How does the process of nucleation work for boiling liquids?

Firstly, the average kinetic energy of the water has to be high enough (water obviously doesn't start to boil at room temp. and pressure the moment it is poured into a pan).

As the water gains more and more energy, small bubbles appear but don't fully rise to the surface as the energy of the gas particles is transferred to the surrounding water particles, eventually losing enough energy and becoming liquid again.

But if the energy of the gas particles is lost too slowly for this to occur they fully rise to the surface. This will only occur at a certain temp. difference i.e. the liquid has to be close to boiling temperature if not it will absorb too much energy and the bubble will become a liquid again.

A single bubble can increase in "size" (more molecules inside the bubble compared to when it started not talking about the diameter) as it rises, similar to the cluster effect you mentioned. This will only occur if the energy difference is high enough to vaporise the surrounding water molecules and not just increase their kinetic energy.

This can also not occur depending on how "violently" the water is being heated, or at which temperature the heating element is, therefore the temperature of the bubble being produced.


Reduction with NH2-NH2 , OH-, and glycol

If you look carefully, the reagent for the first step is $\ce{O3/H2O}$, which is used for Oxidative ozonolysis. If the mechanism of ozonolysis is drawn, you will realise that a molecule of Hydrogen peroxide($\ce{H2O2}$) is released by the oxidation of the alkene to aldehyde and if that $\ce{H2O2}$ is not removed from the reaction mixture, it will oxidize the aldehyde ultimately to acid. That's why during reductive ozonolysis, people use reducing agents like $\ce{Zn}$ or $\ce{Me2S}$ or $\ce{PPh3}$ which can get oxidised themselves by the $\ce{H2O2}$  produced  to $\ce{ZnO}$, $\ce{Me2S=O}$ and $\ce{Ph3P=O}$ respectively and your aldehyde is not oxidized to acid any more. But in this case no such reducing agents is used and thus the product of the first step in the diacid not di-aldehyde. After that heating with $\ce{Ca(OH)2}$ doesn't perform aldol reaction here but eliminates one molecule of $\ce{CaCO3}$ from the compound i.e. essentially decarboxylation and you are left with one carbon less i.e. $\text{Cyclopentanone}$. Then as you recognised correctly is the Wolff-Kisher reduction.

So your final product is $\text{Cyclopentane (B)}$. Here is the schematic of Oxidative ozonolysis releasing $\ce{H2O2}$ responsible for oxidizing the di-aldehyde to di-acid.


What are the products in a reaction between a metal and an acid?

Nitric acid ($\ce{HNO3}$) is a strong oxidizing agent. The hydrogen gas ($\ce{H2}$) produced in the reaction with metal is oxidized to $\ce{H2O}$, in turn evolving nitric oxides like $\ce{NO2}$ and $\ce{NO}$. Even magnesium can be observed evolving nitric oxides when reacted with $\ce{HNO3}$.

Additionally, you know that chlorine gas isn't released by the reaction of $\ce{HCl}$ with magnesium because you would see it evolved from the solution as a pale, yellow-green gas. Instead, only clear hydrogen gas is evolved.


That's a tricky question.

Given your setup, it follows that you will only be sampling water (not simply "fluid", mind you; fluid is a common term for both gas and liquid). But gases might be more or less soluble in liquids. Chances are that your water is saturated with some gas (say, $\ce{CO2}$) and pressurized with the same gas in the pure form. Then, as soon as you take your sample and put it in a sealed (or worse yet, unsealed) container, it will release some gas, and so the composition of the sample will become different from that of the liquid in the pipe.

Whether or not this issue is relevant to your situation, is up to you to assess.


Is there a field of chemistry that doesn't do experiments?

You are correct that it is possible to study theoretical or computational chemistry, however these typically only represent a handful of courses in a normal undergraduate chemistry curriculum. Additionally, most undergraduate chemistry curriculum are designed to instruct students broadly in the core disciplines of chemistry, hence it's usually not possible to specialize.

Non-chemisty degrees may be an alternative. However any science undergraduate program will require laboratory-based coursework. Hands-on experience teaches us how scientific concepts work in the real works, giving us an appreciation of how difficult experimental work can be. It also functions to reinforce content from books and lectures; to teach technical skills (which are usually necessary for future employment); and to develop lab safety habits. So there really is no means to escape it. Moreover, in order to take advanced chemistry electives, prior courses which include lab sessions are generally necessary.

That said, there are other routes to get into chemistry. I've know individuals whom have majored in mathematics, engineering, physics, and biology who either did research in chemistry or who did graduate degrees in chemistry subsequent to obtaining an undergraduate degree in those fields. However in order to obtain undergraduate degrees in those fields and to be able to transition to chemistry programs, lab experience and some senior-level chemistry courses are typically required.

However, once in graduate school, it is possible to avoid experiments altogether: theoretical and computational chemistry graduate degrees typically do not require that the candidate engage in any hands-on experiments. Many however do require that the graduate student work as a teaching assistant in undergraduate chemistry laboratories as a condition of funding. In such cases, laboratory experience is either necessary or a significant advantage.

So while it may be possible to minimize the amount of lab work that you engage in, some will always be required. Additionally I would not recommend avoiding lab work, as it greatly enriches the study of chemistry overall.


Why burn methane to CO2 for isotopic analyis?

There are different types of mass spectrometers with different modes of sample introduction, analyte ionization and ion selection/focusing. You cannot generalize "with a mass spec."
In many cases of isotope ratio mass spectrometry (IRMS), compound specific isotopic analysis (CSIA), or Gas Chromatography Combustion Isotope Ratio Mass Spectrometry organic compounds (not just methane) are burned and oxidized to carbon dioxide and water and analyzed as such to determine the total ratio of carbon-13/carbon-12 and hydrogen-2/hydrogen-1. However, combustion is not the only way to introduce compounds in the mass spectrometer.
Another secondary issue is that there if often more noise at low mass ranges. Methane is a small molecule (m/z =16) and carbon dioxide is relatively large (m/z=44).


How many watt-hours are theoretically necessary to produce 2kg of methane from sea water and air?

Too much for making it of any practical interest

This question cannot be answered easily, there are too many factors. Bytheway it would not be practically convenient making such a process. With $\ce{CO2}$ as your only source of carbon, you would need an extremely energy dispendious air separation apparatus to get some $\ce{CO2}$. You can get then hydrogen from water electrolysis and use  Sabatier reaction (you need a Nickel catalyst) to get finally methane.


Other metabolic reactions aside, what causes beta oxidation to stop?

Different enzymes specialize to work on different size ranges. If the fatty acid is too short, it might not bind to enzymes that work on large fatty acids. An enzyme capable of binding short fatty acids might have most binding determinants targeting the CoA moiety instead of the fatty acid itself.

  At what point would this process stop ? lets say we have a short chain fatty acid like butanoate (4c) would it still transfer an acetyl group leaving an acetate molecule ?

It goes all the way as long as the CoA is attached.


Determine bond length via rotational spectroscopy

The method they are using seems to be looking at the separation between two different transitions to determine $B$.

$J''$ denotes the lower energy state in a transition and $J'$ denotes the higher energy state. Here, you are looking at two different transitions: $$J_1'=4\to J_1''=3 \text{ with } \nu_{4\to 3}=\pu{15.365cm^-1}$$ $$J_2'=9\to J_2''=8 \text{ with } \nu_{9\to 8}=\pu{38.356cm^-1}$$

We can see in the figure below that $\Delta\nu=2B|J_2'-J_1''|$ or that the separation between transitions depends on difference between the $J$ of the higher state of the higher energy transition and the lower state of the lower energy transition. In your case, this gives that $\Delta\nu=2B|9-3|=12B$.

$\hspace{5cm}$

You could also get $B$ from the individual transitions, but you have used the wrong formula. $E_J=BJ(J+1)$ is the energy of the rotational state $J$, but you are given the energy of the transitions, which can be computed as $\nu=2B(J''+1)$. Checking with either transition should give the same result as using the spacing between the transitions.

One reason you might use the spacing to compute $B$, rather than individual transitions, is that if you have any systematic error in measuring transition frequencies, taking the difference will cancel out this error.


Does the iodoform reaction form hydrogen iodide or water?

Your balanced equation is correct, both water and iodide (and iodide in water will be in equilibrium with hydrogen iodide) are formed in the iodoform reaction.

$$ \mathrm{CH_3COR + 3I_2 +4OH^- \rightarrow  CHI_3 + RCOO^- +3I^- +3H_2O}$$

The Wikipedia article that you mention, could have been written a bit clearer.  Specifically, there are two key steps in the iodoform reaction.  First, there is formation of the tri-iodo ketone and water is generated in this step.

In a subsequent step, the carbonyl undergoes an SN2 reaction to generate iodoform.

The Wikipedia drawing in your post seems to focus on the mechanism of the second step (3 to 4b).  The first step (2 to 3) is just illustrated in a cursory fashion and the formation of water is not included in the figure.


Admittedly, I'm having a professional photochemical bias when it comes to the generation of singlet oxygen!

A rather safe and simple approach - even in home experiments - is the use of air, sunlight and a sensitizer (rose bengal, methylene blue, or even chlorophyll from natural resources, such as Spirulina).

Here, the sensitizer is irradiated and excited to the first singlet state, undegoes intersystem crossing (ISC) and transfers the energy to ground state triplet oxygen, upon which singlet oxygen is formed:

\begin{align}
\ce{Sens &->[h\nu]\ ^1Sens^{\ast}}\\
\ce{^1Sens^{\ast} &->[ISC]\ ^3Sens^{\ast}}\\
\ce{^3Sens^{\ast} +\ ^3O2&->\ Sens +\ ^1O2}\\
\end{align}

A prominent reaction of $\ce{^1O2}$ is the concerted, diastereoselective addition to 1,3-dienes in Diels-Alder fashion.

A classical example for this reaction is the synthesis of the natural peroxide ascaridole (2) from $\alpha$-terpinene (1). G. O. Schenck, the former director of the Max Planck institute for Radiation Research, performed the reaction in batteries of flasks in his garden and published the results in 1944 (Naturwissenschaften 1944, 32 (14-26), 157–157). Ascaridole is a natural anthelmintic (= against parasitic worms), hence the name.

Maybe not really suitable for home experiments due to the chromatographic workup envolved but worth to be mentioned is the Schenck ene reaction of allylic alcohols:

The reaction products can be converted to 1,2,4-trioxanes by peroxyacetalization, as shown by Griesbeck and coworkers (Beilstein J. Org. Chem. 2010, 6 (61))

Note that the 1,2,4-trioxane moiety is the key element in the natural anti-malaria drug *artemisinin.


Based on your comments, it sounds like you have some sort of colored sand and you want to immobilize it. My suggestion would be some sort of polymer matrix.

There are multiple possibilities, but one of the easiest is polydimethylsiloxane PDMS which is optically clear and pretty easy to work with (i.e., safe). There's the "base" and a curing agent which creates cross linking.

Probably the easiest how-to comes from U. Wisconsin.

If done right (i.e., no air bubbles) you should be able to pour in the PDMS, cure, and have a polymerized matrix around the sand. Then you could cut it up or do whatever needed to see different regions.


Why doesn't velocity enter into Heisenberg's Uncertainty Principle?

Some problems with your method: you mention taking 100 copies of the same atom of hydrogen and finding the positions at 100 different times. (i) You actually can't make a copy of a quantum system without knowing its quantum state (the no-cloning theorem), (ii) your measurements of position will disturb the "trajectory" of the electron, so your 100 cases are distinct and cannot represent the path of a single electron, and this is especially true the more accurate you make your measurements.

You should realize, however, that the bound placed by the Heisenberg uncertainty principle is really quite small, and on a macroscopic scale we can essentially determine the trajectory of an electron. A bubble chamber or cloud chamber is essentially a major revision of your proposed method: we fill a vessel with supersaturated vapor, wait for charged ions to pass through the vapor and cause it to condense, and observe the resulting tracks. This describes the ion's motion with a resolution on the order of microns.


Yes, this is a known phenomenon. It's a process called electroporation. Here's a paper talking about using it for sterilization. Basically, when you apply an electric field to a cell membrane, you can render it porous. At low field strength, the pores are small and can be closed by the cell, but if you apply a larger potential across the cell membrane, cell lysis is the result, killing the cell. It's actually used for killing cancerous tumour cells in humans.

Unfortunately the trick with doing it with something like milk is that, to get a large enough potential across the cells in the liquid, an enormous potential (up to 1 kV/mm) needs to be applied to electrodes immersed in the liquid, even though the cell membrane is only seeing a volt or two. For this to be manageable, you would probably need a thin flow cell or something. It could be a fair bit more expensive than pasteurization. Even though only a short pulse is used, electrolysis will occur, which could affect the taste. How well it works will also be affected by the electrical properties of the solution, which you can't necessarily play with too much in foods without changing the taste, so it might not be the best technique for foods, but there are certainly other areas where it's useful.


The picture is not a lattice, it is a periodic pattern. In a lattice, all lattice points are related by the translational symmetry given by the unit cell vectors. I can take any two lattice points and take the vector between them. Then, when I add that vector to any other lattice points, I should land on another lattice point.

This is not the case here. If I take the vector between the pair of closest points, and add the vector to the upper right point of the pair, I land on empty space.

So let's paint half of the points a different color. The points of different color are related by two-fold axis (located for instance at the center of two close pairs). The two-fold symmetry applies not only to the two closest points, but to the entire pattern, so it is called a crystallographic symmetry (as opposed to non-crystallographic symmetry).

This crystallographic symmetry imposes no restraints on the dimensions of the lattice. Other symmetry elements would impose restraints, for example in the presence of a four-fold crystallographic symmetry, the lattice would have to be tetragonal.

For the example at hand, the convention is to put the origin on one of the two-fold axes. It looks like you could have one unit cell vector along x and one along y, at approximately right angle (this is not imposed by point group symmetry, so it is "by accident" and you would have to measure carefully to see whether the angle is exactly 90 or maybe 90.0001 degrees).

To summarize, it is monoclinic with an angle that happens to be close to 90 degrees.

For specifics on the lattice types, see e.g. Wikipedia.

Thanks, @Ivan_Neretin, for the helpful comment.


It depends on the priority of the substituents on the ring. If the highest-priority group is the alcohol group, then the numbering of the carbons in the ring starts there - so the 1 is implicit, and you do not need to state it. See 2-(3-Methylbutyl)cyclohexanol as an example.

Note that the alkyl group position is assigned a number relative to the position of the -OH group. However, if there is a higher-priority group than -OH then the position of the alcohol will be assigned a number relative to this higher-priority group; see 2-hydroxy-3-chlorobenzoic acid.

In your example, the 1 is not necessary.


Why is it possible, that an equilibrium shifts?

Let's do a simple generic example. We have a reaction:

$$\ce{2A <=>[k_1][k_{-1}] B}$$

Experimentally we determine that the rates of the forward and reverse reactions are:

$$\text{Rate}_1=k_1[\ce{A}]^2$$
$$\text{Rate}_{-1}=k_{-1}[\ce{B}]$$

At equilibrium:

$$\text{Rate}_1=\text{Rate}_{-1}$$
$$k_1[\ce{A}]^2=k_{-1}[\ce{B}]$$

Since the rate constants are constants we can combine them into the equilibrium constant to generate the law of mass action:

$$K_{eq}=\frac{k_1}{k_{-1}}=\frac{[\ce{B}]_{eq}}{[\ce{A}]_{eq}^2}$$

Now, if $k_1=2k_{-1}$, we have a value of the equilibrium constant, which gives us the relationship between $[\ce{A}]_{eq}$ and $[\ce{B}]_{eq}$ that defines equilibrium:

$$K_{eq}=\frac{k_1}{k_{-1}}=\frac{2k_{-1}}{k_{-1}}=2$$
$$K_{eq}=\frac{[\ce{B}]_{eq}}{[\ce{A}]_{eq}^2}=2$$
$$2{[\ce{A}]_{eq}^2}=[\ce{B}]$$

In this case, only concentrations of $\ce{A}$ and $\ce{B}$ that satisfy $2{[\ce{A}]_{eq}^2}=[\ce{B}]$ constitute equilibrium. Other combinations of concentrations are not at equilibrium and therefore the forward and reverse rates are not equal.

Let's see how this works. An equilibrium situation is $[\ce{A}]=1\text{ M}$ and $[\ce{B}]=2\text{ M}$ because:

$$\frac{[\ce{B}]_{eq}}{[\ce{A}]_{eq}^2}=\frac{2}{1^2}=2$$

The rates are:

$$\text{Rate}_1=k_1[\ce{A}]^2=2k_{-1}[\ce{A}]^2=2k_{-1}(1^2)=2k_{-1}$$
$$\text{Rate}_{-1}=k_{-1}[\ce{B}]=k_{-1}(2)=2k_{-1}$$

However, what if $[\ce{A}]=[\ce{B}]=2\text{ M}$?

$$\frac{[\ce{B}]_{eq}}{[\ce{A}]_{eq}^2}=\frac{2}{2^2}=\frac{1}{2}\ne K_{eq}$$
$$\text{Rate}_1=k_1[\ce{A}]^2=2k_{-1}[\ce{A}]^2=2k_{-1}(2^2)=8k_{-1}$$
$$\text{Rate}_{-1}=k_{-1}[\ce{B}]=k_{-1}(2)=2k_{-1}$$

The concentrations, when plugged into the mass action expression, do not match the relationship defined by the equilibrium constant. More importantly, the rates are no longer equal. In this example, the rate of the forward reaction is now 4 times the rate of the reverse reaction. This is how equilibrium is reestablished. The forward reaction is going faster, but as it does so, $\ce{A}$ is consumed (slowing down the forward reaction) and $\ce{B}$ is produced (speeding up the reverse reaction) until the rates are equal again.

We can even do a bit of algebra to figure out what the concentrations of $\ce{A}$ and $\ce{B}$ will be once equilibrium is reestablished. Let's have $x$ represent the change in concentration of $\ce{B}$, so the change in the concentration of $\ce{A}$ is $-2x$.

$$K_{eq}=2=\frac{[\ce{B}]}{[\ce{A}]^2}=\frac{2+x}{(2-2x)^2}=\frac{2+x}{4-8x+4x^2}$$
$$2(4-8x+4x^2)=2+x$$
$$8-16x+8x^2=2+x$$
$$6-17x+8x^2=0$$
$$x=0.44695..., 1.6781...$$

In our case, we must use $x=0.44695$ since the other value would lead to a negative concentration for $\ce{A}$. Thus:

$$[\ce{A}]_{eq}=2.00-2(0.44695)=1.11$$
$$[\ce{B}]_{eq}=2.00+0.44695=2.45$$


What is the mechanism for electrospray ionization?

How electrospray works has been hotly debated for decades.  Two competing theories have been put forward.

Initial stages: theories agree on droplet collapse due to Rayleigh instability

The two theories agree on the initial stages of droplet formation.  An aerosol is formed by the extrusion of low ("nano", or 25 to 100 nL / min) to high (~1 mL / min) flows of eluent through a capillary charged to ~3000 V.  The volatility of the solvent used in electrospray techniques ensure that the aerosol droplets evaporate.  Charged spheres of liquid experience a self-repulsion due to electrostatic effects.  This repulsion is counterbalanced by surface tension.  As Lord Rayleigh noted in 1887, as long as $T > \frac{Q^2}{16 \pi a_0^3}$, where $Q$ is the droplet charge, $T$ is the surface tension, and $a_0$ is the droplet radius, then the droplet is stable.  But as heat evaporates the droplet, $a_0$ shrinks, while $Q$ stays constant and $T$ nearly so.  Thus the droplet eventually becomes unstable, exploding into a mist of finer droplets, each of which has a smaller charge. What happens after that has been the subject of debate.

The two theories for what happens after droplet collapse

ION EVAPORATION.  This theory holds that as droplets shrink, eventually the field strength at the droplet surface is high enough to force remaining solvated ions into the gas phase, an effect called "field desorption".
CHARGE RESIDUE.  This model holds that charged analytes remain solvated in droplets through many cycles of droplet evaporation / fission.  Eventually, droplets contain on average only one analyte molecule.  Remaining solvent molecules evaporate, leaving the ion in the gas phase.

Current evidence suggests that most small molecules ionize primarily by ion evaporation, but that for large molecules such as proteins, charge residue mechanisms may be important.  In the ion evaporation mechanism, molecules or ions must translocate to the surface of a droplet before they can become gas phase ions.  This  means that molecules which preferentially segregate toward the surface of liquid droplets will ionize more easily.  Thus, the ion evaporation mechanism, but not the charge residue mechanism, predicts that surfactants will have higher ionization efficiencies than non-surface active agents.

An extension of the ion evaporation model

EQUILIBRIUM PARTITIONING MODEL.  This theory divides droplets into an electrically neutral core, where ion pairing maintains local electroneutrality, and a charged shell, from which small molecule ions can move into the gas phase by ion evaporation.  The amount of "excess" charge in the shell available for ionizing neutral molecules from the core is determined by electrospray flow parameters, in particular by the ratio of electrospray current to flow rate, or $[Q] = \frac{4.2}{F} \sqrt{\frac{K \gamma }{V_f \kappa_e}}$, where $F$ is Faraday's constant, $K$ is the conductivity of the medium, $\gamma$ is surface tension, $\kappa_e$ is the dielectric constant,  and $V_f$ is the volumetric flow rate. Conditions where excess charge is limiting could cause severe ion suppression.  For more discussion, see this paper.

Conclusion:

This answer is very long which should be a clue that this topic is complex.  Usually, saying that things are complex is a sign that people don't understand something very well, and that is the case here.  But in general, perhaps the closest to a consensus in the community seems to be that small molecules ionize primarily by ion evaporation, but that for large molecules such as proteins, charge residue mechanisms may be important.


I can only really speak for bromine(VII). Bromine(V) is pretty common and I'm not entirely sure what's the deal with Br(I) readily disproportionating to Br(V) + Br(-1). (The tendency of Br(I) to undergo disproportionation explains why we don't see it very commonly, but why it has this tendency I'm not so sure.)

This is an example of what is sometimes called the "alternation effect", or "d-block contraction" in p-block chemistry.

Due to the intervention of the ten 3d electrons, which are poorly shielding, the 4s and 4p electrons of the elements Ga through Kr are more tightly bound than one might expect. These elements show a distinct reluctance to take on their group valency. In $\ce{BrO4-}$, the higher-than-expected energy cost of ionisation to form Br(VII) is not fully compensated for by covalent bond formation (the strengths of which decrease linearly down the group). A selection of other observations that can, to various extents, be attributed to this are:

the anomalously high electronegativity of germanium (2.01), compared to silicon (1.90) and tin (1.96)
the anomalous instability of $\ce{AsCl5}$ (decomposes above $-50~\mathrm{^\circ C}$) even though $\ce{PCl5}$ and $\ce{SbCl5}$ are well-known,
the fact that nitric acid oxidises sulfur to $\ce{H2SO4}$, but selenium only to $\ce{H2SeO3}$

In a sense, it is exactly the same as the inert pair effect, which arises partly due to the lanthanide contraction - i.e. the intervention of the fourteen 4f electrons, which shield 5d, 6s and 6p electrons poorly. The difference is that the inert pair effect is even more pronounced because of significant relativistic stabilisation of the 6s electrons. Here's an insightful discussion on the topic: Why does tin form tin (II) compounds?


Confused on the peak intensities in this mass spectrometry problem

Such problems in chemistry don't work out to infinite precision like pure math problems. You have to apply some chemical knowledge to the problem too.
The book formula is obviously an approximation to the exact solution. To find the exact values expected you'd need to use combinatorics.
Considering the chemistry:

7 carbons have a mass of 84, so a hydrocarbon molecule can't have that many carbons since there wouldn't be any hydrogen atoms.

5 carbons have a mass of 60, but that would mean at most 12 hydrogen atoms for an aliphatic molecule, so the mass could be at most 72.

Thus there must be 6 carbons in the molecule.


What are some of the more easily manipulated residues on an enzyme?

I'm assuming the OP concerns chemical means of modifying residues. Multiple protein residues other than Lysine may be labeled with an adduct. Here are some examples:

Histidine residues may be labeled using an epoxide attached to a ligand. (Chen et al., 2003)

Cysteine residues may be labeled with a nitrile derivative where the sulhydryl acts as the nucleophile. (Obella et al., 2007)

Arginine residues may labeled using a derivative of the naturally occurring methylglyoxal. (Oya et al., 1999)

Carboxylate functionality such as in Aspartate and Glutamate residues as well as the C-terminus may be labeled via the formation of an amide linker using a carbodiimide such as N-ethyl-3-N',N'-dimethylaminopropylcarbodiimide (EDC). (Gilles et al., 1999)

The above list of residues is not comprehensive, nor are the labeling agents. I recommend checking out the reviews by Basle et al. (2010) and Chen and Wu (2016).
References
Basle, E, Joubert, N, Pucheault, M (2010) "Protein Chemical Modification on Endogenous Amino Acids" Chemistry & Biology 17:213-227. DOI: 10.1016/j.chembiol.2010.02.008
Chen, X and Wu, Y-W (2016) "Selective chemical labeling of proteins" Org. Biomol. Chem. 14:5417-5439. DOI: 10.1039/C6OB00126B
Gilles, MA, Hudson, AQ, Borders, CL (1990) "Stability of water-soluble carbodiimides in aqueous solution" Anal. Biochem. 184:244-248. DOI: 10.1016/0003-2697(90)90675-Y
Gong Chen, Alexander Heim, Doris Riether, Dominic Yee, Yelena Milgrom, Mary Ann Gawinowicz, and Dalibor Sames (2003) "Reactivity of Functional Groups on the Protein Surface: Development of Epoxide Probes for Protein Labeling" J. Am. Chem. Soc. 125:8130–8133. DOI: 10.1021/ja034287m
Oballa, RM, Truchon, JF, Bayly, CI, Chauret, N, Day, S, Crane, S, et al. (2007) "A generally applicable method for assessing the electrophilicity and reactivity of diverse nitrile-containing compounds" Bioorg. Med. Chem. Lett. 17:998–1002. DOI: 10.1016/j.bmcl.2006.11.044
Oya, T, Hattori, N, Mizuno, Y, Miyata, S, Maeda, S, Osawa, T, Uchida, K (1999) "Methylglyoxal modification of protein. Chemical and immunochemical characterization of methylglyoxal-arginine adducts" J. Biol. Chem. 274:18492-18502. DOI: 10.1074/jbc.274.26.18492


Why does ozone disproportionate on reaction with potassium iodide?

The reaction equation does not really give information about the full mechanism, but one possibility is that the ozone is first reduced to ozonide ion, which is known in compounds with heavier alkali metals like potassium,; then the ozonide ion disprportionates to give diatomic oxygen:

$\ce{O3}+\ce{e-}\rightarrow \ce{O3-}$

$\ce{O3-}+\ce{e-}+\ce{H2O}\rightarrow 2 \ce{OH-}+\ce{O2}$

Note that the reducing agent is not specified as iodide ion.  Any mild reducing agent can act similarly.


How to separate KCl from NaCl and MgCl₂?

As I mentioned in the comments, a quantitative separation of the three compounds will probably not be possible outside of a proper laboratory, however a coarse separation is possible. A quick search on the internet yields this link for the separation of $\ce{NaCl}$ and $\ce{KCl}$. You didn't give us the exact composition of the table salt substitute, so I assume $\ce{MgCl2}$ is present in much smaller amounts than the first two compounds.

The simplest strategy is to use the difference between the solubilities of the salts in water or ethanol. Using water is much more convenient, so we'll stick to that. I'll outline the method to obtain the highest purity $\ce{KCl}$. First, add a very large amount of table salt substitute to a container with boiling water (you'll want to add some 200-250 g per 100 g of water) and let the mixture dissolve as much as possible (though it probably won't even come close to dissolving completely). What you have then is a boiling water solution saturated with both $\ce{NaCl}$ and $\ce{KCl}$. Now, while the water is still very hot, quickly filter off the excess salt (might be doable with a fine sieve, or a funnel with some cotton would work better) and reserve the liquid in another container. Let the liquid cool down (using an ice bath is optional but increases yield and purity), and it will start to precipitate the salts. Now you perform another filtration, but this time you keep the solid retained on the filter, not the liquid. This yields purified $\ce{KCl}$.

This strategy works because it turns out that $\ce{NaCl}$ and $\ce{KCl}$ behave rather differently in how their solubility in water is affected by temperature.

$\ce{NaCl}$ soubility in water:

35.6 g/100 g water (0°C)
35.9 g/100 g water (20°C)
39.0 g/100 g water (100°C)

$\ce{KCl}$ soubility in water:

28.0 g/100 g water (0°C)
34.2 g/100 g water (20°C)
56.3 g/100 g water (100°C)

Evidently, as the filtered liquid cools down, $\ce{NaCl}$ remains almost as soluble in cooler water as in hot water, while $\ce{KCl}$ is only about half as soluble. This means that a large amount of $\ce{KCl}$ will precipitate as the filtered liquid cools, while only a bit of $\ce{NaCl}$ will follow. Hence, ideally, from 100 g of water saturated with both salts at 100°C (and containing nothing else), you will be able to extract 25.0 g of a mixture of salts containing 87.6% $\ce{KCl}$ if you perform the second filtration at 20°C. If you filter at 0°C, then the yield increases to 31.7 g of a mixture containing 89.3% $\ce{KCl}$. If for some reason you don't want to do the second filtration, you can just let the water evaporate completely after the first filtration, producing 95.3 g of a mixture containing 59.1% $\ce{KCl}$ from 100 g of saturated boiling water.

In practice your yield and the purity of the $\ce{KCl}$ will likely be somewhat lower due to losses in manipulation and the presence of other impurities. The procedure may be complicated by the very large initial amount of table salt mixture that you have to dissolve in water. You can reduce the amount of the mixture, but at some point the purification yield will start to go down.


The questions is a good one, and answering it winds up being central to a lot of key concepts in thermodynamics. A brief correction before starting, though I don't think it's central to the question,  "When work is done on by the system it means that a part of system kinetic energy is used to do the work..."

The disconnect between your understanding and that used in thermodynamics is that in thermodynamics, the system and its surroundings are separate. So, when our system does work on the surroundings that energy won't come back to heat our system. If we do as you say and use our system to do work on the surroundings, turn that work into heat, and then apply that heat back into our system, the result is just as you say! The kinetic energy in our system isn't lost. Or put in another way when we're allowing heat to flow in and out of our system, all of that heat ultimately winds up as kinetic energy and $ΔU = Q.$

Let's look at a couple examples of putting $\pu{1 J}$ of heat into a system with different surroundings to see how this works out. Our system will be an insulated piston filled with an ideal gas (say helium), which can expand and push on one of 3 things in its surroundings: a mass, a vacuum, or a generator.

Mass: This follows $ΔU = Q - W$.  Of the $\pu{1 J}$ of heat we apply, the portion that goes into warming up our system and increasing internal kinetic energy is $C_v$ (the molar heat capacity is $1.5R),$ whereas the portion of that instead goes into doing work on the surroundings is $R$ (the universal gas constant). So our system gains

$$[\pu{1 J}]\cdot \frac{C_v}{C_v + R} = \pu{\frac{3}{5} J}$$

of thermal energy, and does

$$[\pu{1 J}]\cdot \frac{R}{C_v + R} = \pu{\frac{2}{5} J}$$

of work on the surroundings by expanding and pushing up the mass. This $\pu{\frac{2}{5} J}$ of work energy won't return. These portions $C_v/(C_v + R)$ and $R/(C_v + R)$ show up a lot later on in thermodynamics as their inverses, $k$ and $k/(k - 1)$, respectively.
Vacuum: This follows $ΔU = Q$. There is no surroundings to do work on, so all of the expansion energy goes into slamming the piston against its backstop, and after everything settles this ultimately results in the full $\pu{1 J}$ of heat becoming thermal energy inside our piston.
Generator: This also follows $ΔU = Q$. It's the same situation as the mass, with $\pu{\frac{3}{5} J}$ of thermal energy and $\pu{\frac{2}{5} J}$ of work done on the surroundings, but that $\pu{\frac{2}{5} J}$ of work can be turned into electricity and then be used to heat our piston.

Just like in the vacuum case, all $\pu{1 J}$ of heating ultimately winds up as thermal energy in the system, though it spent some time as work on the surroundings during the process. So, $ΔU = Q - W + W.$ I think this final case is the closest to matching your question.


What is chemistry considered the study of?

While the answers so far are not wrong, I'd like to add a refinement. Chemistry is the study of matter at the level of electron interaction. The basic question in chemistry is about chemical bonding, and as far as we know this is a function of electron interaction. In contrast, radioactivity is also about matter, but is not at all about electron interaction and is not generally considered chemistry (it's considered a part of physics). Also, the chemistry of isotopes (i.e. differing neutron count) is not generally different across different isotopes. This is good, because we need solvents with are deuterium based ($\ce{D=^2H}$) as opposed to hydrogen based for NMR spectroscopy. If we go lower than electrons, we are into quantum mechanics, which is again physics. So electron interaction is a good level at which to characterize chemistry.

EDIT: I realize the original question was for a one-word answer. Frankly, that's not the greatest question, but it's generating answers that make it more interesting, and it's why I answered.

Let me give a counterexample to my own answer to show the difficulty: gases. We learn a lot about gases in chemistry which has nothing to do with chemical bonding; e.g. all the gas equations. I'd argue that most of this is not really chemistry, it's physics. Understanding how temperature, pressure, and volume interact dynamically is a very physical sort of study, not all that different from studying light, or magnetism, or force.  So why do we study gas physics in chemistry? I suspect it's because we use gas physics a lot in chemistry! You need to understand the gas laws to do any real experimentation involving actual gas chemistry; i.e. how would you measure gas as a reactant or product without knowing the gas laws?

So to be clear, if you want one word, I agree with "matter". But I think if you want to get at what chemistry is about, it's not so simple.


If I drink water without sipping, is there a chance of saliva diffusing into my bottle?

This question reminds me of a germaphobic classmate I had back in Middle-school. He would always stay clear of public lavatories/urinals because he was under the impression that urinating at one would result in bacteria from the urinal climbing up the stream of urine and into his urethra...needless to say, he was a very lonely boy.

The situation you described ("Drinking without sipping") is a modification (albeit, a more reasonable one) of the "Bacteria-climbing-up-a-stream-of-urine" theory.

I'm not a chemist myself, and like you, I am faced with... ahem, "issues", while searching for speeds of common ions in water (at a given temperature, and in the absence of an external electric field). But I feel this question can still be satisfactorily approached even without these values.

From the wording of your question, I suppose you assumed that potassium, sodium, magnesium and other ions would diffuse from a region of higher concentration (the saliva in your mouth) to a region of lower concentration of those ions (the stream of water entering your mouth). Now this would seem perfectly plausible if it weren't for a few things:

1) You're dealing with flowing water

This isn't a simple case of spitting into a glass of water and waiting for the contents of your saliva to (passively) diffuse into the water (and eventually form a dilute saliva solution). By drinking it the way you do, the passive diffusion of electrolytes in your saliva won't be able to match the speed of the water flowing into your mouth, much less exceed it in order to enter the bottle.

2) All the electrolytes in your mouth are trapped in a viscous protein-soup

Human saliva contains a variety of proteins (among other things). Common experience tells us that saliva is (a little more) viscous than water, and doesn't "instantly" dissolve in it (<----- Coming from a guy who had to spit in test tubes for his Biology lab assignments). Now not only do the electrolytes have to actively diffuse upstream to get into the bottle, but they also have to "free" themselves from the saliva first.

3) What do you mean by "saliva"?

This is the philosophically inclined conclusion to my answer. You're worried that the entry of electrolytes from your mouth into the bottle (won't happen) counts as contaminating it with your saliva. At first glance, even if the electrolytes seem to be able to make it to the bottle, you're pretty sure that larger molecules (such as proteins) won't be able to do the same.

If you don't count the proteins, epithelial cells and bacteria that normally constitute human saliva...how would this make saliva any different from a simple laboratory-prepared cocktail of sodium, potassium, magnesium, calcium, etc? Besides, ordinary (un-branded) bottled water would already contain these electrolytes; so by that logic, the water's already contaminated with your "saliva" even before you opened it!


Is it possible to obtain pure, precipitated iron with no oxidation by some chemical process?

Very pure iron can be obtained by pyrolysis of iron pentacarbonyl.  Iron metal obtained in this manner is in fact called carbonyl iron and is commercially available.

$$\ce{Fe(CO)5 -> Fe(s) + 5 CO}$$

Doing this yourself is very dangerous, however, because iron pentacarbonyl is volatile and toxic.  The decomposition reaction to produce iron requires high temperatures, and the byproduct is carbon monoxide gas, which is also notoriously toxic.

  Alternatively is it possible to electrically purify iron using an anode and cathode?

This is also possible.  I don't know about other salts of iron, but to form iron from iron oxides, very high temperatures are required.  The other answer mentioned Dr. Sadoway's work.  Let me highlight a recent article of his unveiling new anode materials for electrolytic decomposition of iron oxide.  At temperatures as "low" as 1538°C, molten iron oxide can by electrolyzed to pure oxygen gas and to molten iron metal using electrochemistry.

Water is not an appropriate solvent for electrochemical formation of iron metal.  At potentials low enough to favor $\ce{Fe^0}$ formation from $\ce{Fe^{+2}}$, water is unstable and would be reduced to hydrogen gas.


The name suggested here (via the question comments), the anion-corresponding part, ‘bicyclo[3.3.1]tetraboroxane-3,7-diolate’ ($\ce{[B4O5(O)2]^2-}$) obviously does not match with the tetraborate formula $\ce{[B4O5(OH)4]^2-}$ present in borax, whose structure, confirmed by X-ray crystallography, including hydrogens[1], is

However, the ‘bicyclo[3.3.1]tetraboroxane’ part is correct, for the parent hydride:

This type of names is known e.g. for the ‘cyclotriboroxane’ name for boroxine, or ‘cyclotetrasiloxane’ (in octamethylcyclotetrasiloxane).
So the name would be rather 1,3,5,7-tetrahydroxybicyclo[3.3.1]tetraboroxane, or bicyclo[3.3.1]tetraboroxane-1,3,5,7-tetrol, however it still does not explicitly tell that it's an anion.
Interestingly, this boroxane, siloxane type names (interchanging element/oxygen pattern) is rather marginally mentioned in the inorganic nomenclature, notably more detailed in the organic nomenclature.[2]
Inorganic additive nomenclature is able to constructs names for similar skeletons,[3] but somewhat differently than organic nomenclature.
E.g. bicyclo[x.y.z]Nane (where N is the numerical prefix root, with the exception for 1..4, ‘meth’..‘but’, for the organic names) would be named: Ncarby-[0a.z1,b]dicycle, where a = x+y+2 and b = x+2. If the structure is an ion, it's ‘…cyclate’ followed by the charge in parentheses. One advantage is that the first numeral is the size of the (largest) cycle, not just a branch.
Some examples from the book:

1,7-diazyundecasulfy-[012.11,7]dicycle
(IR-1.5.3.4 Example 6)

cyclo-triphosporic acid
2,4,6-trihydroxido-2,4,6-trioxido-1,3,5-trioxy-2,4,6-triphosphy-[6]cycle
(IR-8.3 Example 5)

tetraaluminy-1-lithy-[05.01,301,402,5]tetracyclate(1−)
(IR-7.4.3 Example 11)

(The leading zero in the first numeral in the square brackets is intentional for the cycles, however in the latter example, it's not used, despite IR-7.4.2 Nodal descriptor, example 3, that states that the nodal descriptor for hexacycle is ‘06’.)
With this nomenclature we can get very descriptive (but almost unusable) names
1,3,5,7-tetrahydroxido-1,3,5,7-tetrabory-2,4,6,8,9-pentaoxy-[08.11,5]dicyclate(2−),
1,3,5,7-tetrahydroxidotetrabory-2,4,6,8,9-pentaoxy-[08.11,5]dicyclate(2−), or
1,3,5,7-tetrahydroxido-1,3,5,7-tetraborypentaoxy-[08.11,5]dicyclate(2−).
References:

Levy H.A., Lisensky G.C. Crystal structures of sodium sulfate decahydrate (Glauber's salt) and sodium tetraborate decahydrate (borax). Redetermination by neutron diffraction Acta Crystallographica Section B 34 (1978): p. 3507
IUPAC Blue Book 2013, P-52.1.6.2
IUPAC Red Book 2005, IR-7.4 Additive nomenclature: Inorganic chains and rings


In molecular orbital theory, why does s-p mixing in the dinitrogen molecule not effect the 1σᵤ orbital?

Rob, if you poke around on the internet you can find many sites that show diagrams for how the $\ce{N2}$ molecular orbitals are formed from two sets of nitrogen ($\ce{N1}$) atomic orbitals.  Many of the diagrams are different one from another - that's a tip off that s-p mixing is not something that can always be predicted in advance.  With that understanding, let's step through your diagram and questions.

It might be useful to take a look at the $\ce{N2}$ MO diagram here.  It's just like your diagram. but they break down the steps and it might be a bit less confusing.

In any case we have a nitrogen atom with 5 valence electrons, 2 in the 2s atomic orbital and 1 each (total of 3) in the 3 degenerate p orbitals.  If we move a second nitrogen atom into the vicinity, these atomic orbitals will hybridize in order to start the bond formation process.

$\ce{N2}$ is analogous to acetylene, they both have a triple bond that is made up of 1 sigma bond and 2 pi bonds.  In acetylene - and nitrogen - the sigma bond is $\ce{sp}$ hybridized.  We form this bond by

Mixing the s and  a p atomic orbitals (the p orbital whose lobe is linear with the s orbital [see the diagrams in the link I supplied above, they call it the $\ce{p_{z}}$ orbital]) on one of the nitrogens creates a bonding and antibonding $\ce{sp}$ orbital on that nitrogen atom.
Same thing happens on the other nitrogen atom
As the 2 nitrogen atoms begin to interact and form bonds, the 2 $\ce{sp}$ orbitals on each nitrogen interact (split further).
The bonding $\ce{sp}$ orbital on each nitrogen interacts to form a lower energy and higher energy pair of bonding molecular orbitals (the $\ce{1\sigma _{g}}$ and $\ce{1\sigma _{u}}$ orbitals in your diagram - BTW, there should be a dashed line from the p orbitals to the $\ce{1\sigma _{g}}$ orbital - it is missing in your diagram, maybe that has led to some of your confusion).
The antibonding $\ce{sp}$ orbital on each nitrogen also interact in the same manner to create 2 antibonding sigma molecular orbitals ($\ce{2\sigma _{g}}$ and $\ce{2\sigma _{u}}$)

Now for pi bonds,

The remaining two p atomic orbitals on each nitrogen are orthogonal to the s and $\ce{p_{z}}$ orbital - they can't mix with those orbitals.
These atomic p orbitals $\ce{p_{x}}$ and $\ce{p_{y}}$ can only mix with their like counterpart on the other nitrogen atom.  This gives rise to a bonding and antibonding $\ce{p_{x}}$ pi molecular orbital and a bonding and antibonding $\ce{p_{y}}$ pi molecular orbital

With that background, let's now look at your questions

  s-p mixing in N 2   causes the 1σ g   orbital to decrease in energy,
  the 2σ g   orbital to increase in energy (to higher than the 1π u  )
  yet the 1σ u   is left unchanged by the interaction. Why is this?

I'd say it a bit differently as I did above. Mixing of the 2 bonding $\ce{sp}$ orbitals, one on each nitrogen atom, causes them to split (one is raised and one is lowered).  The same thing happens to the antibonding $\ce{sp}$ orbital on each nitrogen - one is raised, one is lowered.

  Clearly they are the s and p orbitals but is this within the atom or
  is the s orbital on one atom interacting with the p on the other?

Hopefully I answered this above.  I tried to distinguish between atomic and molecular orbitals.  If I haven't been clear, post back.

  Some clarification as to why the 1σ g   orbital to decreases in
  energy, the 2σ g   orbital to increases in energy would also be much
  appreciated.

Hopefully what I wrote above clarified this as well, let me know if it didn't.


In the current form the question is addressed, and assuming you took a look at a listing of software already in the field like here, I speculate the sheer number of organic reactions (just some) alone renders such a task as a one-man / one-woman project too large.  There is a lot of work associated to create such a database, or software. Organic Synthesis is one open-access reference for organic reactions, yet the normal output is a html / pdf of the reaction procedure, and not a *.sdf (or other chemically relevant *.xml-like) format.  Inorganic Synthesis is a publication with similar intent as the former, yet not openly accessible.

Nevertheless, if you are interested in this topic, I suggest to consult the recent publications Computer-aided synthesis design: 40 years on (DOI 10.1002/wcms.61) as well as the just about a year old Computer-Assisted Synthetic Planning: The End of the Beginning in Angewandte Chemie (English version DOI: 10.1002/anie.201506101).


Can two positive ions be next to each other in a formula?

Yes, there are many examples of this, $\ce{HNO3}$ for instance

  So Arsenic must have a charge of 3+ for all the charges to add up to
  be 0? Correct?

Yes, that's correct

  However, my professor keeps telling me to keep the positive oxidation
  numbers away from each other. Is that a rule when writing a formula?

It's not a rule that I'm familiar with.  Here is a link to a nice, concise listing of the rules for writing formulas for inorganic compounds.

  I thought it only applied when drawing Lewis structures or resonance
  structures.

Yes, it certainly applies in those two cases


Not all the matches have sulfur but only some types in The Preparatory Manual of Black Powder and Pyrotechnics
by Jared Ledgard see page 191, you can find the best matches for your needs, classic matches are okay.

You are right, no reaction is reported between sulfur and ammonia. However if you put in a closed box a compound containing sulfur and a little bit of organic matter Sulfur-reducing bacteria (SRB) reduce sulfur in sulfide anion.

$$\ce{S + 2e- -> S^{2-}}$$
$S^{2-}$ is a base and reacts with water to produce hydrogen sulfide:
$$\ce{S^{2-} +2 H2O -> H2S + 2OH-}$$

$H_{2}S$ is a conjugate acid of $S^{2-}$ so now can react with ammonia with an acid-base reaction:

$$\ce{H2S + NH3 <=> NH4SH}$$
or:
$$\ce{H2S + 2NH3 <=> (NH4)2S}$$

Of course like many ammonium salts it can easily decompose into the two odorous gasses.

Others odorous substances are  skatoles, mercaptans, indoles however I don't think that these are present in great quantities in this case. See this site for further informations.


Determining Charge of Protonated Alkaloids

Are you sure you're trying to calculate the molar mass of the mescaline salt? Isn't that known, depending on the acid you use?

It seems to me the key you need is the mole ratio. For a weak acid-base conjugate pair, you can use the Henderson-Hasselbalch equation:

$pH = pK_a + log(\frac{[B]}{[A]})$

where [B]/[A] is the ratio of concentrations of conjugate base to conjugate acid. If calculated for a single solution containing both forms, the total solution volume cancels and this is the mole ratio. In a more convenient form,

$\frac{B}{A} = 10^{pH - pK_a} = \frac{K_a}{[H^+]}$

(The first equality is more practical, the second is useful for theoretical manipulations.)

Note that when $pH = pK_a$, $B = A$ (the analyte will be 50% protonated and 50% deprotonated).

For mescaline (more precisely, its conjugate acid), $pK_a =$ 9.56, typical of an amine base/alkaloid. So at a pH of 9.56, approximately half of your mescaline should be protonated. You could extract this, measure the weight of the salt, use the known molar mass to convert to moles of the protonated mescaline (I recommend using a monoprotic acid such as acetic if possible to simplify the situation). By the Henderson-Hasselbalch, the moles of protonated mescaline you extracted at its $pK_a$ will equal the moles of freebase mescaline remaining in your organic solvent.


Can we use silver nitrate to distinguish between chloride and carbonate salts?

The test for Chloride ion is silver nitrite in an Acidified solution [with nitric acid]. This removes carbonate (as carbonic acid), cyanide[careful] and sulfide (as hydrogen sulfide). Sulfate must be absent, if present it can be removed with barium nitrate (as barium sulfate).  The silver chloride formed is solubilized by addition of ammonia; insoluble precipitate remaining may be bromide or iodide. The exact procedure to follow is given in reputed indepth Qualitative Analysis lab books.


What's important to note in this situation is that the number of moles of the titrand isn't changing. This matters because all acid/base reactions go to completion, so every drop of titrant that falls into the solution will react with the titrand until there are no moles of titrand left regardless of concentration, at which point you've reached the equivalence point.

Adding a little water here or there doesn't really change the pH at the equivalence point, because ideally there are no moles of acid or base present in solution.

The reason this doesn't mess up your measurement for the concentration is because you measured the volume of titrand before the titration, and then recorded the volume of the known titrant needed to reach the equivalence point. From this information, you can calculate the moles of titrant needed to reach equivalence, which in most cases equals the number of moles of titrand present, and with the original volume of titrand noted previously, you can calculate the original concentration of the solution the sample was taken from. voila!


Colligative properties and sizes of particles

Colligative properties From Wikipedia

  In chemistry, colligative properties are properties of solutions that depend on the ratio of the number of solute particles to the number of solvent molecules in a solution, and not on the nature of the chemical species present. The number ratio can be related to the various units for concentration of solutions. The assumption that solution properties are independent of the nature of solute particles is only exact for ideal solutions, and is approximate for dilute real solutions. In other words, colligative properties are a set of solution properties that can be reasonably approximated by assuming that the solution is ideal.

Emphasis in above text is mine.

So it the assumption that colligative properties are independent of particle size, not the reality. The assumption is very useful, but not perfect.


What is inhibiting polystyrene's ability to dissolve in traditional solvents like ethanol or acetone?

Polystyrene is soluble in acetone, MEK and other ketones, but solubility and the speed with which it dissolves depends, in part, on the molecular mass of the polymer.

If the solvent diffuses very slowly into the polystyrene, only the surface will appear to be affected. Applying a bit of energy with heat or mechanical crushing of the plastic greatly speeds the action of the solvent.

For more technical information, see Solubility Parameters: Theory and Application.


Can I test for lead with household chemicals?

General remark: get a bit of lead (metal or salt - whatever you look for) to have a positive control for your experiments.

More non-chemical tests for lead (metal):

lead is soft: you can cut/scrape it quite easily with a normal knife.
sheets of lead are easily bent, and they stay so.
For bigger pieces: Hammering on it would leave a rather big impression of the hammer. The sound would be dull.
it usually looks grayish, not silvery-metallic, but the cut will be silvery.

If it is about the paint:

what colour is it?
white lead: $\ce{2 PbCO3 \cdot Pb(OH)2}$ I'm more familiar with that in the context of paintings or painting of wooden things (window frames)
$\ce{PbSO4}$ also white. I think it has also been used for things like window frames.
side thought: lead paints for wood may have been popular due to the fact that the toxicity to microorganisms helps very much with the anti-fouling effect. Anti-corrosion paints for metal need different properties.
lead yellow: $\ce{PbCrO4}$ Wikipedia says that was the original pigment for school buses.
$\ce{2PbO \cdot PbO2}$(Mennige) red anti-corrosion pigment that was at least here in Germany the most important lead based priming paint. Only difficulty with recognizing by color: lots of non-toxic replacements are produced with similar colour to have the familiar look...
Very sensitive but but not very distinctive test$^*$ for $\ce{Pb^2+}$: the sulfide is black and very insoluble. Kitchen chemistry: put some egg on and even the surface of the metal should turn black. Possibly some heat is needed (I didn't test, though). Otherwise, use acidic $\ce{Na2S}$ solution from the school lab. If the $\ce{Na2S}$ solution doesn't give a black precipitate, there could in theory still be lead in the paint which is not accessible by the sulfide due to the binding material of the paint. In that case, however, the toxicity will be less of a problem, too (i.e. unless you dissolve the paint somehow, burn it, ...)
$^*$ OTOH it is highly improbable that there are substantial amounts of silver in the paint...
reductive borax bead should be gray, oxidative colourless.
Isolating Pb is quite easy, maybe you can do that in your school lab: precipitate $\ce{PbCl2}$ with cold HCl, filter. The dissolve the chloride in hot water. You can then precipitate PbS (black) or $\ce{PbCrO4}$ (yellow).  Looking at the crystals under the microscope can help further. Qualitative inorganic textbooks often have microscope photos of different precipitates.


What is the reason behind the numbering of nitrogen bases?

ACDLabs maintains a nice summary of the IUPAC nomenclature rules with examples:

Heterocyclics (pyrimidine): http://www.acdlabs.com/iupac/nomenclature/79/r79_702.htm

Fused heterocycles (purine): http://www.acdlabs.com/iupac/nomenclature/79/r79_960.htm

Briefly, apply the following rules:

Number your rings so that the nitrogens end up with the lowest number combination. Thus pyrimidines have (1,3).
Number your rings so that other functional group have lowest possible numbers. Hence pyrimidine is numbered to the exocyclic carbonyl groups are (2,4).
In a fused system, numbering should prefer (in this order): ring with more nitrogens, rings with other heteroatoms, larger rings, nitrogen atom closer to ring junction.
Number toward the farther ring junction from where you started. Do not number junction positions.

Number 3 is fine and all, but purine seems to have a numbering scheme that does not match these rules, and it is supported by IUPAC. Likely, this represents a historical numbering pattern that predates the IUPAC rules. Everyone was using it and so it was adopted. You will likely just need to memorize it (or look it up when you need to).


Magnetic properties of potassium manganate(VII)

Actually it is diamagnetic but due to a phenomenon called charge transfer spectrum (CTS), oxygen transfers one electron to manganese and $\ce{KMnO4}$ as a whole becomes paramagnetic. This is also the reason why it shows a characteristic colour while diamagnetic complexes are colourless due to absence of d-d transitions.


multi state or multistate or multi-state

I prefer option (1), and cite Donald Knuth's reasoning as given here.

Knuth asserts that hyphenated versions of "newly coined" words morph to unhyphenated versions (e-mail becomes email, for example) as they gain acceptance and are more frequently used, which indicates to me that option (2) is not off the table, certainly because the word is not newly coined.

However, I argue that wide acceptance and usage of the wording in the context given here is not demonstrated and fall back to option (1). Finally, as pointed out by Martin in the comment thread, be consistent and use (or don't use) hyphenation regardless of which variation you choose.


Potential energy versus reaction coordinate (graph of Gibbs free energy)

There is a long explanation for that, however in short they are not the same thing. Free energy relates to thermodynamic systems with many particles and is governed by Boltzmann's distribution, a sum of all possible energy configurations, where F is the free energy of the system.

$Z = e^{- \frac{F}{kT}} = \sum_i e^{- \frac{E_i}{kT}}$

Potential energy is non-statistical and well-defined for one particle. Thermodynamic free energy (one type being Gibbs) is a subset of the total potential energy. Even though they may seem similar, they should not be confused and should not be interchanged.


Why doesn't silica gel dissolve in ethyl acetate

Because (among other things) silica is not actually polar. Yes, theoretically a molecule of $\ce{SiO2}$ should have some degree of polarity to it. However, silica as a solid does not exist as single molecules.

Silica exists as a network of $\ce{Si}$ cations bonded together by $\ce{O}$ anions. One way of looking at it would be to say that each small bead of silica gel (or any other form of silica, glass or quartz) is one huge single molecule of silica.


Guess the shape and phase of the MOs of butadiene using LCAO and Hückel's method?

The shape of the molecular orbitals is not guessed, it is determined by the LCAO-MO expansion coefficients which one needs to find out in the first place. The hint is that an eigenvalue equation is usually solved in two steps:

We start by writing down the eigenvalue equation in the matrix form and proceed then by writing down and solving the characteristic equation in which the secular determinant is equated with zero. By solving the characteristic equation we find the eigenvalues (energies).
Then we put eigenvalues (energies) back into the matrix eigenvalue equation and solve it for the coefficients (LCAO-MO expansion coefficients) that determine eigenvectors (molecular orbitals).

As far as I understand, the second step was not simply done yet. Note that any scalar multiple of an eigenvector is also an eigenvector with the
same eigenvalue, so that eigenvectors are not unique. In solving the matrix eigenvalue equation one will find not the actual values of the LCAO-MO expansion coefficients, but rather some relationship between them which could be satisfied by a number of different coefficients. In principle one is free to choose any set of coefficients satisfying the relationship, and usually there is some obvious choice.

For the simplest case in which just two coefficient are related with each other through some ratio the obvious choice is to take some coefficient to be one, then the value of the second coefficient is nothing but the value of the ratio. One could also normalize the wave function then if desired.

Also note that one can use symmetry to simplify the math. OP did not mention were the symmetry adapted linear combinations (SALCs) of atomic orbitals used at the first step, so I mention just in case.

P.S. Fully worked examples could be found in different places, for instance, here (PDF, 150 KB).


In organic chemistry bulkiness isn't just a matter of geometry; one can find the volume of the group using vdW radii, but it's a crude approximation.
Quantitatively steric bulk relies on energy measurements and can be described with a cyclohexane A-value — energy required for a group to switch between axial and equatorial positions in cyclohexane.
From the table of A-values methoxy group is indeed less bulk in comparison to nitro and methyl groups:

$$
\begin{array}{lc}
\hline
\text{Substituent} & \text{A-Value}/\pu{kcal mol-1} \\
\hline
\ce{-OCH3} & 0.6 \\
\ce{-NO2} & 1.1  \\
\ce{-CH3} & 1.7 \\
\hline
\end{array}
$$


This example is probably described quite thoroughly on the internet for the case of $\ce{O2}$ (which is isoelectronic), so it may be worth searching a bit for alternate explanations.

You cannot have a $^3\Delta$ term symbol, because of the Pauli exclusion principle. The π* orbitals come as a set of two: essentially, one of those two is associated with +1 angular momentum, and the other is associated with −1 angular momentum.

Here, to be specific, "angular momentum" refers to the projection of the angular momentum onto the internuclear axis. The "angular momentum" quantum number of each MO is denoted $\lambda$, so in other words, the π* orbitals come as a set of $\lambda = +1$ and $\lambda = -1$.

The quantum number that measures the "angular momentum" of the term is denoted $\Lambda$, and because $\Lambda$ is a projection quantum number, the value of $\Lambda$ is simply the sum of the individual $\lambda$ values. The Greek letter that represents it in the term symbol depends on the absolute value of $\Lambda$: if $\Lambda = 0$ then it is a $\Sigma$ term, if $|\Lambda| = 1$ then it is a $\Pi$ term, and if $|\Lambda| = 2$ it is a $\Delta$ term.

Now, in the ground state configuration of $\ce{S2}$, we can ignore all the orbitals below the π* orbitals because all those $\lambda$'s sum to zero. The only thing of interest is the π* electrons.

If you want to have a triplet state, i.e. the electrons have the same spin, then you cannot put them into the same π* orbital: one will have to go into the $\lambda = +1$ orbital, and the other will have to go into the $\lambda = -1$ orbital. Hence you get $\Lambda = 0$, which leads to a $^3\Sigma$ term symbol, not a $^3\Delta$.

Finding the parity of the term symbol is nothing more than multiplying all the individual parities of the electrons (i.e. for doubly occupied $\mathrm{g}$ MOs you need to multiply by $\mathrm{g}$ twice). This is useful, because $\mathrm{g \otimes g = u \otimes u = g}$, which allows you to immediately discard all doubly occupied MOs. You only need to look at the two π* electrons, which are both $\mathrm{u}$, leading to an overall $\mathrm{g}$ parity and a $^3\Sigma_\mathrm{g}$ term symbol.

For $\Sigma$ term symbols there is an additional detail in that you need to specify the symmetry with respect to reflection in a plane containing the molecule. In this case it turns out that the molecular wavefunction is antisymmetric with respect to this reflection, so the proper term symbol is $^3\Sigma_\mathrm{g}^-$.


The property F in an equation for chemical potential

$F$ is the Faraday constant: $9.64853399(24)\times10^4~\mathrm{C~mol^{-1}}$.

A quick dimensional analysis confirms that this makes sense.

$$\frac{\mathrm{J~K^{-1}~mol^{-1}~K}}{\mathrm{C~mol^{-1}}} = \frac{\mathrm{J}}{\mathrm{C}} = \mathrm{V}$$


Why doesn't a buffer solution change ph(Appreciably?)

Having problems with that concept myself, above answers seem mighty complicated, this is how I think it works (warning- only my understanding, could well be wrong)

Point One:
You have a lot of $\ce{CH3COOH}$ (not much decomposition) and a lot of $\ce{CH3COO−}$ (from the salt) and very little $\ce{H+}$

Point Two:
pH works on the amount of $\ce{OH-}$ and $\ce{H+}$ in the system, there are usually only very very small amounts (when looking at weak acids and bases) present compared to the reactant (in water, there is about 500 million molecules to ever one that dissociates) So the point I'm making is adding a few $\ce{OH-}$ or $\ce{H+}$ ions will have a big impact on the pH.

Conclusion drawn:
As you have so much $\ce{CH3COO−}$ to begin with, removing a very very small amount by adding a few $\ce{H+}$ ions and forming a very very small amount of $\ce{CH3COOH}$, is not going to have a noticeable impact on the concentrations, in comparison to the amount already in the solution (both $\ce{CH3COO−}$ and $\ce{CH3COOH}$). You could even say that it is a negligible amount and can be ignored, therefore can say this will not have an effect on the equilibrium.

As the $\ce{H+}$ is removed, the pH is not changed; however as there is so little $\ce{H+}$ ions to begin with, not removing them will have an impact on the pH (see point two).

As already stated, there are so little $\ce{H+}$ ions in the system, removing them (by adding $\ce{OH-}$) will have an impact on equilibrium, you cannot say the amount of $\ce{H+}$ removed is negligible, you cannot ignore Le Chatelier's principle, the equilibrium will shift right to replace the lost $\ce{H+}$.

Note: I'm an A-level student, don't take this as 100% right or even 1% right, this, at best is a calculated guess as to why equilibrium can be ignored when you add $\ce{H+}$ ions and equilibrium can not be ignored when you add $\ce{OH-}$.


Is the pz orbital wave function three- or four-dimensional?

The wavefunction covers 3-dimensional space. Formally, it maps each point in space to a complex number. That is, it is a function that takes a point in 3-dimensional space as input and returns a complex number as output: $\Psi:\mathbb{R}^{3}\rightarrow \mathbb{C}$.
There are many ways to represent 3-dimensional space: cartesian coordinates $(x,y,z)$, spherical coordinates $(r,\theta,\phi)$, cylindrical coordinates $(z, r, \theta)$, to name the most common. We have these different ways because functions might look "nice" in one coordinate system and not nice in another.
As it turns out, the wave function for the $p_{z}$ orbital looks "nice" in spherical coordinates, that is, $\Psi(r, \theta, \phi)$. It looks less nice in cartesian coordinates, but you can do the replacement for $\theta$ and $\phi$ as noted in the question. However, people do not usually mix coordinates systems together because they are redundant. For example, no one writes $r\sqrt{x^{2}+y^{2}}$. They write either $r^{2}$ or $x^{2}+y^{2}$.
To conclude, the $p_{z}$ orbital you have written is a function of 3 variables: $r, \theta, \phi$. It just happens not to depend on $\phi$, so it looks like just a function of $r$ and $\theta$. Forget this concept of dimensionality because it's not doing you any favors.


If for a chemical reaction the standard gibbs free energy is negative then :

All reactions are in some sense equilibrium reactions. The Gibbs free energy is just a convenient measure of energy that we can use to understand more about our system. In particular you may be familiar with a few equations for the Gibbs energy:
\begin{equation}
\Delta G =\Delta H-T\Delta S,
\end{equation}
\begin{equation}
\Delta G=-RT\ln k
\end{equation}
\begin{equation}
\Delta G =-nFE
\end{equation}
For chemists, these three equations are the most commonly used and depend on what exactly you are investigating, but there are plenty of others too.

If we consider the second of these equations we note that $k$, the equilibrium constant, appears. We can see that as the equilibrium constant increases (and since $x$ follows $\ln x$ in limits) we note that the change in Gibbs energy becomes more negative the larger the equilibrium constant becomes. In other words the more products you have relative to reactants, the more negative the change in the Gibbs energy is, we call this spontaneous.

We could also take the third equation and note that $E$ is the reduction potential for a redox couple ($X^{n+}+e^{-}\rightarrow X^{(n-1)+}$). The more positive the potential for the reduction, the more spontaneous the reaction is.

Hope that helps you, oh and I'm assuming you mean reversible in the reaction sense of equilibria not in the thermodynamic sense of a reversible process?


What open-source software can produce 3D PDF representation of molecules?

Disclaimer

This is not for the faint-hearted.
If you are not interested in using LaTeX, stop reading now, as this uses the media15 or movie9 packages to embed U3D in a pdfLaTeX generated PDF. This answer assumes basic familiarity with LaTeX. I have tested this with pdfLaTeX and it should work fine in XeLaTeX.
One of the tools  I have used (DAZ Studio) is apparently 'free for a limited time'. It is also not available on Linux. FOSS replacement suggestions are welcome.
I'm of the personal opinion that 3D in PDF is not as good as it sounds.

What doesn't work

VMD offers the ability to output as OBJ, which can then be processed with FOSS MeshLab into a U3D file that can be embedded into a PDF. This is sub-optimal for two reasons:

There is no way to tell MeshLab that the VMD-generated mesh is supposed to be Gouraud-shaded. This means that the hard facets of the mesh will be glaringly obvious in the pdf, even if you use an unwieldy degree of subdivision.
Colours are not preserved. Your model will appear in the PDF with a uniform grey texture.

As such, MeshLab is not suitable at this stage.

Preparing a model

VMD offers a number of decent export options to get our geometry out of .PDB or what have you and into a mesh format. I opted to use the VRML format to output a mesh of a ribbon representation of PDB entry 4DM9. I then imported this into Blender.

Blender has no reliable U3D export capabilities, however I can use it to assign Gouraud shading to all of the shapes and assign colours to them. I then use the blender export function to export the scene as a Wavefront .OBJ file, with attendant .MTL file.

Processing with DAZ Studio

I am indebted to the author of this article (John Nyquist), which indicated that DAZ Studio can convert Wavefront to U3D just fine, with preservation of textures. DAZ is a very specific program that seems to be heavily geared towards posing, clothing and rendering humanoid figures in the spirit of programs like Poser. However all we really need is the U3D export function and we can read in the structure just fine.

Sometimes, this is how I see myself:

Okay in all seriousness, don't put a human figure in or on your molecule without good reason. Export as U3D and then we can embed in pdfLaTeX.

pdfLaTeX and movie9

As a proof of concept, I just followed the minimal example given in the media9 manual for getting a 3D model into a pdfLaTeX-generated PDF. Here it is, being viewed in Adobe Reader X (I know, I know!). Bear in mind that very few readers can actually display 3D PDF, which is one of the reasons why I feel that it is not such a good idea for chemical visualisation.

It starts off zoomed way too far in because I did not set up any options for the import (however you can find out all about that in the media9 or movie15 package manuals) and is a bit laggy on account of having over 1 million faces (I went way overboard with subdivision), however the important thing is that it works.

Improvements welcome!

Obviously, this is a pretty Rube Goldbergesque route to getting a molecule into a PDF. There probably exist easier ways which should definitely be suggested.


How can the intense color of potassium permanganate be explained with molecular orbital theory?

Let’s take a look at a qualitative MO scheme for a tetrahedric transition metal complex whose ligands have three p-type orbitals each. On the left of figure 1 you have the metal orbitals ($\mathrm{3d}$, $\mathrm{4s}$ and $\mathrm{4p}$) and on the right the twelve degenerate ligand p-orbitals (transform as $\mathrm{a_1 + e + t_1 + 2t_2}$). Only orbitals of the same symmetry will mix, so the $\mathrm{t_1}$ ones stay where they are originally. I have colour-coded both the $\mathrm{t_2}$ (red) and $\mathrm{e}$ (blue) orbitals.

Figure 1: Qualitative MO scheme of a tetrahedric complex with σ and π bonding between metal and ligands. Double vertical lines represent electron pairs.

Note that the image is qualitative only. The order of energy levels can be different; it is not possible to predict it a priori. The more electronegative a side is, the lower its orbital energies will be. Manganese(VII) is lacking seven electrons so the orbital energies will be very low, will oxygen profits from its high electronegativity. All in all, the energies will be somewhat similar.

We can consider the filled orbitals to be mostly ligand-centred and the empty ones to be mostly metal-centred, in full alignment with the picture of a $\ce{d^0}$ $\ce{Mn^{VII}}$ core and four $\ce{O^2-}$ ligands with electron octets.

The electronic transition that causes the intense purple colour corresponds to an excitation from $\mathrm{t_1}$ to $\mathrm{e}$ — $\mathrm{t_1}$ should always be the highest ligand orbital, while tetrahedric symmetry typically implies that $\mathrm{e}$ lies lower than $\mathrm{t_2}$ on the metal’s side. The higher the central metal’s oxidation state, the lower its orbitals and the closer it is to the ligands’ p-orbitals.

Since the transition is from a ligand-centred to a metal-centred orbital, it is called Ligand to Metal Charge Transfer (LMCT). Oversimplified, one could think of it as a one-electron reduction of manganese by an oxygen ligand to intermediately create $\ce{Mn^{VI}-O^{.}}$.

The transition is obviously spin-allowed (filled to empty orbitals) and the Laporte-rule is invalid in tetrahedric complexes as there is no centre of symmetry. Therefore, the observed colour is very intense.


What's the actual or official definition of activation energy?

Activation Energy is the minimum amount of energy needed extra than the average potential energy of the reactants to cause a reaction.And it does not change with the change of temperature, and for clarification the energy of the reactant molecules don't also change with the reaction. It is just the rate which changes with the reaction according to the arrhenius equation k=Ae^{-\frac{Ea}{RT}}So,the fraction of molecules having energy higher than activation energy changes with change in temperature. I think the easier definition can be stated like this


What is the best way to convert SMILES strings to file names?

I personally find SMILES quite unreadable, even for only small compounds. I think they are even less readable when performing substitutions. Given the vast amount of special characters they use, I do not think they are a good way to create a naming scheme.

  An alternative is using InChIkeys but since they are not human readable, I want to avoid that option.

While that is true, InChI strings are human readable, especially for small molecules. In principle only the layers need to be separated by something other than / (underscores would do that).

Other problematic characters are *;?+() which do not fit in [0-9a-zA-Z-.,_], and you would have to find ways around them when using in scripts.

  I am mainly interested in small organic molecules with less than 20 atoms.

If you are really only interested in molecules with less than 20 atoms, general nomenclature should be the easiest to comprehend, and should yield reasonably short file names. For these you probably only have to worry about spaces.


There are x-rays (Li K-$\alpha$ is 50 eV) and there are x-rays (Uranium K-$\alpha$ 98 keV).

In reality all electromagnetic radiation tries to pass through everything. Let $\rm I_0$ be the intensity of the beam at the surface and $\rm I_x$ be the fraction that gets through to depth $x$, then:

$\rm I_x = \rm I_0 e^{-\mu\rho x}$

where:

$\mu$ is the mass absorption coefficient, units = $\rm{cm}^2/\rm g$
$\rho$ is the density of the material, units $\rm g/\rm{cm}^3$
$x$ depth of radiation, units = $\rm{cm}$

As a gross generality as the energy of the radiation increases (wavelength shorter) the transparency of matter increases. So a thin slice of carbon, be it graphite or diamond,  will stop the vast major of Li K-alpha x-rays, but very few U K-alpha x-rays.

Note with some rearrangements and variable substitution this gives the Beer-Lambert Law used in UV-Vis spectroscopy.


There are multiple different things to consider for phosphate $\ce{PO4^3-}$:

The bonds themselves. Each of the four $\ce{P-O}$ single bonds is polarised towards the oxygen. Each bond is polar.
The molecule as a whole. It is of near-perfect tetrahedral shape thus having the point group $T_\mathrm{d}$. This point group includes the inversion symmetry operation $i$. Therefore, macroscopic properties of a vectorial type (like the overall dipole moment which is a vector more or less pointing from one pole to the other) must, by definition, be $0$. Thus, the entire molecule is not a dipole.
The molecule’s total charge. It has an overall charge of $-3$. This charge is located on the surrounding oxygens but the entire thing can be simplified from far away to a ball with the charge in the centre, i.e. the phosphorus atom. We would consider this molecular ion a monopole.


Can iproniazid be prepared by reacting isoniazid with 2-chloropropane?

The $\ce{NH_2}$ group in given compounds is very similar to regular amine fragment and is more active than acylated $\ce{NH}$. However, alkylated amines are more active than non-substituted ones. So, I'd expect a very significant amount of double-alkylated product.

To avoid purification issues I recommend to consider synthesis of acetone hydrazone followed by reduction with $\ce{Na[BH_3(CN)]}$ or similiar low-active hydrogen donor (hi-active ones will reduce amide fragment)


Is there an equivalent unit cell that can describe a bcc unit cell where all points are the same element?

You certainly can construct what is called a primitive cell containing one atom for a bcc lattice.  There could be more than one way, but here is my method.

Start with the usual cubic representation.  Pick any face and connect the midpoints of its edges to make a smaller square.  Do the same with the opposite face.  This gives a pair of squares which are then the bases of an elongated square prism, whose lateral edges connect the two opposing faces you "contracted".  This cell, which contains just the central atom of the original bcc structure, is a primitive cell.  In this cell the lateral edges, which we did not "contract", are about 1.41 (actually, the square root of 2) times as long as the basal edges.

How does this replicate to make the lattice? If you just scoot the primitive cell over to "catch" a neighboring atom on either side of the lateral faces, you find that you must move the adjacent cell "up" or "down", in the direction perpendicular to the bases, to get the second atom centered properly.  The "up"/"down" move is half the height of the prism.  But once you see this, you can repeat this process and get all the atoms involved in the entire cubic representation.

Last question:  I said the primitive cell is an elongated square prism.  It no longer has cubic symmetry.  Where did the cubic symmetry go?  The answer is you can't make a primitive cell in the bcc lattice with the full cubic symmetry.  Instead, note that when you modified the cube you had the choice of three pairs of opposite faces to "contract".  Thus there is really a combination of three primitive cells with different orientations.  The cubic symmetry emerges from the combination rather than any one primitive cell.


Chemical function of Mg and Conjugated bonds in Chlorophyll

but why do conjugated bonds increase this ability

Adsorption of light results in excitation of an electron, moving from low-level (usually bonding or non-bonding) to a higher (usually antibonding) orbital. (for understanding of what bonding, non-bonding and antiboding orbitals are, see here)

Generally, when two orbitals of same energy partially overlap, two more are formed: one with slightly higher energy and one with slightly lower. When orbitals of conjugated bonds interact, energy of one bonding orbital goes up and of one antibonding orbital goes does, narrowing the gap. When the gap is withing or closely near ~1.6-3.2 eV it can participate in visible light adsorption.

  this molecule does not seem to be have any resonance structures

It does, see carefully.

  what role does this metal complex serve in the molecule

The relevant metal-free molecule do exist in nature. It has, however, different color. $Mg$ ion seems to adjust the gap mentioned above. For a more clear understanding of the matter you might look for data on phtalocianin pigments, that may be easily obtained with different central atom in a very similar structure.


It means very little, at least in terms of stabilization.  Perfectly planar or not, larger aromatic rings bring less stability than smaller ones, so they react more like mere conjugated polyenes than what we think of as "aromatic rings".  Wikipedia:

When the annulene is large enough, [18]annulene for example, there is enough room internally to accommodate hydrogen atoms without significant distortion of bond angles. [18]Annulene possesses several properties that qualify it as aromatic.[1] However, none of the larger annulenes are as stable as benzene, as their reactivity more closely resembles a conjugated polyene than an aromatic hydrocarbon.

Cited reference:
1.
Oth, Jean F. M.; Bünzli, Jean-Claude; De Julien De Zélicourt, Yves (1974-11-06). "The Stabilization Energy of [18] Annulene. A thermochemical determination". Helvetica Chimica Acta. 57 (7): 2276–2288.


The reaction of Alkyl Halides with sodium nitrite in DMF

The $\mathrm{S_N1}$ or $\mathrm{S_N2}$ reaction of alkyl halide with nitrite ion is complicated one since nitrite is an ambident nucleophile. For instance, it was reported that silver nitrite reacts with alkyl halides to give nitrites while sodium nitrite gives more nitroalkane than alkyl nitrite (based on Hard Acid/Hard Base concept; Ref.1):
$$\ce{AgNO2 + R-Br -> R-ONO + AgBr \tag1}$$
$$\ce{NaNO2 + R-Br -> R-NO2 + NaBr \tag2}$$
However, these results were contradicted by work of Kornblum who compared the reactions of iodoalkanes with $\ce{NaNO2}$ and $\ce{AgNO2}$ (Ref.2):
$$\ce{C7H15CH2-I + AgNO2/NaNO3 -> C7H15CH2-NO2 + C7H15CH2-ONO + AgI \tag3}$$
It was shown by the work of Kornblum and coworkers that both primary alkyl bromides and iodides have given excellent yields of pure nitroalkanes (73-83% as major product) when $\ce{AgNO2}$ is used as the nucleophile (Ref.3). In contrast, primary chlorides are completely inert under the conditions $(\pu{0\!-\! 25 ^\circ C}/\ce{Et2O})$:
$$
\begin{array}{l|r|r}
\hline
\text{Heloalkane} & \% \ \ce{R-NO2} & \% \ \ce{R-ONO}  \\
\hline
\text{$n$-Butyl bromide} & 73 & 13 \\
\text{$n$-butyl iodide} & 74 & 12 \\
\text{$n$-Hexyl chloride} & 0 & 0 \\
\text{$n$-Hexyl bromide} & 76 & 10 \\
\text{$n$-Hexyl iodide} & 78 & 13 \\
\text{$n$-Heptyl bromide} & 79 & 11 \\
\text{$n$-Heptyl iodide} & 82 & 10 \\
\text{$n$-Octyl chloride} & 0 & 0 \\
\text{$n$-Octyl bromide} & 80 & 14 \\
\text{$n$-Octyl iodide} & 83 & 11 \\
\hline
\end{array}
$$
Based on the results obtained, the authors had concluded that this is a good method to prepare primary-nitroalkanes (Note that nitroalkanes, $\ce{R-NO2}$, are thermodynamically more stable than alkyl nitrites, $\ce{R-ONO}$; Ref.1). Some of the further work by Kornblum et al. have noted that use of tertiary alkyl halides to prepare corresponding nitro compounds is useless (0-5% yield) while those reactions with secondary alkyl halides have also given only less than 15% yield of expected nitroalkanes (Ref.4). After extensive  work on synthesis of nitroalkanes and alkyl nitrites, declared that:

In the synthesis of saturated primary nitro compounds, silver nitrite gives nitroparaffins in about 80% yield as against about 60% yields obtained with sodium nitrite. While silver nitrite is, therefore, the reagent of choice here, the lower cost and ready availability of sodium nitrite, the not very large disparity in yield, and the shorter reaction time all combine to make sodium nitrite an excellent second choice*" (p.114, Ref.2; and Ref.5).

In their mechanistic studies, Kornblum et al. (Ref.6) have found that yield of $\ce{R-NO2}$ isomer falls progressively when substrate changes from primary to secondary to tertiary halide. In contrast, the yield of $\ce{R-ONO}$ counterpart increases with the same order of primary to secondary to tertiary. Although there is a steric effect in play, the authors conclude that the reaction proceeds via a transition state, which has both $\mathrm{S_N1}$ and $\mathrm{S_N2}$ character in proportions that vary gradually with the structure of the halide. The greater the $\mathrm{S_N1}$ character of the transition state, the greater is the preferences for covalency formation with the atom with higher electronegativity (e.g., $\ce{R-ONO}$). Conversely, the greater the $\mathrm{S_N2}$ contribution to the transition state, the greater the preferences for covalency formation to the atom with lower electronegativity (e.g., $\ce{R-NO2}$).
In this regard, I'd say the reaction, $\ce{R-X + NaNO2 ->[DMF] R-NO2 + NaX}$ would proceed differently under different conditions. Since the solvent is (assuming) anhydrous DMF, the condition is set for $\mathrm{S_N2}$ reactions. Thus, for instance:

If $\ce{X = Cl}$ and $\ce{R-X}$ is a primary alkyl chloride, the reaction would not proceed.
If $\ce{X = Br or I}$ and $\ce{R-X}$ is a primary alkyl halide, the reaction would
proceed rapidly to give nitroalkane, $\ce{R-NO2}$, as the major product but in lower yield $(\approx60\%)$. There should also be some alkyl nitrites, $\ce{R-ONO}$ as well (Ref.3).
If $\ce{X = Br or I}$ and $\ce{R-X}$ is a secondary alkyl halide, the reaction would
proceed to give both nitroalkane, $\ce{R-NO2}$, and alkyl nitrites, $\ce{R-ONO}$, as the product. The yield would be lower $(\approx20\%)$ (Ref.4).

I think that would be enough of discussion since very little information is given in the question. For example, we cannot decide whether reaction goes in ether $\mathrm{S_N1}$ or $\mathrm{S_N2}$ without knowing the nature of $\ce{R}$ group. Knowing the structure of $\ce{R}$ group is extremely important since even some specific primary halides such as neo-pentyl halide do not react under same conditions where other primary halides proceed rapidly.
Finally, it is noteworthy to mention following reaction, which play a significant role in the product ratio of $\ce{R-NO2}$ to $\ce{R-ONO}$:

Reactions of the blue benzhydrylium salts, $\bf{1}$-$\ce{BF4}$, with $\ce{n-Bu4N+NO2-}$ in anhydrous acetonitrile gives colorless adducts, $\bf{1}$-$\ce{NO2}$ (Ref.5):

The monitoring method in Ref.5 does not suggest the product is direct formation of $\bf{1}$-$\ce{NO2}$ or through the kinetic product, $\bf{1}$-$\ce{ONO}$. It was suggest that when use more reactive benzhydrylium salts for the reaction, the monitored reaction process is the formation of $\bf{1}$-$\ce{ONO}$, which successively rearrange to the observed nitro compounds $\bf{1}$-$\ce{NO2}$. To prove this theory, the authors prepared pure $\bf{1}$-$\ce{ONO}$ in different method (Reaction $\bf{A}$), and monitored its rearrangement to thermodynamically stable $\bf{1}$-$\ce{NO2}$ (Reaction $\bf{B}$):

These experiments clearly suggest that preparation of nitroalkane is not simple as the reaction suggests, but rather complicated process!

References:

Ian Fleming, In Frontier Orbitals and Organic Chemical Reactions-Student Edition; John Wiley & Sons, Ltd.: Chichester, West Sussex, United Kingdom, 2009, p. 122–123 (ISBN 978-0-470-74660-8).
Nathan Kornblum, “The Synthesis of Aliphatic and Alicyclic Nitro Compounds,” Organic Reactions 1962, Vol 12, 101 – 156 (DOI: https://doi.org/10.1002/0471264180.or012.03).
Nathan Kornblum, Bernard Taub, and Herbert E. Ungnade, “The Reaction of Silver Nitrite with Primary Alkyl Halides,” J. Am. Chem. Soc. 1954, 76(12), 3209–3211 (DOI: https://doi.org/10.1021/ja01641a029).
Nathan Kornblum, Robert A. Smiley, Herbert E. Ungnade, Alan M. White, Bernard Taub, and Stephen A. Herbert Jr., “The Reaction of Silver Nitrite with Secondary and Tertiary Alkyl Halides,” J. Am. Chem. Soc. 1955, 77(21), 5528–5533 (DOI: https://doi.org/10.1021/ja01626a028).
Alexander A. Tishkov, Uli Schmidhammer, Stefan Roth, Eberhard Riedle, and Herbert Mayr, "Ambident Reactivity of the Nitrite Ion Revisited," Angew. Chem. Int. Ed. 2005, 44(29), 4623 –4626 (DOI: https://doi.org/10.1002/anie.200501274).
Nathan Kornblum, Robert A. Smiley, Robert K. Blackwood, and Don C. Iffland, “The Mechanism of the Reaction of Silver Nitrite with Alkyl Halides. The Contrasting Reactions of Silver and Alkali Metal Salts with Alkyl Halides. The Alkylation of Ambident Anions,” J. Am. Chem. Soc. 1955, 77(23), 6269–6280 (DOI: https://doi.org/10.1021/ja01628a064).


Activation energy and degrees of freedom

This depends upon the theory used to model the reaction; there are two major ones.
Collision Theory
If your molecules may be assumed to be spherical particles with no rotational or vibrational DOF, then collision theory appropriately describes your system. In this case, only translational energy is accounted for. The rate expression may then be derived from the Maxwell-Boltzmann kinetic theory of gases, where the energy of each particle is $\frac{1}{2}mv^{2}$.
I. Filot has a decent explanation of this.
Transition State Theory
Unfortunately, most reaction systems are much more complicated; the reactants experience not only translational, but also rotational and vibrational DOF. This must be accounted for in the rate expression. Again, I. Filot has a good explanation for this.
A general derivation includes the following:
Consider a bimolecular reaction:
$$A+B \rightarrow C$$
We assume there exists a transition state (TST), which defines the highest point along a minimum energy path on the potential energy surface. Moreover, we assume that our reactants are quasi-equilibrated with this TST; the rate of the reaction depends upon the frequency of reactants overcoming this barrier $(k_{f})$. Then:
$$A+B \leftrightarrow C^{\ddagger} \rightarrow C$$
Assuming this reaction occurs in the gas phase under ideal conditions, we can approximate their fugacity using the partial pressure of each reactant. Let us also assume that the reference partial pressures are unity. We derive an equilibrium constant between the TST and the reactants:
$$K^{\ddagger}=\frac{P_{C^{\ddagger}}}{P_{A}P_{B}} = exp(-\Delta G^{\ddagger}/RT) = exp(-\Delta H^{\ddagger}/RT)exp(\Delta S^{\ddagger}/R))$$
Our rate expression becomes:
$$r = K^{\ddagger}k_{f} P_{A}P_{B} = [(k_{f}) exp(\Delta S^{\ddagger}/R))exp(-\Delta H^{\ddagger}/RT)]P_{A}P_{B} = kP_{A}P_{B}$$
Our rate constant is:
$$k = (k_{f}) exp(\Delta S^{\ddagger}/R))exp(-\Delta H^{\ddagger}/RT) = (A)exp(-\Delta H^{\ddagger}/RT) \equiv (A)exp(-E_{a}/RT)$$
Where the partition functions defining the enthalpy and entropy include translational, rotational and vibrational DOF.
Therefore, to answer your questions:
"First, I want to understand if the factor $e^{-E_{a}/RT}$ takes into account only the percentage that have the necessary kinetic energy in translational degrees of freedom or in general the percentage with at least this energy irrespective if this energy is stored in a combination of various degrees of freedom?
No, it takes into account all DOF. In transition state theory, both the activation energy and pre-exponential factor consider the translational, rotational and vibrational DOF of the transition-state irrespective of the DOF providing the energy (This is directly tied to the canonical partition function, which is used in defining the enthalpy and entropy. It only cares about the total microstate energy and not from which DOF it comes from).
"Secondly, if some reactions speed up depending in which type of degrees of freedom (vibrational or translational) are excited, does this mean that the Arrhenius equation even for an elementary bimolecular reaction holds only if the reactants have only translational degrees of freedom?"
Reactions do not speed up depending on which type of DOF are excited. However, certain DOF speed up the reaction much more than other DOF. For example, based on the derivation I have provided, the pre-exponential factor scales exponentially with the transition-state entropy.
$$S_{total} = S_{translations}+S_{rotations}+S_{vibrations}$$
The DOF with the largest contribution to the entropy are translations. In theory, we may increase the rate of the reaction by increasing the vibrational entropy of our transition state. But, we may increase it to a much larger extent by increasing it's translational entropy.


Is the percent error formula you are looking for something like the formula this website service offers? Your lab instructor should have given you the formula, perhaps you missed that. Do you have a reliable lab partner you can ask? The error in mass would probably be something to do with how far each measurement deviates from the average.

If for example I used a 250 ml Erlenmeyer flask which had marks at only 25ml increments. I would then say that I could make a 'guess' as to what the volume is at if I were trying to measure something less than 25ml or between two marks. But the error for that instrument would then be somewhere around +/- 12.5ml error. My general-chem teacher was fine with that, maybe you should ask yours?

I was told that the convention is to expect at least 2% error in the last significant digit if not explicitly defined. When taking a measurement with a device like a graduated cylinder, you always measure down to the most precise "known" value, and then the last significant digit is your own judgement.


Why does salty water make hard boiled egg shells more fragile?

The eggshell is made of calcium carbonate mostly, but the crystal structure is held together by matrix proteins (more references in Wikipedia). Salt creates ionic interactions with the protein molecules, breaking protein-protein interactions, and in this case, probably protein-mineral interactions as well, debilitating the eggshell structure.


Gas chromatography analysis - number of peaks?

The GC is the instrument used (in this case, although for very small amounts, there is no reason GC couldn't be used to separate materials - assuming collection after detection (via, say, refractive index - for obvious reasons flame ionization, etc. wouldn't be usable)) to determine if his primary separation technique was effective. It wasn't.


Why is this reaction spontaneous in standard conditions?

It has been a while since I took electrochemistry so I'm not sure about the formal procedure that you're using.

You wrote the correct overall reaction.

$$\ce{2H3O+ + Fe -> H2 + Fe^{2+} +2H2O}\tag{1}$$

and you almost wrote the correct half cells reactions (should be $\ce{Fe^{2+}}$ in equation 3):

$$reduction:  \ \ \ce{2H3O+ + 2e− ->H2}\tag{2}$$

$$oxidation: \ \ \ce{Fe -> 2e^{-} +Fe^{2+}}\tag{3} $$
Then you wrote the scheme of the cell: $$\ce{Fe|Fe^{2+}||H_{3}O^{+}|H_{2}}\tag{4}$$ And an equation for the overall cell potential.
$$E^{0}_{cell}=E^{0}_{right}-E^{0}_{left}\tag{5}$$

Your problem is between equations 4 and 5. To subtract $E^{0}_{left}$ the potential must be written as the reduction potential for the half cell. So either you're supposed to be:

Adding the values for the right and left reactions in equation 5

-- or--

Writing both half cells as reduction potentials in equation 4.

The gist here is that the tabular values are always written as reduction potentials. So for iron:

$$\ce{2e^{-} +Fe^{2+} -> Fe }\quad\quad E^{0}_{red} = -0.44\tag{6} $$

I was taught to reverse the sign for an oxidation reaction.

$$\ce{Fe -> 2e^{-} +Fe^{2+}}\quad\quad E^{0}_{ox} = 0.44\tag{7} $$

Then add the reduction and oxidation potentials to get the overall cell potential.

$$E^{0}_{cell}=E^{0}_{red} + E^{0}_{ox}\tag{8}$$