$\ce{WF6}$ is not an uncommon gas used in industrial manufacturing of semi-conductors. Commercial gas detectors are readily available, and will detect $\ce{WF6}$ down to low ppm range, which is about an order of magnitude less than the $\mathrm{LC}_{50}$ for rats, apparently.

You could google tungsten hexafluoride detector, but some examples include (no affiliation or endorsement):

http://www.newcosmos-europe.com/gas-by-name/144-tungsten-hexafluoride-detection.html
http://www.toxicgasdetection.co.uk/gas/?gas=Tungsten%20hexafluoride&search=Tungsten%20hexafluoride&form=gas&id=416
https://www.gas-find.com/hexafluorure-de-tungstene.html

The most common type of detector is a diffusion monitor with electrochemical sensor.
Some sensors actually use conversion of $\ce{WF6}$ to $\ce{HF}$, and then detect the $\ce{HF}$. Commercial $\ce{HF}$ detectors at low ppm range are also common.


According to the Pottery Histories web site:

  Petuntse
  Petuntse is a historical Chinese term for rocks with a high content of
  mica and feldspathic minerals that have been used as one of the
  ingredients in the manufacture of Chinese porcelain. It is generally
  similar in composition and use to China stone. Petuntse, with china
  clay, forms the base ingredients for hard-paste porcelains similar to
  those produced in China in historic times. See China stone.

and

  China stone (Cornish stone, Cornwall stone)
  China stone is a
  feldspar-rich mineral derived from partially decomposed granite. China
  stone includes feldspar, mica, fluorospar, quartz and other
  granite-derived minerals such as kaolinite, but lacks iron-bearing
  minerals.  China Stone is used worldwide as a flux in the manufacture
  of ceramic bodies and glazes. In the United Kingdom, the material is
  mined in one small area of Cornwall.

I don't know if this meets the degree of specificity that you were looking for.


How many Ga atoms are connected to one As atom in solid gallium arsenide?

It is quite well established that stoichiometric gallium arsenide (CAS 1303-00-0) forms a zincblende analogous structure. It consists of two stacked face-centered cubic lattice systems (Ioffe Physical Technical Institute database). See also Uncle Al's Answer. In other words, the arsenic forms the lattice and the gallium fills half of the tetrahedral positions (and vice versa). Therefore every gallium has four arsenic nearest neighbours.
This does not necessarily mean that these are bonds. In this case however, the bonds are best described as covalent (having a somewhat ionic character).
The Britannica Online Encyclopedia (Academic Edition) (You might need access provided through an educational institution) states:

Besides the elemental semiconductors, such as silicon and germanium, some binary crystals are covalently bonded. Gallium has three electrons in the outer shell, while arsenic lacks three. Gallium arsenide (GaAs) could be formed as an insulator by transferring three electrons from gallium to arsenic; however, this does not occur. Instead, the bonding is more covalent, and gallium arsenide is a covalent semiconductor. The outer shells of the gallium atoms contribute three electrons, and those of the arsenic atoms contribute five, providing the eight electrons needed for four covalent bonds. The centres of the bonds are not at the midpoint between the ions but are shifted slightly toward the arsenic.
Such bonding is typical of the III–V semiconductors — i.e., those consisting of one element from the third column of the periodic table and one from the fifth column. Elements from the third column (boron, aluminum, gallium, and indium) contribute three electrons, while the fifth-column elements (nitrogen, phosphorus, arsenic, and antimony) contribute five electrons. All III–V semiconductors are covalently bonded and typically have the zincblende structure with four neighbours per atom. Most common semiconductors favour this arrangement.

This is still a fluorishing field for researchers in the field of materials design semiconductors, semiconductor lasers, computational physics/forcefield development, and more.
The book "Electronic Devices and Circuit Theory" clearly gets it very wrong. I had a look into it and noticed they also argue with the Bohr model, which might not be a very good starting point at all. However, the focus of the book is clearly not solid state chemistry, and therefore it might fit the purpose.


If Gibbs energy of activation is positive, how is the change to the transition state possible?

...according to the Second Law of Thermodynamics, entropy of the universe can only increase.

Well, there was a nice similar question in bio.SE about this. A system is not universe. The overall entropy of the universe increases, but that isn't true for only a particular thermodynamic system.

Also, no one said the transition state is stable!

  The transition state of a chemical reaction is a particular configuration along the reaction coordinate. It is defined as the state corresponding to the highest potential energy along this reaction coordinate. Wikipedia

(High potential energy = lesser stability)

This chemwiki.ucdavis page is also good to help reshape your understanding of second law of thermodynamics.


Is it possible to have negative corrorion rate and weight loss, meaning specimen [steel] gained weight after immersion in electroyte solution?

In the data cited in your reference, corrosion is measured by weight loss or gain. In the case of the steel coupon immersed in sodium carbonate, the weight gain is not due to "negative corrosion", but likely due to a tightly-adhering film of oxides that was not removed in cleaning. The oxides are made from the iron of the coupon plus oxygen from water, which is greater than the mass of the iron alone. (Normally, cleaning the coupon removes loosely adhering oxides, so there is a weight loss.)

According to Toyin, "the corrosion inhibition efficiency of $\ce{Na2CO3}$... indicates that a protective film is formed on the metal surface. The passivation of the mild steel may be caused by the formation of $\ce{Fe2O3}$  and $\ce{Fe3O4}$."

That said, it does appear that high carbon concentrations may be effective in greatly reducing the corrosion of mild steel.


Do SN2 reactions always invert absolute configuration?

For the ease of thinking about this, let's imagine $\mathrm{S_N2}$ attack of $\ce{CN-}$ on bromochlorofluoroiodomethane, where iodide is being displaced.

Absolute stereochemistry is retained in this case because the highest priority substituent is becoming the lowest priority substituent. If you were to find the absolute stereochemistry of $\ce{2 ->3->4}$ in (R)-bromochlorofluoroiodomethane, which becomes $\ce{1->2->3}$ after attack, you would see that it is S and after attack becomes R. You can see that here:


Why is LDL more easily oxidized/damaged than HDL?

Just a brief background about lipoproteins:

Lipids and proteins associate noncovalently to form lipoproteins, which function in the blood plasma as transport vehicles for triacylglycerols and cholesterol. Simple, unesterified fatty acids are merely bound to serum albumin and other proteins in blood plasma, but phospholipids, triacylglycerols, cholesterol, and cholesterol esters are all transported in the form of lipoproteins.

In general, lipoproteins are classified according to their densities. The densities are related to the relative amounts of lipid and protein in the complexes.  Thus, there are:

high-density lipoproteins (HDLs),
low-density lipoproteins (LDLs),
Intermediate density lipoproteins (IDLs),
very-low-density lipoproteins (VLDLs), and also
chylomicrons.

Chylomicrons have the lowest protein-to-lipid ratio and thus are the lowest-density lipoproteins. They are also the largest.

There are a couple of things to take into consideration:

How they function

LDL appears to be the major circulatory complex for cholesterol and cholesterol esters. At various target sites, particularly in the capillaries of muscle and adipose cells, these particles are degraded by lipoprotein lipase, which hydrolyses triacylglycerols. Lipase action causes progressive loss of triacylglycerol (and apoprotein) and makes the lipoproteins smaller. This process gradually converts VLDL particles to IDL and then LDL particles, which are either returned to the liver for reprocessing or redirected to adipose tissues.  This results in even more increase in LDL than other lipoproteins.

Environment

Another explanation is based on environment conditions -the LDL binds to specific LDL receptors, which cluster in domains of the plasma membrane which subsequently form coated vesicles which pinch off from the plasma membrane and form endosomes. In the low pH environment of the endosome, the LDL particles dissociate from their receptors. The endosomes then fuse with lysosomes, and the LDLs are degraded by lysosomal acid lipases into amino acids, fatty acids, and cholesterol.

Newly formed HDL contains virtually no cholesterol ester (however, over time, cholesterol esters are accumulated through the action of lecithin:cholesterol acyltransferase).

Another associated protein, cholesterol ester transfer protein, transfers some of these esters to VLDL and LDL. Alternatively, HDLs function to return cholesterol and cholesterol esters to the liver. This latter process apparently explains the correlation between high HDL levels and reduced risk of cardiovascular disease. High LDL levels, on the other hand, are correlated with an increased risk of coronary artery and cardiovascular disease.

Reference

Biochemistry (Grisham)

Biochemistry (Campbell and Farrel)


The electric charge on an electron is fixed, regardless of its speed. The apparent mass of an electron INCREASES with its speed (relative to the observer). So charge/mass will decrease as mass → ∞. The major problem with these statements is the use of "mass" to mean "apparent mass" or "observed mass". More and more (and probably the majority now days) of the Physics community reserve the term "mass" to mean "rest mass" (which is the mass observed at zero velocity relative to the observer). This is a bit more complicated than it seems: consider a gas. The center-of-mass velocity may be zero (relative) but obviously each molecule has its own velocity not dependent on center-of-mass. Fortunately, in the range of temperatures chemists are concerned with, relativistic velocities aren't seen, so no correction is necessary between a molecule's rest mass and its apparent mass. Apparent mass, observed mass, relativistic mass, dynamic mass, are (usually) all different ways of expressing the fact that the energy of an object affects its mass; E= mc² and that energy is the sum of potential and kinetic terms, where the kinetic term is ½mv². Hence the mass you will observe depends on how fast the object is traveling relative to you (or your observational equipment). The Special Relativity equation for mass is m = m0 ÷ √( 1 - (v/c)²) where v is the particle's speed (magnitude of its velocity), c is the speed of light, and m0 is its mass when v=0, i.e. its rest mass. As you can see, as v approaches c, the denominator approaches 0 and so the observed mass gets larger without limit.

I should also note that the generalization I made, that chemists need not "worry" about relativistic velocities, isn't entirely correct. For instance, the color of gold is due to relativistic effects (otherwise it would be silver just like the other more common metals). In general, atoms do not have relativistic velocities, but larger atoms have electrons with relativistic momentum, which can be thought of as relativistic speed. There's no reason why atoms can't have relativistic speeds, other than the obvious problem of how we'd contain and observe them. If there wasn't any "chemistry" at relativistic speeds, we'd have fusion power already - confinement of plasma is "hard" because it interacts with the container.


Why is acetic acid stronger than p-nitrophenol?

The difference in acidity between acetic acid and phenol can be attributed to the electron-withdrawing properties of a carbonyl function versus a phenyl ring. In both cases, the deprotonated oxygen can be depicted in a resonance structure as forming a double bond to the carbon in question and the charge being delocalised within the ring.

It should immediately strike anybody that a phenyl ring primarily provides a resonance body (i.e. a greater volume into which the charge can be delocalised) while a carbonyl group is also actively withdrawing (both inductively and mesomericly). This is reflected when comparing the $\mathrm{p}K_\mathrm{a}$ values of $\ce{CH3CO-CH3}$ (acetone) and $\ce{Ph-CH3}$ (toluene) — $16.5$ logarithmic units in $\mathrm{DMSO}$ according to the Evans table[1] or approximately $24$ logarithmic units in water ($\mathrm{p}K_\mathrm{a}$ of acetone in water taken from the Brodwell data).[2]

Adding a nitro group to the phenyl ring reduces this difference drastically to $-6.1$ logarithmic units (i.e. para-nitrotoluene is more acidic than acetic acid) — in $\ce{DMSO}$. One should always remember that acidity differences in one solvent can be drastically different in another solvent. For example, the difference between $\ce{MsOH}$ and $\ce{TfOH}$ is $11.4$ logarithmic units in water but only $1.3$ logarthmic units in $\ce{DMSO}$ (that is not a typo). One should also note the difference between acetic acid in water ($\mathrm{p}K_\mathrm{a}\approx4.76$) and $\ce{DMSO}$ ($\mathrm{p}K_\mathrm{a}\approx12.3$).

$$\textbf{Table 1: }\text{Acidity constants of various compounds in various media}\\\text{All data taken from the Evans data unless noted otherwise in the text.}$$
\begin{array}{lrc}\hline
\text{compound} & \mathrm{p}K_\mathrm{a}(\ce{H2O}) & \mathrm{p}K_\mathrm{a}(\ce{DMSO})\\ \hline
\ce{CH3CO-CH3} & 19.3\phantom{000} & 26.5\\
\ce{Ph-CH3} & 31.5\phantom{000} & 30.6\\
\ce{$p${-}NO2-C6H4-CH3} & & 20.4\\
\ce{MsOH} & -2.6\phantom{000} & \phantom{0}1.6 \\
\ce{TsOH} & -14\phantom{.0000} & \phantom{0}0.3 \\
\ce{AcOH} & 4.76\phantom{00} & 12.3\\ \hline \end{array}

Given that a $\mathrm{p}K_\mathrm{a}$ value of para-nitrotoluene has not been reported in water (although it’s $\mathrm{p}K_\mathrm{b}$ value has been),[3] we are not in any position to draw definite conclusions. However, water being the solvent it is, I would still assume the electron-withdrawing ability of a nitro group which relies solely on mesomeric effects to be weaker than the effect of a directly neighbouring carbonyl group. Remember that water is excellent at stabilising charges; $\ce{DMSO}$ not so much.

References:

[1]: The Evans $\mathrm{p}K_\mathrm{a}$ table

[2]: The Brodwell $\mathrm{p}K_\mathrm{a}$ data

[3]: J. C. D. Brand, J. Chem. Soc. 1950, 997. DOI: 10.1039/JR9500000997.


Why does pouring carbonated beverages onto fruit cause them to fizz?

All carbonated drinks are oversaturated solutions of $\ce{CO2(g)}$. Pouring over fruits lead to contact with many residual air bubbles, which serve as  fizzing places for $\ce{CO2(aq) -> CO2(g)}$.

It is more or less the same as the ( previously already opened ) drink shaking, where violent fizzing occurs as well.

It is a similar process activation, like if you bend the activation chaff in a pocket heater with molten $\ce{CH3COONa \cdot 3 H2O}$, or if you pour overcooled water from a freezer ( see YouTube ) and watch fast ice growing.


Does a double bond or a triple bond have a higher priority in nomenclature?

The relevant section in the 2013 IUPAC Recommendations (Blue Book) reads:

  P-31.1.1 [...] Locants as low as possible are given to multiple bonds as a set, even though this may at times give 'yne' endings lower locants than 'ene' endings. If a choice remains, preference for low locants is given to the double bonds. [...]

In this case, the two possible numbering schemes lead to the same locant set of (1,4).

However, according to the italicised portion of the quoted text, the double bond is considered to have the higher priority, and the name would correctly be pent-1-en-4-yne.

Note that the final "-e" from "-ene" is dropped if the following suffix begins with a vowel or "y," and is retained if the suffix begins with a
consonant. Additionally, spaces are only used in very specific cases, like in the carboxylic acid ending "-oic acid." Otherwise, dashes are used to separate different functional groups and numbers in the name.


Is there any special rules for checking the aromaticity of polycyclic compounds?

First, Hückel's rule is only valid for monocyclic systems:

Hückel's rule is not valid for many compounds containing more than three fused aromatic nuclei in a cyclic fashion. For example, pyrene contains 16 conjugated electrons (8 bonds), and coronene contains 24 conjugated electrons (12 bonds). Both of these polycyclic molecules are aromatic even though they fail the 4n + 2 rule. Indeed, Hückel's rule can only be theoretically justified for monocyclic systems.

Pyrene is a commonly used counter-example of Hückel's rule. A refinement specific to Polycyclic Aromatic Hydrocarbons (PAH) is Clar's rule.
The case of pyrene is explained in detail in the literature by a 2007 paper in Petroleum Science and Technology:

This representation agrees with Clar’s sextet rule in which the molecule is represented as having two aromatic sextets, one for each of the rings a and d, and two fixed double bonds on rings b and c (Hernández-Trujillo et al., 2005).


Why are carbonyl compunds more reactive towards catalytic hydrogenation than alkynes and alkenes?

That is not true in general: the stronger nature of the C=O bonds (larger bond energy) means that they are not as easily hydrogenated as C=C bonds. However, your compound (c) is not a regular carbonyl group, but an acyl chloride, which is a quite reactive species and thus very easily hydrogenated.

For a bit of reading, see here the section on “Catalytic Hydrogenation”:

  As a rule, the carbonyl group does not add hydrogen as readily as do the carbon-carbon double and triple bonds. Thus, it is fairly easy to reduce an alkene or alkyne function without affecting any carbonyl functions in the same molecule. By using a platinum catalyst and increased temperature and pressure, it is possible to reduce aldehydes and ketones to alcohols, but carboxylic acids, esters and amides are comparatively unreactive. The exceptional reactivity of acyl halides, on the other hand, facilitates their reduction under mild conditions, by using a poisoned palladium catalyst similar to that used for the partial reduction of alkynes to alkenes. This reduction stops at the aldehyde stage, providing us with a useful two-step procedure for converting carboxylic acids to aldehydes, as reaction #1 below demonstrates. Equivalent reductions of anhydrides have not been reported, but we might speculate that they would be reduced more easily than esters. The only other reduction of a carboxylic acid derivative that is widely used is that of nitriles to 1º-amines.


What is the Difference between Menthol Crystal and Menthol Powder?

Unless there is something bizarre going on (which should be mentioned more or less clear on the website/in the catalogue because no chemist would expect it), menthol will be menthol as there is only one compound defined as menthol. This compound is chiral, so I would expect a stereodescriptor. In its absence, I would not be sure whether I am dealing with a racemic sample (more likely in the case of a chemical supplier aimed at practicing chemists) or a natural, chiral sample (more likely in the case of a supplier selling to the general public).

Chiral menthol crystallises easily and nicely, and because crystals are easily separated from contaminants, they are often the preferred form of isolation of compounds for sale; especially since the crystals themselves very rarely contain impurities. While a powder can be the better choice for several purposes (e.g., weighing small sample amounts into a flask), it may not be as simple to produce. In the case of menthol, I expect the powder to derive from powdering the crystals. This is an extra step and thus costs more money.

Depending on how important the quality of the powder is to you, it may make more sense to buy the powder directly or to buy crystals and grind them to powder yourself. However, as I said above, I am very sure that they contain the same chemical in a comparable purity.


Your formula, $\ce{Rb[Pt(NH3)Br2Cl]}$ is correct for the named complex, rubidium amminedibromochloroplatinate(II) (For clear understanding of naming, you may read Naming Coordination Compounds-1 and Naming Coordination Compounds-2).

Many coordination compounds have distinct geometric structures. Two common forms are the square planar, in which four ligands are arranged at the corners of a hypothetical square around the central metal atom, and the octahedral, in which six ligands are arranged, four in a plane and one each above and below the plane. Altering the position of the ligands relative to one another can produce different compounds with the same chemical formula.
Coordination compounds often exist as isomers—i.e., as compounds with the same chemical composition but different structural formulas. Many different kinds of isomerism occur among coordination compounds (Source).

Cis-trans-isomerism  is one of the more common types. Cis-trans (geometric) isomers of coordination compounds differ from one another only in the manner, in which the ligands are distributed spatially. Most of platinum(II) complexes are square planner. Most common example is Cisplatin ($\ce{Pt(NH3)2Cl2]}$) where the each of two ammonia molecules and the two chlorine atoms are situated next to one another (see the diagram in the blue box below), and hence isomer is called the cis (Latin for “on this side”). Thus, coordination name for Cisplatin is  cis-diamminedichloroplatinum(II) (Wikipedia).

When the identical groups are in opposite sides, it is called the trans-isomer (Latin for “on the other side”). To emphasize this nomenclature, I sketched cis- and trans-isomers of diamminedibromoroplatinum(II) (above) and your compound (below) in the maroon box below for better understanding (cis-isomers are in LHS and trans-isomers in RHS).


Acid-base reaction is screwing up my synthesis

Just do a decantation. Simply discard the acid layer and then do the wash.


Can thiols be created using a Grignard reagent?

I've done this with aromatic Grignards, but I have no reference and no database access any longer.

I think when I did it, I was relying on an older colleague's instructions. It was pretty much generate your Grignard from ArBr + Mg in THF, cool in ice bath, stir vigorously and add elemental sulfur. You may need to introduce a reductive step as you can generate polysulfides.

Grignards will also add to disulfides so you could react with dibenzyl disulfide then reduce off the benzyl group.

This reference is the nearest that Google has managed to find Can. J. Chem. 70 (1992) 30


Placed in equivalent freezers, would a liter of water or a liter of lava turn from liquid to solid first? (question from a 6 year old)

You've got the right idea — you want to simplify the problem — but I don't think you're using quite the right simplification.  Take a look at the book's set-up of the situation, and ask yourself how the water is stopping the lava. You'll see the idea is that we're using liquid water, not ice, to solidify the lava.
So your question should really be:  How much heat do we need to absorb from a liter of lava to turn it into a solid, and how much heat can a liter of water at room temperature absorb before it turns to steam?  If the latter is larger than the former, then a liter of water can cool a liter of lava to the point where it solidifies before the water all changes into steam.
[I'm using "heat" when I should really be using "thermal energy", but this is for a 6 yo, so I'm keeping it simple.]
First, let's do the calculation for water.  Here (since it's for a 6 yo), I'm not going to show all the steps in the calculations:
Energy to heat 1 L liquid water at room temp (25C ) to 100C = 75 kcal
Energy to turn 1 L liquid water to steam at 100C = 533 kcal
Total = 608 kcal
According to https://en.wikipedia.org/wiki/Lava, lava is typically 700 to 1200 C, so let's call it 1000 C.
And, using this source (https://ocw.mit.edu/courses/earth-atmospheric-and-planetary-sciences/12-109-petrology-fall-2005/lecture-notes/Nov3notes.pdf), let's assume it melts at 900 C (there's actually a wide range of lava types, and thus a wide range of lava temps and melting points).
Density of lava = 3.1 g/cm^3, so 1 L lava weighs 3100 g
Energy released when 1 L lava cools from 1000 C to 900 C = 93 kcal
Energy released when 1 L lava solidifies at 900 C: 310 kcal
Total = 403 kcal
So, based on the above, 1 L of water is enough to solidify 1.5 L of lava.
Finally, why does the water win?  The main reason is that the water is going between a liquid and a gas, while the lava is going between a solid and a liquid.  And, in general, it takes much more energy to change liquids into gases than solids into liquids, because in the former case you are pulling the molecules completely apart from each other.
But I said it takes "much more energy" to change a liquid to a gas than a solid to a liquid, yet the difference here is only a factor of 1.5. The discrepancy is because  in this case you're not comparing masses of water and lava, you're comparing volumes (and for the same volume, the mass of lava is 3.1 x greater). Thus, if you do it on a mass basis, 1 kg of water is enough to solidify 6 kg of lava (because, while 1 L of water weighs 1 kg, 1.5 L of lava weighs 6 kg).
[For these quantities, mass is typically a more fundamental basis of comparison than volume, which is why heat capacities, heats of fusion, and heats of vaporization are usually quoted on a mass basis. When you see the term "specific", as in "specific heat capacity, that means "per unit mass".  Of course, one can also use a number basis (moles).]


"Reversible" is not binary.  Both the forward and backward reactions always occur and the equilibrium system never has zero reactants or zero products.  Thus, irreversible reactions are called this not because they cannot be reversed - they absolutely can - but because reversal is impractical.  The equilibrium constant may be so skewed toward product that the reactant quantities at equilibrium would be undetectable.  The products may be gases that escape solid reactants, which ensures that the system never reaches equilibrium in the open anyway.  There may be a lot of side reactions on the products.

A "reversible" reaction just lacks factors like this so that we can have measurable quantities of reactant beside measurable quantities of product at equilibrium.  We add some reactant, watch the reaction go forward.  We add some product, it goes in reverse.  See?  It's reversible.  There is no exact cutoff in any measure.


$\ce{H2S}$ has a mostly covalent character. You can decide for it by the crude way textbooks give: Calculate the difference in electronegativity to decide whether it's a non-polar covalent compound, a polar one or has ionic characters.
For the dissociation part (which is usually endothermic overall) it'll need more "ionic character". And it does dissociate in water, resulting in a weak acid named  sulfhydric acid. (The spelling varies with citable sources) Your textbook likes to think about it as not dissociable, only because it's a weak acid.
Addendum:
The "ionic character" plays its role of a rule a good one as explaining acid 'strength'. However, it acts almost flawlessly in the same period, but has exceptions when it gets to groups. As Dave mentions, $\ce{HF}$ is weaker than for instance, $\ce{HCl}$ and that's because the atomic radius changes more dramatically than electronegativity difference. (See here ) But, in the $\ce{H2S}$ molecule, this exception hasn't occurred. Dave pointed out a good example of when the rule isn't working that well, but as I said, we can ignore the other factors in this very case.


How does standardising a secondary standard solution make it suitable as a standard solution?

When talking about standards, there's always the dilemma: How is a standard standardized? Well, you compare it to another standard you already have, and after you consider all the sources of errors, you have a new standard.

In case of the primary standards, you can always promptly trust the label's purity information, as well as your balance's and other measuring equipment (like glassware) accuracy, and at a much lower degree the molar masses, and so on. Having that concentration "certified", anything you calculate using that as a reference will be "certified" as well. However...

If you're going to follow strict analytical rules, than you won't blindly trust those, which is why you cshould calculate the error (or many of the related statistical measures) by considering the accuracy of, like I said, the balance, the glassware, temperature effect on volume (density), etc.

One common way to do this, called the standardized uncertainty, is take the square root of the sum of the squares of each instrument/measure uncertainty divided by it's quantity. You should define a minimal formula which describes the property you want to account the error for, and go for there. Let's take, for example, the concentration of 100 mL of a 0.01 M solution of potassium hydrogen phthalate (KHP), used as a primary standard to standardize sodium hydroxide.

The concentration of the solution is $[KHP]=\frac{n}{vol}$, but we obviously didn't measure molês, so we write

$$[KHP]=\frac{\frac{m}{MM}}{vol}$$

where $m$ is the mass weighed on the balance, and $MM$ the molar mass. Now that we know what influence in your concentration, we have to use some numbers. Let's consider that:

Weighed mass: 0.20422 g, on a balance with uncertainty of ± 0.001 g (1 mg)
Volume measured: 100 mL, in a volumetric flask with uncertainty of ± 0.1 mL

This means that the uncertainty of the prepared solution is

$$\frac{u_{[KHP]}}{0.01} = \sqrt{\left(\frac{0.001}{0.20422}\right)^2 + \left(\frac{0.1}{100}\right)^2} $$

$$u_{[KHP]} \approx 5\times10^{-5}$$

Now note that I didn't take into account the purity of the salt (to make things simpler), and the error on the molar mass (we could assume it's much lower than the others). And what does all this have to do with your question?

My point is, even standards have a variations in them, and when you standardize something, like a HCl or NaOH solution, you could take that into account. The concentration of NaOH, in this case, will have a relative uncertainty higher than the concentration of KHP, because all of the uncertainty of the KHP preparation must be taken into account, as well as the titration variation. The outcome is that a standard based on  a standard of another standard (...) will eventually have a very high uncertainty, which is why we try to sitck to primary and secondary standards.

PS.: The uncertainties of the equipments/glassware are usually available in the manuals or even engraved on them. This values are usually the calibration precision, and should be transformed into uncertainty using some statistical inference, which could be that the variation follows a retangular distribution, resulting in

$$u = \frac{precision}{\sqrt3}$$

Disclaimer: There's a lot more to this than what I summarized above, starting with being able to indentify what should and shouldn't be taken into account, not to mention replicates and a lot of statistics. I don't recommend blindly following this formulae to calculate uncertainties.


Why does carbocation stability increase in the order 1°, 2°, 3°?

One word: hyperconjugation! The more carbons bound to the carbocation, the more bonds can take part in hyperconjugation. This effect happens when a bond between two atoms donates electron density to the electron-deficient carbocation, making it more stable.


Why does ring contraction take place in Wagner–Meerwein rearrangement?

Kornblum1 demonstrated that nitrocyclohexane 3 was the major isolable product from the reaction of iodocyclohexane 1 and silver nitrite. Earlier work by Rosanow2 (1915) claimed the additional formation of tertiary nitro compound 7, a result that Kornblum did not confirm. If indeed nitrocyclopentane 7 did form, a ring contraction is required. Ring contraction of cyclohexyl carbocation 2 sacrifices the greater stability of a secondary carbocation for a primary carbocation (4) and presumably some loss in stability through ring contraction in forming cyclopentyl carbocation 6. There is no evidence that primary nitrocyclopentane 5 is formed. However, a hydride shift in 4 would afford the more stable tertiary carbocation 6 that would lead to 7.

1) N. Kornblum and C. Teitelbaum, J. Am. Chem. Soc., 1952, 71, 3076.
2) See reference in Kornblum paper.


What happens when HCl is added to acetylene in presence of mercuric ions?

I'm not entirely sure but I'll have a go.

As a strong acid, HCl will be dissociated into Chloride and protons in water. I'd suggest that the mercury will first associate with the acetylene forming an Hg- Pi complex which will weaken the triple bond. Then the chloride ion will attack one side of the acetylene and the electron density one of the triple bonds will form a new single bond, first to the mercury ion and then this bond will hydrolyse and pick up a proton. This is similar to an oxymercuration reaction but with an attack from chloride rather than water.

The resulting vinyl chloride is more electron poor than the starting species so should should undergo further 'chloromercuration?' slower than the starting species. I would suggest that the selectivity is then derived from the milder conditions this reaction must be run at. Probably you have to force acetylene to react with HCl without a catalyst by heating the hell out of it (if it happens at all) therefore the addition happens twice.


What criteria have to be met to term a carboxylic acid a 'fatty acid'?

I suggest to consult the IUPAC Compendium of Chemical Terminology, aka the Gold Book for a definition of fatty acids!

In a narrow sense, fatty acids are defined as

  aliphatic monocarboxylic acids derived from or contained in esterified form in an animal or vegetable fat, oil or wax.

The definition continues to mention that

  natural fatty acids commonly have a chain of 4 to 28 carbons […]

The paragraph closes with a wider definition, namely:

  By extension, the term is sometimes used to embrace all acyclic aliphatic carboxylic acids.

Personally, I have used the term fatty acids only for natural fatty acids.


Possible structure if a compound like AB4 is polar?

Given all $\ce{B}$ are equivalent in terms of AXE notation, for $\ce{AB4}$ the following coordination environments are possible [1, p. 177]:

square-planar $(D_\mathrm{4h},$ SP-4);
square-pyramidal $(C_\mathrm{4v},$ SPY-4);
tetrahedral $(T_\mathrm d,$ T-4);
see-saw $(C_\mathrm{2v},$ SS-4).

Table IR-9.3 Polyhedral symbols, geometrical structures and/or polyhedra

Option (A) square planar implies a centrosymmetrical environment and such molecule cannot be polar.
The derivative of (A), option (B) rectangular planar, as well as option (D) tetrahedral also refer to the environments that posses zero dipole moment.
Speaking of (C), I think you are misinterpreting the geometry of square pyramidal coordination.
It's a square non-coplanar arrangement with a non-zero dipole moment commonly seen in period 4 transition metal porphyrins $(\ce{MN4}$ fragment) and also lead(II) oxide and tin(II) oxide $(\ce{MO4}$ fragment).
According to Brecher [2, p. 1944], the structure of bis(N,N-dimethyldithiocarbamato)lead(II) would be yet another example:

Sample depiction of a compound known to exist in square pyramidal configuration:

Since a see-saw option is not present, this leaves only one answer, option (C).
Reference

IUPAC. Nomenclature of Inorganic Chemistry, IUPAC Recommendations 2005 (the “Red Book”), 1st ed.; Connelly, N. G., Damhus, T., Hartshorn, R. M., Hutton, A. T., Eds.; RSC Publishing: Cambridge, UK, 2005.
Brecher, J. Graphical Representation of Stereochemical Configuration (IUPAC Recommendations 2006). Pure and Applied Chemistry 2009, 78 (10), 1897–1970. DOI: 10.1351/pac200678101897.


What is the configuration of dimethyl[tris(trifluoromethyl)]-λ⁵-phosphane?

You can note that $\ce{-CH3}$ shows electron releasing (+I) effect while $\ce{-CF3}$ shows strong electron withdrawing effect. Now, you can treat inductive effect as the organic analogue of electronegativity, giving axial positions to $\ce{-CF3}$  and equatorial ones to $-\ce{CH3}$ in accordance with Bent's rule.


They would be highly unstable due to the inverted tetrahedral geometry of the carbons in the top face. Generally, a polyhedron can form the basis of a stable hydrocarbon if it has at most 3 edges for each corner. A big caveat is that the hydrocarbon will distort the shape if the angle strain is too great. I found that out in my computational research on prismanes.


How can we calculate enthalpy change in a closed tank containing methanol adsorbed by activated carbon?

Enthalpy ($H$) is defined as, $H=U+PV$ where $U$ is the internal energy and $P$ is pressure and $V$ is volume. The change in enthalpy is given by

$$\Delta H = \Delta U + \Delta(PV)$$

And $\Delta U$ is equal to the sum of the heat transferred $q$ and the work done $w=-P\Delta V$.

At constant $V$ (which is what you have in a closed system), $\Delta V =0$ (and thus $w=0$) and $\Delta(PV)=V\Delta P$.

Thus...

$$\Delta H = q+V\Delta P$$

You can determine $q$ based on the amount of methanol that vaporized using the heat of vaporization of methanol. We can assume that the majority of heat absorbed was used to distil methanol.

$$q_{vap}=m\cdot c_{vap}$$

How can we determine the volume of the system? Or... if the change in pressure is very small, perhaps that term can be ignored.


How do pharmaceutical manufacturers ensure 100% purity in their medications?

Most registered specifications will contain the main active pharmaceutical ingredient (API) at an assay of 98-102% to allow for hplc variation(the most common analytical technique for pharmaceuticals). The list of known impurities usually follows with their maximum limit ; this is usually derived from toxicology studies. But not always.

An exception is something like Sativex which has an assay of <90% for the main API, but the other ingredients are not detrimental but generally regarded as safe (GRAS); sterols, triglycerides.

The drug product then has the API in  specified range and can comprise most of the medication e.g. paracetamol tablets, or not very much e.g. acyclovir in ointment form or cisatracurium besylate in water for injection. Excipients are added which could be seen to reduce the "purity" of the medication, but they are present to aid bioavailability, taste etc.

Any technique you can think of can be used industrially... at a cost. So while liquid -liquid extraction is cheapest and easiest,  it rarely gives the desired purity. Recrystallisation is a cost effective technique and is frequently used industrially. Put solid in a reactor, pump in solvent, heat, cool. Extract, distil solvent,  solvent switch, heat, cool is also valid.

Yield loss from industrial scale recrystallisation is usually 5% or greater.  I have seen 30% for achiral material. I've seen 55% for a diastereomeric salt crystallisation (so at least 50% is to be expected).

Sometimes one crystallisation is enough; sometimes three are required. It depends on the chemical transformation efficiency and co-crystallising impurities.

As for recovery ; economic viability depends on the cost of the material. So at 20 dollars/kg, probably not. 15000 dollars/kg, probably yes.

Not all impurities are bad as they can be very similar to the API and simply have a different effect e.g. anti-histamine.  Some could genotoxic and are controlled at the ppm level.

Unknown impurities are tightly controlled to, typically, the total being no more than (NMT) 0.5%. Again, there are exceptions.


Why are the given compounds in a homologous series?

Because they can all be described by the general formula $\ce{CH3-(CH2)_n-OH}$. The only varying parameter in this case is the number of middle $-\ce{CH2}- $ groups.

So while the the number of middle $-\ce{CH2}- $ groups may vary $\left(n = 0,1,2,\ldots\right)$ they all fit in this general formula.

The formula $\ce{CH3-(CH2)_n-OH}$ tells you that:

The chain starts with a $\ce{CH3}-$ group
The chain ends with an $-\ce{OH}$ group
Contains a variable number, $n$, of $-\ce{CH2}-$ groups in between.


Finding out pKa of acid from molar conductivity

You need to add the limiting ionic conductivities for $\ce{H+}$ and $\ce{OH-}$ together to get the limiting ionic conductivity for all the ions in solution ($\Lambda_{0}$, which will replace $\Lambda^{0}_{\mathrm m}$ in your equation).

This arises from a simplification for calculating $\Lambda_{0}$ in weak electrolyte solutions (such as yours) according to Kohlrausch's Law in which it is stated:

  Each ionic species makes a contribution to the conductivity of the solution that depends only on the nature of that particular ion, and is independent of the other ions present.

from which we can then estimate $\Lambda_{0}$ as:

$$\Lambda_{0} = \sum_{i}\lambda_{i,+}^{0} + \sum_{i}\lambda_{i,-}^{0}$$

For your problem:

$$\Lambda_{0} = \underbrace{(34.96 + 19.91)}_{54.87}\ \mathrm{mS\cdot m^{2}\cdot mol^{-1}}$$

we solve (I am omitting units here for clarity, and have confirmed via dimensional analysis that the answer is correct):

$${1\over 4.61} = {1\over 54.87} + {4.61\times 0.025\over K_{\mathrm a}\times (54.87)^{2}}$$

which simplifies to yield the answer (spoiler alert):

  $$K_{\mathrm a} = 1.926\times 10^{-4} \implies \mathrm{p}K_{\mathrm a} = 3.72$$

and can be favorably compared to the value of 3.77 given here.


Molar absorption coefficient for potassium hydrogen phthalate

I would imagine that you could just use one of the spectrophotometers you have available to determine the molar absorptivity constant experimentally. After all, the molar absortivity constant is simply the constant which equates the terms (those being Absorbance and Concentration) which are already proportional. So, using something like the beer-lambert law...

$$A = \epsilon b C$$

...make a solution of a known concentration, measure its absorbance at your wavelength of interest, and solve for $\epsilon$.

$$\epsilon = \frac{A}{b C}$$

You might consider doing this a number of times to get an average value and thus minimize the random error associated with your constant.


Suggest the product from two-substituted benzene ring

On substituting an already substituted benzene ring, you need to know which kind of substitution takes place; It is electrophillic substitution that is taking place.

Then can you now think about the correct choice?
$\ce{-NH2}$ has a greater positive mesomeric effect (+M effect)
in comparison to the Halogen groups.

Since the para position with respect to $\ce{-NH2}$ is already occupied by chlorine the ortho position would be attacked.

What we come to know is that since $\ce{-NH2}$ creates more electron availability in comparison to halogen it leads to "electrophilic substitution" on ortho position with respect to $\ce{-NH2}$.

And moreover $\ce{-NH2}$ group since a very very strong electron donor so once again another substitution of next bromine takes place at the other ortho position with respect to $\ce{-NH2}$.


Cool question! Always stay curious!

A while ago, I had a similar question that started it's journey with silicon, being of similar bonding capability to carbon. However, over the course of my research, I realized that silicon was just too low on the periodic table (and therefore much heavier) to be so extremely stable in such a wide variety of forms such as fullerenes and long, complex hydrocarbons.

However, my search was not in vain. I stumbled upon Boron Nitride, an isoelectronic analog to carbon graphite. Many different compounds have been synthesized from boron nitride, consisting of a wide array of fullerene analogs, such as Boron Nitride Nanotubes, or BNNTs, which by themselves have plenty of applications. For example, paper products made from these nanotubes are highly flame retardant, as is shown in the image below:

The trick is that $\ce{B-N}$ has very similar properties to $\ce{C=C}$, with very little difference in molecular weight, and so many analogous compounds can be built using either monomer. However, boron nitride is different enough to have unique applications in certain areas. Cool, right?


The full name of the compound is 3,4-methylenedioxymethamphetamine, which is typically abbreviated as "MDMA". It is commonly used recreationally, and is casually referred to as "ecstasy", "molly", "E" and many other names.


Copper oxide is not green, but green verdigris, basic copper carbonate (or acetate) forms on copper exposed to air and moisture.

Heating copper in air can form a near-transparent film of copper oxide, which can exhibit all the colors of the spectrum due to interference of light. The same effect cause a "rainbow" in oil slicks on water.


The most fascinating thing about aldol reaction is that it is reversible. So, when the aldol product is kept in a basic medium, it can actually undergo reverse mechanistic pathway to regenerate the two carbonyl compounds between which condensation occured to form the product . Then those two carbonyl compounds can literally undergo condensation in a different way to produce some different aldol product. This phenomenon is called as Retro-aldol reaction.

In this case, also if you observe carefully, the compound (A) is an $\alpha-\beta $ unsaturated ketone (basically a dehydrated aldol product), which by hydrolysis can convert itself into an aldol and thus undergo retro-aldol reaction. Here the aldol condensation was done intramolecularly, so the two ketones produced undergoes intramolecular aldol condensation in a different way to produce another isomer (B).

The mechanism of formation of (B) is as follows,

Thus, option (a) should be the correct answer.


pH Difference Our Taste Buds Can Recognise

First, the pH of water changes with rising temperature because the total number of $\ce{H+}$ ions (well, solvated hydronium ions, really) increases, not because it has become more alkaline. The water is still neutral, having an equal number of hydroxyl ions as hydronium ions. Heat simply increases the ionization.

Second, it depends on the individual taster, but a sample of water at $\mathbf{20~^\circ C}$ with a $\mathrm{pH}$ of $6.63$ (perhaps acidified with citric acid) would probably taste sour in comparison with a sample of water at $\mathbf{20~^\circ C}$ with a $\mathrm{pH}$ of $7.47$ (perhaps with a bit of baking soda, $\ce{NaHCO3}$).

See Temperature Dependent[sic] of the pH of pure Water for a more detailed explanation.


Black Sludge Appearing When Making Iron Oxide Through Electrolysis

The black stuff is your iron oxides. FeO and Fe3O4 are both black, Fe2O3 is red. I would guess you are making some complex iron oxide/ hydroxide/ hydrate. I have seen XRD analyses of many aqueous iron corrosion products , they are typically a complex mix.


There is probably going to be a separate reaction rate law for each phase, assuming that the reaction even occurs in other phases.  Of course, in liquid and solid phases, the pressure is going to have much less of an effect than in a gas phase.  Another consideration is that there may be interphase mass transport between the various phases (at their interfaces), and this may have to be taken into account in the chemical species mass balances for the individual phases.


Would Z and E isomers show different optical activity?

Not necessarily. Although the change you suggest would not alter the absolute stereochemistry (S in both cases) of the asymmetric center starred, the two molecules are diastereomers and therefore will have different physical and optical properties. Even if they did both rotate light the same way, they probably won’t have the same degree of rotation for the same concentration (specific rotation).

The problem boils down to whether we can predict the direction (or degree) of rotation from a drawn absolute stereochemical configuration. The answer to this is still “no”. We don’t currently have a theory that will allow us to say that, for a certain molecule, R gives us d or l, etc.

Some studies have been able to come up with some rules of thumb for small molecules, but there is no comprehensive theory to make this prediction in general. https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2704476/


1.With "salt", does Vogel mean just $\ce{NaCl}$?

Yep, it's cheap and available.

  How do I go about making these ice shards that are used in the lab?

Wrap the ice cubes in a towel and hit the bundle with a hammer until you get whatever size you want.  That's how we did it in the lab.


Is there any electronic component to water conductivity?

I agree with the commenters that electrical conduction is very unlikely, but it's worth going through some possible mechanisms:

actual solvated electrons: As others have noted, free electrons would be expected to react rapidly with protons, even in a basic solution, so this changes quickly to a scenario of sequential electron transfer between protons, so let's do that next.
Sequential electron transfer between $\ce{H.}$ and $\ce{H+}$: Let's assume the solution is strongly acidic, so protons are abundant, and a proton gets reduced at the cathode to a hydrogen atom radical. Based on the bond dissociation energies, abstraction of $\ce{H.}$ from water to form $\ce{H2}$ and $\ce{HO.}$ is slightly unfavorable, the hydrogen radical would be preferred over the hydroxyl radical. (The H-H bond formed has a BDE of ~ $\pu{105 kcal/mol}$, while the $\ce{H-OH}$ bond broken has a BDE of ~$\pu{120 kcal/mol}$.) The problem is the rate of quenching by reaction of two hydrogen atom radicals to form $\ce{H2}$ (which is water electrolysis produces hydrogen gas). I couldn't find a rate constant for that, but there is a published rate constant for recombination of hydroxyl radicals in water that is around $\pu{10^10 M-1 s-1}$. As you might expect, that's essentially diffusion limited, so the rate constant of hydrogen atom recombination is going to be at least as high. If we optimistically assume that transfer of the electron from $\ce{H.}$ to $\ce{H+}$ has a comparable rate constant, you would still have to have a very low concentration of radical and very short path to travel in order for an electron to make it from a cathode to an anode, but it doesn't seem theoretically impossible.
The third possibility would be sequential transfer of electrons from $\ce{HO-}$ to $\ce{HO.}$. In a strongly basic solution, this also seems like a theoretical possibility given a very short path and a very low concentration of radical, assuming there are no other molecules in solution that can quench the radical.

I'm not suggesting that either of these theoretical possibilities actually ever occurs, just that these are the mechanisms that seem most likely to me.


Why do ionic solids change to a structure with a higher coordination number at high pressures?

Lets put some numbers on the phase transition of NaCl from the normal NaCl structure (based on fcc, 6-fold coordinated) at atmospheric pressure, to the CsCl (based on bcc, 8-fold coordinated) structure at high pressure.
From Compounds and Alloys Under High Pressure: A Handbook, by E. Yu Tonkov (google found it) one gets that the transition occurs at 27-29GPa, with a volume change of 5%.
The density of NaCl, taken from Wikipedia, is 2.165 gm/cc, or about 27cm$^3$/mol, or 27E-6 m$^3$/mol.
Multiply on through and you get, very roughly, the PV energy gained by converting from the less dense NaCl structure to the denser CsCl structure is 40kJ/mol - this is not an inconsequential energy gain for the Gibbs free energy.  Thus, the PV energy gained by a denser solid with the higher coordination plays off (and pays off) against the ionic unhappiness.

In the literature there are also indications that the CsCl, and/or higher pressure phases, may be metallic in nature, so the assumption of purely ionic bonding may be in jeopardy.


Reactivity of chlorobenzene and benzene in electrophilic substitutions

considering electrophilic aromatic substitution is chlorobenzene or benzene more reactive?

Let's expand the question to include all of the halobenzenes.

All of the halogens ($\ce{F, Cl, Br, I}$) are more electronegative than hydrogen.  Therefore they will inductively remove electron density from a benzene ring. However, you can also draw resonance structures for the various halobenzenes where the halogen substituent is donating electron density to the ortho and para positions of the benzene ring through resonance.

Therefore, in the case of the halobenzenes these two effects are working in opposite directions.  In order to determine which effect controls the reaction, we need some data.

The following table compares the relative rates of electrophilic aromatic nitration of the halobenzenes to the rate for benzene itself. If we assign benzene a relative rate of 1, we see that all of the halobenzenes react slower than benzene itself. This tells us that the inductive effect is stronger than the resonance effect in the halobenzene series.
\begin{array} \hline
 \text{Relative rates of aromatic electrophilic nitration} \\ \hline
\end{array}
\begin{array}{|c|c|c|c|} \hline
\ce{Ar-X} & \text{Relative rate}  \\
\hline
\ce{Ar-H} & 1.0 \\ \hline
\ce{Ar-F} & 0.11 \\ \hline
\ce{Ar-Cl} & 0.02 \\ \hline
\ce{Ar-Br} & 0.06  \\ \hline
\ce{Ar-I} & 0.13 \\ \hline
\end{array}

(data source: see p. 26 here)

However, when the halobenzenes do undergo electrophilic aromatic substitution, they react preferentially at the ortho and para positions, because, as explained above, the resonance effect donates electron density to the ortho and para positions.  Therefore, these two positions are the least deactivated and react preferentially.


Why does my professor use both the upper, and the lower, case literal of a letter to symbolize the same concept?

Of course, there is a convention. In physics $E$ usually stands for energy, $e$ for elementary charge. You professor is inconsistent, to say the least. In my opinion using lower case $e$ for anything rather than for elementary charge is wrong by itself, and using it for energy is simply ridiculous. Besides, I don't see any reason for doing so.

P.S. During my university years I had a professor who randomly picked letters for physical quantities every lecture. :D Today I think of it with a smile, but at that time I hated both his lectures and him personally.


Oxidation to allylic alcohols without stoichiometric metal reagents

Almost all allylic oxidation reactions have been using metal catalysts such as chromium, selenium, copper, etc. because it is relatively easy to abstract hydrogen radical from allylic position and metals are well known for doing so with or without single electron transfer (SET). This fact is well documented in the review published in 2013 (Ref.1), the abstract of which simply states that:

  Although $\ce{C–H}$ oxidation of hydrocarbons is generally difficult, allylic $\ce{C–H}$ oxidation is relatively simple and predictable, even on a preparative scale, because active species generated at the allylic position are stabilized by the double bond. Therefore, allylic oxidation has been employed in natural product synthesis, and a variety of reagents and conditions for allylic oxidation have been reported...

Some of the specific catalysts the review referred are  selenium Dioxide, diphenyldiselenide–Iodoxybenzene, chromium Trioxide–3,5-Dimethylpyrazole, pyridinium chlorochromate (PCC), chromic acid, etc., all of which are toxic to humans (e.g., selenium and chromium compounds) and fairly expensive (e.g., palladium and rhodium compounds). Recently, a paper published in Nature journal describes an electrochemical allylic $\ce{C–H}$ oxidation strategy that does not need toxic and expensive metal catalysts, but uses inexpensive and readily available materials. The method exhibits broad substrate scope, operational simplicity, and high chemoselectivity. The notable advantage of this electrochemical allylic $\ce{C–H}$ oxidation strategy is It represents a scalable allylic $\ce{C–H}$ oxidation (results have demonstrated on $\pu{100 g}$ scale reactions).

The General Procedure for electrochemical allylic oxidation is described as follows in the paper:

  With no precautions to exclude air or moisture, a test tube is charged with a stir bar, $\ce{Cl4NHPI}$ $(\pu{0.2 equiv})$ and $\ce{LiClO4}$ ($\pu{0.6 equiv}$, $\pu{0.1 M}$). Acetone ($\pu{6 mL/mmol}$ substrate) is added, and the resulting suspension is stirred until most of the solids dissolve. Substrate ($\pu{1 equiv}$) is added, followed by pyridine ($\pu{2 equiv}$) and $70\%$ aqueous tert-butyl hydroperoxide ($\pu{1.5 equiv}$). The anode and cathode (RVC, separated by a glass slide) are placed in the solution, and the reaction mixture is electrolyzed at a current of $\pu{10 mA/mmol}$ substrate until the reaction is complete as judged by TLC or GCMS analysis. The reaction mixture is transferred to a separatory funnel, and the electrodes are washed thoroughly with hexanes, diethyl ether, or ethyl acetate. The organic layer is washed twice with $\ce{H2O}$, once with brine, dried, and concentrated. The resulting crude product was then purified on $\ce{SiO2}$ to furnish the desired enone product. Note: Pyridinium perchlorate can be added to the reaction ($\pu{0.1–0.2 M}$) to improve solubility of the reaction components. This co-electrolyte has negligible effect on product yields and may improve the reaction if the substrate contains any easily reduced functional groups since this increases the concentration of protons available to be reduced at the cathode.

Note that the mediator, $\ce{Cl4NHPI}$ is N-Hydroxytetrachlorophthalimide, which was synthesized from tetrachlorophthalic anhydride and $\ce{H4N-OH/HCl}$ in pyridine using thermal or microwave technique for this research. However, it was available in Sigma-Aldrich for $\pu{62 USD}/\pu{10 g}$.

The schematic diagram of the general procedure is depicted in above diagram (Figure A).  The variety of mediators used during the progress of the process are listed in Figure B. Among them, N-Hydroxyphthalimide ($\ce{NHPI}$) and $\ce{Cl4NHPI}$ have given the best results (have given total yield of $56\%$ and $77\%$, respectively in otherwise optimized conditions). They are drastically efficient than others, e.g., next best has been TEMPO, which has given only $26\%$ total yield under identical conditions. Among the co-oxidants tried, $\ce{t-Bu-OOH}$ has given the optimal results ($51\%$). The next best is $\ce{PhC(CH3)2-OOH}$ with $43\%$ yield. Acetone ($56\%$) and $\ce{CH3CN}$ ($51\%$) have given equally impressive results as choice of solvent, and acetone was chosen for optimal conditions as the solvent for this research. Hence optimized electrochemical parameters were: $(\pu{0.16 M})$ of substrate with $\ce{Cl4NHPI}$ $(\pu{0.2 equiv})$, pyridine $(\pu{2.2 equiv})$, $\ce{t-Bu-OOH}$ $(\pu{1.5 equiv})$, $\ce{LiClO4}$ $(\pu{0.6 equiv})$ in acetone.

The tentative mechanism is illustrated in Figure C. The formation of tert-butyl-peroxy-intermediate (III) was confirmed. In a separated experiment, one of corresponding tert-butyl-peroxy-intermediate was synthesized in a different route and reacted in identical condition used in electrolytic procedure ($\ce{Cl4NHPI/t-BuOOH/LiClO4/pyridine/acetone}/RT$). The reaction has completed both with and without electrolysis within $\pu{6 h}$, confirming requirement of electrolysis during initial tert-butyl-peroxy-intermediate formation steps (Ref.2).

References:

Akihiko Nakamura, Masahisa Nakada, “Allylic Oxidations in Natural Product Synthesis,” Synthesis 2013, 45(11), 1421-1451 (DOI: 10.1055/s-0033-1338426).
Evan J. Horn, Brandon R. Rosen, Yong Chen, Jiaze Tang, Ke Chen, Martin D. Eastgate, Phil S. Baran, “Scalable and sustainable electrochemical allylic C–H oxidation,” Nature 2016, 533, 77–81 (https://doi.org/10.1038/nature17431).


There are actually two ways to answer this question: by lawyering or by trying to capture what a polymer is. I'll try both.
Answer by lawyer
The IUPAC defines the following terms as follows:

monomer: a molecule which can undergo polymerisation thereby contributing constitutional units to the essential structure of a macromolecule
macromolecule: A molecule of high relative molecular mass, the structure of which essentially comprises the multiple repetition of units derived, actually or conceptually, from molecules of low relative molecular mass.
molecule: An electrically neutral entity consisting of more than one atom ($n > 1$). Rigorously, a molecule, in which $n > 1$ must correspond to a depression on the potential energy surface that is deep enough to confine at least one vibrational state.

Using the data from this paper and a harmonic oscillator approximation, I estimate that the zero-point energy of a harmonic oscillator with two waters and a hydrogen bond is just over 45 kJ/mol.
Since the energy contributed by hydrogen bonding in liquid water is estimated to be between 2kJ/mol and 6kJ/mol, I conclude that the hydrogen bond of water cannot confine a vibrational state.  This means liquid water does not qualify as a single molecule.
Since a long chain of waters connected by hydrogen bonds does not qualify as a molecule, it cannot be a polymer, which means that water cannot be a monomer (at least, not in this sense).

A better answer
The real problem here isn't the technical stuff I wrote above (which, by the way, probably has a few holes large enough to sneak trucks through). It's that words have certain meanings, and we can't capture everything in a simple definition. For the same reason that you'd probably feel a bit weird calling "yellow" a "reddish-green", even though it is literally the result of mixing red and green light, it feels strange to call water a polymer.
Here are some common ideas that spring to mind when I see the word "polymer":

Made of organic monomers bonded together covalently
Structure of covalent bonds (e.g. branching) affects physical properties
Intermolecular forces affects physical properties

These are not strict, but they explain why quartz is not referred to as a polymer, even though it's made of covalently-bonded $\ce{SiO2}$ units: everything is covalently bonded and there aren't any intermolecular interactions to speak of! For another example, depending on the exact context, proteins are sometimes rarely thought of as polymers, even though they're made of repeating amino acids, because their structure is fixed (must be linear chain) and their intermolecular forces don't contribute much.
So the more sensible answer to this question that doesn't rely on lawyering is that liqid water just doesn't have the properties that most people think of when they hear the word "polymer" and so it isn't usually referred to as a polymer.
Supplemental Materials
The code I used to compute the harmonic oscillator energies
from math import pi,sqrt
# Compute the harmonic oscillator energy of two water molecules
# using a two-body QHO approximation
kg_per_amu = 1.66054e-27
watermass = 18.0 * kg_per_amu # Mass of water: close enough for now
reducedmass = watermass**2 / (2 * watermass)

hbar = 1.054571800e-34   # J-s
k = 845.4 # From https://www.princeton.edu/~fhs/paper79/paper79.pdf, N-m

E_zeropt = hbar * pi * sqrt(k / reducedmass) # for n = 0
E_zeropt_kjm = E_zeropt * 6.022e23 / 1000  # Scale from J/molecule to kJ/mol
print(E_zeropt_kjm)


When identical groups are present such that neither can be preferred over the other then, such nomenclature is preferred to emphasise on the equivalence of the groups such as in the one we have here. We cannot name in such a way that includes the three groups together. So, we name it that way instead.

As such, the rules of IUPAC Nomenclature are themselves arbitrary therefore, it is just something that is the way it is. For more information you might like to see https://old.iupac.org/publications/books/principles/principles_of_nomenclature.pdf, Page 34 onwards.

Edit: I almost forgot that a compound can have more than one name, even by the IUPAC nomenclature rules. As such, the names used commonly as IUPAC names are the preferred ones.


50-90'C color changing chemical to use as warning marker for hot electronics?

There are commercial  sources for thermochromic permanent-change ink, but you might also try thermal printer paper. You might need to check if there is a time-dependence on the color change, as well as temperature dependence. The printer paper is available at Amazon, Staples and elsewhere for less than US$20 / package, but you might just test a cash-register receipt, many of which are on thermal paper. According to this PDF, most start to darken at 79 C.


First of all, there is no such thing as "alumina oxide". Alumina is aluminium oxide. Melting alumina is pretty hard, its melting point is above 2000 degrees Celsius. It's so high that people use it as a container (crucible) for melting other stuff. Such high temperatures are not easily achieved without melting a bunch of stuff around it.

Let's assume that you did manage to melt it, maybe using one of those fancy levitation laser techniques. Your  mold has to be heat resistant and inert so it will not melt or react with the molten alumina. Graphite is usually a good one.

Now, you were asking about the structure. Assuming that you started with regular crystalline alumina (aka corundum), then yes: the resulting material after cooling in air will be the same corundum. Amorphous alumina (glass) will only form by extremely rapid cooling, for example dropping a very small amount to ice cold water. In all normal circumstances it will be crystalline.


How to read TINKER's force field parameters

Okay I will take a stab at this.

Consider the following bond stretching parameters in the TINKER FF file:

      ##################################
      ##                              ##
      ##  Bond Stretching Parameters  ##
      ##                              ##
      ##################################

bond          1    1          310.00     1.5260

This tells you that this is a bond parameter between atom 1 and atom 1.  The index of atoms is given in the FF file as

   #####################################################
   ##                                                 ##
   ##  TINKER Atom Class Numbers to Amber Atom Types  ##
   ##                                                 ##
   ##    1  CT      11  CN      21  OW      31  HO    ##
   ##    2  C       12  CK      22  OH      32  HS    ##
   ##    3  CA      13  CQ      23  OS      33  HA    ##
   ##    4  CM      14  N       24  O       34  HC    ##
   ##    5  CC      15  NA      25  O2      35  H1    ##
   ##    6  CV      16  NB      26  S       36  H2    ##
   ##    7  CW      17  NC      27  SH      37  H3    ##
   ##    8  CR      18  N*      28  P       38  HP    ##
   ##    9  CB      19  N2      29  H       39  H4    ##
   ##   10  C*      20  N3      30  HW      40  H5    ##
   ##                                                 ##
   #####################################################

So clearly this is a bond parameter between CT and CT.  What do these atom notations stand for?  They are outlined in Table 1 of this paper.  The value 310.00 is $K_r$ (in kcal mol$^{-1}$)  and 1.5260 is $r_{\mathrm{eq}}$ (in Ang).

For angle bends:

      ################################
      ##                            ##
      ##  Angle Bending Parameters  ##
      ##                            ##
      ################################

angle         1    1    1      40.00     109.50

angle denotes what you'd expect.  The last two parameters, 40.00 ($K_{\theta}$ in kcal (mol radian$^2$)$^{-1}$) and 109.50 ($\theta_{\mathrm{eq}}$ in deg.) are for CT-CT-CT (atom1-atom1-atom1).

For torsional angles:

      ############################
      ##                        ##
      ##  Torsional Parameters  ##
      ##                        ##
      ############################

torsion       1    1    1    1            0.156    0.0  3

This is clearly a torsion between CT-CT-CT-CT.  Here is where it gets a little tricky.  If anyone finds this to be erroneous, please let me know and I will correct it.  The parameter, 0.156 is simply determined by the following formula:

$$\frac{\frac{V_n}{2}}{\mathrm{no.~of~paths}}$$

where $V_n/2$ is the magnitude of torsion in kcal mol$^{-1}$.  In Table 14 of the previously cited paper, for Torsional Parameters you find for X-CT-CT-X:

no. of paths = 9
$V_n/2$ = 1.40

Therefore, $1.40/9$ $\approx$ 0.156 which is exactly what is given in the TINKER FF file.  I've done this with a few other torsional parameters and it seems to add up. I believe footnote $f$ in Table 14 explains this.

The 0.0 is simply $\gamma$ (phase offset in deg.) and 3 is $n$ (the periodicity of the torsion).

Functional Form of Amber94 FF

It may be useful (and convenient) to spell out the functional form as presented in the cited paper here.

$$E_{\mathrm{tot}} = \sum_{\mathrm{bonds}} K_r(r-r_{\mathrm{eq}})^2 +$$
$$~~~~~~~~~~~~\sum_{\mathrm{angles}} K_{\theta}(\theta - \theta_{\mathrm{eq}} ) ^2 + $$
$$~~~~~~~~~~~~~~~~~~~~~~~~~~\sum_{\mathrm{dihedrals}} \frac{V_n}{2}[1+\cos(n\phi - \gamma)] + $$
$$~~~~~~~~~~~~~~~~~~~~~~~\sum_{i<j} \left[ \frac{A_{ij}}{R_{ij}^2} - \frac{B_{ij}}{R_{ij}^6} +
\frac{q_i q_j}{\epsilon R_{ij}} \right]$$

Reference: Cornell and co-workers. J. Am. Chem. Soc. 1995, 117, 5179.


How does the partial pressure of gas C change with the change in volume of the container?

The equilibrium constant $K_P$ is independent of pressure but does depend on temperature which we assume to be constant. The equilibrium constant $K_P$ is the ratio of partial pressures $p_i$ of the gaseous substances raised to the stoichiometric powers, $K_P= p_C^4/p_B^4$ and is defined at a standard state of $1$ atm. pressure.

Writing the equilibrium constant for the reaction $\ce{mB <=>lC}$ in terms of the mole fraction $x_i=n_i/n$ for n total moles gives $K_x= x_C^l/x_B^m$ where l and m are constants, both four in your question. The mole fraction i of a species behaving as a perfect gas can also be written  $x_i=p_i/P$ where $p_i$ is the partial pressure and P the total pressure.

Rewriting the equilibrium constant in terms of partial pressures of B and C gives
 $$K_x=\frac{p_C^l}{p_B^m}P^{m-l}= K_PP^{\Delta n}$$
where $\Delta n$ is the change in the number of moles of the gases. Thus unless the change in the number of moles is zero, which it is in your example, the mole fractions of the components  will change with change in total pressure even though $K_P$ is constant. If $\Delta n > 0$ the pressure will reduce the mole fraction of products.


I have quoted the following answers from the UC Davis ChemWiki page on "The Extraction of Iron."

  Why is coke added as well, when the purpose is to decrease the carbon content?

Coke is used primarily as a reducing agent (see below) and comes out as carbon dioxide (and thus does not increase the carbon content of the molten iron).

  What exactly occurs in this process of making cast iron - what reactions happen in the furnace?

The reactions are given below in the narrative of the process.

The Blast Furnace

The common ores of iron are both iron oxides, and these can be reduced to iron by heating them with carbon in the form of coke. Coke is produced by heating coal in the absence of air. Coke is cheap and provides both the reducing agent for the reaction and also the heat source. The most commonly used iron ores are hematite, ($\ce{Fe_2O_3}$), and magnetite, ($\ce{Fe_3O_4}$).

The air blown into the bottom of the furnace is heated using the hot waste gases from the top. Heat energy is valuable, and it is important not to waste any. The coke (essentially impure carbon) burns in the blast of hot air to form carbon dioxide - a strongly exothermic reaction. This reaction is the main source of heat in the furnace.

$\ce{C + O_2 \rightarrow CO_2}$

The Reduction of the Ore

At the high temperature at the bottom of the furnace, carbon dioxide reacts with carbon to produce carbon monoxide:

$\ce{C + CO_2 \rightarrow 2CO}$

It is the carbon monoxide which is the main reducing agent in the furnace:

$\ce{Fe_2O_3 +3CO \rightarrow 2Fe + 3CO_2}$

In the hotter parts of the furnace, the carbon itself also acts as a reducing agent. Notice that at these temperatures, the other product of the reaction is carbon monoxide, not carbon dioxide:

$\ce{Fe_2O_3 + 3C \rightarrow 2Fe + 3CO}$

The temperature of the furnace is hot enough to melt the iron which trickles down to the bottom where it can be tapped off.

The Function of the Limestone

Iron ore is not pure iron oxide - it also contains an assortment of rocky material that would not melt at the temperature of the furnace, and would eventually clog it up. The limestone is added to convert this into slag which melts and runs to the bottom. The heat of the furnace decomposes the limestone to give calcium oxide.

$\ce{CaCO_3 \rightarrow CaO + CO_2}$

This is an endothermic reaction, absorbing heat from the furnace. It is therefore important not to add too much limestone because it would otherwise cool the furnace. Calcium oxide is a basic oxide and reacts with acidic oxides such as silicon dioxide present in the rock. Calcium oxide reacts with silicon dioxide to give calcium silicate.

$\ce{CaO + SiO_2 \rightarrow CaSiO_3}$

The calcium silicate melts and runs down through the furnace to form a layer on top of the molten iron. It can be tapped off from time to time as slag. Slag is used in road making and as "slag cement" - a final ground slag which can be used in cement, often mixed with Portland cement.

Cast Iron

The molten iron from the bottom of the furnace can be used as cast iron. Cast iron is very runny when it is molten and doesn't shrink much when it solidifies. It is therefore ideal for making castings - hence its name. However, it is very impure, containing about 4% of carbon. This carbon makes it very hard, but also very brittle. If you hit it hard, it tends to shatter rather than bend or dent. Cast iron is used for things like manhole covers, guttering and drainpipes, cylinder blocks in car engines, Aga-type cookers, and very expensive and very heavy cookware.


Understanding colour changes of systems in dynamic equilbrium

In the general case of $\ce{A + B <=>C}$ if  there are initially $a_0, b_0,c_0$ amounts present , then if an amount x of A and B react to produce C then the equilibrium constant $K_e$ is given by
$$ K_e=\frac{k_1}{k_{2}}=\frac{c_0+x_e}{(a_0-x_e)(b_0-x_e)}$$
where $k_1, k_2$ are the rate constants for the forward and reverse reaction and $x_e$ is the amount reacted at equilibrium.
Solving the equation and  for (algebraic) simplicity let $c_0=0$, gives the amount at equilibrium
$$x_e=\frac{1}{2}\left(a_0+b_0+K_e^{-1}-\sqrt{(a_0-b_0)^2+2(a_0+b_0)/K_e +K_e^{-2}}\right)$$
Using this the amount of each species can be obtained as for example the equilibrium amount of A is $a_e=a_0-x_e$. However, in your question you ask about the colour not the concentrations and this alters things a little. The quantity that is measured optically as in a spectrophotometer is the optical density of for example species A is $D=\epsilon_\lambda[A]l$ where $\epsilon_\lambda$ is the extinction coefficient of the molecule at wavelength $\lambda $ and is effectively how strongly the molecule absorbs at that wavelength, $\ce{[A]}$ is the concentration and l the path-length of the sample through which the light passes.

The total absorption is the sum of that for all species so depending on the values of $\epsilon$ and how the concentrations change the absorption may go up, down or remain constant at any given wavelength.

Furthermore, if you do not use a spectrophotometer and measure optical density but you are looking through the sample you will observe the transmitted light which is by the Beer-Lambert law given by $I_{trans}=I_0e^{-D}$ for each species present at each wavelength, so what you may see becomes pretty complicated.


Superionic water is water under extreme heat and temperature with the properties of both solids and liquids.
It is where the water only exists as a mixture of $\ce{H+}$ and $\ce{OH-}$ ions. At high pressures, the oxygen crystallises, leaving the hydrogen ions free - thus making superionic water highly conductive.
It's crystal structure is a face-centred cubic lattice, and it is thought that Uranus and Neptune hold a layer of superionic water.


You are treating the rubber band as an isolated system when in fact you are acting upon it.  The entropy increase occurs in your body as glucose is oxidize into water and carbon dioxide to produce the energy needed for your muscles to contract to stretch the rubber band.


How to attach a liebig condenser to a flask for reflux heating?

Maybe I don't understand your question well, but often in the lab when we do reflux heating, we don't really attach the liebig condenser.

You fix the flask using a clamp. And gravity holds the liebig condenser into place. You can put a clamp around it to hold it straight but you'd better not tighten it too much because these things are fragile.

I hope it helps you :)

If it's not clear maybe it would be easier to answer if you posted pictures of your glassware!


The smoky black stuff is probably not tantalum oxide.  The only such oxide that's well characterized is $\ce{Ta2O5}$, which is white.  Tantalum itself is a rare element, so if the jewelry contains tantalum at all it's probably a thin coating or cladding on a base metal substrate.


This molecule is a ketene because it contains the ketene functional group:

Whilst the ketene appears to contain an alkene and a ketone functional group, which as you have suggested is where its name comes from, its chemistry is notably different from both of its 'constituent groups'.


Alkylation of alkenes by nucleophilic substitution

The pi electrons in an alkene or alkyne can act as a nucleophile.  This class of reactions is often taught a bit later than the standard $\ce{S_{N}1}$ and $\ce{S_{N}2}$ reactions and comes up when neighboring group participation or anchimeric assistance is covered.

The pi electrons in a double or triple bond are not strongly nucleophilic but when you arrange things in their favor, pi electrons can act as a nucleophile.  The reactions involving double bonds that I am familiar with involve an $\ce{S_{N}1}$ mechanism.  Here are a few examples.

(image source)

In example (1) notice how the compound with the double bond solvolyzes $10^{11}$ times faster than the saturated model compound!  Note too that only the product with acetate group anti to the double bond is formed.

Triple bonds and benzene rings can participate as well.  Often in cases involving double or triple bonds new carbon-carbon bonds are formed between the unsaturated linkage and the carbon with the leaving group [examples (2) and (3) above]. In cases involving aromatic rings aromaticity is restored so no new carbon-carbon bonds are formed between the aromatic ring and the carbon with the leaving group.


You used a salt/acid mixture and, upon bringing it to a boil to clean a stainless-steel pot, parts turned black.
All stainless steel is not alike, and all are not “stainless” or non-reactive towards oxidation – given the circumstances.
Getting to your specific question:
The “black” you are seeing upon boiling is a Fe(II)/Fe(III) oxide, the same or similar to the black oxidative protection [the seasoning of a skillet] so valued by the fastidious owners of iron skillets.
One of many procedures for “seasoning” an iron skillet is similar to what you did.  I have done the following A) cleaning the skillet down to shinny metal, particularly important if you do not know the history of the skillet. B) rub a mixture of table salt dampened with an acid [vinegar] over the pan until you get the classic yellow-rust of iron we are so familiar with C) now add water to the skillet and bring it to a boil.  This step is critical.  The yellow rust will turn black upon boiling.  This is a protective Fe(II)/Fe(III) oxide coating.
Now repeat with the salt/vinegar rub and boil until you get a uniform blackish coating on the iron skillet.    Finally, fry up some bacon and eggs to give a carbon-based coating to the skillet.  The latter is more of a “paint” but with the black metal oxide coating as a base, your skillet is protected against “rust” and mild oxidation.
The “original Teflon ® coating” if you will.


What is the balanced equation for H2O2 + KI + dish soap

Ah!! The elephant toothpaste experiment.
First thing first, this reaction is actually a decomposition experiment as the following equation.

$$\ce{2H2O2 -> 2H2O + O_2}$$

Second, $\ce{KI}$ acts as a catalyst in this experiment. You also can use $\ce{MnO4}$ too as the catalyst for this experiment.

Third, the dish soap makes the product of the experiment more soapy and bubbly. So, you can see it will create look like a toothpaste. If you didn't put the dish soap, it will only look like gas bubbling out if the solution.


Why does mercury in mercuric acetate have a lone pair?

TLDR:

Mercury in its (+2) salts does not have lone pairs readily accessible for covalent bonding. However, its 5d electrons do affect its properties,  resulting in a relatively high polarisability.

The long story:

Mercury is the last 5d element and has electron configuration $[\ce{Xe}] 4f^{14} 5d^{10} 6s^2$. The $6s$ electrons are the outermost electrons and readily available for covalent bonding. $4f$ electrons are virtually inert since they are very deep. $5d$ electrons belong to the previous shell, but are the outermost of them and once mercury looses its $4s$ electrons, $5d$ electrons become the outermost and define its properties. They are still too strongly bound to the nucleus to participate in covalent bonds under ordinary conditions ($\ce{HgF4}$ was only observed spectroscopically in inert gas matrix). , but are far enough to wiggle around and participate in non-covalent interactions, such as van-der-waals interactions. $6p$ orbitals are also available for covalent bonding, though extent of their use is often debatable and/or unobvious.


why is NaOH used instead of KOH for preparation of K2Cr2O7?

Manufacturers usually prefer sodium hydroxide over potassium hydroxide because it is cheaper. Sometimes, NaOH is preferred because a smaller quantity (as compared to KOH) serves the purpose. And also the product from by KOH is much more soluble than NaOH, make it harder to extract than using NaOH.


The best explanation I was able to find is given by Avery A. Morton , John B. Davidson , Barton L. Hakan, J. Am. Chem. Soc., 1942, 64 (10), 2242–2247. If I understand it correctly, this boils down exactly to that you already propose.

  Disproportionation has been so universally accepted as the criterion of a
  free radical that there has been no attempt to show a derivation from other molecular species. Nevertheless, a mechanism for this phenomenon
  by way of an organometallic intermediate appears exceptionally reasonable. If an alkylsodium meets an alkyl halide in such manner that the two alkyl chains are adjacent to each other while sodium halide is being formed, the two alkyl residues will have unlike charges and a proton will be drawn to the alkyl radical having the two unsaturated electrons. The reaction is merely the conventional prototropic change taking place between adjacent portions of two molecules whose inorganic components are effecting, or have just
  completed, a union.

  If this view be accepted, the olefin produced must come from the alkyl halide rather than at random from both participant alkyl radicals. Though conclusions are to some extent obscured by a metal-halogen interchange (see later), all available facts unquestionably point to the correctness of this deduction.

However, there might be another reason, which I cannot back up with literature. This is probably due to the persistence of the radical species; higher concentration of radicals increases the possibility that they react with each other. In other words, one radical abstracts a hydrogen from another radical, hence creating an alkane and an alkene, in a purely radical reaction.
$$\ce{2 R-CH2-CH2. -> R-CH2-CH3 + R-CH=CH2}.$$

But then again radical mechanisms are never simple and it is quite possible that this does not happen at all, yet it would appear to be at least an explanation for the Wikipedia statement.


Beer's law: UV-Vis absorbance values exceeding 2

Beer's law limitation can be more easily understood if the coefficient $\epsilon$ is considered as a cross-section effect. Let's start with a broad beam or a bundle of rays having a section $A$ (in $m^2$). Its measured intensity is $I_o$. Now suppose there exist plenty of tiny obstacles suspended in the bundle of rays, and that each such particle behave like a dark disk perpendicular to the bean, with a surface or a cross-section $\sigma$. $\sigma$/A is the fraction of $I_o$ that is blocked off by one such particule. If there are $n$ particules in the whole cylindrical bundle, the fraction of $I_o$ that is captured is $\Delta$$I$/$I_o$ = $n·\sigma$/$A$.

Now let l be the length of the bundle where the particules are present. The volume of this bundle filled with dark particles is $V=A·l$. So the fraction of $I_o$ absorbed by the particules is $\Delta$$I$/$I_o$ = $n·\sigma ·l/A·l$ = $(n/V)·\sigma ·l$ = $c·\sigma ·l$, if $c = n/V$ is the concentration of the particules in the volume crossed by the beam.

Going to the limit, this expression becomes : $dI/I_o$ = - $\sigma ·l·dc$.
Integration gives : $ln(I/I_o)$ = - $\sigma·c·l$

This last expression is similar to Beer's law :  $log(I/I_o) = - \epsilon ·c'·l$. I have written $c'$ because in the beginning of my text $c$ is the number of particules per volume unit. And the concentration in Beer's law is in mol/L. $c' = c/N_o$, where $N_o$ is Avogadro's number.

In the end we obtain the final formula : $\epsilon = N_o·\sigma /2.3$

Now if a particule is a disk of $0.2$ $nm$ diameter (dimension of a small molecule), the corresponding $\epsilon$ is about $10^5 L mol^{-1} cm^{-1}$. This is the order of magnitude of $\epsilon$ for the usual dye.

But if $\epsilon$ = $N_o·\sigma /2.3  = 10'400 m^2/mol$. And this is the point where I would like to go today. It means that if one mole of the dye molecules were all to be touched by the beam, they would have to cover a surface of $10'400 m^2$. This is as large as a football ground. It also means that when the concentration becomes too high, Beer's law does not apply any more. The majority of the molecules of the dye are hidden behind others.

You may even calculate that when $c > 1/ (\epsilon · cm), $ then $  : I/I_o < 0.01$, the absorbance is $> 2$, and this is the limit of application of Beer's law, because most of the molecules are hidden behind some other ones at higher values of $c$.

Sorry for the length of the demonstration.


Loading nickel metal hydride battery for the first time

Battery on 0% usually still has some voltage, and battery on 100% usually can be overcharged. However, for many types of batteries it is not healthy, leading to premature degradation, the exact 0% and 100% charges as well as initial load procedure and storage requirements are marked at point that happens to prevent battery degradation and increase its capacity.

While I'm not aware of what exact process is intended to happen in Ni-hydride battery, I'm pretty aware that preparation procedure for battery MAY be required. For example, during initial cycles of lead accumulator, its capacity increases. The reason is in its chemistry, as it actually works on lead oxides transformations and the cycles increase layer of oxides on electrodes.

Since Ni-hydride cell includes electrode with half-reaction

$\ce{Ni(OH)2 + OH- = NiO(OH) + H2O + e-}$

I guess, that during initial charge the manufacturer expects that layer of nickel hydroxide is formed on positive electrode.  But this is just a guess, you'll have to search specific info on this particular type of battery to be sure. Ni-hydride batteries are resistant to overcharge at low currents, so if you use appropriate smart charger, say, shipped with the batteries, the overcharge is safe, so, since the manufacturer requires it, you should follow the manual.


History of octanol-water partition coefficient

The most oft-cited publication is Partition Coefficients and Their Uses. A. Leo, C. Hansch, and D. Elkins. Chemical Reviews 71(6), 525-616 (1971). This manuscript was noted as a "Citation Classic" (see text here) in which A. Leo writes (in part):

  In recognition of the effort of designing
  a method of calculating log P (o:w) from
  structure, the early version of which was reported
  in this article, the Chesapeake section
  of Sigma Xi presented me with its Excellence
  in Science Award for 1980.”

Note bene: In the last article I've linked, mention is made that the phenomena was first investigated by Bertholet in 1872 and "put on a firm basis by Nernst in 1891." That said, the 1971 publication of Leo $\it{et\; al}$ is the first modern/comprehensive listing of octanol-water partition coefficients and the place where the symbology (and definition) were written.


Why is the conductivity of non-metals not equal to zero?

When we think of molecules we think of their molecular orbitals.  If we place these molecules into an ordered solid like a crystal, then the orbitals on adjacent molecules are close enough to interact and "extended" molecular orbitals are formed.  We call these extended molecular orbitals "bands". Typically these bands are "thicker" (spread over a wider range of energies) than molecular orbitals.  That is to say that because the energy of a molecular orbital in a molecule near an edge will be slightly different than the same molecular orbital in a molecule in the middle of the crystal, there is a spread of energies associated with a band.  Ultimately, because of this "thickening" of bands, all of the bands containing electrons overlap and form what is termed the valence band.  Similarly all of the empty bands overlap and form the conduction band.

(image source)

The valence band and the conduction band are analogous to the highest occupied molecular orbital (HOMO) and lowest unoccupied molecular orbital (LUMO) that exist in individual molecules.  The HOMO-LUMO separation is termed the "band gap" in a solid.

In metals the band gap is essentially zero with the valence and conduction bands overlapping; electrons move readily from the valence band to the conduction band. Electrons in the conduction band then move freely throughout the solid and give metals their characteristic conductivity. In insulators the band gap is very large and very few electrons have the thermal energy needed to be promoted to the  conduction band, hence there is basically no conduction.  Semiconductors lie in between these two extremes and thermal energy can promote some electrons (not as much as in a metal, but far more than in an insulator) to the conduction band, enough to have some conduction.

(image source)


How do we identify left handed and right handed crystals?

Yes, you are right that the handedness refers to the molecular 3D arrangement. Such handed molecules are stereoisomers known as enantiomers.

I think that the first part of your question will make more sense following an answer to the second part. So with regard to the part about discerning handedness by looking at it:

(I couldn't get to the article right now due to network issues, so I hope I'm not repeating too much.)

A molecule is chiral when one of the atoms in it has four unique groups attached to it.
Chiral molecules observably rotate plane-polarized light, either clockwise or anti-clockwise (see here for a picture of what plane-polarized light is).

Pasteur observed this along with the cancelling effect of a racemic mixture. As in any scientific endeavor, Pasteur was building upon the efforts of previous work--in this case, that of Jean-Baptiste Biot in studying tartaric acid. Pasteur's careful work enabled him to deduce that the naturally occurring or biologically produced form of tartaric acid was enantiomerically pure because it rotated light and the mirror image crystals also rotated light opposite directions. This was in contrast to the racemic acid which did not rotate light since it was a 50:50 mixture that cancelled the rotation in either direction.

Bryn Mawr College has a helpful explanation of how enantiomers interact with light:

  You have probably [not] learned in lecture that chiral molecules have optical rotations and that enantiomers rotate plane polarized light in equal and opposite directions... This behavior upon exposure to plane polarized light is not some magical property that defies the general principles of enantiomers. The reason enantiomers interact differently with plane polarized light is that plane polarized light is chiral. Though polarized light is a planar phenomenon, it is actually comprised of two superimposed helical forms. You can think of the electrical component as oscillating in such a way that there are two electrical vectors rotating in a cork screw fashion in opposite directions away from the source of the propagation at the same rate. Since a helix is a chiral shape , the two versions of the light constitute an enantiomeric pair. If you resolve the vectors of the two superimposed helices, you end up with oscillation in a single plane...Enantiomers interact differently with the two forms of light. The basic idea is that when plane polarized light encounters enantiomerically pure, chiral molecules, one form of light will slow down more than the other as a consequence of the interaction. This results in one helix coming out of sync with the other. When the vectors are resolved, the resultant plane is rotated with respect to the original position. This change in position is what we mean when we say a compound has an optical rotation.

So looking at a crystal being exposed to polarized light should allow you to observe whether it rotates light to the right or left. I'm not sure how easily detectable to the eye the rotation is, but polarimeters are instruments which are used to measure the angle of rotation.


Why is diazobenzene named so and not phenyl diazobenzene?

Diazo group is $\ce{=N2}$.

Example: diazomethane $\ce{CH2=N2}$
(detailed structure: $\ce{H2C=N+=N- <-> H2C^{-}-N+\bond{#}N}$)

$\ce{-N=N-}$ is azo group.

Example: azomethane $\ce{CH3-N=N-CH3}$
However, preferred names are based on diazene $\ce{HN=NH}$, so better name for azomethane is dimethyldiazene.

So, note that “diazobenzene” does not exist. You obviously mean “azobenzene”, $\ce{C6H5-N=N-C6H5}$, whose preferred name is diphenyldiazene.


yes, they are, all of them are bidirectional.
It is not possible to draw conclusions about the pH value, because of the different abilities to solve in water and because those reaction may cause an increased or a decreased temperature.

(Source: my teacher)


Entropy change and heat transfer in non-ideal heat engines

So apparently the definition of entropy change in a system is $\Delta S = \int_\mathrm{rev} \frac{\mathrm d Q}{T}$ and this is a state variable.

This definition is for the surrounding if it is boring, e.g. a temperature bath. The $_\mathrm{rev}$ subscript is just for the temperature bath, i.e. there could be irreversible processes elsewhere (e.g. in the system).

Yet in my professor's notes it says $$\mathrm dS = \oint \frac{\mathrm dQ}{T} < 0 \text{ for irreversible processes }$$

The inequality is correct. Equating it to $\mathrm dS$ is not. The entropy change for a cyclic process is zero because entropy is a state function.

[OP...] elsewhere, talking about the Sterling process, it also says
$$ \oint_\mathrm{irrev} \frac{\mathrm dQ}{T} < 0 \text{ , so the entropy of the machine (not the total entropy) decreases } $$

This is the Clausius inequality. If a heat engine is not ideal, it will not extract the maximal energy possible from the heat transfer from hot to cold reservoir. The claim about the entropy of the machine is incorrect. Indeed, the entropy of the machine stays the same (use your argument of taking a reversible path from one cycle to the next, i.e. do nothing). Overall, the entropy increases because the entropy loss in the hot reservoir is no longer matched by the entropy gain of the cold reservoir. Thus, we are not extracting the maximal work in the process.
I found this explanation helpful. They used $\delta Q$ instead of $\mathrm d Q$ to remind us that there is a path-dependency.


Does the aluminium chloride assay (Ordonez et al.2006) for total flavonoid content work on coloured extracts?

Depending on the intensity of the yellow color (specifically, the degree of absorption of light at the wavelength of your measurement) you might be able to subtract out the yellow from the extraction in order to differentiate between that and the color arising from the aluminium chloride assay.

In other words, just measure the absorbance of your material before the addition of your reagents, and subtract that value from your final absorbance reading.

Alternatively, since this is a matrix interference issue, you might do well using the method of standard additions. This is described in this Wikipedia article:

  The method of standard addition is a type of quantitative analysis approach often used in analytical chemistry whereby the standard is added directly to the aliquots of analyzed sample. This method is used in situations where sample matrix also contributes to the analytical signal, a situation known as the matrix effect, thus making it impossible to compare the analytical signal between sample and standard using the traditional calibration curve approach.

This article gives a reasonable outline of the procedure that should be sufficient should you decide to try this method.