I quote the Green Book by IUPAC, 2nd printing (2008), section 1.6, enumeration item 2:

  The overall rule is that symbols representing physical quantities or variables are italic, but symbols representing units, mathematical constants, or labels, are roman. [...]

As such, the correct way to write it is $\ce{NO}_x$, because $x$ is a variable. Please note that it is only the sum formula for the compound and not its name. Since the element names still are in roman (upright) font, everything is fine.

The same goes for a formula like $\ce{C_{$n$}H_{$2n+2$}}$, which is typeset exactly like that in the Gold Book (hat tip to @Martin).

Here at StackExchange, you can type it as $\ce{NO_$x$}$ ($\ce{NO_$x$}$) or $\ce{NO_x}$ ($\ce{NO_x}$).


What is the mechanism of formation of Grignard's Reagent?

Looking at the mechanism:

We see that these are not double headed arrows, but single headed arrows. This indicates the movement of single electrons, rather than electron pairs as is common in most mechanisms.

Mechanisms that involve movements of single electrons tend to form free-radicals.

The lewis structure of Mg has the electrons seperated in order to make it clearer to read.

In step 1, the shared electron pair in $\ce{R-X}$ has been split to form two radicals: $\ce{R}{\small{•}}$ and $\ce{X}{\small{•}}$. However the $\ce{X}{\small{•}}$ radical is further attacked by an electron, this time coming from the $\ce{Mg}$. Thus forming a free-radical magnesium halide: $\ce{X^-Mg^+}{\small{•}}$.

After this, the two free-radicals undergo a termination step to form the Grignard reagent.


Concept of enthalpy at a given temperature

The enthalpy change at a fixed temperature is equal to the amount of heat you need to add in order for the final temperature to be equal to the initial temperature.  If the enthalpy change is negative, this means that the reaction is exothermic, and you need to remove heat for the final temperature to equal the initial temperature.

If the reaction is carried out adiabatically (no heat addition or removal), the temperature of the products will be either higher or lower than the temperature of the reactants, depending on whether the reaction is exothermic or endothermic, respectively.  To determine the temperature change, you can follow a two-step procedure:  1.  allow the reaction to take place isothermally, by adding or removing the heat of reaction as necessary; then, 2.  reverse the heat addition or removal that you did in step 1, by applying it to the products of the reaction from step 1.  The net effect will be the adiabatic temperature rise (or fall) produced by the chemical reaction.


Thallium triiodide does exist  and it is formulated as $\ce{Tl+I3-}$ and not $\ce{Tl^3+(I^{-})3}$. Thus $\ce{TlI3}$ is a thallium(I) compound and contains the triiodide ion, $\ce{I3^{−}}$. This is confirmed by considering the standard potential which indicate that $\ce{Tl(III)}$ is rapidly reduced to $\ce{Tl(I)}$ by the iodide.

$$\begin{align}
\ce{Tl^3+ + 2e− &-> Tl+} & E^\circ &= +1.252~\mathrm{V} \\
\ce{I2 + 2e- &-> 2I-} & E^\circ &= +0.5355~\mathrm{V} \\ \hline
\ce{Tl^3+ + 2 I- &-> Tl+ + I2} & E^\circ_\mathrm{cell} &= +0.717~\mathrm{V} > 0 \\ \hline
\end{align}$$

(data from Wikipedia).

$\ce{TlI3}$ can be prepared by the evaporation of stoichiometric quantities of $\ce{TlI}$ and $\ce{I2}$ in concentrated aqueous $\ce{HI}$, or by reacting $\ce{TlI}$ with $\ce{I2}$ in ethanol.


The quantum mechanics behind periodicity of elements

As you surmise, quantum mechanics was developed to help explain observations such as that periodicity, spectral lines and the photoelectric effect.
However, the solution of the Schrödinger equation is difficult, and not complete beyond hydrogen. For example Wikipedia states, "Unlike for hydrogen, a closed-form solution to the Schrödinger equation for the helium atom has not been found." Matrix mechanics can also predict some properties, such as spectral emissions. Neither, approach, though, is likely to give a clear insight into chemical behavior for heavier elements, particularly where relativity affects inner electron behavior for the actinides.
So you've raised a good question, and as mathematics and computational speed increase, exact solutions for other elements will be found.
[To me, it seems a bit odd (no pun intended) that there is one pair of s electrons, three pairs of p, five pairs of d and seven of f, as if it's numerology. But that's my limitation. Sigh... or ψ.]


Chemical Formulae and the Periodic Table of Elements

According to this entry in the IUPAC Gold book

https://goldbook.iupac.org/html/D/D01648.html

the symbol for Deuterium is H together with the isotopes notation.

However deuterium and tritium are routinely indicated by D and T respectively. Their use is dictated by the framework. Besides nuclear reactions, when only straightforward chemical properties* are considered, it is convenient to use a single letter rather than H and super/sub-scripts.

In organic nomenclature, while writing names for molecules, the presence of deuterium is signalled by the symbol (d).

*note that isotopic effect might however affect chemical properties at fine levels. The toxicity of heavy water is a notable striking example.


What does it mean if a catalyst support is doped?

If a substance is doped, it means that some small amount of another substance has been introduced into it. More specifically, a substance is doped by a dopant, the latter of which is the minority component. The dopant can be an additive to or a replacement within the bulk material.

In your case, some fraction of $\ce{Ti}$ atoms are replaced by $\ce{Nb}$ atoms (the replacement scenario is the usual one for crystalline materials such as yours). Thus, the physical properties of your substate will be slightly different that those of pure $\ce{TiO2}$.

See this Wikipedia entry for a more complete summary of what I have written here.


How to separate a racemic mixture of 1,2-diols

The standard method for ketalization is illustrated in the reaction of cyclohexanone 1 with ethylene glycol to form the ketal 2. The reaction is reversible which requires irreversible removal of water shifting the reaction to the right. In this reaction the toluene/water azeotrope may be separated by a Dean-Stark trap. Because of an unfavorable equilibrium, hindered ketones, such as camphor 3, are more difficult to ketalize. Standard conditions may be employed in this instance using more efficient removal of water.1
Dauben, et al.,2 have demonstrated an alternative technique for ketalizing camphor and other hindered ketones by employing high pressure at ambient temperature as illustrated in 3 --> 4.3. To illustrate the resolution technique I have chosen (R)-camphor because it is hindered, inexpensive and recoverable. The full structure of your racemic 1,2-diol was not provided so I will employ racemic butan-1,2-diol to this end. There are four possible diastereomers that can be formed in the ketalization--namely-- 5, 6, 7 and 8. The (R)-diol can form ketals 5 and 6 while the (S)-diol can afford ketals 7 and 8. Because each pair is formed under equilibrating conditions, ideally one would hope that the less hindered of the 5/6 pair and 7/8 pair dominate such that a single ketal would arise for each of the enantiomers of the diol. The success of such a technique requires separation of diastereomers, presumably via chromatography, or, ideally, by selective crystallization. If the latter method were viable, an enantiopure diol can be liberated by acid hydrolysis. The enantiomeric diol would arise by utilizing (S)-camphor, which, unfortunately, is much more expensive than (R)-camphor. continued...

One need not be limited to camphor. Other hindered ketones may be considered. The methodology described here is not a guarantee of success. Conceptually, it is no different from Ladenburg's resolution of racemic coniine with tartaric acid in the 19th-century.
1) Estrone methyl ether was successfully ketalized by user55119 using a spinning band fractionating column to remove water. The Dean-Stark method had failed.
2) W. G. Dauben, J. M. Gerdes and G. C. Cook, J. Org. Chem., 1986, 51, 4964.
3) The role of ethyl orthoformate is to scavenge water thereby forming ethanol and ethyl formate. Ketalization was also accomplished at high pressure using TMSOCH2CH2OTMS/TMSOTf.


Is it possible for a bicyclo compound to have three bridges of equal length?

Why not? The simplest bicyclic bridged compound, bicyclo[1.1.1]pentane, is a well-known example of such a molecule for its use in synthesizing the even more strained propellane. I believe your teacher was trying to prescribe the rules of bicyclic nomenclature, i.e.

  In between the prefix and the suffix, a pair of brackets with numerals denotes the number of carbon atoms between each of the bridgehead atoms. These numbers are arranged in descending order and are separated by periods.


What are the products formed in acidic hydrolysis of a tetrahydropyranyl ester?

The actual mechanism
Waylander has given an accurate reaction since this is a THP ether

The actual mechanism of the de-protection using alcoholysis would be as follows:

The actual mechanism of the de-protection using acidic hydrolysis would be as follows:

A place to start from if you didn't know the mechanism
Seeing that the compound given resembles a carbonyl group that has been protected using ethylene glycol, a mechanism using a reaction similar to that of protection of carbonyl groups was used here to hypothesis what would happen(not knowing the original mechanism) and to decide the products formed.

Hence, taking a reversed approach here, the first question I ask is:

What could have formed this acetal in the first place?

One part of the acetal seems to be connected to the carbonyl group and is part of the main molecule. The other side, however, is a different molecule. So, this is what the acetal formation looks like in the mechanism. Here, I assume $\ce{R}-$ is just an inert part of the molecule and so is equivalent to $\ce{CH3-CH2}-$. This may not be the right mechanism but it would give us a good starting point.
Formation of the hemiacetal takes place as follows:

Formation of acetal using propanol can happen as follows:

Which is the THP ether in the question.

What are the final products?
Therefore, the products of acid catalysed hydrolysis of the acetal would be $\ce{R-CH2-OH}$ and 5-hydroxypentanal which exists in equilibrium with its cyclic form dominating in mildly acidic conditions.

The hydrolysis of the hydroxyderivative of THP would happen as follows. Both alcohols will get protonated but protonating the oxygen in $\ce{-OH}$ results in the same molecule being regenerated.

Therefore, the final products would give $\ce{R-CH2-OH}$ and a hydroxy derivative of THP.
However, if $\ce{HCl}$ or $\ce{NaHSO3}$ is used instead of a weak lewis acid, we see the dominating compound in the equilibrium being 5-hydroxypentanal as shown in this paper here:

[...] As a result of the acid-catalyzed hydration of 1, the 2-hydroxytetrahydropyran (2) was formed primarily which equilibrated with its open chain form, the 5-hydroxypentanal (3). The acid-catalyst (conc. hydrochloric acid), however, promotes not only the addition of water to the C-C double bond of 1, but also that of the OH groups in 1 and 2, thus resulting in the protected compounds 2-(2-tetrahydropyranyloxy) tetrahydropyran 0, and 5-(2-tetrahydropyranyloxy)pentanal (4). The ratio of 4 + 5 to 1 + 2: depends on the amount of acid used.[...]

Reference:

Peter Vinczer , Zoltan Juvancz , Lajos Novak & Csaba Szantay (1989) AN
IMPROVED SYNTHESIS OF 5-HYDROXYPENTANAL, Organic Preparations and Procedures International: The New Journal for Organic Synthesis, 21:3, 344-346, DOI: 10.1080/00304948909356391


I recommend to fetch the crystal structures of inorganic compounds via specific Google search pattern site:crystallography.net <keyword(s)>, e.g. in your case

site:crystallography.net PbWO4

yields in this COD page where you can use JSmol to do basic visualization right in your web browser. Alternatively, you can download CIF and work further offline with numerous free tools such as Mercury or Olex2. Both are free and available for Windows, Linux and MacOS.

Mercury

Olex2


Why do we assume 100% ionization for acetic acid (a weak acid)?

You do not assume 100% ionization of acetic acid during titration. Acetic acid is a weak acid. Look at this equilibrium

$\ce{CH3COOH <=> H+   + CH3COO-}$

The base, $\ce{NaOH}$, keeps consuming $\ce{H+}$ during a titration, as a result acetic acids keeps on providing $\ce{H+}$ to maintain the equilibrium in water until all the acetic acid is converted into acetate ion.


Acidic cleavage of ether with a 2° alkyl group

In each of these cases, the ether will generally be cleaved to produce 2 alkyl halides. Acidic clevage can occur via Sn2 or Sn1 depending on the nature of the substrate. A bulky group, such as tert-butyl, will undergo an Sn1 mechanism in which the protonated ether group leaves before the nucleophile attacks. The same process is generally consistant with secondary substrates. I have drawn a mechanism involving both a primary and tertiary substrate to demonstrate this point; however, it is important to note that in the presence of most concentrated halogen acids and heat, ethers will almost always undergo complete acidic clevage in which 2 alkyl halides are generated. I have also included another mechanism involving a secondary and primary substrate. The primary substrate will undergo an Sn2 reaction due to the lack of steric hindrance; however, it is a little less clear with the secondary substrate. Will it undergo Sn1 since Sn2 would be more sterically hindered or will it undergo  Sn2 since the carbocation intermediate in Sn1 is secondary and thus less stable? Since we are using an aqueous halogen acid in this case, it will most likely undergo Sn1 since it is in the presence of a polar protic solvent, which will thus stabilize the carbocation intermediate in the process of forming.


Why does Si2H4 readily polymerize at RTP but ethene does not (requires catalyst)?

Why does $\ce{Si2H4}$ readily polymerize at RTP but ethene does not?

$\ce{Si2H4}$, the silicon analogue of ethylene, is referred to as disilene, a member of the silene family.

The silicon atoms in disilene are $\ce{sp^2}$ hybridized and the orbitals  overlap to form both a sigma and pi bond, analogous to the hybridization and bonding in ethylene. However, the $\ce{Si=Si}$ bond length is significantly over 2 angstroms, whereas the double bond length in an ethylene is much shorter at 1.34 angstroms. This much larger double bond length in silenes results in extremely poor p-orbital overlap which yields an extremely weak silicon-silicon pi bond.  While it takes energies on the order of 65-70 Kcal/mol to cause rotational cis-trans isomerization about a carbon-carbon double bond, it only takes 10-20 Kcal/mol to effect similar isomerization about the silene double bond.

Since the pi bond in a silene is so weak it doesn't require much energy to disrupt it making the molecule very reactive, much more so than an olefin.  The molecule can polymerize or react with whatever else may be around such as oxygen, water, etc.

Replacing the 4 hydrogen atoms in disilene with bulky groups stabilizes the molecule and allows the molecule to be isolated and characterized.  The bulky groups make it more difficult for another molecule to get close enough to react with the disilene pi bond.

Still, even these sterically protected molecules react with oxygen and water at temperatures not far above room temperature.  Here is a link to a full paper that discusses the preparation and reactivity of these sterically protected silenes.


How to deal with the unit "Coulomb per equivalent"?

For an electrochemical reaction, you count the atoms / ions by mol, and use the coulomb as a counting unit of charge. For a more intuitive explaination of the $n$ factor in the Faraday equation, try this analogy:

The summer olympics include swimming in a pool with lanes $50\,\pu{m}$ long.  Among the typical competitions are runs about $50$, and $100\,\pu{m}$.  Now, for the sake of the analogy, think the swimmers were representatives of ions.  Each of the swimmers could represent

a $\ce{Cl-}$ to yield $\ce{Cl}$, a change by one unit of charge.  So they swimm only one lane in one competition, $n = 1$; the single discharge happens when the swimmers clap the wall with their hand.
a $\ce{Cu^{2+}}$ wanting to yield $\ce{Cu}$. As they start with two charges now, two discharges in this single race are required ($n = 2$). Their first discharge is when they contact and turn at the opposite wall ($50\,\pu{m}$), their second and final discharge when they clap the other wall with their hand ($100\,\pu{m}$).


Why is there no non-bonding molecular orbital in the overlap between H1s and Be2s in BeH2

Well, you need two sigma bonds here, and two anti-bonding ones. And there are actually two non-bonding orbitals here.

First of all: how you write it down doesn't really work, since there are p-orbitals in Be, not only s-orbitals. You need to look at them too, not only the s-orbitals. And I'm not quite sure if you did, but don't confuse Valence Bond and MO theory.

So for Valence Bond Theory:

We know this molecule is linear. Therefore we need two orbitals on the Be which can form bonds with the hydrogens and are in an 180° angle. So we do sp-hybridization and with one sp-orbital from Be and the s-orbital from hydrogen we can form a $\sigma$ and $\sigma^*$ orbital on each side. Two sigma bonds and two empty p-orbitals on Be which are non-bonding.

For MO Theory:

We construct a $\ce{H2}$ first, in this case we could write + + and + -, so we got a bonding and an antibonding orbital. Now we put our Be in the middle. And now it get's quite complicate to write down, so here on page 10 you can find the MO of $\ce{BeH2}$. You see that the bonding "$\ce{H2}$" orbital was combined with the 2s orbital of Be in an bonding and antibonding way, and the antibonding + - orbital of "$\ce{H2}$" was combined with a p-orbital in a binding and antibinding way. The two other p-orbitals aren't used and are non-bonding, since they would have equal constructive and deconstructive interaction with both "$\ce{H2}$" orbitals. In the end same result as with the valence bond theory.


Can bone mineral be obtained in pure, crystalline form?

Apatite is the chemical name for the mineral found in bone and teeth. It is occasionally found in the form of large crystals, such as this:

BTW, just as bone is made of the mineral apatite, there are critters with skeletons of silica, another common mineral, such as Venus' flower basket!


Behavior of NaCl in water with electrodes

$\ce{NaCl}$ once in water dissociates, producing $\ce{Na+}$ and $\ce{Cl-}$ ions. Note: this process is possible only because immediately after dissociation ions are solvated by water, i.e. water molecules bind to ions

This is releasing energy and increasing effective size of ion, decreasing ion charge density. Once relatively free ions are produced, they can move between electrodes, transferring charge between electrodes. On electrodes ions may discharge taking/giving up electrons, and this is what usually happens with $\ce{Cl-}$ ion. However, if this is not an option, as the expected product is no stable enough in water, water molecule may react instead, producing another ion and either hydrogen or oxygen. This is what happens with $\ce{Na+}$: $\ce{Na}$ is too active to be stable in contact with water, so instead of sodium, hydrogen and $\ce{{}^{-}OH}$ are produced instead, resulting in  net reaction (assuming cold solution) $\ce{2NaCl + 2 H2O <=> 2 NaOH + Cl2 + H2}$ with chlorine and hydrogen released on different electrodes.


Can the change in internal energy be nonzero if temperature is constant?

TL;DR: Do not just memorise thermodynamics equations! And if you have an issue with the equations $\Delta U = 0$ or $\Delta H = 0$ for an isothermal process, read on...

The first problem
You said that an exothermic reaction corresponds to $\Delta U < 0$. This is not true. It is defined by $\Delta H < 0$.
However, we should still discuss this issue because $H$ is also a state function and in some cases $H$ is dependent only on temperature.

How to study thermodynamics
This kind of issue in thermodynamics frequently crops up here, and it is a very common mistake amongst students to indiscriminately use equations that they have learnt.
So, you have an equation that says $\Delta U = 0$ for an isothermal process, i.e. one at constant $T$. The important question here is not "what is the equation" or "what is the answer"! Instead, you should be asking yourself "how do I derive this result", and the answer will naturally follow. This should really apply to everything you do - how can you expect to apply a formula that you do not actually understand?
Furthermore, if you are able to actually understand how an equation comes about, then you don't need to memorise it - you can derive it. For example, what's the point of memorising $w = -nRT\ln(V_2/V_1)$, along with the conditions, if you can simply obtain the equation from the very definition of work $đw = -\int p \,\mathrm{d}V$ by substituting in $p = nRT/V$ and integrating? What's more, the fact that you actually substituted in the ideal gas law should tell you something about the conditions that accompany the equations - for one, it's only applicable to ideal gases. What about when you reach the equation $\mathrm{d}U = T\,\mathrm{d}S - p\,\mathrm{d}V$ and happily memorise it, only to find out later that there are actually three more equations that look exactly like it? Are you going to memorise all four? No - you need to know where they come from.

Internal energy
In general, the internal energy $U$ of a substance is a function of at least two variables, for example $T$ and $p$. However, it can be shown using statistical mechanics that for an ideal gas,
$$U = n\left(\frac{3NRT}{2} + U_0\right)$$
Here, $N$ is the number of atoms in one molecule of the gas (for example, $N = 2$ for $\ce{H2}$, and $N = 3$ for $\ce{H2O}$). $U_0$ is the molar internal energy at absolute zero (this contains terms such as electronic energy), and crucially, this depends on the exact identity of the gas. This result is one example of the equipartition theorem in action (although the theorem itself is more powerful).
We do not need to concern ourselves with its derivation now. The point is that: this equation can only be applied to a fixed amount of a fixed ideal gas at a time. For example: if I have one mole of $\ce{H2}$ at $298\mathrm{~K}$, it is going to have a different internal energy from one mole of $\ce{Ar}$ at $298\mathrm{~K}$, even though the amount of substance and the temperature are the same.
What does this mean?
Firstly, it means that for anything that is not an ideal gas, you cannot assume that $U = U(T)$ and therefore you cannot assume that $\Delta T = 0$ implies $\Delta U = 0$. Let's say I compress $10\mathrm{~g}$ of water by reversibly increasing the pressure from $1\mathrm{~atm}$ to $10\mathrm{~atm}$, at a constant temperature of $323\mathrm{~K}$. Is the change in internal energy equal to zero? No, because water is not an ideal gas.
Secondly, it means that if the amount of ideal gas changes in any way, $U$ is going to change. Let's say I have a balloon filled with $1\mathrm{~mol}$ of argon gas, and I add a further $2\mathrm{~mol}$ argon gas into the balloon, all at a fixed temperature of $298\mathrm{~K}$. Is the change in internal energy equal to zero? No, in fact it will triple because you are changing $n$ from $1\mathrm{~mol}$ to $3\mathrm{~mol}$!
Lastly, it means that if the chemical composition is changing in any way, you are automatically not allowed to say that $\Delta T = 0$ implies $\Delta U = 0$. And this is the case even if all reactants and products are ideal gases. Why not? Let's look at this reaction, and let's assume that everything there behaves as an ideal gas, and let's say that the reaction vessel is kept at a constant temperature of $300\mathrm{~K}$.
$$\ce{H2 (g) + Cl2 (g) -> 2HCl (g)}$$
Let's write an expression for the internal energy of the reactants. On the left-hand side, we have:
$$U_\mathrm{reactants} = \left[\frac{3}{2}RT + U_0(\ce{H2})\right] + \left[\frac{3}{2}RT + U_0(\ce{Cl2})\right]$$
On the right-hand side, we have:
$$U_\mathrm{products} = 3RT + 2U_0(\ce{HCl})$$
So, is $\Delta U$ equal to zero? No; even though the multiples of $RT$ cancel out with each other, the values of $U_0$ for the reacting species are different, which makes
$$\Delta U = 2U_0(\ce{HCl}) - U_0(\ce{H2}) - U_0(\ce{Cl2}) \neq 0.$$

Enthalpy
Let's go back to our (closed) system where there is only one ideal gas and it's not undergoing any chemical reaction. Within this system, we can safely say that if $\Delta T = 0$, then $\Delta U = 0$. So:
$$\begin{align}
\Delta H &= \Delta (U + pV) \\
&= \Delta U + \Delta (pV) \\
&= \Delta (pV) \qquad \\
&= \Delta (nRT) \\
&= nR \Delta T \\
&= 0
\end{align}$$
We haven't made any assumptions here apart from the ideality of the gas (the fact that $n$ is constant came from the fact that it is a closed system). So, for an ideal gas, $\Delta T = 0$ implies $\Delta H = 0$.
However, we did use the fact that $\Delta U = 0$. Therefore, what I said about the cases above where $\Delta U \neq 0$ also applies here.
Regarding the terms endothermic and exothermic, they are used to describe chemical reactions. And for chemical reactions, there is absolutely no way you can say that constant temperature implies $\Delta U = 0$ or $\Delta H = 0$. That's why we can actually use the terms endothermic and exothermic. If every single chemical reaction had $\Delta H = 0$, life would be rather plain!


The search for a suitable replacement for lithium ion batteries has been ongoing pretty much since the invention of lithium ion batteries. However, the motivation for this research has primarily been cost, and shorter charge times (along with other advantages) is simply a fortuitous property of the cheaper potassium ion batteries.

Part of the reason that potassium ion batteries are much cheaper that their lithium counterparts is the relative prevalence of potassium in the Earth's crust, potassium being present at levels about 900 times that of lithium.

According to this Wikipedia article on the development of the potassium ion battery:

  The prototype device used a potassium compound Prussian blue as the cathode material for its high electrochemical stability. The prototype was successfully used for more than 500 cycles. A recent review showed currently that several pragmatic materials have been successfully used as the anode and cathode for the new generations of potassium-ion batteries.
  [...]
  The potassium battery has certain advantages over similar lithium batteries (e.g., lithium-ion batteries): the cell design is simple and both the material and the fabrication procedure are cheaper. The key advantage is the abundance and low cost of potassium in comparison with lithium, which makes potassium batteries a promising candidate for large scale batteries such as household energy storage and electric vehicles. Another advantage of potassium-ion battery over lithium-ion battery is the possibility for charging faster. This means that the next generation of mobile phones based on potassium-ion batteries can be charged within a few minutes only.

So, while faster charging times are certainly a desirable characteristic, this was not the driving force for the development of potassium ion batteries.


CuSO4.5H2O >> CuSO4 + 5H2O : chemical or physical reaction?

Copper II forms many types of complexes with water. The aquo ion is $\ce{[Cu(H2O)6]^{2+}}$ where the oxygen ligates to the metal atom. The structure is a distorted octahedral one, two bonds are longer than the other four. (The Jahn-Teller effect$^*$ is behind the fact that two ligands are not bound as strongly as the other four). It is possible with ammonia to form various hydro complexes; $\ce{[Cu(NH3)(H2O)5]^{2+}}$... $\ce{[Cu(NH3)5(H2O)]^{2+}}$. Similar complexes form with chloride ions. Copper sulphate is coloured in solution (due to $\ce{[Cu(H2O)6]^{2+}}$ ion) and in the crystal; the anhydrous crystal is colourless and this seems to indicate that the waters are bound, due to metal to ligand transitions, as in other transition metals. (The reason that sulphate is not significantly bound to the copper in aqueous solution, as it is in the crystal, is due to the vast excess of water (water is 55.5 M) which will displace the sulphate.)

In the crystal of $\ce{CuSO4.5H2O}$ the metal is, as usual, 6 coordinated, with four water oxygens in the plane and one oxygen from a sulphate occupying each axial position. The extra ($5^{\text{th}}$)  water molecule is hydrogen bonded between the second oxygen of the sulphate and a bound water molecule in the plane. This same water's oxygen is also $\ce{H}$-bonded to a sulphate oxygen bound to the next $\ce{Cu}$ atom. The $\ce{H}$ bonds are not quite of equal length  0.287 and 0.294 nm respectively.

(Fifth water oxygen thick black circle, sulphate hashed, $\ce{Cu}$ small circle, other four oxygens, large circles. Source: Wells 'Structural Inorganic Chemistry')

So bonds are broken, $\ce{Cu}$ to oxygen, also oxygen to oxygen and oxygen to sulphur hydrogen bonds as water is removed.

$^*$ The Jahn Teller theorem shows that any non-linear molecule in a degenerate state will undergo a distortion which will lower its symmetry and split the degenerate state. The theorem does not indicate what the distortion will be, or how big it will be. Nevertheless it is extremely important in describing the geometry of metal complexes.


Potassium chlorate from potassium chloride and hydrogen peroxide

Do not attempt! Peroxide is not a strong enough oxidant to oxidize chloride ion to chlorate. Neither is oxygen. However, peroxide can oxidize chloride to poisonous chlorine gas.

Redox reactions are spontaneous if the two half reactions would produce a positive potential in a galvanic cell. Most redox half reactions have measured standard electrode potentials against a known standard so that they can be compared.

The reaction you want to do is:

$$ \ce{3H2O2 + Cl- -> 3H2O + ClO3-}$$

The two half-reactions are:

Oxidation:   $\ce{Cl- +3H2O -> ClO3- +6H+ +6e-}$
Reduction:   $\ce{H2O2 +2H+ +2e- -> 2H2O} \ \ \ \ E^ \circ =  +1.78 \text{ V}$

The reduction potential of hydrogen peroxide is on the table, but the reduction potential for the other reaction is not. However, the other reaction is the sum of two reactions on the table, so we can use Hess's Law.

$$\ce{ 2ClO3- + 12H+ + 10e- ->Cl2 + 6H2O} \ \ \ \ E^\circ = +1.49 \text{ V}$$
$$\ce{Cl2 +2e- -> 2Cl-} \ \ \ \ E^\circ = +1.36 \text{ V}$$
$$\ce{2ClO3- 12H+ + 12e- -> 2Cl- + 6H2O}  \ \ \ \ E^\circ = +2.85 \text{ V}$$

Since standard potentials do not scale by stoichiometric factor (it's in the Nernst equation instead), the value we just found also works for our oxidation reaction. However, since this reaction is reverse of the "reduction half-reaction", we need a negative sign:

Oxidation:   $\ce{Cl- +3H2O -> ClO3- +6H+ +6e-}\ \ \ \ E^ \circ =  -2.85 \text{ V}$
Reduction:   $\ce{H2O2 +2H+ +2e- -> 2H2O} \ \ \ \ E^ \circ =  +1.78 \text{ V}$
Overall: $\ce{3H2O2 + Cl- -> 3H2O + ClO3-}\ \ \ \ E^ \circ =  -1.07 \text{ V}$

This reaction is not spontaneous. Peroxide cannot oxidize chloride, at least not to the chlorate ion. Note that peroxide can and does oxidize chloride to poisonous chlorine gas in the presence of acid. This redox is spontaneous and will happen.  Do not attempt!

Reduction:   $\ce{H2O2 +2H+ +2e- -> 2H2O} \ \ \ \ E^ \circ =  +1.78
   \text{ V}$
Oxidation:   $\ce{2Cl- + 2e- -> Cl2} \ \ \ \ E^ \circ =
   -1.36 \text{ V}$
Overall: $\ce{ 2Cl- + H2O2 + 2H+ -> Cl2 + 2H2O} \ \ \ \ E^ \circ =  +0.42 \text{ V}$

Peroxide will oxidize chlorine gas to chlorate, but who wants to handle chlorine?


What is the "blue spirit" (or "red spirit") in mercury-free thermometers?

As per the Wikipedia page on Spirit Thermometers:

  The liquid used can be pure ethanol, toluene, kerosene or isoamyl acetate, depending on manufacturer and working temperature range. Since these are transparent, the liquid is made more visible by the addition of a red or blue dye.

1,4-bis(butylamino)anthraquinone - commonly known as "Oil Blue 35" - may possibly be used as the blue dye.


Are Na and Cl ions from NaCl in water separate charged particles?

Most of what you wrote is correct. It is difficult to feed many electrons into the salt water because the Coulomb forces are extremely strong, see 1 and 2, but you can feed electrons in if you drain them out elsewhere so there is no net gain or loss of charge. that is called an electric current.

You can use that to electrolyze salt water, and get chlorine as you surmise; NASA has a nice lab. BTW, you would expect to get Cl gas at the positive electrode, and do get it, but you do not get Na metal at the negative, because hydrogen bubbles out before Na. You can electrolyze molten salt, though, to get a bit of Na (or at least you see it form and then catch fire with a yellow flame).


Deriving a proper MO digram for N$_3^-$ is not completely trivial a task (e.g. a random google hit, which seems to me well done: www.chem.uci.edu/~lawm/Azide%20MO%20Diagram.pdf). In the end you will have to show that the degeneracies of the orbitals together with the electron count will not require open shells. Beware of Lewis-structures and mesomery, think of O$_2$, it looks perfectly o.k. to draw O=O, only the MOs and their degeneracies will reveal that its paramagnetic.


For what processes is Mayer's formula applicable?

The former difference is for molar heat capacity in J/K/mol, the latter for specific heat capacity J/K/kg, or J/K/g, depending on M unit.


Neutralization between Ca(OH)2 + H2SO4 in a 30% hydrogen peroxide solution

in your question formulation, you have forgotten to take into account $\ce{H2O2}$ is a weak acid.

The title should rather be:
Neutralisation between calcium hydroxide and 30% hydrogen peroxide"

Unless $\ce{Ca(OH)2}$ was in excess over $\ce{H2O2}$ - and it was said it was not - $\mathrm{pH}$ would be always significantly lower than pH of the hydroxide.

$$\mathrm{pH}=\mathrm{p}K_  \mathrm{a,\ce{H2O2}} + \log \frac{[\ce{HO2-}]}{[\ce{H2O2}]}$$

where $\mathrm{p}K_  \mathrm{a,\ce{H2O2}}=11.75$$ by Wikipedia, but see the links below.

If we consider reaction

$$\ce{Ca(OH)2 +  H2O2 ->  H2O + Ca(OH)(HO2)}$$

we need to neutralize 50% of $\ce{H2O2}$ to reach $\mathrm{pH}=\mathrm{p}K_  \mathrm{a,\ce{H2O2}}$

The hydroxide forms from $\ce{H2O2}$  the $\mathrm{pH}$ buffer solution of a weak acid and it's salt.

$$\begin{align}
\ce{Ca(OH)2  &<=>> CaOH+  + OH-}\\
\ce{CaOH+ &<=>> Ca^2+  + OH- }\\
\ce{H2O2 &<<=> H+ + HO2-}\\
\ce{H+ + OH- &<=>> H2O}\\
\end{align}$$

$\ce{Ca(OH)2}$: $\mathrm{p}K_\mathrm{b1} =1.37$, $\mathrm{p}K_\mathrm{b2} =2.43$
( Wikipedia )

Additionally, $\ce{HO2-}$ is partially eliminated by precipitation, therefore ratio $ \frac{[\ce{HO2-}]}{[\ce{H2O2}]}$ is kept low and so does $\mathrm{pH}$.

$$\ce{CaOH+ + HO2- + 7 H2O <=>> CaO2 \cdot 8 H2O v}$$

Note also the hydrogen peroxide is weakly acidic even without addition of sulphuric acid and that it's $ \mathrm{p}K_  \mathrm{a}$ depends on $\ce{H2O2}$ concentration.

H2O2 pH-and-Ionization-Constant

The solubility constant of calcium peroxide octahydrate in relation to temperature; its influence on radiolysis in cement-based materials


Why is electronegativity increasing with oxidation state increase?

Yes absolutely, electronegativities are hardly static values when you start combining elements and forming ions or molecules.

Electronegativity is simply a measure of an ability of an atom to attract electrons in a covalent framework.

So keeping in mind that like charges repel (electrons repel electrons) ... it follows that elements in a high oxidation state are more electronegative - i.e. more able to attract and stabilize electrons. High oxidation state elements have essentially been "robbed" of their electrons and likely bear some degree of partial positive charge ... this is "good" for attracting and stabilizing nearby electrons.

Conversely consider a lone pair - which is defined to have an electronegativity of 0. Lone pairs cannot stabilize other electrons and don't attract other electrons to themselves ... the electron-electron repulsion would be too great.


Why do reusable heat packs have a limited time frame before the reaction becomes irreversible?

The sodium acetate that shows this behavior is not anhydrous $\ce{NaCH3COO}$ as you write. It is the hydrated salt $\ce{NaCH3COO·3H2O}$. The anhydrous salt does not show the phenomena of surfusion. Only the hydrated form does it. If you keep the solid form of the hydrated salt a long time, I think it will slowly loose water by evaporation through the plastic bag. So it will be partly transformed into anhydrous acetate. And by heating the whole system, it will never melt entirely. So it may never return to its supersaturated form. It is necessarily to add a few drops water to get supersaturation again.


Quantum numbers required to specify the position of an electron

First of all, the question is badly formulated. It asks about "position" but clearly means "orbital", which are two different things. (An orbital is a wave function, which depends on the electrons position: $\psi(\vec r)$.) Although the difference may be difficult to explain at high school level, this choice of words is imprecise and often leads to wrong ideas about the basic concepts of quantum mechanics.

The question is obviously about the 4 quantum number ($n$, $l$, $m_l$ and $m_s$), thus including spin. Whether spin is really required to describe the spatial part of an orbital is arguable. But if we stick with high school level chemistry and consider what is typically called restricted orbitals, then both electrons ($\alpha$ and $\beta$) share the same spatial part of the orbital, and the spin quantum number is unrelated to the "position" of the electron. But the spin quantum number is still required to "distinguish" the two electrons, what seams to be what the question has in mind. (I used quotation marks, because the electrons are actually indistinguishable.)

If we go beyond high school level and into actual quantum chemistry, then it depends on what type of orbitals are considered. I already mentioned restricted orbitals. But one may also use unrestricted orbitals, where $\alpha$ and $\beta$ electrons do not share the same spatial part. Here it is obviously necessary to include the spin quantum number.

One final remark: There are actually 5 quantum numbers that specify an orbital. Analogous to $l$ and $m_l$ of spatial angular momentum, in spin angular momentum there is $s$ and $m_s$. However for electrons we always have $s=\frac{1}{2}$, therefore explicit specification of $s$ is not necessary.


Analysis of multiple spectra to determine an unknown structure

A very quick solution to this problem before my train stop.

Mass Spec gives MW of 122.

1H spectrum gives 5 aromatic, a 1H quartet at 4.8ppm, a 1H exchangeable proton at 2.4ppm and 3H doublet at 1.5ppm. The 4.8/1.5 splitting indicates a substituted ethyl chain with the C1 group having two substituents (CHRR'-CH3). The chemical shift of this 4.8ppm proton indicates some substantial substitution, such as a heteroatom. We also have a labile 1H, which could come from an alcohol or an amine.

13C spectrum shows monosubstituted benzene ring (4 carbon environments with ipso carbon downfield shifted). Also shows two alkyl carbons corresponding to the ethyl chain we have worked out above.

To summarise data to date:

1 monosubstituted benzene (C6H5)
1 ethyl chain, doubly substituted on C1 (CH-CH3)

That gives us a molecular formula of C8H9 which adds up to 105. That leaves 17...... which includes one labile proton, leaving a mass of 16.

As this is clearly a homework question, I'll leave you to join the pieces.


Isomer Identification Using Condensed Structural Formulae

isomer
One of several species (or molecular entities ) that have the same atomic composition (molecular formula) but different line formulae or different stereochemical formulae and hence different physical and/or chemical properties.

Source:
Pure and Applied Chemistry, 1994, 66, 1077 (Glossary of terms used in physical organic chemistry (IUPAC Recommendations 1994))
Pure and Applied Chemistry, 1996, 68, 2193 (Basic terminology of stereochemistry (IUPAC Recommendations 1996))
a)
C5H10
and
C10H20
different chemical formula → not isomeric
b)
CH3(CH2)4CH3 = hexane (C6H14)
and
CH3(CH2)3CH3 = pentane (C5H12)
different chemical formula → not isomeric
c)
CH3CH(CH3)(CH2)2CH3 = 2-methylpentane (C6H14)
and
CH3(CH2)2CH(CH3)2 = 2-methylpentane (C6H14)
same chemical formula and same structural formula → identical, not isomeric
d)
(CH3)3CH = isobutane (C4H10)
and
CH3CH2CH2CH3 = butane (C4H10)
same chemical formula and different structural formula → isomeric


To the best of my knowledge, ordinary alkanes do not typically absorb anywhere in the UV/Vis region. The organic molecules that do typically contain electrons in $\pi$- or non-bonding orbitals, which have energy level transitions corresponding to UV/Vis wavelengths. That said, I did find one paper1 describing absorptions around the far UV ambit, where $\lambda < 200\mathrm{nm}$.

DOI: 10.1021/jp903435c


Water is very unique liquid, because it has a higher density in the liquid state than the solid state. The maximum density of water is found at $\pu{4 ^\circ C}$, which is reported as $\pu{999.9720 kg\:m^{-3}}$ (Temperature Effects on Density). Therefore, molarity of water at $\pu{4 ^\circ C}$ can be calculated as $\frac{\pu{999.9720 g\:L^{-1}}}{\pu{18.015 g\:mol^{-1}}} = \pu{55.51 mol\:L^{-1}}$.

Following chart shows accepted densities of pure water at range of temperatures:

$$\text{Table of Densities of Pure Water at Various temperatures under }\pu{1.0 atm}$$
$$
\begin{array}{ccc}
\hline
\text{Temperature }(\pu{^\circ C }) &\text{Density }(\pu{g\:cm^{-3} }) & \text{Temperature }(\pu{^\circ C }) &\text{Density }(\pu{g\:cm^{-3} }) \\\hline
-30 & 0.98385 & +30 & 0.99565 \\
-20 & 0.99355 & +35 & 0.99406 \\
-10 & 0.99812 & +40 & 0.99225 \\
-5 & 0.99930 & +45 & 0.99025 \\
0 & 0.999840 & +50 & 0.98807 \\
+4 & 0.99997 & +60 & 0.98313 \\
+10 & 0.99970 & +70 & 0.97763 \\
+15 & 0.99910 & +80 & 0.97180 \\
+20 & 0.99821 & +90 & 0.96506 \\
+25 & 0.99705 & +100 & 0.95838 \\\hline
\end{array}
$$

So you may calculate the molarity of pure water at each temperature using the chart. An example for step by step calculations to find the molarity of pure water at particular temperature ($\pu{20 ^\circ C}$):

$$\text{Density of pure water at }\pu{20 ^\circ C } = \pu{0.99821 g\:cm^{-3}}$$
Since molarity $= \pu{mol\:L^{-1} }$, let's consider $\pu{1.0 L}$ of water.
$$\text{Mass of } \pu{1.0 L} \text{ of pure water at }\pu{20 ^\circ C} = \pu{1.0 L}\times \frac{\pu{1000 cm^{-3}}}{\pu{1.0 L}} \times \pu{0.99821 g\:cm^{-3}}= \pu{998.21 g}$$
$$\text{Molar mass of pure water at any temperature} = \pu{18.015 g\:mol^{-1}}$$
$$\therefore \text{ # of moles in } \pu{1.0 L} \text{ of pure water at }\pu{20 ^\circ C} = \frac{\pu{998.21 g}}{\pu{18.015 g\:mol^{-1}}} = \pu{55.41 mol}$$
$$\therefore \text{Molarity of pure water at } \pu{20 ^\circ C} = \frac{\pu{55.41 mol}}{\pu{1.0 L}} = \pu{55.41 mol\:L^{-1}}$$


Why can't electrons be transferred using hydrogen ions?

Firstly, $\ce{H+}$ would be called a hydrogen ion, and $\ce{H-}$ would be called a hydride ion. It would take a lot of energy to get three electrons onto a hydrogen atom, and considering that we are talking about a biological system, this just won't happen.

$\ce{H+}$ would not be employed in the transfer of electrons because it is a lone proton, and has no electrons to give. Whatever electron the hydrogen atom originally had was left behind, and the proton now exists on its own. This, I believe, should explain your homework question.

To answer your question, $\ce{H-}$ ions are common reducing agents in non-biological organic reactions, in the forms of substances like $\ce{NaBH4}$ and $\ce{LiAlH4}$, but these are extremely unstable in aqueous solution.

Instead, cells uses NAD+ to facilitate the transfer of electrons from biomolecules, like those used in cellular respiration, in the form of $\ce{H-}$. During the transfer, one electron from the $\ce{H-}$ is transferred to a nitrogen, and the other electron along with the proton forms a bond to a carbon directly across the ring from the nitrogen, converting NAD+ to NADH, a reaction that is readily reversible.


Calcium ion and sulfate did not form precipitate

While calcium sulfate is usually termed insoluble, it is not the ‘sitting at the bottom like a rock’ type insoluble; rather, it is the ‘there’s no practical way for me to get the two ions into the same solution without precipitation, but I’m still able to identify both ions in solution’ type insoluble. Those sentences don’t really help, so let’s look at numbers.
\begin{array}{lcr}
\hline
\text{salt} & \text{mass solubility} & \text{molar solubility} \\ \hline
\ce{BaSO4} & \pu{2.45e-3 g/l} & \pu{1.05e-5 mol/l}\\
\ce{CaSO4} & \pu{2.1 g/l} & \pu{1.54e-2 mol/l}\\
\ce{CaC2O4} & \pu{6.7e-4 g/l} & \pu{5.23e-6 mol/l}\\
\hline
\end{array}
To calculate your actual experiment, you would have to transform these values into $K_\text{sp}$ values and then calculate how much would precipitate at each given step. Without wanting to do that tedious calculation, you can still arrive at the general conclusion:

addition of sulphate precipitates practically all barium ions

addition of sulphate precipitates a large part of calcium ions
but: a non-neglegible amount of $\ce{CaSO4}$ remains dissolved

addition of oxalate precipitates practically all the remaining calcium.

In fact, the solubilities of $\ce{BaSO4, SrSO4}$ and $\ce{CaSO4}$ are such that a saturated solution of $\ce{SrSO4}$ will always give precipitate when barium is added but never with calcium or strontium; while a saturated solution of $\ce{CaSO4}$ will give a precipitate if either strontium or barium is added but never with calcium. This can be used to identify the heaviest alkaline earth metal in an unknown sample.


Does the charge on the acidic and basic amino acids make them more polar and hydrophilic than the uncharged polar amino acids?

Yes, they are less likely to be on the inside of a protein. Hydrogen bonds with water will be stronger, and interactions with ions as well.

  Moreover, cysteine is classified as an uncharged amino acid, but because it has an ionizable side chain, would it be more polar than serine, asparagine, etc.?

Cysteine and tyrosine are more hydrophilic when deprotonated. So at strongly basic pH, their properties are very different from that of serine. At neutral pH, however, cysteine is less polar than serine.


Reactions between Ca(OH)2 and CO2, and Sr(OH)2 and CO2

Your solid is the respective carbonate. Group II bicarbonates can form aqueous solutions, but do not form stable solids, ergo the solid flakes must be carbonate and formed along the lines of:

$$\ce{Ca(OH)2 (aq)+ 2CO2 (aq/g) <=> Ca^2+ + 2(HCO3)- (aq) <=> CaCO3 (s) + CO2 (aq/g)}$$


1. Retrosynthesis
This proposed route relies on being able to make the tetrahydrofuran via some type of ring closing etherification.
Through disconnection of the C-O bond, a linear diol is revealed which may be accessed in several ways.
Initially, I'd thought about doing a Sharpless dihydroxylation of  (Z)-hex-1,4-diene, however the starting material is surprisingly expensive and there are few literature reports of doing this dihydroxylation (and in those instances where it has been carried out, the e.e. and regioselectivity has been moderate to poor).
A different disconnection, using the chiral pool (less elegant, perhaps, but the starting material is cheap and ensures good selectivity) allowsthe diol can be disconnected across the C-C bond, revealing an aldehyde which is derived from a lactate ester.

2. Forward synthesis

Reagents and conditions: (a) TIPSOTF, 2,6-lutidine (b) DIBAL; allyl-MgBr (c) BnTCA, PPTS (d) Shi epoxidation (e) HF-Py, Py (f) TsCl, Py (g) NaN3 (h) H2, Pd/C (i) MeI
The synthesis starts with protection of methyl lactate with a TIPS group - a silyl group is chosen here as upon deprotection under fluoride conditions later in the synthesis, the THF should spontaneously close.
The TIPS protected methyl lactate may then be treated with DIBAL to afford the lactaldehyde. This must be used immediately to prevent epimerisation. In all likelihood, this could be achieved by adding a solution of allyl magnesium bromide to the same pot as the DIBAL reduction but failing this, there is good precedence for a reagent controlled boron mediated allylation (Tet. Lett. 2003, 44, 1737).
The terminal olefin installed by allylation is now setup to do a Shi epoxidation (Sharpless is terrible at terminal olefins) providing the final stereocentre. Treatment of the substrate with HF-Py should then deprotect the TIPS ether with concomitant formation of the THF ring system.
Finally, functional group transformations provide the desired natural product as shown in the scheme. Some issues may be observed when attempting to methylate the amine in the presence of the alcohol however from experience of similar substrates, the amine is likely to quaternise faster than the alcohol methylates and as such careful control of stoichiometry and conditions should allow the desired product to be isolated as the major component.
Summary
A synthesis of muscarine is proposed in 9 steps (linear) starting from methyl lactate, with the remaining stereocentres being installed using a substrate controlled allylation and a reagent controlled asymmetric epoxidation.
Addition steps could likely be cut out by (1) not protecting the alcohol that is currently protected with a benzyl group and (2) making use of an asymmetric aziridination in place of the Shi epoxidation.


Bond length and bond strength correlation

Decrease in bond length and increase in strength is only a trend, that is observed only between the same pairs of atoms, so its approximately true for C C bonds, single, double triple and similarly for N N bonds. But in general this is not true as can be seen in the following list of bonds with  approximately the same length e.g.
$\ce{H-Br,141 pm 366} \pu{ kJ mol^{-1}}$
$\ce{F-F, ~142 , ~~~~158}$
$\ce{C-O, 143 ,  ~~~~360}$
$\ce{O-O, 148 ,  ~~~145}$
$\ce{N-N, 145 ,  ~~~170}$


What you did wrong was adding the concentration of the hydron and the hydroxide ion. Also, although it hardly matters, the concentration of $\ce{H_2O}$ is considered a constant and ignored in the product.


Reduction of glucose to hexane with hydroiodic acid

Hydroiodic acid is a reducing agent.  As Wikipedia says:

  Although harsh by modern standards, HI was commonly employed as a reducing agent early on in the history of organic chemistry. Chemists in the 19th century attempted to prepare cyclohexane by HI reduction of benzene at high temperatures, but instead isolated the rearranged product, methylcyclopentane (see the article on cyclohexane). As first reported by Kiliani,[10] hydroiodic acid reduction of sugars and other polyols results in the reductive cleavage of several or even all hydroxy groups, although often with poor yield and/or reproducibility.[11] In the case of benzyl alcohols and alcohols with α-carbonyl groups, reduction by HI can provide synthetically useful yields of the corresponding hydrocarbon product (ROH + 2HI → RH + H
  2O + I2).[8] This process can be made catalytic in HI using red phosphorus to reduce the formed I2.[12]

When $\ce{HI}$ acts as a reducing agent, the oxidized product is $\ce{I2}$.  To regenerate $\ce{HI}$ from the $\ce{I2}$, phosphorus can be used.  Thus, phosphorus is not strictly required for the reaction, as long as you are OK with using $\ce{HI}$ in superstoichiometric amounts and/or getting very low yields.
The overall scheme for $\ce{HI}$ reduction of an alcohol to an alkane is first displacement of the (protonated) alcohol by iodide, leading to formation of alkyl iodide.  Then, $\ce{HI}$ catalyzes the reduction of the alkyl iodide to the alkane, leading to $\ce{I2}$ as a byproduct.  The reduction step apparently follows a radical mechanism, at least under some conditions.
I'm not 100% sure how $n$-hexane would form from glucose.  It is unclear to me how $\ce{HI}$ would catalyze the reduction of the glucose aldehydic (or anomeric) carbon to an alcohol.  But, that is only one carbon out of six.  The other carbons have bona fide hydroxyl groups, and as such can be deoxygenated and reduced by $\ce{HI}$ to form alkyl carbons.  For example, Lv et al. report that hydrocarbons can be obtained from the five-carbon sugar alcohol xylitol in yields above 60%.  Unlike glucose, xylitol has no aldehydic/anomeric carbon, and only has alcohol groups.  This process did use phosphorous acid as an in situ iodine recycling agent.  Also, Yang et al. were able to obtain 5-methylfurfural from fructose by using $\ce{HI}$, without using an in situ agent to recycle $\ce{}$ to $\ce{HI}$.


lindane extraction from aqueous solution using hexane

According to this publication, lindane has the following properties:

Practically insoluble in water
Specific gravity = 1.85
Melting point = 113 $\mathrm{^oC}$
Solubility in acetone > 200 g/L
Solubility in n-heptane* = 10-14 g/L
* Used as surrogate for hexane solubility for this argument.

Notes: lindane is by far the most soluble in acetone, though only a small amount could have been used in the 1:9 acetone:water solvent.  If we assume that the reaction mixture were extracted with an equal volume of hexane (conservative I think, probably used more hexane), then the solution would be about 1:10:10 acetone:water:hexane. Given the properties of lindane, I'd expect it to make for a gooey, separate, small layer between the water and hexane.

Prediction:

  Bottom layer - Water, undissolved iron, dissolved iron chlorides.
  Middle layer - Acetone, lindane, partially reacted lindane, some iron chlorides.
  Top layer - hexane,
  Small amounts of lindane and organic byproducts.


Do molecules with polar bond, but with no dipole moment experiences a greater effect from the london dispersion forces?

Yes, an example of a molecule would be CO$_2$, which has no dipole moment but does have two polar bonds. Although CO$_2$ has no dipole moment, its charge distribution gives rise to a quadrupole moment. This quadrupole moment provides a handle for the molecule to interact with other particles (with other quadrupoles, but also with dipoles, induced dipoles or electric charges). As you might know, dipole-dipole interaction scales as $r^{-3}$ where $r$ is the distance between the particles and Van der Waals interaction (induced dipole-induced dipole interaction) scales as $r^{-6}$. It can be shown that quadrupole-quadrupole interaction scales as $r^{-5}$ and therefore represents a stronger interaction than the induced dipole-induced dipole interaction.


Could I settile for galvanized steel over stainless for salt corrosion resistance?

I'm a chemist, but this seems more like a question for an engineer.
Unfortunately I don't have some references to back up my answer.

If you use galvanized steel you have a protective layer on the outside of the steel. As soon as you handle it for attaching, welding... you destroy/scratch/bruise this top layer and the underlying steel gets in contact with the salt solution which is normally aggressive towards metal. This mostly is right for e.g. chloride salts in a mix with oxygen and I'm not quite sure how corrosive MgSO4 is.
Here is a paper about corrosive currents of steel in Saltwater/cement solutions, but that`s not my speciality: http://www.jmst.org/fileup/PDF/pass543.pdf

Stainless steel is more robust, because it's an alloy of steel and chromium. Therefor it's not just covered with a protective layer but saver as a whole. An I guess it's therefor recommended.

(As a side note, some years ago I had a field trip to a big chemistry company with lots and lots of pipelines. We had some time there with the technical stuff in charge of teh pipelines. They explained, that the most annoying pipelines were the ones with salt water, since it was more aggressive towards the pipes than acids, mineral oils, gases etc.)


difference in boiling point of 2-iodopentane and 3-iodopentane

I have spent a quite a time to figure out "so-called" irrational behavior of the boiling points of iodoalkanes compared to relevent othe halpalkane, but I have seen noting really radical, what OP claimed there are. The following chart listed boiling points of all possible long chain halides of pentane (all have been measured at $\pu{760 mmHg}$):
$$
\begin{array}{ccc}
\text{locant} &\text{fluoropentane} & \text{chlororopentane} & \text{bromoropentane} & \text{iodopentane}\\\hline
1 & \pu{62.8 ^\circ C} & \pu{108 ^\circ C} & \pu{130 ^\circ C} & \pu{155 ^\circ C}\\
2 & \pu{58.4 ^\circ C} & \pu{99 ^\circ C} & \pu{117 ^\circ C} & \pu{150 ^\circ C}\\
3 & \pu{59.5 ^\circ C} & \pu{97.5 ^\circ C} & \pu{119 ^\circ C} & \pu{145 ^\circ C}\\
\end{array}
$$

Please be aware that, all of these values are extracted from EPA(USA) website. Also note that values of 2-fluoropentane and 3-fluoropentane are predicted on EPA, but ChemSpider gave experimental values of $\pu{56 ^\circ C}$ and $\pu{56.7 ^\circ C}$, respectively. Also, the boiling point of 3-iodopentane ($\pu{145 ^\circ C}$) has also been predicted, but it is available commercially at Alfa Aesar, which listed the boiling point of it as $\pu{146.6 ^\circ C}$. Thus, these listed values are very reliable. Therefore, I'd say that the values OP has must be not so reliable. I admit that I have to dig deeper to find a reliable source of those predicted values. All three bromo compounds have no problem because all of them are available in Sigma-Aldrich.


Which has a higher lattice energy CaO or NaCl?

Lattice enthalpy of an ionic solid is the minimum thermal energy required to completely convert one mole of that ionic solid to isolated gaseous ions. Here "isolated ions" in the sense that the distance between the separated ions is sufficient to reduce the electrostatic interaction between them to zero practically.

It is also the energy released for the reverse process of  settling of isolated  ions into the solid lattice.

Greater the charge densities of ions higher the electrostatic interaction and so higher the lattice energy by magnitude. A smaller anion/cation with more charges will have higher charge density. Here calcium ion (114pm)is  not only smaller than sodium ion(116pm) but also having twice the charge as that of sodium ion.
Similarly oxide ion(126pm) is much smaller than chloride ion(167pm). Former is doubly charged where as latter has only single charge. For both these reasons CaO has a much high lattice enthalpy (3414kJ/mol) compared that of NaCl (787kJ/mol)


Kinetic energy of molecules in liquid state?

The average translational kinetic energy of a molecule is $3kT/2$ irrespective of whether the molecule is in the gas, liquid, or solid phase.  In the liquid the motion giving rise to kinetic energy is restricted to a narrower range about the potential energy minimum  than it is in the gas phase. The equipartition theorem is quite general. A derivation is given here;  Derivation of mean kinetic energy

For ideal gases, the result may also be derived via kinetic theory. However, the most general derivation comes from the equipartition theorem.


How to dissolve steel without affecting aluminium alloy

The part of the steel insert that is stuck to the aluminum is not accessible to water.  Let's look at linear coefficients of thermal expansion,

http://www.engineeringtoolbox.com/linear-expansion-coefficients-d_95.html

Aluminum is 22.2 ppm/K. Steels are smaller, the largest being 304 austentic at 17.3 ppm.  Ambient temp is around 24 C, a 400 F kitchen oven is around 204 C.  That should not be hot enough to lose temper in the alloys.  If the aluminum hole is an inch in diameter, worst case increasing the temp 180 K times 4.9 ppm/K net gives you 882 ppm or 22 microns of slack.  The steel insert pushes out.


Which phase is more efficient for temperature changes?

A liquid is better than a solid for cooling an object, because it can make more complete contact with the object being cooled, plus it can undergo convection, so the layer of liquid next to the object being cooled is replaced by cooler liquid as the liquid circulates around the object.

A liquid is denser than a gas (giving better contact with the object being cooled) and also will likely have a higher heat capacity, which translates into a higher "cooling" capacity as well. But don't forget about "wind chill factor" - a moving blast of air is a lot more efficient at cooling an object than stationary air. Convection ovens take advantage of this principle too, but for heating rather than cooling. Liquids will probably be more effective than gases for most cooling applications, but there will probably be some exceptions.


Why is mercury much more toxic when inhaled than injected?

I guess the answer is that mercury atoms stick with each other very well (e.g. high surface tension in comparison to most other liquids $\gamma_{293K}^{\ce{Hg}} = 0.47 \text{ N/m}$, where  $\gamma_{293K}^{\ce{H2O}} =  0.073 \text{ N/m}$ [1]) and that this continues to happen in the blood stream. Such clumps of mercury are not able to pass the blood brain barrier (BBB) effectively, especially in comparison to finely divided dissolved vapor, which is transported trough the BBB by the formation of a complex with cysteine.


The parent structure of chlordiazepoxide consist of the heteromonocyclic component 2H-1,4-diazepine that is fused to a benzene ring.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the name of the fused parent structure is generated as follows:

  P-25.2.2.4 Heteromonocyclic components fused to a benzene ring

  Unless listed as a retained name (…), a benzene ring fused to a heteromonocycle of five or more members (a benzoheterocycle) is named by placing the locant(s) indicating the position(s) of the heteroatom(s) at the front of the name consisting of the fusion prefix ‘benzo’ followed by a retained name, a Hantzsch-Widman systematic name, or a name formed by skeletal replacement (‘a’) nomenclature (…). The locants cited correspond to the full bicyclic structure. As in Hantzsch-Widman names, locants are placed in the order corresponding to the order of citation of the heteroatoms in the heterocyclic component. The locant ‘1’ is always assigned to the atom of the heterocyclic component next to a fusion atom. Heteroatoms are allocated lowest locants as a set, without regard to kind; if there is a choice, lowest locants are assigned in accordance with the seniority of the ‘a’ prefixes (…). In general nomenclature locants may be omitted when the name is unambiguous; for preferred IUPAC names locants must be cited. The letter ‘o’ of the ‘benzo’ prefix is elided when followed by a vowel. Indicated hydrogen is placed at the front of the name, when required.
  (…)

Accordingly, the name ‘benzodiazepine’ consists of the prefix ‘benzo’ followed by the name ‘diazepine’. The locant ‘1’ is assigned to the atom of the heterocyclic component next to a fusion atom and the heteroatoms are allocated lowest locants as a set (here: 1,4):

The same numbering of locants applies to the complete structure of 7-chloro-2-(methylamino)-5-phenyl-3H-1,4-benzodiazepine 4-oxide:


What is the mechanism for this Horner-Wadsworth-Emmons (HWE) reaction?

First dil. aq. HCl hydrolyses the starting material to the dialdehyde.
Second the diethylphosphonoacetate anion reacts with one of the aldehyde groups to give the expected intermediate alkoxide. This is in equilibrium with the carbanion shown below. This is not normally depicted in the HWE mechanism as there is usually no pathway forward from it, however in this case there is. This carbanion can, and does, undergo a second HWE with the other aldehyde which proceeds through the usual mechanism to give the products shown in the question


Does the hybridization model gives us any further insight on Molecular Geometry?

I find orbital hybridization of very limited use.  Orbital hybridization is a mathematical manipulation of atomic wavefunctions, but that that is the only relationship to a quantum mechanical description of an atom.  Presently, we have no evidence that supports the claim that orbital hybridization happens.  The reason why I am not in favor of using orbital hybridization to describe chemical phenomena is because it is not a predictive tool.  We only know the hybridization of an atom after we know the structure.

VSEPR, however, is a predictive tool, although it has no significant quantum mechanical basis.  The VSEPR model is based upon the idea that pairs of electrons (whether bonding or not) will repel one another.  Geometrical factors are considered to determine how far away pairs of electrons can be from one another (e.g 180 degrees apart for two pairs, and 120 degrees apart for 3 pairs).  The model accurately predicts the H-C-H angle in methane and also predicts that the H-N-H angle in ammonia is closer to 109 degrees than 120 (as is the case in a molecule such as $BF_3$.  Because VSEPR assumes ideal conditions, the model starts to fall apart when considering complex molecules.  For example, while the VSEPR model correctly predicts that $SF_4$ is seesaw shaped, it can only predict that the structure won't be precisely seesaw (with F-S-F angles of 180, 120 and 90 degrees).  It only allows us to suggest that the lone pair on the sulfur will be a "space hog" in a manner of speaking and repel the bonding electrons.

Now, on to the last part of your question why do we still use it?  (This is somewhat speculative and opinion-based.) We have to blame the organic chemists here.  orbital hybridization terminology ($sp^2, sp^3$, ...) provides a lexicon for describing the bonding environments of carbon atoms that is compact and visually constructive for organic molecules.  If I state that there is an $sp^2$ carbon in a molecule, you can begin to visualize what that part of the molecule looks like, that it is a region of planarity, that it cannot be a chiral center, that it likely has a double bond and therefore may undergo certain types of chemical reactions, etc. So while orbital hybridization doesn't provide any insight into molecular geometry, it provides a basis for terminology that chemists can use to transmit meaningful information.


How are N95 masks different from dust masks?

The short answer to your question "What's the science behind the N95 filter?" is that it is what's called a tortuous path filter, removing particulates from a flow of contaminated air by impact and absorption of the particulate onto the filter material or by trapping particles of sufficient size between the fibers constituting the filter.

The term N95 comes from the NIOSH (National Institute for Occupational Safety and Health, a body of the US government) rating where the 95 means that the material is certified to block 95% of particles 0.3 μm or larger in diameter. The limiting factor toward achieving the stated level of performance during usage is how well the mask seals to the face of the user. Whiskers, dirt and improper fit can all allow a larger portion of particles to circumvent the filter material. The image in the OP shows a mask having a valve to facilitate breathing and maintain a seal by opening during the exhale portion of a breath.


What software can calculate aqueous solution equilibria?

ChemEQL is pretty nice in some instances and free, but it looks like it has been mostly abandoned.

I'm aware of Titrator, but it only runs on Windows and I've never had  a chance to run it. It's also free.

I've heard good things about Visual MINTEQ, but it is Windows only and I've never had a chance to run it. Supposedly well maintained.

There is also PHREEQC, which sort of works on stuff other than Windows. I hear it is harder to use, but probably worth it.


Will the ions in salt bridge be used up?

The ions in the salt bridge are meant as charge carriers, and using a suitable diaphragm intends to keep the mass transport pretty low.

Nethertheless, some of the electrolyte (like $\ce{KCl}$, $\ce{KNO3}$, $\ce{Na2SO4}$ or $\ce{Bu4N+ ClO4-}$ to mention a few) does enter the other half cell.  This is why, as one criterion, the electrolyte in the salt bridge is choosen to not interfer with the electrochemical reaction you want to characterize (analysis like cyclic voltammetry) or execute (electrochemical synthesis).  Likewise for solvents used in cyclic voltammetry, there is an electrochemical window, the intervall of electrode potentials where it is safe to use them so that they are not oxidized, or reduced, respectively.

And if the electrolyte is consumed and charge carriers are absent, provided the electrical potential applied is not altered, yes, the electrical current drops to 0.


Electrons split in Stern-Gerlach experiment but not in Thompson's experiment?

As has already been suggested by some of the comments to your question, there is a fundamental difference between a Stern-Gerlach-type experiment and a Thompson-type experiment, and this difference lies in the type of interaction of the particles.

First of all, as Ivan Neretin points out, a Stern-Gerlach experiment is performed with neutral species (the original experiment involved silver atoms), while Thompson's experiment employed a beam of charged electrons.

The Coulombic or Lorentz interaction between charged particles (or charged particles and fields) is rougly four orders of magnitude larger than the interaction between neutral particles and fields. In the latter case, one can have interaction between a magnetic dipole moment and a magnetic field and/or interaction with a electric dipole moment and an electric field. These interactions are referred to as Zeeman and Stark effects. The important thing to realize is that the sign and magnitude of the interaction depends on the quantum state of the particle (or in other words on the relative orientation of its angular momenta). It is therefore a quantum-state specific interaction. As a consequence particles in one state may be attracted to a region with high fields, while particles in another state may be attracted to regions with low fields, as if you have two different compasses; one that points to the North pole and one that points to the South pole.

Electromagnetic interactions between charged particles and fields are so strong that the quantum-state specific effects are much harder to see (and the Coulomb and Lorentz forces dominate).

Let's go back to the Stern-Gerlach experiment. In their experimental setup, Stern and Gerlach used an inhomogenous magnetic field with a well-defined gradient. The silver atoms they employed have one unpaired electron that could be oriented with respect to the external magnetic field "up" or "down", these different quantum states experience different forces and are therefore separated.

In Thompson's experiment charged electrons are send through a magnetic field where the force they experience only depends on their charge and velocity vector with respect to the magnetic field. This force is identical for all electrons, so no splitting is observed.


How do I determine the average carbon oxidation numbers in ethanol and ethanal?

What you have been asked to calculate is what is called an average oxidation number. This is often used in inorganic chemistry (not because it is rather powerful, but because you can easily get away with it) but lesser in organic chemistry.

To properly determine oxidation states, you should do the calculation on a per-atom basis. And doing that would give you $\mathrm{-III}$ and $\mathrm{-I}$ for ethanol’s two carbons and $\mathrm{-III}$ and $\mathrm{+I}$ for ethanal’s. This method has a number of advantages:

you are always dealing with integers; no need for any fractional states
it is easy to see exactly which carbon’s oxidation state changed.

The downside of this method is that you cannot use it if you don’t know the structural formula. Consider if, for example, you only saw $\ce{C2H6O}$ or $\ce{C2H4O}$ — especially the first one could represent two different structures with different properties. This is were average oxidation states come in handy: you calculate an average of all the atoms of a single element. You may remember that oxygen is typically given $\mathrm{-II}$, hydrogen $\mathrm{+I}$ (except in peroxides and hydrides, respectively; although oxofluorides should get a special mention). You may also remember that the sum of oxidation states of a neutral compound must be $\pm 0$ that allows you to calculate the average oxidation states as follows:

$$\begin{align}\text{For }\ce{C2H6O}\text{:}&\\
0 &= 2 x + 6 \times (+1) + (-2)\\
0 &= 2x + 6 - 2\\
0 &= 2x + 4\\
-4 &= 2x\\
x &= -2\\
\\
\text{For }\ce{C2H4O}\text{:}&\\
0 &= 2 x + 4 \times (+1) + (-2)\\
0 &= 2x + 4 - 2\\
0 &= 2x + 2\\
-2 &= 2x\\
x &= -1\end{align}$$

Since these are averages, you can also arrive at the same conclusion by taking the arithmetic mean of all carbon atoms’ oxidation state; you will obviously arrive at the same result:

$$\begin{align}\ce{C2H6O}\text{:}&\\
\bar{x} &= \frac{-3 + (-1)}{2}\\
\bar{x} &= \frac{-4}{2}\\
\bar{x} &= -2\\
\\
\ce{C2H4O}\text{:}&\\
\bar{x} &= \frac{-3 + 1}{2}\\
\bar{x} &= \frac{-2}{2}\\
\bar{x} &= -1\end{align}$$


Segregation in the lower atmospheric region is prevented by various mechanisms, mainly turbulent mixing of the atmosphere by rising and falling air streams. The main driving force of the large-scale atmospheric circulation is solar heating. The most important circular air flow pattern in the troposphere is the Hadley cell circulation. (Considerable information on the movement and mixing processes in the atmosphere was gained from monitoring following numerous tests of nuclear weapons in the atmosphere mainly during 1951–1962. The radioactive substances served as tracer material.)

Winds mix the atmosphere much faster than gaseous molecules can settle according to their weight. The natural composition of the dry atmosphere remains constant in the lower 80–100 km. Only above 100 km, segregation takes place because the concentration of light gases (e.g. helium) decreases more slowly with altitude than those of the heavy gases.

The same atmospheric mixing processes apply to chlorofluorocarbons. For example, trichlorofluoromethane $(\ce{CCl3F})$ is nearly uniformly distributed throughout the lower 15 km. At higher altititudes, the abundance of trichlorofluoromethane decreases – not because of its density, but because it is decomposed by UV radiation. By way of comparison, the less reactive tetrafluoromethane $(\ce{CF4})$ is uniformly distributed up to at least 50 km.
(Source: Frequently Asked Questions About Ozone (1): How Can Chlorofluorocarbons (CFCs) Get to the Stratosphere If They’re Heavier than Air? in Assessment of Ozone Depletion: 1998 – World Meteorological Organization Global Ozone Research and Monitoring Project – Report No. 44 [html] [pdf])


Relationship between coordination numbers and stoichiometry

Surely there is a logic behind that, and a pretty simple one. Let's take a crystal and count all $\ce{Ca}$ atoms in it. We'll get some huge number N (possibly about as huge as $N_A$). Now how many fluorine atoms are there? Each $\ce F$ is a neighbor of some $\ce{Ca}$, so if we just count all neighbors of all $\ce{Ca}$, we won't miss anything... and will end up with 8N. But wait, each $\ce F$ is a neighbor of four $\ce{Ca}$, so it was counted 4 times! We divide by 4 to compensate for that and happily arrive at 2N, which is consistent with the formula $\ce{CaF2}$.

So yes, it is no coincidence that the ratio of coordination numbers is also the ratio of stoichiometric coefficients.


Easiest way is with an iodide, e.g. $\ce{NaI}$ or $\ce{KI}$, and starch.

An old trick was to sprinkle a bit of $\ce{KI}$ on a (damp) slice of potato, stick a couple of wires into the potato and attach a small cell to the wires. Iodine is liberated at the positive terminal and it reacts with starch in the potato to make a deep blue color.

There are more suggestions from Flinn Scientific, such as using an indicator to show acid or alkali liberated at electrodes.


Is reduction of copper oxide a combustion reaction?

By the question asked, I assume OP is in high school of freshman in college. Thus, I'd use this general description of combustion as a starting point:

  Combustion reactions are common and very important. Combustion means burning, usually in oxygen but sometimes with other oxidants such as fluorine. A combustion reaction happens quickly, producing heat, and usually light and fire. Combustion describes how the reaction happens, not the reactants and products. Chemists as early as Lavoisier suggested that people get their energy from combustion-like reactions, but even though the products and reactants are the same when you burn food in a fire and in your body, the way it happens is different. In a combustion reaction, the thing that burns (the reactant that isn't $\ce{O2}$ or $\ce{F2}$) is called the fuel. Combustion reactions are a type of redox reaction.

As  M. Farooq pointed out a combustion reaction happens quickly, producing heat, and usually light and fire. For example, lets look at combustion reaction of an alkene (a hydrocarbon). If it is a complete combustion, the fire have a blue flame:

$$\ce{C_nH_{2n} + $\frac{3n}{2}$ O2 -> nCO2 + n H2O}$$

If it is a partial combustion, it can have a multiple $\ce{C}$ compounds as products, and have a yellow flame due to presence of elemental $\ce{C}$:

$$\ce{C_nH_{2n} + x O2 -> m C + p CO + $(n-m-p)$CO2 + n H2O}$$

where $x = \frac{2(n-p-m) +p}{2} = \frac{2n-2p-2m +p}{2} = \frac{2n-p-2m)}{2}$. In your reaction would not produce fire and it didn't use either oxygen or other oxidants ($\ce{CuO}$ is not that type of oxidant). It is true that the reaction is a redox reaction.


Nanocar racing competition and the race tracks used

While I can't find specific justification for the surface, the first competition involved both gold and silver "racetracks."

Drivers gear up for world’s first nanocar race
How to build and race a fast nanocar

The competition involves propulsion and imaging using STM so the substrate must be conductive. Gold (and silver to a lesser degree) is particularly useful, since it won't easily oxidize and is easy to purify.

I've seen several comments about having 'tracks' in the gold surface, which suggests a well-defined crystal face with reconstruction, e.g. Au(100):

(Image from Wikipedia)


If numbering from each end of the parent chain gives the same locants, the next criterion stipulates that lowest locants are assigned to the substituent that is cited first as a prefix in the name. This means in effect that alphanumerical order is used to establish the numbering when a choice is possible since alphanumerical order is used to establish the order of citation of substituent prefixes in the name.

On this matter, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

  P-14.4 NUMBERING

  When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:

  (…)

  (f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;

  (g) lowest locants for the substituent cited first as a prefix in the name;

  (…)

Note that Rule f takes precedence over Rule g. If no decision can be attained by application of P-14.4(f), the lowest set of locants is selected according to P-14.4(g) in their order of their citation in the name.

For example, a correct systematic name is 4-ethyl-5-methyloctane (not 5-ethyl-4-methyloctane) since ethyl is cited first as a prefix in the name.

Another correct systematic name is 4-ethyl-5-propyloctane (not 5-ethyl-4-propyloctane). Again, the reason is that ethyl is cited first as a prefix in the name. The size of the substituent is not relevant in this step.


Which of the following alkenes would produce two products when they undergo an addition reaction with hydrogen bromide?

Well as mithoron said.its about symmetry.Let's see how but-1-ene comes out to be the answer

On addition of HBr markownikov rule is followed but lets draw both product.

1.)CH2=CH2 + HBr--->Br-CH2-CH3 + CH3-CH2-Br  (both products are 1-bromoethane)

2)CH2=CH-CH2-CH3 + HBr--->Br-CH2-CH2-CH2-CH3 + CH3-BrCH-CH2-CH3 (notice the                name of both the product. 1st one is 1-bromobutane. 2nd one is 2-bromobutane)

3)CH3-CH=CH-CH3 + HBr---->CH3-BrCH-CH2-CH3 + CH3-CH2-BrCH-CH3 (both products are same i.e both are 2-bromobutane)

4)CH3-CH2-CH=CH-CH2-CH3 + HBr--->CH3-CH2-BrCH-CH2-CH2-CH3 +
                    CH3-CH2-CH2-BrCH-CH2-CH3 (here also both the compounds are same i.e both are 3-bromohexane)

So what distinguishes the 2nd one from these three is that the second molecule  is unsymmetrical.You can find its unsymmetrical by looking at the no:of carbon to both side of double bond.If they are equal then molecule is symmetrical.


Isopropyl alcohol still leaves evaporation residue after distillation

I assume the problem is, like in the watch glass the solutes after IPA evaporates, i.e. higher boiling point solutes.

Making the distillation slower and reduce the amount of aerosol to escape to distillate flask will surely diminish the amount of impurities. More time, less temperature elevation, it should solve the problem.


Identifying if reaction is reduction or oxidation. Is there an error in this question?

Before the reaction, every carbon is attached to to other carbon atoms, one hydrogen atom and one chlorine atom. So oxidation state of each carbon = 0+0+-1(due to C-H bond) +1(due to C-Cl bond)= 0; average oxidation state of carbon is also zero.
In the product only two carbon atoms suffer a change in bondings; now each of these two carbon atoms is attached to one carbon through double bond, another carbon through single bond and a hydrogen atom; so O.S. of this carbon = 0+0+-1 = -1; so to conclude four carbon atoms remain in zero oxidation states, where as the doubly bonded carbon atoms are in -1 oxidation state. so average oxidation state of carbon = -2/6 = -1/3

P.S.:Fractional oxidation state results only when we take average, and that too in some cases


$\mathrm{sp}$ hybridized carbanions are more stable as the lone pair of electrons is placed in an orbital consisting of high s-character. In general, electrons are more stable in s orbitals rather than the p orbitals of the same shell. This can be understood in a simple way, by thinking that the s orbitals are closer to the nucleus, and can also be verified by the n+l rule.
Now, when we say $\mathrm {sp}$ hybridized species are more electronegative than $sp^2$ hybridized species, we mean that the $\mathrm{sp}$ orbital are more accomplished in holding an electron within them when compared to their $\mathrm{sp}^2$ counterparts. As electrons are more stable in an $\mathrm{sp}$ orbital, electrons are pulled more towards them rather than the $\mathrm{sp}^2$ orbitals, which makes us say that $\mathrm{sp}$ orbitals are more electronegative than $\mathrm{sp}^2$ orbitals.


My confusion regarding Hydrogen peroxide

Each oxygen atom is connected to a hydrogen atom(which develops -1 charge on oxygen and +1 on hydrogen) and to another $\textbf{oxygen}$ atom which contributes no charge to both(0 and 0). Similar for other oxygen atom. Hence oxygen state of each oxygen atom is -1 and for each hydrogen atom it is +1.

In general, in peroxide linkage, oxygen has -1 oxidation state


Anions like nitrate forms strong hydrogen bond with water. Here water bridges the nitrate oxygen with hydrogen bonds giving it a very stable arrangement.


How can I easily explain to a layperson the difference between types of "contains"

I would say something like this:

All matter -- that is, all substances -- are made of atoms.  The properties of a given substance are determined by (a) the type and (b) the arrangement of the atoms in that material.

Chemists refer to the "types" of atoms as "elements".  For example, vegetable oil and table sugar (sucrose) are both composed of three types of atoms only: carbon, hydrogen, and oxygen.

The arrangement of the atoms in the two materials differs drastically.  (You could refer to structure diagrams at this point if your friend is interested.)  First, oil has much less oxygen and slightly more hydrogen per carbon atom.  Second, the way the atoms are connected -- their arrangement -- is very different.

Chemists sometimes give a name to a group of similar arrangements of atoms.  For example, "sugars" are compounds that (a) have carbon, hydrogen, and oxygen atoms in ratios that are approximately $\ce{CH2O}$, and (b) often form ring structures, as seen in table sugar, glucose, and corn starch.  But the key words are "approximately" and "often" because some molecules that chemists call sugars don't satisfy these rules.  Biochemists interested in metabolism sometimes refer to molecules that can "easily" be converted into "true" sugars as sugars.

Naming in chemistry is complicated by the fact that chemists use words for different purposes: to indicate metabolic similarity, to indicate structural similarity, or to indicate similarity in reactivity.  Those different purposes sometimes lead situation where chemist B says that what chemist A calls a "sugar"  isn't really a sugar.  But that's just an argument about humans' imperfect way of naming things, not about chemistry.  There is a lot of wiggle room in the words "approxiately", "often", "similar", and "easily".  So if we focus on metabolism only, the glycerol part of the vegetable oil arrangement (i.e. structure) is a bit like a sugar.  But it is dissimilar from sugars in terms of both structure and reactivity.

Again, what ultimately matters is (a) the type and (b) the arrangement of the atoms in a substance.  Those two things are together necessary and sufficient for a complete description of all substances.  Whether to call particular arrangements of particular atom types "sugars" or not is something people can argue about.


How to determine the dimensions of a bixin molecule?

There is a paper on bolaamphiphiles (1) where it is stated that measured surface area corresponding to the area of a bixin molecule lying on the water surface with most methyl groups pointing outside is approx. $\pu{1.0 nm^2/molecule}$ at zero pressure. If the methyl groups were parallel to the water surface, a molecular area of approx. $\pu{1.7 nm^2/molecule}$ would be occupied.

Also, there is a linear size of a single molecule stated as the following:

  Bixin therefore constitutes a valuable starting material for various bolaamphiphiles which can be tied within lipid membranes. Its molecular length is about $\pu{3.1 nm}$ in the all-trans configuration.

I think it might be worth for OP to have a look at the full-text (1) paper as it also discusses various methods of characterization and altering of bolaamphiphiles.

A few years later a crystal structure has been published (2). There is not much to add, as it seems like there is no contradiction with the data obtained with reflection spectroscopy (1). Here are the notable linear dimensions of cis-bixin and overall geometry in 2 projections:

(1) Fuhrhop, J. H.; Krull, M.; Schulz, A.; Moebius, D. Langmuir 1990, 6 (2), 497–505. DOI: 10.1021/la00092a034
(2) Kelly, D. R.; Edwards, A. A.; Parkinson, J. A.; Olovsson, G.; Trotter, J.; Jones, S.; Malik, K. M. A.; Hursthouse, M. B.; Hibbs D. E. 1996, J. Chem. Res., 446, 2640


As I recall, $\ce{Cu}$ is inert to $\ce{HCl}$, however $\ce{Al}$ will react with it to produce $\ce{H2}$ and $\ce{AlCl3}$, resulting in a turbid, dark gray solution that bubbles vigorously as the $\ce{H2}$ gas is evolved. The reaction may not have been instantaneous, especially if the $\ce{Al}$ had a passivation layer of oxides on it.

Once initiated, that reaction would've continued even after removal of the external current, and is probably the source of the black precipitate you describe.


Can ammonia be detected by a 1D-1H-NOESY experiment?

In the 1H,1H-NOESY experiment you are using the nuclear Overhauser effect to transfer magnetization from one proton to another proton in close proximity (< 5 Å).  This only works within a single molecule, unless you have a reasonably stable complex you can't observe signals in the NOESY between protons of different molecules in your solution.

Ammonia has three protons which are at least chemically equivalent, their signals will appear at the same chemical shift. In a 1D NOESY you irradiate at the frequency of a specific proton and then observe the transferred magnetization at other frequencies. You can't observe magnetization transfer between protons at the same frequency that way, as your irradiation will destroy any signal at this specific frequency.

You can ignore the explanation above, as the NOESY experiment you are using is different. In metabolomics the NOESY 1D is used for water suppression, this is a different experiment than the 1D NOESY I explained above.

The common variant of this I found also uses presaturation to suppress water. This presaturation will also affect all signals that exchange quickly with the solvent, and the protons of ammonia are exchanging with water. So I would expect the ammonia signal to be far weaker or even not visible in such an experiment.

If you want to observe ammonia, you have to use a water suppression method that doesn't affect exchanging protons.


How many peaks would there be in H1 NMR of ethene?

You need to consider whether the protons in ethene are chemically equivalent, magnetically equivalent, or both. The lewis structure of ethene:

Because of the symmetry of the planar ethene molecule, all H atoms are chemically and magnetically equivalent. The spin system is therefore A$_4$. The signal is not split because the environment is the same for all H atoms. The signal would therefore be a single peak. Compared to an "average" methyl group, the ethylene protons are less shielded, and would appear at a higher chemical shift than methyl groups.


There is a general principle known as “last in-first out” that I think applies here. If a molecule contains multiple acidic positions with significantly different pKa, you can selectively react on the least acidic position (in your case, the alcohol).

Specifically, I would try treating your molecule with 2 eq. Of strong base such as NaH or LDA. The first equivalent will deprotonate all of your acid to form the quite stable metal acetate. The second equivalent removes the alcohol hydrogen to generate a less stable alkoxide. If you add 1.0 eq. Of the electrophile it ought to react selectively on the more reactive alkoxide nucleophile, leaving the metal acetate unaffected, then you can just neutralise in work up.


Enthalpy of a reaction changes with temperature, does this mean bond strengths change with temperature?

No: bond strengths as such do not change with temperature. The potential energy surface holding the atoms together, i.e. all the bonds, is constant for a given molecule.

However, all molecules have vibrational and rotational energy levels and molecules also have bulk kinetic energy due to thermal motion in a gas or liquid. As the temperature increases the bulk thermal energy increases, i.e. molecules move faster, but also more vibrational and rotational energy levels become excited, which means that the vibrational and rotational quantum numbers increase. This means that less extra energy is needed to break a bond if the temperature is already high, compared to that needed at low temperature, simply because the bond is already has a lot of internal (vibrational/rotational) energy.

Not all molecules have exactly the same energy but instead there is an exponential distribution of energies (the Boltzmann distribution) where the higher the energy is the fewer molecules are there that have this energy.

The subject of  Statistical Mechanics describes these molecular effects and also forms the foundation for Classical Thermodynamics, i.e. what we usually call 'Thermodynamics', and which was developed before atomic/molecular theory.


A useful way to reuse non-recyclable plastics: Turn them to petrochemicals, shall we?

The broad question here is whether "non-recyclable" plastics can be turned into fuel. The short answer to that is yes. More on that below. However, the specific proposal to convert polystyrene to isooctane is not particularly feasible. I'll address that first.
The proposed process begins with depolymerization of polystyrene back to styrene. Although lab-scale processes have been reported that give as high as 80% yield of styrene monomer, there is not yet an economically feasible process for scale-up. In fact, if styrene could be depolymerized practically, it would not be considered "non-recyclable". Furthermore, if the styrene could be recovered, it would be much more valuable as styrene than as gasoline, so there would be no incentive to spend the money on the subsequent steps described by OP.
Nonetheless, if one wanted to make liquid fuel from styrene, there is no need to produce isooctane specifically. Liquid fuels are generally mixtures of a variety of hydrocarbons, and the exact chemical composition does not especially matter as long as the autocombustion temp, vapor pressure and other relevant properties fall within an acceptable range.
So rather than make isooctane, one could simply hydrogenate styrene to ethylbenzene, an already common anti-knock agent or all the way to ethylcyclohexane (a more likely final product than octene).
Going back to the original premise of getting liquid fuel from plastics, just as there is no need to produce a specific compound from the styrene, there is no need to recover styrene specifically from the polymer, nor even restrict one's self to a specific polymer type such as polystyrene. Instead, bulk plastic mixtures can be pyrolyzed to produce a mixture not all that different from some crude oils and gases. These liquids and gases can be used as fuels themselves or mixed with existing petroleum products and refined.
Many challenges remain to make such a process economically viable, however, not least of which is the cost of collecting and transporting plastic waste (generally quite low density) compared to recovering oil from the ground.


The primary candidates for volatilization of odorous sulfur gases are $\ce{SO2}$ and $\ce{H2S}$.  Unless the elemental sulfur has been physically ground and dispersed into the air it is not likely the direct source of the smell.

According to the International Volcanic Health Hazard Network, who of course study gaseous volcanic emissions with respect to safety issues:

  "Sulfur dioxide ($\ce{SO2}$) is a colourless gas with a
  characteristic and irritating smell. This odour is perceptible at
  different levels depending on the individual's sensitivity, but is
  generally perceived between 0.3-1.4 ppm and is easily noticeable at 3
  ppm."

Regarding $\ce{H2S}$, according to the US Occupational Safety and Health Administration (OSHA), $\ce{H2S}$ has a perceptible smell at between 0.01 and 1.5 ppm.

The two gases have different smells, $\ce{H2S}$ being that of rotten eggs and that of $\ce{SO2}$ being described as "pungent and irritating".  So that would be one way to differentiate which gas you are smelling.

The most likely candidate is $\ce{SO2}$ as our air is generally oxidizing, and to me elemental sulfur smells more like $\ce{SO2}$ than rotten eggs.


If I add some KOH, where will the reaction shifts?

The equilibrium will shift to the left, as $\ce{Ni^2+}$ is removed from the equilibrium by the precipitation of $\ce{Ni(OH)2}$:

$$\ce{Ni^2+ + 2 OH- \rightleftharpoons Ni(OH)2\downarrow}$$

The precipitate forms when $[\ce{Ni^2+}]$ and $[\ce{OH-}]$ are sufficiently high to exceed the solubility product of $\ce{Ni(OH)2}$ ($K_{sp}=5.48 \times 10^{-16}$ at 25 °C, source). The concentration of $\ce{Ni^2+}$ before addition of $\ce{KOH}$ depends on the stability constants of the ammin complexes (there are several species in equilibrium with each other, as the ligands are exchanged stepwise, and not all at once). Nickel hydroxide will dissolve again when $\ce{NH3}$ is added, as this shifts the equilibrium of ammin complex formation to the right.


3-Bromo-3-ethylpentane is so bulky that an E2 elimination seems to be unlikely as the $\ce{OH}^{-}$, despite being a strong base, has a very low chance of directly attacking the beta $\ce{H}$. Moreover, proceeding by the E1 mechanism, we have a tertiary carbocation which is quite stable compared to primary and secondary carbocations.
However, as @Zhe has pointed out, looking at the experimental data gives the best answer. A general rule of thumb (although exceptions may exist) is that tertiary alkyl halides tend to follow E1 mechanism rather than E2 even under strong bases.