants composed of original and transformed spin orbitals are related as follows,
$$
    \Phi' = \det(\mat{U}) \Phi \, ,
$$
which by taking the absolute values and squaring both sides leads to
$$
    |\Phi'|^{2} = |\det(\mat{U})|^{2} |\Phi|^{2} \, .
$$
This means that the transformed Slater determinant $\Phi'$ describe exactly the same system as the original Slater determinant $\Phi$ in every observable way. Or, in other words, the Slater determinant is invariant to any linear transformation of spin orbitals (except for normalization). The only advantage of the unitary transformation from that perspective is that $|\det(\mat{U})|^{2} = 1$, so that there is no even need for renormalization.

Transformed canonical orbitals

Now, as it was mentioned earlier, it is convenient to do calculations with canonical orbitals, however, once the Hartree-Fock equations are solved, the resulting canonical spin orbitals can be transformed by any arbitrary transformation making no changes with respect to observable properties of the many electron system in question. And it advantageous to use unitary transformations since there is no need for renormalization of the Slater determinant.

So, other sets of orbitals can be constructed by transforming the canonical orbitals, i.e. by forming linear combinations of them. The way these linear combinations are formed (i.e. the choice of a unitary transformation) is dictated by what is the ultimate goal of the transformation. For instance, canonical orbitals are usually fairly delocalized over a molecule, rather than concentrated in limited spatial regions of it as anticipated from general chemistry models. Thus, of a particular interest for a chemists is a transformation which localizes the orbitals. The way localized orbitals are obtained depends on a particular choice of the unitary transformation and there are in fact few different approaches.

But just localizing the orbitals might not be the end of the story. For instance, one might want to not just localize them, but do so in a way that the maximum-occupancy character in localized 1-center and 2-center regions of the molecule. This procedure results in the most accurate possible natural Lewis structure corresponding to a given Slater determinant and is the basis of the natural bond orbital model.


Helium in stars is an initial product in the fusion reactions that go on to make "more interesting" elements.  In that sense we do need helium to get life going.

Once the more complex atoms are made, however, life appears to have no further use for helium.  As Nicolau mentioned, it seems too hard to see life use helium. The neutral atom shares a spot with neon for the least reactive substance in all of chemistry, whereas the other commonly-found form of helium, the alpha particle (He2+) from radioactive decay is arguably the most reactive substance in all of chemistry. Life can only work with supremely delicate balancing of reactivity in the middle of the scale, not such crazy extremes.


Why so differance in alkali halide's crystal forms?

Perhaps you have heard of the following terminology: "rock salt", "cesium chloride", "zinc blende" structures? If so, you may recall that in this case, the main factor dictating structure is "size" or atomic radius ratio, rather than group. For example, in the case of the rock salt structure, it is most likely that there is "a cation/anion radius ratio of 0.414 to 0.732". As such, you could consider the radius of cesium in comparison to sodium: significantly larger. Thus, it adopts the cesium chloride configuration, which is preferred when "ions are of roughly the same size ".


How to find the concentration of potassium permanganate in this experiment?

I think your experiment has one setup mistake.:

You're mixing $\pu{0.8265 g}$ of oxalate with $\pu{50 mL}$ of water and $15$-$\pu{20 mL}$ of sulfuric acid. So your $\pu{25 mL}$ oxalate solution does not contain $\pu{0.8265 g}$  of oxalate (more or less $\frac{25 \cdot 0.8265}{ 65-\pu{70 mL}}$). So you can't know the exact oxalate concentration of this solution.

I recommend you to use a $\pu{100 mL}$ volumetric flask to prepare the oxalate solution. You can then calculate its concentration and the amount of oxalate you have in $\pu{25 mL}$.


Beside Jon Custer's comment, to be taken with a grain of salt (as the op mentions a heating unit implemented in the ultrasonic bath), if the manual allows, it depends.

You will recognise that by simple use (even if nothing but water is in the trough), that the temperature of the water will increase simply bu use.  Hence, to keep the results reproducible from one batch, to the next, you power on the bath until this temperature stabilises.  (It is better to briefly switch off the sonification to record the temperature with a thermometer.)

In terms of duration of sonification: while one aims to keep the reaction time under sonification short(er) than by conventional heating (alone), like

(source, doi 10.1016/S0008-6215(00)90155-1)

there are examples, where the sonification reportely has to last much longer than $\pu{90 min}$, for example

(source)

I've seen colleagues run reactions from morning till noon, for the sake of selectivity of the reaction intentionally cooling the bath by intermittent addition of water ice into the trough.

To attenuate the cavatational noise the bath may generate, esp. if your flat bottom vessel hovers with some distance just above the resonators (the hot spots), either consider a different room (or at least, keep the sash closed), or have a "noise-proof shoe box" like

(source)

that you may design by yourself, too, covering your bath durin the operation.  Instead of the foam-lining, a wooden box lined with polysterene is even cheaper, but offers less "relief". It is even better to place the bath (if weight permits) at the same time on a sheet of expanded polysterene that you get "for a smile" at at your departemental chemical shop, or appliance shop.

But do not forget the ultrasound bath underneath the box / in the other room, especially if your heating unit is used.

(source)

Beside this, you may interested in periodicals like Ultrasonics and Ultrasonics Sonochemistry, or Luche and Capelo-Martínez as secondary references, too.


What you've performed is, technically, "thermal decomposition of cellulose", or pyrolysis. During heating, there are large number of changes producing many different chemicals, so the smoke you saw was a mixture of different chemicals.

Pyrolysis is used to produce many industrial chemical, such as "wood alcohol" (methanol), and during gasoline shortages, to run internal combustion engines.


Can the structure of an azide ion contain a coordinate bond?

You can’t really tell a dative bond from a ‘normal’ covalent bond until you break the bond. A dative bond will then dissociate heterolyticly while a covalent bond will dissociate homolyticly.

Just using that simplified way of putting it, you could say that there are many reactions of azides wherein $\ce{N2}$ is liberated leaving a nitrene or related structure; just one of many examples would be the Curtius rearrangement. However, that is not what is meant in the definition above; the definition talks about supplying the bond dissociation energy.

The BDE is not typically supplied in chemical reactions except in coordination compound ligand exchange and radical generation. If you were to apply it to $\ce{N3-}$, I predict homolytic dissociation. Thus, the bond between the nitrogens is more of a typical single bond.

Of course, aside from the chemical implications of dative versus covalent bond you may just have decided to use an arrow notation rather than formal charges. Technically nobody can stop you but still you shouldn’t call it a true dative bond.


Differences between ¹³C-decoupled NMR experiment vs a ¹³C-coupled

The answer is a bit complicated because there are a number of ways to apply decoupling (see eg this site which describes different decoupler gating modes), depending on the goal. For simplicity I limit the discussion to acquisition of a 1D $\ce{^{13}C}$ spectrum:

$\ce{^1H}~$ decoupling during the entire experiment: this provides a signal enhancement due to the NOE from $\ce{^1H}$ to $\ce{^{13}C}$, and suppresses couplings, which reduces spectral complexity and also increases carbon s/n due to collapsed multiplets.
Decoupling during acquisition only (inverse gated decoupling): this suppresses couplings simplifying the spectrum and providing some enhancement by collapsing multiplets, but does not give NOE enhancement.
Gated decoupling: this provides the NOE enhancement but gives a coupled spectrum showing multiplicity.

Applying decoupling to obtain a signal enhancement from NOEs allows you to reduce the overall acquisition time (reduce the number of FIDs you need to acquire) relative to no NOE. This typically works for all but quaternary carbons. Decoupling during acquisition also can allow you to reduce the overall acquisition time, even if you do not obtain an NOE enhancement, by collapsing signals from multiplets into single peaks. The longest experiment would be a truly quantitative one, and particularly if you want full coupling information and s/n equivalent to that from a decoupled experiment. Then you would not apply decoupling at all. In that case you would have to wait on the order of $\mathrm{5 \times } \ce{^{13}C} ~\mathrm{T_1}$ and acquire enough scans to get the desired s/n.


Oxidation of sodium thiosulfate by iodine

Whether compound $\ce{X}$ is going to be reduced/oxidized and to what degree is mostly dictated by the corresponding redox potential for the given medium. The thiosulfate anion $\ce{S2O3^2-}$ is used in quantitative analysis and in iodometric titration in particular since it is a reasonably strong reducing agent [1 pp. 714-717]:

$$\ce{S4O6^2- + 2 e- <=> 2 S2O3^2-} \qquad E^\circ = \pu{0.169 V}$$
$$\ce{2 S2O3^2- + I2 -> S4O6^2- + 2 I-}$$

However, stronger oxidants such as chlorine are capable of oxidizing thiosulfate further down to sulfur(VI), e.g. forming a sulfate instead:

$$\ce{S2O3^2- + 4 Cl2 + 5 H2O -> 2 HSO4^- + 8 H+ + 8 Cl-}$$
Bromine, being intermediate between iodine and chlorine, can cause $\ce{S2O3^2-}$ to act either as a 1-electron or an 8-electron reducer
according to conditions. For example, in an amusing and instructive experiment, if concentrated aqueous solutions of $\ce{S2O3^2-}$ and $\ce{Br2}$ are titrated, and the titration is then repeated after having diluted both the $\ce{S2O3^2-}$ and $\ce{Br2}$ solutions 100-fold, then the titre will be found to have increased by a factor of exactly 8.

Note, however, that it is not entirely correct to state that sulfur in thiosulfate has oxidation state $+2$. In fact, there are two nonequivalent $\ce{S}$ atoms with oxidation states -1 and +5. Same goes for tetrathionate anion $\ce{S4O6^2−}$ where two central $\ce{S}$ atoms have oxidation state $0$ and the two terminal ones $+5$.
References

Greenwood, N. N.; Earnshaw, A. Chemistry of the Elements, 2nd ed.; Butterworth-Heinemann: Oxford; Boston, 1997. ISBN 978-0-7506-3365-9.


Why are fluoroalkyl chains hydrophobic/oleophobic?

You have identified a fairly common but counter-intuitive result. At least it's counter-intuitive based on the way the properties of fluorine are described in undergraduate chemistry classes. Specifically, we're told it's very electronegative and hence forms very strong hydrogen bonds, and that it is very reactive. This is, however, a poor description of what tends to dominate the behavior of fluorine when it is bonded to carbon.
Let's start with something simple but illustrative. If I told you that the boiling point of $\ce{CCl4}$ is $\pu{350 K}$ and the boiling point of $\ce{CBr4}$ is $\pu{463 K}$, what would be your guess for the boiling point of $\ce{CF4}$? Probably less based on the trend, but just how much less?

$\ce{CF4}$: b.p.=$\pu{145.5 K}$
$\ce{CH4}$: b.p.=$\pu{111.7 K}$

First time I've used one of these :)

Well hopefully you don't look this far, but you'll notice that the boiling point of $\ce{CF4}$ is much, much lower than that of $\ce{CCl4}$. Importantly, the difference between the fluorine and chlorine compounds is much larger than that of the chlorine and bromine compounds.
Additionally, the fact that $\ce{CF4}$ boils only slightly higher than $\ce{CH4}$ indicates that the molecule basically loses all interactions when going from the chlorine to the fluorine.
What does this actually illustrate? It doesn't tell us anything about hydrogen-bonding or dipole interactions (all have zero dipole), but it tells us about the dispersion forces present in the system. Namely, if this result generalizes, we have learned that a $\ce{C-F}$ bond is essentially non-polarizable. This is very important because polarization is most of what determines the strength of dispersion interactions. Okay, this basically answers the question, but let's be more systematic.
$\textbf{Dipole-Dipole Interactions:}$
The paper you linked is basically looking for a material which won't interact strongly with water and other polar substances. One important feature is to make sure the ligands on this material do not have an appreciable dipole then. It is true that the $\ce{C-F}$ is statically polarized towards the fluorine atom, so there is a dipole along the bond, but the molecular dipole is generally what we will care about. The ligands are basically cylindrical, so any dipole perpendicular to the cylinder ought to roughly cancel out. If anything there would be a dipole along the ligand backbone, but this is probably very small if it exists at all. So, dipole interactions don't particularly matter.
$\textbf{Hydrogen-Bonding:}$
Despite what you might think, it actually is quite possible for water (or other $\ce{O-H}$ containing solvents) to hydrogen-bond with the fluorine in a $\ce{C-F}$ bond. See [1] for some calculations demonstrating this.
So, if you're designing this material, you might have to worry about this a little bit, but leave it up to the experiment to tell if this matters.
$\textbf{Dispersion Forces:}$
Here is the key. In organic chemistry, you're usually sold a lie that the things determining the size of dispersion forces are the surface area and the molecular weight of a molecule. If this were strictly true, then we would expect the increase in boiling point we saw above going from $\ce{CH4}$ to $\ce{CF4}$ to be much more dramatic than it is based on the change going from $\ce{CCl4}$ to $\ce{CBr4}$. The reason this rule works is because generally organic compounds are just a bunch of $\ce{C-H}$ bonds, so most things are roughly as polarizable as the next thing.
Okay, but it's not like the $\ce{C-F}$ bonds are inherently unpolarizable. They are about as polarizable as a $\ce{C-H}$ bond, so for all consideration of physical interactions, we can basically imagine that these ligands are just lipids.
$\textbf{Why Do Lipids Form Membranes?}$
This is actually quite a tricky, but interesting question. Imagine a big fat lipid in a solvent of water. One thing we know is that water likes to hydrogen bond. This lipid disrupts the hydrogen-bonding network. Furthermore, because there has to be some water at the surface of the lipid, it is likely these waters will get stuck aligning their dipoles with whatever dipole is present in the bond it is near.
So, it seems like having the lipid there is both energetically unfavorable and entropically unfavorable. It is unclear without an experiment which of these things will dominate and if clumping all the lipids into a membrane is actually better because that's definitely a decrease in entropy for the lipid.
Well, [2] shows the experiment describing this same idea. Slightly different system but same idea. Basically the conclusion is that the energy plays almost no role ($\Delta H\approx 0$) but the entropy for forming a membrane dominates.
Shamelessly quoting from the abstract:

Possible mechanisms for the entropy increase include: (i) the attraction between discrete oppositely-charged areas, releasing counterions; (ii) the release of loosely-bound water molecules from the inter-membrane gap; (iii) the increased orientational freedom of previously-aligned water dipoles; and (iv) the lateral rearrangement of membrane components.

The lesson to learn here is that the entropy of the solvent can actually be very important when considering both the equilibrium arrangements of heterogeneous systems as well in chemical reactions.
$\textbf{Bringing it All Together:}$
So, I hope you see what I mean when I said that noticing $\ce{C-H}$ and $\ce{C-F}$ bonds have almost identical intermolecular forces behave basically answers the question.
When we consider explicitly the interactions of a substance with the ligands you have asked about (in comparison to a lipid), we see that dipole interactions might be a little stronger between the solvent and the ligand, but the ligand-ligand interactions should also be a bit stronger than in the lipid. So, again, $\Delta H\approx 0$. In contrast, the stronger dipole means that the solvent is more tied up, so the entropy change should increase (not accounting for the effect of the cube thing the ligands are mounted on).
Note that this argument for membrane formation based on entropy is not exclusive to strongly polar solvents such as water. That is, pretty much any solvent gains a lot of entropy when a membrane forms. So, as long as it does not interact strongly the ligand, the ligands should form a membrane.
Finally, based on what I've said, why not just use $\ce{C-H}$ bonds rather than have fluorine present? I only skimmed the paper, and they mention durability which makes sense because a $\ce{C-F}$ bond is much stronger than a $\ce{C-H}$ bond. Plus alkanes have the problem of being very combustible which is probably not good as these materials would likely be used at high temperatures.
Random semi-relevant note. Teflon is just an alkane but with all $\ce{C-F}$ bonds, and it too is quite slippery (it's the stuff on non-stick pans).

$\textbf{Edits:}$
I forgot to address the oleophobic nature of fluorocarbons, which is the most important part of the question, really.
$\textbf{Oleophobicity of Fluorocarbons}$
One interesting trend which I was not aware of, is that the boiling point of fluorocarbons increases more slowly than boiling point of corresponding hydrocarbons. For instance, the boiling point of perfluorooctane is $\pu{103 ^\circ C}$ while the boiling point of octane is $\pu{125 ^\circ C}$. So, this means that the dispersion forces in hydrocarbons grow more quickly than that in fluorocarbons. This makes some sense because the carbon in fluorocarbons will be more positively charged, and the fluorine atoms are comparable in polarizability to hydrogen, so the polarizability of the carbon atoms is diminished.
It seems, however, that the oleophobicity of this particular structure has a lot to do with the fact that the chains are not free to move about in solution. For instance in [3], the authors show that by attaching a fluorocarbon, which is ordinarily not oleophobic, to silica particles, the flurocarbon becomes oleophobic. This indicates to me (and this could be argued) that by restricting the translational motion of the fluorocarbons by mounting them on a large structure, as is done in the example you give, that the favorability of mixing is lost. This assumes that the energetics of having the fluorocarbons dispersed throughout the oil is roughly the same as having the fluorocarbons and oil separated into phases.
So, it seems like entropy is the dominant factor in both the hydrophobic and oleophobic characteristics of these fluorocarbons. The big difference is that basically any nonpolar organic molecule will form its own phase in the presence of a polar solvent like water. This is because the interactions between the polar substances are fairly strong, but there is also still a lot of free motion (hydrogen-bonds, etc. break and form again quite often). With the fluorocarbon, one has to manually restrict the entropy gain due to mixing by forcing the fluorocarbons onto a silica base. These things individually have the same effect of disrupting some of the degrees of freedom of the solvent, so the two separate to maximize entropy.
Obviously, some of the parts about oleophobicity are a little bit hand-wavy because really either effect could dominate, but the fact that putting it on the silica base makes the fluorocarbon oleophobic as ref. 3 points out convinces me this is due to entropy and not enthalpy.

References:
[1]: Andrew V. Razgulin, Sandro Mecozzi, "Binding properties of aromatic carbon-bound fluorine," Journal of medicinal chemistry 2006, 49(26), 7902-7906 (https://doi.org/10.1021/jm0600702).
[2]: Husen Jia, John R. Liggins, Wah Soon Chow, "Entropy and biological systems: Experimentally-investigated entropy-driven stacking of plant photosynthetic membranes," Scientific reports 2014, 4, Article number: 4142 (7 pages)(DOI: https://doi.org/10.1038/srep04142)(PDF).
[3]: H. F. Hoefnagels, D. Wu, G. de With, W. Ming, "Biomimetic superhydrophobic and highly oleophobic cotton textiles," Langmuir 2007, 23(26), 13158-13163 (https://doi.org/10.1021/la702174x).


Depending on the material that the casing is made out of, you could have a mercury or gallium payload.  They both are capable of significantly weakening the structures of some metals (such as aluminum).  See the following video of gallium weakening a coke can.  Mercury can also weaken the structure, especially in conjunction with gallium as seen here.  Mercury will form an amalgam with aluminum and significantly weaken it, as will gallium.  You might conceivably have nanobots bring a payload of the two metals to weaken the structure, especially since they are not nearly as reactive with other things.

Unfortunately, this would only work if the structure you are trying to damage is primarily aluminum, but might give a starting place for non-acidic solutions.


Stern-Volmer plot and calculation of rate constant for quenching

Benzophenone is a notorious case where prompt fluorescence, phosphorescence and delayed fluorescence can mix. Have a look at Modern Molecular Photochemistry by N.J. Turro for some general information and Prompt and delayed fluorescence from benzophenone by
R.E. Brown and L.A. Singer for the original reference.

\[
\begin{align*}
\ce{BP &->[h\nu]\ ^{1}BP^{\ast}}\\
\ce{^{1}BP^{\ast} &-> BP + h\nu'\quad &(prompt\ fluorescence)}\\
\ce{^{1}BP^{\ast} &->[ISC]\ ^{3}BP^{\ast} &(intersystem\ crossing)}\\
\ce{^{3}BP^{\ast} &-> BP + h\nu''\quad &(phosphorscence)}\\
\ce{^{3}BP^{\ast} &->\ ^{1}BP^{\ast} &(repopulation\ of\ S_1)}\\
\ce{^{1}BP^{\ast} &-> BP + h\nu'\quad &(delayed\ fluorescence)}\\
\end{align*}
\]

Let's ignore all this (for a moment).

Fluorescence quenching, i.e the observed decrease of fluorescence intensity in the presence of a quencher, may occcur through two different pathways:

Static fluorescence quenching

In this case, formation of a ground state complex between the fluorophore $\ce{F}$ a quencher $\ce{Q}$ decreases the concentration of $\ce{F}$, hence the concentration of the excited state $\ce{^1F^{\ast}}$ and the fluorescence intensity. This has no impact on the fluorescence lifetimes! These are the same in the absence ($\tau_0$) and presence ($\tau$) of a quencher.

\[
\frac{\tau_0}{\tau} = 1
\]

Dynamic fluorescence quenching

Here, quenching occurs from the excited state $\ce{^1F^{\ast}}$. With other words, the presence of a quencher introduces an additional, non-radiative pathway for the deactivation of the excited singlet state. Consequently, the fluorescence lifetime $\tau$ is shortened!

Let's try to derive the Stern-Volmer equation and see if that leads to some insights:

(You probably know all of this already, but I'm just in the mood...)

In an experiment on dynamic fluorescence quenching, the following processes and their rate constants have to be considered:

\[
\begin{align*}
\ce{F + h\nu &->[k_{abs}]\ ^1F^{\ast}\quad &\mathrm{(excitation)} } \\
\ce{^1F^{\ast} &->[k_{ic}]\ F + heat\quad &(internal\ conversion)}\\
\ce{^1F^{\ast} &->[k_{f}]\ F + h\nu'\quad &(fluorescence)}\\
\ce{^1F^{\ast} + Q &->[k_{q}]\ F + Q'\quad &(quenching)}\\
\end{align*}
\]

The fluorescence quantum yield $\Phi$ is given as the quotient of emitted ($I_{em}$) and absorbed ($I_{abs}$) photons:

\[
\Phi = \frac{I_{em}}{I_{abs}}
\]

The number of emitted photons depends on the concentration of the fluorophore in its excited singlet state $\ce{^1F^{\ast}}$ and the rate constant $k_{f}$:

\[
I_{em} = k_{f}\cdot[^1F^{\ast}]
\]

The number of absorbed photons depends on the concentration of the fluorophore $\ce{F}$ and the rate constant $k_{abs}$:

\[
I_{abs} = k_{abs}\cdot[F]
\]

The rate of absorption processes is difficult to measure, they're bloody fast. Fortunately, we don't have to! Under steady-state conditions, population of the excited singlet state and its deactivation balances out.

Consequently, we can express $I_{abs}$ as:

\[
I_{abs} = k_{f}\cdot[^1F^{\ast}] + k_{ic}\cdot[^1F^{\ast}] +k_{q}\cdot[^1F^{\ast}]\cdot[Q]
\]

The expression for the fluorescence quantum yield thus is:

\[
\begin{align*}
\Phi &= \frac{k_{f}\cdot[^1F^{\ast}]}{k_{f}\cdot[^1F^{\ast}] + k_{ic}\cdot[^1F^{\ast}] +k_{q}\cdot[^1F^{\ast}]\cdot[Q]}\\
&= \color{\red}{\frac{k_{f}}{k_{f} + k_{ic} +k_{q}\cdot[Q]}}
\end{align*}
\]

The latter is pretty convenient since we don't have to think about the pesky $[^1F^{\ast}]$ anymore! YAY!

If we do perform a fluorescence measurement in the absence of the quencher,  $[\ce{Q}] = 0$, $\Phi_0$, the fluorescence quantum yield under this condition is:

\[
\Phi_0= \frac{k_{f}}{k_{f} + k_{ic}}
\]

The effect of a quencher $\ce{Q}$ on the fluorescence is expressed as:

\[
\begin{align*}
\color{red}{\frac{\Phi_0}{\Phi}} &= \frac{k_{f}}{k_{f} + k_{ic}} \cdot \frac{k_{f} + k_{ic} +k_{q}\cdot[Q]}{k_{f}}\\
&= \color{\red}{1+ \frac{k_{q}\cdot[Q]}{k_{f} + k_{ic}}}
\end{align*}
\]

Too bad that we don't know all these rate constants and using just a steady-state spectrometer, we won't be able to measure them! And for a lab chemist, there's too much stuff in the equation anyway! Bummer!

But we can argue about what influences the fluorescence lifetimes, which, btw, can be measured!

Lifetimes are pretty much the inverse of the (sum of the) rate constants.

Without the quencher, the fluorescence lifetime $\tau_0$ is:

\[
\tau_0 = \frac{1}{k_{f} + k_{ic}}
\]

In the presence of a quencher, the fluorescence lifetime $\tau$ is:

\[
\tau = \frac{1}{k_{q}\cdot[Q] + k_{f} + k_{ic}}
\]

If we plug that in again and rearrange a bit, we get:

\[
\begin{align*}
\frac{\Phi_0}{\Phi} &= 1 + \tau_0\cdot k_q\cdot[\ce{Q}]\\
&= 1+ {K_{SV}}\cdot[\ce{Q}]
\end{align*}
\]

We name $K_{SV}$ the Stern-Volmer constant!

Summary

Plotting $\frac{\Phi_0}{\Phi}$ or the quotient of the areas of the fluorescence spectra vs the concentration of the quencher should give a straight line, that cuts the $y$ axis at 1.
The slope of the curve is the Stern-Volmer constant $K_{SV}$.
The real rate constant $k_q$ for the reaction of the excited state $\ce{^1F^{\ast}}$ with the quencher can only be obtained from $K_{SV}$ if the fluorescence lifetime $\tau_0$ has been obtained from a time-resolved fluorescence measurement!


What are the differences between the various household surfactants?

Can I wash my hair with dishwashing agent [...]!

If it's an agent for dishwashing machines: Please DON'T!

Cleaning in dishwashing machines is performed in alkaline medium.

The tabs usually contain (among others):

alkaline compounds, such as sodium carbonate and sodium hydroxide
bleaching agents, such as sodium perborate and sodium percarbonate
pentasodium triphosphate (to keep calcium and magnesium in solution)

The solution of these tabs is not meant to get in contact with skin and eyes.


Can pressure remain constant in isothermal expansion?

$$\Delta H = \Delta U + \Delta PV$$
Here you were wrong you wrote that
$$\Delta H = \Delta U + P\Delta V,$$ thus the product of
 $PV$  remains constant and the $\Delta H$ is zero, so that the enthalpy is constant. And you are wrote that entropy will increase, hence the answer is B.


What is the difference between cyanato, cyanido and cyano? When should i use which?

Cyano- is related to the neutral CN group. Cyanido- is related to the CN- anion. Cyanato- is related to the CNO- anion.


Zwitterions/IEP of glycine at pH 6? (Paradox?)

You need to separate the two concepts.

When dissolving a neutral amino acid in water, a buffered acid-base reaction will take place as you highlighted, giving a certain pH value. In this case, the following equilibrium:

$$\ce{HOOC-CH2-NH3+ <=> ^{-}OOC-CH2-NH3+ <=> ^{-}OOC-CH2-NH2}$$

will be somewhere between the pure second and the pure third species. (The neutral, non-zwitterionic species is neglectable in aquaeous solution.) Superfluous protons liberated into the solution will end up as $\ce{H3O+}$ explaining the overall reduction in pH value when compared to a neutral solution. You won’t notice it unless you perform really accurate experiments, but the solution’s pH value will be a tad higher than the isoelectric point.

If you go on and add external protons, i.e. by carefully adding an acid, you can adjust the overall solution’s pH. Since the glycinide anions are the strongest base present, they will take up the protons first, followed by the zwitterions. At a certain pH value (corresponding to a certain external proton concentration added), the concentrations of glycinide anions and glycinium cations will be identical and we have reached the isoelectric point of glycine. It is, and has to be, slightly different from the pH value obtained by dissolving glycine.


Difference between nickel catalyst and Raney nickel catalyst

Yes, the Raney nickel catalyst is different from simple, fine nickel powder.  Raney nickel is prepared in two steps.  First, the nickel is alloyed with aluminum and then in second step much of the aluminum is removed through a leaching process with sodium hydroxide. As the aluminum is removed two things happen.  The resulting nickel-aluminum alloy becomes very textured and porous with extremely high surface area (sponge-like).  Also hydrogen gas is liberated during the leaching process and some of it remains adsorbed on the alloy surface.  This is the active catalyst that is ready for use.

$$\ce{2 Al + 2 NaOH + 6 H2O -> 2 Na[Al(OH)4] + 3 H2}$$

If you'd like to read more about the initial alloying process or the activation (leaching) step, see this Wikipedia article.

As an aside, Murray Raney discovered this catalyst (US1628190) in the late 1920's and he was the sole assignee.  I'm guessing that W. R. Grace bought the patent rights from him since they hold the trademark on use of the term "Raney nickel"


Validity of the leveling effect of water in nitration reactions

What you have gone wrong in this question is you completely disregard the fact that mentioned stronger acids than water are all in aqueous medium, in other words, dissolved in water. Thus, fast acid-base reaction ($\ce{H2O}$ acts as a base here) happens to give $\ce{H3O+}$ as the only acid in these mixtures (leveling effect of water).

On the other hand, concentration of sulfuric acid is $\approx 98\% (w/w)$. That means, approximately only $\pu{2 g}$ of $\ce{H2O}$ in $\pu{100 g}$ of concentrated solution. Therefore, mole ratio of $\ce{H2O : H2SO4}$ would be about $0.1 : 1$. Thus, ($\ce{H2SO4}$ is not leveling with water. That extra remaining $\ce{H2SO4}$ can be acted on $\ce{HNO3}$ in nitrating mixture as shown in your scheme (vide supra).

Note: As a rule of thumb, you may simply consider $1:1$ mole ratio of water and any stronger acid than water would be the limiting level for the leveling effect.


According to the IUPAC organic chemistry nomenclature 2013, section P-83 ISOTOPICALLY LABELED COMPOUNDS, it has to do with uniformly labeled compounds:

  P-83.5 GENERAL AND UNIFORM LABELING

  P-83.5.1 In the name of a selectively labeled compound in which all positions of the designated element are labeled, but not necessarily in the same isotopic ratio, the symbol ‘G’ is used in place of locants to indicate a ‘general’ labeling.

  Examples:

  Mixture of isotopically substituted compounds (selective labeling) $\ce{CH3-CH2-CH2-^{14}COOH}$, $\ce{CH3-CH2-^{14}CH2-COOH}$, $\ce{CH3-^{14}CH2-CH2-COOH}$, $\ce{^{14}CH3-CH2-CH2-^{14}COOH}$, etc., when added to $\ce{CH3-CH2-CH2-COOH}$, is designated as [G-14C]butanoic acid.
  D-Glucose in which all six positions are labeled with 14C, but not necessarily uniformly, is designated as
  D-[G-14C]glucose.

  P-83.5.2 In the name of a selectively labeled compound in which all positions of the
  designated element are labeled in the same isotopic ratio, the symbol ‘U’ is used in place of locants to denote ‘uniform’ labeling.

  Examples:

  Mixture of isotopically substituted compounds (uniform labeling) $\ce{CH3-CH2-CH2-^{14}COOH}$, $\ce{CH3-CH2-^{14}CH2-COOH}$, $\ce{CH3-^{14}CH2-CH2-COOH}$ and $\ce{^{14}CH3-CH2-CH2-COOH}$, in equal amounts, when added to $\ce{CH3-CH2-CH2-COOH}$, is designated as [U-14C]butanoic acid.
  D-Glucose in which 14C is equally distributed among the six positions is designated as
  D-[U-14C]glucose.

  Note: In the case of radioactive nuclides, ‘same isotopic ratio’ means ‘same specific radioactivity’.

(Please note that the D- stereodescriptor should be in smallcaps)


Should the volume of a solute be factored into the volume of a solvent when preparing a substance of a certain molarity?

The only volume that is important in the denominator of your equation is the total volume of solution; neither the volume of solute or solvent are explicitly relevant.

Pragmatically, there is not likely to be much of a volume change when taking a concentrated solution and diluting it.

So with reference to your question, you should add enough solvent to get the proper final volume.  Explicitly, add the 25 mL of stock solution and 'dilute to mark'-- that is add enough solvent so the total solution is 50 mL.  The added solvent might not be exactly 25 mL to make this final volume.