An introduction to "oxidation state"/"oxidation number"

#TL;DR;

Oxidation States are useful tools used in nomenclature, redox chemistry etc.
There are a few simplistic rules to determine the oxidation state of elements in an atom; but they're not applicable to all the cases.
There are instances where the assignment of oxidation state isn't useful, or where the oxidation state is ambiguous.
We're easier off assigning, rounding off or estimating oxidation states ourselves in some cases, such as intermetallic compounds.

##What is an oxidation state?
Oxidation state (OS), or oxidation number, is an integer or decimal number assigned to an element in a chemical species. This assignment follows certain rules and guidelines, to make it a useful tool whenever we need it. Oxidation states can be negative, positive, or zero.
According to Comprehensive definition of oxidation state
(IUPAC Recommendations 2016), p 833:

The oxidation state of an atom is the charge of this atom after ionic approximation of its heteronuclear
bonds.

All those fancy words up there mean that, when you're assigning OS, you take all of the bonds in the species to be completely ionic. No purely ionic compound exists, and even, say, the $\ce{Cs-F}$ bond in $\ce{CsF}$ has some covalent character. So you have to understand that OS does not happen. It doesn't refer to any real-life phenomenon.
##Why do we assign OS's, then?
Oxidation states are a powerful tool for chemists, and save us a lot of headache. You might question "why do we need to give an element a number anyway?" which arises because you usually only learn how to give elements those numbers, but no real-life application becomes necessary or evident.
OS is a tool, and is not the only tool, wherever it's being used. It's like the shovel that digs the ground. You can also do the digging with a spoon. Why do you choose a shovel?
#Assigning OS
##The Basis
The basis of assigning OS, in simpler cases, is just good old electronegativity. There are exceptions where this would not work, but you don't need to worry about them at this stage.
You're usually given the values of electronegativities, and those aren't what you should worry yourself about. There are different scales for measuring electronegativity, and IUPAC recommends the Allen scale because it's the only scale independent of OS.
##OS Assignment Using the (Molecular) Formula
You're given a formula of a compound you've never met, and you're asked to determine the OS of some, or all of the elements present. This is a classic scare if you don't understand what you're doing.

Determine the OS of all the elements in $\ce{H2C2O4}$.

Like most of the problems you face, there are some "unknown" and some "known" values. You need to identify a starting point, go on from there until everything sorts itself out.
Here's what we know: We need to use the electronegativities to find out what the OS's are. O is the most electronegative, C is next, and H is the least electronegative.
Here's what we want to know: We have three OS's here that we need to determine. C, H and O.
As it happens, the elements in the rightmost and leftmost parts of the periodic table are the most and least electronegative elements of all, respectively. So we expect their stance to be more clear: The ones on the left, the alkaline metals, tend to lose electrons and the ones on the right, tend to gain. As a general rule of thumb, the closer an element is to the middle of periodic table, the fuzzier and harder determining its OS is.

Always try to remember the location of the element in the periodic table, and start from one that's the rightmost and/or the leftmost. Don't try to quantify this rule, since it's merely a guideline. Sometimes, deciding the OS of oxygen becomes difficult!
Thus, we would expect oxygen or hydrogen to be easier to determine here. That is the case: the electronegativities tell us that oxygen must have a negative OS, and hydrogen a positive one.
Time to quantify this. You remember from the intro to electronic configurations that O's electronic configuration ends in $\mathrm{2p^4}$, and that if it ended in $\mathrm{2p^6}$, it would've given oxygen a relative amount of stability. It would've done the same here if it could grab the two electrons from the carbon, but in reality, that hasn't happened. But anyway, when assigning OS, you assume that the bonds are ionic, and the oxygen has already stolen carbon's two electrons.
If a neutral oxygen with an electronic configuration ending in $\mathrm{2p^4}$ gains electrons to become one ending $\mathrm{2p^6}$, how many electrons has it gained? Two. Thus, the OS of oxygen in $\ce{H2C2O4}$ is −2.
You can apply the same process for hydrogen. It has only one electron in its neutral form, and it's about to lose it. How many electrons can it lose?
Determining carbon's OS is more tricky. Carbon is in group 14, and it's not obvious what its OS should be. It's the closest element to the middle. Here's a workaround: Instead of trying to find out which OS's carbon can have, we calculate how much is left as carbon's share.
$$\color{green}{\text{The sum of the OS's in the species is equal to the charge} \\ \text{of the species, be it a molecule, an ion etc.}}$$
This is a rule you have probably also met in your textbook. The charge of the species is an indication of excess or lack of electrons, and should definitely affect what OS's we assign to the elements. Why? Because OS is itself some form of counting electrons.
($\Rightarrow$ if they're just counting electrons, why do we see fractional oxidation states? See OS Assignment Using the Lewis Structure below)
$\ce{H2C2O4}$, oxalic acid, is a neutral molecule. So the sum of OS's must equal zero. Note that we count hydrogen twice, because two hydrogens have two electrons in neutral form, and they will lose two electrons. Likewise, four oxygens will, in a hypothetical purely ionic oxalic acid, gain eight electrons overall.
$$\begin{equation}
2(+1) + 2(\mathrm{OS}_\ce{C}) + 4(-2) = 0 \\
\mathrm{OS}_\ce{C} = +3
\end{equation}$$
##Most Common Patterns in the OS of elements
If you consider all the possible species that can be synthesized, there are great many bizarre trends and anomalies in OS assignment. Refer to the Further Reading section to learn more about them.
However, for most educational purposes, OS's for a number of elements follow simple guidelines worth mentioning:

Fluorine's OS is always −1 (except in $\ce{F2}$ itself).
Halogens below fluorine typically adapt a −1 OS as well, but always check when there are more electronegative elements such as fluorine and oxygen present.
Alkaline metals adopt a +1 OS.
Alkaline earth metals adopt a +2 OS.
You're almost always better off determining the OS of transition metals using the sum rule.
If oxygen is not the leftmost element present, or there are alkaline and alkaline Earth metals present, don't start by assigning an OS to oxygen.
If there are any of the elements in guidelines 1–4; don't start by assigning an OS to hydrogen.

Determine the OS of all the elements in $\ce{NaO3}$.

There are only two possibilities: Start from Na, or start from O. Both are, as said, in left and right parts of the periodic table. So which?
If we take oxygen, and assign the "normal" OS, −2, we see that Na has to be +6 because the sum should equal zero. Something doesn't sound right. As guideline 6 says, we shouldn't start from O when there's Na present in the compound. So we take the other route.
$$\begin{equation}
+1 + 3(\mathrm{OS}_\ce{O}) = 0 \\
(\mathrm{OS}_\ce{O})= \frac{-1}3
\end{equation}$$
###Check your understanding
Find the OS of all the elements in these compounds:

1.$\ce{NaAlF6}$  2. $\ce{C2O2F2}$
3. $\ce{C8O8}$ 4. $\ce{Au2O3}$

##OS's That Don't Exist
The upper bound for OS's that you will meet is +7 for some manganese $(\ce{MnO4-})$, or chlorine $(\ce{ClO4-})$ compounds. OS's more than that are only found in exotic compounds. Thus, if you get a, say, +9 OS, you should start doubting the route you took.

Determine the OS of W in $\ce{H4W2O13}$.

Three possible starting points. We rule out tungsten because it's the nearest to the middle. Of the other two, hydrogen would lose 1 electron, and oxygen would tend to gain two. Right?
Well, if you solve it for tungsten, considering the neutral overall nature of the compound, you'll get an OS of +11! That's definitely not right, since the highest observed OS is +10 (See Further Reading). This could only mean one thing, since hydrogen can't adapt another OS or tungsten would have an even higher OS. This means that some of the oxygens are in the form of peroxides, superoxides etc.
Had we known that the structure of the compound, more elaborately written, is $\ce{(H2O)(O)(O2)2WOW(O2)2(O)(H2O)}$, we wouldn't have made the mistake of thinking all the oxygens have the normal OS of −2.
Hence, it becomes necessary to be able to deduce OS from structures.
##OS Assignment Using the Lewis Structure
At times you need to determine OS in an electron dot diagram, or Lewis structure. This is sometimes useful/needed over determining from the molecular formula because

either the molecular formula doesn't give us the whole picture (see below), or
it gives us impossible answers or no way to go further, like the tungsten compound's case, or
it's already what the examiners gave us!

Determining OS from Lewis structure is even easier than deducing it from the molecular formula, because you already see the bonds. It's the same process in the IUPAC definition above: Ionic approximation

The electrons in a bond go count as the more electronegative atom's. (There are exceptions to this, see Further Reading)
The electrons in a bond between the same atoms are shared equally. So in a $\ce{C-C}$ bond, each C atom gets one electron from the bond.
The OS of each element is the number of valence electrons minus the sum of lone pairs and electrons it gains from bonds.

A good example used to illustrate this is acetone.

Determine the OS of all the elements in acetone, $\ce{C3H6O}$.

The answer is in the spoiler below (no peeking ಠ_ಠ):

 $\mathrm{OS}_\ce{O} = -2, \mathrm{OS}_\ce{C} = \frac {-4}3, \mathrm{OS}_\ce{H} = +1$

There's one oxygen in acetone, and it's bonded to the carbon:

Assigning the rules, we get

$\ce{C1}$ has four valence electrons, but there are "seven electrons around it". Both electrons from the $\ce{C-H}$ bond belong to carbon. Thus, $4 - 7 = -3$.
$\ce{C2}$ has four valence electrons, but only two electrons are around it. One from of each of the two $\ce{C-C}$ bonds. That makes $4 - 2 = +2$.
$\ce{C3}$ is just like $\ce{C1}$, $\mathrm{OS} = -3$.

##Fractional OS's
Didn't we say earlier that OS assignment is some kind of counting electrons? Then how come we encounter fractional OS?
Now you know the answer. Some elements in some compounds show two different oxidation states, and when computing the OS from the molecular formula, you actually compute the average OS. None of the carbons in acetone have an OS of $\frac {-4}3$.
Familiar examples of this 'average' OS are carbon in acetic acid, which is computed to be zero but is in fact −3 and +3, and iron magnetite, which is $\frac 83$.
##Further Reading
The school lessons are over, but the fun is actually just about to start. This is the part where, you know, you see what you've been taught isn't accurate. Read on if you're interested. It might get confusing for first-timers.
###Be Liberal in Thinking
OS is a useful concept. Unfortunately, it's hypothetical. There's this mindset that there should only be one OS for each element in a compound out there. That's simply not the case.
$\ce{S2O3^2-}$, thiosulfate anion, is one such compound where the two approaches we just introduced produce conflicting results. The average $\mathrm{OS}_\ce{S}$ is +2 for sure, but is it +5 and −1, or +6 and −2?
Turns out, studies are conflicting on this matter. The final say here is provided by IUPAC recommendations linked above:

###5.3 Nominal OS
In systematic descriptive chemistry, OS is used to sort out compounds of an element; in electrochemistry, it represents the electrochemically relevant compound or ion in Latimer and Frost diagrams of standard (reduction) potentials. Such purpose-oriented OS may differ from the OS per definition and are termed nominal, or more specifically “systematic” or “electrochemical”.

Consider thiosulfate, the $\ce{S2O3^2−}$ anion: The sulfur–sulfur
bond is practically a single bond, which yields OS = +5 for the central S and −1 for the terminal S atom by both algorithms. The electrochemical OS of sulfur in thiosulfate is their average, +2. On the other hand, a systematic OS of +6 may be chosen for the central S atom, as if the sulfur–sulfur bond were approximated ionic, to emphasize the similarity of the peripheral O and S atoms obtaining OS = −2 as in sulfate.

##Exceptions
These are the cases where the methods above don't reliably work in determining OS:

The most glaring example is when the more electronegative atom is bonded reversibly as a Lewis acid.
We originally introduced electronegativity as the basis to decide OS. Truth is, the real basis is the energy level of molecular orbitals, and this, in most of the cases, corresponds to the more electronegative atom getting all the fish. It's difficult and/or impractical to use complicated MO schemes to determine OS, so we stick to electronegativities.
In the case of Lewis-acid ligands, however, this backfires and gives us the opposite of the correct answer.
Non-innocent ligands are ligands whose OS is ambiguous because they're very redox-active. This article does a nice job of explaining $\ce{H2}$ and nitrosyl non-innocent ligand briefly (§ 10 & 11).
Nominal OS's; like thiosulfate above.
Metallic compounds; there are some obvious cases such as TiO, but it gets more complicated – Source
Intermetallic compounds, such as Zintl phases, where it's safe to assume an OS of 0 for redox purposes or compounds with delocalization over different atoms  – Source (§ 13)
Compounds with a close electronegativity such as Se and C in $\ce{Se-C}$ bonds or H and P in $\ce{H3PO3}$. – Source

##Exotic OS
These are compounds whose certain elements demonstrate unusual OS's:

Alkalides, which are salts containing alkali metal anions with an OS of −1. Lithides are currently not known, but some kalides and natrides are stable in room temperature.
The highest known OS belongs to Ir in $\ce{[IrO4]+}$ which is +9, and potentially +10 for Pt in $\ce{[PtO4]^2+}$. The lowest known OS is −5 for boron in Zintl phases. – Wikipedia
Platinum hexafluoride is a strong oxidizing agent that can oxidize $\ce{O2}$, giving dioxygenyl, with an unusual OS of $+\frac 12$ for O.
Alkaline earth metals are observed to show unusual +1 OS in some complexes. – This article reports the synthesis of complexes with Mg–Mg bonds.

##Useful Links
[1]: List of Oxidation States of Elements
[2]: High-valent Iron
[3]: Using OS to analyze redox reactions; a tutorial
[4]: ChemGuide OS tutorial
<Under construction . . . Probably>
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Depicting a mechanism of peptide bond formation with protonated and deprotonated amino group

In neutral aqueous solution, the amino group is protonated (3 bound hydrogen atoms ) and the carboxilic acid group deprotonated (no bound hydrogen atom). In other solvents, you might have both groups in the neutral form the name amino acid suggests.
The mechanism of peptide bond formation is complex and involves intermediates with tRNA in the cell and starting materials distinct from amino acids for synthesis in the lab.


Can lead-acid batteries release CO? Or can a CO sensor detect gasses other than CO?

You're probably picking up hydrogen gas, which is produced when lead-acid batteries are overcharged at high charging voltages (a danger in its own right). This article details a situation similar to yours: charging a lead acid battery in a golf cart (in a confined space) sets off a $\ce{CO}$ alarm, and typical sensors are activated by $\ce{CO}$ at levels of 150 ppm for 30 minutes and/or 300 ppm $\ce{H2}$ for 30 minutes. Because $\ce{CO}$ is not produced by battery charging, the offending gas is thus identified as hydrogen. After venting, the levels recede.
Also see this post from Hawk Environmental, which details how hydrogen gas can create false positives for carbon monoxide. This post goes into a little more detail about the hydrogen/carbon monoxide sensor issue, and the author states that:

Lead-acid batteries produce Hydrogen when charging.
Carbon Monoxide detectors use something called a "Metal Oxide Semiconductor (MOS)" sensor, which detects a variety of gases including Hydrogen.
A MOS sensor calibrated for CO will give a false positive in the presence of Hydrogen gas at ~10% of the actual value. So, that ~85ppm reported of CO was in reality ~850ppm of Hydrogen gas.

Finally, in an article in the September 1996 issue of Fire Engineering, entitled "Other gases may set off $\ce{CO}$ detectors" (here), it is stated that

CO detectors are electrochemical sensors, which detect a variety of specific gases. In most cases, the gas of interest reacts with the chemical solution in the sensing electrode to produce ions (charged particles) that move through the solution to an electron-accepting electrode, completing an electric circuit. The number of ions produced and the magnitude of the current developed are proportional to the concentration of the gas being measured.

and that

Interfering gases of similar molecular size and chemical reactivity may produce false positive response. Major interferents that can affect $\ce{CO}$ detectors are acetylene, dimethyl sulfide, ethyl alcohol, ethylene, hydrogen cyanide, hydrogen sulfide, isopropyl alcohol, mercaptan, methyl alcohol, propane, nitrogen dioxide, and sulfur dioxide.

I suggest you confirm/deny the above by contacting Fluke and asking them specifically if the sensor does not detect hydrogen gas at all and, if so, at what level. I strongly suspect that the well-documented behavior I've found online and summarized here is the cause of what you have observed in this case.


The test is better suited for $\ce{Fe^3+}$.
$$\ce{Fe^3+ + [Fe(CN)6]^4- -> Fe4[Fe(CN)6]3}$$

The compound $\ce{Fe4[Fe(CN)6]3}$ has a specific name. Its called Prussian Blue.

If you perform the test for $\ce{Fe^2+}$, the precipitate won't be that dark. First partial conversion of $\ce{Fe^2+}$ to $\ce{Fe^3+}$ takes place. Then the above reaction takes place to give the prussian blue precipitate.

There are specific tests for $\ce{Fe^2+}$ as well. You can use Dipyridyl to get a red colouration which is due to the formation of $\ce{[Fe(dipy)]^2+}$.
Or, you can use Dimethylglyoxime to get red colouration again which is due to $\ce{[Fe(DMG)2]}$.


Is it possible to melt diamonds into a liquid?

Liquid carbon does indeed exist, but perhaps surprisingly, relatively little is known about it. It only exists above around $4000\ \mathrm{K}$ and $100\ \mathrm{atm}$, which are not trivial conditions to sustain and probe. There certainly are many theoretical studies into the properties of liquid carbon, though. You can find a phase diagram for carbon here, though it is likely just a rough guide. The specific bonding properties of carbon atoms in the liquid phase still seems uncertain to some extent because carbon is very versatile and can create very dissimilar structures (easily seen comparing its two most common phases, graphite and diamond). This source suggests that bonding in the liquid varies continuously between linear acetylenic chains, graphite-like and diamond-like structures, depending on the exact conditions.

Edit: Now I realize that obtaining liquid carbon and melting diamonds are two slightly different things. Liquid carbon is trivial to obtain compared to melting a diamond. As you mention, heating diamond at relatively low pressures will cause it to convert into graphite before it could melt. The only way to inhibit that would be to exploit the slight difference in density between graphite and diamond; as diamond is denser, the increased pressure will stabilize it slightly, decreasing the tendency for diamond to revert back to graphite.

You'll need a lot of pressure though! Rigorously speaking, according to the phase diagram above, to obtain liquid carbon directly from melting a diamond would require a pressure of at least $10\ \mathrm{GPa}$, which is about $10^5\ \mathrm{atm}$. To put that into perspective, it's about $1/30$ of the pressure at the centre of the Earth. Amazingly, we actually can reach such absurd pressures in the lab by compressing two diamonds against each other. However, if the conditions required conditions for melting diamond (and sustaining it in molten form) are reached, it would probably compromise the integrity of the instrument and cause catastrophic (possibly explosive) failure, so I don't think anyone will be purposefully melting diamonds any time soon! Other techniques could be used (laser inertial confinement, light-gas guns, etc), but I believe most would only transiently produce liquid diamond.


You drew the radical, which is in normal cases very unstable.
The cyanide ion has an ion pair instead of the single electron at the carbon atom.


How do potentiometric cells do not proceed through a spontaneous reaction like in a galvanic cell?

First, I wouldn't classify cells as potentiometric or galvanic.

If you take any electrochemical cell with non-zero total cell potential and connect a wire between the two electrodes it will discharge. This will happen spontaneously in the galvanic direction.

In potentiometric measurements of electrochemical cells, what you do is a potential measurement which ideally happens at zero current(definition of a voltmeter), and practically at picoampere or lower current. Therefore you are not allowing the electrons to flow from the outside circuit and hence your cell doesn't discharge.


Coordination complex - nitrito-N & nitrito-O

Linkage isomerism is of major interest in the field of inorganic chemistry. Linkage isomerism occurs when an ambidentate ligand such as $\ce{NO2, SCN,}$ etc. binds to the metal center. Nitro-nitrito metal complexes can be regarded as good examples of linkage isomerism. It must be noted that the $\ce{NO2}$ group can be coordinated to the metal atom in different ways such as nitro (via the nitrogen), nitrito (via oxygen), chelating via both oxygens (nitrito-O,O′), and bridging nitro, usually via nitrogen and oxygen atoms (Ref.1):

Source of diagram: Modified from Ref.1
Amongst various coordination modes, above diagram shows not only $\ce{NO2}$ is capable of having the wide range of binding modes in mononuclear metal complexes, but also its potential tendency to act as a bidentate ligand in some cases.

Will I get a 50-50 mixture of nitrito-N & nitrito-O co-coordination complexes? Or will there be a chance of getting random combinations of nitrito-N & nitrito-O coordinate bonds to a Co3+ (e.g. 4 nitrito-O + 2 nitrito-N)?

Not necessarily! It is not easy (at least not as easy as OP described it) to synthesize specific linkage isomer. It has been shown that the coordination mode of nitrite ion depends on its nature and the stereochemical environment which surrounds the metal ion (Ref.1 & references there in).

If I just want to get nitrito-N complex, are there ways to prevent the formation of nitrito-O complex? Or the mixture must form and then we need to do some other reactions to get rid of nitrito-O complexes?

These specific compound have been synthesized. Thus, there should be ways to such synthesis. However, I'm not qualified to give you a definitive answer for that question. Inorganic coordination chemistry expert would be the most qualified one. Yet, references in here including Ref.1-3, and references there in would give you better view. For instance, Ref.3 has described a ways to synthesize $\ce{Cu(NO2)2}$ and $\ce{Mg(NO2)2}$ complexes. Both metal showed nitrito-O,O′ (as bidentate) coordination with different stabilities.

References:

Hoda Pasdar, Bahare Hedayati Saghavaz, Naser Foroughifar, and Mehran Davallo, "Synthesis, Characterization and Antibacterial Activity of Novel 1,3-Diethyl-1,3-bis(4-nitrophenyl)urea and Its Metal(II) Complexes," Molecules 2017, 22(12), 2125 (12 pages) (DOI: https://doi.org/10.3390/molecules22122125).
Andrew J. Timmons and Mark D. Symes, "Converting between the oxides of nitrogen using metal–ligand coordination complexes," Chem. Soc. Rev. 2015, 44, 6708-6722 (DOI: 10.1039/C5CS00269A).
Andrew J. Timmons and Mark D. Symes, "Coordination of $\ce{NO2}$ to $\ce{Cu}$ and $\ce{Mg}$ in $\ce{M(NO2)2}$ Complexes. A Theoretical Study," Chem. Soc. Rev. 2015, 44, 6708-6722 (DOI: 10.1039/C5CS00269A).


The notation, upon replacing potassium with sodium, can refer to a water glass composition. The unique notation, in my opinion, is warranted based on the following comment from a paper1:

Despite of years of investigations, there are still questions about
the molecular species configuration of these solutions that need to be
answered to gain clarity over the best conditions for the various
applications
When adding sodium hydroxide ($\ce{NaOH}$) to a water glass, the ratio (molar
or weight) $\ce{SiO2/Na2O}$ is decreased. This ratio is called water glass
modulus (n) and determines various physical and chemical properties
such as the pH and the viscosity of the solution.
[...]
the viscosity is extremely dependent on the modulus and increases as
the solution becomes either more siliceous or more alkaline, i.e. at
both higher and lower $\ce{SiO2/Na2O}$ molar ratios (n). Since the viscosity
of such disperse systems like this is given by the silicate
conformation i.e. extended chain conformation and the degree of
polymerization
[...]
the degree of polymerization of the predominant silicate species in
the alkali-solution [...]

So we have apparently a weakly specified structure characterized by extended chain conformation and degrees of polymerization with its rather unique notation (seems right to me).
Reference

Helén Jansson, Diana Bernin, and Kerstin Ramser , "Silicate species of water glass and insights for alkali-activated green cement", AIP Advances 5, 067167 (2015), DOI: 10.1063/1.4923371


Why are Terminal Alkynes less reactive than internal alkynes?

The mechanism of hydration is an electrophilic addition and requires polarization of the carbon-carbon triple bond. This task is more difficult compared to case of alkene due to it having more %s character
In case of internal alkynes the carbocation intermediate formed ( since this mechanism is electrophilic addition the intermediate is a carbocation) is stabilized by the hyperconjugative effect and the inductive effect of the alpha carbons.
This support is taken away in case of terminal alkynes as their is no alpha-carbon. Now terminal alkynes can still be hydrated but only in presence of $\ce{Hg^2+}$ ion in acidic medium because of $\ce{Hg^2+}$'s high polarizing power(because it has a pseudo inert configuration). This polarization succeeds in catalyzing the reaction.
Ref.: Morrison and Boyd, sec 12.10 (hydration of alkynes)


What is the CNC bond angle in methyl isocyanate?

This is not a good exercise question, because none of the options is actually correct. It basically asks you to blindly apply a model which cannot yield the correct result.
Before continuing any deeper into the matter, let's have a look at the actual structure. I was not able to find a crystal structure, so I ran a quick optimisation on the RI-BP86/def2-SVP level of theory. For this purpose, this is accurate enough.

The bond angle $\angle(\ce{CNC}) = 144^\circ$ is right in between $120^\circ \dots 180^\circ$.
The (pure) VSEPR model must fail for this molecule, as it is  only really working for central atoms with distinct and non-delocalised ligands.
For simple cases it performs very well in describing the approximate shape of a molecule. Once you have gotten this down, you can refine predictions with other tools. Please note, the lone pairs take up more space argument is actually a post-rationalisation and should not be used within the framework of VSEPR theory. More reliable predictions can be made with Bent's rule, which says that more electronegative ligands will concentrate p-orbital contributions towards them, decreasing the bond angle (this is a very, very brief summary).
As far as the VSEPR model goes, one should concentrate on simple coordination patterns:

linear; the bond angles will be close to $180^\circ$
trigonal; the bond angles will be close to $120^\circ$
tetrahedral; the bond angles will be close to $109^\circ$
higher coordinations like trigonal bipyramidal, octahedral, etc.

(TL;DR) From that we see that the central nitrogen has three ligands, so the general coordination around it is trigonal and the bond angle should be close to $120^\circ$. This is the answer, which your book will accept as correct, because it blindly applies the VSEPR model.
Applying a bit of Bent's rule won't help much. Carbon is slightly less electronegative than nitrogen, hence one would expect a slight opening up of the angle (more s-character directed towards carbon), but that's not enough to push it to $144^\circ$.
More importantly, the $\ce{-NCO}$ group is highly delocalised. Another very important concept comes into play: resonance. Ignoring the underlying structure for a moment, these are the most important contributors:
$$\ce{Me-N=C=O <-> Me-N^+\bond{3}C-O^-}\tag1\label{eq:resonance}$$
Within the concept of resonance, the positions of the nuclei will not change, and the structures do not exist on its own. A single structure of a set is therefore always an incomplete description. Please read What is resonance, and are resonance structures real?
The left structure in \eqref{eq:resonance} is basically what we have described above with VSEPR. From the depiction on the right, we would expect a $\angle(\ce{CNC}$ of about $180^\circ$, because triple bonds are coordinated linear. Since these are electronic descriptions of the same structure, we can expect the actual (underlying) structure to be somewhere in between these hypothetical forms.
The observed molecular structure is always the result of a delicate balance of effects. So repulsion of nuclei, electronic repulsion, electronic attraction, dispersion, etc. will all have an effect. Many of these interact quite complicated with each other. Sometimes our used models will yield good approximations, sometimes they don't.
In this particular case, we see that VSEPR alone fails; including resonance will refine the prediction. However, without a more rigorous approach, it is impossible to predict the bond angle correctly, i.e. we do not know the contribution of the right structure without doing some more maths. I used a quantum chemical approach, which is obviously nothing you could do in an exam. This is one of the reasons why I dislike this question.

A quick word on hybridisation:
The concept of hybridisation is independent of VSEPR. Many books will throw them together, but it is actually incorrect. You can make a prediction of the shape of a molecule with VSEPR. After that, you can use the molecular structure to choose the closest matching hybridisation model (if any at all). As such, hybridisation is always a result from a particular structure, never a cause. Therefore, your initial assertion is reverse:

My current assumption is that the $\ce{e-}$ in nitrogen in $\ce{CH3NCO}$ forms sp2 hybridization (and therefore should forms a shape like ethene).

Correct is, because the structure around nitrogen is trigonal, the electronic structure can be approximated with sp2 orbitals.

Why $\angle\ce{CNC} \approx 104^\circ$ cannot be correct
In order for this angle to be achieved, the nitrogen first needs to be surrounded by four ligands (or electron pairs), giving us an angle of about $109^\circ$ to start with. Then more p-character needs to go into the bonds, reducing the angle. Hence the ligands must be more electronegative than the other ligands. A popular example is dichloromethane,  or even water (but that is a bit more complicated). See the Wikipedia article on Bent's rule, our Q&A Utility of Bent's Rule - What can Bent's rule explain that other qualitative considerations cannot?, and multiple other questions with regard to Bent's rule.


DSVA's comment pointed me to a database that contains the information I need. The wikipedia articles contain links to an app called jsmol, but that app appears to be getting its data from a database called ChEBI. You can search the database using search terms such as "formaldehyde" or "sf6." The entry for each molecule contains a link to a .mol file, which avogadro can read.


TL;DR
$\rm pH=6$ is the neutral pH at $\rm 50\,^{\circ} C.$
Long Version
Actually there is nothing wrong with your calculations. It seems that you are confused about the definition of a solution being acidic. If something is acidic, this means that the concentration of $\ce{H+}$ ions (technically $\ce{H3O+}$ ions) is greater than the concentration of $\ce{OH-}$ ions. At room temperature (25 degrees Celsius) a solution is neutral if its pH is 7. This means that for a solution to acidic, its pH must be lower than 7 at 25 degrees.
When we increase the temperature to 50 degrees, the pH of a neutral solution is actually 6, not 7. You can check this as if you were find the total concentration of $\ce{H+}$ ions and $\ce{OH-}$ ions, you will find that they both equal $10^{-6}$. Since there concentration is equal, the solution must be neutral. So for a solution to be acidic at a temperature of 50 degrees, its pH must be lower than 6, not 7.
Explanation
Now you might wonder what the reason for this. Consider the following equation which is the auto-disassociation of water $$\ce{2H2O + heat \leftrightharpoons H3O+ + OH-}$$ As you can see, this reaction is actually endothermic. This means that when you increase the temperature, due to Le Chatelier's principle, the equilibrium will shift to the right; hence the concentration of $\ce{H3O+}$ and $\ce{OH-}$ ions will increase. This is indicated by the increase in the $\rm K_w$ value, as more products are formed.
Therefore since there is a greater of concentration of $\ce{H3O+}$, a solution which is neutral will have more $\ce{H3O+}$ ions at 50 degrees than it would when it is 25 degrees. Therefore its pH will be lower than 7 despite being neutral.


Why do the names of most chemical elements end with -um or -ium?

To expand on @BelieveInvis's answer -- in the early 19th century, when the Royal Society was really in the swing of things, the dominant language of scholarship was still Latin. Since Latin didn't have words for the new metallic elements, new words were coined from the existing terms for the substances and given Latinate endings.

From the OED's entry on -ium:

The Latin names of metals were in -um, e.g. aurum, argentum, ferrum; the names of sodium, potassium, and magnesium, derived from soda, potassa or potash, and magnesia, were given by Davy in 1807, with the derivative form -ium; and although some of the later metals have received names in -um, the general form is in -ium, as in cadmium, iridium, lithium, osmium, palladium, rhodium, titanium, uranium; in conformity with which aluminum has been altered to aluminium.

So, I think after that, other elements were simply given the suffix to fit the generally useful naming scheme, and then, metal names which were already in common use kept their common language names (e.g. gold as opposed to aurum) simply by force of usage.


Which species dissociate completely in ionic equations?

The second equation you’ve given appears to be the right overall ionic equation. Phosphate is a weak acid, $\ce{NaOH}$ is a strong base. We expect NaOH to dissociate at any pH, where the speciation of phosphate ($\ce{H3PO4}$, $\ce{H2PO4-}$, $\ce{HPO4^2-}$, and $\ce{PO4^3-}$) depends on pH.

A “net ionic equation” doesn’t include spectator ions. Spectator ions exist on both sides of the arrow and can be “canceled like” (think as in algebra). For any acid base reaction, it’s really easy to remember!

$$\ce{H+ {(aq)} + OH- {(aq)} <=> H2O {(l)}}$$

Lastly don’t forget the state: $\ce{OH {(aq)}-}$. I was taught to right them as subscripts, but I’ve seen then as regular text too: $\ce{OH- (aq)}$.


In one of his great answers, Martin gave us the calculated molecular orbitals for carbon monoxide I am copying below. Note that while cyanide has a nitrogen atom in place of the oxygen and features a single negative charge, we can still qualitatively consider the MO schemes identical; thus I will base my discussion on what you see in the image.

In this scheme, the HOMO is referred to as MO number 7; since 5 and 6 are π-type MOs, this corresponds to the 5σ you mentioned.

One thing we immediately note is that the carbon-centred lobe of the HOMO is very diffuse and extends far into space; much farther than any other orbital of the molecule. It is also the orbital that an acid (either proton or Lewis acid) would bond to. In Pearson’s HSAB theory, large and diffuse orbitals — often also called polarisable — are associated with soft acids or bases, as the charge per volume value is considered comparatively low. Therefore, just by looking at the size of this orbital, we can identify it as a ‘soft’ one.

It should also be explained why soft bases tend to create more stable complexes with soft acids. This is a direct consequence of the more diffuse nature of the orbital: the overlap between two orbitals (which will later form a bond) is typically best if the orbitals have similar shapes and sizes. Thus, because the HOMO has a rather diffuse lobe on carbon’s side it will prefer binding to other compounds with rather diffuse orbitals in a Lewis acid-base reaction.

That should be all that is required to explain cyanide’s behaviour under Pearson’s theory.


Need to find series of equation products from given reagents

I will assume that all of these reactions are happening in solution. Since we've established that all these reactions are redox reactions, we know the general structure of each reaction--- $$\text{reducing agent} + \text{oxidizing agent} \to \text{oxidized thing} + \text{reduced thing}$$ ---and we generally won't have to worry about other things (like what you've mentioned in your edit) happening. I will work out one reaction, and leave you with the rest. Let's look at reaction 4.

We can ignore $\ce{Na+}$ and $\ce{K+}$, because we know from prior experience that these are relatively inert spectator ions that won't participate in most reactions. Then we have the three species $\ce{S2O5^2-}$, $\ce{NO3^-}$, and $\ce{C12H22O11}$.

$\ce{N}$ has its maximum oxidation state of $+5$ in $\ce{NO3^-}$, so we expect that $\ce{NO3^-}$ will be a good oxidizing agent. $\ce{NO3^-}$ is commonly reduced to $\ce{NO2^-}$ in solution.
On the other hand,  $\ce{S}$ has an oxidation state of $+4$ in $\ce{S2O5^2-}$ and a maximum oxidation state of $+6$, so we might expect $\ce{S2O5^2-}$ to be a reducing agent. Fully oxidized $\ce{S}$ is usually present as $\ce{SO4^2-}$ in solution.
Finally, $\ce{C}$ also has its maximum oxidation state of $\ce{+4}$ in $\ce{C12H22O11}$, so $\ce{C12H22O11}$ is likely to also be a reducing agent. Fully oxidized $\ce{C}$ is commonly present as $\ce{CO2}$.

This gets us the three half-reactions
$$\begin{aligned} \ce{NO3^- + H2O + 2e- &-> NO2^- + 2OH-}\\ \ce{S2O5^2- + 6OH- &-> 2SO4^2- + 3H2O + 4e^-}\\ \ce{C12H22O11 + 48OH- &-> 12CO2 + 35H2O + 48e-} \end{aligned}$$ where we've also assumed basic conditions and hence added $\ce{H2O}$ and $\ce{OH-}$ as required to balance out charges and atoms on both sides of each equation. Now we simply need to take linear combinations of these half-reactions to cancel out electrons on both sides, and we're done.

In response to your comment, everyone starts out bad at chemistry. You'll get better.


The Workings of Pulse-Height Discrimination

Pule Height Analaysis (PHA) is done on the current pulse from the detector. It only works for detectors which have a current pulse which is proportional to the energy of the x-ray (scintillation counters and gas proportional counters). It does not work for Geiger counters where the current pulse doesn't depend on the x-ray energy.

Think of dropping the current through a resistor. The current pulse then has a voltage. Basically you set a low and a high voltage (a window) for the detector pulses which you will count. Any detector signals within the window are counted, those outside the window (too low or too high) are rejected.

Given that the detector current (or voltage really) should be proportional to the energy of the x-ray detected this improves the S/N ratio. So "low" current pulses don't have enough energy to be a Cu K-alpha x-ray, and high energy pulses have too much energy. High pulses could be for example an x-ray with twice the energy.

It is important to note that the rejected pulses are not "background" counts per se. But there is a twist. The rejected pulses do create detector deadtime. So you have to be careful that the overall count rate of the detector is low enough to still be linear, or do a correction for the total count rate.


The calcium amount may change from one fruit to the next one, or from one day to another one, or from one field to another one. 20 % is not a significant difference between such a measurement and another one. And this may happen for any constituant, not specially for calcium.


Product of Friedel-Crafts cyclisation of (3-methylpent-4-en-1-yl)benzene

Though the question is old and pretty much answered in the comments, I ran into this question and draw the mechanism, so I thought about summarizing all that was brought up in an answer.

The carbocation is first formed on the secondary carbon. However, molecules are more stable when the charges are delocalized (this page talks about the stability of carbocations). When the positive charge is in the secondary carbon, it can only "share the burden" of having this charge with 2 neighbor carbons. So the tertiary carbon donates an hydrogen through a sigma to atomic orbital interaction (the sp3 hybdrization allows a π-type interaction) resulting in the more stable tertiary carbocation seen in the mechanism below:

Then the reaction follows analogously to the pathway you deduced for the 6 membered ring: First a π to atomic orbital interaction between the aromatic ring and the tertiary carbocation, which I believe to be the Rate Determining Step since it breaks the ring aromacity (bigger energy gap). Then finally the sigma to atomic interaction between the hydrogen and the ring carbocation, that restores aromaticity and releases the hydrogen.


How can I work out what reaction will happen?

There are no 'free electrons' that I can see in ethene

That's correct, but...  In ethene there are two bonds between the two carbons that hold the two carbons together.  A chemist calls this a "double" bond.  The important thing about a double bond (that's not obvious from the drawing) is that the two bonds are not the same.  One is a strong sigma bond, just like in ethane; the other is a weaker pi bond.  While the electrons in the pi bond are not free, they are not held as tightly

as the electrons in the sigma bond and are, therefore, more available for reaction.  This is why ethene is so much more reactive than ethane.

  Am I supposed to apply some logic... Or, am I just overthinking it?

Yes, there is logic and that body of logic is what organic chemistry is all about.  In the case of bromine reacting with ethene, the reaction belongs to a class of reactions known as electrophilic additions.  Most reactions in organic chemistry involve a molecule that has electrons available to share (a nucleophile) and a molecule that would like more electrons (an electrophile).  These two molecules react (form a chemical bond) by the one donating\sharing its electrons and the other molecule accepting\sharing those electrons.  Here, the pi bond in ethene has electrons that it is willing to share (they're not free, but they're not tightly held either), while the bromine molecule is highly polarizable, much like $\ce{Br^{\delta +}-Br^{\delta -}}$, with the positively polarized bromine atom wanting electrons.  The reaction proceeds as pictured below, to yield 1,2-dibromoethane.

  A side problem to this is why, if we place an alkene in bromine water,
  we don't get the same thing happening with ethane.

I bet you can answer that one now.  Ethene has a pi bond with electrons loosely held and available for reaction.  Ethane, on the other hand, only has a sigma bond where the electrons are more tightly held and less available for reaction.


Diamond is a covalent network solid, like a number of other common materials (quartz, graphite, glass, and a whole bunch of stuff).

Because they are not discrete molecules - there is no 'diamond' molecule the same way there are molecules of caffeine, benzoic acid, citric acid, N,N-dimethylaminopyridine, etc. - network solids form one of the two main classes of macromolecules, the other being polymers. You'll note that the Wikipedia article on macromolecules seems to imply that 'macromolecule' and 'polymer' are synonymous. They are not, at least not to those chemists working in the field.


Why does the fructose monomer in sucrose appear different from isolated fructose?

Ah, a (fairly) common conundrum that assails us Chemistry students when we start Biochem. ;-)

At first glance, the fructose molecule in your first picture, and that in the second picture appear to have different conformations:

Believe me, they're the same. "How?" you ask? Stay with me here (it requires a little imagination):

Draw a vertical line/axis on your first diagram along the plane of the furanose ring, connecting the oxygen atom at the apex, right down to the middle of the C3-C4 bond. Then rotate the molecule 180 degrees counter-clockwise.

Now compare this with the fructose residue in sucrose.

(I lack the necessary Org. Chem. sketching tools to depict this... apologies)

If you weren't able to wrap your head around "rotation" bit, then, let's assign locant numbers to the carbon atoms in the two diagrams and then compare them:

   I've drawn them tiny, my bad

Must've clicked by now ;)


Dalton's Law is only true for ideal compounds, which theoretically have no intermolecular forces between their molecules. Those forces become more and more significant when increasing the concentration of any compound, because all molecules become packed up closer.

This will lower the pressure, because if any molecule is about to hit the container wall, it gets pulled backwards $-$ and thus slowed $-$ due to the attraction of the other molecules behind it. But for high temperature this effect is negligible, because they can't be slowed down so easily.

Now lets come to the calculation using Kay's rule:

Since you only have one liquid component, you don't need Roult's Law to calculate the vapor pressure for a mixture of liquids.

1.) You need to find the values for the critical temperatures and critical pressures for your components (gases and water vapor).

2.) You need to know the relative amounts of each substance:
$$y_i = \frac{n_i}{n_m}$$

$n_i$... amount of a specific component [mol]

$n_m$... total amount of substance of the mixture [mol]

3.) Calculate the pseudo-critical temperature and pressure:
$$T_c' = y_1 \times T_{c1} + y_2 \times T_{c2} + y_3 \times T_{c3} $$

$$P_c' = y_1\times P_{c1} +y_2\times P_{c2} +y_3\times P_{c3}$$

$T_c$... critical temperature of a specific component [K]

$P_c$... critical pressure of a specific component [Pa]

4.) calculate the pseudo-reduced values for T and the volume:
$$T_r = T / T_c'$$
$$v_r' = V / (R \times T_c' / P_c')$$

$T$... temperature in the container

$V$... volume of the gas phase [m$^3$]

$R$... universal gas constant = 8.314 J/(K mol)

5.) Now you need to look up $v_r$ and $T_r$ in a compressibility chart to get the value for a specific $Z$ factor, where their lines intersect.

6.) And finally calculate the pressure: $$P = ZRT / V$$

This will be more accurate than if you would use the standard way for ideal gases, especially for low temperatures

References:
real gases

calculation

compressibility chart examples


Why don't such ions form if half-filled or full-filled configurations are considered stable?

As also pointed out in the comments, there is no such thing as absolute stability. One species is stable only with respect to another species.

You may have come across the fact - that half-filled configurations are "stable" - while studying ionization enthalpy/electron gain enthalpy, or the electronic configuration of $\ce{3d}$ elements. Pay careful attention: the half-filled configuration is "stable", but only with respect to the neighbouring configurations. That is to say, the $\ce{2p^3}$ configuration is much stable than $\ce{2p^4}$ or $\ce{2p^2}$. Yet, $\ce{2p^3}$ is still very less stable than $\ce{2p^6}$. This is why, in nature, nitrogen prefers to exist in diatomic molecules, so as to complete its octet.

Now for the fully-filled $\ce{3s^2}$ configuration (sodium anion, aka "sodide"): again, note that the sodium cation is way more stable than the sodium anion, as the former has a fully filled octet, while the latter doesn't. (Sodide has indeed been formed, but by using special techniques only. See Wikipedia).

Also see: Why are atoms with eight electrons in the outer shell extremely stable? and Why do atoms "want" to have a full outer shell?


Orbital correlation diagram for the cleavage of cyclopropylmethylene carbene

What is the type of the ring opening; is it a concerted or radical
  based C-C bond cleavage?

This is a chelotropic reaction, a subset of cycloaddition\fragmentation reactions.  As to whether it proceeds by a concerted or radical mechanism, well, you haven't presented any experimental data that can help us decide.  For example, if the reaction had been run with the cis-2,3-dimethylcyclopropane analogue and both cis- and trans-but-2-ene were found among the products, we could argue for a biradical, two-step (non-concerted) process.  When we create an orbital symmetry diagram, that just allows us to make a prediction as to whether or not we expect the reaction to be concerted.

  I strongly assume the empty p-orbital to be in hyperconjugation with
  the walsh orbitals

There are a number of different conformations available to the molecule, but I agree with you that having the carbenic p-orbital aligned with the plane of the cyclopropane ring (the "bisected" conformation) should be favored.  Just as the bent carbon-carbon bonds with high p-character in a cyclopropane ring stabilize the empty p-orbital in a cyclopropylcarbinyl cation (your drawing above), the same thing should happen here with the carbene's empty p-orbital.

  I was asked to draw the orbital correlation diagram.

I have difficulty reading your diagram, so I created a new orbital correlation diagram.

Let me say a few things about my figure in order to make it easier to understand.

The rear cyclopropane bond is bisected by the plane of the screen, this plane also contains the cyclopropyl carbon connected to the carbene center.
The carbene p-orbital is perpendicular to the plane of the screen, the carbenic $\ce{sp^2}$ orbital lies in the plane.
The acetylenic sigma bond lies in the plane of the screen just as the sigma bond connecting the cyclopropyl carbon to the carbenic carbon does.
There are two pi bonds in the acetylene, Px is perpendicular to the plane and Pz is contained in the plane of the screen.
This plane of the screen is the plane of symmetry that we will use to determine the symmetry of the various orbitals and bonds involved in this transformation.
I haven't redrawn the Walsh cyclopropane orbitals, we can refer to your drawing up above.
$\mathrm{\sigma_1}$, $\mathrm{\sigma_2}$ and $\mathrm{\sigma_3}$ refer to the 3 bonding Walsh orbitals (cyclopropane) that you've drawn above.
I haven't drawn any of the antibonding orbitals, as we'll see below, we don't need them in this case.
The empty carbenic p-orbital and $\ce{sp^2}$ orbital should be around E=0, but I've drawn them slightly below for clarity.
The "A" and "S" labels tell us whether the orbital is antisymmetric or symmetric with respect to our preserved symmetry element, the plane of the screen.

Now, if you look at the reactant occupied orbitals and their symmetries, we have 3 orbitals with S symmetry and 1 with A symmetry.  In the products we also have 3 occupied orbitals with S symmetry and 1 with A symmetry. So there is a smooth correlation between reactant orbitals and product orbitals.  This analysis suggests that the thermal reaction can proceed in a concerted fashion.

Miscellaneous

Just because a reaction is allowed to proceed in a concerted fashion does not mean that the reaction will occur.  The most common reaction of cyclopropyl carbenes is ring expansion to the corresponding cyclobutene.  Other concerted reactions may be preferred because they happen to be of lower energy.
Of course, if a reaction is permitted in the forward direction, then it is also permitted in the back direction.  But when we heat ethylene and acetylene we don't generate cyclopropyl carbene - why not?  Well, these are all equilibria and the forward reaction is extremely exothermic.  We remove a highly strained cyclopropane and a high energy carbene and replace them with pi bonds.  Therefore we would expect the back reaction (combination of ehtylene and acetylene) to be extremely endothermic.  The products are much more stable than the reactants and we don't expect to see a significant concentration of reactants at equilibria.  It would be interesting to heat ethylene and acetylene to high temperature and see if any cyclobutene (or its ring opened product, buta-1,3-diene) is formed.  If these products are observed they would be suggestive of the presence of cyclopropyl carbene which then underwent the ring expansion reaction to cyclobutene.

Response to OP's comments:

  How can we see in general from an orbital symmetry correlation diagram
  if a reaction is concerted or not?

If all of the reactant bonding MO's correlate with product bonding MO's, then the reaction is allowed to proceed in a concerted fashion.  If some reactant bonding MO's correlate with antibonding product MO's, then the reaction is not allowed to proceed in a concerted fashion.

In the current example, the reactant bonding (occupied) MO's $\sigma_1,~ \sigma_2,~ \sigma_3$ and the carbene $\ce{sp^2}$ correlate with product MO's $\sigma$, ethylene $\pi$, acetylene $\pi_x$ and acetylene $\pi_z$ respectively.  All ground state reactant orbitals correlate with ground state product orbitals.  Therefore the reaction is allowed to proceed in a concerted manner.

  Is it okay to say biradicaloid instead of carbene?

No, not really. "Biradical" is generally used in cases where the electrons are on different atoms.  Like when we break a bond in cyclopropane we have a 1,3-biradical.

  In the orbital correlation diagram I would predict the acetylene a
  little bit lower because of more s character (sp) than ethylene (sp2).
  May be the energy difference is roughly as high as the difference
  between carbene p orbital A and carbene sp2 orbital.

That may be, but in both ethylene and acetylene the pi electrons (which is what we are talking about) are in pure p orbitals.  My guess was that they would have roughly the same energy.

  For the retro cheletropic reaction I assume a photolytical pathway

The retro reaction, like the forward reaction, should be thermally allowed and photochemically forbidden.

  According to the paper, I've found here...

I can only see the abstract, but from what I read, I don't think this is a reaction involving a photochemically produced carbene in an excited state.  I think that they are photochemically decomposing the diazo compound and produce the same carbene they generate thermally at higher temperatures.  The photochemical route just allows them to generate the same ground state carbene at very low temperatures.


How to derive the steady-state solution for simple two-step reaction with differential equations?

Based on the discussion in comments, the OP is interested in the equilibrium concentration of the species ABC in this reaction scheme:

$$\ce{A + B + C <=>[k_1][k_2] AB + C <=>[k_3][k_4] ABC}$$

Equilibrium is reached when the rates of the forward and backward reactions of each step are equal. Thus, we have have two equations describing the equilibrium:

$$k_1\ce{[A][B]}=k_2\ce{[AB]}$$
$$k_3\ce{[AB][C]}=k_4\ce{[ABC]}$$

We also know from mass conservation that
$$\ce{[A]_o}=\ce{[A] + [AB] + [ABC]}$$
$$\ce{[B]_o}=\ce{[B] + [AB] + [ABC]}$$
$$\ce{[C]_o}=\ce{[C] + [ABC]}$$

These five equations can be manipulated to find an expression for $\ce{[ABC]}$ in terms of $\ce{[A]_o}$, $\ce{[B]_o}$, and $\ce{[C]_o}$.


What type of gloves should be used while diluting sulfuric and hydrochloric acids?

There are numerous sources online comparing polymer materials resistivity in hazardous environments, but most of them agree that in your case ordinary nitrile gloves would protect in case of incidental contact, and butyl rubber/neoprene/PVC gloves are more suitable if a prolonged exposure is expected.

Nitrile gloves are disposable and cheap, and always should be changed when contaminated with chemicals as they only offer basic protection.


The index of refraction of $\ce{BaSO4}$ at 589 nm can be found in "Index of Refraction of Inorganic Crystals", in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.

Separate values are given for the indexes of the orthorhombic crystals:

$\begin{align}
n_x &= 1.6362\\
n_y &= 1.6374\\
n_z &= 1.6480
\end{align}$


What is the IUPAC nomenclature for Intermetallic Compounds?

In your example you have given two metals and a bond-indicating symbol.

Grammar

Hyphens, Plus and Minus Signs, 'em'Dashes and Bond Indicators

'em' Dashes (ALT+0151, 'en' dash would be ALT+0150)

a) Used to indicate metal–metal bonds in polynuclear compounds (e.g. "[Mn2(CO)10 bis(pentacarbonylmanganese)(Mn—Mn)").

b) Used to separate the indvidual constituents in names of (formal) addition compounds (e.g. "3CdSO4·8H2O cadmiumsulfate—water")

Alphabetical Order

a) Intermetallic compounds are sorted alphabetically within the group of cations and anions. Deviations are allowed in order to express structural information.

b) In coordination complexes, ligands are sorted alphabetically and the donating atom is put first (e.g. OH2 instead of H2O).

Further Reading

Therefore, I would write your example as "Co—Fe".

There are a lot more rules that you can read here: https://web.archive.org/web/20181005064417/https://www.iupac.org/fileadmin/user_upload/databases/Red_Book_2005.pdf


What is the function of tosylchloride in the synthesis of an ether?

There are probably only two principal mechanisms in organic chemistry, one being electrocyclic reactions (not applicable here) and one being nucleophile/electrophile reactions. In the latter, which is what is happening in your case, you must always look for the group with either the highest electron density or the ‘best’ (I know, a hand-wavery term) nucleophilic properties — this group is the electrophile — and for the group with the lowest electron density or the ‘best’ (same hand-waveryness) electrophilic properties — termed the nucleophile.[1]

The first step to determining this would be to check electronegativities and oxidation states. Groups with high electronegativities and low oxidation states are typically nucleophiles and vice-versa. In your example, all alcoholic oxygens have a much higher electronegativity than all atoms they are connected to (carbon and hydrogen), are in the $\mathrm{-II}$ oxidation state and would thus act as nucleophiles. The oxygens and the chloride of tosyl chloride also have a larger electron density than their surroundings. For reasons outlined above that I cannot go into details with, they will not react nucleophilically, though.

The carbon atoms which are directly attached to the alcoholic oxygens all lack electron density due to the neighboring electronegative oxygen. They are electrophiles. And also the central sulfur of tosyl chloride — which not only features a $\mathrm{+VI}$ oxidation state but should also be drawn with a double positive formal charge to satisfy the octet rule[2] — is an electrophile.

To synthesise the ether from the two alcohols, we would now somehow need either of the two alcoholic nucleophiles to attack the other carbon electrophile and displace the other alcoholic hydroxy group — this reaction is termed $\mathrm{S_N2}$ reaction or bimolecular nucleophilic substitution. However, it is not likely to occur directly: a direct side-product would be the removal of a hydroxide ion, which is termed a bad leaving group.[3] We need to turn it into a better leaving group.

This is where tosyl chloride comes in. But first another little excursion: Notice the little 1. and 2. in your reaction scheme? They stand for first and second, implying that we are dealing with two separate reactions interceded by a workup and typically purification of an intermediate. So first we react the 1-phenylpropan-2-ol with tosyl chloride, collect the product and then react said product with ethanol. As I outlined before, a pretty good electrophile in tosyl chloride is the sulfur atom, and with the adjacent chloride, we have a good leaving group. So under a nucleophilic substitution,[4] this chloride can be displaced (along with a proton) leading to our intermediate 1-phenylprop-2-yl tosylate.

Technically we would now need to make a fresh analysis of nucleophiles and electrophiles. But it may suffice to note the key differences: there is no longer a good leaving group on sulfur (the alcoholate which would be a leaving group again is the conjugate base of a very weak acid), but if ethanol now attacks the carbon adjacent to what is now a tosylate group we have the tosylate anion as a good leaving group. In a second bimolecular nucleophilic substitution, tosylate and a proton are displaced and the ether is formed.

Notes:

[1]: Actually, which group is nucleophilic and which is electrophilic depends on the frontier orbitals, and that can also be used to unify both reaction types into one general scheme, but that analysis is very hard to do a priori.

[2]: Organic chemistry’s preference to write sulfonic sulfur atoms as doubly bonded to two oxygens, creating a dodectet gained popularity probably because of one of Pauling’s not-so-bright moments. However, evidence points towards d-orbitals (which would be required for the dodectet) not taking part in bonding for non-metals and thus the charge separation depiction being a more accurate description of the truth. This is documented a lot across this site.

[3]: Again, I won’t go into detail here when leaving groups are good or bad, but just leave as one principal pointer that conjugate bases of strong acids are bad leaving groups. The conjugate base of $\ce{OH-}$ is $\ce{H2O}$ which is a very weak acid; thus, hydroxide is a very bad leaving group.

[4]: Technically, this is no longer an $\mathrm{S_N2}$ reaction. Since sulfur is participating as the central, electrophilic atom, the transition state is more long-lived involving a four-electron three-centre bond. The difference is marked enough to call this bimolecular nucleophilic heavy atom substitution or something along the lines. It will usually not be referred to as an $\mathrm{S_N2}$ reaction.


The carbonyl oxygen is more nucleophilic. You can draw a resonance structure of the carboxylic acid that places a formal negative charge on that oxygen which rationalizes its greater nucleophilicity. Also, after it adds to the carbocation, the resulting intermediate will be resonance stabilized. If the other oxygen attacked, no resonance structures would be possible.


Why could it be assumed that the standard enthalpy change is independent of temperature down to 25 °C?

Generally speaking you can write that
$$\Delta H^\circ(T) = \Delta H^\circ(T_\mathrm{ref}) + \int_{T_\mathrm{ref}}^T \Delta C_p \,\mathrm dT \tag{1}$$
where $T_\mathrm{ref}$ might be $\pu{25 ^\circ C}$ and the pressure is 1 bar (in your case, approximately 1 bar).
If your temperature window is not too large you may get away with assuming $\Delta H^\circ$ is constant. This amounts to assuming that the second term in the equation, dependent on $\Delta C_p$, is small compared to the first. If it isn't then rather than using the integrated van't Hoff equation

$$\ln\left(\frac{K_2}{K_1}\right)=\frac{-\Delta H^\circ}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right) \label{eqn:2}\tag{2}$$
maybe you should attempt a polynomial fit to $\ln K_\mathrm{eq}$ and taking the derivative of the polynomial wrt $T^{-1}$ to evaluate $\Delta H^\circ$ as a function of $T$.

To test whether \eqref{eqn:2} is appropriate you might evaluate the quality of a fit to that equation or check the linearity of a plot of $\ln K$ versus $T^{-1}$.


Why do Hydrogen Halides(with the exception of HF) form mostly strong acids?

I would like to correct a few points firstly that the halide part is not negatively charged. Also the electronegativity of the halide is the dominating effect in this case.

Hydrofluoric Acid as we know is not a very strong acid as compared to other hydrogen halide acids, had it been the case of electronegativity this would not have been the case.

The dominating factor here is the sizes of the atoms forming the molecule. As the size of halide increases, the bond strength of the H-X bond decreases, making it easier to break the Hydrogen halide bond and act as a stronger acid.

So the acidic strength comes out to be:
              HF < HCl < HBr < HI


is ph at eq point dependent on acid concentration at all

The concentration of both acid and base solution matters - or rather the ratio of concentrations matters. Let's look at some simple examples titrating acetic acid, which has a $K_a = 1.74 \times 10^{-5}$ with sodium hydroxide.

$$\ce{HC2H3O2 + NaOH -> H2O + NaC2H3O2}$$

  Example 1 - 20.0 mL of 0.100 M acetic acid titrated with 0.100 M NaOH

In this example, since the concentrations are the same and the stoichiometric ratio is 1:1, we will need 10.0 mL of NaOH solution to reach the end point. The new volume is 40.0 mL and the concentration of acetate anion is 0.0500 M (volume doubled, concentration halved). Some of that acetate ion reacts with water in the following equilibrium:

$$\ce{C2H3O2- +H2O <=> HC2H3O2 + OH-}\\ K_b=\dfrac{[\ce{HC2H3O2}][\ce{OH-}]}{[\ce{C2H3O2-}]}=5.76\times 10^{-10}$$

To figure out how much hydroxide is formed (and thus eventually the pH), we need to do an ICE problem:

$$\begin{array}{|c|c|c|c|}
 & [\ce{C2H3O2-}] & [\ce{HC2H3O2}] & [\ce{OH-}] \\ \hline
I & 0.0500\ \text{M} & 0\ \text{M} & 1\times 10^{-7}\ \text{M} \\
C & -x\ \text{M} & +x\ \text{M} & +x\ \text{M} \\
E & 0.0500-x\ \text{M} & x\ \text{M} & 1\times 10^{-7}+x\ \text{M} \\
\end{array}$$

Then we plug in the concentrations into the law of mass action:

$$5.76\times 10^{-10} = \dfrac{(x)(1\times 10^{-7}+x)}{(0.0500-x)}$$

Solving for $x$ (since I am not making any approximations, I am letting Wolphram Alpha do the grind:

$$x=5.32\times 10^{-6}\ \text{M} \\
[\ce{OH-}] = 5.42 \times 10^{-6}\ \text{M} \\
[\ce{H+}] = 1.85 \times 10^{-9}\ \text{M} \\
\text{pH} = 8.734$$

  Example 2 - 20.0 mL of 0.100 M acetic acid titrated with 0.200 M NaOH

In this example, the NaOH solution is twice as concentrated, so it only takes 10.0 mL to titrate the acid. The concentration of acetate anion is 0.0667 M.

$$\begin{array}{|c|c|c|c|}
 & [\ce{C2H3O2-}] & [\ce{HC2H3O2}] & [\ce{OH-}] \\ \hline
I & 0.0667\ \text{M} & 0\ \text{M} & 1\times 10^{-7}\ \text{M} \\
C & -x\ \text{M} & +x\ \text{M} & +x\ \text{M} \\
E & 0.0667-x\ \text{M} & x\ \text{M} & 1\times 10^{-7}+x\ \text{M} \\
\end{array}$$

Then we plug in the concentrations into the law of mass action:

$$5.76\times 10^{-10} = \dfrac{(x)(1\times 10^{-7}+x)}{(0.0667-x)}$$

$$x=6.15\times 10^{-6}\ \text{M} \\
[\ce{OH-}] = 6.25 \times 10^{-6}\ \text{M} \\
[\ce{H+}] = 1.60 \times 10^{-9}\ \text{M} \\
\text{pH} = 8.796$$

  Example 3 - 20.0 mL of 0.200 M acetic acid titrated with 0.100 M NaOH

In this example, the NaOH solution is half as concentrated as the acetic acid solution, so it takes 40.0 mL to titrate the acid. The concentration of acetate anion is 0.0333 M.

$$\begin{array}{|c|c|c|c|}
 & [\ce{C2H3O2-}] & [\ce{HC2H3O2}] & [\ce{OH-}] \\ \hline
I & 0.0333\ \text{M} & 0\ \text{M} & 1\times 10^{-7}\ \text{M} \\
C & -x\ \text{M} & +x\ \text{M} & +x\ \text{M} \\
E & 0.0333-x\ \text{M} & x\ \text{M} & 1\times 10^{-7}+x\ \text{M} \\
\end{array}$$

Then we plug in the concentrations into the law of mass action:

$$5.76\times 10^{-10} = \dfrac{(x)(1\times 10^{-7}+x)}{(0.0333-x)}$$

$$x=4.33\times 10^{-6})\ \text{M} \\
[\ce{OH-}] = 4.43 \times 10^{-6}\ \text{M} \\
[\ce{H+}] = 2.26 \times 10^{-9}\ \text{M} \\
\text{pH} = 8.646$$


Does any compound exist with multiple liquid phases?

There's no reason why chemical substances with multiple liquid phases shouldn't exist.

In fact there is evidence for the existence of a least a second form of (surprise) liquid water. If you search the Web for liquid-liquid phase transitions you will find more of these, e.g. cerium.


A compound that absorbs all visible light

Substances which absorb almost all the light falling on them appear black. Therefore you are looking for the blackest known compound.

The record is currently held by Vantablack[1], a substance composed of vertically aligned carbon nanotubes, which absorbs up to 99.965% of visible light incident upon it. As you can see it is really black.

Previously, the record was held by Superblack[2,3], a nickel-phosphorus alloy with an absorbance of up to 99.65%.

For comparison, regular black paint generally absorbs around 97.5% of visible light.

Sources

[1] https://en.wikipedia.org/wiki/Vantablack

[2] https://www.newscientist.com/article/dn3356-mini-craters-key-to-blackest-ever-black

[3] http://pubs.rsc.org/en/Content/ArticleLanding/2002/JM/b204483h#!divAbstract


Regarding the infinitesimal form of the First Law of Thermodynamics

The first law of thermodynamics is a statement of energy conservation thus and defines the internal energy E as an extensive state function. In an infinitesimal transformation, the first then law reduces to
$$\mathrm{d}E = \mathrm{d}Q + \mathrm{d}W$$

where $\mathrm{d}E$ is a total (exact) differential for infinitesimal transformation.

However, $\mathrm{d}Q$ and $\mathrm{d}W$ are not exact ($Q$ and $W$ are not state functions); $Q$ and $W$ in a thermodynamics transformation are process-dependent. All of these are properties of functions of more than one variables


What is physically responsible for different compounds with the same types of atoms?

Okay, I'm sure there's a much more direct, down-to-earth answer that you'd find more helpful, but I'm going for something different. To stop myself from writing a book chapter, there will be some hand-waving and cut out details. With more time and thought it could be unpacked further, but hopefully I'll at least be able to give you a taste.

Right, let's forget chemistry and try to deduce the product for the following simple reaction:

$$\ce{H2 + O2 -> ?}$$

To start off, we establish some basic rules:

Molecules are formed from the combination of integer amounts of atoms
Atoms cannot be formed or destroyed (i.e., the number of atoms is a conserved quantity)
There is a quantity (let's call it potential energy) inherent to each kind of free atom and molecule
The Universe tends towards the lowest possible potential energy

With nothing else to guide us, we are forced to consider every possible product (isomers neglected for simplicity):

$$\ce{H2 + O2 -> H + O + H_2 + HO + O_2 + H_3 + H_2O + HO_2 + O3 +  H4 +...}$$

Now, if the point is to minimize the total potential energy of the system, we can develop a heuristic (which will turn out to be wrong, but entertain the idea for now). First, find the specific atom or molecule among the listed products with the lowest possible energy,$^1$ then make as much of it as allowed by stoichiometry. When no further can be made, look for the next lowest energy molecule, then make as much of that as possible, and iterate the process until no lower energy molecules can be made.

That's nice and all, but without knowing which molecules have lower or higher energy, you don't get anywhere. Here, we either resort to some hefty quantum mechanical calculations, or empirical data. By doing some thermochemical experiments (measuring enthalpies of formation), it turns out that $\ce{H_2O}$ is the molecule with the lowest energy of the list by quite a margin, so it's going to be the main product. Indeed that is what is observed empirically when hydrogen is burnt in oxygen.

So we know that water is the lowest energy product for the reaction, but what happens if we mix $\ce{H2}$ and $\ce{O2}$ with a ratio different from that in water? Say you have two moles of $\ce{H2}$ and two moles of $\ce{O2}$, does that mean you now get $\ce{H2O2}$ or potentially some other products? In fact, this is not the case. The heuristic above still applies to minimize the potential energy of the system, so $\ce{H_2O}$ is preferentially produced. This shouldn't come as too much of a surprise, because the low energy of $\ce{H_2O}$ is an inherent feature of the molecule, and is entirely independent of how much $\ce{H2}$ and $\ce{O2}$ you decide to mix in an experiment.

Then what happens to the excess $\ce{O2}$? Again following the heuristic, it will seek out the lowest energy molecule available. Since there's no more hydrogen left to react, it has a more limited number of possibilities. It could conceivably still react and turn into other molecules (for example, $\ce{O3}$ is perfectly acceptable), but it turns out that of the remaining candidates, $\ce{O2}$ is the molecule with the lowest energy, so the reaction just stops there, and we get a mixture of $\ce{H_2O}$ and left over $\ce{O2}$.

Okay, we've made some progress, but we need to talk about some important complications. In the real world, at any temperature above absolute zero, you can actually retain a small amount of molecules with a higher potential energy. All of those lower energy molecules can randomly gain a bit of thermal energy, and every so often that causes them to react and turn into a higher energy molecule. If the lowest energy molecules are much lower in energy than anything else (as is the case for $\ce{H_2O}$ among the combustion products of $\ce{H2}$ and $\ce{O2}$), then this effect is very small and can be neglected. However, in many cases there isn't such a clear-cut winner, and you end up getting a mixture of different products with similar energies. This is especially true as you get into more complicated chemistry.

Unfortunately, even after all this we're still missing a very big piece of the puzzle. Everything I've mentioned so far is a part of thermodynamics, and that alone cannot always predict the product of a reaction. Allow me to prove it. According to everything I've argued, $\ce{H2}$ and $\ce{O2}$ will always react to form primarily $\ce{H2O}$ since it has the lowest energy, and $\ce{H2O2}$ can never be a significant product. But hydrogen peroxide is made in very large amounts industrially, so how do they do it? A quick search on the internet provides the answer, the anthraquinone process, whose net reaction is simply:

$$\ce{H2 + O2 -> H2O2}$$

What gives? What we've been missing is the concept of kinetics. For many transformations in nature, chemical or otherwise, even if the final state has a lower energy than the initial state, they are often separated by an energy barrier. Even if there can be a net release of energy, the transformation is hindered because to start off it needs a net input of energy.

This is the case even for the reaction $\ce{H2 + O2 -> H2O}$. If you just mix the gasses together at room temperature, they will not react even after a century. Either some initial energy needs to be supplied (a flame, a spark, etc.) or the energy barrier needs to be decreased by using a catalyst (Pt), after which the reaction will proceed in a fraction of a second.

When you consider all possible reactions, and all the energy barriers separating the different molecules, you get the very high-dimensional potential energy surfaces I mentioned in my comment. All our work in Chemistry boils down to finding reliable paths in (or making our own by actively modifying) the crazily complex energy landscapes and exploiting them for our purposes.

$^1$ Technically, the lowest amount of potential energy per amount of atoms the molecule is made of. You can think of it as the lowest potential energy per gram of molecules.

Here's something which might also help. Is there a case where all this potential energy and kinetic stuff doesn't skew the results, and you can tease out the purely statistical effect? Sure, let me present a typical case.

Consider the following transesterification reaction (poorly drawn because I don't currently have access to ChemDraw):

Both sides of the equation have an alcohol and an ester. The alcohols are very similar to each other, and the same can be said for the esters. Because of this, there's barely any difference between the potential energy of the products and the reagents.

Now consider taking a mole of the starting  ester and alcohol. In the beginning, there is only ethanol and no methanol, so statistically you'd expect the reaction to happen in the direction shown. As the reaction progresses, the amount of ethanol available is reduced, disfavouring further generation of products, and the presence of methanol now actually allows the reaction to also go the other way around. By pure statistics, you'd expect the reaction to stop when there's about the same amount of methanol and ethanol, half a mole each, and this is what is observed. In short, simple transesterification reactions tend to have equilibrium constants very close to 1 (e.g. this article).


Some background knowledge
You need to be aware of electronegativities, assignement of s and p character, relation of structure to the donating or accepting nature of the substituent, and most importantly, the overlap efficiency of the donor and acceptor positions (roughly estimated by overlap integral , although for the sake of this discussion, most of the mathematical rigor is not required)
I will be comparing most of the order for the +M effect, and provide some hints for the -M effect.
Group 1: from $\ce{O-}$ to $\ce{OCOR}$
This group of substituents consists of two kinds of donor atoms from the 2nd period, namely $\ce{O}$ and $\ce{N}$. As they are from the same period as $\ce{C}$ , therefore, overlap efficiency will be maximum in this case, for 2p-2p overlap.

Recall that p-p overlap> s-p overlap> s-s overlap as per overlap efficiency

Due to this reason, this group of substituents will have better +M effect as compared to the halogens, because as we move down the group 17, overlap efficiency for 2p-3p,2p-4p overlap and so on decreases.
However, a member of period 2 does exist among the halogens, namely $\ce{F}$ .But since it is more electronegative than both $\ce{O}$ and $\ce{N}$, it will again have lesser donating tendency as compared to this group.
Now, let's get down to the specific cases:
The group leader $\ce{O-}$ :
First,let us look at some enthalpy values:
$\ce{ O(g) + e^- -> O^-(g) 1st EA = -142 kJ /mol }$  (1)
$\ce{ O^-(g) + e^- -> O^2-(g) 2nd EA = +844 kJ /mol }$ (2)
As you can clearly see, adding an electron to $\ce{O-}$ ( or adding a second electron to $\ce{O}$) is an endothermic process, which means energy will have to be provided externally for pushing more electron density inside $\ce{O-}$.
That is pretty obvious as well, because you are forcing an electron into an already negative ion. It's not going to go in willingly! The second electron affinity of oxygen is particularly high because the electron is being forced into a small, very electron-dense space.[source]
Conversely, this means that the reverse reaction for (2) will be pretty favorable, in the sense that $\ce{O-}$ will quite readily give away electron density to a suitable acceptor atom,as is evident from the enthalpy values. Hence, $\ce{O-}$ leads the charge in terms of +M effect

It is to be kept in mind that the factor of compatibility of overlap still holds. Donation from $\ce{O-}$ to, say, $\ce{Si}$ will still not be as effective as that to $\ce{C}$

Subgroup 1:The curious case of $\ce{-NH2}$,$\ce{-NHR}$ and $\ce{-OR}$
First of all, since we have got neutral donor atoms in this subset, it stands to reason that simple electronegativity will be the major governing factor here to judge the overall donation tendency. Hence, both the $\ce{N}$ donor groups should be more donating than the $\ce{O}$ donor group.
The slightly peculiar judgement crops up when you are deciding between $\ce{-NH2}$ and $\ce{-NHR}$ . On a preliminary basis, you might say that since the $\ce{R}$ will most probably be exerting a strong +I effect ,and thereby the latter should have had more +M effect than the former.
But here's the catch: the electron density of the $\ce{N-R}$ bond will mostly shift towards N as compared to the $\ce{N-H}$ bond, but the donation to the benzene ring will ultimately take place by the $\ce{N}$ atom. And hence, it becomes important to look at what the donor nitrogens of these two groups have to say.
We are basically going to compare the bond angles of $\ce{H-N-H}$ and $\ce{H-N-R}$ .Due to the sterically bigger $\ce{R}$ group , the latter should be greater than the former. This means that the s-character will go up on the $\ce{N}$ atom of the latter, and this directly means that electronegativity for it will increase.

(Think of the various hybridization schemes, namely $\ce{sp}$ , $\ce{sp^2}$ and $\ce{sp^3}$, the percentage s character in each, the bond angles for each, and the electronegativity trend of $\ce{sp}$>$\ce{sp^2}$>$\ce{sp^3}$ )

Hence, donation power of $\ce{-NHR}$ should be less than $\ce{-NH2}$

A similar argument will also explain why the +M effect of $\ce{-OR}$ is less than $\ce{-OH}$

Subgroup 2: $\ce{-NHCOR}$ and $\ce{-OCOR}$
Both these group are analogous structurally, and that's the key to their relative comparison for our purpose. On a preliminary basis, we see that the number of highly electronegative atoms( that is, $\ce{O}$, $\ce{N}$ and $\ce{F}$) has increased in both these groups as compared to the previous members of this group. $\ce{-NHCOR}$ has an $\ce{O}$ and $\ce{N}$, while $\ce{-OCOR}$ has two oxygens. This immediately tend to make them less donating than the previous members of the group.
Another reason to see why they would be less donating is because of the distribution of partial charges in the resonance hybrid. The donor atom's lone pairs in both the cases are conjugating between a $\ce{C=O}$ group and the benzene ring. Since negative charge is more stable on $\ce{O}$ than on $\ce{C}$, the oxygen of the $\ce{C=O}$ will end up possessing much of the negative charge which would otherwise be available for donation to the ring.
In other words, we can say that the charge will be dispersed over a competing electron sink, namely, the $\ce{C=O}$ .Hence, their mesomeric effects will decrease considerably as compared to other members of the group.
As for their individual comparisons, you can easily judge that by the electronegativities of their individual donor atoms, $\ce{N}$ and $\ce{O}$
The phenyl group
Wikipedia says:

It is generally considered an inductively withdrawing group (-I), because of the higher electronegativity of sp2 carbon atoms, and a resonance donating group (+M), due to the ability of its π system to donate electron density when conjugation is possible.

Drawing from this, one can say that since the phenyl group is donating by using bond pair of electrons(namely, it's $\ce{C=C}$ π bonds) rather than lone pairs(unlike all the previous groups which were using lone pairs of $\ce{N}$ or $\ce{O}$) therefore it will have lesser donation tendency than the preceding members as lone pairs are generally more "freely available" for donation, being attracted by a single nucleus rather than bond pair of electrons which are held by two nuclei.
Still, as it is a donation from a $\ce{C}$ donor atom, it has better +M effect as compared to the halogens.
Comparison between $\ce{-Ph}$ and $\ce{-F}$ is a tough case, as it may seem that fluorine which is using it's lone pairs should have been a better donor than $\ce{-Ph}$. To this I would say that maybe since fluorine is the most electronegative element, it's electronegativity would dominate the donating tendency by the lone pairs(in technical terms, HOMO-LUMO gap would be more in case of F's non-bonded electrons than the $\ce{C=C}$ pi-bonded electrons)
But again, this reasoning would seem unsatisfactory.  Anyway,the mesomeric effect of the $\ce{-Ph}$ group usually holds little chemical interest. It mainly serves as a control, and we are much more interested in the effect of hetero atoms on the ring as compared to the effect of one benzene ring on another(in case of mesomeric effect)
Group 2:The halogens
I have already given enough idea about them above, more details can be found here
So there you have it. A similar idea can be applied for deducing the logic of the -M ordering as well.

Hints for -M effect

$\ce{-NO2}$ has 3 highly electronegative atoms, and an electron sink in the form of $\ce{-N=O}$, so it should lead the pack.

Just like the previous ordering, we are again interested in the characteristics of the acceptor atom here. While for +M effect, it increased with decrease in s-character(and hence electronegativity) on the donor atom, and more freely available electron density

-M effect will increase with increase in s-character(and hence electronegativity) of the acceptor atom. Since in your list, the remaining substituents are $\ce{C}$ acceptors, then the hybridization of $\ce{C}$ needs to be considered.
Accordingly, $\ce{-CN}$ should fall next with an $\ce{sp}$ hybridized $\ce{C}$

Now most of the items I can see on the list are substituents with $\ce{sp^2}$ carbons, having different combinations of $\ce{O}$ and $\ce{N}$ attached to the acceptor atom. Your job would now be to checking the varying extents of cross conjugation happening at the electron sink directly attached to the acceptor atom(and thereby comparing the electron accepting ability of the same) by looking at the electron donating abilities of the attached heteroatoms

You can ask for further clarification if these hints should not prove to be sufficient.


Activities of Solutes - especially acids and bases

For almost the last 100 years, the definition of pH has been -log of hydrogen ion activity.  Prior to 1920, the definition was -log of hydrogen ion concentration.

Hydrogen ion activity or concentration does not refer to bare protons, but solvated protons, as $\ce{H_3O^+, H_9O_4+}$ etc.

See Determination of Hydrogen Single Ion Activity Coefficients in Aqueous HCl Solutions at 25°C for authoritative data.


Why are there 2 equivalents of Ca in an example stated in Wikipedia?

The Wikipedia article uses a funny defintion of equivalent that I have not heard yet even though I use the term extremely frequently in the lab. But they define it well so here goes:

One equivalent of ions is defined as $1~\mathrm{mol}$ of charge. So if we dissolve $1~\mathrm{mol}$ of $\ce{Na+}$, we have exactly $1~\mathrm{mol}$ of positive charges in solution, so $1~\mathrm{mol} \cdot 1~e = 1~\mathrm{eq}$ ($e$ being the elemental charge). The $1~\mathrm{mol}$ of $\ce{Cl-}$ contributes $1~\mathrm{eq}$ of negative charges, balancing the entire thing.

Calcium, however, has a $\mathrm{+II}$ oxidation state. So $1~\mathrm{mol}$ of $\ce{Ca^2+}$ gives us $1~\mathrm{mol} \cdot 2~e = 2~\mathrm{eq}$ of positive charge.


How much muriatic acid to dissolve cinder block?

Keep in mind that a cinder block is likely to contain silicates, these will not dissolve well in hydrochloric acid. The silicates will either not react or they will form silica.

If you want to dissolve silicates then you need to use something which will complex to the silicon such as fluoride (HF). I have seen methods for the digestion of soil samples to get the plutonium out. To obtain the plutonium from a soil sample it is often important to digest down the silicate minerals. To do this you must boil the soil sample to dryness many times with a mixture of hydrofluric acid and nitric acid. This will convert the silicate minerals into silicon tetraflouride which will boil out of the mixture.


What components of safety should be included in a chemistry laboratory experiment conclusion?

Perhaps your instructors want more specific information about the hazard itself, rather than generic safety advice.  Some examples:

1) HCl: Strong acid, highly corrosive (nitpick: caustic is usually reserved for strong bases, corrosive for acids), emits corrosive and toxic fumes (not all acids do)

2) 29% NH4OH: Caustic, emits strong fumes (lachrymator--makes your eyes water), work in a fume hood, outdoors or in a well-ventilated area

3) Bunsen burner: No open flames in the presence of flammable vapors or liquids with high vapor pressure (much bigger hazard than just burning yourself)

4) Hotplate: Hotplate looks the same whether hot or cold, so assume it is hot, take care with flammable liquids with low flash points

Again, these are just examples, but I tried to tailor the answer to the item in question.


Water is a polar species in which the oxygen atom in $\ce{H2O}$ has a partial negative charge and the two hydrogen atoms have partial positive charges. As you suggested, this is because oxygen is highly electronegative (i.e. it has a high affinity for electrons) and so the negative electrons tend to have a higher density around the oxygen atom than the hydrogen atoms, creating this separation of charges due to water's molecular structure. Take a look at the molecular structure of water, which can be shown below. It has an asymmetry bent geometry and therefore the charge distribution in water is also not symmetric.

The hydrogen in $\ce{HCl}$ does not actually give its electron to the chlorine atom. Rather, the more electronegative chlorine ion is attracted to the partially positive hydrogen atoms in water, and the hydrogen ion in $\ce{HCl}$ is attracted to the partially negative oxygen atom in water. This separates the $\ce{H+}$ from the $\ce{Cl-}$ in $\ce{HCl}$ so that the acid dissociates into its respective ions. You therefore get the equation $\ce{HCl -> H+ + Cl-}$ when in water. If you include water on the reactant side, it can be equivalently written as $\ce{HCl + H2O -> H3O+ + Cl-}$. Here, $\ce{H3O+}$ is simply indicating that the $\ce{H+}$ ion is in an aqueous solution.


Why is the helium wavefunction a linear combination?

First of all, note that
$$
\psi_a(\vec{r}_{1}) \, \psi_b(\vec{r}_{2})
\neq
\psi_a(\vec{r}_{2}) \, \psi_b(\vec{r}_{1}) \, ,
$$
since same functions in these products ($\psi_a$ and $\psi_b$) have different arguments ($\vec{r}_{1}$ and $\vec{r}_{2}$) to the left and to the right of the inequality sign.

Now, to the question, why do we have a linear combination. For a system of two indistinguishable particles (say, electrons) the obvious requirement is that there should be no observable difference between the system in state $\psi(1, 2)$ and the system in state $\psi(2, 1)$,
\begin{equation}
    |\psi(1, 2)|^{2} = |\psi(2, 1)|^{2} \, ,
\end{equation}
where $1$ and $2$ stand for coordinates of particles.
This implies that
\begin{equation*}
    \psi(1, 2) = \pm \psi(2, 1) \, .
\end{equation*}

And in order to satisfy this equality in what is often called the orbital approximation, where the state of a many electron system $\psi$ is represented as a product of a single-particle states $\psi_i$, the wave function $\psi(1, 2)$ should be actually written as not just a product of $\psi_1$ and $\psi_2$ (none of which, neither $\psi_1(1) \psi_2(2)$ nor $\psi_2(1) \psi_1(2)$, is symmetric or antisymmetric), but rather as the following linear combination of these products,
\begin{equation*}
    \psi(1, 2) = \frac{1}{\sqrt{2}} \Big( \psi_1(1) \psi_2(2) \pm \psi_1(2) \psi_2(1) \Big) \, ,
\end{equation*}
where $1 / \sqrt{2}$ is the normalization factor.

The choice of the sign in the expressions above is not arbitrary, but rather determined by the particles spin $s$, namely,
\begin{equation*}
    \psi(1, 2) = (-1)^{2s} \psi(2, 1).
\end{equation*}
Particles with integer spins, called bosons (photons, alpha particles), are described by a symmetric wave function
\begin{equation*}
    \psi_{\mathrm{s}}(1, 2)
    =
    \frac{1}{\sqrt{2}}
    \Big(
        \psi_1(1) \psi_2(2)
        +
        \psi_1(2) \psi_2(1)
    \Big) \, ,
\end{equation*}
while the particles with half-integer spins, called fermions (electrons, protons, neutrons), are described by the anti-symmetric wave function
\begin{equation*}
    \psi_{\mathrm{a}}(1, 2)
    =
    \frac{1}{\sqrt{2}}
    \Big(
        \psi_1(1) \psi_2(2)
        -
        \psi_1(2) \psi_2(1)
    \Big) \, .
\end{equation*}


Potassium Manganate and Acidified Potassium Dichromate colour changes

Although you are asking for the color changes of the reduction of potassium manganate ($\ce{K2MnO4}$; a green-colored salt, (Wikipedia1)) by potassium dichromate ($\ce{K2Cr2O7}$; a red-orange-colored salt, (Wikipedia2)) in acidic medium, the equation you are showing is reduction half reaction of potassium permanganate ($\ce{KMnO4}$; a purple-colored salt, (Wikipedia3)) in acidic medium. According to Wikipedia1, potassium manganate is an intermediate in the industrial synthesis of potassium permanganate. Thus, color change for that specific reaction is green to purple (disregarding other interference such as color change of the oxidizing reagent).

The reduction half reaction of $\ce{K2Cr2O7}$ in acidic medium is:
$$\ce{Cr2O7^2- + 14 H+ + 6e- <=> 2 Cr^3+ + 7 H2O}  \qquad  E^{\circ} = \pu{1.36 V}$$

The oxidation half reaction of $\ce{K2MnO4}$ is:

$$\ce{MnO4^2- <=> MnO4^- + e-}  \qquad  \qquad  E^{\circ} = \pu{-0.558 V}$$

The total redox reaction of $\ce{K2MnO4}$ and $\ce{K2Cr2O7}$ in acidic medium is:
$$\ce{Cr2O7^2- + 14 H+ + 6 MnO4^2- -> 2 Cr^3+ + 6 MnO4^- + 7 H2O}  \qquad  E^{\circ}_{\text{Rxn}} = \pu{0.802 V}$$

Since $E^{\circ}_{\text{Rxn}} \gt 0$, the reaction is spontaneous.


It is important to clearly separate the terms denticity, hapticity and bridging.

An atom or a ligand is bridging is one single atom bonds to two different acceptors.

A ligand (but never a single atom) can be multidentate or multihaptic if two atoms of that ligand can bond to the same acceptor.

A ligand such as the amide anion cannot be multidentate or multihaptic as it only has one coordinating atom: nitrogen. This nitrogen can coordinate to two different metals (in a μ-amido complex) but it still remains a monodentate ligand.
A ligand such as ethylenediamine or en for short ($\ce{H2N-CH2-CH2-NH2}$) can coordinate to the same central atom via each of the two nitrogen atoms. This is a bidentate ligand as it has two ‘teeth’ in the central atom.
A ligand such as ethene ($\ce{C2H4}$) can coordinate to a single central atom in a bihaptic (η2) fashion. Drawing shortest lines from nucleus to nucleus, both carbon atoms would be the same distance from the acceptor atom (in reality, it is better thought of the π system donating). Hapticity is reserved for such cases in which a continuous set of atoms coordinate to a central metal.
A ligand can be both multihaptic and multidentate. An example would be dibenzylideneacetone or dba ($\ce{Ph-CH=CH-C(=O)-CH=CH-Ph}$), in which both of the $\ce{C=C}$ double bonds act as a bihaptic ligand. (Due to geometry constraints, dba usually coordinates to two different acceptors rather than to a single one in a chelate fashion.)
Finally, it is possible for a ligand such as en to be both bidentate and briding, if either of the nitrogen atoms coordinates to two different metals. This requires deprotonation of at least one nitrogen.
The bottom line is: briding ligands can be monodentate or multidentate; they can be monohaptic or multihaptic. Only bridging ligands based on a single donor atom must be monodentate.


Acid Catalysed Ring Expansion – Mechanism?

You are correct about protonating the carbonyl group. [The first of your products is the same as the reactant.] The cyclobutanone 2 was prepared from a mixture of geranic acids. (J. J. Beereboom, J. Org. Chem., 1965, 30, 4230). The structures of 2, 4 and 5 were determined by Beereboom. I am providing the most reasonable mechanism. Cyclobutanone 2 must have a more stable cis ring juncture owing to its method of formation or exposure to p-TsOH. The trans isomer of 2 cannot give the trans isomer of 3 because of strain. Tertiary carbocation 3 arises by addition of the double bond to the protonated carbonyl. Collapse of the carbocation produces bicyclic ketone 4, which in the presence of p-TsOH isomerizes to the exocyclic isomer 5. In general, 1,1-disubstituted double bonds are less stable than tri-substituted ones. In this case, ring strain may be present in the endocyclic alkene 4.


Why are there only two irreducible representations in the C3 group?

You got everything right. It is just that two of these representations would use complex numbers ($\varepsilon=e^{2\pi i\over3}$ and $\varepsilon^2$), and we don't like that, so we stick them together into one representation of dimension 2, which makes the imaginary part vanish. Afterwards we treat this representation as "kinda irreducible". In fact it can be reduced, but we just don't want to go that way.

The same applies to other cyclic groups.


Relationship between normality and molarity

The normality of a solution is the molarity of the solution multiplied by the number of equivalents per mole:

$$\mathrm{N} = \mathrm{{eq\over mol}}\cdot\mathrm{M}= \mathrm{{eq\over mol}}\cdot\mathrm{{mol\over L}}=\mathrm{{eq\over L}}$$

Example: Calculate the normality of a 1 $\mathrm{M}$ solution of $\ce{H2SO4}$. There are 2 equivalents per mole, from the reaction

$$\ce{H2SO4 -> 2H+ + SO4^{2-}}$$

which yields 2 moles of $\ce{H+}$ for every 1 mole of $\ce{H2SO4}$.

Using the relation given above, we find that a 1 $\mathrm{M}$ solution of $\ce{H2SO4}$ is 2 $\mathrm{N}$:

$$\mathrm{{2\;eq\over mol}}\cdot\mathrm{1\;M}=2\;\mathrm{N}$$

One resource you might find useful in order to check your math is Sigma-Aldrich's Normality & Molarity Calculator. Additionally, there are details relevant to the definitions of molarity and normality that you might find clearer that the ones you've conveyed in your question.