Can diamond undergo a self sustaining combustion reaction all the way to carbon dioxide?

Yes, diamond will combust in air. Regardless of the ambient air temperature, e.g. your example of $21\ \mathrm{^\circ C}$, you of course have to heat it to it's ignition temperature somehow, whether in a furnace, a flame, etc.  The autoignition temperature for diamond is around $900\ \mathrm{^\circ C}$ (source 1, source 2), compared to about $730\ \mathrm{^\circ C}$ for graphite.  The autoignition temperature is described in this Wikipedia article as:

  "...the lowest temperature at which it spontaneously ignites in normal
  atmosphere without an external source of ignition, such as a flame or
  spark. This temperature is required to supply the activation energy
  needed for combustion."

So, once you have initiated combustion by some means heating it to about $900\ \mathrm{^\circ C}$, it will continue to burn, primarily to $\ce{CO2}$.  Note that combustion does not require that a flame be produced or maintained.

How efficiently diamond combusts in air, in other words how much of the carbon is converted to $\ce{CO2}$ vs. $\ce{CO}$ and other partially oxidized carbon species is a function of a few different things. The same is true when burning graphite, wood, or most anything else. Is it a single smooth "rock" or a finely ground powder? How much of what sort of impurities does it contain? How efficiently ventilated is the system? These issues and other conditions of the combustion will determine the degree to which $\ce{CO}$ and other partially combusted carbon species are produced.


What is the proton NMR spectrum of p-methoxyphenol?

In theory, this structure is expected to provide four different signals:

a broad singlet, due to the potential exchange of the phenolic OH
a singlet accounting for three protons of the methyl group
and signals for the remaining four aryl protons ... which may be tricky

The chemical shift of the aryl protons is influenced by the two, para standing towards each other, subsitutents; here: OH and methoxy.  With their electronic properties, they influence the electron density of the aromatic ring, and eventually alter the location of the signals of the aryl H's. The centre of theses signals may be predicted by empirical increment rules which many textbooks about NMR include, like Silverstein, for example. Depending on the experimental conditions, accidentally they may simplify to a signal which only looks like a singlet:

(source, recorded in $\ce{CDCl3}$ at a Lamor frequency of 89.56 MHz)

Addition: The distance between the aryl H's and the phenol or the methoxy group (at least three bonds away) and their very similar electronic influence on the aromatic ring may render the aryl H's very similar.  So it takes a spectrometer with higher measurement resolution (measurement points per ppm) and higher magnetic field to discern them.  Rayment et al. published more recently in J. Org. Chem., 2012, 77, 7052–70602012 (doi 10.1021/jo301363h) data recorded again in $\ce{CDCl3}$, on an instrument with stronger magnetic field.  To quote:

  1H NMR (400 MHz, CDCl3) δH 3.78 (3H, s), 4.77 (1H, br s),
  6.76−6.82 (4H, m)

Even here, some discern, yet likely the two doublets overlap a bit, as a prediction by ChemDoodle suggests:

(the other doublet corresponds to the second pair of aryl H's).  Note the spread of the scale, too.  While I imagine even more advanced NMR spectrometer may eventually resolve the doublets (600 MHz is not so uncommon, and a few of about 1 GHz are around, too), I doubt someone will invest the more precious measurement time on these to delve more into detail and publish the then very high resolution 1H NMR, instead of collecting 1H NMR data of proteins and other natural products.

Addendum and correction:  Following the comment by @long, I reread the relevant pages in  Silverstein again.  Stating the for aryl H's would yield two doublets is wrong.

There are three types of coupling each of the aryl protons in p-methoxy phenol participates which are illustrated below:

(Source for the values of coupling constants: Silverstein, loc. cit., 6th edition, 1998, p. 212)

In blue is the proton which signal is of concern.  In the top row, it is the left in ortho position towards the methoxy group.  This proton couples with the close ortho standing proton (only three bonds away, hence $^3J$), typically with a coupling constant of $\pu{9 Hz}$, hence yielding according the $(n + 1)$ rule a doublet.  Simultaneously the proton marked in blue couples with the meta standing one with a typical coupling constant of $\pu{3 Hz}$; hence the previous doublet would become a doublet of doublet (dd).  Even farther distant, the para standing proton, hence coupling with an even lesser coupling constant which not every NMR spectrometer is capable to resolve -- but in principle yielding ddd, a doublet of doublet of doublet -- solely for this proton, up to $2 \times 2 \times 2 = \pu{8 signals}$ may become observable.

Following the theory further, the proton marked in blue of the bottom row is not identical to the one marked in the top row, and consequently, this proton may yield up to 8 observable signals, its own ddd.  This already adds up to 16 lines, which however is not (yet) the end of nitpicking theory as -- according to Silverstein's textbook -- the aryl Hs on the left hand are not equivalent to the ones on the right and a total of 32 lines possibly to observe.

With the graphical example provided by the Silverstein book for p-chloro nitrobenzene

which will serve for a follow-up question by mine, to record the aryl H's in a signal in shape like a singlet was by chance.


Why do electrolytic concentrations affect voltage in electrochemical cells?

Potential is work to move charge and this work is done by the chemical energy stored in the reactants, so if the concentration of reactants increases then there are more reactants which will do more work so this implies that potential increases with comcentration.


Best way to dissolve/emulsify beeswax and carnauba wax in water-based solution

Try Oelic acid and TEA with Waxes.
1. Take 3% Oelic acid
2. Add 3% TEA in Oelic acid and stir it with continuous heating up to 50C.
3. Take 100ml Water in Above solution.
4. Take 20grm carnuaba wax and dissolve it in above solution with continuous mixing and heating up to 60C.
5. Again add 40 to 50ml water in above solution.


How to calculate a reaction's DeltaH using a MM forcefield calculator software?

What you are forgetting is that the reaction will also involve hydrogen.
$\ce{C6H6 + 3 H2 -> C6H12}$
In principle, the $\Delta H$ for $\ce{H2}$ should be 0, but in practice, that may not be true with a force field. The energies from molecular mechanics are seldom accurate enough to compute reaction energies. They are parameterized on a set of molecules, but the errors are usually large, both systematic and random errors.


Comparison of acidity between angelic and tiglic acid

I have some thoughts of my own. However, I am not certain if they are correct.

The key to the question certainly cannot lie in the electronic factor because the cis- and trans- isomers do not differ in this aspect, except for the difference in the overall dipole moment. Thus, I believe we can only rationalise this with steric considerations. Let's first take a look at the chemical structures.

Fig. 1: Structure of tiglic acid

Fig. 2: Structure of angelic acid

The steric interaction I would like to focus on is the repulsion between the methyl groups and the negatively-charged oxygen atom of the carboxylate ion. Due to resonance, the negative charge would be delocalised over the two oxygen atoms of the carboxylate ion conjugate base. In the trans-isomer (angelic acid), the methyl group is located closer to one of the oxygen atoms (i.e. the oxygen atom of the carbonyl group). Thus, there would be a stronger repulsive interaction between the two. In the cis-isomer (tiglic acid), the methyl group is located further away from that oxygen atom, thus the intramolecular repulsion would be lessened, thus it is more stable.

Since a more stable conjugate base implies that the acid is stronger, the cis-isomer (tiglic acid) is more acidic than the trans-isomer (angelic acid). This argument is in full agreement with the information presented in your reference book.


Are any other orthonormalizations schemes besides Lowdin normally used when solving SCF equations?

Molcas developer here, currently implementing symmetric and canonical orthonormalization for the RASSCF module in addition to Gram-Schmidt.

As a rule of thumb you should treat those Orthonormalization schemes (ON-schemes) as different tools for different jobs.

If there are $n$ linear independent atomic orbitals (AOs), of which $n$ molecular orbitals (MOs) are to be constructed, then symmetric ON is probably the best, because the squared deviation of the MOs to the AOs is minimized. This means that your MOs are as close to the AOs as possible, while still being orthonormal.
If there is linear dependency, symmetric orthogonalization can detect, but not cure it.

Canonical and Gram-Schmidt orthonormalization have the advantage, that they can be used to construct only $m$ MOs from $n$ AOs ($n > m$) and to cure linear dependence.

If you compare Gram-Schmidt with Canonical ON, Gram-Schmidt has the disadvantage, that it is not as numerically stable. Besides the AOs that are orthonormalized last, can look completely different after the procedure. If you manually constructed your basis with chemical knowledge, this is not what you want. Modified Gram-Schmidt cures the numerical instability, but still has the disadvantage, that MOs might look completely different to the initial AOs.

PS: Just an addition to the comments: Grahm-Schmidt can be written as linear transformation. Although one would rarely do this.


Ab initio method to calculate C-C bond dissociation energy?

This "experiment" is about determining bond dissociation energy.

The general way to solve this isn't like calculating enthalpies of reaction, as I initially thought, but rather through calculating the energy of two CH3 radicals, and comparing it with ethane.  As another user pointed out, this should be done with unrestricted Hartree Fock and with diffuse basis sets.

Thus, energy(ethane) - 2*energy(methyl radical) is the solution.


Balancing a redox reaction with only one product

You should use the the same product species in both half -reactions. In the example you gave, this is $\ce{N2}$.
$$\ce{NH4NO3-> N2}$$
The half-reactions are constructed from the ammonium and nitrate ions separately, since these feature nitrogen in different oxidation states:
\begin{align}
\ce{NH4+ &-> N2} \tag{1} \\
\ce{NO3- &-> N2} \tag{2}
\end{align}
Balancing the reduction half-equation $(1)$:
\begin{align}
\ce{2NO3- &-> N2} & & \text{(balance N)} \\
\ce{2NO3- &-> N2 \color{red}{+ 6H2O}} & & \text{(add } \ce{H2O} \text{ to balance O)} \\
\ce{2NO3- \color{red}{+ 12H+} &-> N2 + 6H2O} & & \text{(add } \ce{H+} \text{ to balance H)} \\
\ce{2NO3- + 12H+ \color{red}{+ 10e-} &-> N2 + 6H2O} & & \text{(add } \ce{e-} \text{ to balance charge / oxidation state)}
\end{align}
And balancing the oxidation half-reaction:
\begin{align}
\ce{2NH4+ &-> N2} & & \text{(balance N)} \\
\ce{2NH4+ &-> N2 \color{red}{+ 8H+}} & & \text{(add } \ce{H+} \text{ to balance H)} \\
\ce{2NH4+ &-> N2 + 8H+ \color{red}{+ 8e-}} & & \text{(add } \ce{e-} \text{ to balance charge / oxidation state)} \\
\end{align}
In order to combine the two balanced half-equations, we need to multiply the first by $4$ and the second by $5$:
\begin{align}
\ce{8NO3- + 48H+ + 40e- &-> 4N2 + 24H2O} \\
\ce{10NH4+ &-> 5N2 + 40H+ + e-} \\ \hline
\ce{8NO3- + 10NH4+ + 48H+ + 40e- + 10H2O &-> 9N2 + 24H2O + 40H+ + 40e-}
\end{align}
which finally reduces to
$$\ce{8NO3- + 10NH4+ + 8H+ -> 9N2 + 14H2O}.$$


Finding simplest formula of a compound with following specifications


you might as well check the reactivity series.There is no unique and fully consistent way to define the reactivity series.The reactivity series is sometimes quoted in the strict reverse order of standard electrode potentials, where it is also known as the "electrochemical series":according to the series which has been defined there is no reason behind it.It is experimentally defined and according to the reactivity series magnesium is placed above aluminium.You might as well check this link for more clarity https://en.wikipedia.org/wiki/Reactivity_series#Reaction_with_water_and_acids.In your question it is given that magnesium doesn't react with hot water but it does actually vigorously.
 Magnesium  reacts very slowly with cold water, but rapidly
in boiling water, and very vigorously with acids.Aluminium reacts with acids and steam but less vigorously than magnesium.Magnesium DOES react with hot water even more vigorously than aluminium . Please check the link given.
Yes another factor is the formation of oxide layer on the surface of aluminium which reduces its reactivity


How can a methyl cation exist without 8 valence electrons

The way you put it that everything wants to achieve noble gas configuration is a good rule of thumb for predicting molecules that are stable and long-lived. However, not everything is in equilibrium. A methyl cation certainly exists, and it certainly wants to achieve the noble gas configuration, but it still takes time to do so. So we can conclude that the methyl cation is an unstable, short-lived, highly reactive molecule, which is true. And you are totally right to explain this with the tendency of having a closed electronic shell.


First off, welcome to Chemistry.SE! For formatting help visit the help center and for more information about this page, take the tour.

While I do think you should learn some theory, it's definitely possible to learn some practical computational chemistry through experimentation. And yes, there are a lot of abbreviations to follow.

One thing you should track while you explore is how long different calculations take. In general, more accurate quantum chemical methods are also more computationally costly. This means that they take longer, but also as systems get bigger, the cost rises faster. You might hear some people talk about "linear scaling" (i.e., double the size of the system, twice as much time) or $N^2$, $N^3$, etc.

A good resource, IMHO, is Molecular Modeling Basics by Jensen. It's available through CRC Press but Jan also blogs extensively. He uses GAMESS and Avogadro but also Jmol and other programs.

While the Avogadro project is working on some more detailed tutorials, I'd suggest starting some calculations with HF/6-31G* or B3LYP/6-31G* and see how things go for different molecules. I strongly suggest running some calculations and then consider how the results compare to experiment. Benchmarking is important. Computational models all have systematic and other errors, and for real work it's critical to run a set of molecules and carefully choose methods that work well.

Here are a few key terms to consider:

Method: There are several types of methods.

Semiempirical methods: These include AM1, PM3, RM1, (and PM6, PM7, and others that I don't think are in GAMESS-US). Such methods are fast, since they approximate some integrals and use empirical parameters. They also have an implied basis set which is usually very small. (Avogadro doesn't support viewing semiempirical calculations with GAMESS yet.)
Wavefunction methods: These include HF and relatives. These are similar to typical quantum chemistry methods taught in school.

MP2, MP3, MP4: These are perturbation methods that improve on Hartree Fock and introduce some level of electron correlation.
CCSD, CCSD(T), etc.: These are "coupled cluster" methods that are highly accurate, but often slow, and do not handle large molecules.

Density Functional: This includes another large set of methods, including BLYP, B3LYP, and newer methods like M06-X. Rather than approach things through the Schrödinger equation,

Basis: Most methods (except AM1, RM1, PM3, etc.) also require a "basis set," a set of functions to describe the atomic orbitals. Wikipedia article There are a few families, including:

Pople Type: 3-21G(d), 6-31G(d), 6-311+G(d), etc. These are older, more traditional basis sets. Generally 6-31G(d) or 6-31G* is considered minimally acceptable today. The "*" or (p,d,f) included at the end indicate that higher polarization is included on some or all atoms.
Correlation Consistent: cc-pVDZ, aug-cc-pVTZ, etc. These are newer basis sets, designed to function better with methods that treat electron correlation. The "aug-" includes various diffuse functions, which are useful particularly for anions and electron density that is particularly, well, diffuse. (In Pople basis sets, these are indicated with "+".


How do I determine whether it is the *ferrous* or *ferric* ion that is implied in this displacement reaction?

You can determine which reaction takes place by the standard electrode potentials of the half reactions. The relevant standard reduction potentials for the possible reactions (source of the values) are:

$$\ce{Cu^2+ +2e- ->~ Cu} \quad E_0=+0.342\mathrm{V}$$
$$\ce{Fe^2+ + 2e- ->~ Fe} \quad E_0=-0.447\mathrm{V}$$
$$\ce{Fe^3+ + e- ->~ Fe^{2+}} \quad E_0=+0.771\mathrm{V}$$

The overall potential of the electrochemical cell is defined as the difference between the standard reduction potentials of reduction and oxidation reaction.

$$E_{cell}=E_{red}-E_{ox}$$

For a spontaneous reaction to occur, $E_{cell}$ must be positive. With the potentials of copper reduction and oxidation of iron to $\ce{Fe^2+}$, you get

$$E_{cell}=E_{\ce{Cu^2+/Cu}}-E_{\ce{Fe^2+/Fe}}=+0.342\mathrm{V}-(-0.447\mathrm{V})=+0.789\mathrm{V}$$

It is obvious from the values that $\ce{Cu^2+}$ cannot further oxidize $\ce{Fe^2+}$ to $\ce{Fe^3+}$ because $E_{cell}$ is negative for this reaction.

$$E_{cell}=E_{\ce{Cu^2+/Cu}}-E_{\ce{Fe^3+/Fe^2+}}=+0.342\mathrm{V}-(+0.771\mathrm{V})=-0.429\mathrm{V}$$

The balanced equation for the redox reaction between iron and copper(II) sulfate is therefore

$$\ce{Fe + CuSO4 ->~ FeSO4 + Cu}$$


Can I change viscosity of an HDPE polymer?

If you use an extruder to prepare your composite material, try to use HDPE with a melt flow index: MFI of 0.1 g/10 min.


Is Orca wrongly breaking bonds? Why can it only optimize the molecule with COPT activated?

It seems to be a bug with optimization in internal coordinates. As I already mentioned in comments, ORCA couldn't converge at geometry relaxation step when optimization is done in internal coordinates. Switching to optimization in Cartesian coordinates (coordsys cartesian) or old internal ones (coordsys redundant_old) in the %geom section of the input, or using the ! COPT simple input as OP has done, solves the issue.

P.S. OP might report it as a bug at ORCA forum so that developers can take a closer look at this case.


Why do molecular orbitals in solids merge to bands?

Let's explain this with your example of silicon that every atom is sp3 hybridised. when you merge two of these orbitals, then it yields bonding and antibonding MO. But  when a third atom bounds, it yields  3 MO (not four as you mentioned) bonding, antibonding and non bonding MO. The number of MO equals the number of atomic orbitals constituting them, according to LCAO theory.
The energy of the third MO is in between the nonbonding and anti-bonding MOs. When you add a forth, a fifth and so on atoms, you will have more and more MOs very close in energy and you will form a band.
This is a qualitative description about the formation of a band.


When an aqueous extract of a substance is said to be administered at a certain quantity, is the quoted mass analogous to one as original substance?

You probably get the $\mathrm{LD_{50}} = \pu{1585 mg/kg}$ value by reading the abstract of the paper. That's why you got confused by that value. However, it is not $\pu{1585 mg}$ of fresh leaves. It is actually $\pu{1585 mg}$ of dry leave extract. I mean dry residue obtained by removal of water from an aqueous extract of dried leaves. The authors have made a stock solution of that dry residue in  $\pu{200 mg/mL}$ concentration by dissolving in enough distilled water ($\mathrm{pH} = 6.7$). Thus, $\pu{7.9 mL}$ of that solution is lethal to a rodent weighing $\pu{1.00 kg}$. On the other hand, since they have used rodents averaging $\pu{100 g}$ of body weight, each of them need only $\pu{0.79 mL}$ of that solution to be lethal.
Hope this explain what you are looking for.

Reference:
Olufunsho Awodele, Ibrahim Adekunle Oreagba, Saidi Odoma, Jaime A. Teixeira da Silva, Vincent Oluseye Osunkalu, "Toxicological evaluation of the aqueous leaf extract of Moringa oleifera Lam.(Moringaceae)," Journal of Ethnopharmacology 2012, 139(2), 330–336 (DOI: https://doi.org/10.1016/j.jep.2011.10.008).


As noted in the comments, the fallacy is the notion that only a select few atoms can form hydrogen bonds with protic hydrogen.  In reality, the strongest hydrogen bonds involve nitrogen, oxygen, or fluorine, but most nonmetals can form such bonds if they have an electron pair to donate to the bond.  Hydrogen bonding with chlorine is seen in the $\ce{HCl_2^-}$, which is analogous to the more familiar bifluoride ion, as well as chlorine-bearing organic compounds.


Concentration of gas in water: What is the time derivative of concentration toward steady state?

The transient diffusion equation for the oxygen concentration is $$\frac{\partial C}{\partial t}=D\frac{\partial^2 C}{\partial x^2}$$where D is the diffusion coefficient.  Henry's law applies at the interface.


Not all hydrogen atoms in a compound are equally acidic. The acidity of a hydrogen at a certain position in a molecule depends on its chemical environment, i.e. the solvent (if the compound is dissolved), which atoms and functional groups it is attached to, and how well the negative charge of the corresponding conjugated base is stabilized. A quantitative measure for the acidity of a proton is its $pK_a$ value which is defined as follows (reference):

$$pK_a=-\mathrm{log}_{10}K_a$$

with $K_a$ being the equilibrium constant of the dissociation reaction in water $\ce{HA + H2O \rightleftharpoons H3O+ + A-}$, in which the hydrogen is abstracted from the compound by heterolytic bond cleavage. $[\ce{H2O}]$ can be assumed to be constant (approx. 55 mol/l) and is therefore included in $K_a$:

$$K_{a}=\frac{[A^{-}][H^{+}]}{[HA]}$$

Let's take formic acid ($\ce{HCOOH}$) as an example. As the name of the compound suggests, it has at least one acidic proton, and if all hydrogens would be equally acidic and dissociate from the molecule, the reaction would look like this:

$$\ce{HCOOH + 2H2O \rightleftharpoons 2H3O+ + COO^{2-}}$$

However, this does not happen in aqueous solution, only one of the protons is acidic:

$$\ce{HCOOH + H2O \rightleftharpoons H3O+ + HCOO-}$$

The reason for this is that both hydrogens are situated at different positions in the structure:

With a $pK_a$ of 3.77, the hydrogen of the carboxyl group is significantly more acidic than the hydrogen bonded to the carbon atom. The typical $pK_a$ of an aldehyde hydrogen (H bonded to the carbon of a carbonyl group) is in the range of 17 (source), so such a position cannot be deprotonated in water.


No, PET it is definitely not a copolymer. Neither is e.g. Nylon 6,11 (or any other polyamide with two numbers). The monomers alone would not polymerise, or at least only by a very different reaction (e.g. ethylene gylcol gives PEG).

What you can have are alternating copolymers, but there the sequence is a statistical property. It comes when a reactive chain end B* has a higher probability to add an A (to give BA*), and a chain end A* to add B (to give AB*). That probability is sensitive to monomer composition, the temperature, pressure, catalyst, etc. If you want to read more about that, "copolymerisation parameters" is the topic to look up. The monomers you see there are afaik invariably olefines, i.e. they have a double bond that undergoes polymerisation.


Competition between substitution and elimination (uni- or bimolecular)

For an $\mathrm{E1}$ mechanism, we would need to generate a carbocation. Our halide is attached to a secondary carbon so the generated carbocation would be secondary. A quick check of neighbouring stabilising groups is negative. And as a professor of mine once put it (possible slightly oversimplified but with some general truth in it):

  Exactly one secondary carbocation exists and that is the prop-2-yl cation.

Thus, $\mathrm{E1}$ is not possible since the carbocation will not form in any measurable amounts. For the same reason, $\mathrm{S_N1}$ is not an option. For everything else, let’s consider the molecule’s 3D structure:

The tert-butyl group acts as a conformation ‘anchor’ meaning that the other chair form is not viable. Luckily, bromine is pointing upwards, which gives us access both to $\mathrm{S_N2}$ and $\mathrm{E2}$ mechanisms without having to invert or use a boat conformation.

Our base/nucleophile is methanolate which is both a good enough base and a good enough nucleophile, able to access both pathways. The centre for nucleophilic attack is also well accessable. If I had to choose, I would indeed go with your choice of saying the main products will be $\mathrm{S_N2}$ adduct and two enantiomers of elimination product will also be formed. (They stem from either abstracting the proton I drew or the enantiotopic proton on the back half of the six-membered ring.)


What are the resonance structures of 3-aminocyclohex-2-en-1-ylium?

The resonance structures and how to construct them for 3-aminocyclohex-2-en-1-ylium are shown below. The second structure in your question is none of them, as it is a different molecule: cyclohexa-1,3-dien-1-amine. For this molecule it is also possible to construct a resonance structure, as also shown below.

However, the two structures can be interconverted via an acid-base reaction.

  And for the stability of 3-aminocyclohex-2-en-1-ylium, the last structure would be most stable (with the positive charge on the EN nitrogen)?

There is no such thing as a most stable resonance structure. Resonance is only a concept that breaks down a highly complicated electronic structure into imaginable contributions of Lewis structures. All of these contributions are present all of the time. However, there are usually a few, that would describe the overall character best. Sometimes there is even only one predominant one.
The concept of resonance is necessary to describe delocalisation effects, hence the Lewis-like representation below is probably much closer to the actual structure than any of the resonance forms depicted above.


How to clean glassware containing pesticide solutions

I like this question because the cleaning of glassware is an often overlooked science in itself. You want to remove "contaminated glassware" as a possible factor in the interpretation of unexpected results in particular. A lot of brain-power and time can be wasted if you cannot assume that all the glassware (and any other surface coming in contact with your reaction) is clean with respect to the compounds of interest, and of any compounds that could interfere with their reaction.

For the given example, since dimetholate is soluble in methanol, methanol would seem to make a good solvent for rinsing the glassware before it dries.  So long as the solute is still dissolved in the solvent, a thorough solvent rinse should be effective with respect to removing dimetholate.  If the solute has dried, even though equilibrium thermodynamics say methanol should re-solublize  anything from the original methanol solution, the kinetics of this can be limiting/prohibitive. If this is the case (i.e. the solution has dried before rinsing), an aqueous detergent scrub followed by a methanol rinse is probably necessary. As water and methanol are miscible, this should not result in any mixed-solvent induced re-precipitation of anything originally present in the methanol-dimetholate solution.

Keeping to your example, as water an methanol are miscible, even in the case where the solution has not dried you should first rinse with methanol then with DI water. This will ensure any methanol that might precipitate the next standard material will be removed, and any remaining solvent (water in this case) will also not precipitate the water-soluble malathion.

The process is more complicated when you are using the glassware for a solvent in experiment 1 that is not miscible in experiment 2.  Say you a toluene  solution followed by an aqueous solution. As water and toluene are not miscible, any toluene left over from your toluene rinse would form a 2-phase solution in the aqueous solution.  While you always have the option of thoroughly rinsing then thoroughly drying (with the disadvantage of this discussed above), it is generally more desirable to do the following: thoroughly rinse the first solution in toluene, then rinse that with a solvent that is miscible with both toluene and water, like acetone. After the acetone rinse, then you can rinse the acetone with Di water.

If you have a reasonably limited number of solvents, it is good to keep a set of glassware dedicated to each solvent matrix. In my experience, the best procedure, and in fact it was part of an SOP for glassware cleaning, was to keep keep the glassware soaking in the solvent for which it will be used, rather than letting it dry.  Obviously this depends on the use case and there are certainly times that it is necessary to thoroughly dry your glassware. But where that is not necessary, to the degree reasonable, I recommend keeping the glassware separated based on what you use it for (i.e. solvent type), and keeping it soaking in that particular solvent.


The best way to solve this problem is to look at the $\Delta H$ charts found online. Using the simple equation below will give you the answer.

  $$\sum \Delta H_{\text{Products}} - \sum \Delta H_\text{Reactants} = \Delta H_{\text{Reaction}}$$
  If the $\Delta H$ is positive in sign then the reaction is endothermic and if it is negative in sign then it is exothermic.


Can HBr add to a non-sp2 hybridized carbon?

There is a carbocation rearrangement. Basically, the H can produce two secondary carbocations. One of those secondary carbocations can do a 1,2-hydride shift, which produces a tertiary carbocation. This carbocation, being tertiary, will react much faster, resulting in being in the major product. So, yes, HBr can add to a non-sp2 carbon.


Spatial orbitals $\phi_i$ in modern electronic structure calculations are indeed typically expressed as a linear combination of a finite number of basis functions $\chi_k$,
\begin{equation}
    \phi_i(1) = \sum\limits_{k=1}^{m} c_{ki} \chi_k(1) \, .
\end{equation}
In the early days, atomic orbitals were built out of basis functions, while molecular orbitals were built out of atomic orbitals, which is where the name of the approach, linear combination of atomic orbitals (LCAO) originated.
Today both atomic and molecular orbitals are built out of basis functions, and while basis functions for molecular calculations are still typically centered on atoms, they are usually differ from the exact atomic orbitals due to approximations and simplifications.
Besides, basis functions centered on bonds or lone pairs of electrons, or even plane waves, are also used as basis functions.
Nevertheless, the approach is still commonly referred to as linear combination of atomic orbitals.

Now, to the very question of why do we do things this way?

On the one hand, the LCAO technique made its way into quantum chemistry as just another example of a widely used approach of reducing a complicated mathematical problem to the well-researched domain of linear algebra. To my knowledge, this was proposed first by Roothaan.1 However, and as it was mentioned by Roothaan from the beginning, the LCAO approach in present day quantum chemistry is also attractive from the general chemistry point of view: it is tempting to construct molecular orbitals in modern electronic structure theory from their atomic counterparts as it was done by Hund, Mulliken and others already in the early days of quantum theory2 and as it is though in high-school general chemistry courses today.

1) Roothaan, C.C.J. "New developments in molecular orbital theory." Reviews of modern physics 23.2 (1951): 69. DOI: 10.1103/RevModPhys.23.69

2) Pauling and Wilson in Introduction to Quantum Mechanics with Applications to Chemistry refer to the following works in that respect (p. 346):

  F. Hund, Z. f. Phys. 51, 759 (1928); 73, 1 (1931); etc.; R. S.
  Mulliken, Phys.  Rev. 32, 186, 761 (1928); 41, 49 (1932); etc.; M.
  Dunkel, Z. f. phys. Chem. B7, 81; 10, 434 (1930); E. Hückel ,Z.f.
  Phys. 60, 423 (1930); etc.

Hund's papers are unfortunately in German, but Mulliken's ones are quite an interesting read. Especially the second one,

Mulliken, Robert S. "Electronic structures of polyatomic molecules and valence. II. General considerations." Physical Review 41.1 (1932): 49. DOI: 10.1103/PhysRev.41.49

which (again, to my knowledge) introduced the very term "molecular orbital".


Different reactions of clean and rusty iron nails in vinegar (acetic acid)

In your first experiment, the rust ($\ce{Fe2O3. xH2O}$) probably reacted with vinegar to form Iron (III) acetate which makes the solution reddish in color.

In your second experiment, the iron on the clean nail surface would have reacted with vinegar to form Iron (II) acetate. There are two main reasons why the solution might appear to have no change:

Usually, Iron (II) acetate is prepared using concentrated acetic acid and scrap iron (or ferrous oxide/hydroxide). Since vinegar is extremely dilute (around 5-8% acetic acid solution in water), the reaction would be very slow and the conc. of Iron (II) acetate formed will be very low.
Iron (II) acetate happens to be very soluble in water and has a very faint light green color in solution which is not easily detectable in lower concentrations.

Then again, when you added $\ce{H2O2}$ to the second solution, you mentioned that a reddish color appeared. This shows that the dissolved $\ce{Fe^2+}$ ions are oxidized to $\ce{Fe^3+}$. The $\ce{Fe^3+}$ formed will react with the acetate ions to form red Iron (III) acetate.


How do I solve questions pertaining to rate of disappearance and appearance?

In general, if you have a system of elementary reactions, the rate of appearance of a species $\ce{A}$ will be

$$\cfrac{\mathrm{d}\ce{[A]}}{\mathrm{d}t} = \sum\limits_i \nu_{\ce{A},i} r_i$$

where

$i$ is each reaction in the system
$\nu_{\ce{A},i}$ is the stoichiometric coefficient of species $\ce{A}$ in reaction $i$ (positive for products, negative for reagents)
$r_i$ is the rate for reaction $i$, which in turn will be calculated as a product of concentrations for all reagents $j$ times the kinetic coefficient $k_i$:

$$r_i = k_i \prod\limits_{j} [j]^{\nu_{j,i}}$$

In other words, there's a positive contribution to the rate of appearance for each reaction in which $\ce{A}$ is produced, and a negative contribution to the rate of appearance for each reaction in which $\ce{A}$ is consumed, and these contributions are equal to the rate of that reaction times the stoichiometric coefficient.

The rate of disappearance will simply be minus the rate of appearance, so the signs of the contributions will be the opposite.

In your example, we have two elementary reactions:

$$\ce{2NO ->[$k_1$] N2O4} \tag{1}$$

$$\ce{N2O4 ->[$k_2$] 2NO} \tag{2}$$

So, the rate of appearance of $\ce{N2O4}$ would be

$$\cfrac{\mathrm{d}\ce{[N2O4]}}{\mathrm{d}t} = r_1 - r_2 $$

Similarly, the rate of appearance of $\ce{NO}$ would be

$$\cfrac{\mathrm{d}\ce{[NO]}}{\mathrm{d}t} = - 2 r_1 + 2 r_2$$

and the rate of disappearance of $\ce{NO}$ would be minus its rate of appearance:

$$-\cfrac{\mathrm{d}\ce{[NO]}}{\mathrm{d}t} = 2 r_1 - 2 r_2$$

Since the rates for both reactions would be

$$r_1 = k_1 \ce{[NO]}^2$$

$$r_2 = k_2 \ce{[N2O4]}$$

the rate of disappearance for $\ce{NO}$ will be

$$-\cfrac{\mathrm{d}\ce{[NO]}}{\mathrm{d}t} = 2 k_1 \ce{[NO]}^2 - 2 k_2 \ce{[N2O4]}$$


Why can't the reaction of grignard reagent with carboxylic acid give a ketone?

Huy Ngo says it all.  A Grignard reageant will react first with just about any hydrogen that has even a modicum of proton donation capability, before "resorting" to the slower nucleophilic attack.
To get a ketone (in most cases), avoid the dissociable proton by using a nitrile instead, assisted by a copper salt. This forms an intermediate imine which is hydrolysed to a ketone after water/acid workup:
$\ce{CH_3MgI + CH_3COOH -> just a salt}$
$\ce{CH_3MgI + CH_3CN ->[Cu^+] (CH3)2C=N-M ->[H2O] (CH3)2C=O}$


What are the correct equilibrium arrows?

According to IUPAC: Quantities, Units and Symbols in Physical Chemistry (Green Book), the arrows are used differently for composite mechanisms in chemical kinetics.

In case of a simple mechanism that is composed of a forward and a reverse elementary reaction (i.e. reactions that occur at the molecular level in one step), it is conventional to write these in one line using two double-sided arrows $(\rightleftarrows)$.
For example:

$$\begin{align}
\mathrm{A} &\to \mathrm{B + C}\\
\mathrm{B + C} &\to \mathrm{A}\\
\hline
\mathrm{A} &\rightleftarrows \mathrm{B + C}
\end{align}$$

This is distinguished from a net reaction, which is written with two one-sided arrows $(\rightleftharpoons)$.
For example:

$$\begin{align}
\mathrm{A} &\rightleftarrows \mathrm{B + C}\\
\mathrm{B + C} &\rightleftarrows \mathrm{D + E}\\
\hline
\mathrm{A} &\rightleftharpoons \mathrm{D + E}
\end{align}$$

The two-sided arrow $(\leftrightarrow)$, which indicates resonance structures, shall not be used for reactions.

I guess that every elementary reaction that is proceeding in both directions $(\mathrm{A} \rightleftarrows \mathrm{B + C})$ may also be considered a net equilibrium reaction $(\mathrm{A} \rightleftharpoons \mathrm{B + C})$, but not every net equilibrium reaction $(\mathrm{A} \rightleftharpoons \mathrm{D + E})$ is an elementary reaction that occurs at the molecular level in one step $(\mathrm{A} \rightleftarrows \mathrm{D + E})$. Therefore, I assume that one-sided arrows $(\rightleftharpoons)$ may be used for all equilibrium reactions, whereas double-sided arrows $(\rightleftarrows)$ could be wrong when one combines composite mechanisms to obtain net reactions.


Number of stereoisomers of 3-ethylpent-1-ene-1,4-diol

The answer lies in the double bond.  The alcohol on the double bond undergoes keto-enol tautomerism. The greatly more stable aldehyde effectively makes the vinylic alcohol non-existent. This removes a factor for stereoisomers.  six is an odd number of stereo isomers to have since 6 is factored to 2 and 3.  This implies a stereoisomer with 3 forms.  I think your teacher may be thinking of the trans, cis and aldehyde state of the alcohol to come up with six, but this would neglect a carbon stereo center. in this case 12 would be the correct answer. However you should ignore the tautomers and just assume aldehyde which means the answer should be 4.


Why does oxygen not condense onto open dewers of liquid nitrogen?

The boiling point of nitrogen is 77 K, and the boiling point of oxygen is 90 K, so oxygen is liquid when liquid nitrogen is around. Wikipedia repeats the warning you mentioned:

  Liquid nitrogen has a lower boiling point at −196 °C (77 K) than oxygen's −183 °C (90 K), and vessels containing liquid nitrogen can condense oxygen from air: when most of the nitrogen has evaporated from such a vessel there is a risk that liquid oxygen remaining can react violently with organic material.

So it's not because the surface of the nitrogen is too warm.

My guess is that the pressure due to the evaporating nitrogen is pushing air away from the opening. LN2 has an expansion factor of 694, which you can compute by dividing the density of the liquid (0.808 g/mL) by the density of the gas at 20 °C (1.17 g/L). Of that expansion factor, a factor of 303 K/77 K = 3.94 comes from warming the gas from 77 K to 20 °C. The other factor of 176 comes from turning the liquid into a gas. That expansion is pushing air out of the headspace of the dewar.


Naturally Occurring Sulfites or Sulphites form

When collagen is hydrolyzed, sulfite ions ($\ce{SO3^2-}$) are released. Usually manufacturers filter out the residues, but may some still remain behind, which is why they put the warning on the label.


Bond Dissociation Temperatures for Hydrocarbons

When hydrocarbons undergo thermal decomposition carbon-hydrogen and carbon-carbon bonds are broken.
$$\ce{R-CH3 -> R-CH2. + ~H.}$$
$$\ce{R-R' -> R. + ~R'.}$$
Here is a link to a table that shows the energies required to break these bonds (the bond dissociation energy). These reactions are unimolecular decompositions and are governed by first order kinetics.  The rate of the reaction is given as
$\text{reaction rate} = k[\ce{A}]$
where k is the rate constant and $[\ce{A}]$ represents the concentration of the hydrocarbon.  The rate constant can be described using the Arrhenius equation
$k= A e^{-E_a/(R T)}$
where $A$ is the pre-exponential factor, T the absolute temperature in kelvin, ${E_a}$ is the activation energy, and R is the universal gas constant.  In the thermal unimolecular decomposition of a hydrocarbon, $A$ will have a value around $\pu{10^{12}sec^-1}$ and the bond dissociation energy can be used for the activation energy.
The Arrhenius equation tells us that this decomposition will be occurring at all temperatures above absolute zero.  As we increase or decrease the reaction temperature the rate constant will increase or decrease respectively, but the reaction is always occurring as long as we are above absolute zero.
As to pressure, yes it will affect the reaction rate.  Pressure is directly related to concentration, the $\ce{[A]}$ term in our rate equation.  As we increase or decrease the pressure, we are increasing or decreasing the concentration of $\ce{A}$, and thereby increasing or decreasing the rate of the reaction.


Can I use the Henderson–Hasselbalch for calculating the pH of non-buffer systems?

Can I use the Henderson-Hasselbalch equation on reactions that are not buffers?

If you know the equilibrium concentrations of weak acid and conjugate base, you can use the equation. You can also use it when you know the $\mathrm{pH}$ and the $\mathrm pK_\mathrm a$ and want to know the ratio of protonated and deprotonated species (for example for relating color and $\mathrm{pH}$ when working with pH indicators). Written as a formula, the following is always correct for a weak acid and conjugate base at equilibrium:

$$ \mathrm{pH} = \mathrm pK_\mathrm a + \log\frac{[\text{weak base}]_{\text{eq}}}{[\text{weak acid}]_{\text{eq}}}$$

or

$$ \log\frac{[\text{weak base}]_{\text{eq}}}{[\text{weak acid}]_{\text{eq}}} = \mathrm{pH} - \mathrm pK_\mathrm a $$

The equation is especially convenient for a buffer because when you mix high concentrations of weak acid and its conjugate base, the concentrations don't change much when the system reaches equilibrium, so you can approximate the $\mathrm{pH}$ by plugging in the initial concentrations. So for buffers,

$$ \mathrm{pH} = \mathrm pK_\mathrm a + \log\frac{[\text{weak base}]_{\text{eq}}}{[\text{weak acid}]_{\text{eq}}} \approx \mathrm pK_\mathrm a + \log\frac{[\text{weak base}]_{\text{initial}}}{[\text{weak acid}]_{\text{initial}}}$$

For a solution of a weak acid only or weak base only, that does not work (one of the initial concentrations is zero, and you get a nonsense result). The approximation is best for high and similar concentrations of acid and base when the $\mathrm pK_\mathrm a$ is close to $7$.


It helps you when you want to balance a red-ox reaction by the mean of the oxidation number change. For example, if you want to balance the reaction of thiosulfate with iodine to tetrathionate and iodide, you realize that the average oxidation number of sulfur changed from +2 to +2.5. The oxidation half-reaction looks like this:
$$\ce{4S^{+2}-> 4S^{2.5+} +2e-}$$


Why are there different values for enthalpy of combustion, depending on the calculation method?

Average bond enthalpies (note the word "average") are calculated in a different way from formation enthalpies. Formation enthalpies are well-defined and precise, meaning that two different people could measure them the same way and get the same answer. Average bond enthalpies are averages over many different types of bonds.

For example, there is only one way to make $\ce{CH4}$, but there are many different molecules with $\ce{C-H}$ bonds. The enthalpy of dissociation of a $\ce{C-H}$ bond in $\ce{CH4}$ is different from that in $\ce{C2H6}$, and in $\ce{C6H6}$, and really in any other molecule you can think of. To find average bond enthalpies, a selection of a large number of those bonds in many different molecules is taken, and the enthalpy of dissociation for each is measured. Then, the average (mean) is recorded. If you look at some different charts of bond enthalpies, you will notice two things: First, there are some numbers marked "exact" - $\ce{H-H}$, $\ce{Cl-Cl}$, $\ce{F-F}$, for example. This is because there is only one molecule with each of those types of bonds. The other thing you will notice is that depending on which chart you are using, many of the numbers will be slightly different. This is because there is no universal, unchanging standard describing which molecules are used to determine each bond - it depends upon what the people making the chart decided to use.

Because of this difference, when making predictions, average bond enthalpies are less accurate than formation enthalpies. In practice, they are used for quick estimations, or as a last resort when formation enthalpies are not available.

When I taught chemistry, I would have my students measure reaction enthalpies in lab, then calculate them using formation enthalpy tables and average bond dissociation enthalpy tables. The goal was to show them that average bond enthalpies can be reasonably close for rough estimates, but that formation enthalpies are much better in practice. It also showed them that accurately measuring reaction enthalpy in the lab is difficult!


How is silica bead bonded into polystyrene?

You can functionalize silica to add polymer "brushes" to the surface. You can also buy pre-made functionalized silica, although the downside to that is it isn't always easy to get the manufacturer to tell you exactly what is on the surface.

This isn't really going to increase the attractive interactions between the silica and the polystyrene, however. Polystyrene has a bulky aromatic ring hanging off of each monomer, and no sites that can participate in hydrogen bonding. When you add hydrophobic brushes to the surface of the silica, what you are doing is limiting the ability of the silica to aggregate due to hydrophilic interactions with itself. As a result, given enough kinetic energy (temperature for melt casting, or sonication for solvent casting) you can get the particles to disperse. Then, if you solidify the polymer quickly enough, they will be kinetically trapped in a dispersed state.

Depending on what your goal is, this could be enough for your application. However, if you want to get the particles to strongly attract to the polystyrene, you aren't going to be able to get much farther than van der Waals interactions. That's a limitation of the polystyrene, not the silica.

Other particles (gold, or carbon nanotubes, for example) will be more hydrophobic and easier to disperse without functionalizing, but you still are not going to get very strong attractive forces between the particles and the polymer.


Question about relation between two time constants

I think I understood your question now. I will reformulate it in more 'chemical' terms and give you my solution. Not nicely formatted for now, I will do that later if I find the time.

You have a cell with $G_{max}$ gates on it. They can open and close reversibly, depending on a voltage V. Closed gates open with rate constant $\alpha(V)$, and open gates close with rate constant $\beta(V)$.

This implies one differential equation. If $G_o$ is the number of open gates at time t, and $G_c$ is the number of closed gates at time t, then:

$\frac {dG_o}{dt} = \alpha G_c - \beta G_o$

And as the total number of gates is $G_{max} = G_o + G_c$:

$\frac {dG_o}{dt} = \alpha (G_{max} - G_o) - \beta G_o$

At steady-state, i.e. when $G_o$ stops changing, you have:

$\frac {dG_o}{dt} = \alpha (G_{max} - G_o) - \beta G_o = 0$

which, solved for $G_o/G_{max}$, yields:

$\frac {G_o}{G_{max}} = \frac {\alpha}{\alpha + \beta}$

The ratio $G_o/G_{max}$ is the probability that a single gate is open. As you say, a 'channel' is open when 4 of its gates are open. If we consider each gate being open or closed independently, the probability that 4 gates in a channel are open is indeed $(G_o/G_{max})^4$

What you measure is the time it takes, when the voltage is abruptly changed from an initial value to a final value, for the number of open channels to go from an initial value to 'equilibrium'.
This time is in theory infinite, but in practice it is probably the time it takes for the number of open gates to go from its steady-state value at voltage $V_1$, to a value sufficiently close to the new steady-state value at voltage $V_2$.

If so, it is sufficient to solve the above differential equation for $G_o$ with initial conditions $(t=0,G_o=G_{o,V_1})$. The solution is:

$\frac {G_o(t)}{G_{max}} = \frac {G_{o,V_1}}{G_{max}} \cdot e^{-{(\alpha+\beta)t}} + \frac {G_{o,V_2}}{G_{max}} \cdot [1-e^{-{(\alpha+\beta)t}}]$

where:

$\frac {G_{o,V_i}}{G_{max}} = \frac {\alpha(V_i)}{\alpha(V_i) + \beta(V_i)}$

The difficulty I see is the arbitrary nature of the 'time' you consider sufficient to reach steady-state. I would rather fit time points, i.e. number of open gates vs time. And my earlier suggestion to use numerical fitting may still be valid, actually.


You likely had either or both of the following:

Copper had some oxide on it, and this oxidized the alcohol leaving the metal behind.

If there was air in your system, especially on heating copper could have reacted with it to generate the oxide (a relatively fast reaction with heating) and you're back to (1).  The metal would then, of course, be available to react with more air.  In effect this would be a catalytic oxidation.

Copper is actually known in negative oxidation states, but that does not happen or need to be invoked here.


If losing electrons means getting oxidized, then does it mean that ALL metals are reducing agents, since they themselves get oxidized?

Your reasoning is sound. All (pure) metals can act as reducing agents, but they do not have to all the time. There are a few examples of metals in negative oxidation states, where they would act as oxidising agent.

Probably the most prominent examples are Zintl phases, where alkaline or earth alkaline metals reduce post transition metals (or metalloids like antimony). One of the first reactions observed was the sodium lead reactions in liquid ammonia in 1907. Back in the day they proposed the following reaction equation:
$$\ce{Na + excess~Pb ~->[\ce{NH3}][\mathrm{-50^\circ C}]~ Na+ + $“$~Pb2^{-}~$”$}$$
It was later found that the actual composition is $\ce{Na4Pb9}$. With crown ethers and/or cryptants it is possible to isolate discrete anions, too.

There are more examples like this, and a nice overview is given in John E. Ellis, Inorg. Chem., 2006, 45 (8), 3167–3186.


There's several aspects to your question.

Does quantum mechanics play a role in protein folding? Yes. The origin of the van der Waals interaction is ultimately a quantum mechanical one. (At least the induced dipole portions: how the electrons move with respect to each other and an external electrical field is driven by quantum mechanics.) Also, while part of hydrogen bonding can be described by classical dipole-dipole interactions, there's a non-negligible portion which is due to non-classical electron interaction. (How much this last bit contributes, if at all, is a bit controversial, though.)

But that's all electrons. The other part of your question is asking implicitly about the quantum behavior of nuclei.

In molecular modeling we often use a viewpoint termed "the Born-Oppenheimer approximation". That is, while we allow the electrons to delocalize and exchange and be their quantum mechanical self, we treat the nuclei pretty much like classical particles. Using this approximation, we can get pretty decent correspondence with experimental results.

So the best we can tell, the quantum nature of atoms as a whole (that is, nuclei as opposed to electrons) isn't needed for things like protein folding. The nuclei are probably not substantially delocalized, they're probably not interfering with one another, and they're probably not quantum tunneling.

Probably. There's some evidence that quantum tunneling of nuclei in the process of enzyme catalyzed reactions. This is mainly hydrogens, though. In most of these experiments the quantum tunneling drops off dramatically when you go to deuterium or tritium nuclei. I'm unaware of any system where carbon/oxygen/nitrogen tunneling is seriously proposed - they're just too heavy.

So there may be some contribution to the quantum nature of nuclei to protein folding, but all indications at this point indicate that it would be minor. The quantum properties of electrons, though, are definitely important.


Which has a greater dipole: chloroform (CHCl3) or trichlorofluoromethane (CCl3F)?

Imagine a $\ce{CHCl3}$ molecule and a $\ce{CFCl3}$ molecule fixed with the carbon atom of each molecule at the origin of a three-dimensional coordinate system. Select the two coordinate systems such that the hydrogen atom of the $\ce{CHCl3}$ molecule and the fluorine atom of the $\ce{CFCl3}$ molecule both lie on the $z$-axis above the origin, and let $u_z$ be the $z$ unit vector. If the four single covalent bonds of each molecule are symmetrical (here, tetrahedral), then it follows that the net vector sum of any four vectors that are equal in magnitude and are directed in the four directions from the origin to each non-carbon atom will be the zero vector. In both molecules, the net vector dipole moment arising from the three $\ce{C-Cl}$ bonds will equal the dipole moment of a single $\ce{C-Cl}$ bond with the chlorine atom in the direction of $-u_z$ from the carbon atom.

Assuming arguendo that the dipole moment of each bond is proportional to the difference in electronegativity of the two atoms between which that bond lies (on the Pauling Electronegativity Scale, hydrogen rates a $2.1$, carbon rates a $2.5$, chlorine rates a $3.0$, and fluorine rates a $4.0$), then to find the dipole moment of the $\ce{CHCl3}$ molecule, we add the vector dipole moment of the $\ce{C-H}$ bond ($[(2.1 - 2.5) \cdot u_z]$, or $-0.4 u_z$) to the vector dipole moment of the $\ce{C-Cl}$ bond ($[(3.0 - 2.5) \cdot (-u_z)]$, or $-0.5 u_z$), obtaining $[(-0.4 u_z) + (-0.5 u_z)]$ or $-0.9 u_z$.

To find the dipole moment of the $\ce{CFCl3}$ molecule, we add the vector dipole moment of the $\ce{C-F}$ bond ($[(4.0 - 2.5) \cdot u_z]$, or $1.5 u_z$) to the vector dipole moment of the $\ce{C-Cl}$ bond ($[(3.0 - 2.5) \cdot (-u_z)]$, or $-0.5 u_z$), obtaining $[(1.5 u_z) + (-0.5 u_z)]$ or $1.0 u_z$.

One might thus conclude that the dipole moment of $\ce{CFCl3}$ might be some ten percent larger in magnitude (as proportional to 1.0) than is that of $\ce{CHCl3}$ (as proportional to 0.9).


Preface

It is being assumed throughout this post that the concentration c equals the activity a. This is of course not true for concentrations above $\approx 10^{-3} \ldots 10^{-2}~\mathrm{mol\,L^{-1}}$ but it is common practice in schools and universities as long as the exercise is not very specifically about activities.
$$\mathrm{pH} = -\lg a(\ce{H+}) \xrightarrow{a=c} \mathrm{pH} = -\lg [\ce{H+}]$$
The other general assumption is that temperature is $22~^\circ \mathrm{C}$ and therefore $k_\text{W} = 10^{-14}~\mathrm{mol^2\,L^{-2}}$.

Given error contours are not "real measurements vs. calculations" but absolute differences between approximation n and the cubic equation that is not more simplified than using concentrations instead of activities.

General

If you have a calculator that can solve cubic equations, you don't need to simplify. In addition you only would have to remember one equation, no matter which pKa or concentration occurs (see above for real world restrictions).

Solve the following equation for x, with $x=[\ce{H+}]$:
$$x = \underbrace{\frac{k_\mathrm{W}}{x}}_{[\ce{OH-}]} + \underbrace{\frac{[\ce{HA}]_0 k_\mathrm{a}}{x+k_\mathrm{a}}}_{[\ce{A-}]}$$
or as a single equation (that is harder to remember):
$$x^3+x^2 k_\mathrm{a}-x([\ce{HA}]_0 k_\mathrm{a}+k_\mathrm{W})-k_\mathrm{a} k_\mathrm{W} = 0$$

Results from this equation will be the reference pH values for the following comparison of pH approximations.

To calculate the pH for any monoprotic(?) base, simply replace [$\ce{H+}$] with [$\ce{OH-}$], $k_\mathrm{a}$ with $k_b$ and of course [$\ce{HA}$] with [$\ce{B}$].

Approximation 1 and 2

For very strong acids the approximation is that they are completely dissociated, which makes the resulting equation very easy:
\begin{align}
x &= [\ce{HA}]_0\\ \rightarrow \mathrm{pH} &= -\lg [\ce{HA}]_0
\end{align}
When am I able to use this approximation with an error that is smaller than $\Delta \mathrm{pH} \leq 0.1$? It's the blue area in the picture below ($\mathrm{pK_c} = -\lg [\ce{HA}]_0)$.

The simple rule is: $6.7 \gtrsim \mathrm{pK_c} \gtrsim \mathrm{pk_a} + 0.5$.

Including the hydroxide ions from the autoprotolysis of water makes it a little bit more complicated and yields a quadratic equation in x:
\begin{align}
x &= [\ce{OH-}] + [\ce{HA}]_0 = \frac{k_\mathrm{W}}{x} + [\ce{HA}]_0\\ \rightarrow \mathrm{pH} &= -\lg \left(0.5 [\ce{HA}]_0 + \sqrt{0.25 [\ce{HA}]_0^2 + k_\mathrm{W}} \right)
\end{align}

This formula holds within: $(\mathrm{pK_c} \gtrsim \mathrm{pk_a} + 0.5) \vee (\mathrm{pK_c} \gtrsim 7.4)$.

It's not surprising, that the range of usefulness increases over the whole are of diluted concentrations. And that the high pKa values are also covers is only due to the fact that the pH is 7 everywhere upon concentrations that are below $10^{-7}\ldots10^{-8}$.

Approximations 3 & 4

The following approximations are used for weak acids. It's the second easiest to remember, so it's used quite often. The assumption is, that $x \gg k_\mathrm{a}$.
\begin{align}
x &= \frac{[\ce{HA}]_0 k_\mathrm{a}}{x}\\ \rightarrow \mathrm{pH} &= -\lg \sqrt{[\ce{HA}]_0 k_\mathrm{a}} = 0.5 (\mathrm{pK_c} + \mathrm{pk_a})
\end{align}

It's application range is: $(\mathrm{pK_c} < 13.7 - \mathrm{pk_a}) \wedge (\mathrm{pK_c} < \mathrm{pk_a} - 0.6)$.

It is obvious, that the approximation is useful in a small area where the acid is not too diluted and it's pKa value is not too low. Seems to fit well for weak acids.

Again, what happens if $k_\mathrm{W}$ gets involved involved?

\begin{align}
x &= \frac{k_\mathrm{W}}{x} + \frac{[\ce{HA}]_0 k_\mathrm{a}}{x}\\ \rightarrow \mathrm{pH} &= -\lg \sqrt{[\ce{HA}]_0 k_\mathrm{a} + k_\mathrm{W}}
\end{align}

Again, the range expands over very weak acids and strong diluted solutions as expected: $(\mathrm{pK_c} < \mathrm{pk_a} - 0.6) \vee (\mathrm{pK_c} > 14.2 - \mathrm{pk_a})$.

Approximation 5

The last approximation is for strong acids that are not diluted. The amount of protons in solution is equivalent to the amount of dissociated acid but this time with taking $k_\mathrm{a}$ into account. As only the influence of the autoprotolysis of water is neglected, it is "the most accurate" approximation for high to medium concentrations. This can easily be seen as it includes the scope of both approximations, 1 and 2. It is again a quadratic equation in x.

\begin{align}
x &= \frac{[\ce{HA}]_0 k_\mathrm{a}}{x+k_\mathrm{a}}\\
\mathrm{pH} &= -\lg(-0.5 k_\mathrm{a} + \sqrt{0.25 k_\mathrm{a}^2 + [\ce{HA}]_0 k_\mathrm{a}})
\end{align}

It can be used if: $(\mathrm{pK_c} < 6.7) \wedge (\mathrm{pK_c} < 13.7 - \mathrm{pk_a})$.

Now there are all those approximations, with all their respective acceptable ranges but it's quite hard to compare five diagrams with another. I made two further pictures that show 1) which approximation is the best per (pkc,pks)-pair and 2) which approximation is the easiest to remember per (pkc,pks)-pair.

1) The best -- which function yields the lowest error in comparison with the exact equation?

2) The easiest -- which function yields the lowest error in comparison with the exact equation and if there is more than function with $\Delta \mathrm{pH} < 0.1$, which one is the easiest to remember?

So ... the pka ranges kind of resemble when to use which function but in the end it's a little bit more complicated. If $k_\mathrm{W}$ is neglected, the low-error-area (LEA) is quite small, only approx. 5 is able to yield acceptable values for a big range of concentrations and pka values. Approximations 1 and 3 should be used with care, as their LEAs are quite fancy. For approximations 1 and 3 it's often better to switch to their low-concentration counterparts 2 and 4 to be on the safe side.


Which carbocation is more stable, the ethyl- or 1-propyl-carbocation?

The ethyl carbocation ($\ce{C2H5+}$) has a different structure than what you might have anticipated.  It is a non-classical ion with a bridging hydrogen.
For a description of just what a non-classical carbocation is see this answer.
The hallmark of a non-classical ion is 3 atoms with 2 electrons spread over them.  This is called a 3-center 2-electron bond (hypercoordinate bonding). In the case at hand the 3-atom 2-electron bond involves the 2 carbon atoms and the bridged hydrogen atom.

So there is no "methyl" group present in the ethyl carbocation that can inductively release electrons.  Additionally, the hyperconjugated resonance structure you mentioned has actually been replaced by a hydrogen that is partially bonded to the other carbon atom.

The n-propyl carbocation would likely have a similar non-classical structure - if it existed at all.  This carbocation has not been experimentally observed, it rearranges too rapidly to the much more stable 2-propyl carbocation.

So while we can't answer your question about whether inductive or hyperconjugative effects are more important in this series of compounds, we can assess which cation is more stable.

A variety of experiments suggest that the heat of formation of the non-classical ethyl cation is around 215 kcal/mol (reference, see pp. 68-72). This same reference points out that the heat of formation of the classical 1-propyl cation lies approximately 20 kcal/mol above that of the 2-propyl cation.  The heat of formation of the 2-propyl cation is around 193 kcal/mol.  Adding 20 kcal/mol to that value would place the classical 1-propyl cation around 213 kcal/mol. But, assuming the 1-propyl cation would also be a non-classical ion it would be lower in energy than the classical version, its heat of formation would therefore be less than 213 kcal/mol.

This analysis suggests that the 1-propyl cation would likely have a lower heat of formation than the ethyl cation.  In other words, the 1-propyl cation would likely be more stable than the ethyl cation.


Are melting points for mixtures of solids ALWAYS below the melting points of the individual components?

Is this always true?

No. As a general rule of thumb, if two compounds are miscible in each other, that is they exhibit solid solution, then the melting point of a (chemical, not mechanical) mixture will be intermediate. If the two compounds are not miscible and it's only possible to make a mechanical mixture and not a chemical mixture, then you will have a lower melting point.

Let's try to see this with several examples of phase diagrams, taken from FACT.

  ...mixtures of compounds are going to have melting points lower than either of the components pure melting points.

This is called eutectic relationship, for example sodium chloride and sodium sulfate. There is no intermediate compound, and they have a eutectic relationship, where the melting point of any mixture is lower than the melting point of both compounds. Note that if you are not in the pure liquid field, you will have a pure compound together with a liquid that is a mixture.

Silver and gold are a good example for a case when there is complete miscibility. All intermediate compounds can exist, therefore the melting point of the mixture changes continuously with the composition:

The double line is there because there is a very thin temperature range in which a liquid coexists with the solid of a different composition. See the solid solution page for more about that.

These were two simple cases. More complex cases arise when you have discrete compounds rather than a complete solution solution. A rather good example is cerium and nickel:

This system consists of several eutectic and peritectic points.


Is there a temperature where salt doesn't dissolve in water?

Solubility is an equilibrium phenomenon, which means that its temperature dependence is a function only of the enthalpy change of the reaction, not the entropy. See the van't Hoff equation for the mathematical relationship.

If by salt you mean sodium chloride, the dissolution in that case is only very slightly endothermic, so the solubility decreases very slowly with temperature.

That means that if you go low enough in temperature, you will reach a point where the solubility theoretically approaches zero. However, the water will freeze long before the temperature gets that low, and solubility will no longer have meaning.


Why are negatively charged ligands behind the neutral ligands in spectrochemical series?

I think you are confused by what the spectrochemical series actually shows. It is not a rating of stronger or less strong ligands to mean they can bind more or less strongly to the metal. It is only a rating of which ligands induce a high crystal field split in the resulting complex.

The splitting derives from the metal $\mathrm{e_g}$ orbitals — $\mathrm{d}_{x^2 - y^2}$ and $\mathrm{d}_{z^2}$ — actually being antibonding to the metal-ligand interaction and thus higher in energy than the (unmodified in a complex with only σ type interactions) $\mathrm{t_{2g}}$ orbitals.

For systems more complex, i.e. including π interactions, the $\mathrm{t_{2g}}$ orbitals (the other three d-orbitals) interact in a π manner with ligand p-orbitals. If the ligand has populated p-orbitals such as halides have, the $\mathrm{t_{2g}}$ energy is raised as it is turned into a slightly antibonding interaction (the bonding one being on the ligands’ side). Since this brings $\mathrm{t_{2g}}$ closer to $\mathrm{e_g}$, the field split is reduced and the ligands are termed weak-field ligands. In simplified Lewis schemes, one could represent this by a double forwards bond from the ligands to the metal.

In the case of empty p-type orbitals such as in $\ce{CO}$, metal to ligand back-bonding can occur by populated metal orbitals interacting with unfilled ligand orbitals; now it is the metal’s side of things that is stabilised. Since this π backbonding stabilises the $\mathrm{t_{2g}}$ orbitals, their energy is lowered and the field split becomes greater. Hence they are termed strong-field ligands. In simplified Lewis structures one could represent that by drawing a donor-acceptor arrow from the ligand to the metal (σ forwards bond) and one the other way (π back bond).

This all has nothing to do with the overall strength of metal-ligand interaction, as I mentioned earlier. In fact, strong complexes are typically obtained for certain metal-ligand combinations and the same ligand can be very bad in interacting with other metals. Classic example: $\ce{CO}$ ligands are extremely good at stabilising low metal oxidation states such as in $\ce{[Fe(CO)4]^2-}$, while they are rather lousy in stabilising high ones: $\ce{[Fe(CO)6]^2+}$ can only be synthesised in the extreme absence of all other ligands, i.e. in $\ce{SbF5}$ from $\ce{FeF2}$ where the fluorides will be drawn towards antimony to generate the even more stable $\ce{[SbF6]-}$ complex.


How to calculate whether a redox reaction is spontanuous?

You need to approch the question in a systematic manner.  The first step is to write the reduction and oxidation half reactions:

Reduction Half Reaction: (cathode)

$\ce{H2SO3 + 4 (H+) + 4 e- -> S(s) + 3 H2O}$

Oxidation Half Reaction: (anode)

$\ce{Mn -> (Mn^2+) + 2 e-}$

Now find the reduction potentials for both half reactions (only one was on the tables you provided, but a different set of tables might have both).  Remember they are reduction half potentials so to find the reduction potential of the oxidation half reaction you must reverse the reaction direction:

$\ce{(Mn^2+) + 2 e- -> Mn}$

The reduction happens at the cathode, and oxidation happens at the anode.

$$\ce{E_{cell} = E_{cat} - E_{an}}$$

$\ce{E_{cat} = +0.45 V}$

$\ce{E_{an} = -1.18 V}$

$\ce{E_{cell} = (+0.45) - (-1.18)}$

$\ce{E_{cell} = +1.63 V (spontaneous)}$

Just to clarify again, the cathode's reduction potential (+0.45) was NOT on the tables you provided.  I had to search the internet for a different set of tables which did have the cathode's half reaction reduction potential.


Why do cyclic ethers have higher boiling points than their acyclic isomers?

The data I could find suggests that cyclic ethers have higher dipole moments than their acyclic counterparts.

$$\mathbf{Four~carbons}$$
\begin{array}{c @{} c} \hline
\text{THF} & \mathrm{1.63~ D ^1, 1.75~ D ^2} \\
\text{1-butene oxide} & \mathrm{1.89~ D^2} \\
\text{ethyl vinyl ether} & \mathrm{1.26~ D ^2} \\
\text{diethyl ether} & \mathrm{1.15 ~D ^3, 1.15 D ^2} \\
\text{methyl propyl ether} & \mathrm{1.11 ~D ^2} \\\hline
\end{array}

$$\mathbf{Five~carbons}$$
\begin{array}{c @{} c} \hline
\text{Tetrahydropyran} & \mathrm{1.58~D ^2, 1.87~D ^4} \\
\text{2-Methyltetrahydrofuran} & \mathrm{1.38~ D ^5} \\
\text{dipropyl ether} & \mathrm{1.21~D^2, 1.00~D^6} \\\hline
\end{array}

This seems consistent with intuitive expectations based on conformational models where, in the cyclic ethers where the molecular geometry is constrained by the ring, the lone pairs are pointing in the opposite direction from the carbon skeleton. Whereas the lone pairs and carbon chains in the conformationally more mobile acyclic isomers are less geometrically fixed and therefore less "directed" in space.

It seems likely that the higher dipole moments in the cyclic compounds would lead to greater alignment / ordering in the liquid phase, which in turn would lead to higher boiling points.

References:

  1: http://en.wikipedia.org/wiki/Tetrahydrofuran
  2: https://physicalchemistryrosamonte.wordpress.com/material-balances/material-balances-on-a-crystallizer/physical-properties-of-pure-methanol/dipole-moment/
  3: http://en.wikipedia.org/wiki/Diethyl_ether
  4: http://www.drugfuture.com/chemdata/tetrahydropyran.html
  5: http://www.stenutz.eu/chem/solv28.php?s=1&p=20
  6: https://books.google.com/books?id=G6jaBwAAQBAJ&pg=PA50&lpg=PA50&dq=propyl%20ethyl%20ether%20dipole%20moment&source=bl&ots=VA52gqJ0kn&sig=GM8tua_QpXccFW3VXpy9qiNf7rk&hl=en&sa=X&ei=rzRvVZ__Cc32yQT9gIPACQ&ved=0CDsQ6AEwBQ#v=onepage&q=propyl%20ethyl%20ether%20dipole%20moment&f=false


Why does thermodynamics only apply to macroscopic systems at equilibrium?

The mathematics of equilibrium systems are far simpler than others

It isn't strictly true that thermodynamics doesn't apply to systems that are not at equilibrium only that the simple theorems and formulae of equlibrium thermodynamics only apply to systems at equilibrium.

The reason why most discussion and most teaching is about those equilibrium systems is because the mathematical results are far simpler and far easier to teach. They are also widely useful even with systems that are not exactly at equilibrium. Chemists can understand the key features of many reactions, for example, by understanding the equilibrium thermodynamics of the reaction (like the heat of formation of the products). Exceptions can be explained by understanding kinetic barriers in the reaction pathway.

But when details of non-equilibrium behaviours are required, the behaviour of many systems is so complicated there are no easy rules to teach or to talk about.

Non-equilibrium thermodynamics is actually a very large field and key results have won Nobel prizes (Prigogine won the 1977 Nobel in Chemistry for work on key aspects of it). Arguably a great deal of the complexity of the universe and life itself is the result of the complexity and structures that can arise when systems are not at equilibrium. So you could even argue that our ability to understand any thermodynamics is a result of the complexity the can arise from non-equilibrium systems. If the subject is that complex, there are not going to be many simple formulae or generalisations to teach or talk about.

And that is probably the real reason why some people think thermodynamics only applies to systems at equilibrium: anything else is too complicated to talk about so educators and textbook writers tend to avoid the subject.


Reaction of NaOH with methanol and ethanol?

The proposed reaction:

$$\ce{CH3OH(l) + NaOH(s) <<=> CH3O^-Na+(s) + H2O(l)}$$

According to wikipedia:

  The solid hydrolyzes in water to give methanol and sodium hydroxide

The equilibrium is biased to the left, as @vapid has mentioned in the comments. To quote him:

  In the large excess of alcohol, the reaction will be considerably shifted towards alkoxide.


Why is water transferred from a beaker of pure water to a beaker of sugar solution when placed in the a closed container?

Water molecules in gas phase collide and are absorbed by both solutions at the same rate, assuming the same humidity over the solutions.
OTOH, as sugar solution has lower molar fraction of water than water itself,  its rate of water evaporation is lower than pure water evaporation.
Therefore, there is the net transfer of water molecules from the water beaker to the sugar solution beaker.
It can be formulated also by other way: The saturated vapor pressure over sugar solution is lower than over pure water.  Therefore vapour saturated over water is oversaturated over sugar solution. And vice versa - vapor saturated over sugar solution is undersaturated over water.


Waylander is right, many sugars are heterocyclic compounds. While sugars such as glucose can exist in an open chain form, many of them are normally found in the form of six or five membered rings which consist of carbons and an oxygen.

Thus glucose fits the very definition of a heterocylic compound. It is important to keep in mind that any cylic (ring) compound where more than one element type is found in the atoms which make up the ring is a heterocylic compound.

For example S4N4, pyridine, ethylene oxide, glucose, 9-BBN and tetrahydrofuran are all heterocycles. They have very different chemistry to each other but they are all heterocylic.


Supercooling, freezing point depression, and finding the right freezing point

Melting (and boiling) points are only defined for pure substances. Rather than defining them as ‘the temperature when pure substance $\ce{X}$ melts/boils’ they are better defined as:

The melting point is the temperature at which for a certain defined pressure (usually $1~\mathrm{bar} = 10^5~\mathrm{Pa}$) the liquid and solid phases of a pure substance are at equilibrium.
The boiling point is the temperature at which for a certain defined pressure (usually $1~\mathrm{bar} = 10^5~\mathrm{Pa}$) the liquid and gaseous phases of a pure substance are at equilibrium.

(I didn’t copy that from anywhere, I made it up. Apologies to everyone who I was unintentionally quoting without attribuition.)
Thus, you can supercool pure liquids but if you allow the supercooled liquid to reach equilibrium at its supercool temperature (and standard pressure), it will solidify. Similarly, you can overheat solids but if you let the overheated solid reach equilibrium at the overheated temperature, it will liquidify. Thus, neither the supercool liquid’s nor the overheated solid’s temperature can be considered the melting point, because neither phase is stable at equilibrium.


The energy amount used for heating up a material can be calculated according to the following equation:
$$ E = c_\text{s}\, m\, \Delta T$$
where $E$ is the energy, $c_\text{s}$ the specific heat capacity, $m$ the mass of the object and $\Delta T$ the temperature difference that is to be achieved.

Solving the above equation for $\Delta T$ we get:
$$ \Delta T = \frac{E}{c_\text{s}\, m}$$

Plugging in the values provided by you, I get $\Delta T = 22.6\;\mathrm{K}$ (or °C, since we're talking about the difference and the scale is the same, it doesn't matter).

Adding this increase in temperature to the initial temperature gives the final answer:
$$ \vartheta_\text{f} = \vartheta_\text{i} + \Delta \vartheta = 21.0\; \mathrm{°C} + 22.6\; \mathrm{°C} = 43.6\; \mathrm{°C}$$

Note the following:

Number of significant digits: Since the data given consistently had three significant digits, I've rounded to the third digit.
Symbol for temperature: I used $\vartheta$ for temperatures in °C and $T$ for temperatures in K. I strongly suggest you follow this example as to never confuse the two.


What is a good source for large tables on the properties of water?

Such values for various properties of liquid water and steam can be looked up in so-called steam tables. Software versions also allow calculations (ususally interpolation) of additional values. For example, I usually use REFPROP – NIST Standard Reference Database (however, not always the latest version) or some old and simple proprietary steam table software from Siemens/KWU.

If you do not have access to professional steam tables, you may want to consider using the steam tables that are included in WolframAlpha.


As you've noted, $\ce{P4O10}$ (or $\ce{P2O5}$ ) is a powerful dehydrating agent. According to the Chemical Book, it can actually dehydrate concentrated sulfuric acid to $\ce{SO3}$. So, it's not surprising that it can indeed dehydrate aliphatic alcohols. Using ethanol as an example, ethene can be produced by refluxing ethanol over $\ce{P4O10}$:

$$\ce{C2H5OH ->[P4O10] C2H4 + H2O}$$

Note that the reaction will proceed with gentle or no additional heating (the reaction itself is exothermic), it will just do so more slowly.


Can visible light or infrared radiation excite electrons?

Different regions of the electromagnetic spectrum correspond to different atomic and molecular processes, each with one or more associated spectroscopies. Here is a general summary, with decreasing frequency/energy going from left to right:

In order to excite a valence electron, the longest wavelength or lowest energy radiation is usually in the visible region, not longer than ~700 nm. I'm not sure if there is a lower limit to this energy, or what the observed lowest energy electronic transition is, but certainly in this region, there is some overlap with the energy required to excite molecular vibrations. For some nice cases you can perform vibrational-electronic spectroscopy (usually called "vibronic" spectroscopy), where you resolve a number of vibrational transitions (peaks) as fine splitting or fine structure for a single electronic transition:

$\hspace{20ex}$

However, we do not say that "infrared radiation can excite electrons", though there is a range of frequencies where the types of excitation couple together. See also rotational-vibrational (rovibrational) spectroscopy; a nice example is hydrogen chloride in the gas phase.

The reason coordination complexes are often colored is because there are valence electronic excitations that correspond to the absorption of electromagnetic radiation in the visible region. The reason many molecules or complexes are white is because their lowest energy electronic transition is in the ultraviolet region. This is more general than losing degeneracy, and holds true for any macroscopic or microscopic system. It is not different from the color we see for conjugated systems. If the lowest energy available transition of a system is in the visible region, the system will appear colored. If it absorbs red-orange light, the opposite side of a color wheel will show you that it should appear green-blue. What you see as the system's color is the remaining color after subtracting the absorbed color. Yes, the kinds of orbitals involved are different (d-d transitions or metal-to-ligand charge transfer versus $\pi\rightarrow\pi^{*}$), but the transition energy matching is the most important thing.

For the "limit of ionizing radiation", ionization refers not to the excitation of an electron, but its removal:

$$
\ce{X + energy -> X+ + e-}
$$

The atom with the smallest first ionization energy is sodium, given as 496 kJ/mol. Converting this to a wavelength gives roughly 0.013 nm or 13 pm, which is a hard X-ray. Normally X-rays only have enough energy to excite core electrons, so this is the minimum wavelength for atomic ionization. Many atoms will require gamma rays, though I also assume that when looking at whole molecules rather than atoms, the ionization potential may be reduced.

Now to touch on the specific example of rhodopsin. It is a protein that contains retinal, which undergoes photoisomerization:

where A is the 11-cis form, and B is the all-trans form:

If we say that the ultraviolet region spans 10 to 400 nm, then retinal absorption is in the UV, not the visible. However, "the absorbance spectrum of the chromophore depends on its interactions with the opsin protein to which it is bound; different opsins produce different absorbance spectra." From that page:

\begin{array}{lcccll}
\hline
\text{Name} & \text{Abbr.} & \text{Photoreceptor} & \lambda_{\text{max}}~(\text{nm}) & \text{Color} & \text{Human variant} \\
\hline
\text{long-wave sensitive}           & \text{LWS}  & \text{cone} & 500-570  & \text{green, yellow, red}  & \text{OPN1LW "red" / OPN1MN "green"} \\
\text{short-wave sensitive 1}        & \text{SWS1} & \text{cone} & 355-445  & \text{ultraviolet, violet} & \text{OPN1SW "blue" (extinct in monotremes)} \\
\text{short-wave sensitive 2}        & \text{SWS2} & \text{cone} & 400-470  & \text{violet, blue}        & \text{(extinct in therian mammals)} \\
\text{rhodopsin-like 2}              & \text{Rh2}  & \text{cone} & 480-530  & \text{green}               & \text{(extinct in mammals)} \\
\text{rhodopsin-like 1 (vertebrate)} & \text{Rh1}  & \text{rod}  & \sim 500 & \text{blue-green}          & \text{OPN2 = Rho = human rhodopsin} \\
\hline
\end{array}

From photopsin:

  The mean absorbance spectra of outer segments of the four classes of human photoreceptors. Curves labelled as follows: '498', mean of eleven rods; '420', mean of three blue-sensitive cones; '534', mean of eleven green-sensitive cones; '564', mean of nineteen red-sensitive cones.

Regarding the mechanism of opsins as triggered by the photoisomerization,

  The photoisomerization induces a conformational change in the opsin protein, causing the activation of the phototransduction cascade. The opsin remains insensitive to light in the trans form. It is regenerated by the replacement of the all-trans retinal by a newly synthesized 11-cis-retinal provided from the retinal epithelial cells. Opsins are functional while bound to either chromophore, with A2-bound opsin λmax being at a longer wavelength than A1-bound opsin.

The reason for the isomerization is that upon electronic excitation, the system can approach a point on the potential energy surface where two or more geometries are degenerate; this point is called a conical intersection. That is, after exciting 11-cis-retinal, certain internal rotation barriers are lowered in energy, the nuclei begin to respond, the system passes through a conical intersection, and further relaxation of the nuclei leads to all-trans-retinal. In general, photochemical processes are very complicated.

References

Kropf, Allen; Hubbard, Ruth. The photoisomerization of retinal. Photochem. Photobiol. 1970, 12, 249-260.
Bowmaker, J. K.; Dartnall, H. J. Visual pigments of rods and cones in a human retina. J. Physiol. 1980, 298, 501-511.
Send, Robert; Sunholm, Dage. Stairway to the Conical Intersection: A Computational Study of the Retinal Isomerization. J. Phys. Chem. A 2007, 111, 8766-8773.


Variation of current with time in gel electrophoresis

I think this might help you a bit:
https://www.researchgate.net/post/SDS_PAGE_should_be_run_at_constant_current_or_constant_voltage


Probability of finding a molecule in the ground vibrational level

You're on the right track.

Also, using $i$ as an index can be confusing some times because it can be confused with the imaginary number; however, here it should not present a problem. As a matter of habbit however, I like to use $j$ or $n$ or something else..there are only so many letters in the alphabet.

The sum in the denominator is called the partition function, and has the form

$$Z = \sum_{j}e^{-\frac{\epsilon_j}{kT}}$$

For the harmonic, oscillator $\epsilon_j = (\frac{1}{2}+j)\hbar \omega$ for $j \in \{ 0,1,2.. \}$

Note that $\epsilon_0 \neq 0$ there exists a zero point energy.

Let's write out a few terms $$Z = e^{-\frac{\hbar \omega/2}{kT}} + e^{-\frac{\hbar \omega3/2}{kT}} + e^{-\frac{\hbar \omega5/2}{kT}} +..... $$

factoring out $e^{-\frac{\hbar \omega/2}{kT}}$

$$Z = e^{-\frac{\hbar \omega/2}{kT}} \left( 1+ e^{-\frac{\hbar \omega}{kT}} + e^{-\frac{2\hbar \omega}{kT}} +.....\right) $$

The sum in the bracket takes the form of a geometric series whose sum converges as shown below $$1+x+x^2+... = \frac{1}{(1-x)} $$
 herein, $ x \equiv e^{-\frac{\hbar\omega}{kT}} $

Putting all of this together

$$Z = \frac{e^{-\frac{\hbar \omega/2}{kT}}}{(1-e^{-\frac{\hbar\omega}{kT}})}$$

Now, $$p_0 = \frac{e^{-\epsilon_0/kT}}{Z} = \frac{e^{-\frac{\hbar \omega/2}{kT}}}{Z} = \frac{e^{-\frac{\hbar \omega/2}{kT}}}{\frac{e^{-\frac{\hbar \omega/2}{kT}}}{(1-e^{-\frac{\hbar\omega}{kT}})}} = (1-e^{-\frac{\hbar\omega}{kT}}) $$


Effect of impurities on enthalpy of fusion of ice

According to the article An Empirical Correlation between the Enthalpy of Solution of Aqueous Salts and Their Ability to Form Hydrates (Pandelov et al. 2010), your observation of a lower enthalpy is correct:

  We have found a correlation between the enthalpy of solution of the salt and its ability to form a hydrate, namely, that the salt’s enthalpy of solution is lower than the standard enthalpy of fusion of ice (6 kJ/mol)

However, the drop in enthalpy you observed of 25%+ seems excessive, as can be seen in the graph below:

Latent heat of fusion comparison between water, $\ce{NaCl}$ solution and ethanol solution.

Image source: Colligative properties of water cited from the original article Formulation of the latent heat of fusion of ice in aqueous solution (Kumano et al. 2009)

KUmano et al. 2009's methodology centred around the use of

  Differential scanning calorimetry analysis was performed and the measurements and calculated values compared with respect to the specific enthalpy of the ice/solution mixture.

As for the 'mushy ice' that you observed, it is likely to be a mixture of ice and brine, as per the diagram below:

Image source: Wikipedia


Hydroboration/oxidation vs acid-catalyzed hydration of 1-phenylpropene

Hydroboration

With the simplest hydroborating agent $\ce{BH3.THF}$, the hydroboration/oxidation of 1-phenylpropene proceeds to give 1-phenylpropan-1-ol as the major product.[1] The isomer of the substrate used was not reported.

There is some ambiguity in the above source, as the labels 8 and 8′ (which the paper uses to refer to the two regioisomers) are consistently swapped; hence, I have taken the names and not the labels to be accurate. However, as additional proof, there are other examples of similar regioselectivity in the hydroboration of styrenes (here, an E/Z mixture of styrenes was used):[2]

Rationalising this selectivity is not very easy, but in one of Herbert Brown's papers where the same regioselectivity is seen with 9-BBN as the hydroborating agent, he writes:[3]

  The preferred site of hydroboration in 1-phenylpropene, the site to which boron becomes attached, is the 1-position. This is not what one would expect on the basis of steric effects. One possible explanation for this is the combined effects of phenyl conjugation (-K) and methyl hyperconjugation (+K) which act to decrease the amount of electron density at the 2-position and to increase it at the 1-position.

I'm not totally convinced by this, but I also don't have any better explanation. Further evidence for this explanation is the fact that the (Z)-isomer reacts with poorer regioselectivity than the (E)-isomer. This is because there is allylic strain in the (Z)-isomer which forces the phenyl group to twist out of conjugation with the C=C double bond (depicted below). Consequently, the electron-withdrawing effect of the benzene ring (via conjugation) is reduced, and the formation of the 2-hydroxy isomer is less disfavoured.

Acid-catalysed hydration

Examples of these in the literature are harder to find and I cannot find any generic examples of acidic hydration of $\ce{R^{1}-C=C-R^{2}}$, where $\ce{R^1}$ is an aryl group and $\ce{R^2}$ an alkyl. However, I see no particular reason to go against the conventional wisdom: the more stable carbocation intermediate is formed at the benzylic position, which should lead to the 1-hydroxy compound as the major product.

References

Camacho, C.; Uribe, G.; Contreras, R. Diphenylamine Borane, A New Stable Amine .Borane with Remarkable Hydroborating and Reducing Properties. Synthesis 1982, 1982 (12), 1027–1030. DOI: 10.1055/s-1982-30051.
Allegretti, M.; Bertini, R.; Cesta, M. C.; Bizzarri, C.; Di Bitondo, R.; Di Cioccio, V.; Galliera, E.; Berdini, V.; Topai, A.; Zampella, G.; Russo, V.; Di Bello, N.; Nano, G.; Nicolini, L.; Locati, M.; Fantucci, P.; Florio, S.; Colotta, F. 2-Arylpropionic CXC Chemokine Receptor 1 (CXCR1) Ligands as Novel Noncompetitive CXCL8 Inhibitors. J. Med. Chem. 2005, 48 (13), 4312–4331. DOI: 10.1021/jm049082i.
Brown, H. C.; Nelson, D. J.; Scouten, C. G. Hydroboration. 63. Kinetics and regiospecificity of the hydroboration of isomeric cis- and trans-alkenes via 9-borabicyclo[3.3.1]nonane (9-BBN). Effects of 1,1- and 1,2-dialkyl interactions. J. Org. Chem. 1983, 48 (5), 641–643. DOI: 10.1021/jo00153a003.


Would an (NH3)2+ molecule be trigonal planar like BH3 rather than trigonal pyramidal?

Not only is changing shape due to ionization possible, it has been observed.  Not with ammonia as suggested in the question, but with cyclooctatetraene, which is nonplanar in its neutral state but switches to a planar, aromatic ring structure when reduced to its di-anion.


Determine rate equations for entire kinetic scheme

The Biomass-equation is:
$$\frac{\mathrm d B}{\mathrm d t} = -(k_1 + k_2) B$$
The (volatile + gases)$_1$ equation is:
$$\frac{\mathrm d VG_1}{\mathrm d t} = k_1 B$$
The (volatile + gases)$_2$ equation is equal to equation for (Char)$_2$:
$$\frac{\mathrm d VG_2}{\mathrm d t} = \frac{\mathrm d C_2}{\mathrm d t} = k_3 C_1$$
(It might be that it needs to be multiplied by 1/2.)

And the equation for (char)$_1$ is this:
$$\frac{\mathrm d C_1}{\mathrm d t} = k_2 B-k_3 C_1$$

The equations for the complete set of (char)s and (volatile+gases)s are then the sums of the solutions for $C_1$ and $C_2$ resp. $VG_1$ and $VG_2$.

For initial concentration of $B(0)=1$ and everything else $f(0)=0$, as well as $k_1=2$, $k_2=10$ and $k_3=3$, you should get something similar to this:


Is zero dipole moment possible in a square pyramidal complex?

A square pyridimal structure can have a dipole moment of zero, e.g. a squarepyramidal cluster of atoms of the same element would have no dipole moment. However for most cases where the bonds are polarized, the dipole moments will not cancel as they would for other symmetries (e.g. tetrahedron). So basically, most square pyramidal complexes with polarized bonds will have a net dipole.

If you have a complex in mind just calculate the dipole computationally or look at the ir spectrum.


Where can I access raw values for near infra red absorbance of liquid water?

I found a database called HITRAN that offers what I wanted. They support a python API that can be used to pull the relevant data. The code I used is written below (you'll need the hapi.py file from here).

from hapi import *
from matplotlib import pyplot as plt
fetch('H2O',1,1,7500,11500)
nu,coef = absorptionCoefficient_Lorentz(SourceTables='H2O')
plt.plot(nu,coef)
plt.show()


Is an acid (just) a proton donor and if not, what is the correct definition of an acid?

Your teacher said that this is not completely correct as there are (at least) three common definitions for acids and bases (Wiki has a good article for this https://en.wikipedia.org/wiki/Acids_and_Bases):

Arrhenius: In water, acids under this definition protonate the water and form Hydronium (H3O+). Bases in this definition deprotonate the water.
Brønsted–Lowry: This is your proton donor definition. Acids in this definition donate protons to bases.
Lewis: In this definition acids are electron pair acceptors and bases are electron pair donors.

The relationship between these definitions is that as chemistry evolved we ran into more chemicals (acids and bases) with similar reactivities that the previous definitions didn't quite encompass and so chemists broadened the definitions. So, as you go down the list above you will find more and more chemicals that match your new definition of acid or base.


Cyclohexene reduction reaction with hydrogen gas and ethanol

Note: This is a homework question, so I will not give a complete answer.

$\ce{H2/Pd}$ is a reducing agent, and it acts so by adding hydrogen to the double bonds. Alkenes are converted to alkanes. This proceeds via a concerted pathway, so rearrangements due to carbonation shifts do not occur. Now I think you can proceed from here.

As far as those substituents are concerned, think carefully on what would happen.


Is there any reaction between amines and AgNO3?

I found this reaction between amines and silver nitrate:


Sublimation is not boiling but there is a reason for listing it instead. Sublimation is direct vaporisation from the solid. The most familiar is solid dry ice producing $\ce{CO2}$ gas. So if metaldehyde melts at $\pu{246^\circ C}$ then one might suppose that the boiling point is higher. However, look at the generic $P$ vs. $T$ diagram below. If we assume that at $P = 1000$ is normal atmospheric pressure then a compound with this $P$ vs $T$ sublimes. As we define boiling to occur only at normal pressure and be the liquid to vapour (gas) transition then this compound can never boil at normal pressure. To become a liquid high temperature and pressure is needed. So you can see why the table lists sublimation at normal pressure rather than boiling.


Your suggested method seems excessively complicated. Substitution of chloroacetone with phenol under basic conditions should give O-alkylation in reasonably good yield. The phenol will be deprotonated and the nucleophilic phenoxide ion will then undergo an $S_N2$ reaction with the primary alkyl chloride.


The term "generally octahedral" refers to a distorted octahedral arrangement of the ligands referring to the central atom (as ron already stated). In this case the distortion is due to the massively different bond lengths.

In addition to ron's answer I did a little calculation at BP86/cc-pVDZ. I did not use any symmetry constraints at first and the final geometry deviated from ideal $C_\mathrm{4v}$ geometry only slightly due to rounding errors. After symmetry constraint reoptimisation, the following parameters may be found:

\begin{array}{lrl}
\text{parameter}                & \text{value}&\\\hline
\mathbf{d}(\ce{S-Br})           & 2.039 &\mathrm{\mathring{A}}\\
\mathbf{d}(\ce{S-F_{ax}})       & 1.638 &\mathrm{\mathring{A}}\\
\mathbf{d}(\ce{S-F_{eq}})       & 1.650 &\mathrm{\mathring{A}}\\\hline
\angle(\ce{F_{ax}-S-Br})        & 180.0 &{}^{\circ}\\
\angle(\ce{F_{eq}-S-Br})        &  90.7 &{}^{\circ}\\
\angle(\ce{F_{ax}-S-F_{eq}})    &  89.3 &{}^{\circ}\\
\angle(\ce{F_{eq}-S-F'_{eq}})   &  90.0 &{}^{\circ}\\
\angle(\ce{F_{eq}-S-F''_{eq}})  & 178.7 &{}^{\circ}\\\hline
\angle_\mathrm{D}(\ce{F_{eq}-S-F'_{eq}-F''_{eq}})
                                & 178.7 &{}^{\circ}\\\hline
\end{array}

I was unable to determine a interconversion barrier related to the $\angle(\ce{F_{eq}-S-Br})$ bond angle change. I was also not able to find a different stable isomer.

In solid state it is most likely, that the geometric parameters will change due to packing effects and/or more internal entropy.

Addressing the edit:

  Are you sure there are 8 F-S-F bond angles < 90 degrees? I'm can see how the Br pushes down on the fluorines in the equatorial plane, creating four bond angles < 90 degrees. Where's the other four? Or are these from taking the Br to be in the equatorial plane?

Yes, the bromine pushes down the fluorines of the equatorial plane resulting in $\angle(\ce{F_{ax}-S-F_{eq}})=89.3^{\circ}$ bond angle. Due to this, the sulfur is out of plane with the equatorial fluorine atoms. Since the planes expanded by $\ce{Br-S-F_{eq}}$ and $\ce{Br-S-F'_{eq}}$ have to be at $90^\circ$ due to symmetry constraints, this is creating also a slightly smaller angle than $90^\circ$ for the equatorial fluorines. In this very case it can be found to be $\angle(\ce{F_{eq}-S-F'_{eq}})=89.992^{\circ}$, which was rounded by the program to fit the one decimal place restriction. The application of this is just trigonometry and you can derive it as a simple thought experiment - if you like.


What are the applications of quantum field theory to chemistry?

Have a look at this article: "The Physics behind Chemistry and the Periodic Table"
Pekka Pyykkö Chem. Rev. 2012 112 (1), 371-384
(http://pubs.acs.org/doi/abs/10.1021/cr200042e)
I think it answers your question and has most of the references to work done in that particular field. For most questions in chemistry the non-relativistic Schrödinger Equation is good enough. Even for systems with an appreciable amount of relativistic effects, the Dirac Equation is often not solved directly. So, the number of applications where Quantum Electrodynamics is used is really limited.