How can I find the parent chains in these two compounds?

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the first relevant criterion to be considered in choosing a principal chain is the length of the chain.

When there is a choice for the principal chain, the following criteria are applied, in the order listed, until a decision is reached:

  P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. (…)

  The senior ring, ring system, or principal chain:

  (a) has the greater number of multiple bonds (P-44.4.1.1);

  (b) has the greater number of double bonds (P-44.4.1.2);

  (…)

  (h) has the lower locant for an attached group expressed as a suffix (P-44.4.1.8);

  (…)

With regard to numbering of locants within the same compound class, multiple bonds have seniority over simple substituent groups (e.g. methyl, ethyl, etc.) that are used for naming of branched alkanes. When there is a choice for the locants of multiple bonds, low locants are given first to double bonds:

  P-14.4 NUMBERING

  When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:

  (…)

  (e) saturation/unsaturation:

    (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;

    (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);

  (f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;

  (…)

Note that Rule e takes precedence over Rule f.

Furthermore,

  P-31.1.1.1 The presence of one or more double or triple bonds in an otherwise saturated parent hydride (…) is denoted by changing the ending ‘ane’ of the name of a saturated parent hydride to ‘ene’ or ‘yne’. Locants as low as possible are given to multiple bonds as a set, even though this may at times give ‘yne’ endings lower locants than ‘ene’ endings. If a choice remains, preference for low locants is given to the double bonds. In names, the ending ‘ene’ always precedes ‘yne’, with elision of the final letter ‘e’ in ‘ene’. Only the lower locant for a multiple bond is cited, except when the numerical difference between the two locants is greater than one, in which case the higher locant is enclosed in parentheses.
  (…)

Therefore, the preferred IUPAC names (PINs) for the examples given in the question are 3-ethylpent-1-en-4-yne and 3,3-diethynylpenta-1,4-diene, respectively.


The enthalpy is not additive under isobar, adiabatic conditions for non-ideal mixtures.

The change in the enthalpy is called the excess enthalpy (enthalpy of mixing). It can be computed with models such as the Margules, van Laar, Wilson, NRTL, and UNIQUAC.

These model give equations for the excess Gibbs energy. The excess enthalpy can be derived from these as described in 6.2 Fundamental Relations of Excess Functions in Molecular Thermodynamics of Fluid-Phase Equilibria (3rd Edition). (F'x already referred to this excellent book.)

One has to be careful with the conflicting sign convetions in the literature.


Why do only certain acids disassociate in water?

There are many factors that determine the strength of an acid:

electronegativity of the non-metal atom: an electronegative atom can attract more easily  the electrons of the hydrogen an so generate the proton.
atomic radius greater the radius of the conjugate base greater the acidity, this because the electrons are dispersed around the atom and there is a less attraction between the positive proton and the negative anion.
stability of the conjugate base if the conjugate base is stable due to some particular electronic configuration (e.g. stability due the resonance structure) the species are more prone to be dissociated.


If we go back to your earlier question on Frost diagrams (I've reproduced the key figure below), we see why simple molecular orbital theory or the "$4n+2$" rule suggests that benzene is aromatic while cyclobutadiene is antiaromatic.

Forming a planar, conjugated, 6-membered ring and placing 6 π-electrons in it creates a π-system that is energetically more stable than 3 conjugated C=C bonds (e.g. in hexa-1,3,5-triene). On the other hand, forming a planar, conjugated, 4-membered ring and placing 4 π-electrons in it creates a π-system that is energetically less stable than 2 conjugated C=C bonds (e.g. in butadiene).  For these reasons we say that the first system, benzene, is "aromatic", while the second system, cyclobutadiene, is "antiaromatic". Other measurements and physical phenomenon such as reactivity, bond lengths, ring currents, etc. support these conclusions.

Free cyclobutadiene has been observed as a transient intermediate.  Further theoretical analysis suggests that due to its lack of aromaticity it will distort from a square structure to a rectangular one with alternating single and double bonds (Jahn–Teller effect).

  sources claim that this instability can be attributed to other factors such as ring and angle strain rather than antiaromaticity

It turns out that other systems involving the cyclobutadiene skeleton have been prepared and studied.  Derivatives of both the cyclobutadiene dication[1] ($4n+2$, $n=0$) and the cyclobutadiene dianion[2] ($4n+2$, $n=1$) have been prepared and studied.  Their stability is much less than that of benzene, but much more than that of cyclobutadiene. Placing two charges in these small rings results in extremely high coulombic repulsions in both the dianion and dication, and this may be a significant part of the explanation as to  why they are less stable than say, benzene.  The fact that they exist as planar structures, with appropriate ring currents and are more stable than cyclobutadiene suggests that the instability observed with cyclobutadiene is not due to ring strain and angle strain alone.  Rather the aromatic stabilization - antiaromatic destabilization suggested by simple MO theory seems to be involved.

References

Olah, G. A.; Staral, J. S. Novel aromatic systems. 4. cyclobutadiene dications. J. Am. Chem. Soc. 1976, 98 (20), 6290–6304. DOI: 10.1021/ja00436a037.
Takanashi, K.; Inatomi, A.; Lee, V. Y.; Nakamoto, M.; Ichinohe, M.; Sekiguchi, A. Tetrakis(trimethylsilyl)cyclobutadiene dianion alkaline earth metal salts: new members of the 6π-electron aromatics family. Eur. J. Inorg. Chem. 2008, 2008 (11), 1752–1755. DOI: 10.1002/ejic.200800066.


Is a sequence of two Wagner-Meerwein rearrangements possible?

There is no reason why a sequence of two Wagner-Meerwein rearrangements wouldn’t happen — assuming low concentrations and thus the absence of directly capturing nucleophiles. In fact, a professor of mine once said:

  The only secondary carbenium ion is the prop-2-yl ion; all others will rearrange to form tertiary carbenium ions.

(He did explicitly exclude carbenium ions stabilised by other means from that discussion; of course an oxocarbenium ion is well-known as is an allyl cation.)

A very well-known example of this is the biosynthesis of lanosterol, in which the enzyme lanosterol synthase takes squalene-epoxide, orders it in a boat-chair-boat conformation, catalyses a domino sequence of carbocations being captured by double bonds, followed by a sequence of four Wagner-Meerwein rearrangements and finally deprotonation to generate the backbone ring-structure of chloresterols. See the following scheme.

If nature can do it, it will happen in the lab at least to a minute extent.


Classically permitted position values in a quantum harmonic oscillator

Assume for a minute that instead of having a quantum oscillator with total energy $E$ you had a classical one with the same amount of energy. That energy is the sum of potential and kinetic contributions:
$$E = E_\textrm{p}+E_\textrm{k}.\tag{1}$$
$E_\textrm{k}$ is zero and the total energy equal to the potential energy when the extension (amplitude) is at a maximum, when
$$E = E_{\textrm{p,max}} = \frac12 kx_\mathrm{max}^2\tag{2}$$
which, solving for $x_\mathrm{max},$ leads to
$$x_\mathrm{max} = \sqrt{\frac{2E}{k}}.\tag{3}$$
All that is left to do is to substitute in the right value for $E.$ That is dictated by the expression for the quantized system,
$$E_n = h\nu\left(n + \frac12\right),\tag{4}$$
so that the corresponding classical maximum amplitude is
$$x_\mathrm{max} = \sqrt{\frac{h\nu(2n + 1)}{k}}.\tag{5}$$


Why is a face-centred cubic unit cell not regarded as equivalent to a body-centred tetragonal unit cell?

In an FCC lattice  a=b=c while in a body centered tetragonal lattice a=b but c is unconstrained. The first example you gave should be that an end centered tetragonal is equivqlent to a primative tetragonal lattice. There is no end centered cubic lattice as that would violate the minimum cubic symmetry of 3 four fold symmetry axes. however the transition from FCC to a body centered tetragonal lattice would be displacive not transformational.


When do I use the hydronium ion in an acid-base equation?

In both your cases, only H+ actually participates in the reaction. Remember that HCl is a strong acid, and dissociates completely into H+ and Cl-. For example, consider the net ionic form of your second equation:

$\ce{C2H3O2- + HCL -> HC2H3O2 + Cl-}$

$\ce{C2H3O2- + (H+) + Cl- -> HC2H3O2 + Cl-}$

$\ce{C2H3O2- + H+ -> HC2H3O2}$

For your first equation, also note that, similarly, although the entire H30+ molecule participates in the reaction, only the H+ is transferred. (Also, I think you meant NH4+ instead of NH3 as a product.) So:

$\ce{NH3 + (H3O)+ -> NH4+ H2O}$

$\ce{NH3 + H+ -> NH4+}$

So basically, they both use the same compound (H+). Usually water is the main producer of H+ (through self-ionization). You just need to consider factors such as the common ion effect that would make other compounds such as HCl the main producer of H+ instead of water.


What causes hydrogen abstraction in the radical chain mechanism of alkane halogenation?

why does the halogen actually abstract hydrogen from the alkane?

Look at the two tables of bond energies on this webpage.  The second or bottom table shows the $\ce{C-H}$ bond energies for various carbon containing compounds.  In the top table we see that the H-Cl bond strength is 103 kcal/m.

If we start with a chlorine radical, then if we break a C-H bond and make a Cl-H bond, the energetics of the reaction will be

$$\ce{\Delta H=bonds~ broken~-~bonds~made}$$
$$\ce{\Delta H=C-H ~bond~ broken~-~H-Cl ~bond~made}$$
$$\ce{\Delta H=C-H ~bond~ broken~-~103~ Kcal/m}$$

Look at the top row in the bottom table again.  As long as our chlorine radical breaks any C-H bond that has a dissociation energy less than 103 kcal/m, our reaction will be exothermic, and the reaction will continue.

Edit

  As I understand, the first propagation step in the free-radical
  monochlorination of methane is slightly endothermic but is driven to
  completion by the exothermic nature of the second propagation step

The first propagation step involves breaking a $\ce{C-H}$ bond (103 kcal/m) and making an $\ce{H-Cl}$ bond (103 kcal/m), so it is roughly thermoneutral according to the above link.

$$\ce{CH3-H + Cl{.} -> CH3{.} + H-Cl}$$

$$\ce{CH3{.} + Cl-Cl -> CH3-Cl + Cl{.}}$$

The second propagation step involves breaking a $\ce{Cl-Cl}$ bond (58 kcal/m) and making a $\ce{CH3-Cl}$ bond (81 kcal/m), so it is quite exothermic.

These steps are consistent with my earlier argument.  You are correct that using free energies would be better (more accurate) than enthalpies, but the Tables I could find were based on enthalpies - I suspect they provide us with a reasonably accurate "back of the envelope" calculation and view of what is going on.

As to the actual mechanism of the reaction, once we've created a chlorine radical, we've created a reactive species.  It is reactive because it doesn't have an octet of electrons.  By undergoing the above-reactions it can regain its octet and become a stable entity.

Here is a diagram of a reaction coordinate (note we are using free energy as the Y-axis in this diagram).  Generally the more exothermic (exergonic for free energy) a reaction, the lower the activation energy required to reach the transition state and pass over from reactants to products.  So our reactive chlorine radical bounces around probing different parts of the reaction coordinate (e.g. different $\ce{C-H}$ bonds, different angles and approaches to these bonds) until it encounters a bond from the right direction, right angle, etc, such that the collision has enough energy to pass over the top of our reaction coordinate and become a product.


The black one is probably $\ce{ZnFe2O4}$; insoluble compound of very weakly toxic metal ions, or $\ce{Fe3O4}$. The red one is $\ce{Fe2O3}$ In terms of disposal all above counts as non-toxic mineral materials, similar to sand.

It is also possible to have some carbon in this setup, but I doubt it.

Anyway, you have nothing too toxic in this setup, you can safely dump it into ground without any meaningful lasting environmental consequences.

The problem may arise, if the steel used was stainless steel or another iron alloy, but it doesn't seem to be the case from your description (and they are rarely galvanized anyway)


The solid (crystalline) state is defined by structure, for which there can be many. Liquid means loss of long range order, unless you talk about liquid crystalline, where orientational, but not positional long range order, is partially retained. In the gas phase, also short range order is lost.

So, in the liquid or gaseous state, phases can only differ by the molecular structure of the constituents. For water, that leaves only one stable possibility. Sulfur has a few different molecular modifications, but they're hard to separate and they tend to interconvert, so they're not really phases.

There are a few liquids with two phases which differ not in structure but in quantum states. Helium I and II are an example, para and ortho hydrogen another.


Having trouble understanding the iron-carbon phase diagram

For any steel with less than 0.02% carbon we should see only ferrite, and for steel with >0.02% carbon we should see only pearlite.

You have got a small misunderstanding here. The maximum solubility of carbon at $723\ \mathrm{^\circ C}$ is 0.02%. At room temperature it is very low ($0.005%$ at $0\ \mathrm{^\circ C}$). Also for carbon level between 0.02 and 0.8 you wont see only pearlite, you will see a combination of ferrite and pearlite for sure. Only at 0.8% you can observe pure pearlite(100%). Also note that the phase diagram is in such a way that if you cool from the ferrite+ austenite region at $800\ \mathrm{^\circ C}$ with carbon content < 0.02% we can still get a ferrite + pearlite + even austenite/martensite.
Phase diagrams give an ideal picture. In real situations, it is very tough to get very slow cooling rates to get the equilibrium structure.


why does Zn become oxidised in the first place in operation of voltaic cell?

A voltaic cell is an electrochemical cell that uses a chemical reaction to produce electrical energy. In anode compartment you have Zn in contact with ZnSO4. In cathode cell you have two other reagents. For example Cu and CuSO4. activity series of the metals suggest that zinc is OK to lose electrons and turn into zinc ion. Copper ions from CuSO4 are strongly willing to turn into reduced copper metal. So zinc gives away electrons to copper ions because Cu$^{2+}$ has higher affinity to electrons than Zn$^{2+}$.

Why Cu$^{2+}$ has higher affinity to electrons than Zn$^{2+}$ is a separate question and it belongs to a different field (quantum mechanics/molecular orbital theory rather than electrochemistry).

In short - there is Zn metal that holds electrons weakly and Cu$^{2+}$ that will hold electrons strongly when it forms Cu metal. Hence, electrons flow to Cu$^{2+}$ from Zn.


Ketenes can participate in [2+2] cycloadditions with their $\ce{C=O}$ double bond.  Ketene dimerization is the best known example of such a cycloaddition.

(image source)

As you add substituents to the ketene $\ce{C=C}$ double bond, you begin to favor cycloadditon across the carbon-carbon double bond at the expense of cycloaddition across the carbonyl double bond.

Some Diels-Alder cycloadditions also involve adding across the ketene carbonyl.

(image source)

With both the [2+2] and [4+2] cycloadditions often the kinetic product involves cycloaddition across the carbonyl, but on mild heating this cycloadduct can rearrange to the product from cycloaddition across the carbon-carbon double bond, the thermodynamic product.

A very nice review of the synthetic and mechanistic aspects of ketene cycloadditions can be found here.


Yes, in fact this process is similar to the processes in lime stone that form cave stalagmite and stalactites but with calcium carbonates.
$$\ce{M(HCO3)2 <-> MCO3 + H2O + CO2 }$$
$$\ce{M} = (\ce{Mg},\ce{Ca})$$
Though I should note the process in not called dehydration.


A spike recovery value of >100% does not mean that you added too much or the recovery process is too good. All it means is that the analytical method which is being followed "in letter and spirit" gives a positive error. There may be a bias in the method or the precision of the method is poor. We cannot say just by looking at one number. At this stage it is assumed that there was no operator error; only the analytical method is to be blamed. A single number like 115% in analytical chemistry has no meaning. Whoever called it a good recovery process needs to revisit that statement.

Whenever we develop a new method, recovery is done is several matrices and even different types of samples. As a result one gets a range of recovery values in different matrices or for different samples. It is better to say that a method shows gives a recovery between 95% to 115%. This statement is far more meaningful than just stating that a recovery of 115% is good. Certainly, that method must be pretty good.


First of all, let me state the obvious: Phosphorus is awesome.
After we got that out of the way we can focus on why.

There are many different modifications of phosphorus in nature. With increasing thermodynamic stability they are $$\ce{P_{white} -> P_{red} -> P_{violet} -> P_{black}}.$$

Apart from this there are many low molecular different allotropes, like $\ce{P4 (white)}$, $\ce{P6}$, $\ce{P8}$, $\ce{P10}$, $\ce{P12}$. And because that is not enough, there are chain like polymeric allotropes, too.

Apart from this it is possible to formulate many different cations and anions, that are derived from the above molecular structures. Just to name a few, there are $\ce{P3+, P5+, P7+, P9+}$ mainly observed in gaseous phase an $\ce{P^{3-},P2^{4-}, P3^{5-}, P4^{2-}, P7-,...,}$ usually in combination with alkali metals. Most amazingly it may form anionic polymer strains of the general form $\ce{[P7-]_{\infty}, [P15-]_{\infty}}$.

But now for the most important part, phosphorus has stable oxidation states in compounds, ranging from $\mathrm{-III}$ to $\mathrm{+V}$. Here are a few examples:
$$\ce{\overset{-III}{P}H3,\ \overset{-II}{P_2}H4,\ [\overset{-I}{P}H]_{n},\ \overset{\mathrm{\pm0}}{P4},\ H3\overset{\mathrm{+I}}{P}O2,\ H4\overset{\mathrm{+II}}{P2}O4,\ H3\overset{\mathrm{+III}}{P}O3,\ H2\overset{\mathrm{+IV}}{P2}O6,\ H3\overset{\mathrm{+V}}{P}O4}$$
While dealing with these compounds it is usually completely unnecessary to describe bonding with hybrid orbitals.

In case of phosphane $\ce{PH3}$ it would be wrong. Assuming $\ce{{}^{sp^3}P}$ one would expect $\angle(\ce{H-P-H})\approx109^\circ$, while it is found to be $\angle(\ce{H-P-H})=93.5^\circ$, which is almost the same angle as the $\ce{p}$ orbitals are having towards each other.

In general your assumption is correct, that it is possible to form only three covalent bonds to reach a stable configuration. And that will most likely be the case when phosphorus forms compounds with more electropositive elements.

Now dealing with oxygen, means dealing with a much more electronegative element, i.e. $\ce{En(O)}\approx3.4$, $\ce{En(P)}\approx2.2$. That also means that bonds are much more polarised towards the oxygen.

Analysing the phosphate anion $\ce{PO4^{3-}}$ it is crucial to recognise its symmetry, which is tetrahedral $T_\mathrm{d}$. In this arrangement it is perfectly safe (but not at all necessary) to describe phosphorus as $\ce{sp^3}$ hybridised.

A Natural Bond Orbital analysis (BP86/def2-TZVPP) reveals that there are 4 equal $\ce{P-O}~\sigma$ single bonds and each oxygen has three lone pairs. The contribution of the $\ce{d}$ orbitals to bonding is well below $1\%$ and can be interpreted as numerical noise (use as polarisation functions) of the DFT method.
\begin{array}{rlrr}\hline
       &             &\mathrm{\%P (hyb)}  &\mathrm{\%O (hyb)}\\\hline
3\times&\ce{Bd(O-P)} & 24 (\ce{sp^3}) & 76 (\mathrm{sp^{2.3}})\\
       &\ce{Lp(O)}   &                & 100 (\mathrm{sp^{0.4}})\\
2\times&\ce{Lp(O)}   &                & 100 (\ce{p})\\\hline
\end{array}

This is consistent with the partial charges, i.e. $q(\ce{P})=2.2$, $q(\ce{O})=-1.3$. Therefore a more accurate Lewis formula is with charge separation.

The corresponding NBO reflect the bonding picture one would expect given all details from above. It should be noted, that natural bond orbitals are a linear combination of the canonical orbitals and do not have a physical meaningful energy eigenvalue. The two top rows represent the $\ce{p}$ lone pair orbitals of oxygen, the third row represents the $\ce{sp^{0\!.4}}$ lone pair orbitals, the fourth row gives the $\sigma$ bonding orbitals. (The last row is the orientation of the molecule, core orbitals are not displayed.)

The corresponding canonical orbitals which have a physical meaningful eigenvalue are delocalised over the whole molecule, hence they are not providing a simple understandable bonding picture. While NBO fail to respect the symmetry point group, canonical orbitals are constructed to obey this principle. (Here shown from highest energy, top, to lowest energy, bottom, core orbitals not displayed.)


What does (heated) mean in a chemical equation?

In terms of balancing the equation, the fact that heat must be supplied can be ignored.  It is just providing information about the conditions necessary to run the reaction.


Do metallic bonds contain London dispersion forces?

If by "Fe particles" you mean distinct particles (as opposed to the atoms inside a continuous Fe crystal structure), then the answer would be VdW/London forces.

From wikipedia:

This is the only attractive intermolecular force present between neutral atoms

Which, in a way, can be applied to particles or particle surfaces as well, as long as they're not charged.

Whereas, if you are talking about a continuous iron crystal structure, and by particles you mean atoms within that structure, then you have a metal-type bond between them caused by delocalization of electrons into the conductive band. See wikipedia for further details.


Which ligand names have to be parenthesized while naming the Mononuclear Coordination entities

Greek prefixes are used to designate the number of each type of ligand in the complex ion, e.g. di-, tri- and tetra-. If the ligand already contains a Greek prefix (e.g. ethylenediamine) or if it is polydentate ligands (ie. can attach at more than one binding site) the prefixes bis-, tris-, tetrakis-, pentakis-, are used instead.
Bottom line - parenthesize only those ligands which already has a greek prefix in its name so that you can highlight the use of the alternative prefixes such as bis-,tris- etc.
Ps -Diaquabis(ethylene-1,2-diamine)nickel(II) sulphate is the correct naming but you can make it more appropriate by using ethane instead of ethylene


Three stacked arrows, what does it signify?

There is an explicit answer in section 3. Other Horizontal Arrows of prof. Santiago Alvarez' published work [1, p. 594]:

[…]
Finally, multiple arrows (7) are used as an ellipsis for several intermediate steps in a chemical reaction.

Reference

Alvarez, S. Chemistry: A Panoply of Arrows. Angew. Chem. Int. Ed. 2012, 51 (3), 590–600. DOI: 10.1002/anie.201101767.


What does iron (III) get reduced to in a galvanic cell: solid iron or iron (II)?

$$\begin{align}
\ce {Fe^{2+} {(aq)} + 2e^{-} &<=> Fe {(s)}} &E^⦵= \pu{-0.44V}\\
\ce {Fe^{3+} {(aq)} + e^{-} &<=> Fe^{2+} {(aq)}} &E^⦵= \pu{+0.77V}\\
\ce {Mg^{2+} {(aq)} + 2e^{-} &<=> Mg {(s)}} &E^⦵= \pu{-2.37V}\\
\end{align}$$

Thus

$$
\begin{align}
\ce {Mg {(s)} + 2Fe^{3+} {(aq)} &<=> Mg^{2+} {(aq)} +2Fe^{2+} {(aq)}}&&E^⦵_{\text{cell}}= 0.77--2.37=\pu{+3.14V}\\
\ce {Mg {(s)} + Fe^{2+} {(aq)} &<=> Mg^{2+} {(aq)} +Fe {(s)}}&&E^⦵_{\text{cell}}= -0.44--2.37=\pu{+1.93V}\\
\end{align}$$

Both reactions are feasible, and I'd imagine the first would proceed until the concentration of $\ce {Fe^{3+} {(aq)}}$ is very low, and then the second would start at a noticeable rate.
This assumes, of course, the iron solid is able to deposit on the carbon cathode, and I don't see why not.

Hope this was helpful.

(Thank you Dr. J for pointing out my typo)


How can Nernst equation apply in this redox reaction?

The flaw in your reasoning is that you are believing that $E^\circ$ is a constant, when really it is a function of temperature.

See Standard Electrode Potentials and Temperature Coefficients in Water at 298.15K for a linear approximation of the temperature dependence of $E^\circ$.


Eli, you should have read the paper very carefully before asking questions. I think you are very intelligent person, so don't choose easy way out. Now, to answer your question, about mol%. It is right in the paper. As you already figured out and also Orthocresol pointed out in his comment, The amount equals to given mol% is $LR \times \frac{given \ mol\%}{100}$. It is in the General procedure A of given reference:

  Condensations of phenols and naphthols with $\alpha,\beta$-unsaturated ketones. An oven-dried $\pu{25 mL}$ Schlenk tube equipped with a teflon-coated magnetic stir bar was evacuated and purged with argon. The phenol ($\pu{0.2 mmol}$), pentafluorophenylboronic acid ($\pu{8.5 mg}, \ \pu{0.04 mmol}, \ \pu{20 mol\%}$), diphenylphoshinic acid ($\pu{8.7 mg}, \ \pu{0.04 mmol}, \ \pu{20 mol\%}$), $\alpha,\beta$-unsaturated ketone ($\pu{0.4 mmol}$) and heptane ($\pu{1 mL}$) were added to the tube under argon.....

Accordingly, The phenol used: $\pu{0.2 mmol}$ ($LR$), pentafluorophenylboronic acid used: ($\pu{0.04 mmol}$ meaning $0.2 \times \pu{20 mol\%}$), and diphenylphoshinic acid ($\pu{0.04 mmol}$ meaning $0.2 \times \pu{20 mol\%}$).


Interpreting Infrared Spectroscopy (IR) Spectra

In your first spectrum, the band at 3517 $\mathrm{cm^{-1}}$ indeed fits best to the $\ce{N-H}$ stretch of a primary amine (source). However, I would not assign the shoulder of this band to a $\ce{#C-H}$ stretch. Such a band would be sharper and perhaps more intense, and as Greg E. already pointed out in his comment to your question, there is also no corresponding $\ce{C#C}$ stretch band present (reference). The absence of bands in the range between 2500-2000 $\mathrm{cm^{-1}}$ allows you to rule out some other functional groups with double or triple bonds, like nitriles, thiocyanates, isocyanates, carbodiimides or azides. The band at 1715 $\mathrm{cm^{-1}}$ is a typical $\ce{C=O}$ stretch, and while it could be a ketone, the presence of the broad band at 2932 $\mathrm{cm^{-1}}$ (which could actually be a double band) makes it likely to be an aldehyde, since the double band can then be assigned to the $\ce{=C-H}$ stretch of the CHO group (it could also include other $\ce{C-H}$ stretches due to its width).

Your second spectrum appears a bit more complex. The broad, intense band at 3298 $\mathrm{cm^{-1}}$ in combination with the band at approximately 1680 $\mathrm{cm^{-1}}$ could be assigned to an amide, but I would expect a stronger carbonyl band in this case. Nevertheless, the intensity and width of the 3298 $\mathrm{cm^{-1}}$ band suggest the presence of a functional group capable of H-bonding. The bands between 2900-2800 $\mathrm{cm^{-1}}$ could be assigned to alkyl $\ce{C-H}$ stretches (source). For the compound to be aromatic, there should be aromatic $\ce{C-H}$ stretch bands present, slightly above 3000 $\mathrm{cm^{-1}}$ (source).


Which rings of this heterocyclic compound are aromatic?

I think that all the four rings are aromatic.

If the top nitrogen donates it's pair of electrons to the ring, and oxygen takes it's pair, the three fused rings would have 14 electrons in the rings' peripheral, causing it to be aromatic:

Even the 3D drawings of this molecule confirm it's planarity of the 3 fused rings:

I think we can at least suspect all the rings in the molecule to be aromatic.


Which are the differences between resonance Raman and fluorescence?

The main difference between Raman scattering and fluorescence is the excited state lifetime. Fluorescence excited states are longer-lived than the 'virtual' states associated with Raman scattering.

In fluorescence, absorption of light excites an electron to a higher energy state. The lifetimes of these excited states are long enough where the geometry of the molecule can relax to accommodate this new electron configuration. However, this new geometry is not equivalent to the lowest energy structure, so when the molecule returns to the ground state, the energy of the system will be slightly higher than where it started. This means that the fluorescence energy is always lower than the excitation energy (this is true for molecules, not atoms).

For Raman scattering, light absorption promotes the molecule to a very short-lived 'virtual' excited state. When this state relaxes, it can return to a different vibrational or rotational sublevel of the ground state than where the molecule started. This leads to slightly different energies than the elastically scattered light, which is at the same energy as the incoming radiation source.

Resonance Raman is a special case where the excitation energy is tuned to a particular electronic transition of a molecule. This has the advantage that vibrational transitions associated with the electronic change are substantially enhanced compared to other Raman transitions, which simplifies the Raman spectrum. Because you are exciting a specific electronic transition, though, it is possible to get a substantial fluorescence background as well.

This J Chem Ed article has a nice description of Raman versus fluorescence. ChemWiki has a page devoted to the differences between Raman and Resonace Raman.


Derivation of an equality in Michaelis–Menten kinetics

Your equation

  $$\ce{Enzyme + Substrate <=>[k_1] ES complex ->[k_2] Enzyme + Product}$$

contains only two rate constants, $k_1$ and $k_2$, but not $k_{cat}$ you refer to.

The correct schema for the Michaelis-Menten kinetics would be

$$\ce{Enzyme + Substrate <=>[k_1][k_{-1}] ES complex ->[k_2 = k_{cat}] Enzyme + Product}$$

Note that $k_{cat} = k_2$ is the rate constant for the decay of the ES complex in the second step.

From this you can derive the Michaelis-Menten rate law as described in the Wikipedia article you refer to.

$[E]_0$ is the initial concentration of the enzyme. During the reaction the enzyme exists as free enzyme ($[E]$) and bound in the enzyme substrate complex $[ES]$. These concentrations are related by the equation $[E]_0 = [E] + [ES]$.


In what compounds does oxygen exhibit +½ oxidation state?

Yes. The dioxygenyl ion, $\ce{O_{2}^+}$, is a rarely-encountered oxycation in which both oxygen atoms have a formal oxidation state of +1/2. It is formally derived from oxygen by the removal of an electron:

$$\ce{O_{2} → O_{2}^+ + e^{-}}$$

Source


How to rationalise the resonance structures and hybridisation of the nitrogen in a conjugated amine?

First of all, are they correct? ChemBioDraw had some complaints, but as far as I can see there's the same amount of electrons, and no valence orbitals exceeding capacity.

Yes, these are the six most important resonance structures for this compound. The reason ChemDraw complains is that it is trying to act smarter than you, and it most certainly is not. It interprets that negative formal charge on the carbon atom as implying a lone pair, since carbon can only have the negative charge if it also has a lone pair. When you add the lone pair and the charge, ChemDraw is suddenly stupid and thinks you have exceeded the octet on carbon (3 bonds + 1 explicit LP + 1 implied LP from the charge).

  In the first structure, nitrogen is $\mathrm{sp^3}$ hybridised, but on all others it's clearly $\mathrm{sp^2}$ hybridised. So what does that mean? It's somewhere in between, but closer to $\mathrm{sp^3}$-hybridised?

If the lone pair on a nitrogen (or on any atom)) participates in resonance, then that nitrogen (or whatever) atom must be $\mathrm{sp^2}$-hybridized so that the the lone pair is in a $\mathrm{p}$-like orbital to ensure appropriate symmetry for $\pi$-overlap. Put another way, if an atom is $\mathrm{sp^2}$-hybridized in one resonance structure, then it is $\mathrm{sp^2}$-hybridized in all of them. Atoms that are $\mathrm{sp^2}$-hybridized and $\mathrm{sp^3}$-hybridized have differing geometries, which is not permitted in the resonance phenomenon.

In truth, hybridization is an approximation we make to make quantum mechanics jive with molecular geometry. Geometry is real (empirically determinable) and hybridization is fictitious (not empirically determinable), but hybridization makes QM behave better conceptually and mathematically. Hybridization is also a useful predictor of chemical reactivity of various bonds and functional groups in organic chemistry. Experimentally, I would guess that the nitrogen atom is trigonal planar (or very close to it). Trigonal planar implies $\mathrm{sp^2}$-hybridized (not the other way around).

In molecular orbital theory, we do away with the need for both resonance and hybridization. This molecule would have 7 $\pi$ orbitals formed from linear combinations of the p orbitals on the 6 participating carbon atoms and the nitrogen atom. The probability density function plots of these 7 orbitals will be more complex than you might be used to, but they will suggest the same electron density and charge density as a resonance hybrid assembled from your six resonance structures. See the Wikipedia article on conjugated systems, which is not very good but will give you the idea.

  My reasoning is that the first structure contributes more.

The five charge-separated resonance structures are more important than you think, since they imply that the five-membered ring is aromatic like the surprisingly stable cyclopentadienyl carbanion. However, you might not yet be this far along in your studies of organic chemistry. At the introductory level of understanding of resonance in organic molecules, the first structure is most important for the reasons you list.


What happens if a buffer is made up with only half as much NaH2PO4 compared to Na2HPO4?

You are not right, Le Châtelier's principle plays a very important role.

Let's have a look at the reactions involving disodium monohydrogen phosphate, as it is an amphoteric substance and therefore a buffer on its own:
\begin{align}\ce{
Na2HPO4 + H2O &<=> 2Na+ + H3+O + PO4^3- \\
Na2HPO4 + H2O &<=> 2Na+ + {}^{-}OH + H2PO4^- \\
}\end{align}
To a small extent, there will also be the formation of phosphoric acid, but we'll ignore that for the moment.

Let's ignore the counter ion and formulate the equilibrium constants:
\begin{align}
\ce{HPO4^2- + H2O &<=> H3+O + PO4^3-} &
K_1 &=\frac{c(\ce{PO4^3-})\,c(\ce{H3+O})}{c(\ce{HPO4^2-})\, c(\ce{H2O})} &
= K_\mathrm{a}\cdot c(\ce{H2O})\\
\ce{HPO4^2- + H2O &<=> {}^{-}OH + H2PO4^-} &
K_2 &=\frac{c(\ce{H2PO4^-})\,c(\ce{{}^{-}OH})}{c(\ce{HPO4^2-})\, c(\ce{H2O})}&
= K_\mathrm{b}\cdot c(\ce{H2O})\\
\end{align}

The whole reaction can be described with a coupled equilibrium constant:
$$K = K_\mathrm{a}\cdot K_\mathrm{b}\cdot c^2(\ce{H2O})$$

At all times in a pure solution there will be phosphate, monohydrogen phosphate, dihydrogen phosphate and maybe even phosphoric acid in solution. Changing any of the concentrations of these species will change the equilibrium.
The same equations can be applied to the sodium dihydrogen phosphate system. The initial concentrations of course will be different.
Depending on the ratio of the two solutions, the pH will also change.

The buffer capacity is of course not only dependent on the overall concentration, but also if it is near the ideal buffer point.

Overall, polyprotic acids are very complicated systems and it solely depends on your understanding of the word "excellent" how the buffer behaves. There are many fine tuned buffer systems for various purposes, this website has quite a few neat tables.


Why don't gases of elements with negative electron affinities exist as ions in nature?

It totally does not work like that. You can't say that $\ce{2F^-}$ is less stable (or more stable) than $\ce{F2}$; you can't compare them at all. People would say "B is more stable than A" when there is a spontaneous reaction $\ce{A\to B}$, but this is not the case here. A reaction must be balanced in charge and in all elements, otherwise it is just a bunch of letters that makes no sense. There is no reaction $\ce{F2<->2F^-}$ (either way), much like there is no reaction $\ce{F2->Au}$.

There is indeed a reaction $\ce{F2 + 2e- -> 2F^-}$, and it is quite favorable energetically (as you already know), but that's another story. You need the electrons for it, and they are just not there. Fluorine can't wrestle those electrons from another fluorine, as explained by IT Tsoi. And if you have lithium around, why, of course they would react to form $\ce{LiF}$, but that's a chemical compound, not just $\ce F$ by itself; also, at room temperature it is solid.

There is more to it. A gas (as well as a liquid or solid, for that matter) formed of negative ions alone is very much impossible. The ions will repel with such force that no material in the world will hold them even for a second. (I mean macroscopic amount, not a thin trickle of ions inside a mass spectrometer.) So you can't have fluorine in this state even if you have a stash of electrons which you are willing to share. $\ce{F-}$ ions do exist, of course, and there is nothing exotic about it (arguably, almost all fluorine in the world is found in this form), but they exist as a part of various ionic compounds, and not by themselves.


Second order reaction with two reactants

Assuming mass action kinetics, we have the following differential equation

$$\frac{\mathrm d}{\mathrm d t} [\ce{A}] = -\kappa \, [\ce{A}] [\ce{B}]$$

where $\kappa > 0$ is the rate constant. If we denote

$$x (t) := [\ce{A}]_0 - [\ce{A}] = [\ce{B}]_0 - [\ce{B}]$$

then

$$\frac{\mathrm d}{\mathrm d t} [\ce{A}] = \frac{\mathrm d}{\mathrm d t} \left( [\ce{A}]_0 - x \right) = 0 - \dot x = - \dot x = -\kappa \left( [\ce{A}]_0 - x \right) \left( [\ce{B}]_0 - x \right)$$


Differentiating between Elementary and Non Elementary reaction

Reactions that are probably not elementary

Whenever there are more than three reactants, it is unlikely that the reaction is elementary. Tenary elementary reactions are already quite rare. If you have a reaction like

$$\ce{2C8H18 + 25 O2 -> 16 CO2 + 18 H2O}$$

it is impossible for 27 particles to come together at once, so you would expect some intermediates. Even for a reaction with just 3 reactant molecules such as

$$\ce{CH4 + 2O2 -> CO2 + 2H2O}$$

it is highly unlikely that a single collision would be sufficient to break 6 bonds and make 6 new ones.

Conversely, a reaction with a single product such as

$$\ce{O + O2 -> O3}$$

is probably not bimolecular because there is no way of getting rid of the excess energy. Instead, the reaction proceeds like this:

$$\ce{O + O2 + X -> O3 + X}$$

where X is "a third molecule" that leaves the collision chemically unchanged but with higher kinetic energy (source: https://cimss.ssec.wisc.edu/wxwise/ozone/OZONE2.html).

Reactions where you can't tell

Bimolecular simple reactions could well be elementary, but often are not. The reaction

$$\ce{H2 + I2 -> 2 HI}$$

shows two distinct mechanisms, one with one elementary step as written (the iodine has to be vibrationally excited, favored by high temperature) and the other one with iodine first dissociating into iodine atoms (source: https://link.springer.com/chapter/10.1007/978-3-642-80299-7_12).

Is it possible to tell just by looking whether a given is reaction is elementary or non elementary?

No, but sometimes it is obvious that the reaction has to be multi-step. In cases where a mechanism is dependent not only on the net reaction, but also on reaction conditions, it becomes obvious that just knowing the net reaction is not sufficient.


How much water has boiled off when precipitation begins?

One expects 25 g of $\ce{NaCl}$ to saturate 62.5 g of water at 100 °C. However, 25 g of $\ce{NaCl}$, 58.443 g/mol, is 0.4278 mol in 62.5 ml of water at 100 °C. Colligative property boiling point elevation is +0.52 (°C kg)/mol. 0.8556 mol of particles in 62.5 ml of water is 13.69 mol/kg or 7.1 °C elevated.

$\ce{NaCl}$ solubility/temp increases from about 36 g/100 g at 0 °C to 40 g/100 g at 100 °C or 0.04 g/°C. This gives first approximation 0.2847 g more solubility. "25.28" g $\ce{NaCl}$ in 62.5 g water at 107 °C is then 25 g $\ce{NaCl}$ in 61.8 g water. 88.2 g water boiled off (with some decimal trim for Zeno’s paradox).

Repeat the problem with ammonium cyanate.


How does sulbutiamine work in the same way as thiamine, even if it doesn't have a thiazole ring?

To get started, here are the structures of some compounds that are relevant.

Thiamine

Sulbutiamine

Allithiamine

Looking at these structures we see that you are of course correct, sulbutiamine does not contain a 5-membered thiazole ring like thiamine.

Thiamine (vitamin B1) was an early focus of interest because sailors at sea for long periods of time developed beriberi due to a nutritional thiamine deficiency.  Since those early days, thiamine substitiutes were of medicinal interest.  One of the early substitutes was allithiamine.  It is a natural product that can be extracted from garlic.

You can see the structural similarities between allithiamine and sulbutiamine.  Sulbutiamine has the advantage that both sides of the disulfide (S-S) bond contain the same active ingredient, so one would expect increased pharmacological properties from sulbutiamine.  The advantage that both allithiamine and sulbutiamine possess over thiamine is that they are more soluble in lipids (the positively charged thiazole ring in thiamine makes it very soluble in water, but not so soluble in lipids).  This enhanced solubility in lipids means that they can cross the blood-brain barrier much more effectively than thiamine, this improves their bioavalabilty.

With that background we can now answer your question, "how does sulbutiamine work in the same way as thiamine."

Once in the brain, sulbutiamine can be easily metabolized to thiamine.  Look at the structure of sulbutiamine and here is what happens:

reductively break the disulfide bond and form two identical mercaptans
$$\ce{RS-SR ->C[{[H]}]\   2 ~RS-H}$$
protonate the acetyl (that's on the nitrogen that will become part of our thiazole ring) oxygen to make it more attractive to nucleophilic substitution
Let the mercaptan sulfur nucleophilically attack the acetyl carbonyl carbon
eliminate water and form the 5-membered thiazole ring
and we now have thiamine

But a much larger dosage in the brain than had we simply ingested thiamine.


Is the Beckmann Rearrangement technically an "SN2" reaction at an "sp2" center?

When you were taught that we don't see $S_{\mathrm{N}}2$ reactions at $sp^{2}$ centers, what evidence or rationale was given? In organic chemistry, very few things are definite. When they are, it's usually because of being forbidden by quantum mechanics (e.g., symmetry in pericyclic reactions). Otherwise, they are just very unlikely, unless you somehow magically make the conditions right.

Greg Fu used to give this example: J. Am. Chem. Soc. 1998, 120, 2275-2282.

In it, a trans-substituted alkene undergoes nucleophilic substitution to become a cis-substituted alkene. Using that fact and other data, the authors conclude that the mechanism most consistent with the observed reaction data is a vinylic $S_{\mathrm{N}}2$. If you have a better mechanism, consider writing a paper and submitting it to JACS.


C-H Bond Dissociation Energy varies with substitution

Nature always try to acheive least potential energy for stability . Any organic reaction which makes the product less stable is not favourable .

If the C-H bonds are cleaved homolytically in all cases , the resulting stability of the free radicals will tell us about the feasibility of the reaction .

The primary free radical is the most unstable and to bring about its C-H bond breaking would mean increasing its energy comparitively . Hence its bond energy is the most and so its the stronger bond of the three .

Tertiary free radical is the most stable while secondary lies in between them . The same explanation can be used to draw out their comparitive strength of the C-H bonds .


How can transition metals form so many bonds with ligands?

Copper is  in +2 oxidation state so it has configuration $\ce{[Ar] 3d^9}$. Since 4s, 4p and 4d orbitals are vacant, Cu will show $\ce{sp^3d^2}$ hybridisation and will accept these elctron pairs in its 6 $\ce{sp^3d^2}$ hybridised orbitals. Read more about VBT here http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch12/valence.php


Colour changes in coordination complexes

Many things can happen when you heat a coordination complex: ligand loss, reactions in the ligand structure, loss of hydration water, isomerization, oxidation/reduction of the metal center etc.

In the specific case of potassium iron(III) oxalate trihydrate, as reported in this paper, the thermogravimetric analysis shows that the first structural change happens at 113 ºC with the loss of three water molecules (from the trihydrate salt), yielding potassium iron(III) oxalate, a yellow-orange solid.
When the temperature reaches 296 ºC (in air), iron(III) is reduced to iron(II) to form iron(II) oxalate, an orange solid.

Further heating leads to the decomposition of iron oxalate into FeO at around 470 ºC. FeO is a brown/red solid (probably the compound you mentioned in your question).


What do chemical constituent measurement units 'g%g' and 'µg g⁻¹' mean?

$\pu{g\%g}$ literally means grams per cent grams or grams of mineral per $\pu{100g}$ of sample. So, in population 1 there is $\pu{2.33g}$ of $\ce{K}$ per $\pu{100g}$ of leaves. Not bad when you consider cooked spinach only has about$\pu{0.05\%}$ $\ce{K}$! Eat your greens, kids. About $\pu{2\%}$ $\ce{K}$ in plants is average across a broad spectrum of plants.

$\pu{ug g-1}$ is micrograms of mineral per gram of sample. Population 1 has 23 micrograms of $\ce{Na}$ per gram of leaves.

You should be able to convert these to percents or ppm, as needed, and check in the text as to whether this refers to dry weight of plant or not. I'm guessing with this multivariate array of mineral analyses, the origin of particular cannabis crops could be determined.


How do I calculate the change in energy of an electron transition?

When an electron transitions from a higher energy state to a lower energy state, the energy has to go somewhere. Usually it is emitted as light of a given wavelength which corresponds directly to the change in energy of the electron.

The energy of light at a given frequency $\nu$ is
$$ E = h\nu\; , $$
where $h$ is Planck's constant.
The frequency of light is directly related to the wavelength $\lambda$ and the speed of light $c$ via
$$ c = \nu\lambda\; . $$

Combining these two equations we get
$$ E = \frac{hc}{\lambda}\; , $$
with which you should be able to calculate the $\Delta E$ values for the transitions.


Mercury is toxic, but you need to carefully define what you mean by toxic or you draw incorrect conclusions

Toxic is a broad term. It means a lot of different things. The timescale matters. Some toxic things take years to exhibit their effects; others act instantly.

A binary distinction between toxic and not-toxic is pretty meaningless: you need to define the context and the timescale of the toxicity.

Mercury metal and mercury compounds are usually considered toxic. But their effects are varied in time and degree. Mercury metal is pernicious but only if you are exposed to it over a long time period. In fact you could probably drink it with few ill effects. The body just doesn't absorb it quickly. What is dangerous about mercury is not short term exposure to the metal but long term exposure to the vapour. This is why people don't suffer immediate ill effects when handling the metal even without skin protection.

Mercury vapour is readily absorbed in the body and will accumulate in tissue causing a variety of long term effects. This was discovered by mercury miners who often developed long term problems from their exposure. And it was documented for science by some chemists who started to suffer effects after working with the metal over long periods of time and managed to document their own decline (see Stock's work, for example). Mercury metal is often widely used in laboratories to provide a limited overpressure for gas distribution (you allow the gas to bubble through a mercury manometer).

Since the toxicity was recognised, chemists have been a lot more careful and always avoid vapour buildup by working in well ventilated spaces and making sure that manometers containing mercury are vented safely to the outside (via scrubbing filters) along with other potentially toxic vapours.

There is little immediate risk when working with metallic mercury as long as you don't spill it somewhere where it will collect and allow vapour to build up in the atmosphere.

Mercury compounds are a bigger risk. Some are readily absorbed into the body. The worst sort are mercury organometallic which are both volatile and penetrate the skin quickly. If you work with those you need to take extreme precautions. Even experienced chemists have been killed by accidents involving things like methyl mercury (see this tragic story).


How can I filter dissolved PVA from water

Rig up a column that you can fill with granular activated carbon (GAC). Throw batting on either end to prevent carbon fines from entering your dissolution chamber. Replace the carbon when you reckon it is no longer working. If you can afford the heat, throw a cooling coil in front of the filter.

Another method is indicated in this patent: US4078129. They add 0.25g sodium borate and 0.6g sodium sulfate per liter to precipitate between 0.03 and 3% PVA.
This may not be a good solution if you are prone to overdosing, as it would say, goop your soup, when you go to do the next batch.

A third method seems to be already popular at the large scale: ultrafiltration. There are many papers discussing its usage for waste water discharge applications. Use the smallest pore UF you can find (50kDa is what I've seen in the literature). The pressures required to operate it are pretty low. Be sure the temperature of the solution is <50 degrees C prior to entering the UF (almost all membrane applications fail >50 degrees C). Run it outside-in. Don't dead-head it. You will generate a small concentrate stream. Perhaps try method 2 on that to reduce waste generation.


You're almost there!

For 8:  For every mole of acid added, it reacts with a mole of the base included in the Pt(NH3)xCly, which is the ammonia.  So you can figure out how many moles of ammonia have reacted, and compare this with the moles of salt you started with.

For 9: For every mole of AgCl you have formed, you have used up one mole of Cl from the Pt(NH3)xCly.  So you can figure out how many moles of Cl you have captured, and compare this with the moles of salt you started with.


What determines the high degree of selectivity in free radical brominations?

First, let’s take a look at the radical chain reaction that leads to the substitution in question:

$$\begin{align}\ce{Br^. + H-R &-> HBr + R^.}\tag{1}\\[0.4em]
\ce{R^. + Br-Br &-> R-Br + Br^.}\tag{2}\end{align}$$

Obviously, the selectivity is determined in the first step as long as no radical recombinations take place. If we break down the process in the first step into sub-processes, we get:

cleavage of the $\ce{C-H}$ bond;
formation of the $\ce{H-Br}$ bond.

Again, the second is the same no matter which radical is formed; therefore, it must be the cleavage of the $\ce{C-H}$ bond that leads to selectivity. Indeed, you can imagine radicals as being electron-deficient compounds and therefore behave similar to cations. Being similar to cations means that you can assume hyperconjugation to be a viable stabilisation mechanism for radicals; therefore, the energy lost by homolytically cleaving a tertiary $\ce{C-H}$ bond is slightly less (a less unstable product) than that lost cleaving a primary $\ce{C-H}$ bond. Exploiting this energy difference is the basis of selectivity.

This selectivity is higher if bromine is used than if chlorine is used. The reason is precisely the Hammond postulate you cited: the reaction with bromine is predicted to be more endothermic and thus the transition state more product-sided. A layman’s explanation would be that bromine can approach slowly and try out whether it feels happy about reacting with this hydrogen — think of playing Jenga and carefully testing if a brick is moveable or not. Chlorine on the other hand, is faster to react and thus has a less product-sided transition state.

There is a second factor that reduces selectivity. Assume 2-methylbutane; this compound conveniently has primary, secondary and tertiary hydrogens. Probabilities of bromine radicals to react with each of these hydrogens have been calculated; since I don’t have actual numbers on me right now, I am going to ballpark the following: assume, that a secondary hydrogen has a five times as high probability of reacting with bromine when compared to a primary one and that the tertiary hydrogen has a 15 times as high probability. Naïvely, we could assume that the reaction yields products with a ratio of $1:5:15$ — but we need to remember how many hydrogens of a given type there is.

The possible products are 1-bromo-3-methylbutane, 1-bromo-2-methylbutane (both primary attacks), 2-bromo-3-methylbutane (secondary attack) and 2-bromo-2-methylbutane (tertiary attack). For each of these, the following number of hydrogens can react to form the product in question:

1-bromo-3-methylbutane: 3
1-bromo-2-methylbutane: 6
2-bromo-3-methylbutane: 2
2-bromo-2-methylbutane: 1

Thus, to accurately calculate the ratios, we need to multiply the preference factor with the number of hydrogens, resulting in a product ratio of $3:6:10:15$ in the same order as presented — a lower fraction of tertiary bromide than presumed.


An electrochemical cell works on the principle that the Gibbs free energy gets converted into electrical energy by making the electron transfer indirect. This is done by itself to make the 'heat energy' that would normally be released in a redox reaction into 'electrical energy' so ideally heat shouldn't be lost.

Also you say you want to charge the cell by heating it so as to make the reaction go in the opposite direction. But for a spontaneous cell reaction, increasing the temperature makes the Gibbs free energy more negative and $\Delta G^\circ = -nFE^\circ$ ($E^\circ$ is standard cell potential).

So the $E^\circ$ value increases, and so does the concentration (because of backwards reaction), so the cell doesn't actually get charged up.

Ideally just charge it by passing current in the other direction.


experimentally determine oxidation state of Iron compound

In all the (physical) experimental literature I've read, at least for biomolecules (usually enzymes), the technique of choice is Mossbauer spectroscopy. Here is a good starting point.


How to interpret TD-DFT results for finding λmax and amplitude

You are looking at the wrong numbers.

You should look for excitation  that has non-zero oscillator strength. In the text file "State # 1,2 .." and in Chemissian on the left, list with columns "# Energy Strength" The transitions with zero or very small oscillator strength has a very low intensity in the spectrum, so you don't see them.
In your case the relevant excitation are #2 (Energy: 6.691 eV, Strength: 0.027) and #4 (Energy: 8.353 eV, Strength: 0.042), and you can see them as small peaks on the graph down.

The "amplitude .. 1.41" has nothing to do with the oscillator strength. It tells you which orbital contributions are the most important ones for the given excitation.


How does the smell of a compound come about, and is it possible to define a smell?

Smell is what you feel when molecules of some compound (and not any kind of radiation, mind you) touch the olfactory nerves deep inside your nose. It is just another property of a compound, like molecular weight, or melting point, or color, only different.

Unlike molecular weight, it can't be calculated from molecular formula by simple rules.
Unlike molecular weight or melting point, it can't even be adequately described with a single number.
Unlike all of the above, it can't be reliably predicted for a new compound, though much can be done by the way of analogy.

$\ce{H2Se}$ smells worse than $\ce{H2S}$, and $\ce{H2Te}$ is worse still. A disturbing characteristic of the latter, given by an anonymous expert in the field, I'd better keep hidden so as not to offend the public taste:

   Smell it, and you'll vomit farther than you can see.


How does the steric effect affects the electronic density of the central atom?

Both methyl and ethyl groups have +I effect and its obvious that the ethyl groups have more electrons than the methyl groups, so in this case the present of more electron density on nitrogen atom with ethyl groups lessen the -I effect of the nitrogen atom toward the ethyl groups compared to that on the methyl groups. I think this should be the reason.


What is the equation for solving the concentration of the unknown from a Single-point Standard Addition Calibration for Potentiometry?

The little problem with the standard addition in potentiometry is, that the response to the addition is not linear, but logarithmic.

If we consider the simplified Nernst equation in context of parameters $A, B$,

and if we consider $c, c_\mathrm{0}$ as unknown concentration and  concentration increment, respectively,

we can follow this derivation:

$$\begin{align}
E_\mathrm{1} &= A + B \cdot \log c\\
E_\mathrm{2} &= A + B \cdot \log ( c + c_\mathrm{0} )\\
E_\mathrm{2}- E_\mathrm{1} &= B . \log( 1 + \frac {c_\mathrm{0}}{c} )\\
10^{\frac{E_\mathrm{2}- E_\mathrm{1}}{B}} &= 1 + \frac {c_\mathrm{0}}{c}\\
c &= \frac {c_\mathrm{0}}{10^{\frac{E_\mathrm{2}- E_\mathrm{1}}{B}} -1}
\end{align}$$

where $B = \frac{RT}{nF} \cdot \ln(10)$


why is tertiary carbocation more stable than primary allylic carbocation?

The decreasing order of stablity of carbocations is:

aromatic (tropylium, cyclopropenyl) > benzyl > allyl > tertiary > secondary > primary

Allylic carbocations are resonance stabilized whereas tertiary alkyl groups are stabilized by hyperconjugation (and inductive effect). The hyperconjugation (and inductive effect), in general, is quite weaker than resonance (and hyperconjugation effect). The only exceptions to the previous statement I am aware of are halogens (inductive effect is stronger than resonance).


What is the chemical equation for the digestion/decomposition of cellulose?

Biologically, methane is produced by methanogens, which are Archeabacteria. They do not utilize carbohydrates such as cellulose or glucose directly. Instead, they typically consume acetate or a mixture of hydrogen gas and carbon dioxide. These stoichiometry of these two processes is

$\ce{4H2 + CO2 -> CH4 + 2H2O}$ or

$\ce{CH3C(O)OH -> CH4 + CO2}$.

To get overall conversion of cellulose to methane, the cellulose is first fermented by the enteric bacteria you read about, which can produce acetate, hydrogen and carbon dioxide by a variety of pathways. These products are then utilized by the methanogens. One example of an enteric fermentation would be

$\ce{C6H10O5 + 3H2O -> 2CH3C(O)OH + 4H2 + 2CO2}$.

Combining this with the methanogenesis reactions, we can write the net equation of

$\ce{C6H10O5 + H2O -> 3CH4 + 3CO2}$.


Can salicylic acid be synthesized from salicylaldehyde?

There are a large number of reagents out there that will transform aldehydes into the corresponding acids (benzaldehydes and benzoic acids in this case) and $\ce{KMnO4}$ is indeed one of them.

Oxidation of the phenolic hydroxide is only possible if at the same time the aromatic system is destroyed. However, it is rather hard to come up with a good product of this oxidation since a simple quinone isn’t possible. It therefore takes much harsher conditions than the oxidation of an aldehyde.

Things would change if a second hydroxy group were attached to the same aromatic system either ortho or para to the first. Then the oxidation to the corresponding quinone is much easier.


Gas spectra in the 400 nm - 1000 nm range

The answer is that you can detect all types of gases if you have the correct light source and use Raman spectroscopy. Raman is more adaptable than absorption or emission spectroscopy as it is a scattering process and the laser does not have to operate at the wavelengths that the molecules absorb, thus one laser can be used for all samples. A suitable source is going to have to be a narrow band laser (although when Raman discovered this effect he used sunlight). You will also need notch -filters to remove the laser wavelength and prevent it from saturating your detector. Not cheap.


To what microscopic quantity does Gibbs energy correspond to?

As others have said, the Gibbs function (or Gibbs energy) makes sense at a macroscopic, thermodynamic level (instead of a microscopic level), but if you need to picture it, you can elaborate what it means from the deconstruction of its mathematical definition.

Internal energy

Let's start with internal energy, U. Internal energy is the mechanical energy (potential + kinetic) of all the particles in the system. Take into account that it isn't reducible to the energy of separate particles - these particles are interacting (for chemical systems, through intermolecular forces) and the potential and kinetic energy are intrinsically related (though quantum mechanics). The only energy that you aren't considering is the kinetic energy of the system as a whole - if your chemical system is moving through space or rotating around its centre of mass, that energy isn't relevant here (it's also trivially easy to calculate via Newtonian mechanics) - and potential energies that arise from external factors that affect the system uniformly - so if your whole system is subject to gravity, that potential energy isn't computed into the internal energy (also trivially easy to calculate via classical physics). In other words, the internal energy discounts anything that you could calculate if you picture your system as a point mass (as you do in high school mechanics); it deals with whatever is happening inside that point mass.

This mechanical energy is conserved, but the balance between its components will depend on the exact positions of all atoms; there would be constant transferences of energy from kinetic to potential, from inter- to intramolecular interactions, etc. It's easier if we start picturing the system in statistical terms - as a huge amount of molecules, moving in random directions at random speeds (following known statistical distributions), of which thermodynamic functions (such as Gibbs') capture an average.

So, in the internal energy you are taking into account: 1) the kinetic energy of molecules in your system, 2) the potential energy of intermolecular interactions, 3) the kinetic energy of intramolecular movements (rotations, vibrations), 4) intramolecular potentials (arising from the electronic structure of molecules).

Enthalpy

Since we are considering the movement of molecules relative to the centre of mass of the system, we know that the total movement of the system is zero - the centre of mass remains fixed. If you add all the momenta of the molecules, you get a zero.

However, there's a possible collective movement that, while averaging to zero, implies a macroscopic physical change: movements towards the system or away from it. It is possible, if you add all the momenta, to get a total momentum of zero and, at the same time, that more molecules are moving away from the centre of mass than towards it - if you picture it in macroscopic terms, that means that the system as a whole remains at the same position, but is expanding (think of a balloon). And, likewise, if more molecules move towards the centre than away from it, the system is immobile but contracting. Both these collective movements can, potentially, change the volume of the system - they are a microscopic interpretation of pressure.

Note that, ff the system actually changes volume, the intermolecular potential energy is affected - if the system expands, its molecules are further apart and its interactions are weaker; for a contracting system, the opposite is true. The energy involved in that process is purely a mechanical work - captured in the term $pV$. When we work with solids or liquids, changes in volume are usually negligible; when we work with gasses, we usually store them in rigid containers - so changes in volume are also usually negligible. Therefore, it's convenient for chemists to define a function that discounts this mechanical work from the internal energy - that's exactly what enthalpy, H, is.

So, in the enthalpy you are taking into account: 1) molecular movement (except expansion and contraction), 2) intermolecular potentials, 3) intramolecular movement, 4) intramolecular potentials.

Entropy

The molecular interpretation of entropy, S, is always contentious (disregard any explanation that includes the words "chaos" or "disorder"), but we can discuss it simplistically. Entropy is more complex to visualise because it's an intrinsically statistical magnitude, and it ultimately relates to how energy is transferred between the different forms (inter- and intramolecular, kinetic and potential). What entropy implies is that when more than one configuration produces equivalent states, that makes the state more likely - effectively stabilising it, lowering its energy.

If you took microscopic photos of your system at different points, you'd get random positions and states for its molecules - but some situations would be more likely, from a purely random viewpoint, than others. If you have a two-component system, it's very very unlikely that their molecules will be spontaneously segregated in different parts of the system - if you place molecules randomly, it's far more likely that they will be well mixed throughout the system, so, more entropy for the latter. Molecules with many internal movements (like large molecules with many vibrations) also produce more equivalent states - you could get many photos in which the only difference is what particular bond vibrates more strongly, while that's impossible for, say, diatomic compounds. An ideal crystal at T=0K, where atoms are fixed exactly in periodic positions, without defects, and without vibration because they don't have any kinetic energy, has only one way to be arranged - so it has minimum entropy.

The ability to populate these configurations is closely tied to temperature. Temperature provides the energy that produces these movements and allows the system to explore more configurations - through Boltzmann's distribution. Since entropy indicates how many configurations there are for a given state and temperature provides the energy to do so, the "entropic stabilisation" term depends on temperature, and, conversely, systems with higher entropy are more sensitive to entropic stabilisation. That's what the term $TS$ measures.

This is important because that "entropic stabilisation" means that a fraction of the energy of the system (precisely $TS$) is tied up in providing these alternative configurations, and cannot be extracted directly from the system. Unless chemical reactions or other structural changes take place, that energy is inaccessible.

Gibbs function

We can then describe a function that computes 1) the internal mechanical energy of the system, 2) discounts the energy involved in expansion/compression work, and 3) discounts the energy that is inaccessible due to entropy. This is what the Gibbs function, G, means at a microscopic level. Note that this doesn't make for a clear-cut distinction between what microscopic forms of energy are included in the Gibbs function - it isn't as simple as "kinetic out, potential in", because some things (mainly entropy) are statistically defined and only make sense for a collective of (interacting) particles.


In both cases the alkoxide will be in an equilibrium with the alcohol. This will be occurring on a faster timescale than the substitution but in the first case it's irrelevant as  the same species are present on either side of the equilibrium.

Methanol is more acidic than ethanol, however, so the reaction lies in favour of methoxide and ethanol.


From the abstract of the article in question:

  The results suggest that the pressure stability of a protein in solution is not directly affected by the presence of these proposed piezolytes, and so they cannot be granted this distinction.

So the answer is "no", the supposed piezolytes do not appear to have the purported function. They do stabilize proteins against thermal denaturation, but not against pressure-induced destabilization.


These are common names, and predate our knowledge about bonds. My guess is that the "ylene" comes from reduction -- reducing ethyl alcohol gives ethylene, reducing acetic acid  gives acetylene.


There is no need to use ideal gas law in $pV = nRT$ form or find molar fraction $X$. The process is isothermal ($T_1 = T_2 = \pu{25 ^\circ C}$, hence $p_1V_1 = p_2V_2$), so it's a matter of finding the sum of partial pressures of each gaseous component $i$ in the system:

$$p = \sum_i{p_{2i}} = \sum_i{\frac{p_{1i}V_{1i}}{V_{2i}}}$$

Also, converting the pressure back and forth between $\pu{torr}$ and $\pu{atm}$ is counterproductive as you are explicitly asked to provide an answer in $\pu{torr}$.

   $$p = p_2(\ce{N2}) + p_2(\ce{O2}) = \frac{\pu{250 torr} \cdot \pu{200 mL}}{\pu{300 mL}} + \frac{\pu{350 torr} \cdot \pu{350 mL}}{\pu{300 mL}} = \pu{575 torr}$$


How do I determine the number and order of ligand orbitals?

T refers to triply degenerate set, E refers to doubly degenerate set, A and B are singly degenerate. The total number of orbitals should add up to the total number of atomic orbitals on your ligands. 1,2,g,u are symmetry labels (look up Mulliken symbols). One thing to remember when constructing you MO diagram is that only orbitals of similar energy and the exact same type and symmetry can mix. So, you can only mix A with A, B with B, E with E and T with T. Also, only g with g, u with u, 1 with 1 and 2 with 2.


Ni-Mh battery makes fizzy sounds while and after charging - why and do I need t replace it?

It's not possible to give a proper answer to (a) without examining the battery but it is likely due to a compromised casing.  The noise is likely hydrogen gas leaking out as it's produced. Nickel metal hydride (NiMH) batteries use a metal hydride (MH) for energy storage.  When this Hydride is exposed to moisture in air it reacts to form hydrogen gas: $$\ce{MH_x + xH2O \rightarrow M(OH)_x + x H2 }$$

For (b) I would say that your battery is probably not usable—safely or otherwise—anymore and may present a fire hazard. You should replace the battery as rechargeables don't cost near as much as the device they can leak inside of and destroy or the cost of fire damage to a house.


Please stay away from sodium hydroxide, it's way too nasty to use in non-lab setup. You can handle it safely in dry conditions, but in contact with water, it becomes sticky and causes bad burns.

As a substitute, I suggest $\ce{Ca(OH)_2}$, readily available as slaked lime. You can structure it into relatively high-surface sculptures and it solidifies into common plaster upon binding enough $\ce{CO_2}$. It is often used for this purpose due to its low price, low toxicity and ease of handling.

In your situation, I would just put some vertically fixed mesh soaked in very viscous suspension of calcium hydroxide. In the high humidity conditions you'd have no problems with normal drying, so once it solidifies it has done its job and you can just hit the mesh against the ground to crumble the plaster away and put on new layer. It is advisable to use gloves and goggles, but it is not as necessary as it would be in the case of sodium hydroxide (look at all the masons around).


Is Nitroglycerine a Nitro Compound or an Ester?

The Oxford dictionary is correct.

A nitro compound is defined by the $\ce{-NO2}$ functional group, but only if that group is attached to a carbon atom. Once an additional oxygen is added, the molecule is basically inorganic nitrate and thus belongs to a wildly different compound class.

You think that is confusing? Well, an alcohol is defined by a hydroxy group — but only if that hydroxy group is attached to a carbon, which in turn is not attached to a different oxygen, nitrogen or sulphur atom, and which is not part of an aromatic system. Even more constraints in there.


How do metals split the oxygen double bond during corrosion?

The corrosion of metals like iron is essentially an electrochemical process.

In corrosion, a metal is oxidised by loss of electrons to oxygen and formation of oxides. Corrosion of $\ce{Fe}$ (commonly known as rusting) occurs in presence of water and air. The chemistry of corrosion is quite complex, but it may be considered essentially as an electrochemical phenomenon. At a particular spot of an object made of iron, oxidation takes place and that spot behaves as anode and we can write the reaction.

\begin{align}
\tag{Anode}
  \ce{2Fe (s) &-> 2 Fe^{2+} + 4 e-} &
  E^\circ_{\ce{Fe^{2+}/Fe}} &= \pu{– 0.44 V}
\end{align}

Electrons released at the anodic spot move through the metal and go to another spot on the metal and reduce oxygen in presence of $\ce{H+}$ (which is believed to be available from $\ce{H2CO3}$ formed due to dissolution of carbon dioxide from air into water. Hydrogen ions in water may also be available due to dissolution of other acidic oxides from the atmosphere). This spot behaves as cathode with the reaction

\begin{align}
\tag{Cathode}
  \ce{O2(g) + 4 H+(aq) + 4 e- &-> 2 H2O (l)} &
  E^\circ_{\ce{H^+/O2/H2O}} &= \pu{1.23 V}
\end{align}

The overall reaction being:

\begin{align}
  \ce{2Fe(s) + O2(g) + 4H+(aq) &-> 2Fe^{+2}(aq) + 2 H2O (l)} &
  E^\circ_{\text{cell}} &= \pu{1.67 V}
\end{align}

The ferrous ions are further oxidised by atmospheric oxygen to ferric ions which come out as rust in the form of hydrated ferric oxide $\ce{(Fe2O3. x H2O)}$ and with further production of hydrogen ions.


In metal complex electron counting, is the neutral or ionic formalism more correct?

I think, the answer to number 1 will remove the need to answer the subsequent questions:

1) Which method gives the true d-electron count?

Neither!
You cannot, and I want to stress that, derive the oxidation state of a metal in a complex solely by looking at its ligands. You can make educated guesses and in some cases only one oxidation state is plausible but there are many cases—most notably those denoted $\ce{\{FeNO\}^n}$ where such an assingment is meaningless to impossible. Without consulting spectroscopic data, it is absolutely impossible to tell whether a $\ce{\{FeNO\}^7}$ compound is $\ce{[\overset{+II}{Fe}(NO^.)]}$ or a $\ce{[\overset{+III}{Fe}(NO^-)]}$. In fact, entire papers have been published arguing one way or the other.
This essentially renders your follow-up questions entirely moot, but I do wish to take the time to point out that the metal’s oxidation state along with the corresponding oxidation state of the ligands is theoretically important information that can predict the electrochemical behaviour of a complex. The only problem is that it cannot always be neatly predicted as shown above. It is worth noting, though, that certain complexes have been assigned metal oxidation states beyond reasonable doubts (e.g. $\ce{[Cu(NH3)4(H2O)2]}$), and in such cases the predictions of reactivity are absolutely adequate.


By how many pH units does rain vs distilled water change a soil pH test?

Pure water (rain as well as distilled water) in equilibrium with the atmosphere ($p_{\ce{O2}}=10^{-3.5}\ \mathrm{atm}$) can be calculated to contain about
$$\begin{align}
\mathrm{pH}=-\log[\ce{H+}]&=5.65\\
-\log[\ce{HCO3-}]&=5.65\\
-\log[\ce{CO3^2-}]&=10.3\\
-\log[\ce{H2CO3^*}]&=5.0\\
-\log[\ce{CO2}]&=5.0\\
-\log[\ce{H2CO3}]&=7.8\\
\end{align}$$

(The calculation can be found in: Stumm, W.; Morgan, J. J. Aquatic Chemistry, Third Edition; John Wiley & Sons: New York, NY, 1996; pp 159–161.)

Note that $[\ce{H2CO3^*}]$ is the total analytical activity of dissolved $\ce{CO2}$, i.e. $[\ce{H2CO3^*}]=[\ce{CO2(aq)}]+[\ce{H2CO3}]$. $[\ce{H2CO3}]$ is the concentration of true $\ce{H2CO3}$.

Thus, $[\ce{H2CO3^*}]=10^{-5}$ and $c(\ce{H2CO3^*})=10^{-5}\ \mathrm{mol/l}$; i.e. a few µmol/l.


What's the configuration of chromium(III) in aqueous medium according to crystal field theory?

You nailed it, this is almost a chicken-and-egg problem. As long as we don’t know which complex will form, we cannot say which orbitals are populated. This is why it is often and regularly repeated that you need to know the geometry before you can make a call.

In an octahedral complex, the $\mathrm{t_{2g}}$ orbitals will always be populated first as they are lower in energy. (Look up any introduction on crystal field theory or check out the answers on this site.) In a tetrahedral complex, $\mathrm{e}$ — not $\mathrm{e_g}$, as there is no centre of symmetry, meaning the the $\mathrm{u/g}$ designators do not make sense, even though the orbitals are still often called $\mathrm{e_g}$ at introductory level — are the orbitals of choice to begin with. Again, just look at the scheme.

Thankfully, however, we can make an educated guess based on various factors.

octahedral complexes are more common especially at introductory levels and for the ‘top middle’ of the d block elements (roughly from titanium to nickel)
the LFSE (ligand field stabilisation enthalpy) can be calculated to $\pu{-12Dq}$ in the octahedral case and $\pu{-8Dq}$ in the tetrahedral case which twice prefers octahedral
there is no compelling reason (size or charge balance) to consider tetrahedral to be preferable to octahedral.

Thus, the complex you are looking at will most likely be octahedral — and so it is: it is hexaaquachromium(III) $\ce{[Cr(H2O)6]^3+}$, which is octahedral.


Does the term 'silver nitrate' express all of its constituent chemical elements?

You should know that nitrate is a polyatomic ion with the molecular formula $\ce{NO3^-}$. For example:

Sulfate is a polyatomic ion with the molecular formula $\ce{SO4^{2-}}$;
Sulfite is a polyatomic ion with the molecular formula $\ce{SO3^{2-}}$;

Nitrate is a polyatomic ion with the molecular formula $\ce{NO3^-}$;
 Nitrite is a polyatomic ion with the molecular formula $\ce{NO2^{-}}$

At first glance, the nomenclature of the polyatomic negative ions seems difficult.
When the name of the ion  ends in either -ite or -ate:

The -ite ending indicates a low oxidation state.
The -ate ending indicates a high oxidation state.

I agree with you that at the beginning, it seems difficult. But by practicing, you find this quite easy.


Calcium oxide is double charged, meaning the force required to separate ion pairs is much larger than Sodium chloride.

Melting an ionic solid requires you give an ion sufficient energy to become free of the ionic bonding of its neighbors.  In a simplified view, each ion has 6 neighbors in bulk (both are fcc, I believe) or 5 at surface face, 3 on corner and 4 on edge.  You need to add at least 3-5 times as much thermal energy to liberate the ions in calcium oxide as sodium chloride due to the doubling of charge from each neighbor.


Why note the solvent level on chromatography paper when it is removed from the chamber?

Building on my comment, marking the distance that the solvent traveled allows us to calculate the retention factor $R_f$ (or apparently retardation factor, according to IUPAC meddlers who need to rename things with perfectly good names that everyone else uses). The retardation factor is the ratio of the distance traveled by the spot to the distance traveled by the solvent:

$$R_f=\dfrac{d_{spot}}{d_{solvent}}$$

If you are only performing one thin-layer chromatography experiment (TLC) ever, then you may be okay not marking the solvent distance. Since everything is on one piece of paper or one plate, you can compare readily:

From the above TLC picture, we can easily tell that unknowns A and B are probably the same compound as the standard, but unknown C is something else.

Now, what if we have two TLC plates/papers?

We can draw the same conclusions about the relationships between standard 1 and unknowns A, B, and C. Likewise, we can make some conclusions about the relationship between standard 2 and unknowns D, E, and F. However, we cannot easily determine if unknown F is the same compound as unknown A.

Determining the $R_F$ value lets us do that. Using the same solvent system, the distance traveled by the solvent and the distance traveled by a compound are both proportionally related to the length/height of the plate/paper. The solvent can travel twice as far on a paper that is twice as long. It turns out that the compound being analyzed also travels twice as far on a paper that is twice as long. The the ratio, $R_f$ is constant.

Let's drop in some numbers:

Now we can calculate the $R_f$ values for each spot:

After calculating the $R_f$ values, we can make some conclusions (assuming there are only two compounds in the entire experiment):

Unknowns A, B, D, and E are the same compound as Standard 1.
Unknowns C and F are the same compound as Standard 2.


Kinetic energy of electrons emitted in Photoelectron Spectroscopy

From Wikipedia (emphasis mine):

  Because the energy of an X-ray with particular wavelength is known
  (for aluminum Kα X-rays, Ephoton = 1486.7 eV), and because the emitted
  electrons' kinetic energies are measured, the electron binding energy
  of each of the emitted electrons can be determined by using an
  equation that is based on the work of Ernest Rutherford (1914):

  \begin{equation}
     E_\text{binding} = E_\text{photon} - \left(E_\text{kinetic} + \phi\right)   \end{equation}

  where $E_\text{binding}$ is the binding energy (BE) of the electron, $E_\text{photon}$ is
  the energy of the X-ray photons being used, $E_\text{kinetic}$ is the kinetic
  energy of the electron as measured by the instrument and $\phi$ is the
  work function dependent on both the spectrometer & the material. This
  equation is essentially a conservation of energy equation. The work
  function term $\phi$ is an adjustable instrumental correction factor
  that accounts for the few eV of kinetic energy given up by the
  photoelectron as it becomes absorbed by the instrument's detector. It
  is a constant that rarely needs to be adjusted in practice.

So, it is the absorbtion of the photoelectron by the detector of the spectrometer that causes the work function term. The work function of the measured sample should be part of $E_\text{binding}$ as this is the ionisation energy of the electron. However, I think the equation you quote might have it slightly wrong because if $B.E._F$ is the energy difference between the energy level of the electron and the Fermi energy then there is indeed some portion of the work function of the sample missing since $B.E._F$ is not equal to the ionisation energy.


Aqueous ammonia and hydrochloric acid are solutions of gases.

In a fume hood place two beakers with aliquots of these two solutions (one in each beaker, don't mix) side to side.

Did I mention the fume hood?

You'll see the formation of solid ammonium chloride from the reaction of the two gases.

$$\ce{HCl(g) + NH3(g) -> NH4Cl(s)}$$

Don't forget to clean up the mess!


How to use Lewis theory to differentiate between an acid and a base?

First of all, within the Lewis framework, you may have substances that can act as both acid and base (though most of the time as one or the other, not both at the same time).$^{1}$ The adjective for such substances is "amphoteric".

It comes down to a bit of chemical intuition to decide which is which. It is true that almost anything that has a lone pair can act as a base - though it may take a strong acid to form an actual Lewis pair. For acids, one should look for an obvious gap in the electron cloud ($\ce{AlCl3}$) or candidates for complex formation ($\ce{FeCl3}$).$^2$

For nitric acid, the situation is pretty clear: the hindrance provided by three oxygen atoms around the central nitrogen prevents access, so it will hardly act as a Lewis acid. It can act as a base, however, weak Brønsted bases are usually also weak Lewis bases because they are content on their own.

For sodium hydroxide, one must realize that in an aqueous medium, the two ions will fully dissociate. The hydroxide anion will act as a Lewis base, but the sodium cation with its full outer electron shell will hardly act as an acid.$^3$

$^1$ Adducts of e.g. $\ce{CO2}$ in frustrated Lewis pairs, wherein the $\ce{CO2}$ acts as a sort-of bridge between the two Lewis sites, are an example of when "both at the same time" can occur.

$^2$ I am aware that these compounds will form dimers or more complex structures in reality. If you are able to spot this behavior and understand why, you are already able to spot Lewis acids.

$^3$ Note that the sodium cation in water will be solvated, i.e. some water molecules will orient their lone pairs towards the ion. The border between solvation and Lewis adduct formation can be blurry.


Elimination of OH from spirocyclic compounds

This question has been unanswered for quite long, and meanwhile, I have found an answer, so I wanted to share it.

On treating the compound with sulfuric acid, the $\ce{OH}$ groups will be protonated to $\ce{OH_2^+}$.

The $\ce{OH_2^+}$ groups are good leaving groups, so they will leave as $\ce{H_2O}$, forming carbocation.(one by one)

As cyclopropane is under significant ring strain, it will undergo ring expansion to form cyclobutane ring, which is somewhat less strained. At the same time, $\ce{H^+}$ ions will be expelled, forming double bonds. (It should also take place step by step.)

In this case I don't think that the double bond will not cause any significant barrier to the formation of the molecules, as the double bond is here the bridging bond, not at the bridgehead[formation of double bonds at bridgehead is not possible for smaller rings(<8 carbon rings), as it causes very high strain: Bredt's rule]

I have considered only elimination in this reaction scheme. Other compounds may also be formed. Feel free to post answers if you think other compounds will be formed or if you think the above mechanism is wrong in any way.


What does the commercial chemical composition notation "Percent dry, wet with water"mean?

I believe there is no general rule.
In your case (assuming you mean this) it is a solid. materials can have 25% and even more water and still be solids.


A homogeneous reaction mixture can hardly show macroscopical chaotic behaviour. Well known cyclic counterexamples exist, and practically any reaction which proceeds faster than diffusion or mechanical mixing can homogenise it again (e.g. any reaction that generates a lot of heat!) is liable to show some intermediate chaotic concentration gradients. Its usually hard to notice, and most reactions are perfectly reproducible, if you set them up in the same way (educt impurities!).

If your reaction vessel contains inflows of different chemical composition, then you easily get chaotic behaviour. Think about a tube reactor. I believe chemical engineers are spending a lot of their time trying to control and avoid chaos.


Gaussian scan function help for constructing input file

This is (probably) not going to work the way you intend it to work. Constructing a meaningful modredundant calculation is quite a difficult task, as you sometimes cannot anticipate what the program actually does. Using this interface you will perform a relaxed scan, i.e. you will step one variable, fix the values of the scanned coordinates, and run a full optimisation at this point. This allows you to systematically, brute force a potential energy surface. In you case you specify 12 coordinates to scan with 20 points each, resulting in a whopping $N=\frac{20!}{12!}\approx5\times10^8$ optimisations.

Here are a few pointers when constructing a modredundant input:

don't scan more than two parameters at a time, that roughly equates to a 3D graph of a PES
don't scan xyz coordinates, scan bond lengths and angles instead; scanning xyz coordinates results in fixing the coordinate system in parts disallowing the algorithm to rotate the molecule/ use coordination transformations to speed up the calculation
keep constant what you need not optimised; fixing things will result in a faster calculation

Let's make this a little bit more visual. As an example I'll choose $\ce{(CO2)2}$ and I am using the ridiculous low level of theory HF/3-21G.

First up optimise the equilibrium geometry.

%chk=opt.chk
#P HF/3-21G opt

title required

0 1
C       -1.345383226      0.000000000     -1.107741226
O       -0.793543854     -0.126832008     -0.084830947
O       -1.897222598      0.126832008     -2.130651504
C        1.651522395      0.000000000     -1.243964208
O        2.203361766     -0.126832008     -0.221053930
O        1.099683023      0.126832008     -2.266874487

The result will be the following geometry:

    6
scf done:  -373.128185
 C    -1.546398     0.008833    -1.171593
 O    -0.884661    -0.109039    -0.227710
 O    -2.221473     0.124768    -2.098858
 C     1.852537    -0.008833    -1.180113
 O     2.527612    -0.124768    -0.252847
 O     1.190800     0.109039    -2.123995

Now we want to move the two molecules closer together. If we would use the approach you have come up with, this would be really tedious. First we need to transform the coordinates in a way that we can understand which molecule to move where. Luckily this is not necessary at all. We can simply scan over the "bond length" of the carbon carbon distance. I chose five steps with an increment of -0.2 angstrom.

%chk=modered1.chk
#P HF/3-21G
opt(modredundant)

title

0 1
 C    -1.546398     0.008833    -1.171593
 O    -0.884661    -0.109039    -0.227710
 O    -2.221473     0.124768    -2.098858
 C     1.852537    -0.008833    -1.180113
 O     2.527612    -0.124768    -0.252847
 O     1.190800     0.109039    -2.123995

B 1 4 S 5 -0.2

That will give us the following graph and animation:

You can see that the OCO bond angles changes. The CO bond lengths changes, too, but that is not as visible. This is expected behaviour in a relaxed scan. Only one coordinate is fixed, which is the distance between the carbons. If you are interested in keeping at least the bond angles constant, freeze them:

%chk=modered2.chk
#P HF/3-21G
opt(modredundant)

title

0 1
 C    -1.546398     0.008833    -1.171593
 O    -0.884661    -0.109039    -0.227710
 O    -2.221473     0.124768    -2.098858
 C     1.852537    -0.008833    -1.180113
 O     2.527612    -0.124768    -0.252847
 O     1.190800     0.109039    -2.123995

B 1 4 S 5 -0.2
A 2 1 3 F
A 5 4 6 F

With the result, we'll see again something unexpected happening. There is a sudden rotation at 2.8 angstrom. For the given set of parameters, this is the lowest energy. This is a side effect of a relaxed scan. not everything will run smoothly and sometimes the calculation turns rogue and fails. It really is a delicate matter.

Just for shits and giggles, let's do one that will make us cringe. We will scan also both of the oxygen oxygen distances. Theoretically this should give us 20 data points. Spoiler alert: This will fail at step 8. Try it yourself:

%chk=modered3.chk
#P HF/3-21G
opt(modredundant)

title

0 1
 C    -1.546398     0.008833    -1.171593
 O    -0.884661    -0.109039    -0.227710
 O    -2.221473     0.124768    -2.098858
 C     1.852537    -0.008833    -1.180113
 O     2.527612    -0.124768    -0.252847
 O     1.190800     0.109039    -2.123995

B 1 4 S 5 -0.2
B 2 5 S 5 -0.2
B 3 6 S 5 -0.2

You can see, that with these small molecules it is already hard to predict what happens when you run the calculation. There are traps and pitfalls lining your way. More often than not it is trial and error.

Thankfully a full relaxed scan is not necessary in most cases. Oftentimes a nice little rigid scan will give you a decent result. Have a look at the scan keyword. The only downside to this approach is, that you need a z-matrix to make it work. And putting your brain to that task is also a bit tedious.

I have tried to convert the specifications to a z-matrix putting constants where needed and introducing the scan variable. I'm not sure I succeeded in the way I wanted to, but here we go:

%chk=scan.chk
%nproc=2
%mem=8000MB
#P HF/3-21G scan

title required

0 1
 c
xx   1  1.40000
xx    1  scan    2   90.000
 o    1 oc4         3   90.000      2  88.981
 o    1 oc5         3   90.000      2  -90.00
 c    3  1.93591    1   89.994      4 178.983
xx    6  1.40000    3   90.000      1  90.000
 o    6 oc8         7   90.000      3 180.000
 o    6 oc9         7   90.000      3   0.000

scan        2.794000  5  -0.20
oc4         1.152818
oc5         1.158751
oc8         1.152817
oc9         1.158750

The result is two rods, moving towards each other:

I expect that since you have more than 100 atoms, accomplishing a meaningful scan calculation is a very, very tedious task. It might be much simpler creating the points you want to scan by hand with a molecule editor. But without more information about the actual system, sound advice is difficult.


What causes photovoltaic (solar) cells to degrade?

Why do they degrade? What exactly is reducing their efficiency?

This link outlines several modes of solar panel degradation, and this report by the National Renewable Energy lab is a very detailed review of studies on solar panel degradation worldwide.

To summarize:

Internal resistance of the cell can increase due to infiltration of contaminants (usually water vapor) when the encapsulating material cracks due to long-term UV exposure or temperature cycling.
The shunt resistance (resistance of the path to ground) can decrease when metal ions migrate (diffuse) through the cell. This happens naturally, with the rate depending on temperature and the type of cell. In general, higher temperatures lead to faster diffusion, and amorphous silicon has higher diffusion rates.
The anti-reflective coating can deteriorate. Anti-reflective coatings can be made of lots of different types of materials, but in general, heat, UV exposure, and exposure to contaminants are what cause them to degrade.

These are all degradation mechanisms of the cells themselves - any of the other components of the panel can also degrade. In all cases, the mechanisms are related to thermal stress or mechanical damage due to things like wind or hail, or corrosion from humidity.

Given all of the above, the main causes of solar panel degradation that can be (sort of) controlled by you are:

Temperature
UV exposure
Mechanical damage

  How can this process of degrading be slowed down? E.g. if you don’t need your PV panels, is it possible to conserve them by covering them with a black, opaque coverage like tarp? If not, what else?

Anything you can do to reduce these three factors would help. However, covering a solar panel with an opaque tarp would only reduce UV exposure. The temperature (and cyclical thermal stress) would likely increase.

Unfortunately, both 1 and 2 increase with exposure to sunlight - and exposure to sunlight is the whole point of solar panels. Mechanical damage can be reduced by installing panels in places and orientations that make it unlikely that they will be stressed by winds, or have things hitting them very often.

However, it looks like the only effective option would be to completely enclose the panels in a temperature-controlled low humidity environment when they are not in use. Whether or not the benefit of doing so is worth the cost will depend on the degradation rate and cost of the particular panels.


Just like electrons nuclei can exist in a number of quantised states. We see evidence of this when, during radioactive decay, a gamma ray is produced. This is a nucleus in an excited state decaying to the ground state, and the excess energy being radiated away as a gamma ray photon. This is precisely analogous to an electron in an atom falling from a high energy state to a lower energy one, and thus producing radiation which we can see in an atomic spectrum. So an isomer of a nucleus is simply that nucleus in a state which is not the ground (lowest energy) state.

Most isomers are very short lived, and thus the gamma radiation appears to be more or less coincident with any other radiation that is produced during a radioactive decay. For instance the $\ce{^{18m}F}$ isomer you mention has a lifetime of 162ns according to https://en.wikipedia.org/wiki/Isotopes_of_fluorine . However the odd one is much longer lived, and for a remarkable case of one that is observationally stable take a look at $\ce{^{180m}Ta}$. 0.012% of natural Tantalum consists of this isomer, and it has never been observed to decay to the ground state - as is usual in such cases symmetry forbids the transition. And what I find even more remarkable is that the ground state itself is unstable with respect to both electron capture and beta emission, decaying to isotopes of Hafnium and Tungsten with a half life of just over 8 hours! See https://en.wikipedia.org/wiki/Isotopes_of_tantalum for details.

Sorry - slightly belated edit, but it occurred to me that isomers do have one very direct use in Chemistry, namely Mossabuer spectroscopy, see https://en.wikipedia.org/wiki/M%C3%B6ssbauer_spectroscopy. In essence very small changes in the energy of the gamma ray that is emitted when the isomer decays to the ground state can be detected by this technique, and can be related to changes in the chemical environment within which the atom finds itself.