The name (1s,4s)-1-ethyl-4-methylcyclohexane is correct; it is even the preferred IUPAC name (PIN). It refers to the cis isomer of 1-ethyl-4-methylcyclohexane.

The preferred IUPAC name for the corresponding trans isomer is (1r,4r)-1-ethyl-4-methylcyclohexane.

In general nomenclature, according to Subsection P-93.5.1.2 of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the stereodescriptors 'cis' and 'trans' may be used to denote the relative configuration of substituted monocyclic compounds.

  P-93.5.1.2 Relative configuration: the stereodescriptors 'cis' and 'trans'

  The stereodescriptors 'cis' and 'trans' are used to show the relationship between two ligands (atoms or groups) attached to separate atoms that are contained in a ring or a ring system. The two ligands are said to be located 'cis' to each other if they lie on the same side of a plane. If they are on opposite sides, their relative position is described as 'trans'. The appropriate reference plane of a ring or ring system (the ring being in a conformation, real or assumed, without reentrant angles at the two substituted atoms) is the mean plane of the ring(s). The stereodescriptors denote relative configuration; the absolute configuration must be described by CIP stereodescriptors such as 'R' and 'S'.

  (…)

In preferred IUPAC names, all stereogenic units shall be specified (unless an omission is explicitly allowed). The presence of stereogenic units does not require the molecule to be chiral.

According to Subsection P-93.5.1.1.1, the absolute configuration in substituted monocyclic compounds shall be described by Cahn-Ingold-Prelog (CIP) stereodescriptors.

  P-93.5.1.1.1 Absolute configuration

  In substituted monocyclic compounds the absolute configuration is specified in preferred IUPAC names by the CIP stereodescriptors such as 'R', 'S', 'r', and 's'.

This rule explicitly applies to achiral cyclic compounds, too.

  P-93.5.1.1.2 Achiral cyclic compounds

  The configuration of achiral cyclic molecules is also specified by CIP stereodescriptors in preferred IUPAC names.

  (…)

Therefore, this rule also applies to (1s,4s)-1-ethyl-4-methylcyclohexane, which is given in the question, and its isomer (1r,4r)-1-ethyl-4-methylcyclohexane. (The examples mentioned in the Blue Book include the similar compounds (1s,4s)-cyclohexane-1,4-diol, (1r,4r)-cyclohexane-1,4-diol, and (1r,4r)-4-chloro-4-methylcyclohexan-1-ol.)

The lower-case stereodescriptors 'r' and 's' are CIP stereodescriptors like 'R' and 'S'. The stereodescriptor 'r' or 's' is chosen by applying the CIP sequence rules as for the stereodescriptors 'R' and 'S'. However, the stereodescriptors 'r' and 's' are used to indicate the absolute configuration of pseudoasymmetric centers, whereas 'R' and 'S' are used to indicate the absolute configuration of real chirality centers.

  P-91.2.1.1 Cahn-Ingold-Prelog (CIP) stereodescriptors

  Some stereodescriptors described in the Cahn-Ingold-Prelog (CIP) priority system, called 'CIP stereodescriptors', are recommended to specify the configuration of organic compounds, (…), and in the nomenclature of natural products (…). The following stereodescriptors are used as preferred stereodescriptors (…):

  (a) 'R' and 'S', to designate the absolute configuration of tetracoordinate (quadriligant) chirality centers;

  (b) 'r' and 's', to designate the absolute configuration of pseudoasymmetric centers;

  (…)

The 'pseudoasymmetric' is defined in Subsection P-92.1.1 (d) as follows.

  Stereogenic units are called pseudoasymmetric (center, axis or plane) when they have distinguishable ligands 'a', 'b', 'c', 'd', two and only two of which are nonsuperposable mirror images of each other (enantiomorphic). (…). The 'r/s' (…) stereodescriptors describing a pseudoasymmetric stereogenic unit are invariant on reflection in a mirror (for example 'r' remains 'r', and 's' remains 's'), but are reversed by the exchange of any two ligands ('r' becomes 's', and 's' becomes 'r'). Lower case stereodescriptors are used to describe pseudoasymmetric stereogenic units.

For example, (2R,3r,4S)-2,3,4-trichloropentane contains two chirality centers (labeled 'R' and 'S', respectively) and one pseudoasymmetric center (labeled 'r').

The two 1-chloroethyl ligands of the pseudoasymmetric center are nonsuperposable mirror images of each other. On reflection of the entire molecule in a mirror, the configurations of the chirality centers are reversed ('R' becomes 'S', and 'S' becomes 'R'), but the configuration of the pseudoasymmetric center is invariant ('r' remains 'r').

The same principle applies to (1s,4s)-1-ethyl-4-methylcyclohexane, which is given in the question.

The two halves of the 1,4-disubstituted cyclohexane are nonsuperposable mirror images of each other. Therefore, the compound contains two pseudoasymmetric centers, which are invariant on reflection in a mirror ('1s,4s' remains '1s,4s').

Although this case is explicitly mentioned in the Blue Book, it may not be obvious how to choose the stereodescriptors 'r' or 's' for 1,4-disubstituted cyclohexanes. The ranking is a complex process and requires the use of auxiliary descriptors.

Generally, the stereodescriptors 'r' and 's' are assigned by applying the Cahn-Ingold-Prelog (CIP) Priority System as for the stereodescriptors 'R' and 'S'. The order of precedence is reached by the construction of a hierarchical digraph and the application of the Sequence Rules.

The constitutional formula of a cyclic molecule is transformed into an acyclic digraph according to the methodology that is described in Subsection P-92.1.4.3 and which has been recommended by Prelog and Helmchen.

In order to obtain the acyclic digraph of a stereogenic unit, the didentate ligand (i.e. the ring) is transformed into two monodentate ligands by leaving intact in each case one bond with the core of the stereogenic unit and cleaving the other bond. At the end of each of the two branches thus obtained, a duplicate atom of the core is attached. (Note that this methodology is similar to the case of multiple bonds.) For example, for the pseudoasymmetric stereogenic unit at the atom numbered '1' of (1s,4s)-1-ethyl-4-methylcyclohexane:

where $\ce{(C)}$ is a duplicate representation of the core of the pseudoasymmetric stereogenic unit.

As the digraph is acyclic, it is now possible to specify the stereodescriptor 'R' or 'S' for the stereogenic unit at the atom numbered '4', which is carrying the methyl group, in each branch. These are auxiliary descriptors that are different in each branch.

Since the auxiliary descriptors are different, they allow the branches to be ranked using Sequence Rule 5, which states that 'R' has priority over 'S'.

  P-92.1.3.5 Sequence Rule 5

  An atom or group with descriptor 'R', 'M', and 'seqCis' has priority over its enantiomorph 'S', 'P' or 'seq Trans'.

Therefore, the configuration of the pseudoasymmetric stereogenic unit at the atom numbered '1' of (1s,4s)-1-ethyl-4-methylcyclohexane can be determined to be 's'.

In the same way, the configuration of the pseudoasymmetric stereogenic unit at the atom numbered '4' of (1s,4s)-1-ethyl-4-methylcyclohexane can be determined as 's'.

Therefore, the complete configuration of (1s,4s)-1-ethyl-4-methylcyclohexane, which is given in the question, is indeed (1s,4s).


How to create a simple do-it-yourself procedure to measure diffusion?

There are way more sensitive ways to measure dextrose (glucose) concentration than polarimetry. For example, there are reducing sugar tests like the Dinitrosalicylic Acid test or (better) the 2,2'-Biconchoninic acid test. Just measure the dextrose concentrations inside and outside of the bag. I have protocols for these tests if you want; it involves getting ahold of the reagents and a spectrophotometer (unless you just want to do it qualitatively, look at the color and see if they're different), but presumably that would be no more difficult than getting hold of a polarimeter.

As far as the second solute S goes, you want something that's a lot larger than glucose. I'm only familiar with dialysis tubing that separates small molecules from proteins, so I would go with that. Cheap regenerated cellulose dialysis tubing has a molecular weight cutoff of 10,000, so use glucose (MW 180) and cheap (and soluble) Bovine Serum Albumin (MW 66,000). You always want something at least five times less than or five times greater than the cutoff for these tubes.

Be careful, there are reducing tests for sugars and reducing tests for proteins, they often use the same techniques but in different conditions, so with one protocol they'll be very sensitive for one and much less sensitive for the other. As long as you have blanks, it shouldn't be a problem, but make sure you're following the right protocol.

Or if you really want to be fancy, there's a fluorimetric test that is extremely specific to glucose, using glucose oxidase.

Or if you want to be DIY, just use a diabetes glucose meter and strips.


How can 30 ml of water be heated in less than 10 seconds?

Well, let's do some math:

Assuming 30 mL of water is 30 g, and we want to heat our water from 20 °C to 90 °C, the energy required is:
$$\begin{align}E&=C_Pm\Delta K \\
&=\left(4.18 \mathrm{\frac{ J}{gK}}\right)(30\mathrm{\ g})(70\mathrm{\ K})\\
&=8.778\mathrm{\ kJ}\end{align}$$

So how much power do we need to do this in a given time? "Instant" doesn't really mean anything, so let's go with 10 seconds:

$$\begin{align}P&=\frac{E}{t}\\
&=\frac{8778\mathrm{\ J}}{10\mathrm{\ s}}\\
&=877.8 \mathrm{\ W}\end{align}$$

This is not an enormous amount of power, but the trick is that it all has to go into heating the water. A good microwave outputs a fair bit more power than this, but it generally doesn't all get absorbed by such a such a small amount of water in only 10 seconds. Your best bet is probably an electric heating element directly inserted into the liquid, though I don't know if you can get a ~1000 W one small enough to sit in that much water.

As Jon Custer notes, it's not necessary to produce all the heat at once. If you heat some kind of thermal reservoir and flow the liquid past/through it, it reduces the demands on your heat source.

Edit: Also, I just tried this with a 1200 W microwave and it only took 15 seconds. How fast do you really need this coffee?


What are the IUPAC names for singly- and doubly-protonated piperazine?

The relevant rule in the current (corrected) version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) is:

P-73.1.1.2 General rule for systematically naming cationic centers in parent hydrides
A cation derived formally by adding one or more hydrons to any position of a neutral parent hydride (listed in Chapter P-2), or whose degree of hydrogenation has been modified (see P-31) is named by replacing the final letter ‘e’ of the parent hydride name, if any, by the suffix ‘ium’, preceded by multiplying prefixes ‘di’, ‘tri’, etc. to denote the multiplicity of identical cationic centers. (…)

Therefore, the preferred IUPAC names are piperazin-1-ium and piperazine-1,4-diium, respectively.


Why is 11-oxabicyclo(4.4.1)undeca-1,3,5,7,9-pentaene aromatic?

The aromatic system should be planar. The oxygen is not part of the aromatic system.
Please note, of course, that your drawing does not capture the 3-dimensional geometry of the system. You would still have to analyze the carbon skeleton to see if it satisfied the planarity constraint enough to be aromatic.


Are crystallographic omit maps only used with molecular replacement?

Experimental phasing and molecular replacement provide initial phases used to start model building. Later on (refinement, validation) phases are calculated from the model.

If the phases are what determine the final electron density irrespective of the amplitudes used

Both amplitudes and phases are used to determine the electron density.
If you calculate the phases omitting part of the model, the electron density in this place won't be biased by that part of the model. (It may be biased by the absence of the model in this place).


Does activation energy depend on steric factors?

The Arrhenius model of a rate rate constant is very basic so has to be modified in many cases. One of these is when steric factors (generally undefined exactly) are used as an empirical  multiplier to A and used as a rationale to fit experimental data.


Dominant contribution of v = 0 → v = 1 transition in vibrational spectroscopy

The reference should really be to the Boltzmann distribution where the  population is proportional to $\exp(-E_n/(k_\mathrm BT))$  where $E_n$ is the energy of the $n^{th}$ energy level. This shows that the population is greatest when $n$ is small. Also you need to know that absorption is proportional to the population difference between any two levels. As the vibrational frequency is generally a few hundred wavenumbers, or higher, and $k_\mathrm BT$ at room temperature $\approx 210$ wavenumbers you can see that the population of $n= 0$ levels can be very large compared to that when $n\gt 0$.


Reaction Kinetics and Stoichiometry -- Mass conservation

Reaction rates should be defined in terms of the Extent of reaction ($\xi$) that corresponds to the number of moles (or the molarity for reactions in solution) of specie $i$ divided by the the stoichiometric number, $\nu_i$:
$$\xi = \frac{[i]}{\nu_i}$$
Therefore, considering that the first reaction should be:
$$\ce{A + A -> B ~=~ 2A -> B}$$
The corresponding reaction rates are:
$$\frac{\mathrm{d}\xi}{\mathrm{d}t} = -\frac{1}{2} \frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = \frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t}$$
In this way, the reaction rate is always the same (positive) number, independently on the specie we are referring to.
The global rate equations are:
\begin{align}
\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} &= −2k_1[\ce{A}]^2−k_2[\ce{A}][\ce{B}]−k_3[\ce{A}][\ce{C}]−k_4[\ce{A}][\ce{D}]\\
\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} &= 2k_1[\ce{A}]^2−k_2[\ce{A}][\ce{B}]+k_4[\ce{A}][D]\\
\frac{\mathrm{d}[\ce{C}]}{\mathrm{d}t} &= k_2[\ce{A}][\ce{B}] -k_3[\ce{A}][\ce{C}] +k_4[\ce{A}][\ce{D}]\\
\frac{\mathrm{d}[\ce{D}]}{\mathrm{d}t} &= k_3[\ce{A}][\ce{C}] -k_4[\ce{A}][\ce{D}]\\
\end{align}

It is conceptually wrong to assume that

  the products of $\ce{A + D}$ should be split evenly between $\ce{B}$ and $\ce{C}$.

The reaction scheme you described is the "parallel reactions" model. The amount of $\ce{A}$ consumed by each reaction depends on the specific rate. The stoichiometry of the first step says that every time 1 molecule of $\ce{B}$ is produced 2 molecules of $\ce{A}$ are consumed. But maybe $k_1$ is very small and so all the other process will consume $\ce{A}$ much more rapidly than this one.

You cannot predict what will happen in a reaction system just looking at the stoichiometry of each step. If you have a large amount of $\ce{C}$, it will probably consume a lot of $\ce{A}$ in the third step, producing a lot of $\ce{D}$ that, in turn, will produce a lot of $\ce{B}$, etc. Stoichiometry is just one piece of information. But kinetics is much more that this.


It's called deposition or desublimation. It's for example how snow is created in clouds or how frost forms on leaf. When leaf is very cold, then vapour surrounding leaf looses so much energy that it changes into solid state.


Why is the bond angle H-P-H smaller than H-N-H?

Starting point: 2s orbitals are lower in energy than 2p orbitals.

The $\ce{H-N-H}$ bond angle in ammonia is around 107 degrees.  Therefore, the nitrogen atom in ammonia is roughly $\ce{sp^3}$ hybridized and the 4 orbitals emanating from nitrogen (the orbitals used for the 3 bonds to hydrogen and for the lone pair of electrons to reside in) point generally towards the corners of a tetrahedron.

In the analogous case for phosphorus (phosphine, $\ce{PH_3}$), the $\ce{H-P-H}$ bond angle is 93.5 degrees.  This angle indicates that the phosphorus atom is almost unhybridized (the bond angle would be 90 degrees if it were completely unhybridized). The 3 bonds from phosphorus to hydrogen roughly involve the three 3p orbitals on phosphorus and the phosphorus lone pair of electrons resides in the 3s orbital of phosphorus.

So the question becomes, why does the nitrogen atom in ammonia choose to hybridize, while the phosphorus atom in phosphine does not? Let's start by listing the factors that will stabilize or destabilize geometries in these compounds.

There are two choices for the central atom:

remain unhybridized: [$\ce{p}$ orbital - H1s] bonds will form and they will be arranged 90 degrees with respect to one another.  As a result,

substituents will be arranged closer together and destabilizing steric interactions will be increased
due to the absence of s-character in the $\ce{X-H}$ bonds emanating from $\ce{X}$, the electrons in these orbitals will be higher in energy
the lone pair electrons will be highly stabilized since they will reside in a low energy, pure s orbital

or

hybridize: [$\ce{sp^3}$ orbital - H1s] bonds will form and they will be arranged 109 degrees with respect to one another.  As a result,

steric interactions will be reduced because the tetrahedral orbital arrangement will space the attached substituents further apart
due to the presence of s-character in the $\ce{X-H}$ bonds emanating from $\ce{X}$, the electrons in these orbitals will be lower in energy
the lone pair electrons will be less stabilized since they will reside in a higher energy orbital that contains significant p-character

An example:

Let's now consider the example of $\ce{NH_3}$ and $\ce{NF_3}$.  Fluorine is much more electronegative than hydrogen, therefore we would expect electron density in the $\ce{N-F}$ bond to be shifted away from nitrogen towards fluorine.  Because of this electron redistribution, the $\ce{sp^3}$ orbital on nitrogen involved in this bond will contain less electron density.  Consequently it will rehybridize - if there is less electron density in the orbital, there is less of a need for lower energy, electron stabilizing s-character in this orbital.  The orbital will wind up with higher p-character and the s-character that has been "saved" can be used to stabilize other electrons (the lone pair!).  Our prediction would be that $\ce{NF_3}$ should have more p-character in its $\ce{N-F}$ bonds than $\ce{NH_3}$ has in its $\ce{N-H}$ bonds.  As a result we would expect the $\ce{F-N-F}$ bond angle in $\ce{NF_3}$ to be smaller than the $\ce{H-N-H}$ bond angle in $\ce{NH_3}$.  Indeed the bond angle in $\ce{NF_3}$ is 102 degrees compared to the 107 degrees observed in ammonia!

Back to our problem:

Nitrogen (3.04) is more electronegative than phosphorus (2.19), which is about the same as hydrogen(2.2).  In our $\ce{X-H}$ bonds, we would therefore expect more electron density around the central atom when $\ce{X~=~N}$ than when $\ce{X~=~P}$.  Using the same reasoning used in our example, we would then expect the $\ce{N-H}$ bonds in ammonia to have higher s-character (and a larger $\ce{H-N-H}$ angle) than the analogous bonds in phosphine, just as observed.  The fact that phosphorus, being a second row element, has longer $\ce{P-H}$ bonds (142 pm) than ammonia (102 pm) lessens steric problems in the unhybridized geometry and further lowers the energy of the unhybridized configuration for phosphine.

In the case of ammonia, the shorter $\ce{N-H}$ bond lengths (increased steric interactions) and the increased electron density in the $\ce{N-H}$ bonds makes the  hybridized case the lowest energy.  Whereas in the case of phosphine, steric interactions are of less consequence because of the longer bond lengths and the decreased electron density in the bonds around phosphorus make the energetics of the (nearly) unhybridized geometry more favorable.


Find the concentration of the HCl solution when mixed with 1.265g of ACES-K+

Let us call the potassium salt "A-" and its corresponding acid for "HA".
The total volume of the final solution is 88.42 + 27.59 = 116.01 ml.
The final concentration of [HA] + [A-] = 1.265/(220.29 *0.11601) = 0.049499 (M).
[K+] = [HA] + [A-] = 0.049499 (M).

The equilibrium equation gives:
6.54 = 6.85 + log ([A-]/[HA]) => [A-]/[HA] = 0.48977 => [A-] = 0.48977 * [HA].
=> [HA] + 0.48977 * [HA] = 0.049499 => [HA] = 0.03322 (M) and [A-] = 0.016279 (M).

The charge balance is:
[H3O+] + [K+] = [OH-] + [Cl-] + [A-]
At pH = 6,54 [H3O+] = 10^-6.54 and [OH-] can be neglected in comparison with [A-].
=> [Cl-] = [H3O+] + [K+] - [A-] = 0.03322 (M).

The amount of chloride ions in the final solution will be 0.03322 * 0.11601 = 0.003853 mole. Thus, the concentration of the hydrochloric acid used will be:
(0.003853 * 1000) /27.59 = 0.1396 or about 0.140 M.


Spontaneous Reaction and Gibbs Free Energy

I think you are confusing two related concepts here. Spontaneity of a reaction is determined by the net change in free energy, $\Delta G$ - notice the capital Delta.

Which direction you will move along a reaction coordinate is determined by $dG$ - note the little $d$.

$\Delta G$ is the overall change in free energy when comparing products to reactants. It tells you whether the reactants are "uphill" or "downhill" of the products.

$dG$ is the slope of the free energy surface at your current position on the reaction coordinate. In the equation you gave, that's given by Q/K. K basically tells you what the equilibrium ratio between reactants and products is, and Q tells you what the current ratio is. The ratio of Q/K is an indication of how far to the left or right you are of equilibrium.

This picture might help explain it a little better:

Image the free energy surface as a gravitational potential energy surface (we can get away with this because free energy is the thermodynamic potential for systems at constant T & P). The ball represents the reacting system, and the x-axis is a reaction coordinate.

$dG$ (the differential) tells us the slope at a given point - it tells us where the system is going to go right now. $\Delta G$ tells us the overall change - it tells us where the system is going to eventually end up.

When you look up standard state $\Delta G$ values for a reaction, you are determining whether an overall reaction is spontaneous or not - you are figuring out which side of the equilibrium has more stuff - reactants or products.

When you look at Q, you are figuring out which direction things need to go to get back to equilibrium, based on where you currently are.


How high in (mmH2O) is the column of water in the tube?

Consider the air pressure in mm of mercury given. Is water more or less dense? Would an amount of water of the same mass as the mercury in the column take up more or less room?

So what would you need to do to change the 735 mm of mercury to an equivalent mass of water, given the densities cited in the problem?

The diameter of the column is red herring; don't go off on the kipper.


What is the importance of iron in complexes and in the process of electron transport?

The Fe-S proteins are part of the electron-transport chain (together with other complex molecules such as other proteins and coenzymes etc.) In my opinion, I think it would be unfair just to describe the Fe-S complex without understanding bits of the processes and other important molecules involved in this system since it involves a chain of interlinked events.

Components of the Electron-Transport Chain

  Many of the proteins embedded in the inner mitochondrial membrane are
  organized into the four respiratory complexes of the
  electron-transport chain. Each complex consists of several protein
  components that are associated with a variety of redox-active
  prosthetic groups with successively increasing reduction potentials.

The mitochondrial respiratory chain consists of a series of sequentially acting electron carriers, most of which are integral proteins with prosthetic groups capable of accepting and donating either one or two electrons. Three types of electron transfers occur in oxidative phosphorylation:

direct transfer of electrons, as in the reduction of $\ce{Fe^3+}$ to $\ce{Fe^2+}$;
transfer as a hydrogen atom ($\ce{H^+ + e_}$); and
transfer as a hydride ion ($\ce{:H^-}$), which bears two electrons.

In addition to NAD and flavoproteins, three other types of electron-carrying molecules function in the respiratory chain: a hydrophobic quinone (ubiquinone) and two different types of iron-containing proteins (cytochromes and iron-sulfur proteins). Ubiquinone (also called coenzyme Q, or simply Q).

The electron-transport chain involves several different molecular species, including:

Flavoproteins, which contain tightly bound FMN or FAD as prosthetic groups
and which may participate in one- or two-electron transfer events.
Coenzyme Q, also called ubiquinone (and abbreviated CoQ or UQ), which can function in either one- or two-electron transfer reactions.
Several cytochromes (proteins containing heme prosthetic groups), which function by carrying or transferring electrons), including cytochromes b, c, c 1, a, and a3. Cytochromes are one-electron transfer agents in which the heme iron is converted from $\ce{Fe^3+}$ to $\ce{Fe^2+}$ and back.
A number of iron–sulfur proteins, which participate in one-electron transfers involving the $\ce{Fe^3+}$ and $\ce{Fe^2+}$ states.
Protein-bound copper, a one-electron transfer site that converts between $\ce{Cu^+}$ and $\ce{Cu^2+}$.

The Electron-Transport Chain Can Be Isolated in Four Complexes

1. Complex I (NADH Dehydrogenase or NADH–Coenzyme Q Oxidoreductase)

Complex I (also called NADH dehydrogenase) passes electrons from NADH to CoQ. Situated in the inner mitochondrial membrane,e contains one molecule of flavin mononucleotide (FMN; a redox-active prosthetic group that differs from FAD only by the absence of the AMP group) and eight or nine iron–sulfur clusters that participate in the electron transport process.

a.    Iron–Sulfur Clusters

These are redox active. They commonly occur as prosthetic groups in iron–sulfur proteins.
There are four common types of iron–sulfur clusters: Those designated [2Fe–2S] and [4Fe–4S]. Clusters consist of equal numbers of iron and sulfide ions and are both coordinated to four protein Cys sulfhydryl groups. The [3Fe–4S] cluster is essentially a [4Fe–4S] cluster that lacks one Fe. One means of identifying these clusters
utilizes the fact that their sulfide ions are acid labile.
The oxidized and reduced states of all iron–sulfur clusters differ by one formal charge regardless of their number of Fe ions. This is because the Fe ions in each cluster form a conjugated system and thus can have oxidation states between the +2 and +3 values possible for individual Fe ions.

The overall equation for the reaction is:

$$\ce{NADH +  H+ +  CoQ -> NAD+  +  CoQH2}$$

b. The Coenzymes of Complex I

FMN and ubiquinone (CoQ), the coenzymes of Complex I, can each adopt three oxidation states. Although NADH can only participate in a two-electron transfer, both FMN and CoQ are capable of accepting and donating either one or two electrons because their semiquinone forms are stable. In contrast, the cytochromes of Complex III, to which reduced CoQ passes its electrons, are only capable of one-electron reductions.
FMN and CoQ, therefore, provide an electron conduit between the two-electron donor NADH and the one-electron acceptors, the cytochromes.

2. Complex II [Succinate Dehydrogenase or Succinate–Coenzyme Q Oxidoreductase (SQR)]

Complex II, which is also the citric acid cycle enzyme succinate dehydrogenase, passes electrons from succinate to CoQ to yield fumarate and CoQH2. It does so with the participation of a covalently bound FAD, a [2Fe–2S] cluster, a [4Fe–4S] cluster, a [3Fe–4S] cluster, and one heme b560.

Complex II is important because it injects these relatively high-potential electrons into the electron transport chain.  It catalyzes the transfer of electrons to coenzyme Q.

However, its source of electrons (in other words, the substance being oxidized) differs from the oxidizable substrate (NADH) acted on by NADH-CoQ oxidoreductase. In this case, the substrate is succinate from the citric acid cycle, which is oxidized to fumarate by a flavin enzyme;

$$\ce{Succinate + E-FAD -> Fumarate +  E-FADH2}$$

The notation E-FAD indicates that the flavin portion is covalently bonded to the enzyme. The flavin group is reoxidized in the next stage of the reaction as another iron–sulfur protein is reduced:

$$\ce{E-FADH2  +  FeOSoxidized -> E-FAD _ Fe-Sreduced}$$

This reduced iron–sulfur protein then donates its electrons to oxidized coenzyme Q, and coenzyme Q is reduced;

$$\ce{Fe-Sreduced  +  CoQ  + 2 H+ ->  Fe-Soxidized  +  CoQH2}$$

The overall reaction is:

$$\ce{Succinate  + CoQ -> Fumarate  + CoQH2}$$

a. Cytochromes

  Cytochromes are redox-active proteins that contain heme groups that
  reversibly alternate between their Fe(II) and Fe(III) oxidation states
  during electron transport.

3. Complex III (Coenzyme Q–Cytochrome c Oxidoreductase or Cytochrome bc1 Complex)

Complex III passes electrons from reduced CoQ to cytochrome c. It contains four redox cofactors: two b-type hemes, a c-type heme, and one [2Fe–2S] cluster. This complex -CoQH2-cytochrome c oxidoreductase (also called cytochrome reductase), catalyzes the oxidation of reduced coenzymeQ (CoQH2). The electrons produced by this oxidation reaction are passed along to cytochrome c in a multistep process. The overall reaction is;

$$\ce{CoQH2  +  2 Cyt c[Fe(III)]  -> CoQ  + 2 Cyt c[Fe(II)]  +  2 H+}$$

Cytochrome c

Cytochrome c is a peripheral membrane protein that is loosely bound to the outer surface of the inner mitochondrial membrane. It alternately binds to cytochrome c1 (of Complex III) and to cytochrome c oxidase (Complex IV) and thereby functions to shuttle electrons between them.

Complex IV (Cytochrome c Oxidase)

Cytochrome c oxidase (COX or CcO), the terminal enzyme of the electron transport chain, catalyzes the one-electron oxidations of four consecutive reduced cytochrome c molecules and the concomitant four-electron reduction of one  $\ce{O2}$ molecule to yield $\ce{H2O}$:

$$\ce{4cytochrome c^2+  +   4H^ + O2   -> 4 cytochrome c3+ + 2H2O}$$

Overview:

Summarised version:

The mitochondrial electron-transport chain. The pathways of electron transfer (black) and proton pumping (red) are indicated. Electrons are transferred between Complexes I and III by membrane-soluble CoQ (Q) and between Complexes III and IV by the peripheral membrane protein cytochrome c (Cyt c). Complex II (not shown) transfers electrons from succinate to CoQ

These Fe-S clusters are present in the complexes described above and play a big part in the electron transport chain, as they are redox active hence allow the transport of electrons along the chain through a series of redox reaction yielding $\ce{H2O}$ in the final reduction of $\ce{O2}$ molecule (producing the ATP molecules along the way). Although this is not exhaustive of the whole process it gives the "main" idea of molecules (including the so-called iron-sulfur proteins you mentioned in the question) and processes involved in the ETC. It was just impossible for me to explain only iron-sulfur proteins without getting into detailed processes

Hope this helps you

References

Biochemistry (Campbell and Farrel) 5th ed (Chapter 20)

Biochemistry (Voet and Voet) (Chapter 22)

Biochemistry (Grisham) 4th ed(Chapter 20)

Lehninger  Principles of Biochemistry (Chapter 19)


How strict is the "to excite electrons the energy must equal the energy state difference" fact?

In fact there are many factors to consider, so the answer to your question depends on what type of material you are analyzing.

First of all you should know that even if the absorption level are discrete the absorption and the emission of a material are never a single value but a peak. If you consider the absorption or the emission of a gas you can assume them discrete value (even is not completely true), and so you can call them lines (see e.g. Lyman series), but if you condense the matter the interaction between the atoms/molecules broad the absorption/emission range and so you have peaks. If you are talking about electrons you are dealing with UV-VIS spectra in this case if you look at solid or a liquid UV-VIS spectra you can see that you don't have a well define peak but a continuous absorption, in this case you have bands. So it really depends of the state of the material you are analyzing.
There are other factors that determine the broadening of the absorption to multiple wavelengths. Form the previous explanation you can derive that more an atom is isolated, less interaction occurs, and the absorption is more limited to determinate wavelength, because you have similar "oscillators". But if you increase temperature and pressure you increase the probability that some atoms/molecules interact, and have different energy levels. The effect is the same: you are broadening the absorption band, respectively you will have a _temperature and a pressure broadening.
So it seems that if you have a single atom this absorbs at a determinate wavelength. Even in this case is not true, the explanation is not so intuitive as the previous ones because you have to understand Heisenberg uncertainty principle but roughly it states that you can't determine precisely the energy states of an atoms. This last broadening is called Natural broadening.

For your specific case you should look at the the UV-VIS spectrum, you will not see a sudden step next to the absorption band but a gently curve. This means that there are less and less atoms absorbing in this region because statistically is less probable that atoms can be perturbed so much to absorb that wavelengths but there are still chances that some atoms absorbs even at wavelength far from the mean.

Have a look to wikipedia spectral lines page for more infos.


$$\ce{2NaN3 -> 2Na + 3N2}$$

As you already found the amount of $\ce{N2}$ required, I'll just give you a hint on how to get the amount of sodium azide required to produce that much nitrogen.

We know that 2 moles of sodium azide gives us 3 moles of nitrogen. So $x$ moles of sodium azide gives us $2$ moles of nitrogen. Solve for $x$. That's all!


Rate determining step for parallel reaction?

Let's suppose that $\ce{A -> C}$ is the fast step in the parallel reaction scheme.  This implies that $k_2 \gg k_1$.  Thus, $k_1 + k_2 \approx k_2$, it is therefore (approximately) true to write the rate of consumption of $\ce{A}$ as $A_0 e^{-k_2t}$.  This is what is meant by the fast reaction "controlling" rate for parallel reactions.


My assumptions:

'ul' Refers to microlitres, be careful that the addition of B is in millilitres.
You are trying to calculate the % v/v that is being removed from the total solution at the end?

So before incubation the total volume is 2ul + 4ul + 2.2ul + 1.8 ul = 10ul (0.01ml)
You are adding in 0.95ml, totalling 0.96ml.
You are then taking 150ul (0.15ml)
This equates to (0.15/0.96)*100 %
= 15.6%


Enthalpy of reaction between ammonia and oxygen

The heat of reaction of a mixture is defined as the extent of reaction, $\xi$ times the enthaply of reaction, $\Delta H_{rxn}$,
$$Q_{rxn} =\xi\Delta H_{rxn} = \sum^n_{i=1}\nu_iH_{f,i}^{o} =\sum^{n,products}_{i=1}\|\nu_i\|H_{f,i}^{o} - \sum^{m,reactants}_{i=1}\|\nu_i\|H_{f,i}^{o} $$
For a reaction system,
$$\ce{4NH_3(g) + 5O_2(g) -> 6H_2O(g) + 4NO(g)}$$
In the case of the example you've provided, you are assuming this reaction is taking place at standard conditions. If we assume for a stiociometic amount of reactants, then the extend of reaction, $\xi$, should be equal to 1. Therefore,
$$Q_{rxn} = 1*\Delta H_{rxn} = 6\Delta H^o_{f,H_2O} + 4\Delta H^o_{f,NO} - 5\Delta H^o_{f,O_2} - 4\Delta H^o_{f,NH_3}$$
Since, $\Delta H^o_{f,O_2} = 0 \pu{kJ//mol}$, the equation reduces to,
$$\Delta H_{rxn} = 6\pu{mol}(-241.8 \pu{kJ//mol}) + 4\pu{mol}(90.3 \pu{kJ//mol}) - 4\pu{mol}(-46.1 \pu{kJ//mol}) $$
$$\Delta H_{rxn}= -905.2\pu{kJ}$$
Now, per gram of $\ce{NO}$, assuming the basis of 4 mol $\ce{NO}$ from the reaction equation,
$$\frac{Q_{rxn}}{m_{NO}} = \frac{\Delta H_{rxn}}{m_{NO}} = \frac{-905.2\pu{kJ}}{4_{mol, NO}(30.01\pu{g//mol})}=-7.541\pu{kJ//g_{,NO}}$$


In synthesizing PbS nanocrystals will there be any problems if I use lead IV acetate instead of lead II acetate?

You might run into problems. $\ce{Pb(OAc)4}$ is a pretty strong oxidant!

What will be your sulfide source? Are you sure that it isn't oxidized these conditions?

Edit 1

At least N,N'-diarylthioureas are efficiently desulfurized by lead tetraacetate and furnish ureas (and sulfur) via a carbodiimide intermediate, see Bull.Chem. Soc. Jpn., 1984, 57, 315-316.

Edit 2
If you don't mind a personal advice: I wouldn't want to add complexity to the protocol just to maneuver around problems (and create others) I wouldn't have if I were using the right starting materials in the first. (The German term for that is verschlimmbessern, which translates to something like disimprove).


Unknown notation relating to stoichiometry

It is probably supposed to be the 'corresponds to' sign[ref, p 105]; in this case simply representing that for every two moles of $\ce{NaOH}$ one mole of $\ce{H2SO4}$ is converted by reaction stoichiometry.

This sign is difficult to typeset in LaTeX which is why I presume the authors used something else instead. Creating the sign is even more annoying in MathJax. The closest I [got at first] were

$$\overset{\lower.5em\wedge}{=}\ \text{or}\ \hat{=}$$

which are $\overset{\lower.5em\wedge}{=}$ and $\hat{=}$, respectively.

Edit. After some testing, this option $\lower.1em\lower.5em\widehat{\raise.5em{=}}$ looks best in my opinion.

$$\ce{2 mol NaOH $\lower.1em\lower.5em\widehat{\raise.5em{=}}$ 1 mol H2SO4}$$

full code: $$\ce{2 mol NaOH $\lower.1em\lower.5em\widehat{\raise.5em{=}}$ 1 mol H2SO4}$$

Just the bare $\widehat{=}$ leads to

$$\ce{2 mol NaOH \widehat{=} 1 mol H2SO4}.$$

ref: E. R. Cohen, T. Cvitas, J. G. Frey, B. Holmström, K. Kuchitsu,
R. Marquardt, I. Mills, F. Pavese, M. Quack, J. Stohner, H. L. Strauss, M. Takami, A. J. Thor. (2007). Quantities, Units, and Symbols in Physical Chemistry. ('IUPAC Green Book'). Third Edition. International Union of Pure and Applied Chemistry. (2008 IUPAC & RSC reprint). ISBN: 978-0-85404-433-7. [free copy]

Here is a picture of how it looks in the 'Green Book':


should charge balance be based on activities?

No, charge is not really a activity based property.  It is a number.  Just as the activity coefficients don't change the number (or moles) of potassium atoms or phosphate anions, they don't change the charge either.

If you told me that one liter of a 1.0 molar potassium chloride solution had activity coefficients of 21 for potassium and 186 for chloride, then the amount of charge in the solution is still one mole of potassium cations and one mole chloride anions.


What makes formaldehyde 'safe' or 'dangerous'?

The toxicity of formaldehyde is dose-dependent. The famous saying "the dose makes the poison" is one of the rules of thumb about toxicology. Your body can effectively detoxify small amounts of formaldehyde in a given amount of time, however if you are exposed to concentrations that result in you ingesting/inhaling more than what your body can metabolize, then toxicity will begin to occur.

Most consumer products with detectable levels of formaldehyde (I'm thinking of shampoos and body washes particularly) do not directly have formaldehyde added to them. As formaldehyde is a fairly reactive gas that will polymerize & evaporate easily, formulation chemists will use compounds that decompose slowly to formaldehyde plus other mostly inert material. What this does is provide a constant, low level of formaldehyde present to halt microbial growth. (Formaldehyde is a wonderful disinfectant - few things if any will survive treatment with concentrated formaldehyde - and it turns out you need a LOT of formaldehyde to kill a human as compared to a microbe!)

Formaldehyde is also a natural trace compound formed in fermentation/decomposition of other organics, so it's detectable in the biosphere if you want to go looking for it - I'm sure you can find detectable amounts in fruits, wood products, even the air around a garden.


Fluorescence quenching by palladium(II) vs. platinum(II) complexes

The theory of this quenching effect was developed about $70$ years ago by Förster, who showed that the quenching is not due to diffusion or collision. It is due to a resonance phenomena. The fluorescence yield is quenched by a solute whose mirror image of the adsorption spectrum overlaps the emission spectrum of fluorescence. It is is proportional to the overlap integral of these two spectra.
The quenching effect due to an agent X on the fluorescence is given by Stern-Volmer equation : ${I_0/I = 1 + \gamma [X]}$. Now T. Förster has shown that the quenching constant $\gamma$ is equal to $\gamma = \frac{2\pi ^2}{3} \frac{N}{1000} R_0^3$, where $N$ is Avogadro number, and $R_0$ is the critical Förster distance, where $R_0^6$ is proportional to the overlap integral $J(\nu )$ of the two spectra, and to the sum of all energy transfer rate constants (solvent to solutes and solute to solute).
This theory will not be developed here.
But, as you are looking for a piece of literature, I recommend the following references, where this theory is explained.
See : T. Förster, Discuss. Faraday Soc. $27, 1, (1959)$ and : T. Förster, Ann. Physik $2, 55, (1947)$.


Sodium aluminate is produced. The reaction is slower than the usual $NaOH + Al$ and liberates $CO_2$.


A Spectrophotometric technique to evaluate Vitamin C concentration in Citrus Fruit

It is not quite correct of the statement that "ascorbic acid cannot be determined by Beer Lambert's law, because it does not absorb visible light." Although the second part of the statement is true, the highlighted part is incorrect since Beer Lambert's law applies for ultra-violet range of the spectrum. According to Ref.1, ascorbic acid (Vitamin C) absorbs light in UV range of the spectrum:

The result revealed that $\lambda_\mathrm{max}$ (peak wavelength of spectra) of L-ascorbic acid at $\mathrm{pH} \ 2.0$ was $\pu{289.0 nm}$ whereas at neutral $\mathrm{pH} \ 7.0$, $\lambda_\mathrm{max}$ was $\pu{295.4 nm}$. In alkaline $\mathrm{pH} \ 8.6$, $\lambda_\mathrm{max}$ was $\pu{295.4 nm}$ and at $\mathrm{pH} \ 7.4$ the $\lambda_\mathrm{max}$ of L-ascorbic acid remained the same as $\pu{295.4 nm}$.

Thus, if OP has access to a UV/vis spectrometer, assuming no interference from other components in the fruit extract in the UV range, OP can try to determine ascorbic acid percentage in fruits of interest using a direct method with calibration curve of known ascorbic acid concentrations.
In the event of having interferences, there is a better method of getting absorbance in visible region of the spectrum (Ref.2). This is a well-established method, which is called the 2,4-dinitrophenyl hydrazine method (Ref.3).
The method involves oxidation of ascorbic acid to dehydroascorbic acid by bromine water followed by preparation of the corresponding osazone using 2,4-dinitrophenylhydeazine (2,4-DNPH). The osazone upon treatment with 85% $\ce{H2SO4}$ forms red colored solution, which absorbs light at $\pu{521 nm}$. The procedure was successfully used to determine vitamin C content in various fruits and vegetables (Ref.3). The calibration curve of the red dye prepared from standard ascorbic acid samples is shown in the following diagram (LHS; the plots are from Ref.3):

This method need you to prepare a working solution (5% metaphosphoric acid-10% acetic acid): Fifteen grams of solid metaphosphoric acid were dissolved in mixture of $\pu{40 mL}$ of glacial acetic acid and $\pu{450 mL}$ of distilled water in a $\pu{500 mL}$ volumetric flask.
The standard vitamin C (ascorbic acid) solutions for calibration curve: First, prepare the $\pu{500 ppm}$ ($\pu{500 mg/L}$) standard stock solution by dissolving $\pu{0.05 g }$ of standard crystalline ascorbic acid in $\pu{100 mL}$ of working solution. From that stock solution, you can prepare $5, 10, 15, 20$, and $\pu{25 ppm}$ standard test samples by serial dilution (using working solution to dilute) to prepare the calibration curve.
In addition, you need 10% thiourea solution, 2,4-DNPH solution, and 85% $\ce{H2SO4}$ solution in hand.
Procedure: To the equal volumes of filtered fruit solution (prepared by using working solution to extract and dilute) or standard ascorbic acid solutions, add few drops of bromine water until the solution became colored (bromine water color; this is to confirm the completion of the oxidation of ascorbic acid to dehydroascorbic acid). Then few drops of thiourea was added to it to remove the excess bromine and the clear solution was obtained as result. Then 2,4-DNPH solution was added equal volume to all these oxidized ascorbic acid samples (both standards and fruit extract samples) and they are kept in $\pu{37 ^\circ C}$ water bath for $\pu{3 h}$ (Ref.4). When the coupling reaction of 2,4-DNPH dye with dehydroascorbic acid has been completed, add equal volumes of 85% $\ce{H2SO4}$ solution ($5\times$ volume of 2,4-DNPH) followed by spectrophotometric determination.
It should be noted that any other compound in the extract also reacts with 2,4-DNPH and 85% $\ce{H2SO4}$ solution, if they contain carbonyl groups. For example, common component in fruit is glucose (and other sugar), dye from which are giving absorbance at around $\pu{450 nm}$ (see the complete scan of dye prepared by D-glucose on the RHS of the above diagram). As indicated in the diagram, these interference is minimal at $\pu{521 nm}$.
Note: This method is good to determine total ascorbic acid content (ascorbic acid + dehydroascorbic acid). The full text of the relevant paper is attached for the convenience.
References:

Shaheen A. Maniyar, Jameel G. Jargar, Swastika N. Das, Salim A. Dhundasi, Kusal K. Das, "Alteration of chemical behavior of L-ascorbic acid in combination with nickel sulfate at different $\mathrm{pH}$ solutions in vitro," Asian Pacific Journal of Tropical Biomedicine 2012, 2(3), 220–222 (https://doi.org/10.1016/S2221-1691(12)60045-8).
R. Riemschneider, M. Z. Abedin, R. P. Mocellin, “Qualitäts und stabilisierung prüfung hitze konservierter Nahrungsmittel unter verwendung von Vit C als kriterium-Mitt 1 (Quality and stability evaluation of food preserved by heat utilizing vitamin C as a criterion),” Alimenta 1976, 15, 171-174.
M. Rahman Khan, M. M. Rahman, M. S. Islam, S. A. Begum, “A Simple UV-spectrophotometric Method for the Determination of Vitamin C Content in Various Fruits and Vegetables at Sylhet Area in Bangladesh,” Journal of Biological Sciences 2006, 6(2), 388-392 (DOI: 10.3923/jbs.2006.388.392)(Full Text).
Anal Parimal Desai, Shuchi Desai, "UV Spectroscopic Method for Determination of Vitamin C (Ascorbic Acid) Content in Different Fruits in South Gujarat Region," Int. J. Environ. Sci. Nat. Res. 2019, 21(2),  IJESNR.MS.ID.556056 (4 pages)(DOI: 10.19080/IJESNR.2019.21.556056)(PDF).


If you labeled the oxygen atoms in carbon dioxide in green and the oxgyen atoms in dioxygen in red (because they derived from photosynthesis, while $\ce{CO2}$ is used up in the Calvin cycle), you would have to add some water molecules to be able to write an equation like this:

$$\ce{6C\color{limegreen}{\ce{O2}} + 12H2\color{red}{\ce{O}} -> C6H12\color{limegreen}{\ce{O6}} + 6\color{red}{\ce{O2}} + 6 H2\color{limegreen}{\ce{O}}}$$

This is problematic because oxygens in carbonyls can interchange with oxygens in water fairly rapidly (e.g. hydration of carbonyl to diol, and reverse reaction losing the "original" oxygen on the carbonyl). It would be fine to write the net reaction of the Calvin cycle separately from the net reaction of photosynthesis. This would also nicely show that hydrolysis of ATP is required for some of the steps, and that NADPH is involved.

On the other hand, if you are talking about the relationship between photosynthetic organisms vs other organisms, it is nice to have a simple net reaction for the entire organism (which also ignores that plant cells do break down glucose as well).

  Clearly there is much more going on as the formula most of us learned in school suggests, so my question is: what is the correct and complete equation for photosynthesis?

There is no official complete equation. You can add as much or as little detail you want. It really depends on where you want to go with the chemical equation. The links in the comments are great starting points.

  [OP, from comments:] Found the answer as to why i was confused by the simplified equation: "Hexose (six-carbon) sugars are not a product of the Calvin cycle. Although many texts list a product of photosynthesis as C6H12O6, this is mainly a convenience to counter the equation of respiration, where six-carbon sugars are oxidized in mitochondria. The carbohydrate products of the Calvin cycle are three-carbon sugar phosphate molecules, or "triose phosphates", namely, glyceraldehyde-3-phosphate (G3P)."

Great, it feels good to figure things out! In biochemical pathways, it can be difficult pinpointing products. The Calvin cycle and pentose phosphate pathway can be tweaked to make a variety of products, and those tweak affect what the "real" net reaction is.


In general, orthogonal wavefunctions are much easier to treat. In some cases they appear naturally, but usually, the orthogonality is imposed as a constrain while constructing the wavefunction.

For example, if you construct electronic wavefunction in the atomic orbital basis, you try to construct the orthogonal basis. This guarantees that the AOs are linearly independent. (Implication, not equivalence). Would you fail to fulfill this, the solution might still be possible, but much more difficult.

If you manage to solve the eigenvalue - eigenvector problem, the solutions are by definition orthogonal to each other. This is the case for the examples you provided.


The melting point of diamond is certainly an exotic topic. Diamond is slightly less stabile than its allotrope, graphite, and will in most circumstances turn to graphite way before its melting point. If you have enough pressure it will stay diamondlike and it will at some point melt.

A diamond is not a single molecule unless you stretch the definition of molecule. It is usually defined as the smallest fundamental unit, and for diamond I'd call the tetragonal carbon unit cell the smallest fundamental unit, or maybe even a carbon atom. Diamond, having one of the highest values for heat transfer among substances would certainly melt almost like a single molecule though. But in theory, one and one carbon decides to leave behind it's friends bound by the diamond structure boundaries behind to seek their fortune in the hippie camp that is called liquid.


What conditions would make SN1 products the major products over E1 products, and vice versa?

In a practical way, you would have to consider each particular case, there are tons of reasons that could make a reaction to yield mostly E1 or SN1 product (it will be always a mixture, that can vary from 50:50 to 0.01:99,9, I hope I'm making myself clear, they are always competing reactions).

But there are several factors that contribute to make one of the two more favorable than the other, for example:

The kind of the reactant employed:

If you use hydrogen halides, ($\ce{HCl, HBr, HI}$) you will mostly obtain SN1 products.

If you use acids like $\ce{H2SO4, H3PO4~or~TsOH}$ you will mostly obtain E1.

Why is this? The reagents/acids of the first case, are very good nucleophiles. The reagents lisetd on the second case, are awful nucleophiles.
A good nucleophile is key for a substitution. But for elimination you only need something that can abstract a product in the alpha position to the leaving group.

On the other hand, very strong and small bases, can favour an elimination through E2 mechanism, with almost no substitution taking place. E1 mechanisms is usually observed only under mild basic conditions.

Solvents, like you say, also can favour one type or another:

In some cases, solvent acts a base for E1 reactions. (i.e. $\ce{EtOH -> EtOH2+}$). Bulkiness of the solvent can cause both E1 and SN1 (in case that the solvent itself is the nucleophile, which usually happens) to be slower: it has to approach the compound that undergoes the reaction anyway. Actually is easier to approach a hydrogen than a carbon, so probably will favour SN1 over E1, BUT, do NOT take this as a determining factor at all. The alpha carbon from which you need to abstract the proton might be more or less hindered.

Solvents with low nucleophility also decreases the % of substitution reaction.

Both E1 and SN1 are favoured in protic solvents, in general.

To sum up, you will always have both products, but yes, there are factors that can be modified in order to maximize one of the two reactions. There are some general considerations, like the ones I posted, but in practice, each reaction is a world.


Is the phenyl cation or ethynylium more stable?

Preamble
As trivial as this question might seem, it is not. It should certainly never be asked in at an exam level lower than basic quantum chemistry.
Any of the tempting explanation schemes will certainly fail. A bonding picture in the Lewis formalism is almost impossible; hybridisation descriptions in these compounds is extremely complex.
As a general tip: The smaller the molecule, the more complicated is the bonding.
What is stability?
This question seems so simple, yet is incredibly complex. As a quantum chemist, I see any molecule as stable if it is an energy well (stationary point) on the potential energy surface (and this might already be a simplification). This concept does not help in any way when you want to compare stabilities, because it is just one property of a particular arrangement of nuclei in a particular state.
There is no such thing as absolute stability. We cannot measure it. It is an abstract concept, that we like to think that we understand it. An example from the real world: ice is stable below 0 °C and vapour is stable above 100 °C. Yet ice is not more stable than vapour.
Comparing stability, we always need a framework of multiple variables or constants and a reference point.
It is therefore often assumed, that the boundary conditions are the same (exactly). Even on a purely theoretical level this it is extremely difficult to find these exact boundary conditions. Sooner or later you have to introduce approximations, which eventually crash your system, especially regarding the reference point.
It is difficult, but not impossible.
In order to compare the relative stability of those two cations we can assume a hypothetical hydride transfer between them, hence building a closed reference frame. In computational chemistry we use the concept of isodesmic reactions for such purposes. It is obviously not free of any bias and a complete hypothetical, but it can give us some more insight.
In this case we can propose the following hypothetical isodesmic reaction:

Note that the number of bonds of the same type are the same on each side are equal. Close enough at least. One could argue, that we break a hydrogen sp²-carbon bond and form a hydrogen sp-carbon bond, and vice versa, therefore not having the same types of bonds on the left and right side. But, this is probably as close as we can get.
Having set up a concise set of parameters, we can now go ahead and perform a few quantum chemical calculations. I decided to use my personal standard: DF-BP86(D3BJ)/def2-SVP level of theory.
And we run into more problems
The singlet, closed shell molecules benzene (despite aromaticity) and acetylene are no problem. But when we come to the cations, things get interesting.
The phenyl-cation does not propose a big threat either, it is big enough to compensate the empty orbital in the σ- and π-system.
The high symmetry of the acetylene molecule make it impossible to treat on this level of theory. The linear form is - by my definition - not stable. At least not at this level of theory, at least not in the singlet state. What we do need is a multi-reference method like CASSCF at least. And then we still don't know which state is preferred and if we can still use our proposed hypothetical reaction. Because if the spin state is different on either side, the whole thing falls apart.
Results
Nevertheless while we're already at it, let us look at some of the results and clear up some misconceptions.
Benzene and the Phenyl-cation
Upon ionisation the system needs to compensate the change with a change in geometry. This is a purely electronic effect. In benzene all carbons are trigonal coordinated, quite nicely having three sp² hybrid orbitals; it is one of the very few examples, where symmetry dictates that. Removing a hydrid ion also means removing electrons from such an orbital, freeing up more stable s-character. As a result the bond angle widens to accommodate more s-character in the carbon-carbon bonds. Now this leads to a visibly distorted geometry - a delicate equilibrium of multiple factors.

The π-system remains largely undisturbed. The σ-bonds compensate almost completely for the change. The cationic carbon is now better described as being between linear and trigonal, hence having hybridorbitals between sp and sp².
Acetylene and the Ethynyl carbocation
Acetylene is a wonderful molecule by itself. It has extremely high symmetry, which is quite rare for four atoms. With that comes a large number of degenerate orbitals. And a highly complicated bonding situation.
You might want to regard the carbons as having sp-hybrid orbitals, and that is fine as an approximation, but the deeper you dig, the creepier it gets. (Also see Bent's rule.)
Disturbing this molecule in any way, results in breaking the symmetry. The linear version of HCC+ still has high symmetry and now we have a problem. Upon removal of the hydrid ion, the remaining electron density will try to use as much s-character as possible. As a result part of the s-character is tried to be redistributed into the π-system. These orbitals now do not match the spacial symmetry anymore and the molecule is not stable in its linear form in the singlet state. This is not the whole truth though, but it gets much more complicated and this will certainly be too much for this answer.
A stable structure on this level of theory in the singlet state is an almost three-membered ring. It is probably best to think of it as a protonated dicarbon. All bonds are elongated compared with acetylene.

Since we have a significant different bonding situation now, the previous assumptions of the same type of bonds left and right break down and we can no longer compare these two molecules in the given reference framework.
(Without respect for actual science, we could still do it. Then we would know that the left side of the equation is about 600 kJ/mol more stable than the right.)
Conclusion
The phenyl-cation might be stable, it is a local minimum at DF-BP86/def2-SVP, while the acetylene-cation does not exist. The HCC-cation is better thought of as a protonated dicarbon molecule.
If I understood you correctly, then your assumption, that the "strain" on the ring is bigger than the instability of the orbitals in the acetylene cation, is indeed wrong.


Why is current non-zero when applied potential is still below E$^0$ in cyclic voltammetry?

I think you may be coming at this question from the wrong angle. Since I'm not entirely knowledgeable about electrochemistry, I'll give a somewhat qualitative answer, the general idea of which I hope to be understandable.

Fundamentally, what you are doing is taking a system containing a species, say $\ce{A}$, and its oxidized form $\ce{A+}$ (of course this same argument applies for the reduced form $\ce{A-}$ if you sweep in the other direction of voltage). In the absence of an applied voltage, the system is at equilibrium with a certain concentration of $\ce{A}$ and $\ce{A+}$. Let us consider that $\ce{[A]}$ $\gg $ $\ce{[A+]}$. This implies that the equilibrium constant for $\ce{A <=> A+ + e-}$ is much smaller than 1, and therefore that the oxidation reaction has a high positive standard reaction free energy change $\Delta _r G^{o} \gg 0$ (loosely speaking, the reaction is "not spontaneous" in standard conditions, though this sort of thinking is misleading, as I explain later).

As you start to sweep the voltage, you begin adding electrical free energy into the system, such that the reaction free energy (in non-standard conditions) now becomes a sum of the initial chemical free energy change and the electrical free energy $\Delta _r G = \Delta _r G^{o} + G_{electrical} = \Delta _r G^{o} + nFE$ (I think that you can argue that by altering the voltage, you are altering the free energy of the electron in the equilibrium). If you sweep in the right direction, then the electrical free energy term can contribute to decrease the reaction free energy. At some defined voltage, the two factors cancel out, giving $\Delta _r G = 0\ \ \rightarrow \ \ \ \Delta _r G^{o} = -nFE^*$, where to $E^*$ we give the special name "oxidation potential", $E^{ox}$.

The important thing to notice is that all this time, the potential sweep is altering the reaction free energy incrementally. Since the free energy is shifting, the associated equilibrium constant is also changing. At voltages much than lower $E^{ox}$,  $\Delta _r G = \Delta _r G^{o} + nFE \gg 0$, and so even though the reaction free energy is lowering, it is still quite positive and the associated equilibrium constant is still much smaller than 1. Thus the reaction is only slowly shifting towards products, and therefore there are few electrons being passed into the circuit, and the current is close to zero. However, when the voltage applied starts to get close to the oxidation potential, even if $\Delta _r G < 0$, the equilibrium constant will be smaller than but somewhat close to 1, and there is some conversion of $\ce{A}$ into $\ce{A+}$, creating a current below the formal oxidation potential.

The whole problem comes from thinking that as long as $\Delta _r G < 0$, a reaction "happens", while for $\Delta _r G > 0$, the reaction "doesn't happen". In truth, there is a gradual change of equilibrium concentrations as a reaction goes from one regime to the other. Even if $\Delta _r G > 0$, if it sufficiently close to zero then the equilibrium constant is only a bit smaller than 1, and you can expect appreciable amounts of products. Notice that in the voltammogram, the current is practically zero until about 0.1 V away from the oxidation potential, and it only really starts to pick up around a mere 0.05 V away. Even though these values mean $\Delta _r G > 0$, they are sufficiently close to the oxidation potential that the equilibrium constants aren't so small, and so you can actually measure a significant amount of species being oxidized.


What is the relationship between dipole moment and solubility?

It's all about the energetics (assuming an isolated system--happy to elaborate in the comments): as acetonitrile has a large dipole moment, it can readily stabilize the individual ions. See this discussion of solvation energy.

The more subtle case is in the case of nonpolar compounds. In this case, the large dipole can create an induced dipole in the nonpolar molecule. Then the dipole-induced dipole attraction would be larger in magnitude than the spontaneous dipole-spontaneous dipole interactions present in nonpolar-nonpolar interactions (which are also still present as small fluctuations in the polar-nonpolar case). See this discussion of how charges/dipoles/multipoles induce dipoles and then subsequently attractively interact with said dipoles.


What reactions create a lot of gas really quickly?

The reaction used for car airbags involves sodium azide (NaN3). A spark initiates the reaction and it generates gaseous nitrogen really fast.

I don't know much more about it but you can certainly find some good informations on google!


How to find the empirical formula for an nitrogen oxygen compound from the given mass of nitrogen and the product?

$\ce{x/2 N2 +y/2 O2 \rightarrow N$_x$O$_y$}$

We have two unknowns $x$ and $y$. So, to determine them we need two independent  equations:

1 - The conservation of nitrogen mass: $5=14x$. So,  $x= 0.35714$.

2- According to the stoichiometry of the above balanced chemical equation: The number of moles of  $\ce{ N$_x$O$_y$}$ equals $n= \frac{5}{28}\times\frac{2}{x}$.

On the other hand, the number of moles of  $\ce{ N$_x$O$_y$}$ equals   $n= \frac{19.3}{14x+16y}=\frac{10}{28x} $.
By substituting $x= 0.35714 $ into the last equation, we find $y=0.89375$.
We find $\ce{ N_{0.35714}O_{0.89375}}$. We divide by $ 0.35714$, We obtain  $\ce{ NO_{2.5}}$.i.e. the formula of nitrogen oxide is $\ce{ N_2O_5}$.


Are all lift-off resist negative photoresists?

There are definitely positive photo resists, but there tends to be little incentive to use them. Usually you use a negative mask with a negative resist to create a positive part. The pattern to be made tends to be a relatively small area of the photo resist on the mask and thus it takes less time for the lithography machine to produce a negative mask which can be a bottle neck in an R&D setting. Also Positive resists tend to be perminate as they have chemically inter-reacted to stay in place not be dissolved later (like during lift-off), thus another reason positive resists are not used frequently.


DNA pairs (adenine-thymine, guanine-cytosine)

In Principles of Nucleic Acid Structure, W. Saenger argues that hydrogen-bonded bases contain at least two hydrogen bonds (forming a "cyclic" pattern). Often, there is a tautomeric form possible that also makes two hydrogen bonds, with two covalent bonds turning into hydrogen bonds, two hydrogen bonds turning into covalent bonds, and double bonds moving around so that each atom still has an octet. In the OPs first example (A:T pair), for example, the ketone would turn into an enol, and the protonation of the amino and imino groups would change.

Saenger goes on to say that 28 pairs are possible.

Many of these are observed (the ones that are labeled are so common that they were given names), in structures of DNA, RNA, and in crystal structures of dinucleotides.

  I was wondering why adenine pairs with thymine and cytosine pairs with guanine.

They pair because they make strong hydrogen bonds with a geometry of the base pairs that allows base stacking and is compatible with base pairing of the other nucleotides in a DNA strand. As the discussion above shows, many other base pairs are theoretically possible, and some of them are actually observed in nature.

  Thymine has the lowest acidity and adenine has the biggest acidity. So it is logical that the intermolecular bond will be strong.

I'm not sure about the premise, and the conclusion is incorrect, as other base pairs also form. As long as a group has the right protonation state, the pKa does not directly determine hydrogen bond strength.


MolView is unable to generate correct 3D structure for hydroxypropadiene

I suspect MolView simply fails to parse the compound's name you are trying to input. "Hydroxypropadiene" isn't the proper name, at least not by IUPAC standards. According to Google services, this name doesn't appear in any textbooks and is mentioned only in three research articles from 1980s.

The PIN propa‐1,2‐dien‐1‐ol appears to work correctly with MolView:

When you are trying to get a 3D image from the 2D drawing, there is the following issue which MolView usually reports (not sure why you haven't received the message):

  Failed to load structure from sketcher. This structure may contain allenes, which cannot be represented in the SMILES notation.

MolView relies on SMILES in order to fetch the 3D structure.
The problem is likely that Ketcher generates wrong SMILES because PubChem indeed receives C=CC=O instead of the proper one, C=C=CO.


Phenol vs 4-hydroxypyridine - Enol Content

Your solution manual correctly predict that phenol has a higher enol content due to aromaticity. For example, you won't see strong carbonyl peak around $\pu{1700 cm-1}$ in FT-IR, which proves keto-form does not exists. It is 100% enol-form as phenol.
On the other hand, 4-pyridone is a different story. The dominance form, whether it is 4-pyridone or 4-hydroxypyridine, would be depend on which phase the compound is in. For example, the Ref.1 states that:

In the gas phase, there is a preference of the hydroxy form in both 2- and 4-hydroxypyridine, while in a nonpolar solvents such as cyclohexane and chloroform, the two tautomers exist in comparable amounts. However, the two equilibria are shifted entirely to the right in favor of the oxo-form, 2- and 4-pyridone, respectively in polar solvents as well as in the crystalline state.

Thus, it is safe to say, it is only in the gas phase, 4-hydroxypyridine dominate. However, in liquid phases, both form exist where predominance is depending on the solvent. In solid phase, 4-pyridone dominate. Since, both phenol and 4-pyridone are solid at ambient conditions, phenol has more enol form than 4-pyridone, which exist in keto-form as a solid.
Note that a public database set like Tautobase, implemented in DataWarrior, may be used to train yourself, too.
Reference:

Jiali Gao, Lei Shao, "Polarization Effects on the Tautomeric Equilibria of 2- and 4-Hydroxypyridine in Aqueous and Organic Solution," J. Phys. Chem. 1994, 98(51), 13772–13779 (https://doi.org/10.1021/j100102a051).
Oya Wahl, Thomas Sander, "Tautobase: An Open Tautomer Database", Chem. Inf. Model. 2020, 60, 1085–1089 (doi 10.1021/acs.jcim.0c00035).


Does nitrogen deaeration need to be performed in a fume hood?

As implied in the comments, you have nothing to fear by sparging your solution with nitrogen in the classroom. Even a good size tank only contains a few cubic meters of compressed gas, while your classroom likely contains a few hundred cubic meters of air. Thus, even if you could instantly empty the tank the effect would be negligible. And you will be using low enough flow rates that there will be no fear of a "localized" depletion of oxygen.

The safety issue that I need to mention, though you are almost certainly already aware, is regarding the sodium hydroxide solution. Even at $\pu{0.1 M}$ it's not something that you want in anyone's eyes, so protection against splatter from the sparging solution would be prudent.


Why does greater orbital overlap mean a stronger bond?

An intuitive explanation is given in the comment above. Better orbital overlap generally indicates more constructive, favorable interaction between the two atoms.

Let's consider a homonuclear diatomic bond (e.g. $\ce{H2}$). The qualitative understanding is similar for two different atoms, but the math from quantum mechanics is a bit easier with a homonuclear bond.

So there's one bonding orbital ($\sigma$) and one anti bonding orbital ($\sigma^*$).

The energy of the orbitals will be:

$e_1 = \frac{H_{11} + H_{12}}{1 + S_{12}}$
$e_2 = \frac{H_{11} - H_{12}}{1 - S_{12}}$

You can put in some numbers and see what happens (e.g., as the overlap $S_{12}$ increases from 0 to 1, the bonding interaction becomes more and more energetically favorable).

N.B. There's a slight asymmetry - the anti bonding orbital is higher in energy than the bonding orbital is lower in energy.


Were gold(II) complexes synthesized for the first time in 2017?

You are correct in noting that there are a number of Au (II) small molecule species synthesized prior to the report of Professor Katja Heinze in 2017.
The heart of the Au (II) claim by Professor Katja Heinze and team as stated in the article you referenced, see reference 1 given below, has to do with the a) relative stability of the mono-nuclear gold-porphyrin complexes compared to other Au (II) systems and b) that the molecule does not have a Au-Au bond.  The latter point being the more important.
Au (II) is a very stable oxidation state for molecules containing Au-Au bonds.  We synthesized and characterized a number of such species years ago, see references 2 and 3 below for two papers describing the work.
For those working in the field, the nature of the claim by Heinze et al. is understandable.  For those outside the niche field of Au (II) chemistry I can see how the distinction is less than clear.
References
https://www.sciencedaily.com/releases/2017/08/170808145949.htm
https://pubs.rsc.org/en/content/articlelanding/1985/c3/c39850001278#!divAbstract
https://pubs.rsc.org/en/content/articlelanding/1986/C3/c39860000321#!divAbstract


The reaction can be thermodynamically unstable (favorable) not only due to positive entropy change but also negative enthalpy change, both result in negative free energy. The thermodynamic stability is determined by the difference between the potential energy of products and reactants. If the result is negative the reaction is said to be favorable and vice versa.

Yet this doesn't say anything about the rate of the reaction. The rate of the reaction or reaction kinetics is determined by taking into account an intermediate step in the way from reactants to products, the transition state. If the difference between the potential energy of the transition state and the reactants is too large, the reaction will be too slow. conversely if the difference is small the reaction takes place so fast.

Here arises a question: From where will reactants get the energy to reach and pass the transition state ?

The answer is that indeed a number of molecules has this energy but their population may be too low. This population is determined from the Boltzmann distribution. Heating may be used to supply some energy that increase this population of molecules.

Returning to your question: "Does that mean that the rxn is taking place but we don't see it because of how slow it is , or that the reaction will not take place in the first place?"

Answer: Although the term "kinetically inert" is only used for complexes, the same rules hold as in the other cases. The reaction does take place but we don't notice it. An example for this is turning from diamond to graphite or burning of paper (oxidation of a cellulose).

please, Don't hesitate to ask any question in the comments if something is unclear to you.

This is a useful link:
https://chem.libretexts.org/Textbook_Maps/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Principles_of_Chemical_Equilibria/Kinetically_vs_Thermodynamically_Stable

Hope this helps.


Placement of Electrodes for Electrochemistry?

It really depends on the process and how
homogeneous electric filed is needed. The more are electrodes close together, the inhomogeneity will be greater.

The second problem will be with the homogenity of the solution. You can assume, that electrolyte is kinda homogeneous in a bulk. But more close you get to the electrode, the more is the electrolyte separated.

Ions, as a charged particles, are transferred to near electrode surface.


The dust in the house mostly contains sand, dirt and dead skin cells.

First, let's start with the easiest, sand. The chemical formulae of sand is $\ce{SiO2}$, so it contains silicon and oxygen.

Next, dirt has rock, sand, and clay. Sand is made of silicon and oxygen. Rock is mostly made up of graphite (carbon) and some minerals like phosphorus and silicon. Clay is made up of silicon and oxygen and some minor minerals like aluminium.

Lastly, dead skin, like other organic compounds, contains carbon, hydrogen and oxygen, maybe some nitrogen as it may have protein in it.

So basically, dust is made up of carbon, hydrogen, oxygen, nitrogen, silicon, phosphorus and some minor minerals.


What is incorrect about this depiction of the hydrolysis of an ester?

There are indeed two mechanisms for ester hydrolysis, one using base catalysis and one using acid catalysis. The mechanisms of both are provided below (as an aside, you mention that your textbook isn't great- you should consider getting a different one, even if your teacher gives assignments out of the other book, it'll give you a second opinion and something to cross check your book against).

Base catalysed ester hydrolysis

Source: Organic Chemistry (Wothers and Clayden)
Acid catalysed ester hydrolysis

Source: Organic Chemistry (Wothers and Clayden)

Is the product really an acid? It doesn't even have an H+.
Yes, it is an acid. The acid is actually initially formed however because of the basic conditions of the base catalysed mechanism it gets deprotonated to give the carboxylate. Upon workup the carboxylate salt (in your example the sodium salt of the carboxylate) would protonate to give the carboxylic acid (COOH).
It's stated that hydrolysis can happen when esters are treated with an acid as well as a base. But acids only donate H+ so where would the extra O come from to produce an alcohol as a product?
The product is a gain a carboxylic acid, and as you can see from the mechanism above, water is involved, which is where the additional oxygen comes from.  Its worth noting that with the acid catalysed mechanism the whole thing is reversible.
It's stated that a molecule of water is produced (in the base catalysed mechanism). Is this statement incorrect, or is the diagram incorrect?
It could be water or an alcohol depending on whether hydroxide or the alkoxide does the deprotonation. Water is most likely however, as shown above. In the acid catalysed mechanism it is the alcohol that is formed.
In general, how would we change the diagram/information below to make it correct?
See the mechanisms above, taken from Clayden and Wothers. In general the diagram in your textbook looks essentially 'fine', though I dont entirely like the way the structures are drawn (its a very american textbook way of doing it.)


Starch in water might work, perhaps with table salt or other bacterial/fungal growth inhibitor. Follow directions in most recipes: mix a bit of starch in cold water and then add gradually to hot water to make a more stable "soup".

BTW, blasting a colloid with ultrasound (e.g. from an ultrasonic humidifier) homogenizes it and may delay separation.


Hydrogenation at atmospheric pressure in amateur environment

I've adddressed a part of your question here.

As far as the catalyst is concerned, 3-5% Pd/C is fine.

  What is an effective method for generating the hydrogen to fill the balloon with? I suppose that normally a pressurized hydrogen tank would be used for this, but I find this overkill for the reaction I'm planning. Furthermore, the use of pressurized hydrogen tanks is not quite without danger and I'm not trained to handle them safely.

In the lab, I only used pressurized tanks. While commercial hydrogen generators are available - they are mostly used to generate hydrogen gas as a carrier gas for GC/MS - these things cost a fortune and really would be overkill.

Actually, pressurized tanks are pretty safe if you follow some rules:

Obviously, you can't attach the pressure regulator and operate the whole thing unless the steel cap over the main valve is removed. Don't start anything unless the tank is secured to the lab bench or the wall with a clamp, a chain or another sturdy contraption that prevents the tank from falling over! You want hydrogenation, not a rocket ;)
Get familiar with the pressure regulator.
When you're done, remove the regulator again and put the cap over the main valve. Keep the regulator at a dry and clean place.
No grease on the threads!

When operated with a bit of common sense, a small pressurized tank is the most convenient option.

  What is the function of Celite when filtering the catalyst?

When you filter off the catalyst, you'll typically use a suction filter with a base of sintered glass (fritted glass). Palladium on charcoal is a very fine powder that will clog the fritted glass. A pad of Celite in the suction will prevent this. Celite is cheap, suction filters are not!


Is there any bad side effects for using calcium chloride to replace a running dehumidifier?

As for safety, calcium chloride, $\ce{CaCl2}$, is in sea water and added to foods, such as beans and pickles. It's about as dangerous as table salt, sodium chloride, $\ce{NaCl}$ (actually, $\ce{CaCl2}$ is safer for concrete).

However, it absorbs only so much water until it dissolves and stops being effective. What would you do with the puddle of $\ce{CaCl2}$ solution? It can be dried with heat and reused, but then your just moving energy usage from a continually running electric dehumidifier to a batch process using $\ce{CaCl2}$... though if you can get heat for free, it would be cost effective! If you don't have a solar concentrator or access to geothermal energy, there would probably be no energy savings, though.


What are the products of [Ti(H2O)6]Cl3 + H2SO4 + K2Cr2O7 --->

I know potassium dichromate is a strong oxidizing agent

You have already identified one partner and its role in a redox reaction. Good start!

So which component will be oxidized to what?

Can we assume that the reaction is carried out in water? If so, let's have a look at the ions in solution.

We'll have $\ce{H3O+}$, $\ce{K+}$, $\ce{Cr2O7^{2-}}$, and "that complex".

Let's have a look at $\ce{[Ti(H2O)6]Cl3}$.

Apparently, it's a complex of titanium with 6 neutral molecules of water as the ligands and 3 $\ce{Cl-}$ as the counter ions. So, what is the oxidation state of titanium in the complex when it is neutral with 3 $\ce{Cl-}$ as the counter ions?

Let's look again: $\ce{[Ti(H2O)6]^{3+}}$. OK, it is obviously $\ce{Ti(III)}$ and it seems that we have found the victim of our stong oxidant in the system.

We remember that most titanium compounds in nature are $\ce{Ti(IV)}$ oxides and conclude that we will have to balance our redox reaction to account for a $\ce{Ti(III) -> Ti(IV)}$ oxidation.

We could now continue to conclude that the reduction of $\ce{Cr(VI)}$ to $\ce{Cr(III)}$ is accompanied by $\ce{[Ti(H2O)6]^{3+} -> [Ti(H2O)6]^{4+}}$ but then we remember that $\ce{Ti(IV)}$ - tiny and higly charged - is a hard Lewis acid and will not exist as a naked cation, barely covered with just six neutral ligands in a hexaquo complex!

Instead, the likely $\ce{Ti(IV)}$ species is the titanyl cation, $\ce{TiO}^{2+}.$

Time to collect the puzzle pieces and write the redox reaction.


What is the hybridization of the carbonyl oxygen in a carboxylic acid?

I already pointed it out in the comments, but I believe it is time to give it some more thought and explanation. Let's deal with some conceptual issues first.

Hybridisation is a model, that can be used to describe a bonding situation. It is never cause for a certain geometric arrangement, it is always the result of a bonding arrangement. Strictly speaking it is not even necessary to be invoked, but it can be quite handy. Quoting from Aditya's answer - I could not have said it better:

  First of all, 'hybridisation' is a hypothetical concept, i.e. orbitals don't really mix to form new orbitals, but 'hybridisations' are very successful in explaining structures of molecules and their physical properties. So knowing that a compound has $\ce{sp^2}$ hybridisation will just help you arrange the constituent atoms around the central atom. Physically it has no other meaning.
  Note: Only orbitals can be hybridised, never compounds.

Having understood that, it is easy to see, that atoms in a molecule could use many different hybridisations schemes. There are some concepts, based on quantum chemical calculations, which would find the most appropriate fit between a Lewis structure and the molecular orbital approach. See for example this fine question: What is natural bond orbital theory used for? There is also a more quantitative approach, which is called Valence Bond theory, but that is beyond the scope of the answer.

Since this is just a mathematical concept, an equilibration between $\ce{sp^2}$ and $\ce{sp^3}$ would not change a thing. We can decide which way we want to look at it. Ron consequently raised the question in the comments:

  I don't understand how "6" can be correct. How can the carbonyl oxygen be $\ce{sp^2}$ (1) and also be rapidly equilibrating between $\ce{sp^3}$ and $\ce{sp^2}$ (4)?

As I pointed out, (4) should not even be an option. The premise of the question is clearly flawed. Please do not take hybridisation too seriously.

This is how I come to my next point. Are the oxygen's orbitals in carbon acids really $\ce{sp^2}$ (or even $\ce{sp^3}$) hybridised? Well.

It is often appropriate to describe oxygen's orbitals as $\ce{sp^2}$, when it is bonded to two other elements. Two $\ce{sp^2}$ lobes are needed for the bonds, the remaining $\ce{sp^2}$ orbital hosts one lone pair, the remaining $\ce{p}$ orbital hosts the other lone pair.
More often the case is that oxygen has only one bonding partner. In these cases it is more appropriate to describe the oxygen with $\ce{sp}$ orbitals due to symmetry reasons. Consider the local symmetry to have an infinite rotational axis $C_\infty$, hence there have to be two degenerate, perpendicular orbitals. In most of these cases you will also have (partial) π-bonding and/or hyperconjugation.
There are very few exceptions, like $\ce{H3O+}$, where it is appropriate to describe the oxygen with $\ce{sp^3}$ orbitals.

In all of this it should be considered that hybridisation can be fractional. See the last note here and more consequently "What is Bent's rule?" and "Utility of Bent's Rule - What can Bent's rule explain that other qualitative considerations cannot?".

An important point to remember is, that when there is a π-bond, the hybridisation can be $\ce{sp^2}$ at maximum.

I believe that answer 6 is really about a wrong interpretation of resonance stabilisation of the molecule. (Please see Why is there a need for resonance? for more information)
Carbon acids can be described in the resonance scheme below. Now it appears, that the oxygen employs $\ce{sp^2}$ orbitals in the first structure (A) and $\ce{sp^3}$ in the others (B, C). It is important to recognise, that the bonding situation in this molecule is a superposition of all (possible) resonance structures. Theses structures are not in equilibrium.

In conclusion there are several issues with this question and you might want to address this with your teacher.


What material most closely resembles/emulates dog's excrements?

You'll never be able to get to 2-3m accuracy with this. With a reproducible sample, you could make a few dozen tests and create a convincing statistics.

But dog excrements? Even if they always used the same dog, which they kept on a constant diet, and which you had impounded, it'd be impossible. The sample is much, much too complex, and, unfortunately, very much alive. (I don't mean the dog.)

I say stick to old-school policework: Keep interrogating your pranksters until one of them sings. ;-)


What is the substance turning salty water green when electrolysed?

My speculation is based on the words specifying the use of 'tap water' during the electrolysis of aqueous sodium chloride.
In particular, the interaction of formed chlorine and existing ferrous bicarbonate (found in tap water) could form small amounts of aqueous FeCl2. [EDIT TO QUOTE SOURCE]

Bacterial Iron in Well Water – Iron bacteria is usually identified by slime in places such as toilet reservoirs or by the presence of a slimy mass fouling softeners or filters.
Ferric Iron in Well Water – Also known as red water iron, ferric iron in well water is essentially clear water iron that has been exposed to oxygen – usually from the air, thereby oxidizing.  Carbon dioxide leaves the water and the oxygen combines with the iron to form ferric ions (Fe+++).  This gives the water a red rust coloring.
Ferrous Iron in Well Water – Often called clear-water iron because it is clear when poured, this substance is found in water that contains no oxygen. Typically, it comes from deeper wells and groundwater sources. Carbon dioxide acts on iron in the ground to form soluble ferrous bicarbonate. In water this forms ferrous ions (Fe++)

Per Wikipedia on Ferrous chloride, to quote:

The compound is highly soluble in water, giving pale green solutions.

Repeat the electrolysis experiment with distilled water and see if you get the same coloration issue.
[EDIT][EDIT] Interestingly, low solubility Fe(OH)2 is clear and only acquires a greenish tint in the presence of oxygen (per Wikipedia, to quote: "Iron(II) hydroxide is a white solid, but even traces of oxygen impart a greenish tinge.). Further, well-water is associated with high levels of bicarbonate and low O2 levels (see this significant work). To claim that Fe(HCO3)2 is at all green is without basis, as I also could not find any source anywhere asserting a coloration, other than that it is, in effect, 'clear' as was claimed above per the comment: 'Often called clear-water iron because it is clear when poured'. Theoretical formation of the usual hydration sphere for Fe(OH)2 interacting with bicarbonate in low solubility conditions with high HCO3-, low O2, significant CO2 presence (and its associated partial pressure) suggest reasons why it may not be a valid argument. If I am wrong, please provide a single source citing color, else it should be likely assumed that my supposition is correct and Fe(HCO3)2 is clear.  By the way, Atomistry on Ferrous bicarbonate, to quote:

Ferrous hydrogen carbonate or ferrous bicarbonate, FeH2(CO3)2, exists in solutions obtained by dissolving ferrous oxide or carbonate in water charged with excess of carbon dioxide. On exposure to air carbon dioxide is evolved, and a precipitate is obtained consisting essentially of ferric hydroxide.

which one could interpret as that clear-water iron, upon exiting the pipe, liberates CO2 and forms Fe(OH)2, which, with further oxygen exposure, may display a green tint (or, per the introduction of electrolysis formed chlorine).


The reason for adding strong acids or bases to water rather than the other way around is that the dissolution/reaction of these compounds with water tends to be very exothermic and can result in splattering or even boiling of the strong acid or base, particularly if water is added to the concentrated acid or base.

On the other hand, if you slowly add the acid or base to water, you will never have a concentrated acid or base present to splatter as it will be rapidly diluted by the much larger amount of water. Even if you accidentally add the acid or base too quickly, at least the splattering will primarily be of water or dilute acid/base rather than splattering concentrated strong acid or base. It is still important to do the addition slowly so that you only slowly increase the acidity/basicity of the solution. This way, the rate of reaction (and thus also the heat evolution) will happen at a safe level and greatly reduce the risk of splattering.

As a side note, this concept does not only apply to the addition of strong acids and bases to water. While mixing any chemical reagents (unless otherwise specified of course), you should add the more reactive or concentrated solution to the less reactive or more dilute solution, for the same reasons as described for the addition of strong acids and bases to water.


What easily available chemicals do I use to turn oak light grey?

Graying is symptomatic of bleaching (oxidation) caused by air and sunshine. You can use chemical bleach for a similar effect.
See for instance: https://www.wikihow.com/Bleach-Wood.
There are a lot of guides online on the subject. This one seems pretty useful, discussing different bleaches, their safe handling, and treatment steps. Even the US Department of Agriculture has a guide on the subject.
The appropriate bleach for your application appears to be the "two-part (A/B) wood bleach" in which a mix of peroxide and sodium hydroxide reacts with the wood. Commercial kits are available. The USDA guide has the following to say:

One of the most powerful, satisfactory, and widely used bleaches in recent years is based on  a  strong  solution  of  hydrogen  peroxide  (30  or  35  percent  strength).  The  wood  is  first coated evenly with a solution of sodium hydroxide (caustic soda, or ordinary lye) containing pounds of the solid per 10 gallons of solution.  After this has dried for about 30 minutes, an even  coat  of  the  hydrogen  peroxide  is  applied.  The  sodium  hydroxide  applied  as  the  first coat  renders  the  hydrogen  peroxide  unstable,  whereupon  it  decomposes  and  liberates  a  large amount of oxygen in a chemically active condition, which does the bleaching. The liquid on the  surface  should  be  allowed  sufficient  time  to  dry,  during  which  time  the  bleaching  occurs; then  the  wood  should  be  rinsed  thoroughly  with  cold  water.  To  neutralize  any  remaining traces  of  sodium  hydroxide,  the  surface  should  then  be  washed  or  sprayed  with  a  solution containing 1 pound of acetic acid (or 1 pound of oxalic acid) per 10 gallons of water, allowed to stand for 15 minutes, then rinsed thoroughly with water and allowed to dry. After the wood is sanded lightly to remove raised grain the new finish may be applied.

See also the section labeled "precautions".
Regarding the chemistry of the bleaching process, Wikipedia provides the following summary:

Hydrogen peroxide
Using hydrogen peroxide to delignify chemical pulp requires more vigorous conditions than for brightening mechanical pulp. Both pH and temperature are higher when treating chemical pulp. The chemistry is very similar to that involved in oxygen delignification, in terms of the radical species involved and the products produced.
Oxygen
Oxygen exists as a ground state triplet state, which is relatively unreactive and needs free radicals or very electron-rich substrates such as deprotonated lignin phenolic groups. The production of these phenoxide groups requires that delignification with oxygen be carried out under very basic conditions (pH >12). The reactions involved are primarily single electron (radical) reactions.[14][15] Oxygen opens rings and cleaves sidechains giving a complex mixture of small oxygenated molecules. [...] While the radical reactions are largely responsible for delignification, they are detrimental to cellulose. Oxygen-based radicals, especially hydroxyl radicals, HO•, can oxidize hydroxyl groups in the cellulose chains to ketones, and under the strongly basic conditions used in oxygen delignification, these compounds undergo reverse aldol reactions leading to cleavage of cellulose chains.

Lignin is a polyphenolic compound that lends wood much of its color, so it is the main target of bleaching. The cellulose is best left alone, so a relatively mild bleaching protocol is best for wood.
Another important factor affecting the aged appearance, evident in the posted image, is more difficult to replicate with a chemical treatment. That is water loss, which is responsible for cracking. A quick surface treatment might not provide the highly aged appearance in the picture$^\dagger$.
Update (and somewhat off-topic)
$\dagger$ Oak is apparently among the most difficult woods to dry without it developing cracks. If the wood is already dry then it is unlikely to crack, but if it is fresh and is allowed to dry quickly (for instance by setting in the sun for a few weeks) than it is almost certain to crack.


What is the final molar concentration (molarity) of the nitrate ions in the resulting solution?

First, when solving problems like this one, you have to make sure that both salts possess good solubility and that no double salt is precipitating:

$$
\begin{align}
\ce{NaNO3 &<=>> Na+ + NO3-}\label{rxn:R1}\tag{R1}\\
\ce{Ca(NO3)2 &<=>> Ca^2+ + 2NO3-}\label{rxn:R2}\tag{R2}
\end{align}
$$

Both nitrates are indeed well soluble in water (equilibrium is shifted to the far right), and no poorly soluble side-product is formed, e.g. we are likely to have a true solution consisting of hydrated cations $\ce{Na+}$, $\ce{Ca^2+}$ as well as $\ce{NO3-}$ anions.

As you already suggested, one can indeed use total amounts of substances $n_i$ and the preservation of volumes $V_i$ to find the final molar concentration on nitrate (here indices $1$ and $2$ refer to initial sodium and calcium nitrate solutions, respectively):

$$c(\ce{NO3-}) = \frac{n_1(\ce{NO3-}) + n_2(\ce{NO3-})}{V_1 + V_2}\label{eqn:1}\tag{1}$$

According to the stoichiometry (see \eqref{rxn:R1} and \eqref{rxn:R2}):

$$n_1(\ce{NO3-}) = n(\ce{NaNO3}) = c(\ce{NaNO3})\cdot V_1\tag{2.1}$$

$$n_2(\ce{NO3-}) = 2\cdot n(\ce{Ca(NO3)2}) = 2\cdot c(\ce{Ca(NO3)2})\cdot V_2\tag{2.2}$$

Finally, we can plug all the known quantities back to \eqref{eqn:1}:

$$
\begin{align}
c(\ce{NO3-}) &= \frac{c(\ce{NaNO3})\cdot V_1 + 2\cdot c(\ce{Ca(NO3)2})\cdot V_2}{V_1 + V_2}\tag{3}\\
  &= \frac{\pu{1.23 mol L-1}\cdot\pu{16.3 mL} + 2\cdot\pu{0.823 mol L-1}\cdot\pu{6.28 mL}}{\pu{16.3 mL} + \pu{6.28 mL}}\\
  &= \pu{1.35 mol L-1}
\end{align}
$$


Which substance has the highest temperature range between melting and boiling point

Gallium melts at 30 °C but doesn't boil until 2200 °C.  If 30 °C is a bit too warm to count as "room temperature" or "normally" for you, I found an old paper that recommends tetralkyl silanes such as tetradodecyl silane as lubricants that are liquid over very wide temperatures.

Addendum:

Dowtherm A is a eutectic mixture of biphenyl and diphenyl ether.  According to its manufacturer:

  These compounds have practically
  the same vapor pressures, so the
  mixture can be handled as if it were
  a single compound. DOWTHERM A
  fluid may be used in systems employing
  either liquid phase or vapor phase
  heating. Its normal application range
  is 60°F to 750°F (15°C to 400°C),
  and its pressure range is from
  atmospheric to 152.5 psig (10.6 bar).

[...]

  The viscosity of DOWTHERM A
  fluid is low and changes only
  slightly between the melting point
  of the product and its top operating
  temperature.


There is a minimal volume you need to fill into the NMR tube to be able to get a good shim, this is around 500 microliters with regular 5mm tubes. Not using a solvent would require a lot more of your actual sample just to fill the tube. In many cases you might not even have enough of your product to fill an NMR tube.

The solvent does have some effect on the spectra, it could be harder to compare spectra of pure product to the more typical spectra measured in a common solvent. You're measuring in a different environment, product in product is different than product in solvent.

Modern NMR spectrometers use the deuterium signal from the solvent for the lock, which corrects a drifting magnetic field. You could measure without the lock if your experiment is short or your signals are very broad anyway. But this is not something that every NMR facility has just set up for all users, you might have to manually set this up at the spectrometer.

I suspect that the automatic shimming methods typically used now would not be able to deal with a sample without solvent. But I never tried that.