How do chemical reactions taking place in an alkaline battery produce electrons?

An alkaline battery is a specific type of galvanic cell.  As I am sure you found in your research, it involves an oxidation-reduction reaction between zinc($\ce{Zn}$), and manganese(IV) oxide ($\ce{MnO2}$.)  They are called alkaline batteries because these chemicals are surrounded by an alkaline (basic) solution of potassium hydroxide, $\ce{KOH}$.

Your question is really one that is more about galvanic cells in general.  When an oxidation-reduction (a.k.a. redox) reaction occurs, electrons are transferred between species.  This sort of reaction is exceedingly common.  That's all that happens in a battery, it's just that people have figured out how to catch the electrons on their way from species A to species B and use their energy to do work.

If I took a piece of zinc and placed it in a slurry of $\ce{MnO2}$ and $\ce{KOH}$, it would react according to the following balanced chemical equation and produce a lot of heat.

$\ce{Zn + 2MnO2 <-> ZnO + Mn2O3}$

Suppose that I separated the two pieces of the reaction into their own jars.  A piece of zinc placed in a solution of $\ce{KOH}$ in one jar and some $\ce{MnO2}$ mixed with $\ce{KOH}$ in another.  The reactants are all present, but they have no way of interacting, so nothing would happen - there's no way for electrons to move from the zinc to the $\ce{MnO2}$.

Now I attach a piece of wire to the zinc on one end and into the $\ce{MnO2 - KOH}$ slurry at the other. Now electrons can move from one jar to the other.  If I have a voltmeter that can measure a very brief signal, I can actually see a quick pulse of electrons flowing through the wire, from the zinc to the $\ce{MnO2}$.  The pulse ends almost immediately though, because the movement of electrons away from the zinc and toward the $\ce{MnO2}$ creates a charge separation.  The former becomes positively charged and the latter negatively charged.  Continuing to move electrons against this charge separation is progressively more difficult, and it doesn't take long for the electrostatic force of the charge separation to balance out the electrochemical potential of the reaction and the entire thing just stops.

To fix this problem, I need a way for the charges to balance back out without allowing the reactants to mix directly.  This is done with some sort of salt bridge.  A salt bridge allows positive and negative ions to move between the jars, but it still keeps the jars separate from one another.  This could be as fancy as an actual salt bridge (a glass tube filled with a gel containing some sort of salt) or as simple as a piece of paper soaked in the $\ce{KOH}$ solution.  Now when I connect the wire, electrons flow from the zinc to the $\ce{MnO2}$, negatively charged $\ce{OH-}$ ions move through the salt bridge toward the zinc, and positively charged $\ce{K+}$ ions move through the salt bridge toward the $\ce{MnO2}$.  The charge stays balanced in both jars, so there is no electrostatic force preventing electrons from moving, and they flow freely through the wire.  If I put a motor in the middle of the wire, it turns.

Specifics of Alkaline Batteries

That all was regarding galvanic cells in general.  A battery has additional requirements.  It needs to be compact and safe, in particular.  The construction of an alkaline battery involves a zinc core with a paste of manganese(IV) oxide and potassium hydroxide wrapped around it.  The paste and zinc are separated by a paper sheet (the salt bridge in this galvanic cell.)  Because all of the reactants and products are solid, there is no concern that they will diffuse through the cardboard; it just keeps the reactants in their places.

The reaction does not start until the battery is put into a device and the device is turned on.  Flipping that switch gives the electrons from the zinc a pathway to reach the manganese(IV) oxide.  Until the circuit is complete, there is no buildup of electrons at one end of the battery.

Electrical potential (voltage) is the ratio of the energy carried by an object to that object's charge.  One electron in this system has a potential of 1.43V, two electrons carry twice as much energy, but they also have twice the charge, so they also have a potential of 1.43V.  Electrical potential is sometimes called electromotive force, and that idea, that we're measuring how hard the electrons are being pushed through the wire, may help you understand the battery's behavior better.

Why 1.43V though?  It's because of the properties of the chemical reactions the battery is based on.  You probably know that if you burn some amount of propane it will produce a given amount of heat, and if you burn an equal amount of acetylene, it will produce a different amount of heat.  The same thing is true here.  The energy associated with the half reactions in each part of the battery depends on the chemicals involved.  Rather than list a total amount of energy available, batteries list the energy per charge, voltage.  This battery's half reactions are:

$\ce{Zn + 2OH- -> ZnO + H2O + 2e-}$  E°=1.28V
and
$\ce{2MnO2 + H2O + 2e- -> Mn2O3 + 2OH-}$ E°=0.15V

Each half reaction gives a specific amount of energy to the electrons involved in it.  Since the energy:charge ratio (E° above, called a reduction or oxidation potential) is constant, we can use that to find the energy:charge ratio of the entire battery just by adding them together.  Why doesn't it go above 1.43V?  For the same reason that a fire doesn't burn hotter.  The chemicals involved only release so much energy per reaction.  Note: The potential can be changed in some systems by playing around with the concentrations of the chemicals in the cell.

As the battery is used up, the zinc slowly but surely turns into zinc oxide and the manganese(IV) oxide turns into manganese(III) oxide.  As their amounts fall to zero, the battery's potential falls and it quits working.


Why do compounds with similar properties have different Stitch IDs?

Bethamethasone and dexamethasone are different chemicals, hence they behave different, therefore they have different identification numbers. The numbers you cite are not STITCH IDs, and as far as my research goes these do not exist. Instead they are CID - PubChem Compound Identification numbers.

As you can see the molecules are diastereomers. The stereo centre of the steroid differs at carbon-16. When you look them up with STICH 4.0 you will see differences with their predicted functional partners. While the first one is nearly identical, the second partner for betamethasone is proopiomelanocortin, while for dexamethasone it is ABCG2.

For more information on STITCH (‘search tool for interactions of chemicals’) I recommend: Michael Kuhn, Christian von Mering, Monica Campillos, Lars Juhl Jensen, and Peer Bork, Nucl. Acids Res. 2008, 36 (Database issue): D684–D688. (Open Access); PMCID: PMC2238848.

Image sources:

National Center for Biotechnology Information. PubChem Compound Database; CID=9782, https://pubchem.ncbi.nlm.nih.gov/compound/9782 (accessed Apr. 20, 2016).
National Center for Biotechnology Information. PubChem Compound Database; CID=5743, https://pubchem.ncbi.nlm.nih.gov/compound/5743 (accessed Apr. 20, 2016).


The semi-reaction of reduction for Zn is :

$Zn^{2+} +2e^- \rightarrow  Zn \quad E^\circ =-0.763$

The semi-reaction of reduction for Cu is :

$Cu^{2+} +2e^- \rightarrow  Cu \quad E^\circ= 0.337$

The standard redox potential for Cu is more positive thus it is reduced and Zn is oxidized.
Electrons flow from a negative to a positive potential. They flow towards the cathode because their potential V, at the anode is more negative, thus they have an higher potential energy: $U_e = V*e$

note that $e$ the charge of the electron is negative as well as V.


What properties does a solvent with a high dielectric constant have?

Before I start, please note that I am not a chemist so be sure to verify what I say...

The electric conductivity of water depends on what is dissolved in it. For example, salt water will be far more conductive than distilled water. So if you're professor was talking about sea water, the answer could actually be different from purified water.

Due to the dissolved salt ions, sea water is a better conductor of electricity and so by definition it will be a poor dielectric, since according to this Wikipedia article: to be a good dielectric a substance must be both a good insulator and able to be polarized by an electric field.

Purified water on the other hand will be a better dielectric because it does not contain any significant amount of ions and is therefor not very conductive (i.e. is a better insulator than sea water).

I think (but don't quote me on an exam) that the reason alcohol is a poor dielectric is because it is not nearly as polar as water is, and according to the above, the second requirement for a dielectric is that it could be polarized by an electric field.

(This might also have to do with the size of the molecules, since water molecules are generally smaller than many kinds of alcohols, and can therefore re-orient themselves more easily in the presence of an electric field, but this is a guess.)

The Solvent Classifications section of this Wikipedia page also mentions that the polarity of a solution can be roughly gauged by the solution's dielectric constant. So this too seems to support the idea that polarity and dielectric properties are related.

If anyone has any improvements or corrections, please let me know in the comments and I'll update the answer accordingly.


How to determine which gas is more soluble

You could answer this question by looking at the possible interactions between a gas molecule and water molecules. The only I can see are van der Waals interactions (because these gas molecules cannot participate in hydrogen bonding).

$\ce{O_2}$ and $\ce{Br_2}$ have no permanent dipole so the only interactions are Debye forces (interaction between water permanent dipole and a gas molecule induced dipole) and London forces (between instantaneous dipoles).

The induced/instantaneous dipoles strength increases with the size of the atom because the bigger the electron cloud the easier it is to modify its shape.It happens because the electrons far away from the nucleus are less attracted by it and can move around a bit more easily than electrons very close to the nucleus.

This is the reason why van der waals forces are stronger for bigger atoms.
This would explain why $\ce{Br_2}$ has a better solubility than $\ce{O_2}$ in water.

As a side note:

London interactions are very important if you look at liquefaction temperature of gases. For instance, it explains why $\ce{Ar}$ becomes a liquid below 87K while $\ce{He}$ become a liquid only below 4K.


Comparing the stabilities of allylic, benzylic, and tertiary radicals

This is a tough question.  I think it might even be unfair to ask such a question on a test in non-advanced classes.  In advanced classes it could make an interesting topic of discussion, but I'm still not sure that the "real" answer is known.

What can be said is that due to resonance, both the allylic and benzylic radicals are more stable than the t-butyl or ethyl radicals which are not resonance stabilized.  You can see from the following diagram that you can draw two resonance structures for the allyl radical.

For the benzyl radical you can draw even more resonance structures.

Superficially, this might suggest that the benzylic radical is more stable than the allyl radical. This was the "answer" and reasoning provided here.

However, if you do some thermochemical calculations (as done here), you arrive at the opposite conclusion that the allyl radical is roughly 2 kcal/mol more stable than the benzyl radical. The number of resonance structures alone is not a perfect indicator of stability.

For me, that is a pretty small difference in energy, and I think it would be fair to answer that the allyl and benzyl radicals have comparable stabilities.


Carbon dioxide sequestration in Limestone

This process does not rely on catalysts, but it addresses the essence of the question. The process described is to inject liquefied carbon dioxide into basalt formations, even if the basalt is very old. This makes the it plausible that we could scrub the excess $\mathrm{CO}_2$ from the atmosphere by:

growing plants,
burning those plants, capturing the CO2, and
burying that carbon using, partly, the energy from the burned plants.


Removing PTFE marks on glass bottles after using magnetic stir bars

Stir bar has a very rough grinding action. It is quite well known among silica manufacturers that if you stir micron sized silica with a stir bar for a long time in a reaction mixture, the spherical particles are damaged to some extent. In such cases, overhead stirring is preferred. This is not applicable to your case , I guess. If you had soil as well, the scenario is no different. It is a grinding action of the stir bar on glass. The "frosting" of glass is like rubbing a sand paper on glass for a long time. Nothing can heal this, except a glass-blower's flame treatment. Of course this is not worth it, as it is not going to affect the analysis of ammonia. I don't think that this is PTFE on glass. It will not stick to glass and secondly it is so hydrophobic that it would be floating on your solution rather than firmly sticking to glass.


How to solve constants out of the internal energy equation?

You need to use the equation $$\left(\frac{\partial U}{\partial V}\right)_T=-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]$$


Major product in Reimer-Tiemann Carbonylation

The product of the Reimer-Tiemann reaction actually also depends on the counter cation of the base we use. It is found that if we use $\ce{NaOH}$ as the base used for formation of carbene electrophile, ortho product is major. But surprisingly, if we use the base $\ce{KOH}$, the para product is produced in major amount.  The reason behind this is the stability of the transition state via formation of an ordered cyclic arrangement.Due to the smaller size of $\ce{Na+}$, this can get fit between the Oxygen and Chlorine atom in ortho positioned electrophile via electrostatic attraction and give rise to a very ordered transition state which governs the major formation ortho product. See for details, https://chemistry.stackexchange.com/a/98197/58497
On the other hand, $\ce{K+}$ is huge in size related to $\ce{Na+}$, which obviously can't increase the stability of the ortho product by forming a cyclic orderd Transition state. That's why, the para product generally gets preference in the case of $\ce{KOH}$.


Why only ketone forms enolate in Claisen condensation between ester and ketone?

The pKa of a ketone alpha hydrogen is of the order of 20, the pKa of an ester alpha hydrogen is of the order of 25 source. Bear in mind that this is a log scale, meaning the proportion of ketone enolate to ester enolate is 100000:1.

In answer to the comment by @orthocresol: the ketone does self-condense, but this is reversible. The condensation with the ester is irreversible due to formation of the anion of the product B-diketone.


Will borosilicate glass absorb 808nm laser light?

Borosilicate glasses are a family that has many members with differing properties. You can look up the provider for your glass, and check whether it is transparent in the near infrared. I think it would be in most cases (see here for one example).


Thermal reduction of other metal oxides beside Fe2O3

One candidate is manganese dioxide, $\ce{MnO2}$.  This decomposes at 535°C, releasing oxygen.  As the phase diagram from Ref. [1] shows below, the reduced manganese fuses with excess silica between 1250 and 1300°C, forming rhodonite ($\ce{MnSiO3}$).  You would see this as a pink color.  But, of course, I'm a chemist rather than a potter, so I make no guarantees on whether or how it works in pottery!

Reference:

1.
Y.-B. Kang, I.-H. Jung, H.-G. Lee, "Critical thermodynamic evaluation and optimization of the MnO–SiO2–TiO2–Ti2O3" system, Calphad
Volume 30, Issue 3, September 2006, Pages 226-234.  https://doi.org/10.1016/j.calphad.2006.05.002


Well yes you can if you assume the Bohr model is correct. However in reality, it is a little bit more complicated than that. In reality everything have both particle and wave like properties, which builds the foundation of quantum mechanics. The lighter the object, the more wave like properties it exhibits and the heavier the object, the more particle like properties it exhibits. So a ball actually has wave light properties but it is too heavy for it to be notices. Meanwhile a light wave actually has particle properties, however it is too light to be noticed. Electrons on the other hand turn out to a mass between these two extremes, hence both its particle and wave like properties are able to be noticed.Hence it is important to stop thinking electrons as small point particles.

Another important concept is the Heisenberg Uncertainty Principle which is the following formula: $$\Delta x \Delta p \geq \hbar$$ This means that the uncertainty of an object's position multiplied by the uncertainity in an object's momentum at best can only be equal to the constant value: $\hbar$ which equals $1.054*10^{-31}$ This equation can be applied to electrons. Hence, the important thing that this equation tells us about electrons in an atom is that rather electrons moving in a predictable circular orbit, they actually travel in any path that they want and we actually can't precisely determine position of the electron in an atom. Hence the quantum mechanical model of the atom actually consists of orbitals, rather than orbits.

Orbitals is the area which an electron is in 90% of the time. There are 4 main type of oribitals, s, p, d and f shown below.

So, finding the velocity of an electron is little more complicated than using simple classical mechanic equations. However its velocity can be determined in simplified conditions using quantum mechanics.


Is it considered safe to sonicate with flammable solvents?

In general it should be relatively safe considering you take all the precautions required when working with volatile flammable liquids.
Specifically for sonication I can think of two safety aspects:

Sonification promotes formation of the aerosol/fog consisting of the atmospheric gases and solvent vapors.
Depending on its composition, it might or might not be fire hazardous or even explosive.
That's why keeping an ultrasonic setup in a ventilated area — a fume hood, ideally — is a good idea.
Keep the power supply and all controller circuits outside, if possible, in order to prevent spark vapors ignition.
Be careful when sonicating heterogeneous solutions.
Some materials would absorb the ultrasonic energy and this can result in local overheating.
To prevent this, use mechanical stirring.


As mentioned in the comments, safety is the main priority.  Take particular heed to the warning labels and information on the chemical label - if unsure, look up the chemical details MSDS (or SDS) = (Materials) Safety Data Sheet.

One of the most important thing to do (and is often overlooked) is to keep an inventory of what chemicals you have.  The safety data sheets can be obtained from the manufacturers or by looking up the chemical SDS itself, or from general websites such as MSDS.com - an example is this one for mineral turpentine - these contain considerable information about storage, disposal and first aid.  (It is always a good idea to have this readily available and updated).

Personal protective gear is a must - gloves, safety glasses etc - usually the label or the SDS will have this information.

According to the document NERC Guidance on Safe Storage of Laboratory Chemicals (2010), has information that is pertinent to a workshop environment.  There are 3 main principles in regard to storage of chemicals explained in the document:

Segregation, from the article:

  The key incompatibles to segregate from each other are strong acids from
  strong bases and strong oxidisers from organic or flammable materials.

The NERC document has a list in its appendix (too long for this reply)

Separation, related to the segregation,

  In a laboratory situation adequate separation can be achieved by means of
  storage cupboards which physically divide incompatible classes of hazardous
  chemicals. The cupboards may need specific properties or provide
  separation by means of distance. They will also need to provide secondary
  containment (eg spill trays or bunded shelves) and security (eg locks / bolted
  to wall).

Ventilation (from personal experience, this one is often overlooked).

  Ventilation is often an essential requirement for safe storage of hazardous
  chemicals. Its main function is to allow dilution and extraction of vapours or
  gases that may escape / seep out from containers during storage so they no
  longer present problems from the viewpoint of noxious smell, hazardous
  personal exposure or creation of an explosive atmosphere.

Also consider an appropriate fire extinguisher, note that water is probably the worst one.  Here is an example of a fire extinguisher type guide - this guide suggests that powder based extinguishers may be best (but, make sure you study it and seek advice if unsure).

Some of this may seem 'overkill', but it is better to have some preparedness for if anything goes wrong, than be unpleasantly surprised by it (from personal experience), also, depending on where you are, there could be a legal requirement for a degree of proper storage, handling and protection.


Yes, a reaction known as "hydrogen exchange" will take place.  When $\ce{H2O}$ and $\ce{D2O}$ (heavy water) are mixed, hydrogen exchange will take place rapidly to form $\ce{HDO}$ as a third component of the mixture.  Depending on the initial amounts of $\ce{H2O}$ and $\ce{D2O}$ added, a statistical mixture of the 3 compounds will result. So if you mixed equal amounts of the $\ce{H2O}$ and $\ce{D2O}$, then you would wind up with something around a 1:2:1 (it won't be exactly 1:2:1 due to the effect of the slightly different stabilities of the 3 compounds - see LDC3's comment) mixture of $\ce{H2O}$, $\ce{HDO}$ and $\ce{D2O}$.


Tl;DR: Wikipedia mentions some of the ways in which elemental sulfur is produced naturally:

Through volcanic emissions, including emissions from hydrothermal vents. Some of them found near hot springs and near volcanic deposits. Also, lakes of molten sulfur up to ~200 m in diameter have been found on the sea floor due to submarine volcanoes.
As minerals: sulfide minerals, such as pyrite (iron sulfide), cinnabar (mercury sulfide), galena (lead sulfide), sphalerite (zinc sulfide), and stibnite (antimony sulfide); and the sulfate minerals, such as gypsum (calcium sulfate), alunite (potassium aluminum sulfate), and barite (barium sulfate).
Action of anaerobic bacteria on sulfate/sulfide minerals.

Long answer:
I found a paper1 which discussed #3 point in details. It said that although sulfide minerals can be oxidized to zero-valent sulfur by sulfate reducing bacteria in presence of molecular oxygen, it may pose the following problems:

exposure to oxygen would drastically decrease growth of microbes thereby slowing down sulfide production
on geologic timescales, excess supply with oxygen would convert
sulfide into sulfate rather than native sulfur
enormous amounts of oxygenated water would be needed to produce sulfur in large amounts

So, in order to overcome these problems, scientists proposed four possible mechanisms as a means to form native sulfur:

a modified sulfate reduction process that produces sulfur compounds
with an intermediate oxidation state,
coupling of sulfide oxidation to methanogenesis that utilizes
methylated compounds, acetate or carbon dioxide
ammonium oxidation coupled to sulfate reduction
sulfur comproportionation of sulfate and sulfide.

These reactions are found to be thermodynamically favorable and especially useful in environments containing dissolved sulfide. This provide evidence that microbial species functioning in such environments can produce native sulfur in absence of light and external oxidants such as $\ce{O2}$, nitrate, and metal oxides.
Reference

Labrado, A. L., Brunner, B., Bernasconi, S. M., & Peckmann, J. (2019). Formation of Large Native Sulfur Deposits Does Not Require Molecular Oxygen. Frontiers in microbiology, 10, 24. https://doi.org/10.3389/fmicb.2019.00024


How much additional hydrochloric acid should be added to arrive at ‘excess acid’?

The volume you've established seems correct, as well as your assumption that extra $\pu{40 mL}$ of $\ce{HCl}$ would be a good excess:

$$\ce{NaHCO3 + HCl -> NaCl + H2O + CO2}$$

$$V(\ce{HCl}) = \frac{n(\ce{HCl})}{C(\ce{HCl})} = \frac{m(\ce{NaHCO3})}{M(\ce{NaHCO3}) \cdot C(\ce{HCl})} = \frac{\pu{0.5 g}}{\pu{84.0 g mol-1} \cdot \pu{0.1 mol L-1}} = \pu{59.5 mL}$$

I suspect your teacher either prepared a series of various antacids, or this one contains additional components, such as $\ce{CaCO3}$, $\ce{Mg(OH)2}$ of $\ce{Al(OH)3}$. Note that for the neutralization of the same mass of, say, aluminium(III) hydroxide, over than 3 times more $\ce{HCl}$ is required:

$$\ce{Al(OH)3 + 3 HCl -> AlCl3 + 3 H2O}$$

$$V(\ce{HCl}) = \frac{n(\ce{HCl})}{C(\ce{HCl})} = \frac{3m(\ce{Al(OH)3})}{M(\ce{Al(OH)3}) \cdot C(\ce{HCl})} = \frac{3 \cdot \pu{0.5 g}}{\pu{78.0 g mol-1} \cdot \pu{0.1 mol L-1}} = \pu{192 mL}$$

Either way, it looks like you understand how to properly find out the amount of acid for the neutralization. I think if you demonstrate your detailed attempt to the teacher, it should clarify the situation.


Can fluorine act as the central atom in interhalogen compounds?

Why can't fluorine be the central atom in inter-halogen compounds?

First off, fluorine can be at the "center of things."  Examples would include the strongly hydrogen bonded hydrofluoric acid

and the very relevant example of the trifluoride anion

$$\ce{[F-F-F]^-}$$

The trifluoride anion example is critical as it demonstrates that fluorine can be hypercoordinated (hypervalent).  Of course, in this last example we've "tricked" the fluorine into occupying the middle position, so your question remains (slightly altered), why doesn't fluorine occupy a central position more often?

In order to answer this question, two general cases need to be considered,

The anionic situation $\ce{[X-F-X]^-}$ vs.  $\ce{[X-X-F]^-}$
The cationic situation $\ce{[X-F-X]^+}$ vs. $\ce{[X-X-F]^+}$

where X is Cl, Br or I.

In the case of the anion, Lewis dot structures suggest a negative charge on the central atom.

However, computational analysis of the electron distribution in the triiodide anion (see p. 156) suggests that the central atom is positively charged (+0.27 electron), while the terminal atoms are negatively charged (-0.63 electrons on each terminal atom).  Assuming that the same general trend would be applicable to mixed trihalide anions, then in the case of $\ce{[X-F-X]^-}$ we would be placing positive charge on highly electronegative fluorine atom instead of the larger, more polarizable (e.g. more capable of bearing a positive charge) bromine, chlorine or iodine atom.

In the case of the cation, similar arguments come into play.  Again, calculations for a variety of trihalide cations (mixed and homogenous) suggest substantial positive charge on the central atom (this time in agreement with Lewis Dot structures). The same arguments made above would operate here to destabilize structures with a central fluorine atom. This analysis also found extremely long (less stabilizing) bonds resulted when fluorine (or the most electronegative of the 3 atoms) occupied the central position.

Arguments that don't explain the infrequent occurrence of fluorine at a central position in inter-halogen compounds include:

the lack of "d-orbitals"; as other's have commented, this is an old, invalid argument; the existence of $\ce{F3^-}$ demonstrates that fluorine is capable of hypercoordinate bonding without the need for d-orbital involvement
inefficient 2p-3p overlap; $\ce{FCl}$ has a higher bond strength and lower heat of formation than $\ce{Cl2}$


Minimizing Gibbs free energy with Helmholtz free energy expressions

I don't think you can use the approach you are outlining without an equation of state.

$G = \sum_i \mu_i n_i$ is valid only at constant $T$ and $P$.

$A = \sum_i \mu_i n_i$ is valid only at constant $T$ and $V$.

If $P$ and $V$ are both constant, then you can equate the two and use $G=A= \sum_i \mu_i n_i$.  Actually, that assumption is probably not horrible for most liquids at "normal" pressures: I think it amounts to assuming the liquid is incompressible.


To make a solution that is 70% solute by weight with $\ce{H2O}$ as the solvent, take 70 $\mathrm{g}$ of solute and add 30 $\mathrm{mL}$ of water. The resulting solution will have a mass of 100 $\mathrm{g}$, 70 $\mathrm{g}$ of which is the solute; hence, 70 wt. % in $\ce{H2O}$.


How is latent heat of vapourisation utilized in changing state?

It is raising the energy level of the system by breaking the bonds that hold it together as a liquid. Once the bonds are all broken within a certain volume, the liquid will vaporize.

The energy seems to 'disappear' or 'hide' because we expect to see a change in the properties of the chemical as energy is added, but the breaking and reforming of intermolecular bonds is happening constantly within a liquid.

When the liquid has enough energy, random bond-less distributions will occur within the liquid that are large enough to overcome the pressure of the liquid and its container. This is how water boils. If enough of the water molecules become unbound, they will form a bubble. Inside the bubble, the water vapor will be exactly equal to the pressure of it's environment.

This gives us our definition of boiling point: vapor pressure equals atmospheric pressure. The water or other liquid can't stay in the liquid state because there's not enough surrounding pressure to hold it together.


Finding the number of cyclic isomers algorithmically

If I remember correctly, several years ago I stumbled over the SMOG program addressing this question (description here).  The software requestested sum formula and a selection of allowed structural elements, cyclic substructures were among them.  Maybe the algorithms implemented are not the first ones to answer such a question, yet publications like this showcase continuing interest in this field, too.


What is the compound I made from KNO3 and table sugar which subsequently exploded when I exposed it to a flame?

It's not a new compound - it's just an intimate mix of fuel (sugar) plus oxidiser (nitrate). It's not the most stable of situations to have...


Extracting sulphur from hot springs with medieval technology?

Sulfur does naturally occur around hot springs and volcanos. It would be very reasonable to assume that a hot spring could have natural deposits of sulfur near it that could be collected without any special effort.


Do molecules change shape when they become vibrationally excited?

Yes they change 'shape'. A diatomic only  in the sense that the bond extends and contracts, but for  a triatomic or polyatomic molecule there are bond stretches and bends. The number of different ways these vibrations change the 'static' shape of a molecule, such as may be measured by x-ray diffraction, depends on the symmetry and this is described by the point group. Only certain combinations of bond stretches and bends are allowed and these are called 'normal modes'.

Some other points.

A molecule will vibrate even when it has zero point energy, i.e. when $v=0$; by the Heisenberg principle this must happen. If it were stationary then it would have infinite energy because its position is determined exactly. The electron density, if it were to be measured would thus be an average of that at the zero point energy. Calculations of this made at zero potential energy would thus be slightly different. Slightly because the bond extension is only a few percent of the bond length at this energy, and also most nuclear wavefunction density is located at the minimum energy.

We assume that the Born-Oppenheimer condition applies, briefly that electrons being far lighter than nuclei instantaneously adjust to the slow nuclear motion as a molecule vibrates. This allows us to draw the potential energy profile, for example a Harmonic or Morse potential. When we draw the familiar wavefunctions for vibrational motion these are the nuclear wavefunctions at each energy and are the stationary solutions to the Schroedinger equation and we interpret these as meaning that the molecule has a probability of being at each position given by the square of the wavefunction.

In a long time spectroscopic experiment (e.g. using a spectrophotometer) only the spectrum is measured, i.e. energy gaps between levels.

In experiments using a femtosecond laser pulse$^*$ it is possible to excite several vibrational wavefunctions almost instantaneously, i.e. in a time shorter than a vibrational period, in this case a 'wavepacket'$^{**}$ is produced and this 'particle' moves left and right inside the potential. This motion has been observed by using a second femtosecond laser pulse as a probe. As the energies of the vibrational levels are not all the same the wavepacket, which starts off in phase, will become out of phase then 'in phase' again and so goes on repeatedly recurring until it is disturbed, say by a collision with another molecule. (In these time resolved experiments Fourier transforming the data gives the spectrum.)

$^*$ As a femtosecond pulse is short it has a wide spectral width and can therefore excite several energy levels.

$^{**}$ A wavepacket is the direct summation of the vibrational wavefunctions multiplied by a time dependent term related to the vibrational frequencies. Thus the wavepacket evolves in time. The wavepacket recurs at regular time intervals due to the time dependent terms coming into phase just as they were at time zero. This is a property of a collection of oscillators; children on swings would show the same effect.


Determining (approximately) the amount of a strong base needed to change a buffer's pH by a certain number of units

You can calculate the quantity of sodium hydroxide, required to change the $\mathrm{pH}$ of your system by two unites,  by calculating the concentration of sodium ion in the charge balance equation:
$$\ce{[Na+] +[H+]=[OH- ] +[H2PO4^{-}]} \,\,\,\,(1)$$
In fact, you know the initial concentration of sodium ion in the buffer system before adding sodium hydroxide (from the known concentration of $\ce{NaH2PO4}$), and you calculate the final concentration of sodium ion  after adding sodium hydroxide by the above equation. So you can deduce the required quantity  of sodium hydroxide.
As for the  other ions in the charge balance equation, they are calculated as follows:
At first the  $\mathrm{pH}$ of the buffer system $\mathrm{pH=pK}_a=2.15$.

If you want to change the $\mathrm{pH}$ of your system by two unites, then the final $\mathrm{pH}$: $$\mathrm{pH}= 4.15\,\,\,\,(2)$$
Now you can estimate the concentrations of ion hydronium and ion hydroxide.

The use of Henderson-Hasselbalch equation gives: $$\ce{[H2PO4^- ]= 100[H3PO4 ]}\,\,\,\,(3)$$

The equation of matter conservation gives: $$\ce{[H2PO4^- ] + [H3PO4 ]= C_0=1 \mathrm{M}}$$ As the buffer is $25\,\mathrm{mL}$ of $1\,\mathrm{M}$ of each of the components, so there's a total of $1\,\mathrm{M}$ in the $50\,\mathrm{mL}$ solution due to dilution (not as you mentioned in your comment).
Now, you can calculate the concentration of ion $\ce{H2PO4^{-}}$
from equation (3) and (4):$$\ce{[H2PO4^- ] = \frac{100}{101}\mathrm{M}}$$

If we substitute the numerical values in equation (1):
$$\ce{[Na+] +10^{-4.15}=\frac{10^{-14}}{10^{-4.15}} +\frac{100}{101}} $$
$$\ce{[Na+]_0 + [Na+]_{\mathrm{added}} \approx 10^{-4.15} + \frac{10^{-14}}{10^{-4.15}} +\frac{100}{101}} $$
$$0.50 +\ce{ [Na+]_{\mathrm{added}} \approx 10^{-4.15} + \frac{10^{-14}}{10^{-4.15}} +\frac{100}{101}} $$
$$\ce{ [Na+]_{\mathrm{added}}} \approx 0.4902 \mathrm{M} $$
$$n=\ce{  [Na+]_{\mathrm{added}}} \times V_{\mathrm{sol}}= 0.4902\times 50\times 10^{-3} =24.51 \times 10^{-3} \mathrm{mol} $$
The volume of sodium hydroxide is $$V_{\mathrm{added}} = 24.51 \times 10^{-3}\times 1= 24.51 \times 10^{-3} \mathrm{L} $$
I suggest to use a  solution of sodium hydroxide more concentrated, so as to neglect the effect of the added volume on the initial concentrations of the buffer system.


Entropy change in mixing a gas at constant temperature and pressure

The reference you gave involves mixing two different chemical species together at constant temperature and pressure.  The problem you seem to be solving is mixing the same chemical species with itself at constant temperature and pressure.  For that process, the change in entropy is zero.  If you really intended to determine the entropy of mixing two different chemical species, then the path you identified is not a reversible path for transitioning between the initial and final states of the system.


Sodium compounds:

$\ce{Na2C2}$ is sodium carbide. It reacts violently with water. $\ce{Na2CO3}$ is sodium carbonate. $\ce{NaO}$ is not a valid formula for any of the oxides of sodium: $\ce{Na2O},\ \ce{Na2O2},$ and $\ce{NaO2}$ are all real compounds, and they all react with water to produce $\ce{NaOH}$, which is probably what is formed.

  "Hydrocarbons"

If indeed you had $\ce{Na2C2}$, then you would form $\ce{H2C2}$ and $\ce{NaOH}$ when it was placed in water. However, you would not need electrolysis to do so. This is an acid-base reaction that is spontaneous under normal circumstances. $\ce{H2C2}$ cannot be the brown sludge on your cathode. It is a gas.

$$\ce{Na2C2 + 2H2O -> 2NaOH + H2C2 ^}$$

  What actually happened?

In an electrolytic cell, the process that occurs (at least under standard conditions) is the one with the least negative standard potential $E^o_{cell}$.

The possible oxidations are either the oxidation of copper or the oxidation of water. Sodium carbonate is the supporting electrolyte.

$\ce{2Cu + 2OH- -> Cu2O + H2O +2e-}\ \ \ E^o=+0.36 \ \text{V}$
$\ce{Cu -> Cu^{2+} + 2e-} \ \ \ E^o=-0.34 \ \text{V}$
$\ce{4OH- ->O2 + 2H2O +4e-}\ \ \ E^o=-0.40 \ \text{V}$
$\ce{2H2O -> O2 + 4H+ +4e-} \ \ \ E^o=-1.23\ \text{V}$

Since the oxidation of copper to $\ce{Cu^{2+}}$ has the least negative $E^o$, it is the most likely oxidation. $\ce{Cu^{2+}}$ also explains the color, since aqueous solutions of $\ce{Cu^{2+}}$ are blue. The first equation is also likely, since it has a positive $E^o$, however it does not explain the color.

The possible reductions are the reduction of water/hydroxide and the reduction of $\ce{Cu^{2+}}$, since we know that it is being formed.

$\ce{2H2O + 2e- -> H2 + 2OH-} \ \ \ E^o=-0.83 \ \text{V}$
$\ce{Cu^{2+} + e- -> Cu+} \ \ \ E^o=+0.16 \ \text{V}$

The reduction of $\ce{Cu^{2+}}$ to $\ce{Cu^{+}}$ is most likely, but not initially. Initially, the is no $\ce{Cu^{2+}}$, so the only reduction that can happen at the cathode is the first one, which produces hydrogen gas. The redox then might be: $$\ce{Cu + 2H2O -> H2 + Cu^{2+} + 2OH-} \ \ \ E^o =-1.19 \ \text{V}$$

Once the $\ce{Cu^{2+}}$ diffuses over to the cathode, a different reduction starts to compete, with a different net redox reaction: $$\ce{2Cu + H2O -> H2 + Cu2O} \ \ \ E^o=-0.47 \ \text{V}$$

This redox reaction still produces hydrogen, but it has a much less negative $E^o_{cell}$. The other peculiarity is that $\ce{Cu2O}$ is produced at the cathode after $\ce{Cu^{2+}}$ has diffused through the cell. $\ce{Cu2O}$ is your brown sludge, since like rust, it is both brown and completely insoluble in water.


Classifying some molecular orbitals after Hartree-Fock calculation

Natural bond orbital (NBO) methods may help you to localized the orbitals of your system after a post-HF calculation. I'm not familiar with Molcas but I know that several computational chemistry softwares have NBO methods implemented. Take a look at this simple explanation that uses MolPro:

https://sites.lsa.umich.edu/zimmerman-lab/tutorial/electronic-structure-calculation/casscf/


Reduction of amino acids to corresponding amino alcohols

Chiral amino alcohols such as L-valinol are generally prepared by the reduction of corresponding α-amino acids and suitable reducing reagent. Most commonly used reducing agent is $\ce{LiAlH4}$ since most common reducing reagent, $\ce{NaBH4}$ does not reduce carboxylic acid to corresponding alcohols. However, use of $\ce{LiAlH4}$ has notable disadvantages such as being expensive and highly flammable. To avoid these disadvantages on high scale syntheses, some novel methods have been developed. One such method is reduction of carboxylic acid using $\ce{NaBH4/I2}$ reducing sysrem (Ref.1). The most remarkable feature of this reagent is its selectivity over other known strong reducing agents. For example, if $\ce{NaBH4/I2}$ is used to reduce bicarboxylic acid, it would reduce both carboxylic groups to $100\%$. However, if one carboxylic acid i
 group was protected (say, ester-carboxylic acid combo), $\ce{NaBH4/I2}$ would reduce only carboxylic acid group leaving ester group alone (Ref.1).

Two years after this important discovery, McKennon, et al. (1993) found out that this same method can be used to reduce N-protected or naked amino acids to their corresponding N-protected amino alcohol or amino alcohol, respectively (Ref.2). A typical procedure is as follows:

  Preparation of (S)-tert-Leucinol: A dry 5-L three-necked round-bottom flask equipped with a mechanical stirrer and a 250-mL addition funnel was charged with sodium borohydride ($\pu{55.6 g}, \pu{1.47 mol}, \pu{2.41 equiv}$) and $\pu{1.6 L}$ of anhydrous THF. The solution was stirred while (S)-tert-leucine ($\pu{80.0 g}, \pu{610 mmol}$) was added in one portion. The flask was cooled to $\pu{0 ^\circ C}$ and fitted with a reflux condenser. The addition funnel was charged with a solution of $\ce{I2}$ ($\pu{155.0 g}, \pu{610 mmol}, \pu{1 equiv}$) in $\pu{200 mL}$ of THF, which was added dropwise to the flask over a $\pu{1.5 h}$ period with considerable gas evolution. The solution was allowed to warm to room temperature. When the brown color had dissipated to give a cloudy white solution, the reaction was brought to reflux for $\pu{19 h}$. The cloudy white suspension was cooled to room temperature with the aid of a water bath. The addition funnel was charged with $\pu{150 mL}$ of $\ce{MeOH}$, which was added drop-wise with rapid stirring. Vigorous gas evolution was observed. Small aliquots of $\ce{MeOH}$ were then added until all of the solid white material had dissolved. The solution was concentrated by rotary evaporation to give a white pasty oil that was dissolved in $\pu{1.20 L}$ of $20\% (w/w)$ aqueous $\ce{KOH}$ and mechanically stirred for $\pu{6 h}$ at room temperature. The light green solution was extracted with $\ce{CH2Cl2}$ ($\pu{3 \times 1.50 L}, \pu{1 \times 600 mL}$). To minimize the emulsion, $\pu{200 mL}$ of brine was added to the aqueous layer after the first extraction. The combined organic extracts were dried over $\ce{Na2SO4}$, filtered through glass wool, and concentrated in vacuo to give $\pu{74.1 g}$ of a white crystalline solid in a thick, slightly yellow oil. Distillation of (S)-tert-leucinol, bp $70$-$\pu{73 ^\circ C}$/$\pu{2 mmHg}$ (Lit. $117$-$\pu{120 ^\circ C}$/$\pu{57 mmHg}$) from a $\pu{30.0 g}$ reduction reaction performed in the above manner yielded $\pu{21.3 g}$ of product ($79\%$ yield) as a clear oil, which solidified upon cooling to room temperature.

References:

J. V. Bhaskar Kanth, Mariappan Periasamy, “Selective reduction of carboxylic acids into alcohols using sodium borohydride and iodine,” J. Org. Chem. 1991, 56(20), 5964-5965 (https://doi.org/10.1021/jo00020a052).
Marc J. McKennon, A. I. Meyers, Karlheinz Drauz, Michael Schwarm, “A convenient reduction of amino acids and their derivatives,” J. Org. Chem. 1993, 58(13), 3568-3571 (https://doi.org/10.1021/jo00065a020).


Quoting the Introduction (P-10) of the IUPAC Blue Book (Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013):

[...] We have developed rules for assigning 'preferred IUPAC names', while continuing to allow alternative names in order to preserve the diversity and adaptability of the nomenclature to daily activities in chemistry [...]
Thus the existence of preferred IUPAC names does not prevent the use of other names to take into account a specific context or to emphasize structural features common to a series of compounds. Preferred IUPAC names (PINs) belong to a 'preferred IUPAC nomenclature'. Any name other than a preferred IUPAC name, as long as it is unambiguous and follows the principles of the IUPAC recommendations herein, is acceptable as a 'general IUPAC name', in the context of a 'general IUPAC nomenclature'.

Therefore, there may be multiple names for a compound which are allowable in IUPAC nomenclature, one or more of which may be a trivial name (or derived from one). Note that there is no stipulation that the PIN must be the systematically obtained name. So, for a compound with a trivial name, either the systematic name or the trivial name may be chosen as the PIN. How does IUPAC choose which one is the PIN?
It might not be the answer we are looking for, but the end of the day, I strongly suspect it is simply arbitrary. The examples you have cited are very illustrative. Here is one more inconsistency which, to the best of my knowledge, is arbitrary.
"Acetic acid" is a PIN that may be modified by substitution (P-65-1.1.1), but only to a limited degree (P-15.1.8.2.1). Therefore "bromoacetic acid" is also a PIN. On the other hand, acetylene is also a PIN; however, substitution of any kind of not allowed (P-31.1.2.1). So, instead of bromoacetylene, the PIN is bromoethyne. (Bromoacetylene, however, remains acceptable within general use. All the names listed below that aren't followed by "(PIN)" are acceptable within general use.)

What makes IUPAC decide that - for PIN purposes - one trivial name is substitutable, whereas another is not? Or in the examples you have cited, why is one trivial name the PIN whereas another isn't? Of course, the rules state that it must be so, but to the best of my knowledge, there isn't any explanation of the philosophy behind the choice of PIN. Maybe I'll be proven wrong, though, and I'd be glad to see it.
To be honest, if you could come up with one, I'd be surprised. It's very tough to draw a line. Maybe we can agree that "acetone" should be a PIN, but should "neopentane" be a PIN as well? How about "estrogen"? Someone has to just arbitrarily choose where to draw a line, and it just so happens that IUPAC has gone with something stricter that doesn't use "acetone" as a PIN.
Anyway, just to reiterate - many of these names are perfectly valid IUPAC names. It is just that, for one reason or another, IUPAC has not selected them as preferred IUPAC names.


Can the guanidinium ion give stable N-nitrosamines under mild metalworking conditions?

By definition, N-nitrosamines are aliphatic or aromatic derivatives of secondary amines, which have a nitroso group ($\ce{–NO}$) attached to nitrogen. In general, N-nitroso compounds are usually formed by the interaction of a nitrosating agent ($\ce{NOX}$) derived from either nitrite salts or nitrogen oxides with an amino substance such as secondary and tertiary amines, secondary and tertiary amides, peptides, substituted ureas, guanidines, and urethanes. Some important N-nitroso compounds are shown below:

As structures indicated, resonance in the framework has nothing to do with $\ce{-N-N=O}$ function (e.g., 1-Methyl-3-nitro-1-nitrosoguanidine is stable at room temperature and commercially available). Nonetheless, guanidine derivatives interact with $\ce{NOX}$ to give stable nitroso-guanidine derivatives (which are type of N-nitrosamides). With $\ce{NOX}$ derived from nitrite salts, the formation of N-nitroso derivatives of guanidine rarely occurs above $\mathrm{pH \ 4}$, but increases steadily with increasing acidity i.e. decreasing $\mathrm{pH}$, a maximum at  $\mathrm{pH \ 2.5-3.5}$ where the yield is inversely proportional to the basicity of the amino substance  (Ref.1).

On the other hand, guanidines do not react with $\ce{NOX}$ derived from nitrogen oxides in aqueous media (Ref.2,3). Note that $\ce{NOX}$ derived from nitrogen oxides worked best with strongly basic amino substances (e.g., aliphatic amines and heterocyclic amines) under non-aqueous or neutral and alkaline aqueous conditions (Ref.2,3).

References:

Sidney S. Mirvish, "Formation of N-nitroso compounds: Chemistry, kinetics, and in vivo occurrence," Toxicology and Applied Pharmacology 1975, 31(3), 325-351 (DOI: https://doi.org/10.1016/0041-008X(75)90255-0).
B. C. Challis, "Nutrition and nitrosamine formation," Proceedings of the Nutrition Society 1985, 44(1), 95-100 (DOI: https://doi.org/10.1079/PNS19850015) (PDF).
Brian C. Challis,  Soterios A. Kyrtopoulos, "The chemistry of nitroso-compounds. Part 11. Nitrosation of amines by the two-phase interaction of amines in solution with gaseous oxides of nitrogen," J. Chem. Soc., Perkin Trans. 1 1979, 299-304 (DOI: https://doi.org/10.1039/P19790000299).


Identifying a quadruply charged anion containing three carbon atoms

Sesquicarbide

A quick Google search suggests that $\ce{Li4C3}$ is a real compound, but based on what I saw in the links, it probably is not ionic in the same way that lithium acetylide $\ce{Li2C2}$ is.

$\ce{Mg2C3}$ appears to be the formula for magnesium carbide, or at least a magnesium carbide.

A carbide is a binary compound of carbon and a less electronegative element, usually a metal or metalloid. Carbide is also a generic name for mono- and polyatomic anions containing only carbon. The following ions would all carbide ions: $\ce{C^{4-}, C2^2-, C2^{4-}, C2^{6-}, C3^{4-}, C3^{6-1}, C3^{8-}}$, etc.

The special name of three of these carbide ions are:

$\ce{C^{4-}}$ - Methide
$\ce{C2^{2-}}$ - Acetylide
$\ce{C3^{4-}}$ - Sesquicarbide


How to isolate and determine an unknown molecule in a known heterogeneous mixture using HNMR

The software that comes with the instrument (Topspin for a Bruker machine,but Varian has a similar product made by them) will certainly be able to do this, as can third party software such as MestreNova or the free ACD/NMR.  You should check with the NMR service in your department as to what software they have licenses for.

The method you describe of 'subtracting' two NMR spectra, whilst possible, is not however very good in practice. For two spectra (more accurately, for two FIDs) to be subtracted, they must be IDENTICAL. That is, the peaks for aspirin in both the pure sample, and the contaminated one, must be perfectly aligned (same peak shapes, intensity, shift etc), which is practically different due to slight differences during each acquisition.

The easiest thing for you to do is probably to purify the sample, for instance by running some chromatography, or perhaps re-crystallising the aspirin, which would hopefully leave your impurity behind, which you could then analyse.

Failing this, and if you have access/inclination, some 2D NMR experiments would be useful to you. A simple COSY would allow you to see which peaks correspond to which peaks (thus allowing you to see which peaks aren't aspirin). More complex methods (HSQC, HMBC, DOSY) would also be of great help, but are probably beyond your needs for a second year lab course.


Intensive and Extensive Properties--Resistance

To the extent that it's appropriate to treat it as a thermodynamic quantity, given that it's generally a description of a specific, macroscopic system (and thus violates the statistical-mechanical large-numbers assumption of thermo), resistance is extensive, always.

Resistance of a wire: The resistance of a wire depends on area and length. This implies it depends on the "quantity" of the wire taken. Does this mean it is extensive?

Yep! You're spot on on this.

Resistance of a cell: As we know, Resistance of a cell is $R=\dot G/\sigma$ where $\dot G$ is the cell constant and the property of the cell, and $\sigma$ is the conductivity. Now, this won't depend on the amount of electrolyte taken in the cell. Does this imply resistance is an intensive quantity?

Actually, this resistance measure will depend on the amount of electrolyte in the cell. Per here, $\dot G$ is a function of the dimensions of the active portion of the cell, in particular the sizes and shapes of the electrodes, and how far they are from one another. These geometrical characteristics then affect how much electrolyte must be used in order to fill the cell. Thus, the resistance $R=\dot G/\sigma$ is still extensive.


Is H2CO3 (carbonic acid) soluble or insoluble in water?

Yes, it is soluble.

For example, the oceans are constantly getting $\ce{CO2}$ from the atmosphere that gets converted into carbonic acid and thus decreasing pH on a global scale. Or the soft drinks we drink contain carbonic acid in them in dissolved form only due to mixing of $\ce{CO2}$ gas.

Also, not all carbonates are insoluble as Ben Norris has commented.

Hard and fast 'rules' in chemistry tend to be rare.  Consider them guidelines.


I found a paper$\ce{^{[1]}}$ regarding the formation of diiodide anion species($\ce{I2^-}$). It is assumed to be an unstable species which forms during the formation of triiodide anion (excess of iodide in solution which makes it brown). This is a part of the paper which describes the formation of the species. For more information, read the full paper. Also, you can check the abstract which @andselisk linked.

The $\ce{I2^-}$ species is formed by the 248-nm laser photolysis of
iodide through the following reactions:
$$\ce{I- + hν -> I + e_s-}$$
$$\ce{I + I- ->[k2]I2^-}$$
$$\ce{I2- + I2- ->[k3]I3- + I-}$$
As discussed later, we found that photodetachment of electrons from
$\ce{I-}$ ions produces solvated electrons ($\ce{e_s-}$) in ionic
liquids as well as in molecular solvents. After photodetachment,
iodine atoms react with $\ce{I-}$ to form diiodide anion radicals. In
aqueous solution, the transient absorption maxima of $\ce{I2-}$ are
located around 400 and 720 nm. The extinction coefficients of
$\ce{I2-}$ in aqueous solution at 385 and 725 nm are 10000 and 2560
M-1cm-1, respectively.[...]

$\ce{^{[1]}}$: J. Phys. Chem. B, 2007, 111 (18), pp 4807–4811


If the gases in a mixture are ideal then their diffusion is independent*.
The relationship described by Graham's law is transitive, therefore the relative effusion rate of multiple substances with respect to a reference substance also provides a ranking of their relative diffusion coefficients (an ordered list).
*Outside of condensed or supercritical phases it seems unlikely that two fluid substances would display sufficiently strong interactions to alter the relative order of their diffusion coefficients.


Reporting reaction energies calculated using electronic energies rather than enthalpies

In the case of atoms or molecules in a vacuum, at $0~\mathrm{K}$, and infinitely apart, the "reaction energy" of

$$
\ce{A + B -> A \bond{....}\bond{....}\bond{....} B},
$$

where $\ce{A \bond{....}\bond{....}\bond{....} B}$ is effectively at infinite separation, is

\begin{align*}
E_{\text{rxn}} &= \sum_{i}^{\text{products}} E_{i} - \sum_{j}^{\text{reactants}} E_{j} \\
&= E_{\ce{A \bond{....}\bond{....}\bond{....} B}} - (E_{\ce{A}} + E_{\ce{B}}) \\
&= E_{\ce{A}} + E_{\ce{B}} - (E_{\ce{A}} + E_{\ce{B}}) \\
&= 0.
\end{align*}

The reasoning is that energies are extensive, and there is no change in the total energy of the system. The total energy of infinitely separated particles is their sum, as there is no attractive or repulsive interaction between them. They're effectively completely isolated. In that sense, there is no reaction, which is I put reaction energy in scare quotes.

In practice, when you have a finite interaction such as a chemical reaction, the total interaction energy should include electronic, vibrational, rotational, and translational components, of which the latter 3 will grow as temperature increases. Most likely the thermal vibrational contribution is the only significant one.

One missing term that is not dependent on temperature and should be added is the zero-point (vibrational) energy,
$$
E_{\text{ZPVE}} = \frac{1}{2} \hbar \sum_{c}^{\text{normal}\\\text{modes}} \omega_{c},
$$
which is the non-zero amount of non-electronic energy that molecules have even at zero kelvin and is purely quantum mechanical in nature. The ZPVE or ZPE can alter energy barriers non-linearly, since vibrations are not necessarily localized, and is the most significant non-electronic contribution to gas-phase reaction energies.


What reaction takes place when potassium iodide is added to a tetraammine copper(II) complex?

Copper(II) is a hard Lewis acid that would be octahedrally coordinated but for Jahn-Teller distortion.  Instead, it is square-planar coordinated with two distant weakly bound apical ligands.  The lovely deep blue color, hard Lewis base ammonia versus hard Lewis acid water coordination, illustrates the nephelauxetic effect.  What would obtain with triethylamine coordination of copper?  Cu(II) is substitution labile.

Jahn-Teller effect
Nephelauxetic effect
Kinetics (and thermodynamics) of substitution reactions

Adding acid protonates the ammonia, its lone pair now in a dative bond with a proton and unavailable for coordination.  Tetraamminecopper(II) (note the spelling of "ammine" when coordinated) becomes tetraaquocopper(II) with fast kinetics.

Aqueous Cu(II) is a modest oxidizing agent; iodide is a modest reducing agent.  Cu(I) is a soft Lewis acid, iodide is a soft Lewis base, $\ce{CuI}$ is very insoluble (look up its $\ce{K_{sp}}$ -- see here.)

Iodine reacts with excess iodide to give triiodide anion.  However, isolate the "$\ce{CuI}$," wash, and titrate with thiosulfate.  $\ce{CuI}$ strongly adsorbs elemental iodine.  It is also modesty light sensitive.  Cyanide, being a pseudohalide, follows the same path to afford exceptionally insoluble CuCN plus cyanogen.

Remember who earned the "A," or do your chair parade for better references, then understanding for the next question.


What does delta letter mean in some ceramics chemical material formula?

Exactly. In your example, it is related to oxygen vacancy (oxygen deficiency) in the crystal lattice.
It can affect the physical properties of the ceramic material. Take the example of this piezoelectric ceramic material:$\ce{Eu_{0.5}Ba_{0.5}TiO_{3-\delta}} $ Oxygen-vacancy induces antiferromagnetism to ferromagnetism transformation.

In general, $\delta$ denotes deviation from the stoichiometry (exces or deficiency) in a chemical formula


Energy of eclipsed and staggered conformations of ethane and propane

That diagram is horribly drawn.  The way it is drawn, the 13.6 kJ/mol number is meaningless - it measures from the maximum of one curve to the minimum of the other curve.

I think that you are supposed to assume that for one of the molecules, the one for the lower curve, the torsional strain is 12.6 kJ/mol and for the other molecule the torsional strain is 13.6 kJ/mol.

Do you have some reason to think that the barrier to free rotation for propane would be higher or lower than for ethane?  Perhaps your book gives the value for the torsional strain of ethane/propane? (through a quick google search I was able to find the torsional strain for both of these)


Are there any papers linking the bad taste of orange juice after toothpaste to sulfates?

There may be more, but this is one paper examining the effects of membrane-modifying compounds on taste receptors. It suggests that substances like sodium lauryl sulfate and gymnemic acid affect taste response by altering taste receptors' lipid membranes. Though many anionic surfactants have sulfate functional groups, not all compounds with sulfate groups act as surfactants (nor does the sulfate anion on its own). It sounds like this is something that any compound that can perturb the structure of a taste receptor's membrane can do, not something that necessarily applies to all sulfates.


Why are regular plastic bags not smell-proof?

Well, you basically answered your own question nicely with your "obvious answer" discussion. On a molecular scale, smallish molecules, like the volatile compounds that evaporate and make way to your nose, are able to permeate the plastic. This happens in part for the reason you said; on such a small scale, the plastic that appears fully "solid" and impermeable on a macro scale is actually a mesh of molecules with paths large enough for these small, smelly, molecules to escape.

There are factors at play other than molecular size, like the chemical composition of the plastic and the escaping compounds. If the compounds have a strong affinity for the plastic, it will take less time for them to permeate as they will more quickly absorb through the plastic.

Regarding your observation that the plastic seems to hold liquids, but not smells; this is probably an illusion. I suspect that the liquids you've placed in the bag actually have little or no smell (i.e. water, food oils). For example, in an analytical laboratory setting, water samples or water-based standards stored in various plastic containers have a limited shelf life because the water slowly permeates the plastic, evaporating into it's surroundings. So the water is escaping the plastic just like a smelly compound might do, but of course water has no smell.


Not quite, an isotope has same number of protons ($ A- N = Z = \mathrm{constant}$), but a different number of neutrons ($\mathrm N$ varies; e.g. $\ce{^3_\color{red}{1}H}$ and $\ce{^2_\color{red}{1}H}$, or $\ce{^235_\color{red}{92}U}$ and $\ce{^238_\color{red}{92}U}$ are isotopes).

An isobar has a fixed number of total nucleons ($Z + N = A = \mathrm{constant}$; e.g. $\ce{^\color{red}{40}_19K}$ and $\ce{^\color{red}{40}_20Ca}$, or $\ce{^\color{red}{3}_2He}$ and $\ce{^\color{red}{3}_1H}$ are isobars). Not nearly as mainstream as isotopes, but isobars are important to consider when doing mass spectroscopy.

Extra fact: For nuclei of the same number of neutrons ($A - Z = N = \mathrm{constant}$), the term is isotones.


I find the following picture from Chemistry Libretexts. I don't own the picture, so the credit goes to respective authors.

I believe the d orbitals illustrated on the diagram are only the $d_{z^2}$ orbitals. For the f orbital, I believe it is $f_{z^3}$. Please let me know of any problems/suggestions.


What are the reaction products in my dishwasher from a steel colander?

I am confident both are aluminum ; Steel would be magnetic, austenitic stainless is non-magnetic ,but normal relatively shiny or bright in consumer products. All steels are about 3 X more dense than aluminum if you could judge the density. Aluminum is subject to corrosion in solutions of salts, acids ,and bases. But I couldn't pick one of your components as the most likely cause. That is, I expect it is corrosion of aluminum you are seeing. I have had a very similar colander  but washed it by hand because it was fairly large to put in a dish washer.


What structural property creates stickiness?

Long story short, no, sticky substances don't have to be nonpolar at all. Pour a little wine or beer on a table, and let it dry almost completely: it will get sticky, but still be water-soluble and hence polar. Also, there are certain water-based glues, which are surely sticky and polar. On top of all that, go read about the hagfish, only make sure you're not eating at the moment, for fear of adverse reaction.

As for the structural properties that make a substance sticky (or viscous, which is basically the same thing), these are not so easy to point out. Many sticky things are polymers, i.e., very long molecules with lots of van der Waals interactions, just like you said, and probably lots of interactions of other sorts, too. But this is hardly a rule, as exceptions both ways are plenty. Polyethylene is a polymer, but is not sticky at all. On the other hand, sugar syrup can be pretty sticky, despite containing no polymers.

As for the sticky solids, I would claim that there are none. Of the possible examples, some are actually highly viscous liquids, some are sticky because they are covered with such liquids, and some would stick to things because of electrostatic effects.


How to prepare potassium cyanide from potassium ferrocyanide

Until about 1900, potassium cyanide was produced by melting dry potassium carbonate with dry potassium hexacyanidoferrate(II):
$$\ce{K4[Fe(CN)6] + K2CO3 -> 6KCN + FeO + CO2}$$
However, the product contains potassium cyanate, which is approximately formed as follows:
$$\ce{K4[Fe(CN)6] + K2CO3 -> 5KCN + KOCN + Fe + CO2}$$

In a small scale, dry potassium hexacyanidoferrate(II) is melted in a covered porcelain crucible. It is heated as long as nitrogen escapes.
$$\ce{K4[Fe(CN)6] -> 4KCN + N2 +FeC2}$$
The iron carbide sinks to the bottom and the molten potassium cyanide can be poured off.


How to calculate the electrochemical potential of Iron(II/III) and saturated calomel electrode?

First of all, you have to use reduction potentials as IUPAC considers the reduction potentials to be called the standard potentials.
It is not clear in your question whether you have used reduction or oxidation potentials.

Secondly, your equations don't make sense. The first equation shows Fe in +2 state getting reduced to Fe in +3 state.
Your second equation shows oxidation of mercury from pure state to +1 state.

And you are right that stoichiometry should not affect the EMF of cell.

The correct cell reactions will be iron(II) getting oxidised to iron(III) and Mercury(I) being reduced to Mercury(0).

At Cathode, Reduction of mercury takes place.

At Anode , Oxidation of Iron takes place. But we always use only reduction potential to calculate EMF. Don’t use oxidation potential of one with reduction potential of the other.

And yes EMF= E(Cathode) - E(Anode) provided both are reduction potentials.

But with the given electrode potentials I see no way to get 1.16V as an answer for the EMF of cell.

I would suggest that you check the question and electrode potentials again. And check the answer key again. Maybe there is an error there. It would be best to check about that with your course instructor.


Do (simple, inorganic) salts ionize completely in water when they dissolve?

When a salt, no matter which salt, or any other polar compound is exposed to enough water, then its polarized molecule parts, depending on its polarity, become attracted to the oxygen or hydrogens of the water, which will result, starting with the outermost molecule layer, in the complete dissolution of the compound. Within this process both the positively and the negatively polarized sections will get surrounded by water molecules (= hydration shell) and thus shielded from one another. In this state the original compound(s) have (theoretically) disappeared completely and the two containers will indeed be identical. Possible back reactions will (mainly) result in the most energy efficient compound for the particular experimental conditions and have the same chance in both containers.

When you now evaporate all the water you may get various compounds and mixtures depending on what was dissolved.

PS: I'd say "of equal molarity" or "they are equimolar", but my mother tongue isn't English, so there might be a more elegant way.


Do acyl fluorides exhibit resonance donation or inductive withdrawal?

Yes, acid fluorides exist.  You can purchase acetyl fluoride from Sigma-Aldrich.

Look back at these two questions for background on how the increased contribution from the triple-bonded carbonyl resonance structure (see below) will lead to an increase in the carbonyl ir stretching frequency (e.g. the triple-bonded carbonyl resonance structure will lead to a stiffening of the carbonyl bond and consequently shift it absorption to higher energy).

Ring Strain and C=O Stretching Frequency
Triple Bond Character in Acyl Chlorides

Here are the resonance structures we considered in the case of an acid chloride.

Resonance structure IV is what we're calling the "triple-bonded" carbonyl structure.

The carbonyl stretch for a acetyl chloride (I, R=Methyl) occurs at 1818 ${cm^{-1}}$.  In the case of acetyl fluoride (replace Cl with F in the above structures) this stretch occurs at 1869 ${cm^{-1}}$, significantly higher.  This suggests that resonance structure IV plays a larger role in the case of acid fluorides, despite resonance structure III presumably also playing a larger role compared to the acid chloride case.  The stability of $\ce{F-}$ vs. $\ce{Cl-}$ seems to be a very significant factor.


Basic question about an expression for Gibbs free energy

In the equation

$$G = \sum \mu_i N_i$$

$G$ is the absolute value of the Gibbs free energy at some point in state space. While not explicitly stated, the values of the chemical potentials $\mu_i$ are in general functions of composition and other state variables (e.g. T and p), rendering $G$ a function of these same state variables.

On the other hand, the equation

$$dG = -SdT + VdP + \sum \mu_i dN_i$$

describes an infinitesimal difference in the free energy between two (infinitesimally close) points in that space.

Neither equation puts constraints on T and p, that is, both are general.


Which software to use to output molecular orbital diagrams?

If your interest is a qualitative reprepresentation, you may reach out for an editor like inkscape and use lines, squares and Unicode to create a picture like the following one

In text fields you may enter Ctrl + U followed by an unicode code like
2191 (for $\uparrow$), 2193 (for $\downarrow$), 21C5 (for $\uparrow\downarrow$).*)
If you have access to LaTeX, you may be interested in modiagram, especially if you are interested in an output as .[e]ps or .pdf.  Only a few parameters are needed for an illustration like

(credit)
(It is possible to construct more advanced orbital diagrams like this and this, too.)

For the quantitative, and more programmatic approach you seek, especially with Python, there are solutions with the matplotlib library which might require some adaptation by you (example).  One of them is PyEnergyDiagrams once shown here on ChemSE:

the other EnergyLeveller.py:

Addition:  With the special chars used here, maybe a couple like Ipe + LaTeX is more handy to use, than inkscape.


How does ammonium chloride increase the solubility of magnesium hydroxide in water?

There is a strong reason. But it is the chemical reaction, not just a better dissolution.

$\ce{Mg(OH)2}$ is a base with the limited solubility, defined by $K_\mathrm{sp}=[\ce{Mg^2+}][\ce{OH-}]^2$

$\ce{NH4+}$ ion, created by $\ce{NH4Cl}$ dissolution, acts as a  weak acid:

$$\ce{NH4+ + H2O <<=> NH3 + H3O+}$$

with $\mathrm{p}K_\mathrm{a}=9.25$

$\ce{OH-}$ ions formed by dissolution of $\ce{Mg(OH)2}$ are eliminated by recombination

$$\ce{OH- + H3O+ <=>> 2 H2O}$$

what supports the dissolution by keeping the product of ion concentrations below the $K_\mathrm{sp}$.

Effectively, there is ongoing equilibrium:

$$\ce{Mg(OH)2 v + 2 NH4+ <=> Mg^2+ + 2 NH3 + 2 H2O}$$


The spin of atomic nuclei can be predicted based on the knowing the number of protons and neutrons in the nucleus:

Even/Even. Nuclei that contain an even numbers of protons and an even number of neutrons have $I = 0$ and are NMR silent. Examples include $\ce{^{12}C}$, $\ce{^{16}O}$ and $\ce{^{32}S}$.

All other nuclei are NMR active:

Odd/Odd. Nuclei that contain an odd number of protons and an odd number of neutrons have I that are positive integers. Examples include $\ce{^2H}$ $(I=1)$, $\ce{^{14}N}$ $(I=1)$ and $\ce{^{10}B}$ $(I=3)$.
Odd/Even & Even/Odd. All other nuclei (odd/even and even/odd) have spins that are half integral. Examples include $\ce{^1H}$ $(I=1/2)$, $\ce{^{11}B}$ $(I=3/2)$, $\ce{^{13}C}$ $(I=1/2)$, $\ce{^{17}O}$ $(I=5/2)$, $\ce{^{19}F}$ $(I=1/2)$ and $\ce{^{31}P}$ $(I=1/2)$.

A precise formulation is far more complex, and arises from the fact that the key components of the nucleus (proton and neutron) are made up of quarks. The neutron, although neutral in charge, is actually made of 3 quarks (udd), which themselves are charged, and not uniformly distributed about the neutron 'sphere'. Protons also consist of 3 quarks (duu). The nuclear spin comes from the total angular momentum of all these quark components. To delve deeper is probably outside the realm of chemistry, and is probably better suited to a physics forum (at least it is outside my ability to explain).


Why is phosphorus pentachloride not used in the synthesis of chlorophenols?

Well you see, phenol do not react with PCl5 to give chlorobenzene as it undergoes EAS mechanism.In order to make the reaction possible you will have to activate phenol with nitro group.


As electronegativity increases the holding capacity of electron by the atom increases. So it is less likely to donate electrons. So we can generalise that access a period in periodic table nucleophilicity decreases.


Optimized General Contractions using the EMSL Basis Set Exchange

Those basis functions are removed because they are considered redundant. There is a completely uncontracted shell with the same exponent and angular momentum, so they can be removed without much loss of accuracy.

N     0
S   9   1.00
   9046.0000000              0.0007000
   1357.0000000              0.0053890
    309.3000000              0.0274060
     87.7300000              0.1032070
     28.5600000              0.2787230
     10.2100000              0.4485400
      3.8380000              0.2782380
      0.7466000              0.0154400
      0.2248000             -0.0028640        <=====
S   9   1.00
   9046.0000000             -0.0001530
   1357.0000000             -0.0012080
    309.3000000             -0.0059920
     87.7300000             -0.0245440
     28.5600000             -0.0674590
     10.2100000             -0.1580780
      3.8380000             -0.1218310
      0.7466000              0.5490030
      0.2248000              0.5788150        <=====
S   1   1.00
      0.2248000              1.0000000        ***Same exponent

Unfortunately, this is not documented on the old BSE. Here is the reference:

T. Hashimoto, K. Hirao, H. Tatewaki
'Comment on Dunning's correlation-consistent basis set'
Chemical Physics Letters v243, Issues 1-2, pp, 190-192 (1995)
https://doi.org/10.1016/0009-2614(95)00807-G

There was a period of time when the option was not working on the new BSE website (https://www.basissetexchange.org) but has been fixed (I hope).

Note: I am the maintainer of the new Basis Set Exchange :)


According to original paper (1), KIT-6 is a highly ordered cubic $Ia3d$ silica ($\ce{SiO2}$) templated by Pluronic P123 ($\ce{[EO20PO70EO20]}$) block-copolymer surfactant. As a source of $\ce{Si}$, tetraethoxysilane (TEOS) or sodium silicate are primarily used.

(1) Kleitz, F.; Choi, S. H.; Ryoo, R.; Chem. Commun. 2003, 0 (17), 2136–2137. DOI: 10.1039/B306504A


In the Cannizarro reaction an aldehyde with no alpha-hydrogens is treated with $\ce{OH^-}$.  Through a series of equilibria the aldehyde goes on to disproportionate into the corresponding carboxylate and alcohol as shown in the following mechanism.

Again, the only requirement for the Cannizzaro reaction to occur is that the aldehyde not have any alpha-hydrogens.  Formaldehyde and benzaldehyde are two very common aldehydes that undergo the reaction when treated with $\ce{OH^-}$.  Chloral ($\ce{CCl3CHO}$), having no alpha-hydrogens also undergoes the Cannizzaro reaction (reference)

Edit: Response to OP's comment

A key part of the Cannizzaro mechanism is shown in the second line of the above mechanism.  Here a hydrogen is transferred in a concerted step to the second molecule of aldehyde.  Free $\ce{H^-}$ is not generated in the process, rather it is a smooth, concerted transfer of a hydrogen atom.

[Note: This has been proven experimentally, when the reaction is run in $\ce{D2O}$, no deuterium winds up attached to the alcohol carbon.  This shows that free $\ce{H^-}$ was never involved, but rather a concerted transfer of the aldeyhdic hydrogen takes place].

Thanks to Mithoron for noting that, with chloral, the haloform reaction can apparently compete with the Cannizarro reaction.  Presumably, the pathway followed depends upon the reaction conditions.


From the properties block on the Wikipedia page for $\ce{NbF4}$:

Niobium (V) fluoride is hygroscopic - it absorbs water from the atmosphere. This hygroscopicity is your main problem. When the powder absorbs water, it probably is becoming sticky.

The added water is probably also providing the extra mobility the ions need to crystallize from the powder. Niobium (V) fluoride forms monoclinic crystals, which on a very small scale could look like hair or needles.


Why is covalent bonding stronger than electrostatic attraction?

Ionic bonds can easily be as strong as covalent ones.

First off, let's be clear that almost everything has some ionic and some covalent character. Moreover, it's not an either-or situation. Some bonds have both strong ionic and strong covalent character.

For now, I'll stick to simple cases.

Let's consider $\ce{LiF}$, which seems a prototypical ionic bond. The interatomic distance is $\pu{156 pm}$ according to NIST. That gives a potential energy of $\pu{903 kJ/mol}$. Now, that assumes a full electron is transferred from $\ce{Li+}$ to $\ce{F-}$, which isn't the case (more like $0.8 - 0.9$). This simple calculation also ignores Born repulsion which is similar to the short-range Van der Waals repulsion between atoms.

A more complete analysis J. Chem. Educ. 2000, 77 (8), p 1076 gives a much better agreement. The real bond dissociation energy is $\pu{577 kJ/mol}$ (Source, PDF via the Internet Archive). A simple spherical ion model from the paper gives the bond enthalpy as $\pu{608 kJ/mol}$.

  calculations based on a simple ionic model reproduce the bond energies of $\ce{LiF(g)}$ and $\pu{NaF(g)}$ to within $5\%$.

If we look up covalent bond dissociation energies, the triple bond in dinitrogen is $\pu{945 kJ/mol}$. Ok, that's perhaps stronger, but $\ce{LiF}$ only has a "single" charge and clearly some ionic compounds have greater charge transfer.

More "prototypical" covalent bonds include $\ce{C-H}$ ($\pu{410 kJ/mol}$) and $\ce{O=O}$ ($\pu{497 kJ/mol}$), both less than $\ce{LiF}$.

Now, as I said earlier, it's not the case of a bond being only covalent or only ionic.

To quote from R. Gillespie J. Chem. Educ. 2001, 78 (12), 1688:

  The large atomic charges in $\ce{BF3}$ and $\ce{SiF4}$ are important to understanding the great strength of these bonds—which are the two strongest known single bonds, with bond dissociation enthalpies of $613$ and $\pu{565 kJ/mol}$, respectively. In comparison, the $\ce{C–C}$ bond in ethane has a bond dissociation enthalpy of only $\pu{345 kJ/mol}$. The great strength of the $\ce{BF}$ and $\ce{SiF}$ bonds arises from both the relatively large amount of shared density and the large atomic charges. So it is not strictly correct to say that these bonds are “predominately ionic”; if we are forced to use these terms we have to say that they are both strongly covalent and strongly ionic.

Finally, you can of course treat both covalent and ionic bonds using MO theory. For example, an approximate picture of $\ce{LiF}$ under MO theory might be something like this:

We readily see $\ce{LiF}$ has very high ionic character. For all intents and purposes, the "bonding" $\sigma$ orbital is localized on the $\ce{F}$ atom, there's very little energy difference between the $\pi$ and $\sigma$ here.


Atmospheric Nitrogen gas and human body pH buffers

One chemical equation that can be written for nitrogenase is this one:

$\ce{N_2 + 8 \ H^+ + 8 \ e^- + 16 \ ATP \longrightarrow 2 \ NH_3 + H_2 + 16 \ ADP + 16 P_{i}}$

The equation shows that not only does nitrogenase generate the basic molecule $\ce{NH3}$, it also consumes protons (i.e. acid) to form hydrogen as an inescapable byproduct of its mechanism.

So yes, in theory nitrogenase could perturb the human body's ability to buffer pH.

However, the relative rates are always important.  Human requirements for nitrogen are on the order of 80 grams of protein per day.  This translates to about 12 g of (fixed) nitrogen atoms per day, which corresponds to 0.85 mol of N.  If all of that nitrogen were to come from nitrogenase, then 0.85 × 8 ÷ 2 = 3.4 mol of acidity would be lost via proton consumption in addition to the 0.85 mol of base generated as $\ce{NH3}$, or about 5.2 mol of "base" in total.

But human pH homeostasis works even though humans breathe out about 900 g of an acid equivalent ($\ce{CO2}$) per day, or about 20 mol.

So even if nitrogenase completely replaced human nitrogen requirements in some hypothetic bioengineered human, the demands on pH homeostasis would only change by about 25%, maximum.  And considering that pH homeostasis still works even when humans are exercising vigorously, when "instantaneous" $\ce{CO2}$ production rates are far higher, I doubt even a 25% change would be too much to handle.

Thus in conclusion I don't think nitrogenase would wreck human pH homeostasis, unless its activity in the human body was increased to a level far beyond what is necessary to replace dietary N requirements.


accurate measurement from inaccurate tools

Generally, improvement of measurement process has much better positive impact on results, than sophisticated mathematical processing of low quality data. But, sometimes we are forced to data what we have.

There are 3 quantities:

resolution = the smallest difference of directly measured value, distinguishable by the tool,

precision = Level of uncertainty of repeated measurement of the same value, due random fluctuances. It is affected by the resolution as well, if comparable with precision.

accuracy = Level of agreement of the estimated and real value, due the tool/method bias.

There can be a very precise but inaccurate tool/method and vice versa,
even if one usually comes with the other.

Precision can be improved by measurement repetition. The arithmetic mean of $n^2$ measurements has $n$ times better precision than a single measurement.

The estimation of the standard deviation of normal Gauss distribution of measurements is

$$s=\sqrt{\frac{\Sigma ( x - \overset{-}x )^2}{n-1}}$$

The estimation of the standard deviation of normal Gauss distribution of real value estimation by arithmetic mean is:

$$s_{\overset{-}x}=\sqrt{\frac{\Sigma ( x - \overset{-}x )^2}{n(n-1)}}$$

For a method consisting of multiple measurement steps, each with  their own error, the error propagation rule has to be followed.

If 2 quantities, subjected to random errors, are added or subtracted, the squares of their standard deviation estimations are additive

$$(s_{A\pm B})^2=(s_{A})^2+(s_{B})^2$$

If quantities are multiplied or divided, their relative standard deviation estimations are propagated:

$$\left(\frac{s_{A\cdot B}}{A\cdot B}\right)^2 = \left(\frac{s_{A}}{A}\right)^2 + \left(\frac{s_{B}}{B}\right)^2$$
$$\left(\frac{s_{A/B}}{A/ B}\right)^2 = \left(\frac{s_{A}}{A}\right)^2 + \left(\frac{s_{B}}{B}\right)^2$$

If we e.g. determine the difference of the standard solution volumes $A - B$,
where $A=20.00 \pm 0.03$, $B=10.00 \pm 0.04$,

then $A-B=10.00 \pm 0.05$,

as $\sqrt{0.03^2 + 0.04^2}=0.05$.

Accuracy is harder to improve by mathematical means. It can be improved by measurement bias analysis, by calibration of tools or methods and by combination of more tools / methods / labs.

There is an additional decrease of the measurement method precision due statistical error of calibration via rules above.

$$(s_\mathrm{rel,method})^2 =  (s_\mathrm{rel, meas})^2+(s_\mathrm{rel, cal})^2$$

More tools/methods/labs lead to the  "mean of the means" and partially decreases the possible bias,
similarly as averaging of measurements improve precision.

But it is more costly.


Why is ammonia so much more basic than water?

When it comes to comparing ammonia and water, the acidity or basicity can be explained because of the central atoms electronegativity. When comparing atoms of the same periodic row, the more electronegative an atom is, the less it will want to donate an electron pair.

The electronegativity values from the Pauling Scale for $\ce{N}$ and $\ce{O}$ are

$$\ce{N}=3.04$$
$$\ce{O}=3.44$$

So based off of these values we can see that nitrogen is less electronegative, it less strongly it holds onto its electrons, and thus it is a better electron donor. This allows the addition of a $\ce{H}$ to be more stable.

The atomic reasoning behind this is that, as a general periodic trend, as you move across the periodic table, from left to right, the nucleus is increasing, which increases it's charge, producing a stronger attraction to its valence electrons.


What is the difference between a mixture and a composed substance?

From the definitions in your textbook, it seems that a mixture is a combination of two or more substances that--though combined--maintain their chemical identifiability. Examples of mixtures might be:

Water/Alcohol Mixtures
Salt/Pepper Mixture
In pharmacy, pills, pastes/creams, and suppositories are mixtures

(there are many example of mixtures)

The consituent compounds might be separated with somewhat mild techniques, including dissolving in certain solvents, distillation, and crystallization.

Composed Substances (according to your textbook) are mixtures of two or more species that under go some chemical reaction. The species involved could lose most of their identifiability, becoming completely different compound(s).  Loosely speaking, chemical reactions can not be reversed by simple separation techniques.

(Of course, there is some arbitrariness here: particularly in what comprises "most" of their identifiability and what defines "simple" separation techniques.  Also, mixtures are actually held together with bonds, just like in composed substances.  They may be weaker, but even "weakness" of the interaction is a subjective concept.)


Rate of reaction involving different stoichiometric coefficients

My doubt is why rate of reaction is taken as the rate of consumption of N2?

It's done that way because rates have units, and most people don't pay enough attention to the units.  Usually in chemistry, rates are units of moles per unit time (per unit volume or per unit catalyst mass or per unit surface area or something).  But you need to know moles of what?

By writing the reaction as $\ce{N2 + 3 H2 <=> 2 NH3}$, you are implicitly defining that a mole of "reaction" corresponds to one mole of $\ce{N2}$, two moles of $\ce{NH3}$, and three moles of $\ce{H2}$.  If you wrote the reaction as $\ce{\frac{1}{2}N2 + \frac{3}{2}H2 <=> NH3}$, then one mole of this new reation would correspond to one mole of $\ce{NH3}$, but only half a mole of $\ce{N2}$, etc.

  Why can't it be the rate of production of NH3? Why can't we say that the rate of reaction is the rate of production of NH3 or half the rate of consumption of N2 or one and half times that of rate of consumption of H2?

It could be any of those things.  The choice is up to you.  You just have to be consistent.  You might this related chem.SE question and answer helpful.

When you say "the rate", you have to specify what rate you are referring to.  And if you talk about rates "of the reaction", the agreed upon reference is that you are counting moles of species that have stoichiometric coefficients of 1.


I found the way to convert the hexagonal parameters:

I have a,b,c and $\alpha$, $\beta$, $\gamma$. The solution in basis vectors is:

basis1 = (a, 0, 0)
basis2 = (xy, yx, 0)
basis3 = (xz, yz, c)

where
\begin{align}
xy &= b\cdot\cos(\gamma)\\
yx &= b\cdot\sin(\gamma)\\
xz &= c\cdot\cos(\beta)\\
yz &= \frac{b\cdot c\cdot \cos(\alpha) - xy \cdot xz}{\sqrt{b^2 - xy^2}}\\
\end{align}

Thus my solution was:

basis1 = (27.1979,  0,       0     )
basis2 = (-13.5990, 23.5541, 0     )
basis3 = (0,        0,       15.499)


Are elemental compounds composed of different isotopes polarised?

When taken isotopes into account, properties of compounds change. The polarisation of the compound included. Your hunch is correct, this probably only matters for very light compounds, or more precise compounds where the mass difference between the isotopes is large. The most extreme case is probably $\ce{HT}$, which has roughly double the mass of $\ce{H2}$.

I found an article by M. Trefler and H. P. Gush on the electric dipole moment of $\ce{HD}$ which explains the presence of the dipole moment as follows:[1]

  The center of positive charge oscillates along the axis of the molecule during a molecular vibration because the excursions of the proton are greater than those of the deuteron; since the electrons do not follow exactly the motion of the positive charge, a weak, vibrating dipole moment results.

There are more articles of the dipole moment of $\ce{HD}$, I simply did not have the time to go through all of them. I also think the above already makes the point you were interested in.[2]

A very thorough theory is presented by W. R. Thorson et. al.,[3] which essentially explains the dipole moment of HD as follows:

  [...] ; it arises directly from asymmetry of the electronic Hamiltonian, provided that we cast. that Hamiltonian in an appropriate set of canonical coordinates. The model of this Hamiltonian as having a pair of nuclei with slightly different effective charges is particularly pleasing, since we can link that directly to the orbital-size and binding-energy effects which we know must be associated with the difference in electronic reduced masses for D and for H.

M. Trefler and H. P. Gush found the dipole moment to be very small, i.e. $\mathbf{p}=(5.85\pm0.17)\times 10^{-4}~\mathrm{D}$. Hence to speak of deuterium protide or protium deuteride is a stretch.

Since the effect is quite small in $\ce{HD}$ already, one can probably assume it will not have a significant effect on the polarisation in other than symmetric diatomic molecules.

However, with isotopes a few notable applications can be achieved. One of the prime examples is “chiral acetate”, (R)-deuterotritioacetic acid $\ce{HDTC-COOH}$, which helped elucidate biosynthetic transformations. For a quite comprehensive, educational article see [4].

References

M. Trefler and H. P. Gush; Phys. Rev. Lett. 1968, 20, 703.
W. R. Thorson, J. H. Choi, and S. K. Knudson; Phys. Rev. A 1985, 31, 22.
(a) S. M. Blinder; J. Chem. Phys. 1960, 32, 105. (b) S. M. Blinder: The dipole moment of HD. http://hdl.handle.net/1811/7809 (c) J. Bradley Nelson and G. C. Tabisz;
Phys. Rev. Lett. 1982, 48, 1393. (d) G. Herzberg; Nature 1950, 166, 563. (e) etc. see google scholar
Addison Ault; J. Chem. Educ. 2003, 80 (3), 333.


What should you couple with the reference material in testing galvanic corrosion?

Wikipedia states:"Galvanic corrosion (also called bimetallic corrosion) is an electrochemical process in which one metal corrodes preferentially when it is in electrical contact with another, in the presence of an electrolyte." What are the two metals? How would we know??!! The general answer is:"They are the two metals which form the galvanic cell (in contact with the electrolyte)." Depending on your exact circumstances, this might be the metallic coating and the base metal, or it could be the metallic coating and the metallic coating,or even some other metal component in electrical contact. Of course, if the metallic coating is perfectly homogeneous and completely encloses the base metal, then there will be no galvanic corrosion (at least, not until other corrosion mechanisms have changed that situation).


How to effectively wash the precipitate or crystals on the filter and why?

Every time the precipitation on filter is washed, there is a certain constant volume $V_\mathrm{ret}$ of washing fluid retained on the filter. If the initial concentration of impurity is $c_0$, then initial amount of impurity on filter is
$$n_0 = c_0V_\mathrm{ret}$$
and after washing with a new portion of the fluid with the volume $V_\mathrm{add}$ concentration of impurity is reduced to
$$c_1 = \frac{V_\mathrm{ret}}{V_\mathrm{ret} + V_\mathrm{add}}c_0.$$
Analogously, after $n$-th washing act
$$
c_n = \left(\frac{V_\mathrm{ret}}{V_\mathrm{ret} + V_\mathrm{add}}\right)^n c_0
\qquad\text{or}\qquad
c_n = \left(\frac{1}{1 + \frac{V_\mathrm{add}}{V_\mathrm{ret}}}\right)^n c_0
$$
and, assuming we have a total volume of washing liquid of $V_\mathrm{tot} = nV_\mathrm{add}$, the former formula becomes
$$c_n = \left(\frac{1}{1 + \frac{V_\mathrm{tot}}{V_\mathrm{ret}}\frac{1}{n}}\right)^n c_0$$
Now, lets assume we have a moderate amount of impurities (e.g. $c_0 =\pu{0.1 mol L-1}$) that we are trying to get rid of by washing with different amounts of the same volume of the same solvent (say, $V_\mathrm{tot} = \pu{100 mL}$). However, we use different filters, for instance:

a large one (retains $1/10$ of the washing liquid, or $V_\mathrm{ret} = \pu{10 mL}$);
a middle one (retains $1/20$ of the washing liquid, or $V_\mathrm{ret} = \pu{5 mL}$);
a small one (retains $1/100$ of the washing liquid, or $V_\mathrm{ret} = \pu{1 mL}$),

and, of course, different portions of the liquid (various $n$). Lets plot the concentration of the impurity for each case as a function of washing frequency:

First and foremost, it's obvious that it is always better to wash with small portions many times, rather than two or three times with large quantities of washing liquid. Second, the smaller the filter, the less impurity will be left after the same number $n$ of washing acts.
Now, can we get rid of the impurity once and for all by washing with an extremely large number of small portions? In fact, lets use an infinite number and take the following limit
$$\lim_{n\to\infty}c_n = \lim_{n\to\infty}{\left(\frac{1}{1 + \frac{V_\mathrm{tot}}{V_\mathrm{ret}}\frac{1}{n}}\right)^n c_0},$$
which is, in fact, one of the notable special limits, sometimes called the second remarkable limit:
$$\lim_{x\to\infty} \left(\frac{1}{1+\frac{a}{x}}\right)^x = \frac{1}{e^a}$$
so that in our case we can conclude it's impossible to completely get rid of the impurity only by washing:
$$\lim_{n\to\infty}{\left(\frac{1}{1 + \frac{V_\mathrm{tot}}{V_\mathrm{ret}}\frac{1}{n}}\right)^n c_0} = c_0\exp{\left(-\frac{V_\mathrm{tot}}{V_\mathrm{ret}}\right)} \neq 0$$
Conclusions

It is impossible to completely remove all impurities by washing the crystals/precipitate on the filter even by washing it infinite number of times.
It is more efficient to wash with small amounts of liquid many times than only a couple with large ones.
Use filter of the right size: not too big to retain the traces of impurities, but big enough to maintain efficient liquid transport through its surface. When possible, use pleated filter paper.

Appendix
The figure have been plotted with pgflots. Feel free to play around with it or reuse for your lab manual. Source code:
\documentclass{standalone}
\usepackage{pgfplots}
\usepackage{siunitx}

\begin{document}

\begin{tikzpicture}
    \begin{axis}[
        title={Effectiveness of washing methods ($c_0 = \SI{0.1}{\mole\per\liter}$)},
        xmin=0, xmax=20,
        ymin=0, ymax=0.001,
        xlabel = $n$,
        ylabel = {$c_n$/\si{\mole\per\liter}},
        minor tick num=4,
        grid,
        legend cell align={left},
    ]

    \addplot [
        domain=1:20,
        samples=200,
        solid,
        thick,
    ]
    {0.1*(1/(1 + 10/x))^x};
    \addlegendentry{$V_\mathrm{tot}:V_\mathrm{ret} = 1:0.10$}

    \addplot [
        domain=1:20,
        samples=200,
        dashed,
        thick,
    ]
    {0.1*(1/(1 + 20/x))^x};
    \addlegendentry{$V_\mathrm{tot}:V_\mathrm{ret} = 1:0.05$}

    \addplot [
        domain=1:20,
        samples=200,
        dotted,
        thick,
    ]
    {0.1*(1/(1 + 100/x))^x};
    \addlegendentry{$V_\mathrm{tot}:V_\mathrm{ret} = 1:0.01$}

    \end{axis}
\end{tikzpicture}

\end{document}


How does pH affect the degradation of ascorbic acid (vitamin C)?

Ascorbic Acid is a known antioxidant molecule and its stability has always been an issue especially for pharma industry which often makes use of AA in formulation. Now as much as pH is a culprit in AA degradation, its only one among other factors like solvent, exposure to light/uv, and the nature of solution (viscous/non-viscous). There are quite a lot of studies on stability of Ascorbic Acid like this one

I normally prepare Ascorbic Acid freshly for each experiment, but it would be an interesting experiment to check the stability of AA for an extended time period at different pH values.

Here is an interesting simulation demonstrating degradation of Ascorbic Acid.


What is the electrochemical reaction of this battery?

Thus we can see that the Al is oxidised but there are no Cu2+ ions to
  be reduced also Na+ reduction would lead to a negative Ecell standard
  and a positive Gibbs. Is it an air cathode where O2 is being reduced?

Well, there is no free lunch in this world. This is not a conventional cell in which each electrode is sitting in its own ions and one can easily calculate their Nernst potential. What we are probably seeing here is the very concept of contact potential. When two dissimilar metals are brought into contact via some electrolyte, a potential difference appears between the two. Recall the first experiment of Volta. He made a long series of
-silver[wet cardboard]zinc-silver[wet cardboard]zinc-silver[wet cardboard]zinc- cells, where the cardboard was wetted with salt-water. He was able to generate large potential differences this way by making very long series of Ag-Zn couple. Is this set-up good for drawing current-probably not.

These experiments were done in the 17th century by Volta. Atoms and ions did not exist then. These are contact potentials and you do not need a redox system to begin the experiment (metal dipping in its salt). Let us say you start with Al, Al will start to dissolve in a corrosive medium (salt is very corrosive) to form Al ions. You have a valid point, there are no copper ions to reduce? What else do we have in the system? Water, it has some ionized $\ce{H+}$, they start to reduce, one should see hydrogen bubbles at the copper electrode. Certainly you can draw small currents to power a LED from such batteries but not for long.

I do have some doubt about the design of the experiment. All soda cans are lined with a very very thin polymer layer. So the contact of Al is not that good with the copper coil. If one scrubs the can very well, only then that polymer layer would come off. The claim of seven LEDs is a stretch.