Why is N-acetyl alanine methyl amide considered a dipeptide?

The nomenclature here is based on counting the number of backbone peptide bonds rather than the number of aminoacids. This does not apply the usual naming convention.
If you look at the wikipedia page on "peptide (amide) bond", it states

A peptide bond is an amide type of covalent chemical bond linking two consecutive alpha-amino acids from C1 (carbon number one) of one alpha-amino acid and N2 (nitrogen number two) of another along a peptide or protein chain.
It can also be called an eupeptide bond to separate it from an isopeptide bond, a different type of amide bond between two amino acids.

Furthermore,

When two amino acids form a dipeptide through a peptide bond it is type of condensation reaction.
[...] The two joined amino acids are called a dipeptide.

Note that you can have aminoacids connected through their side chains. Glutathione (GSH) comes to mind as a good example of this:

It contains three aminoacids (Gly, Cys and Glu). Would you call this a "tripeptide"?
The wikipedia does:

It is a tripeptide with a gamma peptide linkage between the carboxyl group of the glutamate side chain and the amine group of cysteine, and the carboxyl group of cysteine is attached by normal peptide linkage to a glycine.

What about the bonds in GSH? The wikipedia goes on:

[...] the tripeptide glutathione is synthesized in two steps from free amino acids, by two enzymes: glutamate–cysteine ligase (forms an isopeptide bond [emphasis mine], which is not a peptide bond) and glutathione synthetase (forms a peptide bond).

So, here they call GSH a tripeptide based on the number of aminoacids, not the number or even type of "peptide bond".
Hmm, pretty confusing. Perhaps then it is best to go to the original literature on the subject, linked in the wikipedia:

"Nomenclature and Symbolism for Amino Acids and Peptides. Recommendations 1983". European Journal of Biochemistry. 138 (1): 9–37. 1984.

"Glossary of terms used in physical organic chemistry (IUPAC Recommendations 1994)". Pure and Applied Chemistry. 66 (5): 1077–1184.

According to reference 1:

Part 1, Section C : PEPTIDE NOMENCLATURE
3AA-11. DEFINITION OF PEPTIDES
A peptide is any compound produced by amide formation between a carboxyl group of one amino acid and an amino group
of another. The amide bonds in peptides may be called peptide bonds. The word peptide usually applies to compounds whose
amide bonds are formed between C-1 of one amino acid and N-2 of another (sometimes called eupeptide bonds), but it includes
compounds with residues linked by other amide bonds (sometimes called isopeptide bonds). Peptides with fewer than about 10-
20 residues may also be called oligopeptides ; those with more, polypeptides. Polypeptides of specific sequence of more than about
50 residues are usually known as proteins, but authors differ greatly on where they start using this term.

There is some degree of ambiguity regarding the numerical prefixes to the term "peptide", which are not discussed explicitly. But the above article provides a solid guideline about what you should call a peptide, namely two or more aminoacids linked through amide bonds.
So what about AcAlaNHMe (the capped alanine peptide in question)? One reason it is called a dipeptide (and not a peptide or tripeptide) is to emphasize that it contains two peptide (or amide) plains with intervening $\phi$ and $\psi$ dihedral angles which confer the potential conformational flexibility of a regular tripeptide. This is discussed for instance in this article.


What units do I use for energy in the Boltzmann distribution?

Any quantity that is used as a power has to be dimensionless (that is, a pure number with no units). This means that $kT$ has to have the same units as $E_i-E_j$, so the terms in the exponent cancel. If you're measuring temperature in $\pu{K}$ and Boltzmann's constant in $\pu{J/K}$, then the units of energy you have to use are joules, $\pu{J}$.


Are all reactions with ΔG < 0 spontaneous?

No, Gibbs energy is only the relevant state function in the special case of constant temperature and pressure.

In the wording of IUPAC:

  partial derivative of Gibbs energy with respect to extent of reaction at constant pressure and temperature [is negative] for spontaneous reactions

In the special case of constant temperature and volume, Helmholtz energy is the relevant state function.

Most generally, a process is spontaneous if the entropy of the universe (system + surroundings) increases.


Is adding the word "solution" necessary when listing the materials used in a synthesis?

No, the repeated addtion of the word "solution" is not needed.

All you want here is to provide a measured quantity (e.g., volume), concentration of your reagent in the solvent, and inferred quantity (e.g., mmol).  This may be reported in the following pattern, too:

  "1-Cinnamyl-4-methylbenzene (4a): The compound 4a was synthesized from
  1-methoxy-4-{[(2 E)-3-phenylprop-2-en-1-yl]oxy}benzene 2b ($\pu{240.9 mg}$,
  $\pu{1.00 mmol}$) with p-tolylmagnesium bromide 3a ($\pu{1.4 mL}$, $\pu{1.1 M}$ in THF, $\pu{1.5 mmol}$) in the presence of 1b ($\pu{4.1 mg}$, $\pu{1 mol\%}$) in $\ce{Et2O}$ ($\pu{5 mL}$) at room temperature for $\pu{24 h}$ ($\pu{198.0 mg}$, $\pu{95\%}$ yield, colorless liquid)."

(source: Hashimoto et al. in Molecules 2019, 24, 2296 (doi.org/10.3390/molecules24122296), section 3.2; open access publication.)

This pattern is seen for inorganic reagents like $\ce{HCl}$ dissolved in water, yet equally available in methanol, 1,4-dioxane, etc. (example); and organometallic reagents, e.g. butyllithium, methylmagnesium bromide.


Non-halogen intermediate between alkane and alcohol under basic conditions?

I think that you're asking if there is a way to convert an alkane to an alcohol without using halogens in the process.

Alkanes are relatively unreactive and so high energy species (like free radicals) are most often used to carry out reactions on alkanes. The  reaction of alkanes with halogens that you've mentioned is an example of this free radical process.

Other high energy molecules that can be used to convert alkanes to alcohols include ozone and (photo-) chemically generated singlet oxygen.  Just like with the halogens, these reactive molecules show little selectivity, so if primary, secondary and tertiary hydrogens co-exist in your alkane, then a mixture of primary, secondary and tertiary alcohols result.  However in molecules such as cyclohexane, where there is only one type of hydrogen, somewhat useful results can be obtained.  I say "somewhat useful" because ozone and singlet oxygen can oxidize the resultant alcohol further to a carbonyl.  Here is a reference to an abstract on the use of ozone to functionalize alkanes.

Reagents such as potassium permanganate and chromic acid also oxidize alkanes by a free radical mechanism to produce alcohols.  Again the issues of selectivity and further oxidation remain.

Here is a link to a nice review entitled "Alkane Hydroxylation." It covers a wide variety of reagents.  Most of the molecules depicted contain functional groups which can help improve selectivity, but there are a number of hydrocarbon examples as well.  Perhaps you'll find an example that is similar to your molecule of interest.


Plotting the data using Beer-Lambert law

You need to be aware of the dilution formula in order to plot concentration on the x-axis and absorbance on the y-axis:
$C_iV_i = C_fV_f$
So your initial concentration is your stock concentration and your initial volume is listed in the table. Your final volume is $\pu{100 mL}$. Can you determine $C_f$? With that you would be able to calculate all 5 different concentrations.
Keep in mind, that you need to fix the wavelength of measurement. Since the spectrum is very broad, $\pu{1 nm}$ will not make a huge difference. You may not have perfect linearity.


How does mass affect bond stretching in UV spectroscopy?

In a very crude approximation, every bond can be considered to be a quantum harmonic oscillator. As you might know, the energies of such an oscillator are given by

$$ E_\nu = hc\omega_e \left ( \nu+\tfrac{1}{2}\right )$$

where $h$ is Planck's constant, $c$ the speed of light and $\omega_e=\sqrt{k/\mu}$. The constant $k$ is the force constant of the bond and $\mu$ is the reduced mass
$$ \mu=\frac{m_1m_2}{m_1+m_2}$$
So if you increase the mass of one of the particles, while $k$ stays roughly the same, $\omega_e$ becomes smaller and the energy spacing between the vibrational levels decreases.


According to Wikipedia, "When dissolved in a large amount of neutral or slightly alkaline water, aluminium sulfate produces a gelatinous precipitate of aluminium hydroxide, Al(OH)3."


Is flour or baking soda really suitable for extinguishing grease fires?

Do not use flour, it may explode!

Use the pot lid to smother the fire.
Use a fire blanket - read the instructions, watch a you tube demo. People who think they know what they're doing often spread it or end up burned.
Salt, will put out small fires, such as a pan fire, you won't have enough salt to smother a pot or absorb the heat.
Good for putting out small fat flare ups on a barbie plate.


Molar absorption coefficient of H2SO4(aq) for UV-vis spectrophotometry

I never ran into uv absorption data for (aqueous) sulfuric acid and I doubt that there is any absorption at $\lambda$ = 508 nm.

There is an old paper by K. F. Palmer and D. Williams in Appl. Optics, 1975,14, 208-219 (DOI) on the optical properties of sulfuric acid droplets with respects to the composition of clouds on Venus and a more recent one from the group of J. B. Burkholder in Phys. Chem. A, 2003, 107, 1112-1118
(DOI) on Vibrational and Electronic Spectroscopy of Sulfuric Acid Vapor. Both, while interesting and therefore mentioned here, are not relevant for your experiment.

My suggestion is simple:

Measure your ferroin sample in aqueous sulfuric acid (or whatever solvent is required) and use the same solvent as a blank. Most spectrometers will automatically perform the subtraction of the blank from your sample!

Concerning a (free) source of UV-vis absorption data, there is the NIST Chemistry WebBook.


Carbene mesomeric structure of vinyl carbocation?

The alternative answer to what @levineds said is that you're asking about a carbene. Carbenes are actually reasonably common in chemistry though we generally have them coordinated to metals for stability. There are a few reactions that proceed through putative naked carbenes.


Does deuterium exhibit hydrogen bonding?

(Based on ron's answer here on the inductive effect accorded by deuterium - do give that a read, and consider upvoting it)

Since deuterium has a higher mass than protium, simple Bohr theory tells us that the deuterium 1s electron will have a smaller orbital radius than the 1s electron orbiting the protium nucleus.
The smaller orbital radius for the deuterium electron translates into a shorter (and stronger) $\ce{C-D}$ bond length.
 See the bottom half of ron's answer for the proof

In your case, that would mean a shorter $\ce{N/O/F-D}$ bond.

A shorter bond has less volume over which to spread the electron density (of the 1 electron contributed by $\ce{H}$ or $\ce{D}$) resulting in a higher electron density throughout the bond, and, consequently, more electron density at the carbon end of the bond. Therefore, the shorter $\ce{C-D}$ bond will have more electron density around the carbon end of the bond, than the longer $\ce{C-H}$ bond.
The net effect is that the shorter bond with deuterium increases the electron density at carbon, e.g. deuterium is inductively more electron donating than protium towards carbon.

So we can expect the $\ce{N/O/F-D}$ bond to be smaller and more polar than the corresponding bond with protium.
Karl very astutely pointed out that the boiling point of heavy water is higher than that of normal water. In light of the inferences drawn earlier, I took the liberty of interpreting this fact as being (at least somewhat) indicative of deuterium permitting a stronger hydrogen bond than protium.
As Linear Christmas pointed out, the variations in mass across different isotopes is known to affect boiling point (as in the case of helium). This is likely to contribute to some degree, to the increase in boiling point in heavy water. However, I'm still of the opinion that the higher dipole moment of $\ce{D2O}$ over $\ce{H2O}$ is the main cause for the higher boiling point.

Finis
Hydrogen bonding with deutrium does occur and should be, in theory (drawn from ron's answer), stronger than that with protium. The fact that heavy water $\ce{D-O-D}$ is more polar (and hence has a higher boiling point) than regular water $\ce{H-O-H}$, appears to be a consequence of this.


What is the difference between an element and an atom?

The wording in your quote is a bit unclear, I agree. A very nice definition is given by the Jefferson Lab page What is the simplest way of explaining what atoms, elements, compounds and mixtures are?

Atoms

  are the smallest bits of ordinary matter

Whereas an element is

  a substance that is made entirely from one type of atom

Atoms themselves can be 'broken down' into smaller sub-atomic particles (protons, neutrons and electrons) - the amounts of these (particularly the number of protons) define each element.

For example, in their un-ionised form, the element carbon is made entirely from carbon atoms, which have 6 protons and 6 electrons each, and cadmium atoms, which have 48 protons and 48 electrons, make up the element cadmium.

The amount of neutrons can vary with atoms of a particular element, forming isotopes, which (from the link):

  For example, carbon-12, carbon-13 and carbon-14 are three isotopes of the element carbon with mass numbers 12, 13 and 14 respectively. The atomic number of carbon is 6, which means that every carbon atom has 6 protons, so that the neutron numbers of these isotopes are 6, 7 and 8 respectively.

Different arrangements of atoms of single elements can form allotropes, such as for carbon, with diamond and graphite (among others - shown below). They both are composed of atoms of the element carbon, but in different configurations.

Below are diagrams of the allotropes of carbon featured in the article Work Function Engineering of Graphene (Garg et al. 2014)


Use of pipette during experiments for determination of oxygen by Winkler's method

[...] the pipette tip should be well below the surface of the solution [...]

Usually, one would indeed keep the pipette above the solution to avoid that a (minor) part of the solution forms a drop (or film) on the tip of the pipette and is carried away.

Here, when the amount of oxygen dissilved in the solution is to be determined, another effect prevails.

If you add your reagents from greater height, you might create a splash, just like throwing a rock into a lake. As a result, you will mix some oxygen from the gas phase into the solution, which might taint the measurement.


Does a covalent bond constitute of 1 or 2 electrons?

The bond in your picture is made up of two electrons, and in the Lewis formalism so is every single bond. Yet there is no negative charge. We can think of this in two ways:

Remember that the bond electrons are shared. So each atom contributed one electron and, upon dissociation, would usually take back that one electron. Counting only that one electron will give a neutral hydrogen atom, not a hydride ion.

Charges on an atomic or molecular level are due to a mismatch of the numbers of protons and electrons. In the hydrogen molecule, you have two nuclei consisting of one proton each, meaning an overall $2+$ charge. Furthermore, you have two electrons that get shared in the covalent bond giving an overall $2-$ charge. $+2 + -2 = 0$, so the overall molecule is neutral.
If you wanted to place both electrons on one atom to form a hydride anion, that would mean the other atom is left with zero. Instead of a hydrogen molecule $\ce{H-H}$ you would have a separated ion pair $\ce{\overset{-}{H}\bond{...}\overset{+}{H}}$.

Once you leave the very simplified Lewis formalism and turn to theories such as molecular orbital theory which explain molecules much more accurately, this way of thinking will be lost altogether since there is no longer an obvious localisation of two electrons to a single bond.


What is the hybridisation in HCN molecule?

$\ce{HCN}$ and $\ce{HC#CH}$ are linear, triple bonded, with a $π$ system consisting of two perpendicular $π$ bonds. They would be symmetrical in $\ce{HC#CH}$, and slightly distorted in $\ce{HCN}$, and they leave two orbitals for the sigma system.
In $\ce{HCN}$, we hybridize/combine the two remaining orbitals on the carbon atom to form two bonding orbitals, one to the hydrogen, another to the atom on the other side of the carbon (a $\ce{C}$ or an $\ce{N}$). The natural first combination is a $50$–$50$ split to form two $\mathrm{sp}$ orbitals, one directed to the $\ce{H}$ and the other directed toward the $\ce{N}$.
That's enough most of the time, but if you get picky, you could point out that the electronegativities of $\ce{H}$ and $\ce{N}$ are quite different ($\ce{H}$ $2.1$, $\ce{C}$ $2.5$, $\ce{N}$ $3.0$), so the nitrogen will be pulling on its $\mathrm{sp}$ bond more than hydrogen pulls on its $\ce{sp}$ bond, so the $50$-$50$ split readjusts to maybe $70$-$30$ (did your teacher say $~100\%$-$0\%$?), meaning that the hydrogen gets less of the carbon orbital (i.e., less $\mathrm p$-character, more $\mathrm s$-character from the carbon orbitals), while the nitrogen gets more of carbon's $\mathrm p$-orbital.
If this is so, the $\ce{H}$ atom in $\ce{HCN}$ should be more easily removed than a H atom in $\ce{HC#CH}$. This appears to be correct: the $pK_\mathrm a$ of acetylene is $24$ (which is considered to be quite acidic for a hydrocarbon), whereas the $pK_\mathrm a$ of hydrocyanic acid is $9.21$, much more acidic (although a weak acid by any other measure).
Your teacher pointed out something interesting, but the exact split ratio remains to be calculated.


What is the mechanism of the reaction between hexane-2,5-dione and nitrous acid?

As you may know from diazotisation reactions, $\ce{HNO2}$ is a generator of the nitrosonium cation, $\ce{NO+}$. Electrophiles like these can react with the enol tautomer of hexanedione:

Following this, you tautomerise the nitroso compound to the oxime. The mechanism is exactly analogous to keto-enol tautomerism:

Note that I've elected to draw the oxime with a (E)-configuration. Both (E) and (Z) oximes can form, but the (Z)-configured oxime cannot cyclise in the following step to form an isoxazole:

Following this, to get to the compound in your question, you supposedly need to oxidise the acetyl substituent ($\ce{COCH3}$) to a carboxyl substituent ($\ce{COOH}$). Nitrous acid isn't an oxidant, which casts doubts on whether your source is veritable (perhaps it's just a typo).

In the literature, I find that this oxidation is accomplished using nitric acid, $\ce{HNO3}$, which is a perfectly fine oxidant. In one case (J. Heterocycl. Chem. 2009, 46, 909), the diketone is first treated with nitrous acid to get the 3-acetylisoxazole, and is then treated with nitric acid to form the 3-carboxylisoxazole. In a separate case (J. Med. Chem. 2004, 47, 3642), the diketone is simply refluxed with nitric acid to form the 3-carboxylisoxazole. I suspect that this works because nitrous acid is present in the nitric acid, so the mechanistic pathway is exactly the same (a similar effect has been observed with the nitration of phenols using dilute nitric acid).

Exactly how nitric acid oxidises an acetyl group to a carboxyl group, though, I don't know (and I'm not sure if it is known).

Edit: There is, in fact, a mechanism by which nitrous acid may oxidise the acetyl group to a carboxyl; see user55119's answer.


Let's be positive. The edited proposal has merit. You can pass a current thru a solution of a carboxylic acid and measure it with an ammeter (a more sensitive milliammeter would be better).
How about this: check for conductivity. Commercial conductivity meters use AC current and an AC ammeter. Use two inert, identical electrodes for this, and use DC if you can't arrange an AC apparatus. Do a conductivity measurement (i.e., measure the ohms resistance of the solution with the two electrodes taped to a certain distance separation (e.g., two copper wires (or stainless steel) 1 cm apart, 1-2 cm immersion, taped to a plastic ruler) so you can do the identical measurement on another solution.
Then, for a second test, examine the conductivity of that same solution diluted by a factor of 10. Now here's where you are on your own: the conductivity will be less, but not exactly 10 times less, because as the carboxylic acid is diluted, it will ionize to a greater percentage, and therefore carry more current at the same voltage than its concentration reduction predicts.
I say you are on your own, because the experimental results will not be as clear as you might wish (lots of opportunity for scatter), but this will enable you to explore the physical chemistry idea of the ionization of carboxylic acids and dilution and the question of voltage vs current, DC vs AC, electrode composition (reactivity); also bring up pH.
Above all, keep it simple! Consider two variables and plot one against another. You would be surprised how difficult that will be and how there are many more variables just waiting to hop into your experiment!
But you will have fun learning about all the variables and writing up the results and impressing your teacher with your comprehension. Good luck!


Why does hydrogen form on such a long time scale?

Why is the time-scale so long?

The time-scale is so long because the density is not just low, it's virtually nonexistent.

The particle density of any gas can be determined at any temperature and pressure using the ideal gas law ($PV=nRT$). Let's use air as an example. At $\pu{25 o^\circ C}$ ($\pu{298 K}$), and 1 atmosphere, the particle density of air ($n/V$) is:

$$ \frac{n}V =\frac{RT}P $$
$$ \frac{n}V =\frac{(\pu{0.08026 L atm (K mol)^{-1})(\pu{298  K})}}{{\pu{1 atm}}} = \pu{23.9 mol L^-1} $$

Converting to molecules per $\pu{cm^3}$:

$$23.9 \frac{\mathrm{mol}}{\mathrm{L}} \times 6.022\times10^{23}\frac{\text{molecules}}{\text{mol}}\times\frac{1 \text{ L}}{1000 \text{ cm}^3}=1.44\times10^{22}\frac{\text{molecules}}{\text{cm}^3}$$

The density of gasses we are used to dealing with is 18 orders of magnitude higher than the density you quoted. If we have a gas of hydrogen atoms at air density, then the timescale for formation of H2 is:

$$ 10^{15} \frac{\text{years}\times\text{cm}^3}{\text{particles}} \times  \frac{1 \text{ cm}^3}{1.44\times10^{22}\text{ molecules}}=6.94\times 10^{-8}\text{ years} $$

$$6.94\times 10^{-8}\text{ years}= 2.54\times 10^{-5}\text{ days}=6.08\times 10^{-4} \text{ hours}=2.19 \text{ seconds}$$

At "normal" densities, the timescale is much shorter.

The second part of your question deals with the quantum mechanics of the activation barrier to this reaction, which I cannot do justice to at this time.


Is this heterocyclic boron molecule aromatic?

That molecule is indeed aromatic.

It satisfies Huckel's rule ($4n+2$) with $n=0$. It's also planar.

Lastly, the boron atom has three electron domains, meaning it can exist with $sp^2$ hybridization. The left-over p-orbital, whilst not actively forming $\pi$  bonds, can still complete a stable $\pi$ molecular orbital.


How can you predict a double displacement reaction with no precipitate?

If there is no precipitate, then there is nothing to drive the reaction. Both possible products would be soluble, and so there is no net reaction (it's just the reactant ions in solution).


How does haloform reaction take place if the reactants NaOH and X2 react with each other?

The disproportionation of halogens $(1)$ is an equilibrium reaction; depending on the concentrations of reactants and products it can go both ways. Since hydroxide (or more generally: a base) is among the reactants, increasing pH will drive the reaction towards the product side.

$$\ce{3 Br2 + 6 OH- <=>> 5 Br- + BrO3- + 3 H2O}\tag{1}$$

The first step of the haloform reaction is to enolise a ketone to the corresponding enole $(2)$. This is also an equilibrium lying on the reactants’ side this time.

$$\ce{RC(=O)-CH3 + OH- <<=> RC(-O^-)=CH2 + H2O}\tag{2}$$

So you are correct, both equilibria should be away from what is required for the haloform reaction to proceed. However, both are equilibria. For both, we can measure an equilibrium constant and use it to calculate the concentrations of both the enolate and the free dihalogen. I don’t have all the numbers here, but the result will be that both reactants are present in sufficient concentrations to drive the reaction forward. Remember that the full haloform reaction is no longer an equilibrium process; each bromination and the final separation is a non-reversible step; see $(3)$.

$$\begin{gathered}\ce{RC(-O^-)=CX2 + Br2 -> RC(=O)-CX2Br + Br-}\\[0.8em]
\ce{RC(=O)-CBr3 + OH- <<=> RC(-O^-)(-OH)-CBr3} \\[0.4em] \ce{-> RC(=O)-OH + CBr3- -> RCOO- + HCBr3}\end{gathered}\tag{3}$$

The mere presence of non-reversible steps will drive any reaction forwards.


What chemical reaction happens on worn out metal handles?

It looks like a brass ( a class of alloys, containing mostly copper and zinc, being golden-like ), which surface was covered by protective plating , possibly by chrome or nickel(*). It was probably worn out mechanically and/or by corrosion due acidic sweat.
(*)Nickel plating could be worn out easier than hard and resistant chrome.


Why is ligand substitution only partial with copper(II) ions and ammonia?

Metal ion complexes have stepwise stability constants:
\begin{align}
\ce{[Cu(H2O)6]^2+ + NH3 &<=>[$K_1$] [Cu(NH3)(H2O)5]^2+ + H2O}\\
\ce{[Cu(NH3)(H2O)5]^2+ + NH3 &<=>[$K_2$] [Cu(NH3)2(H2O)4]^2+ + H2O}\\
\ce{[Cu(NH3)2(H2O)4]^2+ + NH3 &<=>[$K_3$] [Cu(NH3)3(H2O)3]^2+ + H2O}\\
\ce{[Cu(NH3)3(H2O)3]^2+ + NH3 &<=>[$K_4$] [Cu(NH3)4(H2O)2]^2+ + H2O}\\
\ce{[Cu(NH3)4(H2O)2]^2+ + NH3 &<=>[$K_5$] [Cu(NH3)5(H2O)]^2+ + H2O}\\
\ce{[Cu(NH3)5(H2O)]^2+ + NH3 &<=>[$K_6$] [Cu(NH3)6]^2+ + H2O}\\
\end{align}
with

\begin{align}
K_1 &= \frac{{\left[ \ce{[Cu(NH3)(H2O)5]^2+} \right]}}{{\left[ \ce{[Cu(H2O)6]^2+} \right]\left[ \ce{NH3} \right]}}\\
K_2 &= \frac{{\left[ \ce{[Cu(NH3)2(H2O)4]^2+} \right]}}{{\left[ \ce{[Cu(NH3)(H2O)5]^2+} \right]\left[ \ce{NH3} \right]}}\\
K_3 &= \frac{{\left[ \ce{[Cu(NH3)3(H2O)3]^2+} \right]}}{{\left[ \ce{[Cu(NH3)2(H2O)4]^2+} \right]\left[ \ce{NH3} \right]}}\\
K_4 &= \frac{{\left[ \ce{[Cu(NH3)4(H2O)2]^2+} \right]}}{{\left[ \ce{[Cu(NH3)3(H2O)3]^2+} \right]\left[ \ce{NH3} \right]}}\\
K_5 &= \frac{{\left[ \ce{[Cu(NH3)5(H2O)]^2+} \right]}}{{\left[ \ce{[Cu(NH3)4(H2O)2]^2+} \right]\left[ \ce{NH3} \right]}}\\
K_6 &= \frac{{\left[ \ce{[Cu(NH3)6]^2+} \right]}}{{\left[ \ce{[Cu(NH3)5(H2O)]^2+} \right]\left[ \ce{NH3} \right]}}\\
\end{align}

The overall stability constant is:

$$K_\text{B} = \frac{{\left[ \ce{[Cu(NH3)6]^2+} \right]}}{{\left[ \ce{[Cu(H2O)6]^2+} \right]\left[ \ce{NH3} \right]^6}} = K_1 \cdot K_2 \cdot K_3 \cdot K_4 \cdot K_5 \cdot K_6$$

(In particular, note the exponent 6.)

Furthermore, usually $K_i/K_{i+1}>1$; thus, $K_1 > K_2 > K_3 > K_4 > K_5 > K_6$. Consequently, it gets more and more difficult to exchange further ligands.

Therefore, the predominating complex strongly depends on the concentration of $\ce{NH3}$.
And you probably work with dilute aqueous solutions.

Simply speaking, the missing $\ce{[Cu(NH3)6]^2+}$ in aqueous solution can be attributed to the high concentration of water which is competing with ammonia for the coordination sites.


Find molar concentration of sodium carbonate dissolved in water

Use units! Consistently, properly, all the time!

You have units all over the place sometimes fitting the quantity you are calculating, most often not. If you approached the problem with correct use of units, there would be no question.

First, the molar mass.

$$M(\ce{Na2CO3}) = 105.99~\mathbf{\frac{g}{mol}}$$

Grams are not equal to grams per mole. Then, let’s do the rest without losing or inventing units as we go along. (I bolded the units for you.)

$$n(\ce{Na2CO3}) = \frac{m(\ce{Na2CO3})}{M(\ce{Na2CO3})} = \frac{5.3~\mathbf{g}}{105.99~\mathbf{\frac{g}{mol}}} = 0.05~\mathbf{mol}$$

$$c(\ce{Na2CO3}) = \frac{n(\ce{Na2CO3})}{V} = \frac{0.05~\mathbf{mol}}{250~\mathbf{ml}} = \frac{0.05~\mathbf{mol}}{0.250~\mathbf{l}} = 0.2~\mathbf{\frac{mol}{l}}$$

Different versions of jumbling around the powers of 10 (especially in the millilitre part) will get you the same result if done correctly. It really bears repeating.

Use units! Consistently, properly, all the time!


Solubility of n-butanol and n-butanamine

Out of butanamine and butanol, butanol is more soluble in water due to higher electronegativity of oxygen . And as butanol  have more polar oxygen so it will be more attracted towards water and form H bond.so it is soluble but amine are miscible due to enthalpies that will release out because of electron donating nature of nitrogen.


How to obtain pure alcohol from mixture of alcohol & carbonyl? (Grignard experiment)

I rather doubt that there is a better option in general. Other possible methods are

distillation
To give an example, acetone has a boiling point of 56 °C, while tert-butanol boils at 82 °C.
column chromatography on silica

But in many cases, your suggestion to separate the compounds by crystallization is a good idea. This is in particular true when your product has a high boiling point and/or you have to purify large amounts of substance.


What is the difference between a phosphoryl group and a phosphate group?

The difference is basically one oxygen atom. If you consider monoethyl phosphate as an example, the ethyl alpha-carbon bears a phosphate group, whereas the ethoxy oxygen bears a phosphoryl group.


Is there always a limiting reagent in every equation?

An interesting philosophical question. We are given a chemical equation: $$\ce{CaCl2 + Na2CO3 -> 2NaCl  + CaCO3}$$ and that the same number of moles of the reactants are used. Now the OP poses the question "Is there a limiting reagent?"

As chemists we can say that 1.00 moles of $\ce{CaCl2}$ were reacted with 1.00 moles of $\ce{Na2CO3}$. We'd pragmatically say that the moles were equal and that there was no limiting regent.

However this is a significant figure argument, not reality. Avogadro's number is about $6.022\times10^{23}$. There is no lab in the world that can weight out precisely one mole of $\ce{CaCl2}$ accurate to 23 decimal places. We don't even know Avogadro's number accurate to 23 decimal places! The best value is $6.022140857(74)\times10^{23}$.

So in reality, despite our best efforts as chemists, one reagent or the other will be in some small excess which we'd consider to be insignificant given the overall precision of the experiment.


Calculate pH of a weak acid and weak base neutralization reaction

You have a system of an acid, an ampholyte and a base. As you are interested in the pH value, your main question is what does "produce" or "consume" them. The proton concentration for your specific system is (citric Acid, sodium Bicarbonate, sodium Carbonate):
$$\ce{[H+] = [OH-] + \underbrace{[H2A-] + 2[HA^2-] + 3[A^3-]}_{citric acid's bases} \underbrace{+ [B^2-] - [H2B]}_{bicarbonate's deprotonated\\ and protonated form} \underbrace{- [CH+] - 2 [CH2^2+]}_{carbonate's sinlgy and\\ doubly protonated form}}$$

The multipliers arise from the fact, that a n-times deprotonated/protonated ion yields/consumes n protons.

Now the question is, where you get these concentrations from. It is by solving a linear equation system for each subsystem.

Citric acid, to be deprotonated three times:
\begin{align}
\ce{[H3A] + [H2O] &<=> [H2A-] + [H3O+]\\
[H2A-] + [H2O] &<=> [HA^2-] + [H3O+]\\
[HA^2-] + [H2O] &<=> [A^3-] + [H3O+]}
\end{align}
This gives you the following linear equation system
\begin{align}
k_{a1} &= \ce{\frac{[H2A-][H3O+]}{[H3A]}} \\
k_{a1} \ce{[H3A]} &- \ce{[H2A-][H3O+]} = 0\\
k_{a2} &= \ce{\frac{[HA^2-][H3O+]}{[H2A-]}} \\
k_{a2} \ce{[H2A-]} &- \ce{[HA^2-][H3O+]} = 0\\
k_{a3} &= \ce{\frac{[A^3-][H3O+]}{[HA-]}} \\
k_{a3} \ce{[HA-]} &- \ce{[A^3-][H3O+]} = 0\\
\ce{[H3A] + [H2A-] &+ [HA2-] + [A3-] = [H3A]0} = c_1\\
\end{align}
Which solves to:
\begin{align}
\ce{[H3A]} &= \frac{c_1 x^3}{k_{a1} k_{a2} k_{a3} + k_{a1} k_{a2} x + k_{a1} x^2 + x^3} \\
\ce{[H2A-]} &= \frac{c_1 k_{a1} x^2}{k_{a1} k_{a2} k_{a3} + k_{a1} k_{a2} x + k_{a1} x^2 + x^3} \\
\ce{[HA^2-]} &= \frac{c_1 k_{a1} k_{a2} x}{k_{a1} k_{a2} k_{a3} + k_{a1} k_{a2} x + k_{a1} x^2 + x^3} \\
\ce{[A^3-]} &= \frac{c_1 k_{a1} k_{a2} k_{a3}}{k_{a1} k_{a2} k_{a3} + k_{a1} k_{a2} x + k_{a1} x^2 + x^3}
\end{align}

Sodium bicarbonate, to be protonated or deprotonated:
$$
\ce{[HB^-] + [H2O] <=> [B^2-] + [H3O+]\\
[HB^-] + [H2O] <=> [H2B] + [OH-]}
$$
\begin{align}
k_{a4} &= \ce{\frac{[B^2-][H3O+]}{[HB-]}} \\
k_{a4} \ce{[HB-]} &- \ce{[B^2-][H3O+]} = 0\\
k_{b1} &= \ce{\frac{[H2B][OH-]}{[HB-]}} \\
k_{b1} \ce{[HB-]} &- \ce{[H2B][OH-]} = 0\\
\ce{[H2B] &+ \ce{[HB-] + [B^2-] = [B]0}} = c_2\\
\end{align}
Which solves to:
\begin{align}
\ce{[H2B]} &= \frac{c_2 x k_{b1}}{x k_{b1} + x k_W/x + k_{a4} k_W/x}
       = \frac{c_2 x k_{b1}}{x k_{b1} + k_W + k_{a4} k_W/x} \\
\ce{[HB-]} &= \frac{c_2 x k_W/x}{x k_{b1} + x k_W/x + k_{a4} k_W/x}
       = \frac{c_2 k_W}{x k_{b1} + k_W + k_{a4} k_W/x} \\
\ce{[B^2-]} &= \frac{c_2 k_{a2} k_W/x}{x k_{b1} + x k_W/x + k_{a4} k_W/x}
        = \frac{c_2 k_{a2} k_W/x}{x k_{b1} + k_W + k_{a4} k_W/x}\\
\end{align}

Sodium carbonate, to be doubly protonated:
$$
\ce{[C] + [H2O] <=> [CH+] + [OH-]\\
[CH+] + [H2O] <=> [CH2^2+] + [OH-]}
$$
\begin{align}
k_{b2} &= \ce{\frac{[CH+][OH-]}{[C]}} \\
k_{b2} \ce{[C]} &- \ce{[CH+][OH-]} = 0\\
k_{b3} &= \ce{\frac{[CH2^2+][OH-]}{[CH+]}} \\
k_{b3} \ce{[CH+]} &- \ce{[CH2^2+][OH-]} = 0\\
\ce{[C] &+ [CH+] + [CH2^2+] = [C]0} = c_3\\
\end{align}
Which solves to:
\begin{align}
\ce{[B]} &= \frac{c_3 (k_W/x)^2}{k_{b2} k_{b3} + k_{b2} k_W/x + (k_W/x)^2} \\
\ce{[BH+]} &= \frac{c_3 k_{b2} k_W/x}{k_{b2} k_{b3} + k_{b2} k_W/x + (k_W/x)^2} \\
\ce{[BH^2+]} &= \frac{c_3 k_{b2} k_{b3}}{k_{b2} k_{b3} + k_{b2} k_W/x + (k_W/x)^2}\\
\end{align}

Putting it all together (it looks a bit different, though, as it is copied from a simplified form of Mathematica):
$$x=\frac{k_W}{x} +
c_1 \frac{3 k_{a1} k_{a2} k_{a3}+2 x k_{a1} k_{a2}+x^2 k_{a1}}{k_{a1} k_{a2} k_{a3}+x k_{a1} k_{a2}+x^2 k_{a1}+x^3} +
c_2 \left(\frac{k_{a4} k_W}{k_{a4} k_W+x^2 k_{b1}+x k_W}-\frac{x^2 k_{b1}}{k_{a4} k_W+x^2 k_{b1}+x k_W}\right) -
c_3 \frac{2 x^2 k_{b2} k_{b3}+k_W x k_{b2}}{x^2 k_{b2} k_{b3}+k_W x k_{b2}+k_w^2}$$

Solving this little monster for $x$ and calculating its negative decadic logarithm gives you the wanted pH of your system, which is 5.65.

As you mention, further thinking about the problem makes it possible to simplify it a litle bit.
\begin{align}
\ce{[H+]  &= [OH-] + [H2A-] + 2[HA^2-] + 3[A^3-] + [B^2-] - [H2B] - [CH+] - 2 [CH2^2+]} \\
\ce{[H+] &= [OH-] + [H2A-] + 2[HA^2-] + 3[A^3-] + [CO3^2-]_B - [H2CO3]_B - [HCO3-]_C - 2 [H2CO3]_C} \\
x &= \frac{k_W}{x} + c_1 \frac{x^2 k_{a1} + 2 x k_{a1} k_{a2} + 3 k_{a1} k_{a2} k_{a3}}{x^3 + x^2 k_{a1} + x k_{a1} k_{a2} + k_{a1} k_{a2} k_{a3}} + \frac{-(c_2 + 2 c_3) x^2 - c_3 x k_{a4} + 2 c_2 k_{a4} k_{a5}}{x^2 + x k_{a4} + k_{a4} k_{a5}} \\
\end{align}

This does not change the resulting pH value and thus it is not more accurate in my eyes, it only simplifies (admittedly quite a bit) the equation to be solved.

If we assume your initial pH value to be exactly 6 and that is caused by CO2, then we can add it to the equation.
$$x = \frac{k_W}{x} + c_1 \frac{x^2 k_{a1} + 2 x k_{a1} k_{a2} + 3 k_{a1} k_{a2} k_{a3}}{x^3 + x^2 k_{a1} + x k_{a1} k_{a2} + k_{a1} k_{a2} k_{a3}} + \frac{-(c_2 + 2 c_3) x^2 + (c_4 - c_3) x k_{a4} + 2 (c_2 + c_4) k_{a4} k_{a5}}{x^2 + x k_{a4} + k_{a4} k_{a5}} $$

Where $c_4$ can be calculated by rearranging the proton concentration equation for solely carbonic acid:
\begin{align}
x &= \frac{k_W}{x} + c_4 \frac{x k_{a4} + 2 k_{a4} k_{a5}}{x^2 + x k_{a4} + k_{a4} k_{a5}} \\
c_4 &= \left(x - \frac{k_W}{x}\right) \left(\frac{x^2 + x k_{a4} + k_{a4} k_{a5}}{x k_{a4} + 2 k_{a4} k_{a5}}\right) \\
c_4 &= 3.21119 \cdot 10^{-6}
\end{align}

Without initial pH 6: 5.6467
With initial pH 6: 5.6465


What are the rate laws for each elementary step?

For elementary reactions, the rate law can be directly read from the chemical equation. In your last equation this is a bimolecular reaction, i.e.
\begin{align}
\ce{N + NO &-> N2O} & r &= k\cdot c(\ce{N}) \cdot c(\ce{NO})
\end{align}

To be more precise here, one molecule of $\ce{NO}$ reacts with on atom $\ce{N}$, and in a wider sense, that is two (bi) molecules reacting. The order of the reaction is two.

That being said, there are a couple of things you need to pay close attention to. An elementary reaction is always a reaction between the number of molecules given as reactants, with exactly one transition state leading to the products. If this number or reactants is greater than three it is highly unlikely to happen this way, since a concerted transition state is statistically very unlikely. An elementary reaction therefore also has to be balanced, so your first reaction needs to be:
$$\ce{2NO -> NO2 + N}$$
In this case we could see $[\ce{ON\bond{~}O\bond{~}N}]^\ddagger$ as a transition state.

I very highly doubt that your second reaction as an elementary reaction, from my point of view, there are too many bond that would need to be broken. Instead, the following reactions might be possible and/or necessary decompositions:
\begin{align}
\ce{NO2 + H2 &-> NO + OHH}\\
\ce{OHH &-> OH + H}\\
\ce{NO2 + H &-> NO + OH}\\
\ce{2OH &-> H2O + O}\\
\ce{NO + O &-> NO2}\\\hline
\ce{NO2 + H2 &-> NO + H2O}
\end{align}
And there are more possibilities.

In all but one the elementary reactions are bimolecular - guess which one would be unimolecular.

As for the mechanism itself, I think these are reactions that occur, just not all of them, maybe not even the majority of them.


According to Aufbau principle, for multi-electron atoms, in the ground state of an atom, an electron enters the orbital of lowest energy first and subsequent electrons are fed in the order of increasing energies.

For multi-electron atoms, the following sequence is observed for orbitals in the increasing order of energy: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s.

Here the numbers 1, 2, 3, 4.. indicates shell number. It may be remembered that, though filling of electrons takes place according energy of orbitals, when electrons are removed from the neutral atom, electrons with highest value of n are removed first. So, electrons from 5p or 5s is removed first than from 4f.

To understand lantanide contraction, lets consider an example. If you consider Pd (of 5th period) and Pt (of 6th period), you would expect Pt to have relatively large radii than Pd. In practice, they have roughly same radii. This is because, Pd has 4d electrons and Pt has 5d and 4f electrons. Due to diffused shaped of 4f electrons, they can't shield 5d electrons of Pt. Thus, contracting the size of Pt roughly same as Pd.

Considering your last question, no 4f subshell lies closer to the nuclei than  5s and 5p. For more info, I would suggest you to read this page: Lanthanide Contraction


Why aren't all bond lengths equal in cyclobutadiene?

Anti-aromatic compounds have no delocalization of electrons hence no resonance. To check aromatic character we use huckel's rule of (4n+2)π e- present in a conjugated closed loop. An anti aromatic compound would have only 4nπ electrons which is the case with above mentioned compound . Hence the bond lengths are different.


The electronegativity difference between carbon and the halogen used for Grignard reagents is actually not all that high.  Alkyl fluorides are generally un-reactive towards magnesium, and other halogen whose alkyl compounds are more reactive have electronegativity ranging from only a few tenths above carbon to a few tenths below carbon.

In any event the neutral alkyl halide molecule does not directly cleave into either radicals or ions.  According to Wikipedia, the first stage of the reaction involves a one-electron reduction of the alkyl halide to a radical anion; the magnesium is oxidized to a radical cation.  The alkyl halide anion, not the original molecule, cleaves off its halide ion to form an alkyl radical.  Thus, in effect, after the initial electron transfer you get "a little bit of both" -- the alkyl group is a radical but the halogen (even with its modest electronegativity) is an anion.  The alkyl radical, the magnesium radical cation, and the previously displaced halide ion then combine to get the Grignard reagent molecule (the combination process finishes off the oxidation of magnesium to the +2 state).  At no point in the given mechanism do you have a free radical halogen atom.


How do I scale up to semi-prep from an analytical HPLC

Best practice is to try and maintain the same column length between scales.  Then the scale up factor is simply the ratio of the square of the radii.

If the lengths are different,  try and keep the aspect ratio in the same ballpark e.g. if the length to diameter ratio is not the same but between 4 to 5 for both, you're in a better place. Then simply scale by the ratio of the two column volumes, using the radii calculated from the internal diameters.

Using the ratio of column volumes should work in other circumstances but scaling from a long, thin column to a pancake may cause some issues.

Make sure to keep your particle size of your stationary phase the same too so you have similar packing.
All things being scaled appropriately,  the back pressure should be the same too.


What should be the pKa and IUPAC name of this compound?

I know this isn't exactly an answer, but I think it might help in reaching a qualitative understanding. These are the geometry optimized structures of the protonated and singly deprotonated form of the molecule (I could not obtain a theoretical pKa; it is a much more involved calculation than I anticipated). I would expect the pKa to be fairly low, as the deprotonated structure would seem to be fairly stable due to hydrogen bonding.

(Images were obtained using GaussView. Optimized structures were from Gaussian Optimization calculation using APFD, Basis set 6-311G+(2d,p)).


Electrolytic cell chemical composition during/after charging?

Voltage determines the kind of reactions, amperage determines the amounts.  The positive graphite electrode is the oxidizing anode.  It will evolve oxygen and chlorine and be eroded to carbon dioxide, carbon monoxide, and chloro-organics.  The negative aluminum electrode is the reducing cathode.  It will reduce Na+ to elemental sodium that immediately reacts with water to make lye and hydrogen.  The lye will erode the aluminum into more hydrogen and aluminum hydroxide gel, plus some black slime for about 1% alloy-reinforcing precipitated intermetallics in the aluminum foil (usually 8111 alloy).  Lye plus chlorine gives hypochlorite bleach that will erode both electrodes, and do a little disproportionation into chloride plus chlorite, then that perhaps into chloride and chlorate.

Sodium carbonate (washing soda) is a more benign electrolyte, but it is an alkali hazard to eyes and skin.  A glass is a more inert container.  Do not short the electrodes, do it outdoors and stay upwind.  Avoid electrolyte or product contact with your eyes and skin.


Calculating the concentration of copper(II) from a non-standard cell potential

You are on the right track with that equation. There is a problem with the question, though: you do not know which direction the reaction is proceeding in. Therefore, you get two possible answers.

One of the two possible ways to write the reaction that's occurring is

$$\ce{Cu^2+(aq) + H2(g) -> Cu(s) + 2H+ (aq)}$$

For this reaction, the value of $E_\mathrm{cell}^\circ$ is given by $E^\circ(\ce{Cu^2+}/\ce{Cu}) - E^\circ(\ce{H+}/\ce{H2}) = \pu{+0.34 V}$. This seems to be the more logical direction for the reaction to proceed in, as we have a positive value of $E_\mathrm{cell}^\circ$. Consequently, to achieve the smaller potential difference of $\pu{0.195 V}$, all we need to do is to reduce the concentration of $\ce{Cu^2+}$ from the standard of $\pu{1 mol dm-3}$, so that the forward reaction becomes slightly less favourable. The corresponding value of $n$ is $2$. The reaction quotient can be approximated by

$$Q = \frac{([\ce{H+}]/c^\circ)^2}{([\ce{Cu^2+}]/c^\circ) \cdot (p(\ce{H2})/p^\circ)} = \frac{c^\circ}{[\ce{Cu^2+}]}$$

since the $\ce{H+}/\ce{H2}$ electrode is operating under standard conditions (and where $c^\circ = \pu{1 mol dm-3}$). The temperature is not specified, but we could assume it to be $\pu{298 K}$.

Therefore:

$$\begin{align}
E_\mathrm{cell} &= E_\mathrm{cell}^\circ - \frac{RT}{nF}\ln Q \\
\pu{0.195 V} &= \pu{0.34 V} - \frac{(\pu{8.314 J K-1 mol-1})(\pu{298 K})}{(2)(\pu{9.649 \times 10^4 C mol-1})} \ln\left(\frac{c^\circ}{[\ce{Cu^2+}]}\right) \\
[\ce{Cu^2+}] &= \pu{1.246 \times 10^-5 mol dm-3}
\end{align}$$

If the reaction is actually running in the reverse direction, then you have the reaction

$$\ce{Cu(s) + 2H+(aq) -> Cu^2+(aq) + H2(g)}$$

with $E^\circ_\mathrm{cell} = \pu{-0.34 V}$. Going through exactly the same steps as above (but taking care to invert the expression for $Q$, since the reactants and products are swapped), you get to

$$\begin{align}
E_\mathrm{cell} &= E_\mathrm{cell}^\circ - \frac{RT}{nF}\ln Q \\
\pu{0.195 V} &= \pu{-0.34 V} - \frac{(\pu{8.314 J K-1 mol-1})(\pu{298 K})}{(2)(\pu{9.649 \times 10^4 C mol-1})} \ln\left(\frac{[\ce{Cu^2+}]}{c^\circ}\right) \\
[\ce{Cu^2+}] &= \pu{8.019 \times 10^-19 mol dm-3}
\end{align}$$

What's happening here? It turns out that as you reduce the concentration of $\ce{Cu^2+}$ further, $E_\mathrm{cell}$ drops more and more, until it actually becomes thermodynamically favourable for the reverse reaction to occur. This concentration isn't particularly sensible, but you can certainly achieve it by serial dilution.

How do we resolve this dilemma? The most straightforward way is to find out which electrode is the anode and which is the cathode, which you can do with the voltmeter. For voltmeters to display a positive reading, as is suggested by the question, you need to have electrons flowing from the black wire to the red wire through the voltmeter. This means that the black wire must be connected to the electrode which provides it with electrons, i.e. the anode, since the anode is the site of oxidation (the chemical species gives up its electrons to the electrode and is itself oxidised in the process).

So, if your voltmeter displays a positive reading and the black wire is connected to the standard hydrogen electrode, it means that $\ce{H2}$ is being oxidised to $\ce{H+}$. That means that we are in the first situation above, where the forward reaction $\ce{Cu^2+(aq) + H2(g) -> Cu(s) + 2H+ (aq)}$ is occurring.


Have Ionic Liquids been used to refine aluminum on an industrial scale?

I think you are misunderstanding the article linked in the question.

In the process described in the article, the starting material is A360 aluminum metal alloy.

The goal is to make very pure aluminum metal from impure aluminum metal.

The process is an electrolysis process.

$$\ce{Al(anode) + 7AlCl_4^{−} -> 4Al2Cl_7^{−} + 3e-}$$

$$\ce{4Al2Cl_7^{−} + 3e- -> Al(cathode) + 7AlCl_4^{−}}$$

The existing industry techniques referred to by the article as requiring ~15kw-hr/kg are also for refining impure aluminum metal to yield high purity aluminum.  This number is not referring to purification from bauxite.  The "three layer process" is one of the existing refining techniques, and it does involve ionic liquids, but the electrolyte is liquid only at high temperature.


Base Peak Intensity different than higher value in Mass Intensity Map

One possibility for the discrepancy is that different peak centroiding / integration algorithms were used at different stages in the data.

For example, the base peak intensity recorded in the plain-text (uncompressed) scan data is probably calculated by the instrument in real-time during acquisition.  The intensities of all detected peaks are also recorded by the instrument, of course, but those intensities may be affected by processes such as centroiding that are not applied to the plain-text recorded base peak intensity.

An exact answer to your question will depend on many factors, such as whether spectra were recorded in profile mode or centroid mode, whether spectra were compressed during conversion to mzXML format, and also details of the (possible) centroiding algorithms and BPC algorithms used by the instrument firmware during data acquistion.  (Good luck trying to find the exact code for vendor firmware...it's probably proprietary.)


How to calculate the start concentration for a first order reaction?

For a first order reaction $\ce{A ->[k]P}$, the rate constant is given by

$$k[A] = - \frac{d[A]}{dt} $$

So, at any time $t$, the concentation $[A]_t$ is

$$[A]_t = [A]_0 \cdot e^{-kt}$$

At the half-life time $t_{1/2}$, $[A]_t = \frac{1}{2}[A]_0$, isn't it?

Would you agree that we can calculate $k$ using $k = \frac{ln 2}{t_{1/2}}\ $?

Knowing $k$ and $[A]_t$, calculation of $[A]_0$ doesn't seem impossible now ;).


When two containers at different T and P are kept together, will pressure become same or temperature?

The situation that is being covered in your class is where the two containers are thermally insulated from one another.  In this case, the only place where heat can be transferred from one container to the other is through the gas in the valve (by conduction).  But the assumption is being made in your class problem that heat conduction through the gas in the valve is negligible.  In any event, even if the valve were left open until the pressures effectively equilibrated, the temperatures in the two containers would require a much larger time to equilibrate.  So, if the assumption is made that, once the pressures effectively equilibrate, the valve is closed again (and does not allow heat transfer between the containers), then at final thermodynamic equilibrium, the temperatures of the gas in the two containers will be different.


Can vinyl alcohol react as a dienophile in Diels-Alder reactions?

Firstly, your Diels-Alder products are missing the double bond in the ring (formed between what was formerly C-2 and C-3 of the diene). You could always simply hydrogenate it down to the alkane, but I feel like this would present some selectivity issues in your professor's suggested synthetic route, especially since a terminal double bond is more likely to be hydrogenated than an internal double bond (sterics), and an internal double bond more prone to ozonolysis than a terminal double bond (more electron-rich).

Anyway, ron pretty much said what there is to say about "vinyl alcohol". It doesn't exist in any appreciable quantity, since the keto form acetaldehyde is much more stable (see another of ron's answers here):

Apart from what ron suggested (vinyl acetate), another alternative is to use 2-chloroacrylonitrile as the dienophile. The cyano group is electron-withdrawing, so the the usual requirement for an electron-poor dienophile is met. Treatment with hydroxide will unmask the ketone functionality, see for example the second and third steps in Corey's prostaglandin synthesis. (For a possible mechanism, see Mechanism for basic hydrolysis of α-chloronitrile to ketone?.) You can then hydrogenate the C=C bond chemoselectively to get the desired product, cyclohexanone.

2-Chloroacrylonitrile is what is called a "masked ketene". Ketenes don't undergo [4+2] cycloadditions such as the Diels-Alder reaction because they have a much greater propensity to react in a [2+2] fashion (even cyclopentadiene, an excellent diene in the Diels-Alder, reacts with ketenes via a [2+2]). So, these reagents are essentially surrogate ketenes: with one additional step (hydrolysis in this case), you can access what is formally a ketene+diene Diels-Alder adduct. There's a nice review of masked ketene reagents here.


Calculating singlet-triplet gap and open-shell singlet character

There is a related discussion on which method to use for a singlet triplet guess, which I encourage you to also read:
U- or RO-method for Singlet-Triplet Gap?

This post is going to be a little longer, so I'll present the key points upfront. It is reasonable to calculate (and/or optimise) the wave function with a restricted method, then test it for instabilities. The same applies to to triplet (etc.) wave functions, where I would also try to start with a restricted open shell approach. To answer 1, it is a reasonable first approach for a singlet-triplet gap. If you are more interested in the electronic excitations a time-dependent treatment is necessary.

As a general recommendation I would advise to split the calculations up and read the respective wave functions from the checkpoint file to perform additional calculations and checks, so that you always have a fallback.
Hence the answer to 2, once you have created and converged a solution to the wave function you can read it in again. Do not generate a new initial guess.

You should also always keep an eye on spin contamination, because this is essentially telling you if the method you are using is suitable in the first place. When it comes to broken symmetry solutions, you should always consider to run multi-reference calculations, too.

(Note that guess(always) applies only to geometry optimisations; see below.)

I'm going to exercise a couple of calculations on dioxygen, because it's a nice study case. I will be using the BP86 functional with the def2-SVP basis set and density fitting. I have done the calculations with Gaussian 09 rev. D.01.
It is somewhat an expansion on the example in Exploring Chemistry with Electronic Structure Methods, 3rd Edition (p.71).

I am going to present the full input of the calculations and at the end the absolute energy to compare to, in case you'd like to run the calculations yourself. Please note that I have not run any geometry optimisations and assumed a equilibrium bond distance of 120 pm. I think that is a reasonable choice since this is just an example and the optimisations can be considered as additional exercise.

I labelled the the used methods RHF, ROHF, and UHF even though I am using density functional approximations.

Because the main focus is the use of the stable keyword, we start with a wave function that is already know to be not stable. The RHF solution of dioxygen, filename bp86svp.rhf.sing.gjf:

%chk=bp86svp.rhf.sing.chk
%nproc=2
%mem=10000MB
#p RBP86/Def2SVP/W06
DenFit
gfinput gfoldprint iop(6/7=3)
symmetry(loose)
pop=full pop=nbo
scf(xqc)

title card unused

0 1
O  0.6 0.0 0.0
O -0.6 0.0 0.0

This input uses a few keywords, that are not strictly necessary to complete the calculations. For example the route section #p requests detailed output. DenFit is already implied with the choice of W06 as an additional basis set. I like to write it in nevertheless, to exactly know what I am doing. The same goes for symmetry(loose); occasionally I like to turn symmetry of, so I know this time I really wanted it. The line gfinput gfoldprint iop(6/7=3) is included so that molden (❤) can read (and display) the orbitals. There is no shame in always requesting population analyses.

In this very case even scf(MaxCycle=9999) will not lead to convergence; with scf(xqc) we are allowing the algorithm to switch to quadratic convergence if conventionally the convergence cannot be achieved. Note that this may converge to a different solution than originally attempted.

For publication type calculations I would also include int(ultrafine) and scf(tight); but I forgot it here.

From here I went on to calculate the stability of the wave function, i.e. bp86svp.rhf.sing.stab.gjf:

%chk=bp86svp.rhf.sing.stab.chk
%nproc=2
%mem=10000MB
%oldchk=bp86svp.rhf.sing.chk
#p RBP86/Def2SVP/W06
DenFit
geom=check guess=read
gfinput gfoldprint iop(6/7=3)
scf(xqc)
stable

title card unused

0 1

Note that in this case the SCF will converge within the conventional cycles and hence the energy is different. You can additionally check this by reading in the checkpoint file and perform a new SCF cycle.

However, in the stability analysis you will find the following output:

 ***********************************************************************
 Stability analysis using  singles matrix:
 ***********************************************************************

 Eigenvectors of the stability matrix:

 Excited state symmetry could not be determined.
 Eigenvector   1:      Triplet-?Sym  Eigenvalue=-0.0560505  =2.000
       8 ->  9         0.70645

 Excited state symmetry could not be determined.
 Eigenvector   2:      Singlet-?Sym  Eigenvalue= 0.0372059  =0.000
       8 ->  9         0.70689

 Excited state symmetry could not be determined.
 Eigenvector   3:      Triplet-?Sym  Eigenvalue= 0.1233031  =2.000
       7 ->  9         0.70493

 Excited state symmetry could not be determined.
 Eigenvector   4:      Triplet-?Sym  Eigenvalue= 0.2035093  =2.000
       5 ->  9         0.70373

 Excited state symmetry could not be determined.
 Eigenvector   5:      Singlet-?Sym  Eigenvalue= 0.2902800  =0.000
       6 ->  9         0.70685

 Excited state symmetry could not be determined.
 Eigenvector   6:      Singlet-?Sym  Eigenvalue= 0.3353297  =0.000
       5 ->  9         0.70254
 SavETr:  write IOETrn=   770 NScale= 10 NData=  16 NLR=1 NState=    6 LETran=     118.
 The wavefunction has an RHF -> UHF instability.

As expected we see a couple of instabilities. At this point we can run another calculation with stable=opt to find such a wave function. For reasons I don't quite understand why it finds the the broken symmetry solution instead of the triplet solution.

For exercise purposes we'll manually generate the broken symmetry solution; i.e. bp86svp.uhf.sing.gjf:

%chk=bp86svp.uhf.sing.chk
%nproc=2
%mem=10000MB
%oldchk=bp86svp.rhf.sing.chk
#p UBP86/Def2SVP/W06
DenFit
scf(xqc)
gfinput gfoldprint iop(6/7=3)
symmetry(loose)
pop=full pop=nbo
geom=check guess=(read,mix)

title card unused

0 1

The keyword guess(mix) generates a UHF guess by mixing the HOMO with the LUMO. You only need to do this once. Note that the keyword guess(always) only applies to optimisations and will perform a new initial guess at every point along the optimisation. I personally have no idea why this option exists or what it is used for.

Now we can go ahead and test for stability; i.e. bp86svp.uhf.sing.stab:

%chk=bp86svp.uhf.sing.stab.chk
%nproc=2
%mem=10000MB
%oldchk=bp86svp.uhf.sing.chk
#p UBP86/Def2SVP/W06
DenFit
scf(xqc)
gfinput gfoldprint iop(6/7=3)
symmetry(loose)
pop=full pop=nbo
geom=check guess=read
stable

title card unused

0 1

Note that we do not need to mix the orbitals in the wave function again. The stability section will now read like this:

 ***********************************************************************
 Stability analysis using  singles matrix:
 ***********************************************************************

 Eigenvectors of the stability matrix:

 Excited state symmetry could not be determined.
 Eigenvector   1:  1.012-?Sym  Eigenvalue=-0.0000016  =0.006
      8A ->  9A        0.70640
      8B ->  9B       -0.70640

 Excited state symmetry could not be determined.
 Eigenvector   2:  1.006-?Sym  Eigenvalue= 0.0930085  =0.003
      8A ->  9A        0.70688
      8B ->  9B        0.70688

 Excited state symmetry could not be determined.
 Eigenvector   3:  2.245-?Sym  Eigenvalue= 0.2031898  =1.010
      7A ->  9A       -0.70381
      7B ->  9B        0.70381
 SavETr:  write IOETrn=   770 NScale= 10 NData=  16 NLR=1 NState=    3 LETran=      64.
 The wavefunction is stable under the perturbations considered.

We already know that the triplet state should be more stable. As I hinted, I like to start with a ROHF calculation. In principle these calculations are always instable with respect to an UHF solution. However, usually this leads to faster results, especially if you have already a converged RHF solution. In this case we could readily work from the broken symmetry solution and directly calculate the triplet from there. Choose whatever is more convenient for you.
For this illustrative purpose I chose to generate a new initial guess; i.e. bp86svp.rohf.trip.com:

%chk=bp86svp.rohf.trip.chk
%nproc=2
%mem=10000MB
#p ROBP86/Def2SVP/W06
DenFit
gfinput gfoldprint iop(6/7=3)
symmetry(loose)
pop=full pop=nbo

title card unused

0 3
O  0.6 0.0 0.0
O -0.6 0.0 0.0

And get the UHF triplet from there; i.e. bp86svp.uhf.trip.gjf:

%chk=bp86svp.uhf.trip.chk
%nproc=2
%mem=10000MB
%oldchk=bp86svp.rohf.trip.chk
#p UBP86/Def2SVP/W06
DenFit
gfinput gfoldprint iop(6/7=3)
symmetry(loose)
pop=full pop=nbo
scf(xqc)
geom=check guess=read

title card unused

0 3

And then obviously check for instabilities; i.e. bp86svp.uhf.trip.stab.gjf:

%chk=bp86svp.uhf.trip.stab.chk
%nproc=2
%mem=10000MB
%oldchk=bp86svp.uhf.trip.chk
#p UBP86/Def2SVP/W06
DenFit
gfinput gfoldprint iop(6/7=3)
symmetry(loose)
pop=full pop=nbo
scf(xqc)
geom=check guess=read
stable

title card unused

0 3

We will find the following stability section:

 ***********************************************************************
 Stability analysis using  singles matrix:
 ***********************************************************************

 Eigenvectors of the stability matrix:

 Excited state symmetry could not be determined.
 Eigenvector   1:  3.003-?Sym  Eigenvalue= 0.2313384  =2.004
      6B ->  8B        0.70567
      7B ->  9B        0.70567

 Excited state symmetry could not be determined.
 Eigenvector   2:  3.003-?Sym  Eigenvalue= 0.2313396  =2.004
      6B ->  9B       -0.70568
      7B ->  8B        0.70568

 LASSW sees vector with all elements less than  0.100000000000E-03
 LASSW sees vector with all elements less than  0.100000000000E-03
 Excited state symmetry could not be determined.
 Eigenvector   3:  3.002-?Sym  Eigenvalue= 0.2704717  =2.003
      6B ->  8B        0.70711
      7B ->  9B       -0.70711
 SavETr:  write IOETrn=   770 NScale= 10 NData=  16 NLR=1 NState=    3 LETran=      64.
 The wavefunction is stable under the perturbations considered.

A time-dependent calculation will look for excitations based on the reference wave function. Therefore the solution will not be different. If you don't need to perform this calculation because you are interested in for example the UV/Vis spectrum, then there is no reason to do it.

I guess we have run the necessary scenarios now, so here are the results:

Command file              Functional            Energy / Hartree ( cycles )
bp86svp.rhf.sing          E(RB-P86)       =       -150.155033509 (      3 )
bp86svp.rhf.sing.maxc     No energy statement found.
bp86svp.rhf.sing.scf      E(RB-P86)       =       -150.154053425 (     16 )
bp86svp.rhf.sing.stab     E(RB-P86)       =       -150.154053425 (     16 )
bp86svp.rhf.sing.stab.opt E(UB-P86)       =       -150.201668045 (      4 )
bp86svp.rohf.trip         E(ROB-P86)      =       -150.213780181 (      9 )
bp86svp.uhf.quin.stable   E(UB-P86)       =       -149.651924155 (      8 )
bp86svp.uhf.sing          E(UB-P86)       =       -150.201668044 (      8 )
bp86svp.uhf.sing.stab     E(UB-P86)       =       -150.201668044 (      1 )
bp86svp.uhf.trip          E(UB-P86)       =       -150.215687576 (      8 )
bp86svp.uhf.trip.stab     E(UB-P86)       =       -150.215687574 (      1 )
bp86svp.uhf.trip.td       E(UB-P86)       =       -150.215687575 (      9 )

Lastly and maybe most important: with DFA it is never enough to just use one method. For more on this: DFT Functional Selection Criteria. Winter is coming!


Modeling 3-center-2-electron bonds in Avogadro

First of all, I am not an Avogadro developer, just an occasional user. My view of this matter is the following:

As 3D molecular drawing/editing software, Avogradro, does not need to "understand" B-H-B bonds. If it allows you to draw them in a sensible way, it is already fulfilling its function as a drawing/editing program.
Avogadro is also an interface to a number of codes, either internal or external, that allow you to perform tasks such as geometry optimisations and conformational sampling. Now, there becomes relevant whether or not those methods can "understand" B-H-B bonding. Ab initio and other quantum mechanical methods should afford a reasonable treatment of such molecules, provided that an appropriate level of theory is used, whereas for semi-empirical and molecular mechanics methods it will depend on whether or not a given parametrisation model or a given the force field (version), respectively, have been parametrised to take such situations into account.

In summary, if you are able to draw and edit the structure in Avogadro, the real question is whether or not you are going to do something else with it within Avogadro and whether or not your method of choice is appropriate to treat your molecular systems.


Predicting sign of enthalpy of allotropic transition of phosphorus

You can use the fundamental relation between energy and enthalpy. At constant temperature and pressure:
$$\Delta H_m = \Delta U_m + P\Delta V_m = \Delta U_m + PM (\frac{1}{\rho_{red}}-\frac{1}{\rho_{white}})$$
where the subindex m labels molar properties and M is the molar mass. Taking values of the density posted in the Wikipedia indicates that red phosphorus has a significantly higher density than white (~2.25 vs 1.82 g/mL). This means the volume decreases and the surroundings do work on the system during the reaction. From this data one can approximate the value of $P\Delta V$ at standard pressure (1 bar):
$$\Delta H_m=   \Delta U_m -\pu{0.32 J/mol} $$
In addition the difference in strain would indicate that $\Delta U_m<0$. This would mean that the enthalpy change should be negative, as can be verified by looking up thermodynamic tables. The Wikipedia page describing the allotropes of P (or alternately the CRC handbook) provides a value  of $\Delta H_m=\pu{-17.6 kJ/mol}$, so the correction for the volume contraction can be seen to be negligible (and does not influence the conclusion that $\Delta H<0$). Interestingly, even though white phosphorus is less thermodynamically stable it is considered the standard state of P.


How do I perform thin-layer chromatography to identify alkaloid content in fruits?

TLC in alkaloid research

Both analytical and preparative TLC have been used to separate and identify alkaloids from plant material.

Pretreatment of TLC plates and adsorbents

If you are using silica plates, it is often advantageous to wave the plate above an open bottle with aqueous ammonia or pretreat the plates with an organic solvent containing some triethylamine. Otherwise, the $\mathrm{R_f}$ values of the alkaloids might be too small.

Alkaloid detection on a TLC

As pointed out by Zhe in a comment, this is crucial for preparative TLC. Usually, you run an analytical TLC (same material and mobile phase) in parallel and stain the analytical TLC using Dragendorff's reagent.

General hints on the extraction

Process different parts of a plant separately

Different parts of a plants have different functions. The distribution of particular compounds or compounds classes, such as

chlorophylls and carotenoids
quinones
polyphenols
carbohydrates
terpenoids
alkaloids

will vary in flowers, fruits, seeds, leaves, stems (and bark), roots, and tubers.

It is therefore reasonable to process the different parts of a plant separately.

Use dry plant material

The water that you remove in the beginning doesn't disturb the extraction in a later stage. Dry the plant material. If you assume that the compounds you are looking for are air-sensitive, dry the material in vacuum.

Use powdered plant material

Extraction is more efficient on solid materials with a larger surface.

Use the right solvent

Two different strategies are conceivable:

Extract all soluble compounds with a polar protic solvent (methanol, ethanol) at once and partition between solvents of different polarity later
Leach the dry plant material stepwise with solvents of increasing polarity:

cyclohexane
ethyl acetate
chloroform
methanol

Use the right technique

You can

stir/shake the dried plant material with a solvent at room temperature
boil plant the material in a solvent
use a Soxhlet extractor

If necessary, perform the extraction under a nitrogen or argon atmosphere

Specific advice on extraction

Rambutan (Nephelium lappaceum)

In Antioxidants, 2014, 3, 371-386 (Open Access article), the authors write

  The fruit samples were washed and air-dried, followed by drying in an oven at 40 °C for 24 h. All the dried samples were ground to fine powder with a grinder, after which they were vacuum packed into a nylon-linear low density polyethylene film before analysis. An extraction with 80%
  ethanol was carried out, and the mixture was shaken at 50 °C for 120 min using an orbital shaking incubator as optimized in a previous study. The residue was subjected to another round of extraction under similar conditions to obtain maximal yield of antioxidant activities. The extracts were combined and centrifuged, followed by filtration and concentration of the supernatant using a rotary evaporator at 45 °C. The concentrated extract was freeze-dried, wrapped with aluminum foil and kept at −20 °C. The crude extract as then partially fractionated using silica packed open column chromatography and using increasing polarity of solvent, ethyl acetate, chloroform and methanol (100:0:0 to 0:0:100 v/v).

In the article cited above, alkaloids were not mentioned.

Ambarella (Spondias dulcis)

In Asian Pac J Trop Biomed., 2013, 3, 682-691, the authors give the following procedure

  The pulverized plant materials were soaked in 3.5 L of methanol in separate containers and kept for 1 week with occasional shaking. The whole mixtures were then filtered through sterilized cotton followed by Whatman No. 1 filter paper and concentrated using a rotary evaporator (IKA , Germany). The weight of crude extracts of fruits and leaves were 135 g and 101 g, respectively. A total of 40 g crude extract of each plant materials was then partitioned with chloroform and dichloromethane. The percentage
  yields of fruit and leaves were methanol ( 21.3 % , 15.8 %), chloroform ( 19.5 % , 23.1 %) and dichloromethane (20.9 % , 25.4 %).

The authors mention that extracts both from the fruits and the leaves of S. sulcis gave a positive test with Dragendorff's reagent, but do not provide further information on the amount and/or nature of the alkaloids.


I really like your process of trying to work it out! Trying to find patterns is a great way to look at this (and all science really). I think all you're really missing is some nomenclature, as this appears to be primarily a question of understanding of what's actually going on.

The 3+ alumnium ion ($\ce{Al^{3+}}$) and 2- oxygen ion ($\ce{O^{2-}}$) form a crystal lattice in solid form.

(A crystal lattice is a $3D$ arrangement of ions in a repeating pattern, in which each ion is interacting with more than one other ion)

The total charges need to balance out per "unit cell" (an intermediate-level chemistry concept that means the "smallest repeating subunit of a crystal lattice.")
So this means that you need to balance the total charges by changing the number of ($\ce{Al^{3+}}$) and ($\ce{O^{2-}}$) ions to get the correct chemical formula.

In this case: $(3+) *(2) = (6+)$ balances $(2-) *(3) = (6-)$, giving $\ce{2 Al^{3+}}$ and $\ce{3 O^{2-}}$.

Ionic compounds that have an overall neutral charge are written without the charge superscripts,  thus, $\ce{Al^{3+}_2O^{2-}_3}$ is written as $\ce{Al_2O_3}$

If you're actually looking at precipitation reactions (where the solid salt forms from a solution), you have to balance the equations.

Thus to form $\ce{Al_2O_3}$ the correct ratio would be: $\ce{2 Al^{3+} + 3 O^{2-}} -> Al_2O_3$

Similarly for $\ce{CaBr_2}$, the correct ratio would be: $\ce{Ca^{2+} + 2 Br^-} -> CaBr_2$.

For $\ce{Na_2CO_3}$ you're applying the exact same concept, except you're dealing with an anionic (negative charged) molecule instead of an atom.

Thus, to form $\ce{Na_2CO_3}$ the correct ratio would be: $\ce{2 Na^+ + CO^{2-}_3} -> Na_2CO_3$.


Does PEO particle size decrease when dissolved in water?

PEO is very soluble in water. At 1 wt% concentration, you should expect it to be individual molecules throughout the solution. A PEO molecule with MW of 600,000 has ~13,600 repeating units. You can imagine them to be roughly ~200 nm blobs calculated based on the radius of gyration of a linear polymer in a good solvent.


The micellization releases water molecules which initially solvate the hydrophilic head groups. The release of these waters greatly increases the entropy.

Reference: Dong et al., Chem. Rev. 2010, 110, 4978-5022.


Openbabel is not converting Gaussian Z-matrix

Turns out that OpenBabel is sensitive to whether blank lines are really blank lines or lines with space characters only. In my case, this was the blank line between the route and the title line. One can see them by pressing the button with a paragraph sign (¶) on it in Your favourite text editor (Notepad++, or even MS Word or  LibreOffice Writer), or by entering :set list in vim or M-x whitespace-mode in Emacs.

What does OpenBabel require for conversion:

First line of the file must begin with a hash sign (#) – a substitute for the Route line
Next line must be absolutely empty
Next line can contain anything You want
Next line again must be absolutely empty
Then comes the charge and multiplicity
And then comes the Gaussian Z-matrix in either compact or full form.

What doesn't OpenBabel require for conversion:

the file to have the extension GZMAT, it can also be GJF
title line not to begin with the hash sign (#) or any other character
route line to contain actual keywords (if You are not trying to convert them to any other input file format, that is)


Why is the standard enthalpy of formation of diamond not zero?

You are on the right track - diamond is not the thermodynamically stable carbon phase at STP. Taking two figures from A.T. Dinsdale, 'SGTE Data for Pure Elements', CALPHAD 15(4) 317-425 (1991) one sees:

and

Since graphite is the thermodynamically stable phase of carbon at STP, it is usually selected as the reference phase so it has $\Delta H^0_f = 0$. In the reference above, both absolute Gibbs free energies, as well as free energies with respect to the stable phase at STP, are given for most elements.


Determining if reaction is endothermic or exothermic using equilbrium constant

Think about it this way.  In an exothermic reaction that is at equilibrium the forward reaction has a lower activation energy.  So, if you increase T the forward reaction rate increases but the reverse rate increases even more.  That's because the rates of reactions with higher activation energy are more sensitive to temperature.  With the reverse rate higher more reactants will be made until the concentration reactants reaches a point where the rate of the forward reaction equals the reverse rate.

In an endothermic reaction increasing T increases the forward rate more than the reverse, producing more product until the increased product concentration causes the reverse rate to equal the forward rate.

The equilibrium constant can be thought of as a ratio of products to reactants needed to equalize the rates of two reactions with different activation energies.


Is 98%+ purity Reagent Grade Ca-EDTA powder as safe to use as USP Grade in the body?

While normally health-related questions such as this one are considered off-topic, this one is sufficiently cut-and-dried that it probably merits an answer.

NO!!!! In general, reagent-grade chemicals are NOT safe to use internally like USP grades.  The entire point of the USP specifications is to ensure, to the extent practical, that the relevant chemicals are prepared from suitable starting materials and using suitable methods such that toxic or hazardous materials are absent from the final product. This is a big reason why the USP grades are often more expensive than the lab grades.

It's certainly possible that a given reagent-grade chemical may actually be safe to ingest, but there is no real way to know for sure without conducting extensive analysis of the impurities.


Why do acetylene and chlorine react to yield carbon and HCl, but acetylene and bromine react less intensely?

Bromine is not as strong of an oxidizing agent as chlorine.  In essence the reaction of acetylene with chlorine is the same as burning, just with elemental halogen instead of elemental oxygen.

The higher electronegativity of chlorine explains why it is a good oxidant as compared to bromine.

Other notes:

Describing something as 'radically' (this phrasing removed  from original question edit) when you mean 'intensely' will
cause much confusion around here (Stack) since 'radically' refers to reactions involving single electron mechanisms.


Is the specific heat of salt the same as Na + CL when dissolved?

Salt does not dissolve to to $\ce{Na}$ and $\ce{Cl}$ atoms, but to hydrated $\ce{Na+}$ and $\ce{Cl-}$ ions, so it's heat capacity is very different to the solid $\ce{NaCl}$.

And yes it looses vibration modes of the crystal lattice, less the rotation, as there is nothing to rotate in the lattice.

There are also vibrations of coordination bonds to water. Dissolved ions also have translational degrees of freedom, what is missing in solids.

Even for the identical substance, its heat capacity in solid and liquid state differ. E.g. liquid water has about 2x bigger heat capacity than ice ( 4.2 versus 2.1 J/g/K ).


Is it really? Always?

In an older article (J. Chem. Educ., 1988, 65, 367), the authors described an experiment that demonstrates the outcome of the alkylation of chlorobenzene (1) with 2-chloropropane (2) under kinetic control.

They first added 25 mL of 1 and 2 mL of nitromethane to 1.2 g anhydrous aluminium chloride, and stirred until the aluminium chloride was fully dissolved. To the cooled mixture (3-5 °C), a solution of 3 mL 2 in 20 mL of 1 was dropwise added within 10 min, while keeping the temperature of the reaction mixture below 10 °C. The reaction was stirred for another hour, allowed to come to room temperature.

After workup, gas chromatographic analysis showed that the o/m/p ratio for the resulting isopropyl chlorobenzenes was 63/5/32.

To cut the long story short, under the conditions describe above, the ortho isomer prevailed.


Can Henderson-Hasselbalch equation be used to determine the pH of a solution when a strong base and a strong salt of that base stay together?

The pH of the solution after $\mathrm{25\ mL}$ of $\mathrm{0.2\ M}\ \ce{ NaOH}$ are added to $\mathrm{5\ mL}$ of $\mathrm{0.2\ M}\ \ce{CH3CO2H}$ is $13.12$. Your book or solution manual is wrong. The $\mathrm{pH} = 8.87$ at the equivalence point.

The Henderson Hasselbalch equation only applies in the buffer region - the part of the titration curve before the equivalence point. I have worked out the pH for a number of volumes of titrant and graphed the curve. The buffer region is labeled. You can see that past the equivalence point, the pH jumps to above 12 and stays there. You determined the pH after 25 mL correctly.

There is a point where the $\mathrm{pH=8.87}$ - the equivalence point (labeled with its pH). After $\mathrm{5\ mL}$ of $\ce{NaoH}$ solution is added, all acetic acid has been converted into the acetate ion. We can solve an equilibrium problem to determine the $\mathrm{pOH}$ and $\mathrm{pH}$ at that point. The following equation applies:

$$\ce{CH3CO2- + H2O <=> CH3CO2H + OH-}\ \ \mathrm{K_b=10^{-9.26}}$$

Since all acetic acid was converted into acetate, and the volume has doubled, the initial concentration of acetate is $\mathrm{0.1\ M}$. The initial volume of acetic acid is $\mathrm{0\ M}$ and the initial concentration of hydroxide is $\mathrm{10^{-7}\ M}\approx 0$.

$$\begin{array}{|c|c|c|c|}\hline
\ & [\ce{CH3CO2-}] & [\ce{CH3CO2H}] & [\ce{OH-}] \\ \hline
I & 0.1 & 0 & 0\\
C & -x & +x & +x \\
E & 0.1 -x & +x & +x\\ \hline
\end{array}$$

If we plug in the values into the law of mass action, we can solve for $[\ce{OH}]$ and determine $\mathrm{pOH}$.

$$\begin{align}
K_b &= \dfrac{[\ce{CH3CO2H}][\ce{OH-}]}{[\ce{CH3CO2-}]}\\
10^{-9.26} &= \dfrac{x^2}{0.1-x}\end{align}$$

Solving for x gives

$$\begin{align}
[\ce{OH-}] &=x=7.41\times 10^{-6}\\
\mathrm{pOH} &=5.13 \\
\mathrm{pH} &=8.87
\end{align}$$

Even if I had not made the assumption that the initial hydroxide concentration was zero, I would get the same $\mathrm{pH}$.

  Your book has a typographical error. Either you were meant to determine the pH after $\mathrm{5\ mL}$ of $\mathrm{0.2\ M}\ \ce{ NaOH}$ are added to $\mathrm{5\ mL}$ of $\mathrm{0.2\ M}\ \ce{CH3CO2H}$ or the pH after $\mathrm{25\ mL}$ of $\mathrm{0.2\ M}\ \ce{ NaOH}$ are added to $\mathrm{25\ mL}$ of $\mathrm{0.2\ M}\ \ce{CH3CO2H}$. Both would yield a $\mathrm{pH} =8.87$ at the equivalence point.


How to choose the best method and basis set for a calculation in computational chemistry?

The experimental equivalent of your question would be: what kind and level of impurity am I going to accept in my experiment?

In terms of basis sets: in principle you keep increasing the complexity of your basis set until you reach convergence in terms of electronic energy. What you consider to be converged depends of course on your preferences.

With regard to the choice of method. Again, this depends on what you consider to be accurate enough. In heterogeneous catalysis, typically density functional theory calculations are performed using a PBE functional. This is accurate to about 5-15 kJ/mol which is fine in this field. This type of accuracy would be completely unacceptable in other fields. If high accuracy is required, people typically use post-Hartree-Fock methods such as MP2 or higher or CI.

My recommendation in picking a method and a basis set is the following:

Decide what you consider accurate enough.
Look into the literature if there are calculations being performed on the same or equivalent type of system and analyse their results.
Experiment using different methods on a model system. Look how the energy converges with respect to basis set and methodology. Pick the method that you think is best, both in terms of computational speed and accuracy. (it is OK to pick a couple of methods)
Perform the bulk of your calculations using this method.
If you are going to report your calculation. Clearly explain the reasoning why you picked a certain method. The potential readers of your report really appreciate to hear your motivation.


Fluoride has a high affinity for $Fe^{3+}$, and $Fe^{3+}$ forms deeply coloured complexes with phenols.  The decomposition of the latter by fluoride ion is the basis for colorimetric determination of fluoride.

Ligands used in this kind of assay include 5-phenylsalicylaldehyde and resacetophenone.  For occasional testing of well water I'd recommend any analytical lab.  I'd be very surprised if Merck didn't offer any fluoride test strips, at their price of course, although whether they go down to the 5 ppm range is a different question.  For those that really like to spend money, there's fluoride-sensitive electrodes.


Why can unsymmetrical tert-butyl ethers be prepared by dehydration of the constituent alcohols?

In order to understand why this is OK, you have to understand the mechanism by which this occurs. It is essentially an $\mathrm{S_N1}$ reaction that is most favoured by the conditions (acidic + polar protic solvent).

So, for the formation of an ether $\ce{^tBu-O-R}$ where $\ce{R}$ is a primary alkyl group:

competing formation of $\ce{R-O-R}$ is slow because primary carbocations are very unstable, and
competing formation of $\ce{^tBu-O-^tBu}$ is slow because $\ce{^tBuOH}$ is a poor nucleophile (tertiary alcohols are very bulky).


Anomalous oxidation states of Transition Metals

Regarding the maximum oxidation state achievable, there are two competing trends. Both have to do with ionisation energies.

In going to a higher oxidation state, the additional IEs are always balanced against the release in energy in the formation of stronger ionic/covalent bonds. However, if the additional IE is too large, then the favourable bonding will not be sufficient to bring out the higher oxidation state. A simple example is sodium, which cannot adopt the +2 oxidation state because its second IE is simply too large, even though the lattice energy of a hypothetical $\ce{NaCl2}$ crystal is larger than that of $\ce{NaCl}$.

Firstly, the maximum oxidation state is limited by the number of valence electrons available. This is exactly analogous to the case of sodium; therefore, manganese does not exhibit the +8 oxidation state, because its eighth IE involves ionisation from the 3p subshell, much lower in energy than either the 3d or 4s. This therefore accounts for the increase in the maximum oxidation state going from Sc to Mn.

Secondly, the ionisation energies generally increase going across the 3d block, because of an increase in effective nuclear charge. This is the factor which leads to the decrease in the maximum oxidation state after Mn. For example, the seventh IE of Fe is larger than that of Mn; therefore, Fe does not exhibit the +7 oxidation state.

You can look up the IEs in any relevant textbook. However, note that since the +2 oxidation state is very common in the later half of the 3d block, it is usually the third and higher IEs that are more important in this discussion. So, looking at the trend of first/second IEs won't get you anywhere.


Vorlander's rule discuss about the directing powers of substituent present in benzene ring. Taken from here:

  Saturated groups (containing single bonds) direct predominately 2- and 4-positions.
  Unsaturated groups (containing multiple bonds) direct predominately at 3-positions.

Here are some examples:


Why is sodium cyanide an inorganic compound?

Sodium cyanide is classified as an inorganic compound for historic reasons, much like the entire organic-inorganic classification system is historical in reasoning.

The basic initial idea was that forming organic compounds required a vis vitalis, a life force, and could therefore only occur in living systems. On the other side, inorganic compounds were easily synthesised in vitro and have been synthesised even before alchemy was renamed to chemistry. These included most ionic compounds then known.

Naturally, sodium cyanaide is ionic. Most notably, early chemists would have been able to precipitate cyanide from other cyanide sources, redissolve it and add a sodium source (most likely sodium hydroxide) to regain sodium cyanide; thereby classifying it as inorganic.

In 1828, Friedrich Wöhler disproved the vis vitalis assumption and laid the foundation for synthetic organic chemistry. Yet the distinction between inorganic and organic chemistry was still upheld. Rather than defining it by synthesisability, it was said that any molecule containing carbon be organic — or one of many other similar definitions. This could exclude the cyanide salts but does not have to. The only salts and carbon-containing molecules that are always and consistently classified as inorganic are carbon monoxide, carbon dioxide and (hydrogen)carbonates.

By now, the distinction between organic and inorganic chemistry has lost most of its relevance and is only there for historic reasons. In principle, the distinction is that organic chemists are more interested in the carbon-containing compounds while inorganic chemists lean towards other elements more. But then there are numerous in-between fields that don’t really fit anywhere such as organometallic chemistry, bioinorganic chemistry etc.

Therefore, treat the organic/inorganic distinction as a historical residue and don’t read anything into it.


Why does sodium sulfate initially harden instead of dissolving?

Sodium sulfate forms various hydrates including a decahydrate. The dissolution is exothermic which saturates the solution. When the warm solution reaches the relatively cold frit the supersaturated solutions precipitates. In the glass frit the pores would clog and it would take time to dissolve the sodium sulfate out.


How dangerous is taking a bath in Coca Cola?

Cola and soft drink products are generally in the pH=2.5-3.5 range.
This is the same pH range as vinegar, lemon juice and most fruit juices. Note apple itself is around 3.0-3.5, so it is not about some magical trick of food industry to kill us all.

Does Cola dissolve your body traceless? No.
Do you want to soak your body in it for hours? Would you soak your body in vinegar for hours? Or even water...


Why is Promethium the only radioactive Lanthanide?

Lanthanoids are not generally supposed to be radioactive, with exception of mentioned very long lived radioisotopes.

The longest half-life radioactive nuclides of lanthanoids ( in years )

  $\ce{^{150}Nd}:\ \pu{6.7e18}$
  $\ce{^{151}Eu}:\  \pu{5e18}$
  $\ce{^{148}Sm}:\  \pu{7e15}$
  $\ce{^{144}Nd}:\   \pu{2.29e15}$
  $\ce{^{152}Gd}:\  \pu{ 108e12}$
  $\ce{^{147}Sm}:\ \pu{106e9}$
  $\ce{^{138}La}:\ \pu{ 102e9}$
  $\ce{^{176}Lu}:\ \pu{ 38.5e9}$

Promethium is rather a victim of circumstances there was no free position to have at least one stable isotope. It is related also to the Liquid drop model.

There are few semi-empirical rules, related to proton and neutron fermionic pairing within the nucleus model:

Elements with the odd proton number have maximally 2 stable isotopes.

2 Isotopes of adjacent elements with the same nucleon number, like $\ce{^N_{A}X, ^N_{A+1}Y}$, are not both stable. ( But some of these unstable ones  have very long half-life as seen above.)

Isotopes not close enough to the Valley of stability are not stable, as they have too little or too many neutrons.

The stability of isotopes of similar nucleon number generally decreases in order

both proton/neutron numbers are even
one number is odd, the other is even.
both numbers are odd.

For Pm and Tc, both have odd proton numbers what gives them disadvantage.

All their possible stable isotope candidates are already taken by their  neighbors as energetically preferred variants. Therefore all isotopes of given 2 elements beta decay to nuclei of adjacent elements with less energy.


Unfortunately, although the answer given by bon provides a very simplistic answer to a fairly common NMR-101 problem, it is not quite correct. It is fine for the propane case, but falls short for butane. While a very simple molecule, butane has a complicated spectrum because, though the two methylene groups are chemically equivalent, they are in fact magnetically non-equivalent. It serves as an example of how all too often NMR is poorly taught at the introductory level.

First, let's clarify a minor point of confusion that is widely taught at introductory levels - that equivalent protons do not couple to each other. They do; the reason that the coupling is not seen is that the transitions are degenerate and so sit directly on top of each other (ie not observable). Remember that the observed lines in a spectrum are the energy transitions between stable spin states - if the energy difference between two states is equivalent, then the lines in the spectrum are degenerate.

So, a better generalisation for interactions giving rise to splitting in a spectrum is that:

Coupling between magnetically equivalent nuclei cannot be detected
and does not affect the appearance of the spectrum
Coupling between
chemically equivalent, but magnetically non-equivalent nuclei will
usually affect the appearance of the NMR spectrum; commonly seen in
second order effects like that of AA'BB' systems.

Second, then, to see what a 1H spectrum of butane really looks like (well, a more realistic simulation at least):

Below is a more realistic simulated spectrum of butane (at 400MHz), which takes into account the fact that the two methylene groups couple to each other, with an average coupling of about 6.6Hz. Complicated isn't it? Have a read of Glenn Facey's page on his NMR blog, (an excellent NMR resource) where he discusses this very problem.

So why is the case? Well, we need to look at what gives rise to magnetic non-equivalence in molecules. There is an excellent on-line resource to explain about symmetry and non-equivalence at Hans Reich's NMR page. But for butane, let us consider the Newman projections:

At no time is the coupling between H2 and H3 the same as H2' and H3. Either it is psuedo-trans or pseudo-cis, but never the same. So, H2 and H2' are non-equivalent, and this will result in non-degenerate transitions, giving rise to complicated splitting patterns in the NMR spectrum. This is the same for other straight chain alkanes, with decreasing complexity as the chain length grows.

And just to show what a real spectrum of butane looks like: reproduced from Tynkkynen et al. (Magn. Reson. Chem. 2012, 50, 598–607), showing the real and calculated (at 500MHz):


Reaction of 2,3-dichlorobutanoate with a decarboxylating agent followed by hydrochloric acid and then aq. potassium hydroxide

Considering the follow up doubt, why geminal and not vicinal?

the carbocation formed is more stable because of $\ce{Cl}$ +M.


Which will be the major monochlorination product?

c is the least likely.

Alkane reactivity order is tertiary > secondary > primary > methyl reference here

I would be interested to see an explanation of why b predominates over a


How do flame tests give off color if they only emit the colors when the electrons return to ground state?

The flame area is a violent place due to high temperature favoring ionization. The color appears due to the dynamic process of the formation of short-lived species formed briefly in the flame. As we are burning hydrocarbons, they get split into unstable radicals easily loosing unpaired electrons. In the area of the flame enriched with electrons metal ions from the probe such as $\ce{Na+}$ are reduced to the metal atoms:

$$\ce{Na+ + e- -> Na^0}$$

which are emitting yellow light in a form of observable doublet of spectral lines ($λ_1 = \pu{589.0 nm}$, $λ_2 = \pu{589.5 nm}$) arising from the electronic transitions $\ce{3s^1 -> 3p^1}$. Note that it is the metal atom, not the ion, that is responsible for the emission of light of the given color.

So, as long as we keep the test probe (metal salt) in the flame, we can observe spectral lines of the given element as both conditions are fulfilled (energy supply (heat) and reducing atmosphere (free electrons from the fuel)). By taking the sample out of the flame, electron transitions are no longer possible as there is neither metal atoms capable of any, nor enough energy to promote the electron to the higher level.


They are isomers. Both of them have the same molecular formula ($\ce{C6H12O6}$) but different structures.

Structure of β-D-fructopyranose

Structure of β-D-fructofuranose


Why is the peak-to-peak separation of cyclic voltammetry is 57 mV?

The derivation of equation for cyclic voltammetry requires a lot of advanced calculus and differential equations based on the Fick's law of diffusion. This is not to discourage you but to show that the teacher is oversimplifying it. The derivation are provided in Allan J. Bard's Electroanalytical Chemistry. Why are they teaching cyclic voltammetry in sophomore classes?


You could oxidize some hydroxide ions, to wit:

$\ce{OH^- + O3 -> HO2^- + O2}$

Such an oxygen transfer is favored by strongly basic solutions because $\ce{HO2^-}$ is a weaker base ($\mathrm{p}K_\mathrm{a}$ hydrogen peroxide = 11.75) than $\ce{OH^-}$, as well as ozone being a more powerful oxidizer than diatomic oxygen.


Calculating the enthalpy of polymerisation of ethylene given the bond strengths

The mistake I was making was not noticing that every time a double bond is broken, i.e.; completely broken, 2 single bonds are formed in its place.

One bond between the 2 carbons of the same molecule and another between two molecules, the former of which I forgot to consider.

Thus, the enthalpy is in fact $y-2x$


Multiday lab courses in classic chemistry

Beside asking upfront the research groups, there are summer schools in normal (non-covid 19) circumstances.  For example Oxford's Chemistry Summer School, or ACS' Green Chemistry Summer School.  Given your account name, I speculate you reside in Europe; if so, a suitable directories page not only about core-chemistry may be  www.summerschoolsineurope.eu.  Sadly, because of the current pandemic, many scientific events are cancelled, though (example).  Still, references of courses in the past are retained (example), so you could find help by the former organizers as well.
Maybe you are an outlier by age in comparison to the «typical attendee».  However, the experience you gained after your master thesis, both in terms of chemistry under the hood as well as e.g., project and data management, may be of great benefit to the class providing additional perspectives.  In particular, I think about planning the experiments in the smaller groups, or discussing the results.  Equally, you may become a role model for younger participants in the way you choose to stay continuously engaged in learning.


Same Activation Energy for Forward and Backward Reactions

The van't Hoff equation describes how the equilibrium constant changes with temperature:

$$\ln \frac{K_2}{K_1} = −\Delta H_R^\circ (\frac{1}{T_2} − \frac{1}{T_1})$$

So if the reaction enthalpy is zero for a given temperature interval, the equilibrium constant will not change. This also means that the activation energy in either direction is the same, at least if entropy is not considered (the typical diagram explaining activation energy is for a single molecule with potential energy as the y-axis, enthalpy as the difference in energy between reactant and product, and activation energies as the difference between reactant or product and the activated complex).

  If so, please provide an example.

Take the transfer of protons across a membrane. Equilibrium is reached when the concentrations on either side are equal. A change in temperature will not change the equilibrium constant (which is 1, with a reaction enthalpy of zero).

Another example would be a reaction that turns one enantiomer into the other, e.g. D-alanine to L-alanine. In the absence of chiral co-solvents or solvents or other unusual circumstances, again K = 1 and enthalpy = zero over a range of temperatures.