Is oxytocin's mirror image still oxytocin?

Oxytocin is composed of $\alpha$-amino acids, which with the exception of one (glycine), are all chiral. A chiral molecule's mirror image is called its enatiomer, and has identical chemical properties except in the presence of other chiral molecules, i.e. in the body. This means enantiomer of oxytocin does not have the same biochemical activity as oxytocin, so you were wise to question whether or not this would be the same molecule. In addition to flipping the molecule, you need to change the wedges to dashes and dashes to wedges for it to still be the same molecule:


Why are dipoles "permanent/induced dipole permanent/induced dipole" and not just "permanent/induced dipole" once?

Because it takes two to tango.

Dipoles interact with each other. A Lone dipole has nothing to interact with (other than an electric field, but if we ignore some externally applied macro field, there is nothing for a lone dipole to interact with).

So molecules with an inherent dipole (like water or chloroform) interact with each other. One molecule's dipole interacts with the other molecule's dipole. So you would never say "dipole interaction" only "dipole-dipole interaction". The repeated word is because it takes two dipoles to interact.

The same pattern applies to non-polar molecules with little or no inherent dipole. For example, benzene. Benzene has no built-in dipole, but the electrons in its bonds are fairly polarisable (which basically means it is easy to induce a dipole in them). So benzene molecules do interact but via London or van der Waals forces which are much weaker than the reactions of molecules with inherent dipoles. But one way to describe those weaker interactions is (simplifying a lot) to see them as forces created by fluctuation and temporary quantum-mechanical induced dipoles. Imagine the electrons in one benzene molecule stray from perfect symmetry in their distribution and create a temporary dipole; this may interact with the electron clouds in other benzene molecules an induce other temporary dipoles (or interact with other existing temporary dipoles). Either way the interaction is an "induced dipole-induced dipole" interaction.

So however you understand the interactions, to describe them properly you have to repeat the phrase twice.


Why should the temperature be maintained at 0–5 °C in a diazotisation?

The temperature at which many reactions are carried out is often arbitrary, with certain values being favoured due to the ease at which cooling baths can be prepared. As an example  -78°C is highly prevalent, and a question about this 'magic temperature' has been asked here previously.

In the case of reactions forming diazonium salts, there has been a common rule of thumb that they must be kept somewhere between 0 - 5°C, which is even quoted on Wikipedia:

  In aqueous solution diazonium salts are unstable at temperatures above 5 °C  (Ref: Wikipedia)

At temperatures above 5 °C, diazonium salts (in aqueous solution) have been known to decompose, often explosively, to liberate nitrogen gas, temperatures too low however can cause crystallisation, which causes issues later on (diazonium salts should never be dried upon isolation - they explode).

Although people often keep to this rule of thumb, theres no real body of evidence for its generality (i.e. different diazonium species will have vastly different decomposition temperatures).

Often the only time people carry out in-depth studies of a particular reaction is when the reaction needs to be scaled up, such as in the the pharmaceutical industry. Process chemists go to extraordinary lengths to study their reactions to avoid any possible safety issues (or even unintended side reactions).  In Organic Process Research & Development 2004, 8, 1059−1064, a team of process chemists at Lundbeck carried out extensive thermal studies on a diasotisation reaction they were doing as part of a Sandmeyer reaction, and they found no issues well above 5 °C (though as a precaution they still kept the reaction below 15 °C).


How many covalent bonds are there in carbon suboxide?

Widely used terminology might not match what an examination wants
It is possible that your question wants the answer 8 in this context, but whether this answer would be widely accepted by anyone familiar with bonding theory is doubtful.
In the early days of theories about bonding Lewis described bonds as consisting of two shared electrons. You are probably familiar with "Lewis Structures" which are simple pictures of bonding where electrons are represented with two dots between atoms. This is also how the theory of bonding is initially introduced to students. In this view the number of bonds in the molecule is, indeed, 8. But this–vastly over simplistic theory–is not good for explaining the bonds in many real compounds. Hence the dispute about the correct terminology.
A good theory of bonding needs to take quantum mechanics (and the shape of electron orbitals) into account and those were not understood in the early 20th century when Lewis developed his explanation of bonding.
But we don't need any quantum obfuscation to see why Lewis theory doesn't work well and why it is better not to pretend that bonds consist of pairs of electrons. Take benzene as an example.  You could claim that the bonds in benzene contain 18 electrons so make up 9 bonds. But this means alternating double and single bonds. But that can't be right as all the bonds in benzene are the same length. It is better to say than benzene consists of 6 bonds each of which have somewhere between the character of a single and double bond.
This inability of the Lewis model of bonding to explain real world structures is very common. Nitrate ions are one example where all the bonds have characteristics of more than a "single" bond but less than a "double" bond. Many boranes have bonds between 3 atoms but only two electrons to share among them.
So equating bonds to shared pairs of electrons is not helpful. Nor is using terminology that defines a bond that way. So real chemists should be reluctant to count bonds that way. Whether this is the answer your question is looking for is less clear as you might still be at the stage where they have taught you Lewis structures and nothing else.


Are sulfonyl-carboxylates stable in water?

In addition to the procedures described by Ian Fang, these mixed anhydrides (with $\ce{R_1, R_2 \ne H}$) have been used to prepare symmetric anhydrides of carboxylic acid in solid state reactions. If that's interesting to anyone, I can look up the reference.

As far as the effect of water is concerned, I do remember a procedure where the mixed anhydrides in question were actually prepared in a biphasic system of water and dichloromethane in the presence of a phase-transfer catalyst.


What results from the reaction of 4-phenylbutanoyl chloride in presence of AlCl3?

My guess was that an other molecule of 4-phenylbutanoyl chloride would react with the phenyl [...]

Your guess is not that bad and it seems that you have identified the type of reaction.

   Apparently, it is a special type of electrophilic aromatic substutution, known as the Friedel-Crafts acylation. In your guess, one molecule of the starting material would serve as the acylating agent (through the acyl chloride), while another is acylated at the phenyl.

What if you only had one molecule of the staring material? Can you imagine a similar reaction?

   The alkyl chain of $\ce{CH2}$ that links the acyl chloride unit to the phenyl is flexible! There's no reason why the Friedel-Crafts acylation could not happen as an intramolecular reaction, where acylating agent and the attacked phenyl group are part of the same molecule.

What would the the outcome?

   The intramolecular reaction would furnish tetralone.


If absorbance is 0, what is the equilibrium constant?

Are we talking about uv-vis absorption?

Supposed that you are monitoring the right wavelength and that the extinction coefficients are not ridiculously small, you are outside the limits of experimental error.

If you still do not see any absorption, this means that the species isn't there (anymore).

If it isn't there anymore, it can't particiate in the equilibrium any longer - there is no measurable equilibrium anymore.

If you still need to measure something in the case of an irreversible and fast reaction, your experimental setup needs to change:

You might need to perform time-resolved  uv spectroscopy and use a stopped-flow setup for the controlled mixing of your reactants.

If you don't have anything but a uv-vis spectrometer, a cuvette and a microliter syringe, do it the old way using the standard addition procedure

Prepare the cuvette with a solution of one reactant with a decent concentration
Prepare a solution of the second reactant. This might be highly concentrated it you want to avoid dilution effects. If the reaction is fast, you can only measure the absorption of the first reactant and the product anyway
Measure the initial absorption of the first reactant
Add some volume of the second solution to your cuvette
Measure the uv absorption of your first reactant. Do you see any change?
Can you measure the absorption of the product too?

Repeat steps 4-6.


My lab has a lot of experience with ITO-coated PET, but unfortunately I don't think anyone's tried etching it.
We typically etch ITO-coated glass with a mixture of hydrochloric acid/water/nitric acid (4:2:1 by volume) which is fairly fast at room temperature, going through a few µm in about 10 minutes and mask with a positive photoresist.

To your questions:

Will dry film work well for this application?

Because the resolution you need isn't great, it should work fine, so long as the film adheres properly to the ITO.
Underetching is a possibility, but if the features are big enough it should be okay.
The big issue here is trying to use a normal fluorescent light to expose the film.
This is probably hopeless; you're better off finding a proper UV lamp or even using the sun.
How can I determine the thickness of the ITO given its sheet resistance of 50 ohms?

The mixture I mention above is fairly quick, but either way, I wouldn't worry about the precise film thickness (which may actually vary somewhat depending on the batch/supplier).
The easy way to deal with this is to simply pull the sample out of the etchant occasionally and test the resistance with a multimeter in a region where the ITO should be etched away.
On PET, the conductivity should be negligible when the ITO is gone.
Am I going to destroy the PET during this process?

The datasheet you linked refers to the use of PET containers for storage of chemicals—it may actually last a few hours without significant damage, but I haven't tried it.
Is my plan fundamentally flawed?

I think it's worth a shot.
You will need to handle the ITO sheets fairly delicately—the ITO can crack if the sheet is flexed too much, but I don't see this method as being totally unworkable, though you may have issues with the quality of etching you'll get.

As a final note, I wouldn't worry about getting special glassware.
For measuring simple ratios and etching/developing baths, kitchen quality Pyrex dishes and measuring cups work just fine for this kind of thing.
We keep stacks of them in the cleanroom because they're nice and cheap.


Why Molecular Solids can be treated as Monoatomic in Reactions

Generally, other than the seven diatomic gases at STP ($\ce{H2}$, $\ce{O2}$, $\ce{N2}$, $\ce{F2}$, $\ce{Cl2}$, $\ce{Br2}$, $\ce{I2}$... though  $\ce{Br2}$ & $\ce{I2}$ have fairly low vapor pressure at room temperature), for convenience, other elements are considered as if they reacted as individual atoms, though, as you state, that is not actually the case. Phosphorus, for example, can be in allotropes with formulas of $\ce{P2}$, $\ce{P4}$, or even a graphitic form, but in a reaction, it's usually considered to be monatomic (unless one must take into account the energy to break the bonds of a specific form).

BTW, iodine vapor is diatomic only at moderate temperatures. From Wikipedia: "1% of a sample of gaseous iodine at atmospheric pressure is dissociated into iodine atoms at 575 °C." Furthermore, at STP, iodine solid behaves as a semimetal, so perhaps it shouldn't be considered diatomic... but it's easier to just say "all the halogens are diatomic", ignoring iodine and fleeting astatine.

In brief, then, this is the conventional list.


What are the conditions for liquid metal to be corrosive to another, different solid metal?

There are no general conditions- indeed there no general conditions for the any of the forms of corrosion. Solubility is a factor and many other factors but it just comes down the actual interactions between a given liquid metal and another metal and the service conditions such as temperature, velocity, impurities.

Liquid-metal attack has the following categories:

simple solution,
alloying between liquid metal and solid metal,
intergranular penetration,
impurity reactions,
temperature-gradient mass transfer,
concentration-gradient mass transfer, or dissimilar-metal
mass transfer.

With that many means- determining what liquid metal damages what solid metal  is very difficult.

http://web.ornl.gov/info/reports/1956/3445603500959.pdf


Why is the dipole moment of chloromethane larger than the dipole moment of fluoromethane?

Dipole moment is not just about charges, it also has $L$ term. Bond length of $\ce{C-Cl}$ is greater than $\ce{C-F}$ and in this case, that is more dominating factor.

The dipole moment is in order

$$\ce{CH3Cl} \gt \ce{CH3F} \gt \ce{CH3Br} \gt \ce{CH3I}$$

You can see that electronegativity plays a more dominating role in $\ce{CH3X}$ when $\ce{X}$ is $\ce{Br}$/$\ce{I}$.


How do I determine stereochemistry of the product

The reaction doesn't change the stereo configuration for either reactant, so we can assume that the product is enantiomerically pure with respect to the carbon attached to the hydroxyl group (in orange below), and a racemic mixture with respect to the tertiary carbon (in blue).

Since there are multiple stereocenters, you can either have both the same (same molecule), both different (enantiomers), or have some different and some the same (diastereomer).  The products would be diastereomers.  In particular they would be epimers, which differ in only one stereocenter.


Significance of Law of reciprocal proportions

The comments certainly clarify the law of reciprocal proportions and are an excellent tutorial on how to write a good question but sheds little light on your question “Is there any intuition/specific reason that Jeremias Richter came up with it?”
From reading a number of short bibliographic sketches about Richter it appears his primary interest was mathematics.  The title of his doctoral thesis was, roughly translated, The use of Mathematics in Chemistry  [https://www.encyclopedia.com/science/dictionaries-thesauruses-pictures-and-press-releases/richter-jeremias-benjamin].  His study of quantitative relationships in the neutralization of acids by bases lead to the law of equivalents. [https://encyclopedia2.thefreedictionary.com/Jeremias+Benjamin+Richter ]
Based on these and similar biographical notes one might argue that he arrived at the law of reciprocal proportions not through “intuition” at all but rather through an empirically based and numbers driven approach.  When a mathematician sees a pattern or consistency in “the numbers” they then might be seek to understand the natural forces giving rise to their observations.
A modern and well written example of “math driving science” can be found here. https://www.quantamagazine.org/geometry-reveals-how-the-world-is-assembled-from-cubes-20201119/ “Scientists Uncover the Universal Geometry of Geology”


Is there a cheap, safe liquid that evaporates EXOthermically around room temperature?

No such liquid, safe or otherwise, can exist. Evaporation is a strictly endothermic process in all cases.

The change in state from liquid to gas is marked by the individual particles gaining enough translational kinetic energy to overcome the mutual attractions present in the liquid phase to "fly free" in the gas phase. It is logically inconsistent for a substance to increase its internal energy and release energy to the surroundings as heat in the same process.

In order to achieve both evaporation and a release of energy, one would have to find a liquid that reacts to (a) release heat and (b) form gaseous products.  The energy required to move from the liquid to the gas phase is substantial; more than likely the only reactions exothermic enough to provide a net release of heat are combustion reactions.

I somehow doubt dumping a flammable liquid into your thermostat, inserting a wick, and lighting it on fire is a satisfactory solution for you.


Why is quinoline less basic than pyridine?

This is not due to the inductive effect, but rather, due to solvation. The positive charge of quinoline on the nitrogen is harder to solvate due to steric repulsion of the nearby ring, whereas in pyridine, the small hydrogen atoms in place do not pose a strong steric hindrance for the solvation of the ion.


Relative acidity of acetic acid derivatives with strongly withdrawing substituents

Some cheminformatic descriptors for the puzzling compounds can be found on the PubChem website:

SMILES    N+(=O)O ;
Nitroacetic acid; 625-75-2; alfa-Nitro
acetic acid; 2-nitroacetic acid; ALFA-NITROACETICACID;
CID: 43581;
MF: C2H3NO4;
MW: 105.05g/mol;
InChIKey: RGHXWDVNBYKJQH-UHFFFAOYSA-N   ;
IUPAC Name: 2-nitroacetic acid   ;

and

SMILES N+(C)(C)C;
TRIMETHYL GLYCINE;
N,N,N-trimethylglycinium; Cystadane;
CHEBI:41139;
carboxy-N,N,N-trimethylmethanaminium;
CID: 248   ;
MF: C5H12NO2+   ;
MW: 118.15g/mol;
InChIKey: KWIUHFFTVRNATP-UHFFFAOYSA-O  ;
IUPAC Name: carboxymethyl(trimethyl)azanium

With glycine betaine it is usually meant the chloride salt, which would result in chloride counterions. For dilute solutions presumably those chloride ions don't matter, but they would at higher concentration.
Nitroacetic acid even has its own Wikipedia page. The acidity reported there is pKa 1.68, which is similar to that found in other sites such as in ChEMBL, pKa 1.75.
Glycine betaine or trimethylglycine also has Wikipedia and ChEMBL entries. From Wikipedia, acidity (pKa) is 1.84. From ChEMBL, 2.26.
So from this data alone it would certainly seem like the book is correct. Still, those pKa values are very very similar.


short/extensive training on computational chemistry

Did you check the offerings of the national chemical societies and their special interest groups?

In Germany, that would be the CIC - Chemie Information Computer group of the GDCh - Gesellschaft Deutscher Chemiker.

Cosmologic provides TURBOMOLE workshops.

In 2014, the Max Planck Institute for Chemical Energy Conversion organized a Summer School during which a course on QM/MM with Frank Neese's ORCA was given. I don't know if they plan something similar this year.

The NSCCS has some upcoming events.


How far-reaching is the effect of conjugation?

1,6-additions are indeed possible (1) but since you have competing 1,4-addition and 1,2-addition it becomes more difficult. To push it to the 1,6-addition bulky organometallic catalysts are used. Reaction pathways are also determined by the electrostatic interactions, and these diminish the further away you get from the carbonyl.

(1) https://onlinelibrary.wiley.com/doi/epdf/10.1002/adsc.201601342, https://www.organic-chemistry.org/abstracts/lit1/353.shtm


but I can't find anything because flame AES seems to be quite outdated.

Please don't say that you cannot find anything, this is the reason (i) why physical libraries still exist and (ii) right key words are needed for proper literature search and (iii) not everything is available on the web free of cost.
Most of the metals are detectable by flame atomic emission spectroscopy. All you need is the right flame. The instrument, called the atomic absorption spectrophotometer, works perfectly in atomic emission mode. There is no separate instrument for Flame-AES.
The flames are quite "cold" for non-metals like sulfur, carbon, nitrogen, and noble gases. ICP also works poorly for non-metals.


Be aware that $\ce{H+}$ is nothing else then a proton $^1_1\mathrm{p}$. Protons have an extreme charge density, typically $10^{15}$ higher than  any other ion ( but bare electrons or atomic nuclei).
Protons can exist free in vacuum or plasma, but even in gases they quickly react with many gaseous molecules, forming molecular ions, like dioxidanylium $\ce{O2 + p -> HO2+}$.
Protons act in aquaeous environment as extremely strong Lewis acid with an electron pair of $\ce{H2O}$(Lewis base), forming a ion called hydronium, a particular type of oxonium ions.:
$$\ce{p + H2O -> H3O+(aq)}$$
For convenience, we often use notation $\ce{H+}$ or $\ce{H+(aq)}$, supposing it is at least $\ce{H3O+(aq)}$.
Experiments mentioned in the above Wikipedia link confirm hydronium is usually coordinated with 6 water molecules as $\ce{[(H3O)(H2O)6]+}$, but multiple various hydration structures have been observed or assumed, like
Zundel ion $\ce{H5O2+}$,
$\ce{H9O4+}$($\ce{H3O+ . 3H2O}$)
or a cluster $\ce{H3O+ . 20 H2O}$
Note that the hydronium ion forms solid salts with acids with dissociation constant $K_\mathrm{a} \gt 10^{9}$.
E.g. a crystallic monohydrate of perchloric acid is hydronium perchlorate $\ce{[H3O+][ClO4-]}$


Consider the auto-ionization of water :

$\ce{ 2H_2O->H_3O+ + OH-}$

The first oxygen has three bonds, the second only has one.

You can think of the reaction taking place by a lone pair on the oxygen of one water molecule ripping off the proton only of the hydrogen of another water molecule to form a covalent bond between them using just the lone pair. The electron of the hydrogen is left behind and stays with the oxygen of the other molecule.

If you calculate the formal charges on each oxygen you will see the first one has a positive charge and the second one has a negative. The formal charge is just the valence number of electrons minus the number of bonds minus non-bonding electrons (using the lewis structure) and is a useful book keeping method to think about where the electrons go/are and what are the most stable structures.

Formal Charge Calculation


Why does the gauche conformation of ethane-1,2-diol show hydrogen bonding?

Cyclopentane has a relatively low strain energy, something on the order of 6.5 kcal/mol more than cyclohexane (reference).  In the case of ethane-1,2-diol, the strain energy in the 5-membered ring formed by hydrogen bonding will be even less because some of the $\ce{H-C-C-H}$ and $\ce{C-C-C-C}$ interactions that produce the strain in cyclopentane are no longer present; the oxygens don't have the hydrogens attached like a methylene group does and the hydrogen used in the hydrogen bond is also much smaller than a methylene group.
A typical $\ce{O-H..H}$ hydrogen bond can stabilize a system by ~7 kcal/mol (reference).  So to form a 5-membered ring in ethane-1,2-diol and create a hydrogen bond it will cost something less than 6.5 kcal/mol to overcome the strain energy in the ring, but we get 7 kcal/mol back due to hydrogen bond formation - a net stabilization.
$%edit$


When writing out synthetic reaction schemes, why are some reactions combined in one arrow?

I want to start off by saying that your example is a particularly bad one. Usually, the arrows in a scheme should at least state a list of reagents and steps. Instead, I would love to discuss this using scheme 7 of Zhang et al.’s Beilstein J. Org. Chem. article.[1] (Beilstein J. Org. Chem. is a open access journal, the full text should be accessible from the DOI link provided below.)

The main reason for combining reactions like this is brevity. For example, every practicing chemist will immediately understand that the homoallyl Grignard under 1. will attack one of the carbonyl functions, that triethylsilane will reduce an oxygen and that TBAF will remove the silyl protecting group on the hydroxy function. None of these steps really warrant any discussion and most people would be able to draw out the missing structures in a few minutes. Likewise, I am not entirely sure why they did not include the preceeding protection into that same reaction scheme because there too there is noting interesting going on.

In more advanced synthetic schemes, the steps that are shortened and written together on a single arrow are often typical name reactions or very easily performed ones:

protections and deprotections
oxidations of alcohols to aldehydes with standard reagents
reductions with DIBAL, LAH or $\ce{NaBH4}$
ozonolyses
double bond isomerisations
simple Wittig/HWE reactions
etc.

They are typically chosen in such a way that the path is obvious and that the structure on the end of the arrow can be drawn up quickly by going through the steps. They will practically never include key steps or steps whose stereochemical outcome is uncommon. Cross-couplings of more elaborate structures or lesser known name reactions are also less common.

However, there is no rule as to what can be combined and what cannot. If anything, the reviewers might mark a scheme as too elaborate (i.e. can be shortened) or too brief (needs elaboration).

Reference:

[1]: J. Zhang, H.-K. Zhang, P.-Q. Huang, Beilstein J. Org. Chem. 2013, 9, 2358–2366. DOI: 10.3762/bjoc.9.271.


How to explain Born repulsion between ions in gas phase?

The term $+b/r^n$ is a power law repulsion term between the ions, where $b$ is the bare ion radius (twice the van der Waals radius)  and $r$ the separation.

The hard sphere model considers the ions as 'billiard balls' which means that $n=\infty$ since $+b/r^n$ is effectively zero when $r \gt b$ and infinite when $r \lt b$.

Two other types of potentials are used (a) the power law potential when $n$ is some integer value, usually between 9 and 16  and (b) the exponential potential $+a_0exp(-r/a_1)$ where $a_0$ and $a_1$ are adjustable parameters with $a_1 $ of the order of 0.02 nm. The justification  for using these potentials is weak (the power law particularly so) and they are used as they represent a better approximation than the hard sphere potential and are mathematically convenient. The exponential form is presumably justified since the combination of all orbitals decay exponentially with distance.


You may pour them down the sink. NaOH will be soon neutralized and destroyed by the CO2 from the atmosphere, or by the bicarbonate ions present in all drinkable calcareous waters.


Why aren't soldering tips coated with platinum?

Sigh - as pointed out by @user83678 some 3+ years after first writing this, I answered for Pd, not Pt. That will teach me to pay attention. But, the general answer remains the same. There are enough interactions of platinum with either lead, tin, or silver (a common component in lead-free solders) such that a platinum tip does not make metallurgical sense.

Lets look first at Pt-Pb, with the CALPHAD parameters from Z.H. Long et al., J. Phase Equil. Diff 30 318-322 (2009):

While Pt does not have a high solubility in Pt, the presence of multiple intermetallics indicates problems: either Pb reacts with the Pt time, or Pt dissolves into the solder melt and forms an intermetallic there. Neither is desired.

For Pt-Sn, one gets (with CALPHAD parameters from V. Grolier and R. Schmid-Fetzer, J. Alloys and Compounds 450 264-271 (2008)):

Here the Pt happily sucks up lots of Sn, and then there are all those intermetallics.

For Pt-Ag (from I. Karakaya and W.T. Thompson, Bull. Alloys Phase Diagrams 8(4) 334-340 (1987)) one gets:

The fcc Ag and Pt have large solubilities for each other (but not complete solid solubility, somewhat surprisingly). Again, you get lots of interactions between a Pt tip and a lead-free solder containing Ag.

Aside from my comment that it is not hard to keep a soldering tip in tip-top shape, I will point out several materials issues.

For this purpose, I will stick to the interactions of Pd with classic Pb-Sn solder. A peek at the Pb-Pd phase diagram (ASM International, or G. Ghosh, Met. Trans. 30A, 5-18 (1999)) one readily sees that palladium has a large solubility of lead, more than 10% at room temperature. Since the solubility rises with temperature, cooling a Pb-saturated Pd tip from soldering temperature to room temperature would result in Pb precipitates forming, likely disrupting the mechanical stability of the Pd tip.

A glance at the Pd-Sn phase diagram (same paper referenced above) shows a fairly similar situation (not so surprising since Sn and Pb are in the same column, but that isn't a given in phase diagrams). Again, even at room temperature the solubility of Sn in Pd is in excess of 10%.

For both systems there is a binary phase, Pd$_{3}$Sn or Pd$_{3}$Pb, very close to the fcc palladium phase field, as well as a range of other phases stretching to higher Pb or Sn contents. The ternary diagrams included in the Ghosh paper show many phases present. This means that reactions between molten Pb-Sn solder and a platinum tip won't stop at just saturating the fcc Pd with Pb or Sn - the reaction will continue, making compounds across the phase diagram and rapidly ruining your fancy Pd tip.

For lead-free solders, note that Sn remains a primary component in many of them, so the same things apply.

I have included a (calculated from the CALPHAD parameters in Ghosh's paper) the Pb-Pd phase diagram.


Why is Diiron nonacarbonyl so exceptional?

Wikipedia (primary reference) suggests a possible reason for $\ce{Fe2(CO)9}$ dissolving preferentially in THF versus nonpolar solvents:  it reacts according to the scheme

$\ce{Fe2(CO)9 + THF <=> Fe(CO)5 + Fe(CO)4 \cdot THF}$

Such a reaction is invoked to account for the dinuclear complex giving mononuclear products with various ligands in THF.  A similar reaction with the corresponding cobalt and manganese species would have to produce radical or ionic products, due to the odd atomic number of the metal, and so would be less favored.

This would not, of course, explain the lack of solubility in nonpolar solvents or nonvolatility of $\ce{Fe2(CO)9}$.  It is possible, however, that the same $\ce{Fe2(CO)9}$ molecules that bind with the oxygen in THF could also bind intermolecularly through the oxygen atoms, in effect delocalizing the covalent bonding and creating more intermolecular cohesion.

We can represent this type of interaction using the SMILES structural notation.  Fr a single $\ce{Fe2(CO)9}$ molecule it would look like this, taken from Wikipedia:

$\ce{O=C1[Fe]2(=C=O)(=C=O)(=C=O)C(=O)[Fe]1(=C=O)(=C=O)(=C=O)C2=O}$

For two molecules we would have

$\ce{[O=C1[Fe]2(=C=O)(=C=O)(=C=O)C(=O)[Fe]1(=C=O)(=C=O)(=C=O)C2=O][O=C1[Fe]2(=C=O)(=C=O)(=C=O)C(=O)[Fe]1(=C=O)(=C=O)(=C=O)C2=O]}$

Then we have a contributing structure where an oxygen atom from the left molecule could combine with the right one, displacing an iron pentacarbonyl molecule:

$\ce{[O=C1[Fe]2(=C=O)(=C=O)(=C=O)C(=O)[Fe]1(=C=O)(=C=O)(=C=O)C2=\color{blue}{O[Fe]}(=C=O)(=C=O)(=C=O)C=O][[Fe](=C=O)(=C=O)(=C=O)(=C=O)C=O]}$

where the blue atoms form an intermolecular bond in this contributing structure.  The displaced iron pentacarbonyl molecule is in the last set of brackets.

The Case of Hydrogen Bonding

The delocalization of covalent bonds, hypothesized above, may be seen in a more familiar context:  hydrogen bonding.  Usually thus is rendered as an electrostatic attraction between oppositely charged atoms.  But it may also be considered as a molecular-orbital interaction:  a nonbonding electron pair overlaps an adjacent molecule's antibonding orbital in exchange for creating an "intermolecular" covalent bond.  With water it would look like this

$\ce{[HOH][HOH]}$

with the hydrogen-bonded, "intermolecular bonded" contribution

$\ce{[H\color{blue}{O}(H)\color{blue}{H}^+][OH^-]}$

where the intermolecular bond is between the blue atoms.  This is also the autoionized structure for water.


Freeze distillation: Does efficiency depend on solubility?

Solids have solubility, just as liquids do. This is easily seen in metal alloys, where different amounts of different compounds can cause different crystal structure, low-level phase separation, and large-scale phase separation (if the solid solution is not stable). The changing of these nanostructures of the material can even occur with a material fully in the solid state, such as tempering of steel.

This relates to your question in that liquids freezing will behave very similarly with respect to phase composition as vapors condensing. The composition of the solid will be proportional to the composition of the liquid and affected by differing freezing points, just like you would expect from a liquid-vapor Txy diagram.

There are some differences, however. Molecules in solids can, as I said above, move around within the solid. However, this diffusion is substantially slower in solids than in liquids, and in non-metal solids, slow enough to be largely insignificant within the process of freezing. This means that the freezing process will create regions of differing composition within the solid. Further and related to this phenomena, the speed of freezing will impact the concentration of the solid, as faster freezing will push it even further from equilibrium.

The slow diffusion in solids also means that once trapped in the crystal structure, compounds with higher freezing points will not as easily exit the solid as they might from a liquid. This means that melting can behave very differently from vaporization, and solids might become a slush in the process of melting.

An explanation as to why this all occurs could go as follows: solubility is a factor of the strength of the interaction between different types of molecules. Those interactions must be stronger than interactions within the individual types to avoid phase separation. The process of freezing tends to not fundamentally shift the strength of these interactions. It is possible for materials partially soluble in liquid to be mostly insoluble in solid, but pairs of miscible liquids have such intense cross-compound interactions that freezing typically can only impact concentration of the separate phases.

The really important chemistry in this case takes place at the phase boundary between the solid and liquid, which behaves almost identically to that between a liquid and a gas. Molecules with weaker interactions with the solid phase (weaker interactions with other molecules in that phase) will tend to leave the solid phase more often - but there is an equilibrium where molecules will enter their less-favored phase. Essentially, everything you would have learned about liquids and vapors applies here, except diffusion is too slow to allow good mixing within the more condensed phase.

Of course, there are, in specific cases, more complicated explanations. One is of interstitial compounds, where a second compound might occupy spaces within a solid. While this happens to a degree in all solids, interstitials that are small enough to fit fully in a crystal structure without deforming it might exhibit unusual solid solution behavior, such as a compound with a lower freezing point showing a greater affinity for solid phase than the other compound.

To answer the title question, yes, separation depends on solubility. Unlike vapor distillation, though, it can also depend on temperature difference (thus speed) and mixing.


You're right--it's got to do with them being transition metals (usually). Transition metal ions form coordination complexes. Their empty $d$ orbitals accept lone pairs from other molecules (called "ligands") and form larger molecules (though we don't call them that--we call them "complexes"). When put in water, the ligand is $\ce{H2O}$, and you get complexes like $\ce{Cu[(H2O)6]^{2+}}$ (this is what you get when you add water to Copper Sulphate). Thus, metals with no empty $d$ orbitals of comparable energy become colourless.

OK, but where do coordination compounds get their color from? Well, $d$ orbitals look like this:

Even though the last one looks different, these are all identical in many respects, the most important being that they have the same energy. So if an electron jumps from one orbital to another, it on't make any difference. In fact, since this is a quantum mechanical system, it (sort of) doesn't even make sense to talk about electrons jumping from here to there.

All this changes when you have some ligands around. In coordination complexes, the $d$ orbitals are hybridised with the $s$ and $p$ orbitals of the next or previous shell. If we're talking about octahedral complexes, all the orbital lobes have to be aligned along the coordinate axes (since the $p$ orbitals are). This can be said to come from the fact that we're "mixing" these orbitals to get orthogonal hybrid orbitals, but that may not be that good an analogy. Now, the ligands approach along the coordinate axes since their final position will be in hybrid orbitals along the axes. So, they approach the $d_{x^2-y^2}$ and $d_{z^2}$ orbitals (as well as the $s$ and $p$ orbitals). The repulsion generated makes these orbitals of higher energy than the other three $d$ orbitals. This is called "d-orbital splitting" {*}

Now, electrons may "jump" between the $d$ orbitals (the energy gap isn't so high that the jump is too improbable). Due to the energy gap, some energy is absorbed/released in the form of a photon and it just happens to fall within the visible spectrum (at least, for most transition metals, it does). This jump is called a $d-d$ transition.

Using this, we can easily show why $\ce{Zn}$ salts are colourless. In $\ce{Zn}$, the $3d$ orbitals are full, so they're useless from the ligand point of view. The $4d$ orbitals are, on the other hand, empty. Ideal for forming a coordination complex with the $4s$ and $4p$, but there won't be any $d-d$ transitions as there are no electrons in $4d$ to jump around. (Remember, this is before the ligand bonds)

*For tetrahedral complexes, the $d$ orbitals take no active part in the bonding, and the ligands approach between the axes. This leads to an opposite gap between the top three(known as $t_{2g}$) and bottom two($e_g$) orbitals, with the $t_{2g}$ orbitals at a higher energy. So we still get $d-d$ transitions and color, though IIRC many tetrahedral compounds do not have a visible color(the energy gap is about half that of the corresponding ocahedral gap--and may not give a photon with a visible frequency)


Can wedge and dash bonds on 1, 2 position of cyclohexane ring be used to signify trans orientation of the substituents?

So if you drew a wedge in position 1 and a dash in position 2, it would mean that they are on the same plane - either both axial or both equatorial - assuming that the ring was in a chair conformation.

However, if the objective is to show the structural isomers very clearly, it may be better to use either the chair conformation sketch:

... or the Newman Projection:


Strong acids disassociate completely, weak acids do less so. When they do disassociate in water, they form $\ce{H3O+}$. This pulls the electrons from the thing it is reacting with. Lets take this to be Fe. ($\ce{H3O+}$ rips the electrons from the substance). So the reaction would go something like: $\ce{Fe + 2H3O+ -> Fe^2+ + 2H2O + H2}$. The $\ce{Fe^2+}$ formed is soluble in water. This also reacts with the other ion from the acid. This is what makes the acids corrosive.


Since you have zero knowledge of chemistry, I will keep explanation as simple as possible. I have to emphasize that this answer is not fully correct and can be misleading. It is suited just for interested layman, but not someone trying to grasp deeper concepts of chemistry.

Nitrogen atom has 5 electrons in its outer shell. The most stable electronic configuration, by octet rule, corresponds to 8 electrons in outer shell. Nitrogen needs three electrons to make that shell full. It can achieve that configuration by bonding with another nitrogen. One nitrogen has his own 5 electrons and another lends him his 3 - now first nitrogen has 8 electrons, e.g. full shell. First nitrogen lends his 3 electrons to second nitrogen which also had 5 electrons, so now both of them have closed shell. Since both of them lended three electrons, and accepted three - they have three shared electron pairs, which corresponds to triple covalent bond. Closed shell configuration, which they both now achieved, corresponds to a state of the lowest energy. Since everything in nature strives to a state of lowest energy, you can see why $\ce{N}$ atoms prefer to bond, instead of being each on it's own. The molecule is inert, because it achieved what it strived to: energy minimum. By adding another (third) nitrogen in a game, you would necessary have to "break" closed shell configuration, which necessary leads to increase of energy, which is not favorable. That's the reason why you can't create another bond and keep inertness of $\ce{N2}$ molecule.


Should chemistry students worry about quarks and such?

Is an understanding of chemistry fundamentals likely to become out of date when the textbooks start to include quarks, leptons and the dozens of other sub-atomic particles?

Well, the general chemisty books do already include leptons, at least electrons and usually positrons.

Other than protons, neutrons, electrons and positrons (and photons), I think that when discussing beta decay and electron capture, neutrinos and antineutrinos should be included, otherwise the nuclear reactions aren't really balanced (lepton number must be conserved).  Alternatively, the general chemistry books could omit all discussion of nuclear reactions, as being physics.

Other than that, I think the rest of the particles can wait for physics class, rather than general chemistry class.

Beyond general chemistry, other particles are sometimes discussed in more advanced chemistry books, such as quantum chemistry texts, where examples involving muons are common.

For some chemistry topics, such as isotope differences in rotational-vibrational spectra of small molecules distinction between fermions and bosons is critical.  Trying to explain the difference between fermions and bosons (and Fermi-Dirac statisitics vs Bose-Einstein statistics vs Boltzmann statistics) naturally leads to discussion of the standard model and the particles in the standard model.  So by the time a student gets to physical chemistry, around the third year of college, they should have some understanding of the standard model.


Is there a shortcut method to calculate the Steric Number?

Short answer: no.

The steric number is a property of an atom, not a compound. You need to know what an atom connected to a given atom to know its steric number. For simple compounds, you can usually determine these connections because the formula suggests a central atom and surrounding groups. For hydrocarbons and other organic compounds, you need to consider isomerism. Given the capability of carbon to form complicated bonding patterns, even simple formulas can produce a fair number of isomers with different bonding patterns and steric numbers.

  Let's look at some examples.

$\ce{C4H10}$

This formula corresponds to two compounds with the structures shown:

In this case, both compounds have all four carbon atoms with steric number of 4.

it is not always true that a set of hydrocarbon isomers will always have the same steric number for all carbon atoms or even the same set of steric numbers.

$\ce{C4H8}$

This formula corresponds to six isomers:

Note that four of these structures have two carbon atoms with steric number 4 and two carbon atoms with steric number 3. The other two have all four carbon atoms with steric number 4.

Any method to calculate steric number for carbon atoms in an organic compound using just the formula will fail. You must examine the structure.


Variation of Arrhenius equation with concentration

The Arrhenius equation gives you a rate constant $k$.

For a second order reaction, $k$ has units of $1/Ms$.

For a first order reaction, $k$ has units of $1/s$.

The Arrhenius prefactor $A$ has the same units as $k$.

The rate, as opposed to the rate constant will depend upon concentration. For the first order reaction, the rate constant is multiplied by one concentration to get the rate, and for a second order reaction the rate constant is multiplied by two concentrations to get the rate.

If $k$ and $A$ do depend upon concentration, that would imply that the reaction is not accurately modeled by the rate constant (for example not really first order when you are modeling as first order or not really second order when you are modeling as second order).

As far as activation energy $E_a$, the equation can be written in terms of energy per mole divided by $RT$ or energy per molecule divided by $k_BT$ (where $k_B$ is Boltzmann constant).  $E_a$ does not depend upon concentration.


Why is carbon monoxide stable(compare with other flammable gas) in room temperature?

What gives you the impression that carbon monoxide does not burn easily?

Below I will link to a youtube video of an accidental ignition of a large amount of carbon monoxide.  But be warned!  The video is a highly uncontrolled, dangerous experiment.  The person who performed it did not have appropriate safety equipment and acted very recklessly.  DO NOT TRY TO PERFORM A SIMILAR EXPERIMENT YOURSELF unless you know what you are doing and have

A CO detector / alarm
A buddy in the room with you
A fire extinguisher
Lots of experience synthesizing gases using laboratory apparatus

...

OK, all that said, here is the video.  As you can see, carbon monoxide is quite flammable.  And that guy is an idiot, at least in my book.

The autoignition temperature of $\ce{CO}$ is ~609 °C, only 100 °C or so higher than the $\ce{H2}$ autoignition temperature, and lower than the autoignition temperature for solid carbon. The LEL and UEL are 12.5% and 74%, respectively, which is a very wide range compared to most compounds.

Compared to other gases, $\ce{CO}$ explosions may be comparatively less violent because of the relatively smaller change in moles during the combustion reaction:

$$\ce{2CO(g) + O2(g) -> 2CO2(g)}$$

Thus, two moles of gas are formed from three moles of reactants, which means that the shock waves and pressure forces produced by a sudden $\ce{CO}$ combustion reaction / explosion are not as large as some gases such as say propane.


Mechanism of a gold catalyzed addition to triple bond

In their 2005 The Journal of the American Chemical Society publication, Genêt, et al. have reported a gold-promoted cyclization reaction of bis-homopropagylic diols (see Graphical Abstract below), which afforded bicyclic ketals at room temperature in very good to excellent yields (ref. 1). This cycloisomerization reaction was catalyzed by either Au(I) or Au(II) chloride in methanol:

Abstract: A highly efficient gold-catalyzed cycloisomerization reaction of bis-homopropargylic diols is described. The cyclizations are conducted in the presence of either $\ce{Au^I}$ or $\ce{Au^{II}}$ catalysts in $\ce{MeOH}$ at room temperature in a very short time. The reaction conditions are compatible with functional groups, such as n-butyl, phenyl, allyl, benzyl, and alcohol groups, leading to original strained dioxabicyclo[2.2.1], -[2.2.2], or -[3.2.1] ketals in good to excellent yields.

There was a sequence of compounds has been used as starting material, which consist of different R- groups, one of them been Ph- and $n = n' = 1$ as in this example. A plausible mechanism for the cyclization is given in the paper, which is shown in  following scheme:

In there, 1a and 2a represent the appropriate structures with $n = n' = 1$ and R- = cinnamyl- group. The authors have given the following explanation for the mechanism:

The reaction may be initiated by the formation of the π-alkynyl complex A through the complexation of the unsaturated triple bond to the $\ce{Au}$ catalyst. The coordination of the triple bond, therefore, enhances the electrophilicity of the alkyne as usually advocated for metal-catalyzed reactions of alkynes (ref. 2). The addition of one alcohol may be favored by an intramolecular complexation of the $\ce{OH}$ group to the gold catalyst, as already envisaged by Teles and co-workers (ref. 3) for the addition of methanol to alkynes. The enol vinylgold intermediate B may then be protonolyzed leading to an enol ether, which then undergoes another intramolecular addition of the remaining alcohol leading to the cyclic ketal 2a. This latter step may also be catalyzed by gold, considering the proposed mechanism for the $\ce{Au(II)}$-catalyzed cyclization of α-hydroxyallenes (ref. 4).

In reference 2, it is stated that that the activation of the carbon−carbon triple bond toward nucleophilic attack can be performed by metals with high Lewis acidity (alkyne activation), as illustrated in Equation 1 of the following scheme. It is also easy to realize that an external nucleophile could make an anti-addition as a result and thus, the product should be obtained with trans-stereochemistry (Equation 1). Yet, as envisioned by Teles and co-workers (ref. 3), syn-addition is also possible by an external nucleophile, through complexation with metal, first by alkyne and then by the external nucleophile (see Last Scheme below). However, in many other cases (specifically where a nucleophile would be available intramolecularly as the case in here) the picture is quite different (Equation 2). Here, metal can activate the nucleophile first (nucleophile activation) and the reactions may proceed as a result of the insertion of the alkyne moiety into M−H or M−Nu bonds of “H−M−Nu” complexes, typically with syn-addition as illustrated in Equation 2 of the scheme (This is the case illustrated in A and B).

Last Scheme:

References:

S. Antoniotti, E. Genin, V. Michelet, J.-P. Genêt, “Highly Efficient Access to Strained Bicyclic Ketals via Gold-Catalyzed Cycloisomerization of Bis-homopropargylic Diols,” J. Am. Chem. Soc. 2005, 127(28), 9976–9977 (DOI: 10.1021/ja0530671).
F. Alonso, I. P. Beletskaya, M. Yus, “ransition-Metal-Catalyzed Addition of Heteroatom−Hydrogen Bonds to Alkynes,” Chem. Rev. 2004, 104(6), 3079–3160 (DOI: 10.1021/cr0201068).
J. H. Teles, S. Brode, M. Chabanas, “Cationic Gold(I) Complexes: Highly Efficient Catalysts for the Addition of Alcohols to Alkynes,” Angew. Chem. Int. Ed. 1998, 37(10), 1515–1418 (https://doi.org/10.1002/(SICI)1521-3773(19980605)37:10<1415::AID-ANIE1415>3.0.CO;2-N).
A. Hoffmann-Röder, N. Krause, “Gold(III) Chloride Catalyzed Cyclization of α-Hydroxyallenes to 2,5-Dihydrofurans,” Org. Lett. 2001, 3(16), 2537–2538 (DOI: 10.1021/ol016205+).


HCl synthesis from sulfuric acid and sodium salt

Ideally, that's the reaction. Realistically, there are a few speed bumps.

First, it occurs stepwise, with the acid salt sodium bisulfate as an intermediate. The first stage happens at room temp between equal proportions of salt and sulfuric acid. The second stage requires temperatures of 200*C (in turn requiring water-free concentrations of acid and dry salt) and an additional equivalent of salt that reacts with the bisulfate.

You did say you were using excess sulfuric acid; if true, you'll react all the salt but what'll be left in the flask is sodium bisulfate, not sodium sulfate. Since you're interested in the HCl gas produced and not the sulfate salt, this isn't a big deal, but I just wanted you to be clear that one mole of sulfuric acid to two moles of salt isn't going to get you the expected 2 moles of HCl unless you're heating a "water-free" combination of the reactants.

Second, if you have too much water in the reaction, the HCl won't bubble out; it'll happily sit in solution. So, dry salt and 10M or higher concentration of sulfuric acid are recommended (as in Nicolau's YouTube video, 18M is plenty strong enough to do the job, and they make and sell it up to 98% concentrations). If you're stuck with 1M, you can heat the flask to boiling, which evaporates some water and increases the acid concentration, liberating the HCl gas (sulfuric acid won't boil below 300*C, so you're fine distilling the HCl like this until you're out of water and the thermometer starts to climb).

Following these steps and bubbling the gas produced through water, what you produce should definitely be hydrochloric acid. Obviously the molar strength of it (and thus its ability to react with test materials as was also indicated by Nicolau) will depend on the relative proportions and amount of reactants, the volume of your trap water, and how much gas is actually absorbed by the water.


Fast way to isolate 4-nitrophenol from 4-aminophenol

I would use solvent extraction. Dissolve the mixture of phenols in an organic solvent. Next shake the organic solution with dilute hydrochloric acid in a separation funnel. Do this several times to wash all of the 4-hydroxyaniline out of the organic phase.

Next shake the organic phase with concentrated sodium chloride solution (brine) and then add anhydrous sodium sulfate. Shake the mixture up in a sealed conical flask. Now decant or filter it to give you a solution of 4-nitrophenol which you can now evapourate down to give you the phenol.

I would like to know how the nitrophenol was reduced into the amine over the years it was in storage. This is a more interesting question to my mind is how did the nitro compound get converted on storage into the amine.


Effect of substituents on arene oxide-phenol rearrangement

What your book says is correct, and the issue arose due to your premature judgement of the intermediates' stability without actually working it out.

First of all, [1,2] Wagner-Meerwin shifts are intramolecular rearrangements which are primarily driven by thermodynamic concerns. Therefore, if their is a possibility of a reasonable cyclic transition state being formed, then the presence of the deuterium isotope hardly makes much of a change for the anionotropic shift, especially since the $\ce{C-D}$ is only marginally stronger than the $\ce{C-H}$ bond as stated here. So, if their is enough motivation for the rearrangement to occur, it can happen via either of the pathways.

Now, if you actually try drawing out the resonance structures of 1(a) and 2(a), you will actually find that the cation on the former will receive stabilisation by the +H effect, and not the second. The positive charge will only be able to align suitably for hyperconjugation in 1(a) as shown below:

Hence, mechanism 1 should be the preferred pathway

I think the rule you are stating refers to the increment in electron density at the ortho and para positions of an aromatic ring by the virtue of a methyl substituent as shown below:

However, this thing is only valid when the methylation has been done at a completely aromatic ring and not at some broken fragment of the same.

Only when their is complete conjugation and the ring itself is aromatic does it even make sense to use ortho and para descriptors as per definition. When you are considering a part of the same with incomplete conjugation, then it's pretty much up to the bonding schematic of the current molecule to see where the positive charge is getting delocalised.

o-/p-/m- descriptors lose their value in such cases, and along with them, their delocalisation rules as well

Clarification

I guess the OP had some confusion regarding the mechanism after the formation of these intermediates as well. The competition of each pathway occurs stepwise:

First, the competition will be between the cleavage of either one of the two $\ce{C-O}$ bonds, one adjacent to the hydrogen and the other adjacent to the deuterium. As the intermediate 1(a) is more stable than 2(a) due to reasons stated above, the cleavage will occur as shown in the second step of Reaction Mechanism-1

Now, the intermediate 1(a) will try go for rearrangements that will generate species which are more energetically favorable. Due to the electron donating(+M effect) of the hydroxyl present adjacent to the cationic part in 1(a), a 1,2 deuterium shift will occur leading to the formation of 1(b) as shown in the third step of Reaction Mechanism-1

Note that as soon as the cleavage step occurred as described above, all considerations of Reaction Mechanism -2 and it's intermediates after 2(a) were rendered moot. 2(a) lost the stability battle against 1(a), and after that, none of the intermediates or products of Mechanism-2 would matter at all


Conversion of barbituric acid to veronal

This paper from Tetrahedron Letters here details synthetic procedures using carbonyl compounds in solution with barbituric acid and Pt or Pd hydrogenation catalysyts to give reductive di-C alklyation. This includes examples of N-unsubstituted barbituric acids R1 = R2 = H. Using barbituric acid plus acetaldehyde should give you veronal in a single step.


"Cage-like" Formazine particles, how are molecules arranged?

First, note that

formazine structure attached from PubChem is a different compound;
note that there are also, related, formazan $\ce{HN=N-CH=N-NH2}$ and “formaldazine” $\ce{CH2=N-N=CH2}$ (dimethylidenehydrazine);
formazin (polymer) does not necessarily has to have cage-like structure, as this characterization belongs to hexamethylenetetramine in the quoted citation.

Hexamethylenetetramine (urotropine, $\ce{(CH2)6N4}$) can be prepared from formaldehyde and ammonia. In water it also somewhat decomposes to its precursors:

Formazin (also called formalazine) structure might be still unknown, as a related publication from 1976 states:[1]

  … This substance, an insoluble condensation product of uncertain composition, is prepared by mixing solutions of hexamethylenetetramine and hydrazine sulfate. …

Sometimes empirical-like formula $\ce{(C2H4N2)_n}$ is claimed.[2]

Sometimes, claimed structure like following one is presented:[3,4]

It was (with several other structures, less nice or likely), hypothesized (well, rather more complicated, with some ether linkages $\ce{-CH2-O-CH2\bond{-}}$ and $\ce{-NH\bond{-}}$ groups present), based on spectroscopic studies.[5,6]

References:

Rice, E. W. The preparation of formazin standards for nephelometry. Analytica Chimica Acta 1976, 87 (1), 251–253.
Ziegler, A. Issues Related to Use of Turbidity Measurements as a Surrogate for Suspended Sediment. Turbidity and Other Sediment Surrogates Workshop; Reno, NV, USA, 2002.
Kaur, N.; Kishore, D. An Insight into Hexamethylenetetramine: A Versatile Reagent in Organic Synthesis. J. Iran. Chem. Soc. 2013, 10 (6), 1193–1228.
Sadar, M. J. Stabilized Formazin Composition. US Patent 5,777,011, July 1998.
Mashima, M. The Infrared Absorption Spectra of the Condensation Products of Formaldehyde with Hydrazine. Bull. Chem. Soc. Jpn. 1966, 39 (3), 504–506.
Bondybey, V. E.; Nibler, J. W. Infrared and Raman Spectra of Formaldazine. Spectrochimica Acta Part A: Molecular Spectroscopy 1973, 29 (4), 645–658.


What is the qualitative definition of enthalpy in beginner terms?

In my judgment, it's not worth spending much of your valuable time trying to assign physical significance to the physical property we call Enthalpy.  Enthalpy is just a convenient parameter to work with when we are doing certain types of thermodynamic analysis.  One situation which it is particularly useful for is in applying the first law to open systems in which material is entering and/or leaving the system.  Another is where changes in a closed system are occurring at constant pressure.


Can organic acids be concentrated and used like concentrated sulfuric acid and similar mineral acids?

They can. Vinegar is typically 4-18% acetic acid by mass, but of course concentrated acetic acid also exists with a concentration >90% by mass. It is corrosive and flammable, but it surely exists.


What is the difference between these wave functions?

The first function you have there

$\psi(x)=A\sin(\frac{2\pi x}{\lambda})$,

is very similar to the function of a particle in a monodimensional box.

This function is a very helpful example in order to understand how does the quantum mechanics works.

The other function is

$\psi_{(n,m,l)}(r,\theta,\phi)=R_{n,l}(r) Y_{m,l}(\theta,\phi)$.

This function represents how an electron moves in a hydrogen atom and the full expression is quite different to the first function. Just see the representations:

Ok. The functions do not mean the same concept. So... Why do we use "$\psi$" in both?

In math we use $f(x)$ to speak about any function. In quantum mechanics we use $\psi(x)$ for the same: represent a function easily.

Yeah, but... Why do i have to study the first function if it's just a lie and not the full story?

Remember it's only a very good example. However, the particle in a box function can be used to determine the energy of an electron in a conjugated system as beta-carotene and it's energy for the first excited level. If you subtract the second energy to the first one it gives you the energy for a photon in the visible region: the carrot's orange color.

This means that the particle in a box is a good example if you start to study quantum chemistry and it even works as a good approximation for some systems.

Ok. Too much information but I want to know more: How can I get the second function from the first?

Well, that's... Impossible...

To obtain the functions we use the Hamiltonian operator, which contains the kinetic energy operator and the potential energy operator.

For the particle in a box, we only use the kinetic energy operator because we don't have any potential into the box. This leads to the first function you have.

For the hydrogen atom we must use both operators (all the Hamiltonian) which means to include a function for the potential and we must work in 3D. The potential function can be obtained using the angular momentum. The spherical coordinates makes easier to work in 3D. This leads to radium and angular functions.

Summarizing, we have a different wave function for every different situation. This situation is what determines the function and it's conditions:

The particle in a box → Quantization.
The free particle → No quantization.
The electron in a hydrogen atom → Quantization + quantum numbers + rotation +... → Atomic orbitals.

Ok, seems legit, but it's hard to understand

I recommend you the books "Physical Chemistry" from Atkins, Levine or Engel to understand better the ideas and see the complete development.

In case you hate textbooks you can visit hyperphysics web page for a general looking.


How to name compounds with more than one spiro atom and cyclic compounds with more than two fused rings?

Bicyclic compounds

Locate the bridgehead carbons, the carbons part of two rings.
Now, identify the three connections between the 2 bridgeheads, called bridges: arrange these in decreasing order of number of carbon atoms (excluding the bridgehead). Say these three numbers are m, n, o [m > n > o].
Write the name of your compound as: bicyclo[m.n.o]"m+n+o+2"alkane.
In case there is a substituent or a functional group is present, you need to number your carbons: start from of the bridge-head, through the longest bridge to the shorter and then the shortest (if any).
Other rules of IUPAC apply as such.

Spirocyclic compounds

Locate the bridgehead carbon, the only carbon part of the two rings.
Now, spot the two ring systems having having 2 or more carbons: arrange these in increasing order of number of carbon atoms (excluding the bridgehead, again).
Say these three numbers are p,q [p < q].
Write the name of your compound as: spiro[p.q]"p+q+1"alkane.
In case there is a substituent or a functional group is present, you need to number your carbons: start from of the atom next to bridgehead in the smaller ring, through the bridgehead carbon to the bigger ring.
Other rules of IUPAC apply as such.


The "-oic" ending means to convert the carbon into a COOH group, not that there's an additional COOH group added to that carbon.

Look at your main chain. You have five carbons, of which carbons 1 and 5 are fully oxidized, hence "pentane-1,5-dioic acid". If the "COOH" group was counted as an independent entity adding an additional carbon, then it would be a three carbon chain, to which was added two COOH groups.

That's why it can't be a 3-oic acid - that would imply that the third carbon of main chain has been oxidized to a COOH group (which is preposterous), not that there's a COOH group that's been added. Instead, the COOH group (carboxy) has been appended to the three position, and the molecule is named accordingly.


I can only answer your question about the relation between refractive indices and other properties with two examples.

When looking at dispersion interactions between particles in solution the Lifschitz Theory connects the refractive indices with the Hamaker constants.
The information about refractive indices is even used to choose materials in a way to cancel out dispersion interactions between particles of different kind. (I will attach a paper URL in the next days, just heard it from our professor in the lecture)

Another application of refractive indices is X-Ray NEXAFS spectroscopy and  many more spectroscopic methods involving electron or high energy light radiation. By the Kramers-Kronig relation it is possible to connect absorption constants and refractive indices. Especially one can find for example absorption edges.

Then here I found some probably interesting papers regarding your question. At least their references will be a good starting point. Unfortanetly I have no access and could only read the abstract:

http://www.sciencedirect.com/science/article/pii/0020089165900205

http://www.sciencedirect.com/science/article/pii/0032395083902836


Keto Enol Equilibrium for Claisen Condensation Products

The Claisen reaction yields a product with an acidified proton in $\alpha$-position of carbonyl and ester, right at the same carbon atom of R depicted in blue.

In presence of a base, yes, $\beta$-dicarbonyl compounds undergo deprotonation at their methylene unit to yield a keto-enolate, stabilized by mesomerism with the secound carbonyl group in conjugation. Compare the $pK_a$-values of $\beta{}$-dicarbonyl compounds with ketones, for example (one reference), to estimate if the equilibrium is on the left, or on the right.
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I think cations of the right size and charge equally favour the formation of the keto-enolate.  If you synthesize $\beta$-dicarbonyl compounds, the risk to loose product due to chelation  is lowered if initial work-up is concluded by drying the washed organic phase with sodium sulphate ($\ce{Na2SO4}$), rather than by magnesium sulphate ($\ce{MgSO4}$).


Can ZnO act as a reducing agent for noble metals like Pd?

Zinc chemistry is overwhelmingly that of the +2 oxidation state, it does not go higher. For ZnO to reduce Pd+2, the Zn has to go up from +2 which it cannot. If you are getting reduction of Pd+2, Zn is not responsible.


Position of nodal planes in molecular orbitals

First a descriptive explanation:

The Molecule is symmetric. Thus the resulting mocular orbitals are symmetric too. If you are constructing different MOs where the nodal planes are different than in the picture above they wouldnt be symmetric.

Mathematical Explanation:

These nodal planes result from a LCAO-MO consideration with hückel approximation. You have to form the secular determinant for such a problem at which the coulomb integrals are Alpha and the resonance is only dissimilar zero when the two considered atoms are neighbors. After solving this hückel-matrix you can determinate the coefficents of the several molecular orbitals. Doing that you will see that the nodal planes exist as you drew them above.
(More Information about the Hückel-Method: here)


You need to draw the structure and identify the most acid and most basic site(s).
Notice how I didn't say "the most acid molecule".
Boron tribromide ($ \ce{BBr3} $) contains both a super-strong acidic site (the trivalent boron atom) and potentially basic sites (the lone pairs of the three bromine atoms).

Whether a molecule (e.g. $ \ce{H2S} $) will act as an acid or as a base depends on the reagents.


Calculating the electrode potential from thermochemical data

As stated above by @orthocresol, these values of $\Delta H$ alone are not sufficient to find the electrode potential. To do so, you would need $\Delta G$ for each reaction, so you would either need to find a table of $\Delta G$ for these reactions, or you would need a table of $\Delta S$ along with the table given.

But your question is why do we care about those particular values to begin with. The reason for this is that those reactions summed together give the reduction reaction for that metal. For example, I will look at the reactions of iron.

The reduction reaction for iron: $$\ce{Fe(s) \to Fe^2+(aq) + 2e^-}$$

All the other reactions in the order they are listed in the original post:
\begin{align}
\ce{Fe(s) &\to Fe(g)} \\
\ce{Fe(g) &\to Fe^+(g) + e^-} \\
\ce{Fe^+(g) &\to Fe^2+(g) + e^-} \\
\ce{Fe^2+(g) &\to Fe^2+(aq)}
\end{align}

If you add together these reactions, you should obtain the reduction reaction. Consequently, by Hess's law for Gibb's free energy, the sum of the $\Delta G$'s for these reactions will give you the $ΔG$ for the reduction reaction.

Then, it is just a matter of converting the $\Delta G$ to a reduction potential using the equation given in the original question:$$\Delta G^{\circ}=-nFE^\circ$$ where $n$ is the moles of electrons, $F$ is Faraday's constant, and $E^\circ$ is the standard reduction potential.


Possible nonclassical ion from a bicyclic system

While there are examples of 1,2-methide shifts, this question has an uncanny resemblance to the dienone-phenol rearrangement whose mechanism was first elucidated by Woodward and Singh in 1950. Dienone 1 under acidic conditions undergoes rearrangement to phenol 6 and not, based on earlier speculation, to phenol 4. The direct 1,2-methide shift (2b --> 3) does not occur but rather the reaction proceeds through the spiro carbocation 5. continued

Applying this approach to the carbocation 7 generated from the alcohol in this question, spiro carbocation 8 forms the predicted tetrahydronaphthalene 10. One way to distinguish between the two mechanisms is via a labeling experiment. All of the carbon label in 7 will retain its location as the red star in 10 if the 1,2-methyl shift mechanism applies. The spiro mechanism will partition the label ~50:50 between the two ring benzylic carbons. For related studies, see reference 2.
1) R. B. Woodward and T. Singh, J. Am. Chem. Soc., 1950, 72, 494.
2) A. J. Waring, J. H. Zaidi and J. W. Pilkington, J. Chem. Soc, Perkin Transactions I, 1981, 1454.


We can do this question in two steps, finding the rate law with respect to both A and $\ce{H+}$ individually.
$$\text{Rate}=[A]^y[\ce{H+}]^x$$

The first part is simple, they've given that the half life is independent of $[A]$, therefore it is first order with respect to $\ce{A}$.
Therefore as you've found,
$$\text{Rate}=[A][\ce{H+}]^x$$

Now, onto finding the order with respect to $\ce{H+}$.
From what I've seen on the contradictory answers on the net that OP has mentioned, they state that since the rate is inversely proportional to concentration of $\ce{H+}$, it is second order with respect to $\ce{H+}$ by the relation:
$$t_{1/2}\, \alpha\, [A]^{1-n}$$
where $n$ is the order with respect to A in a reaction.
Is this true? Yes. Is it correct in this context? No. This is because this is true only when the reactant in question actually decreases in the mixture, i.e.
$$-\frac{\mathrm d[A]}{\mathrm dt}=k[A]^n$$
Here, it is specifically given that it is in a buffer of $\mathrm{pH} = 4$. This means that there is no change in the actual concentration of $\ce{H+}$.
Therefore, what you've done is conceptually sound and is the correct method.
$$k'=k[\ce{H+}]^x = \text{constant}$$
\begin{align}
\frac{t_{1/2}(\mathrm{pH}=5)}{t_{1/2}(\mathrm{pH}=4)} &= \frac{k'(\mathrm{pH}=4)}{k'(\mathrm{pH}=5)} = \frac{k[\ce{H+}]^x(\mathrm{pH}=4)}{k[\ce{H+}]^x(\mathrm{pH}=5)} \\
\implies \frac{100}{10}&=\left(\frac{10^{-4}}{10^{-5}}\right)^x = 10^x \\
\implies x &=1
\end{align}
Therefore, the rate would be:
$$\text{Rate}=k[A][\ce{H+}]$$


What is the product formed after nitrosation of ketones (HNO2 + ketone)?

Hydrolysis of the ketone oxime gives the 1,2-diketone as reported in this Synlett reference:

  Treatment of alpha-methylene ketones with excess sodium nitrite and
  aqueous HCl in THF at reduced temperature provieds an effective tool
  for the preparation of a variety of 1,2-diketones


Is it possible to compress an atom to infinite density?

We do not know. Physicists THINK that there ought to be a fundamental limit in scale for space-time that occurs near $10^{-33}$ centimeters and $10^{-43}$ seconds; often called the Planck Scale. There is also a unit of mass associated with this scale which is about $10^{-5}$ GRAMS or $10^{19}$ Billion Electron Volts (BeV or GeV). It is a simple matter to construct a 'Planck density' by computing $10^{-5}$ grams / $\text{(} 10^{-33} \text{centimeters)}^{3} = 10^{94}$ grams/cc. Many physicists think that this represents the highest density for any natural compression mechanism, because at higher densities, space-time itself transmutes into some type of quantum mechanical system for which 'density' is not a meaningful concept. So, if fundamental particles are truly without any structure larger than the Planck scale, then they can be packed together into a 'system' with a density of $10^{94}$ grams; the highest achievable density for energy/mass in this physical universe. Even within black holes, the famous 'Singularity' may not have infinite density, but merely a finite density near the Planck value. This is how 'quantum gravity' is thought to regulate, or eliminate real physical singularities; something which ordinary general relativity is not able to do.

I read it in a book Elegant Universe, by Brian Greene. This book has a whole seperate chapter for this.


Calculating the volume of 1 mole of liquid water

The molecule length value you are using is the problem. It is more like 300 pm, which is the van der Vaals diameter and includes regions where electron density is significant. You have calculated an internuclear distance, but the nuclei of one molecule won't come close to those of another molecule because of electron-electron repulsion.

This will change your answer by a factor of 8.

See Finney, J. L. The water molecule and its interactions: the interaction between theory, modelling, and experiment. J. Mol. Liq. 2001, 90 (1-3), 303–312. DOI: 10.1016/S0167-7322(01)00134-9.


Why is the oxidation state of Co -1 in [Co(CO)4]-?

Ignore list of "allowed oxidation states" for metal complexes. "allowed oxidation states" might be the most common, but definitely not the only ones the atom is allowed to be in. For example, for $\ce{Fe}$ all oxidation state from -2 to +6 are known, for $\ce{Mn}$ -3 and from -1 to +7 are known, for $\ce{Ir}$ it is -3 and from -1 to +8  with tentative +9 reported for gas-phase monocation.

  In $\ce{[Co2(CO)8]}$, most people say the charge of $\ce{Co}$ is 0. How does it become $\ce{[Co(CO)4]-}$?

It is traditionally assumed that ligands do not change oxidation state of the central atom and all atoms in them have same oxidation state as in a free molecule. At the same time, the total sum of all oxidation states of the particle must be equal its charge. Assuming total charge of the particle $-1$ and neutral ligands, the charge by necessity means that oxidation state of  $\ce{Co}$ in $\ce{[Co(CO)4]-}$ is -1.

This is not the only case, there are numerous carbonyl anions with negative charges, like $\ce{[Mn(CO)5]-}$, $\ce{[Fe(CO)4]^{2-}}$. Compounds with even higher formal negative oxidation state of the central metal atom were reported, with anions like $\ce{[Mn(CO)4]^{3-}}$ and $\ce{[Cr(CO)4]^{4-}}$.

It would be a gross misunderstanding to assume that extra charge is located on the metal atom. It is siphoned onto $\ce{CO}$ ligands via $\pi$-backbonding. However, again, metal-ligand interactions are traditionally ingnored in calculation of oxidation state.

Truly negatively charged metal atoms are much rarer, but still happen. Example would be salt $\ce{[Na(15-crown-5)]Na}$ with sodium anion. Such salts are extremely powerful reducers.


What is the white powder that I have extracted from cinnamon?

"6 grams of cinnamon can contain from to 42 to 189 mg cinnamaldehyde." http://herbpedia.wikidot.com/cinnamaldehyde  That's not very much.

"Cinnamon consists of a variety of resinous compounds, including cinnamaldehyde, cinnamate, cinnamic acid, and numerous essential oils." https://www.hindawi.com/journals/ecam/2014/642942/

The extraction method starts with 10 grams of cinnamon and salt is used to salt out the cinnamaldehyde from the steam distillate. Could the solid be salt carried over from the separation funnel? If the solid you obtained is more than about 400 mg, it might well be contaminated with NaCl. You could add a milliliter or two of H2O to see if you get a (saturated) salt solution plus the cinnamaldehyde oil on top. If the cinnamon you started with was very old or had very little cinnamaldehyde to begin with, the amount of oil you obtain might well be overshadowed by a little NaCl contamination.


Actually the copper and silver aren't bonded. The silver that precipitates is pure. Also, a bond between two metals could not be ionic. Ionic bonds are always been metals and non-metals. Looking at the picture again, I think what may have thrown you off is that it looks like the silver at the end is Ag+. However, the silver is actually Ag+ in the beginning of the reaction when it is in the silver nitrate. That's how it is bonded with the nitrate. When the copper donates an electron to the silver, the silver goes from having a +1 charge to having no charge. It becomes neutral, and hence it doesn't bond with the copper.


Correct Electron Pushing with HCl and Water

1) Nucleophiles often have excess electron density while electrophiles are characterized by a lack thereof. The electrons flow in the direction of lower electron density (and not the other way around), because they are "attracted" by the positive partial or formal charge of the electrophile. This is the reason why the first scheme from your book is wrong, because it shows a nucleophile attacking another nucleophile.

2), 3) and the reaction scheme you have drawn are correct.


How to control the purity and quality of fatty acids or their CoA thioesters?

Short answer: Both methods are (technically) suitable but it depends on what impurities you want to detect. But in your case, I would probably suggest mass spectroscopy in most cases.

Long answer:

It depends on how many different compounds you’re getting, how much you trust the company, which impurities you are expecting and how important it is to have exactly the one compound you need.

Mass spectroscopy, ideally coupled with HPLC beforehand, will give a specific retention time for a specific compound, and of course you will get the compound’s mass. That alone is not sufficient for a priori identification! When one detects purity by mass spectroscopy, one would usually have a standard run of one’s desired compounds and compare it to the unknown sample. To identify, compare the retention time (ideally standardised, if the project goes on for more than, say, a year) and the compound masses.

The different fatty acids will have similar, but not identical retention times. The cis-double bond should allow for easy detection via UV absorption (good for quantitising known impurities) and the mass will, of course, differ only by the number of carbon atoms. Impurities that are not fatty acids will have completely different retention times, different UV spectra and different masses which should allow an easy attribution of ‘good peak, bad peak’. If it is not of paramount importance to have only exactly one fatty acid, one can even rest happily with quantitising the amounts of the different ones, too.

Note that it will take some fine-tuning of the HPLC gradient, since most gradients are designed to fit a wide range of polarities with average accuracies, while what you want is a high accuracy in a low polarity range.

Everything I said about HPLC-MS of fatty acids also holds true for the CoA-thioesters, although it is possible that the HPLC solvent hydrolyses the thioester. Work-arounds would be buffered solvents and similar.

$\ce{^1H}$-NMR is best used on pure substances, but can also be used to detect impurities. It should allow for an easy identification on which bond is a double bond by the characteristic shifts and/or multiplets of the two vinylic protons (shift range $6-7~\mathrm{ppm}$). However, you will most likely get unseperable signals at around $1-2~\mathrm{ppm}$ — the aliphatic region where all the saturated carbons’ protons will resonate. Thus it will not be easy to detect saturated impurities, say for example the corresponding unsaturated fatty acid. However, it will be easy to detect other impurities stemming from the synthesis such as remaining reactands (which will depend on the synthetic route).

If you go to the CoA-thioesters, it will be a lot harder to estimate purity just by an NMR spectrum, because CoA has a lot more signals in the $2-5~\mathrm{ppm}$ range. But it is still possible.

NMR would allow the quantification of impurities if you can identify them just by integrating the proton signals and dividing them by the amount of protons giving identical signals.

On a time scale, a typical HPLC-MS run will be around $40~\mathrm{min}$ while a typical proton NMR can be recorded within $5~\mathrm{min}$.

I would suggest NMR in situations where you have a really high number of samples that need to be analysed very timely, where it does not matter if the saturated acid is present or where identifying the stereochemistry and regiochemistry of the double bond is of prime importance. In any case where saturated fatty acids would impede your results, I would strongly suggest HPLC-MS. In either case, I would suggest acquiring pure samples (at least of the fatty acids) first to run a series of standards.


What is the difference between an angular and linear skeletal formula?

Those are $sp^2$ carbons and should have a triangular planar 120° geometry. The angles you've drawn are clearly not 120° .

I've not exactly come across the terminology before, but I believe this is the distinctions:

In a "linear" formula, you disregard the actual bond angles and just try to keep everything in a line.
In an "angular" formula, you draw everything according to the bond angles.

Not sure of this, though.


What is the difference between a Chemical Formula and Formula Unit?

Molecular compounds such as water exist as discrete particles, molecules. This is due to the forming of covalent bonds where each atom has a specific partner to which it is bonded. Each molecule of water contains one atom of oxygen and two atoms of hydrogen. So, H2O is its formula. This can be more specifically called a molecular formula.

In ionic compounds such as table salt, NaCl, the atoms (as ions) do not bond to specific neighbors. Surrounding each chloride ion in a salt crystal are six sodium ions. Likewise, six chloride ions surround each sodium ion. This attraction of oppositely charged ions extends throughout the entire crystal. There is no discrete bonding of a particular sodium ion to a specific chloride ion. So the formula for sodium chloride is expressed as the smallest whole number ratio between the Na and Cl ions which is 1:1 for sodium chloride. So, NaCl is also a formula but not a molecular formula.

The reason for the term "formula unit" is that it is useful when we talk about how much of one substance is required to combine with a particular amount of another substance. For example: to make the smallest amount of hydrogen carbonate we combine one molecule of water with one molecule of carbon dioxide.

But now suppose we want to combine silver nitrate with sodium chloride. Both of these are ionic compounds, they do not form molecules. In this case you would say that one formula unit (FU) of sodium chloride combines with one FU of silver nitrate.

The one-to-one combining in both these examples is simply coincidence. It may take 3 FUs of substance A to combine exactly with 2 FUs of substance B in a different example.


Rule to determine the total number of elements in a period

The total number of elements formed due to their last electron entering the $n$-th shell will be equal to the total number of electrons that shell can fill. For each electron added or removed (along with a proton), you'll get a new element, since a unique element is itself defined by a unique count of protons.

Thus, this problem is reduced to finding the total number of electrons in $n$-th shell. And, simply enough, the total number of electrons in $n$-th shell = number of orbitals in $n$-th shell $\times$ number of electrons each orbital can hold.

For example, for $n=3$, we have $\ell=0,1,2$. Thus, the number of orbitals here is $=1+3+5=9$ (recall that s-subshell has 1 orbital, p-orbital has 3 orbitals, and so on[1]). Since $s=\pm 1/2$, each orbital can fit two electrons, thus total count of electrons is $=2\times9=18$, which is also the total number of elements in the third shell.

In your question, the only change is that each orbital can now hold three electrons (by some phenomenon we don't know yet!) So, in the final step, just multiply the total number of orbitals by three instead!

To calculate the number of elements in a particular period, you need to note the individual subshells that belong to that period, and then sum up the number of the elements from those subshells.

In my solution above, I have shown how to get the total number of elements formed due to their last electron entering the $n$-th shell, which is not equivalent to the number of elements in $n$-th period.

For example, in the fourth period of the periodic table, you have the $\ce{4s, 3d}$, and the $\ce{4p}$ subshells. This is the calculation that your solution manual has shown in the second row.

[1]: In general, the total number of orbitals $k$ in shell $n$ is:

$$k = \sum_{\ell=0}^{n-1}(2\ell+1)$$

since: a) each $m$ value stands for a unique orbital b) $-\ell\leq m\leq\ell$, thus $m$ has a total of $2\ell+1$ values for a particular $\ell$ value


How to get the first step of the derivation of the Maxwell-Boltzmann distribution

KT comes in from requiring agreement with the ideal gas law.

See "section 1.5.3 $\langle v^2\rangle$ should agree with the ideal gas law" and section 1.3 "The functional form of the dependence of T on $\langle\varepsilon\rangle$ cannot be determined solely from kinetic theory, since the temperature scale may be chosen in many possible ways. In fact, one way to define the temperature is through the ideal gas law: T = pV/(nR)" in Kinetic Theory of Gasses


Are atomic mass unit (amu), unified mass unit (u) and Dalton (Da) the same unit but with different names?

It is a simple question that requires a rather complicated answer.
Origins

For technical reasons, in 1898, chemist Wilhelm Ostwald and others proposed to redefine the unit of atomic mass as 1/16 of the mass of an oxygen atom. That proposal was formally adopted by the International Committee on Atomic Weights (ICAW) in 1903. ... This suggestion was made before the discovery of the existence of elemental isotopes, which occurred in 1912. ... This definition remained unchanged until 1961.$^1$

I think the Wikipedia article is wrong about the name. I think the notion was initially conceived as "atomic weight," not "atomic mass."
atomic mass unit(amu)

The discovery of isotopes of oxygen in 1929 required a more precise definition of the unit. Unfortunately, two distinct definitions came into use. Chemists choose to define the amu as 1/16 of the average mass of an oxygen atom as found in nature; that is, the average of the masses of the known isotopes, weighted by their natural abundance. Physicists, on the other hand, defined it as 1/16 of the mass of an atom of the isotope oxygen-16 ($\ce{^{16}O}$).$^1$

unified atomic mass unit(u)

...in 1961 the International Union of Pure and Applied Chemistry (IUPAC), which had absorbed the ICAW, adopted a new definition of the atomic mass unit for use in both physics and chemistry; namely, 1/12 of the mass of a carbon-12 atom. ... The new unit was named the "unified atomic mass unit" and given a new symbol "u", to replace the old "amu" that had been used for the oxygen-based units. However, the old symbol "amu" has sometimes been used, after 1961, to refer to the new unit, particularly in lay and preparatory contexts.$^1$

I'll be less diplomatic. Since 1961 "atomic mass unit" has been conflated with the "unified atomic mass unit."
dalton

In 1993, the IUPAC proposed the shorter name "dalton" (with symbol "Da") for the unified atomic mass unit.$^1$

"And now..., for the rest of the story" as Paul Harvey used to say.
Until the 2019 redefinitions of SI units, the number of "entities" in a mole was determined experimentally. Since 1889 the kilogram was the mass of a particular platinum/iridium cylinder. With the 1961 definition of the unified atomic mass unit, exactly 12 grams of $\ce{^{12}C}$ had exactly 1 mole of carbon atoms. So the experimental measure of a mole was based on how many $\ce{^{12}C}$ atoms would be in exactly 12/1000 of a kilogram.
However in 2019 both the kilogram and the mole were redefined to give them particular values that would no longer depend on experimental measurements. This was in a continuance of an effort by scientists to define a set of primary physical constants that didn't depend on particular physical standards such as scratches on a bar to determine the meter or a blob of metal to determine the kilogram.

The mole is now defined as exactly $6.02214076\times10^{23}$ entities.
The kilogram "is defined in terms of three fundamental physical constants: The speed of light c, a specific atomic transition frequency ΔνCs, and the Planck constant h."$^2$

The point of this is that with the mole and kilogram now being given definite fixed values, exactly one mole of carbon-12 atoms no longer has a mass of exactly 12 grams by definition. Thus now $u≈\pu{1.66053906660(50)\times 10^{−27} kg}$.

To chemists any difference between the pre and post 2019 values will be infinitesimally small. This is because the average atomic masses of the elements depend on the relative abundances of the isotopes of the element which vary. Thus the isotopic variation greatly limits the precision of the average atomic mass of the elements.
To physicists the difference in measurements will make a difference since the mass of a particular isotope is known with much greater precision that the average atomic mass of the element.

tl;dr For practical purposes the three terms are synonymous as user Maurice noted. It would seem that the preferred term is now the dalton.
$^1$ Wikipedia's article on the dalton (Da).
$^2$ Wikipedia's article on the kilogram.


At what point does Avogadro's law not apply?

My feeling is that the size of the molecule matters much less than you would think. For instance, from a purely ideal gas perspective (as Avogadro's law assumes), the expected value for the speed of a particle is $$v=\sqrt{\frac{8kT}{\pi m}}$$
where $k$ is Boltzmann's constant, $T$ is the temperature, and $m$ is the mass of the particle.

So, that means that as the mass of your particle doubles, the average speed of the particle is decreased by a factor of $1/\sqrt{2}$. But, since pressure, from a kinetic molecular theory perspective treats pressure as being the average force which the particles apply to a wall (over a certain area), and force is the time derivative of momentum $p=mv$, we see that $p$ increases by a factor of $\sqrt{2}$ when the mass is doubled (because $v$ decreased by $1/\sqrt{2}$.) That means, the average force on the wall will be greater per collision, but not very much greater.

For instance, I went ahead and calculated the pressure which a mole of hexane exerts on $22.42\ \mathrm{L}$ according to the van der Waals (VDW) equation and found it was $1.05\ \mathrm{atm}$. So, as expected, the value is larger, but not much larger. That's because VDW actually accounts for a parameter (b) which is the volume excluded per mole (meaning the volume that particle actually takes up, sort of).

So, hexane, which is significantly larger than $\ce{CO2}$ or anything else mentioned, only increases pressure by 5% when confined to the same volume. That also means the volume should be 5% larger.

Be wary of scaling that to larger molecules though! As hydrocarbons get larger (and being linear helps) they have a much greater polarizable volume which results in (how about this for a name) greater van der Waals attractions. This extra attraction counteracts the increase in size of the molecule a pretty significant amount.

All that to say, the deviation from Avogadro's Law at SATP is actually quite small.

You're only going to get noticeable deviations at SATP when you have very polar molecules, even then, however, it's not too bad. Water for instance takes up $22.45\ \mathrm{L}$ at SATP which isn't even deviating in the direction we would expect, but if you increase the pressure to $1.1\ \mathrm{bar}$, then the volume occupied for water is $22.33\ \mathrm{L}$.

Intermolecular forces are more significant than the particle's size, particularly at lower temperatures and higher pressures.


What does the number of orbitals in a subshell represent?

Short answer

Number of orbitals does not have direct relation to either probability of  electrons occupying them, either the space they are describing.

The related probabibility is to be understood as:

IF the electron is in the quantum state, belonging to the particular orbital, THEN the wave function of the orbital describes spatial distribution of the electron  occurence probability, what is geometrically illustrated by the orbital 3D shape.

Longer answer
There are 3 meanings of the term orbital:

A complex wave function $\Psi$ being a particular solution of the quantum wave equation
A quantum state of an electron, following 1., described by particular quantum numbers $n$, $l$ and $m$, representing discrete values of electron energy, orbital angular momentum and one of component of angular momentum, respectively.
3D shape, following 1. and 2., representing statistical  probability of the electron presence at given point by$\Psi^2$, determined by the conventional value by residual external probability and by the isoprobability surface.

If there is 1 orbital 2s and 3 orbitals 2p, it means,
for  $n=2$  and $l=0$, there is 1 value $m=0$.
But for $n=2$  and $l=1$,
there are 3 values $m=-1,0,+1$

Electrons are fermions with half integer 4th spin quantum number $m_s$, and every electron in the atom must have unique set of quantum numbers.

Therefore there can be 2 electrons in the orbital 2s, having $n=2$, $l=0$.

But there can be up to 6 electrons in 3 2p orbitals, having $n=2$, $l=1$.


Why does increasing the temperature of a *solid* increase its entropy?

All right, someone bearing the standard of thermodynamics will give you the equations shortly... From a layman to another, here goes my attempt at a simpler explanation.

Entropy may be seen as the "disorderlyness" in some settings, but that is not a very useful way of seeing it. The metaphor is often used, but creates the wrong conclusions when looking closer at it. To me, is best to look at it as a quantity closely related to the number of possible (different) states available to a certain system. It is then obvious that if you increase the temperature, more modes of vibration are available, thus more states, ergo more entropy. The more stuff is in your room,  the less probable is the state where everything is ordered - if all states are equal in probability ;)


What is the largest non-polymeric heterostructured molecule possible?

We know of no inherent limitation concerning the size of a molecule. As you pointed out, certain macroscopic entities like diamonds, graphite or charcoal can be considered fully macroscopic molecules to a certain extent (especially diamond). If chemists let their imagination run loose, they can draw whatever size a molecule you wish to create and with the desired complexity.

Whether it can be synthesised is another matter. Even very small molecules, especially natural products, can be very resistant against attempts to synthesise them. The larger a molecule gets, the more mass a rather small sample will have. It will get increasingly tedious but by no means impossible. However, unless you have oil money, Silicon Valley money or a billionaire funding you, you will likely not get the chance to finish what you want.

One of the most well-known large and not inherently repeating (or so Nicolaou says) molecules is maitotoxin whose total synthesis is famously still being attempted by the Nicolaou group.


Why isn't 'chemically-strengthened glass' made with potassium carbonate to begin with?

The potassium is not added first because the potassium does not intrinsically make stronger glass, it is the substitution of a larger ion for a smaller one that does.

To understand why ion exchange strengthens glass, you have to understand why the ion exchange makes the glass harder. The process of ion exchange hardening is done at a temperature that allows ion diffusion, but disallows a reconfiguration of the glass structure or relaxation of bonds (i.e. below $T_{g}$, the glass transition temperature). When sodium atoms are replaced by potassium in glass, the potassium ions occupy a site that is sized for sodium; this mismatch creates the compression that in turn gives the glass its scratch resistance. Thus the only way to chemically harden the glass is to substitute larger ions into the glass.

If you make a glass with potassium, you have potassium ions occupying potassium sites, thus no strengthening occurs.


Is there a database on chemical reactions, similar to NIST, but far more complete?

The best known databases are Beilstein and Chemical Abstracts with around 22 millions reactions each since the end of the 19th century but they are commercial.

Organic Syntheses has made available its database for free and it covers 6000 reactions. One of the greatest advantages is that each step is described in depth and each reaction has been thoroughly tested prior to publication.

Have a look here for a list of databases.