Calculate ozone layer thickness at a given temperature and pressure

Final conditions ($\pu{1.0 atm}$) give a hint at what altitude the thickness should be estimated – at the Earth's surface. Let's denote the unknown thickness as $h_x$, radius of the Earth as $r_\mathrm{E} = \pu{6371 km}$; also take $h_1 = \pu{15 km}$ and $h_2 = \pu{35 km}$ and put these dimensions on a rough 2D sketch:

Fig. 1. Schematic representation of two ozone layers at given altitudes.

To find $h_x$, we can use the volume of the new hypothetical ozone layer $V_x$ as the difference in volumes of two spheres with corresponding radii:

$$V_x = \frac{4 \pi (r_\mathrm{E} + h_x)^3}{3} - \frac{4 \pi r_\mathrm{E}^3}{3} \tag{1}$$

and express $h_x$ from (1) as follows:

$$h_x = \left(\frac{3V_x}{4 \pi} +r_\mathrm{E}^3\right)^{\frac{1}{3}} - r_\mathrm{E} \tag{2}$$

From ideal gas law for a relatively thin gas layer without temperature and pressure gradients:

$$V_x = \frac{nRT_x}{p_x} \tag{3}$$

where $n$ is the amount of gas in ozone layer, $T_x = \pu{20 ^\circ C}$ and $p_x = \pu{1 atm}$ (new conditions). Let's also denote ozone concentration as $N = \pu{3.0e13 cm-3}$, then total amount of ozone is to be found via initial volume $V_0$ and Avogadro constant $N_\mathrm{A}$ as

$$n = V_0 \times \frac{N}{N_\mathrm{A}} \tag{4}$$

Initial volume can also be found from Fig. 1 using the difference in volumes of the corresponding spheres with radii $r_\mathrm{E} + h_1$ and $r_\mathrm{E} + h_2$:

$$V_0 =  \frac{4 \pi (r_\mathrm{E} + h_2)^3}{3} - \frac{4 \pi (r_\mathrm{E} + h_1)^3}{3} = \frac{4 \pi}{3}((r_\mathrm{E} + h_2)^3 - (r_\mathrm{E} + h_1)^3) \tag{5}$$

Now all we have to do is carefully plug in the values with the corresponding units carefully and find all the unknown variables. From (5):

$$V_0 = \frac{4 \pi}{3}((\pu{6371 km} + \pu{35 km})^3 - (\pu{6371 km} + \pu{15 km})^3) = \pu{1.03e10 km3} = \pu{1.03e25 cm3} \tag{6}$$

Now let's find the amount of ozone using $V_0$ and equation (4):

$$n = \pu{3.0e13 cm-3} \times \frac{\pu{1.03e25 cm3}}{\pu{6.023e23 mol-1}} = \pu{5.12e14 mol} \tag{7}$$

and, subsequently, the unknown volume $V_x$ using (3):

$$V_x = \frac{\pu{5.12e14 mol} \cdot \pu{8.314 J mol-1 K-1} \cdot \pu{293 K}}{\pu{101325 Pa}} = \pu{1.23e13 m3} = \pu{1.23e4 km3} \tag{8}$$

Finally, plugging all the values in (2), we can find the thickness:

$$h_x = \left(\frac{3 \cdot \pu{1.23e4 km3}}{4 \pi} + (\pu{6371 km})^3\right)^{\frac{1}{3}} - \pu{6371 km} = \pu{2.4e-5 km} = \pu{24 mm} \tag{9}$$

The result is somewhat close to the fact posted by NASA:

  Around the world, the ozone layer averages about 3 millimeters (1/8 inch) thick, approximately the same as two pennies stacked one on top of the other.


In summary: IE and EA are just partial terms in the formation enthalpy of solid ionic compounds, that includes also LE and atomization enthalpy of elements. Their values alone can give a hint, but do not provide the final word.

Notice that the text says nothing explicit about ion pairs.
IE + EA relate to free gaseous unbound ions. Like:
$$\ce{M(g) -> M+(g) + e-(g)}\tag{IE}$$
resp.
$$\ce{X(g) + e-(g) -> X-(g)}\tag{EA}$$
The lattice energy (LE) is the energy released by forming solid ionic compounds from free ions:
$$\ce{ M+(g) + X-(g) -> MX(s)}\tag{LE}$$
Gaseous ion pairs have already released a part of the lattice energy.
Note that for ion pairs, the pairing energy may be smaller than the sum of IE+EA, especially for big ions, so forming pairs may not be favoured.


How do proton exchange membrane electrolysers work?

PEM process allows to increase the density of current. the distance between the cathode and anode is relatively small. The polymer has acidic residues that can transfer protons. You can use pure liquid water or even water vapor (Energy Environ. Sci., 2011, 4, 2993).

Cathode and anode remain wet. They touch the PEM.

Pt and Ir are not required, but they worth the money. PEM itself is relatively expensive, so there is not much point to save on Pt or Ir. However they can be replaced with a cheaper material. But then efficiency drop to alkaline electrolysis method.


The functional group in question is a vinyl ether, and it is a mistake to consider these groups as isolated.  The oxygen is in conjugation with the vinyl group, meaning that resonance favors the entire system attacking.  There is no way to leverage conjugation as an energetic driving force if the oxygen attacks by itself.


Discrepancies in calculating free energy values listed in Stumm and Morgan

TL;DR: It appears that OP is right and has estimated standard Gibbs free energy correctly. The latter book accurately fetched data from the original source, but the calculations were carried out based on a false premise that $\Delta G^\circ$ refers to $\pu{25^\circ C}$ exclusively.

First, a general disclaimer. The uppercase circle ($^\circ$) generally denotes standard state. According to IUPAC, temperature is not declared within, but many authors attribute (negligently) $\pu{298.15 K}$ to Gibbs standard free energy $\Delta G^\circ$. Also check out What is the difference between ∆G and ∆G°?

For the given equilibrium one can use

$$\Delta G_T^\circ = -2.303RT\log K_T, \tag{1}$$

and at $\pu{298.15 K}$ this equation can be rewritten as follows, assuming $[\Delta G_{298.15}^\circ] = \pu{kJ mol-1}$:

$$\Delta G_{298.15}^\circ = -5.71\log K_{298.15}. \tag{2}$$

Now to the references. I didn't find the 3rd edition of The Physical Chemistry of Electrolyte Solutions by Harned and Owen (1958), but I do have an earlier version from 1943 [1]. In chapter 15 "The Ionization and Thermodynamic Properties of Weak Electrolytes" [1, p. 485], table 15-1-1 accumulates values for $K_\mathrm{w}$ determined electrometrically:

  In subsequent investigations, the cells without liquid junction, were employed in certain combinations for the determination of the ionization constant itself, as well as $m_\ce{H}m_\ce{OH}$ and $\gamma_\ce{H}\gamma_\ce{OH}/a_\ce{H2O}$ salt solutions.
  \begin{align}
&\ce{H2 | HX($m$) | AgX-Ag} \tag{I}\\
&\ce{H2 | MOH($m_1$), MX($m_2$) | AgX-Ag} \tag{II}\\
&\ce{H2 | HX($m_1$), MX($m_2$) | AgX-Ag} \tag{III}
\end{align}

  $$-\log K'_\mathrm{w} \equiv \frac{(\ce{E_{II} - \ce{E_{III}}})F}{2.303RT} -
 \log\frac{(m_\ce{H}m_\ce{X})_\mathrm{III}}{(m_\ce{X}/m_\ce{OH})_\mathrm{II}} + \\ + \frac{3.629 \times 10^6}{(DT)^{3/2}}\sqrt{\mu} = -\log K_\mathrm{w} + f(\mu) \tag{15-1-10}$$

So, Stumm and Morgan [2, p. 57] correctly adapted the data from table 15-1-1:

\begin{array}{rr|rr}
T (\pu{^\circ C}) &K_\mathrm{w} \times 10^{-14} {^{[1]}} &\log K_\mathrm{w}{^\text{calcd. [1]}} &\log K_\mathrm{w} {^{[2]}}\\
\hline
0 & 0.1139 &-14.9435 &-14.93\\
10 & 0.292 &-14.5346 &-14.53\\
20 & 0.6809 &-14.1669 &-14.17\\
30 & 1.469 &-13.8330 &-13.83\\
50 & 5.474 &-13.2617 &-13.26
\end{array}

But, what Stumm and Morgan did is they calculated the standard free energy exclusively at $\pu{25^\circ C}$ using the $K_\mathrm{w}$ values for other temperatures; disregarding the fact that equation (2) is only valid for $K_{298.15}$, but it was applied to all values as follows:

$$\Delta G_{298.15}^\circ {^{[2]}} = -5.71\log K_T, \tag{2a}$$

which is $\color{red}{\text{incorrect}}$. The latter column is populated with appropriate values calculated with (1), as it was done by OP:

\begin{array}{rr|rr}
T (\pu{^\circ C}) &\log K_\mathrm{w}{^\text{calcd. [1]}} &\color{red}{\Delta G_{298.15}^\circ {^{[2], (2a)}} (\pu{kJ mol-1})} &\Delta G_T^\circ {^{(1)}} (\pu{kJ mol-1})\\
\hline
0 & 0.1139 &\color{red}{85.29} &78.14\\
10 & 0.292 &\color{red}{82.96} &78.79\\
20 & 0.6809 &\color{red}{80.86} &79.50\\
30 & 1.469 &\color{red}{78.95} &80.28\\
50 & 5.474 &\color{red}{75.69} &82.04
\end{array}

Bibliography

Harned, H. S.; Owen, B. B. The Physical Chemistry of Electrolyte Solutions; American Chemical Society Monograph Series; Reinhold Publishing Corporation: New York, 1943; Vol. 95.
Stumm, W.; Morgan, J. J. Aquatic Chemistry: Chemical Equilibria and Rates in Natural Waters, 3rd ed.; Environmental science and technology; Wiley: New York, 1996.


Gibbs free energy ΔG=ΔG°+RTln(Kp) different expressions for ΔG

$K_p$ is normally reserved for equilibrium thus only apples when $\Delta G=0$ and conventionally $Q$ is used otherwise. In a plot of free energy $G$ vs extent of reaction $\zeta$, the slope is $(\partial G/\partial \zeta)_{T,p} =\Delta G$, and equilibrium occurs when the slope is zero then $\Delta G^\mathrm{o} =-RT\ln(K_p)$.  (The extent of reaction is zero with only reactants present, and 1 when one mole of reactants has become product.)

Thus to speak of 'distance' does not mean the energy difference from equilibrium as $\Delta G$ is the slope not the value of the energy $G$ vs $\zeta$.

In the sums you ask about, always calculate products - reactants. The only experimental quantities you have are the concentrations or pressures so it would be possible to calculate $\Delta G$, using your first equation, if the values are known at equilibrium and at some other point. This seems to be what you ask about.


What is lowest energy state, how does an atom gain it and what does it have to do with hybridization?

VSEPR stands for Valence Shell Electron Pair Repulsion.  It's a complicated acronym, but it means something that's not difficult to understand. Basically, the idea is that covalent bonds and lone pair electrons like to stay as far apart from each other as possible under all conditions (to minimize electrostatic repulsions of different pairs of electrons) .  This is because covalent bonds consist of electrons, and electrons don't like to hang around next to each other much because they have the same charge. This, of course, allows the molecule to be in its lowest energy state.

This VSEPR thing explains why molecules have their shapes.  If carbon has four atoms stuck to it (as in methane), these four atoms want to get as far away from each other as they can.  This isn't because the atoms necessarily hate each other, it's because the electrons in the bonds hate each other.  That's the idea behind VSEPR.

Now, one problem with the whole VSEPR thing is that if you have four things stuck to carbon, for example, there are no orbitals that want to get 109.5 degrees apart from each other (109.5 degrees corresponds to the geometric maximum distance the atoms can get apart).  After all, s-orbitals go in a complete sphere (360 degrees) and p-orbitals are 90 degrees apart.

What happens instead of using s- or p- orbitals is that when covalent bonds are formed, the s- and p- orbitals mix to form something called hybrid orbitals.  "Hybrid" just means "mixture of two different things", and that's exactly what a hybrid orbital is.  When three p-orbitals with 90 degree angles combine with one s-orbital with 360 degrees, they average to form four sp3 orbitals with 109.5 degree bond angles.  Depending on the numbers of s- and p-orbitals that mix, you can get a bunch of different bond angles.


Why do spectra appear continuous despite their allowed transitions being discrete?

You are right that absorption and emission are quantized processes, but UV-visible peaks represent a broad range of wavelengths because there are multiple vibrational states of their representative ground and excited electronic states.

In the case on my diagram, you have six different quantized electronic transitions with slightly different ΔE values. In reality, there are more energy levels than that, plus the rotations and vibrations are constantly changing the orbital energy levels as a result of interactions with the matrix, so that helps fill in the peaks. The peak shape can actually be changed by altering the temperature. This gives rise to high-resolution UV-Visible spectrophotometric techniques, which have a few specific applications.


The nuclei absorb the rf radiation and the spin population in the upper level increases at the expense of that in the lower level. There is now a chance that the upper levels will radiate away their extra energy and so return to equilibrium. However, in an nmr experiment the excitation pulses used in most if not all nmr machines are very short, just a few microseconds (and so spectrally wide) but the natural radiative lifetime of the excited levels is some seconds. Thus it is possible to excite the nuclear spins and build up a relatively large population before any significant decay occurs.

(This is one of the reasons why nmr works so well considering the tiny population difference that is produced because the energy levels are so close in energy and so all levels are significantly thermally populated.  The long lifetime of upper levels means that it is possible to coherently build up a significant macroscopic magnetisation in the sample and this what is detected.)


Crystal Field Splitting of d-Orbitals in Octahedral and Tetrahedral Ligand Fields

The labels you are talking about refer to the orbitals’ irreducible representations. Which irreducible representations can occur is dictated by the point group of the entire compound. For octahedral compounds, the point group is the very symmetric $O_\mathrm{h}$, which basically includes every element of symmetry you can have in a cube. Tetrahedral complexes belong to the $T_\mathrm{d}$ point group with a considerably lower symmetry (i.e. a smaller number of symmetry operations).

Octahedral complexes are probably more present in the average non-coordination chemist’s brain. Since $O_\mathrm{h}$ includes a centre of symmetry, all irreducible representations have a subscripted g or u to indicate whether equal phases of an orbital are mapped onto each other upon inversion (g for gerade) or whether phases are mapped onto their opposites upon inversion (u for ungerade). Due to the d-orbitals having two nodal planes, they are automatically g, which is why they as a whole transform as $\mathrm{e_g + t_{2g}}$. Since, as I said, this is the more prevalent case in general, this is what a lot of chemists have in mind.

The tetrahedral point group $T_\mathrm{d}$ does not contain a centre of symmetry. This is easily seen when looking at a tetrahedron, which is not inversion-symmetric. Thus, the irreducible representations do not gain a subscripted g and u. The d-orbitals thus transform as $\mathrm{e + t_2}$. Adding a g to these is wrong, but common (see above).

Thus, you were correct in assuming: ‘just because everyone does it does not mean it is correct.’ The same thing can be said for $\ce{S=O}$ double bonds in sulfate.


We define Arrhenius acidity with respect to water

That definition is new to me, whatever it is. The definition I know is:

An Arrhenius acid is a substance that dissociates in water to form $\ce{H+}$ ions.
An Arrhenius base is a substance that dissociates in water to form $\ce{OH-}$ ions.

The key concept to answer the question is the self-ionisation of water given in equation $(1)$.
$$\ce{H2O <<=> H+ + OH-}\tag{1}$$
This ionisation happens in water, and the ion product as written in equation $(2)$ is $10^{-14}$.
$$K_\mathrm{w} = [\ce{H+}] [\ce{OH-}] = 10^{-14}\tag{2}$$
Thus, water qualifies as a substance that dissociates in water to form $\ce{H+}$ ions. It also qualifies as a substance that dissociates in water to form $\ce{OH-}$ ions. It is both an Arrhenius acid and an Arrhenius base and thus the only Arrhenius amphoteric compound.


How to compare acidity in the following aniline derivatives?

Taking the case of ortho and meta in which you have confusion:

The acidity of aniline will increase if there is a electron donating group in the compound to stabilise the conjugate positive acid formed.

Since OMe is shows +R and -I effect, it is a net electron donating group as +R effect generally dominates over -I.So in case of ortho it is able to show electron donating nature due to +R effect which makes it more basic than the meta isomer which shows electron withdrawing effect due to only -I effect as +R effect will not affect from meta position.


Would gold have the negative charge in gold phosphides?

Because their electronegativity difference is less than 0.4, I think the bond would have a lot of covalent character. I personally don't think it is a good idea to treat this substance as an ionic compound and assign a positively charged ion and a negatively charged.
According to a chemical supply company, $\ce{Au}$ is +3 and $\ce{P}$ is -3.  We typically write the cation first, so from the formula $\ce{AuP}$, you should get another clue that the gold ion is positively charged (sort of if you have to assign a positively charged ion).
I don't think you can assign charge signs solely on electronegativities. Atomic radius and position on the periodic table also play roles in determining which atom is more positively charged.
For electronegativity, more reference here.


Why is lemon mixed with water considered alkalizing water?

The claim that lemon water is alkaline is simply wrong.  However, that hasn't stopped people from claiming that it's true:

  Lemons are an incredibly alkaline food, believe it or not. Yes, they are acidic on their own, but inside our bodies they're alkaline (the citric acid does not create acidity in the body once metabolized). As you wellness warriors know, an alkaline body is really the key to good health.

Every sentence above is wrong.  Lemons do not turn alkaline inside the body.  They are not an alkaline food.  Citric acid creates neither alkalinity nor acidity in the body when it is metabolized.  An alkaline body is not the key to good health.  In fact, if your body were more alkaline by only 0.5 pH units from its usual state, you would probably die from it.

  I wanted to know the specific reactions that occur when mixing water and fresh lemon slices [...]

There aren't really many reactions that occur.  Lemon juice is mostly water.  Mixing the lemon juice present in lemon slices with more water just dilutes it.  Lemon juice is acidic, due to the presence of citric acid and other acids.  Diluting will very slightly lower the acidity.  None of that has any relevance to the pH inside a mammalian body, which is exquisitely regulated by other mechanisms.

I don't know why people make up weird claims that lemons are an "alkaline" food, but those claims, sadly, are out there everywhere.


Do processes involving synthesis need heat?

Definitely. Although combination processes may be exothermic, they do have activational energy. As an example, consider the reaction:

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

While the formation of two $\ce{H-I}$ bonds is quite energy releasing, you still have to go through breaking a $\ce{H-H}$ and a $\ce{I-I}$ bond. You need to supply heat to give thermal agitation to allow the molecules with sufficient energy to break the bonds. A typical reaction like this one would have this energy-reaction coordinate graph:

The reaction proceeds into a high-energy transition state, and then comes down to the low energy products. To reach the transition state however, you must supply heat.

Edit: As @porphyrin mentioned, there are some reactions that have essentially zero activational energy. Combination of radicals may be an example:

$$\ce{Cl. + Cl. -> Cl2}$$

This one has no activational energy because there is no unstable transition state involved. Another popular example is the basic neutralization reaction:

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

Here again, there isn't any highly unstable transition state.


Why is quickly cooling a a hot pan bad for it?

Specifically it depends on material the pan is made of. Just like how heated glass can crack once immediately expose to a source of the opposite temperature. For example, a cast iron skillet has the potential to crack when at a high temperature it is exposed to a much lower temperature. To answer your second question, the reason why reheating the pan wouldn't return it to its original shape is first, you would have melt the metals to reform it and second of all if a crack has occurred in the situation, it means the elastic limit has been reached. The elastic limit is the greatest stress something can undergo and return to its original shape. Beyond the elastic limit, it is impossible to return something to its original shape. I hope this helps!


How exactly do you measure the free energy of a reaction?

How would you go about measuring the energy (in joules) of a reaction
  using the heat of the reaction?

You can measure the heat of a reaction (the enthalpy of reaction, $\ce{\Delta H}$) using a calorimeter, but the enthalpy of the reaction is only one part of the free energy of the reaction ($\ce{\Delta G}$). $\ce{\Delta G}$ is related to $\ce{\Delta H}$ through the following equation $$\ce{\Delta G = \Delta H} - T \Delta S$$ where $\ce{T}$ is the temperature and $\ce{\Delta S}$ is the change in entropy for the reaction.

To use a calorimeter you put your reactants into the calorimeter and let the reaction proceed.  The calorimeter is thermally isolated from its surroundings, so any heat given off (exothermic) or absorbed (endothermic) during the course of the reaction can be accurately measured.  Knowing the moles of reactant you put into the calorimeter and the heat change in joules (using the measured $\ce{\Delta T}$ and the known heat capacity of the calorimeter) you can determine $\ce{\Delta H}$ in joules/mole.  Here is a link that provides more detail about calorimeters along with some problems that are worked out in detail.


Is it possible to create a solution with superoxide?

Potassium superoxide (Aldrich) and an equimolar
  amount of 18-crown-6 ether used to increase solubility were
  dissolved in dry DMSO.

Role of Hydrogen Bonding in the Active Site of Human Manganese Superoxide
Dismutase Biochemistry 2004, 43, 7038-7045


When are the terms diphosphate and pyrophosphate used?

Great question, but unfortunately without a great answer.  Chemistry is very inconsistent in its naming of phosphates.  In addition to the terms diphosphate and pyrophosphate, the term bisphosphate is also used.

Someone will probably put in an answer that cites "official" IUPAC or IUBMB rules.  In many cases (but not all) these rules are more sensical and well thought-out than the terms in common usage in the field, but unfortunately, they are not in universal (or even wide) use.

Here is a rough breakdown of my sense of how these three terms are used:

Inorganic chemistry

Pyrophosphate indicates a phosphate anhydride.
Diphosphate indicates a salt that contains more than one (ortho)-phosphate anion.
Bisphosphate is not widely used.

Biochemistry

Pyrophosphate: used inconsistently to name an organic derivative of a phosphate anhydride.  This is what you see in geranyl pyrophosphate (GPP).  It is also used to refer to "inorganic" pyrophosphate anion, often abbreviated as PPi.
Diphosphate: an unfortunate term that is ambiguous, although usage is shifting towards synonymity with pyrophosphate.  It is often used to mean "pyrophosphate" and but sometimes still used to mean "bisphosphate".  Biochemically, the most important phosphate anhydrides are undoubtedly ATP, ADP, etc.  These are never referred to as pyrophosphates even though that term would be more accurate.  History has chosen "triphosphate" and "diphosphate" as the terms to use for adenosine-substituted phosphate anhydrides, icky though the term may be.
Older literature will also use "diphosphate" to mean "bisphosphate".  The name "fructose diphosphate was formerly in wide usage for what is now known as "fructose-1,6-bisphosphate", for example.  That compound contains no phosphate anhydrides.
Geranyl diphosphate is still a fairly widely used synonym for geranyl pyrophosphate.  You see abbreviations for this molecule of both GPP and GDP.  I prefer the name GPP, but unfortunately not everyone uses it.
bisphosphate: Now in wide usage as the preferred term for an organic molecule with multiple ortho-phosphate esters, such as fructose 1,6-bisphosphate.  Biochemists have done a relatively good job of adopting this usage in modern times.

A common confusion that stems from this nomenclature concerns the stability of organophosphate compounds. Phosphate esters, e.g. glucose-1-phosphate or glucose-6-phosphate or adenosine monophosphate, are much more stable and less reactive than phosphate anhydrides such as acetyl phosphate (a mixed anhydride of acetic acid and phosphoric acid) or the phosphoanhydride bonds in adenosine triphosphate (ATP).


How do I interpret the results of this chromatogram and does it tell me about the compounds' polarities?

You have several queries. First of let us look at the gradient method used by the authors.

The HPLC gradient for synthetic colorants was 0 min A 100 p. cent, 10
min A 0 p. cent, 14 min A 100 p. cent. The HPLC gradient for natural
carmine was 0 min A 80 p. cent, 20 min A 50 p. cent, 23 min A 30 p.
cent, 25 min A 30 p. cent

Factually, notice the trend in the mobile phase for synthetic ones. It is a linear gradient elution that they started with 40% MeOH at t=0 min, and at t=14 min, you have 70% MeOH. In the second case, it is a purely empirical "step" gradient. Try to draw a graph on a piece of paper.
Regardless, the authors use a water-rich mobile phase in the beginning and go to methanol rich mobile phase. Note I am avoiding the word polarity.

Polarity is an umbrella term in chromatography. For elementary purposes, it may be okay for serious work predicting retention on the ill-defined polarity that carries little or no meaning. You linked a GC webpage for an HPLC separation; those are different worlds.
Qualitatively, in top figure (a), it is clear that amaranth spends more time in the water-rich mobile phase, followed by azo ruby and then ponceau 4R if we use a C18 phase.

All you can say is that amaranth has a higher solubility in the water-rich mobile phase than ponceau 4R. One should not over-interpret retention behavior with this limited data.
In short, polarity is not a great terminology for describing the retention order. If you are interested, see the "Hydrophobic subtraction model" of Snyder et al. This one describes the retention in a better way of C18 molecules.
2)

In the chromatogram above (specifically chromatogram "a") and the
associated table, the elution time for the three compounds ranged from
6.60 minutes to 9.76 minutes. Since solvent A and solvent B were used, how do I know which compounds were eluted by which solvent?

This is an interesting question. As you noticed, the composition of the mobile phase changing with time, so the k, the retention factor is changing with time. People have worked out the theory of simple linear gradients. You know the retention time of elution, and you have the gradient composition chart. Correlate it, but this is a very, very rough estimate. The actual treatment is way more involved. Check "Theory of Gradient Elution Liquid Chromatography with Linear Solvent Strength: Part 1. Migration and Elution Parameters of a Solute Band" by Leonid Blumberg to see the complexity of the problem of gradient elution.
One of the best books is  "Introduction to Modern Liquid Chromatography By Lloyd R. Snyder, Joseph J. Kirkland, John W. Dolan, 2011."


Has QTPIE ever been applied to 1D metals such as polyacetylene?

How flattering that someone asked this question ;)

The short answer is yes, I've tried QTPIE on series of conjugated molecules like the oligoacetylenes and oligoacenes. The results, however, are fairly dismal, as can be seen in the preprint. The problem is that in these empirical charge models, the polarizability grows faster than the size of the molecule, which means that these models predict infinite polarizabilities and charge transport behavior for bulk materials, which is obviously nonsense.

The problem is quite intimately related to that of self-interaction error in approximate density functional theories, which also exhibit similar problems.


Which temprature is taken in Gibbs free energy equation, of the system or the surrounding?

The Gibbs free energy is a property of a system, so you should use that system's temperature when calculating it.  The Gibbs free energy is the amount of work that could be extracted from the system in a isothermal, isobaric process: i.e. the temperature and pressure of the system won't change.  Thus Gibbs energy has to do only with the temperature of the system, not of the surroundings.

There is a related but distinct variable called exergy or sometimes "availability".  This variable is given the symbol $B$, and is also a thermodynamic state function.  It is defined as $\Delta B = \Delta H - T_0 \Delta S$, where $T_0$ is the temperature of the surroundings.

Exergy represents the maximum amount of work that could be extracted from a system when that system is brought into thermal equilibrium with the surroundings, for example using Carnot engines.  Thus exergy indicates work obtainable through a combined (a) isothermal change in Gibbs free energy at temperature $T \neq T_0$ and (b) a temperature change occurs when Carnot engines are used to transfer heat from the system at $T$ to the surroundings at $T_0$ until thermal equilibrium is obtained.


Can compounds found in concrete leach into soil against gravity?

Gravity is a small force on ions when compared to water flow and diffusion; lead ions won't sink in a graduated cylinder. But rather than ponder the drift upwards of ions from the concrete, consider the drift downwards of ions from the soil. For instance, the soil may be acidic, and will almost always have more water in it than the concrete has, so the acids in the soil will be much more mobile than the ions in the concrete, and will fairly easily drift into the concrete, neutralizing $Ca(OH)_2$ or even dissolving $CaCO_3$. Whether these calcium ions will travel upwards... well, yes, but  calcium compounds may adsorb on soil particles and travel even slower. So the soil may be as much a barrier as the concrete is a poor source of ions.
Concrete may lose some ions (especially $Ca(OH)_2$) in contact with flowing water, but even more, it tends to pick up $CO_2$ and the $Ca(OH)_2$ becomes $CaCO_3$, which is less soluble. So the soil right next to the concrete may be a little more alkaline than the soil at the top surface.


Would the carbocation intermediate undergo rearrangement?

Unfortunately, I cannot back this with references at the moment but in my opinion there should be no carbocation formed in the first place and therefore no tendency to undergo any Wagner-Meerwein rearrangements.

Secondary carbocations, while being less unstable than primary ones, are still rather unstable. There are no stabilisation mechanisms that suggest themselves in the reaction conditions (concentrated sulphuric acid) in which the nitrogen will be protonated and therefore unable to stabilise a carbocation in α position. I would assume that the elimination proceeds much more $\mathrm{E2}$-like: likely with no intermediate carbocation at all but if there is one it should be extremely short lived and the neighbouring proton should dissociate practically instantly.


How would one carboxylate CBG into its acid form, CBGA?

The substrate is a resorcinol derivative: a strongly activated aromatic ring, which upon reaction with a base delocalizes its negative charge on the ring, making it reactive towards acids (Eg. CO2).

Here (entries 130-132) good conversion is claimed (85%) in a high pressure vessel with magnesium methoxide and dry ice, or a lower conversion (40%) is reported at ambient pressure with magnesium methyl carbonate


What is the ground state electron configuration of isoprene?

The ground state is a singlet, when excited by UV light a $\pi\pi^*$ singlet excited state can be produced. (Singlet has electron spin paired with one electron in $\pi$ HOMO orbital one in LUMO [or higher orbital for other excited states]). The singlet may then form a triplet or decay to the ground state by emission/ internal conversion/reaction as per normal. In some polyenes the lowest excited state may be symmetry forbidden from the ground state, you will have to look into the details for each molecule.


More than one way of balancing a chemical equation

This happens when you have a sum of two (or more) independent reactions. You can balance each one, and then add them together in an arbitrary proportion. People usually run into this when trying to come up with an equation for combustion of gunpowder, where the oxidation of carbon and oxidation of sulfur are pretty much independent.

Now, in your case the reactions are:
$$\ce{2O3 -> 3O2}\tag A$$
and
$$\ce{2NO + O2 -> 2NO2}\tag B$$
Your first reaction is a combination of ${1\over2}A+{1\over2}B$, the second is $A+2B$, and the third is ${3\over2}A+{1\over2}B$. There are infinitely many possible combinations.

Ozone, being an extremely powerful oxidant, is known to participate in certain reactions where it spends one atom of oxygen and the rest falls off as the less reactive $\ce {O2}$, but this is not the case here. $\ce {O2}$ reacts with $\ce {NO}$ instantly, as well as $\ce {O3}$.


Reactivity and transport safety of substituted methyl benzoates

I’ve already addressed this in a comment, but now I went the lengths to check out the material safety data sheets of these compounds from Sigma Aldrich. My suspicions have proven correct: methyl benzoate is not a dangerous compound to transport so any recommendations to reduce its reactivity by adding a substituent must be moot.

In short, section 14 of the MSDSs as I access them contains transport information; and this entry for the compounds methyl benzoate (CAS 93-58-3), methyl 3-nitrobenzoate (CAS 618-95-1), methyl 2-methylbenzoate (CAS 89-71-4) and methyl 2-bromobenzoate (CAS 610-94-6) contains nothing but ‘not dangerous goods’ and the required iterations and variations of that theme to the different categories there.

Therefore, the question must be rejected as totally pointless: the starting compound is not considered a dangerous good for transportation and neither is any of the three possible substituted ones (one generic placeholder for each; I’m not going to check an additional six MSDS sheets).

Naturally, the only answer you can remove by deduction is b, because a hydrogen ‘substituent’ does not change the properties of the benzoate residue since hydrogen is already present there five times.

Considering we have found out that the compound is not dangerous to transport, how can we approach this question? Well, it’s hard. Concerning the reactivity of electrophilic aromatic substitutions — which is low anyway because of the deactivating ester substituent — of course an electron-withdrawing substituent would help; the nitro group suggests itself. On the other hand, concerning the reactivity of the ester group’s carbonyl function, only electron-donating substituents would help; the strongest being the inductive effect of the methyl group. So we are left with two possibilities and it is unclear which one we should follow.

Of course, neither the reaction with electrophiles nor the reaction with nucleophiles is inherently more dangerous than the other. We have again reached a vicious circle in which we cannot proceed at the risk of taking the wrong path.

Therefore, the best advice I can give is to ignore the question in the book. If it is given in an exam, challenge the exam for there is no reasonable answer to it.


Is the salt of a weak acid and a weak base also a weak electrolyte?

When a weak acid such as acetic acid is added to pure water, it does not ionize much, but sufficiently to make the solution acidic. On the other hand, when small amounts of weak acids are added to buffered solution with a pH near neutral, acidic acid dissociates almost quantitatively. So the van't Hoff factor depends on the pH.

The same goes for weak bases used as ammonia - in combination with pure water, you get a basic solution and very little ammonium, but when dissolved in a solution maintaining a neutral pH, ammonia forms the ammonium ion almost quantitatively.

  I have also heard blunt statements that all salts are strong electrolytes, but never with satisfactory explanation or even some weak-weak examples.

If we combine ammonia and acetic acid in aqueous solution and remove the water, we get the salt ammonium acetate. If we dissolve ammonium acetate in pure water, the pH will be roughly neutral, and both ions will remain ions. Therefor, this is a strong electrolyte solution.

  So are there any salts that are weak electrolytes?

In the pure salt, the components are ions by definition. Water is very good at dissolving ions, so the dissolved ions stay ions. To get a salt that is also a weak electrolyte, you would need an uncharged acid and an uncharged base that  would transfer a proton upon forming a salt, but not while in aqueous solution. I am not aware of any such system.


Is there a common name to refer to the groups 13 and 14?

It seems that the carbon group is known as either the tetrels or the crystallogens; while the boron group is known as triels or icosagens. These names are very rarely used.

Source: Wikipedia articles on Carbon family (also, crystallogen) and Boron family. Although other (seemingly) respectable sites use them.


What makes the cyclic structure of glucose more stable than the open chain structure?

You get to form a C-O $\mathrm{\sigma}$ bond at the expense of a C-O $\mathrm{\pi}$ bond. The single bond has a higher bond energy, even though it is somewhat de-stabilized by the anomeric effect. It's still an overall win, despite taking the entropic hit to confine your molecule into a ring.


Self aldol condensation of butanal under basic conditions

Conceptually, we can think of this reaction as occurring in three stages. I've provided diagrams for the self-condensation of acetaldehyde, however you should be able to draw an analogous mechanism for the self-condensation of butanal (your aldehyde).
Your mechanism is essentially correct, except for the final elimination step, which as you've drawn it gives the wrong product.
1. Enolisation of the aldehyde

Source: Organic Chemistry, 2 ed., Warren, Clayden, Wothers & Greeves

As you guessed, the hydroxide is basic enough to cause enolisation of the aldehyde. Using KOH/MeOH, we form the enolate under thermodynamic control, essentially meaning the enolate is in equilibrium with the aldehyde.
2. Addition of the enolate into another molecule of aldehyde

Source: Organic Chemistry, 2 ed., Warren, Clayden, Wothers & Greeves

The newly formed enolate may then attack an un-enolised molecule of the aldehyde (this is okay, since we've established that the aldehyde and the enolate are in equilibrium under these conditions). This nucleophilic attack leads to a beta-alkoxy aldehyde (the structure on the right), which will quickly pick up a proton to form a beta-hydroxy aldehyde.
Under these basic conditions (KOH, MeOH), we may be able to isolate the hydroxyaldehyde, however, they are prone to elimination (see below), which is what must happen to form the product you require.
3. Elimination of hydroxide

Source: Organic Chemistry, 2 ed., Warren, Clayden, Wothers & Greeves

The beta-hydroxyaldehyde formed (see above) is itself able to undergo enolisation, to form a new enolate. Rather than attacking another molecule of aldehyde intermolecularly, it may instead do an intramolecular E1cb type elimination, kicking out hydroxide (which is okay under these basic conditions), and giving you the product as drawn.
Your elimination step isn't the most favourable pathway to get rid of that alcohol, and infact the elimination only occurs under these conditions due to assistance form the carbonyl. The quote below puts this rather well.

You cannot normally eliminate water from an alcohol in basic solution as hydroxide is a bad leaving group. It is the carbonyl group that allows elimination here: these are E1cB reactions.
Source: Organic Chemistry, 2 ed., Warren, Clayden, Wothers & Greeves


How to store bulk gas products and reagents?

It probably depends on local regulations and the properties of the intermediates!

Gas bags might indeed be a viable solution.

In the biogas community, see SolarCities for an example, PVC bags from PUXIN, a chinese supplier, are quite popular.


Concentrated sulphuric acid is a highly acidic medium and thus it very exclusively prefers the $E_1$ mechanism over $E_2$ mechanism and thus a carbocation is surely formed in this conditions.
The example you provide proceeds through the following mechanism:

Now the mechanism you provide, other than following the wrong pathway(very minor pathway) also has a major mistake. During the protonation of 1-methylenecyclopentane the tertiary carbocation is not more stable than the primary carbocation. The tertiary carbocation is $\mathrm{sp^2}$ hybridized and thus prefers a bond angle of 120° which cannot be provided by the cyclopentane ring as it is not flexible enough.
It prefers the very less angle strain in case of $\mathrm{sp^3}$ hybridized orbitlas but with less positive inductive effect over higher angle strain but more positive inductive effect. That's the reason the below equilibrium is highly backward favoured.


Does the Location of Line-Dash-Wedge Notations Matter?

Yes, both inner and outer placements are possible with some nuances.

From Graphical representation of stereochemical configuration (IUPAC Recommendations 2006) [1, p. 1926]:

  ST-1.3.2: Stereogenic centers at ring fusion atoms

  Stereogenic centers at ring fusion atoms should be drawn with hashed wedged or solid wedged bonds to the exocyclic substituent at the fusion atom whenever possible. If necessary, an implicit hydrogen should be made explicit in order to provide an exocyclic substituent to bear the hashed wedged or solid wedged bond. When one of the fusion bonds is oriented vertically and the exocyclic substituent is graphically small (such as a hydrogen or a methyl group, or even such physically larger substituents as a phenyl group represented with a graphically small “Ph” group label), the exocyclic substituent is also
  preferentially oriented vertically and opposite to the vertical fusion bond. This orientation is preferred even if it results in the substituent being placed within the ring system, and is particularly common in the depiction of steroids and other natural products.

From Graphical representation standards for chemical structure diagrams (IUPAC Recommendations 2008) [2, p. 354–357]:

  GR-4.2.2 Fusion atoms with one substituent drawn

  Ring fusion atoms already have three bonds within the ring system. For external ring fusion atoms, the substituent should preferentially be positioned outside the ring system and oriented so that it bisects the angle between the adjacent bonds.

  [...]

  When two substituted fusion atoms are adjacent, placing both substituents outside the ring system will typically cause them to overlap. In those cases, it is better to draw one of the substituents within the ring system. A substituent drawn in this way should preferentially be drawn exactly vertically or horizontally, whichever direction would minimize overlap with other atoms and bonds.

  If a substituted fusion atom is adjacent to an unsubstituted fusion atom, there typically is enough room to orient the substituent either inside or outside the ring system. That substituent should be oriented outside the ring system if all substituents can be oriented outwards, but should preferably be drawn exactly vertically or horizontally if any other substituent is placed within the ring system to avoid
  conflict with an adjacent substituent.

  Small substituents on the C-8 and C-9 atoms of steroids should be drawn within the ring system in this fashion even if there are no substituents on the adjacent fusion atoms C-10 and C-14. Certain other natural products [23] also have preferred orientations that place substituents within the ring system.

  A substituent should not be oriented within a ring system when attached to a fusion atom that lacks any vertical or horizontal fusion bond. The substituent should be oriented outside the ring system, minimizing overlap as best as is possible. Sometimes, it might be possible to reorient the ring system so that the fusion bonds are vertical; the substituents should certainly be placed vertically if so.

  It is not acceptable to orient two substituents within the same ring if the ring has fewer than eight atoms. One or both of the substituents should be oriented outside the ring system, minimizing overlap as best as is possible.

  Substituents on interior fusion atoms have no option other than being drawn within the ring system. As above, such substituents should preferentially be drawn exactly vertically or horizontally, according to the direction that would minimize overlap with other atoms and bonds.

References

Brecher, J. Graphical Representation of Stereochemical Configuration (IUPAC Recommendations 2006). Pure and Applied Chemistry 2009, 78 (10), 1897–1970. https://doi.org/10.1351/pac200678101897. (Free Access)
Brecher, J. Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008). Pure and Applied Chemistry 2009, 80 (2), 277–410. https://doi.org/10.1351/pac200880020277. (Free Access)


Synthesis of 4-amino-3,5-dimethylphenol from N-(2,6-dimethylphenyl)hydroxylamine using perchloric acid?

N-Phenylhydroxylamines undergo what is known as the Bamberger rearrangement when subjected to aqueous acid. The mechanism is a bit strange, as it involves divalent positively charged nitrogen (very rare - in fact, it is so rare that ChemDraw keeps trying to correct my $\ce{NH}$ to $\ce{NH3}$!), but that's just how it is. Wikipedia already has what I consider to be a very good explanation, but I'll try.

The mechanism begins with protonation at oxygen. Ordinarily, one would expect the nitrogen to be the more basic heteroatom, but protonation at nitrogen is unproductive - you can protonate nitrogen, but it's a dead end. On the other hand, if you protonate on oxygen, you can lose a water molecule.

The cationic intermediate on the right is rather like a benzyl cation, but with a nitrogen instead of a carbon. It has several resonance forms:

and the only decent nucleophile in the system, water, has three choices as to where it wants to attack.

Attack at nitrogen just goes back the way we came.
Attack at the ortho positions in this case would lead to a loss of aromaticity because you can't deprotonate at the end (see the mechanism below).
Attack at the para position. This completes the mechanism; after a proton exchange between N and O, a deprotonation leads to rearomatisation.


How do "basic" amino acids (lysine, arginine) act as base under physiological conditions?

I will paraphrase the question:

  Why are the amino acids arginine and lysine called basic? Why are the amino acids glutamic acid and aspartic acid called acidic?

This nomenclature makes sense for situations where the amino acids are dissolved in a non-aqueous solvent. In that case, the side chains will be neutral and have nitrogenous base groups and carboxylic acid groups, respectively.

When the amino acids are dissolved in water at neutral or physiological pH, however, the name does not make sense because they have already acted as base (Arg and Lys) or as acid (Glu and Asp), and are now the conjugate acid and conjugate base, respectively. For glutamic and aspartic acid, it is common to refer to them as glutamate and aspartate (the name of the conjugate base). For arginine and lysine, even when protonated, those names are usually used nonetheless.

  [OP in comments:] So, in physiological situations (aka in the human body), how do arginine, lysine, and histidine act as bases, if they exist in their protonated form, considering how they are often used in enzymatic catalysis (same question asked of acidic amino acids in their deprotonated forms)?

Histidine

Histidine is a common acid/base catalyst. Its pKa is close to 7, so an appreciable fraction is protonated and an appreciable fraction is deprotonated. Often, this amino acid will act as acid and as base in different steps of the same mechanism (e.g. RNase).

Lysine

Lysine is not a typical acid base catalyst. It has roles in binding negatively charged ligands (such as ATP), acting as a "positive charge on a stick". It also acts as a nucleophile, forming e.g. Schiff's base with aldehydes and ketones. Lysine's pKa in the context of a folded protein varies considerably (see e.g. https://www.pnas.org/content/108/13/5260), so in some contexts, it will be an efficient acid/base catalyst.

Arginine

Arginine is not a typical acid base catalyst. Like lysine it has roles in binding negatively charged ligands but different from lysine, it can act as a bidendate hydrogen bond donor (e.g. making a nice salt bridge with aspartate or glutamate with two hydrogen bonds). It also has a role in cation-pi interactions, stacking its planar guanidinium group on aromatic rings. This web page has more on the functional role of arginine: http://www.russelllab.org/aas/Arg.html (and there are similar pages for the other amino acids on that site)


What is the definition of thermodynamic stability?

The full phrase should be thermodynamic stability with respect to ____, where the dash indicates a process, or a chemical reaction.
A mixture of hydrogen and oxygen is thermodynamically unstable with respect to water formation.
Similarly, a diamond is not forever (which may not please De Beers and ladies). It is thermodynamically unstable with respect to conversion to graphite.
Also, thermodynamic stability is a relative term which is often contrasted with reactivity or kinetic stability.
Diamond is kinetically stable at room temperature for the same process (lucky ladies can smile again).


Finding the mystery ester strucuture using NMR

Prediction software always has its limitations, and there is always a degree of error in the calculation. For the ChemDraw predictions, you will see that for the 3 aromatic environments, it has done 3 independent calculations, and has happened to arrive at the same chemical shift. This simply means that these shifts are coincident, not equivalent.

Remember, prediction software is like any tool — only as good as the person using it, and should not replace a proper assessment of the data.

Your assessment here should first look at your proton environments; all 10 protons are accounted for. We have a $\ce{CH3}$, a $\ce{CH2}$, and a mono-substituted benzene. The $\ce{CH3}$ and $\ce{CH2}$ are not directly connected to any other group that is causing obvious splitting. Secondly, the $\ce{^13C}$ spectrum shows 9 peaks, consistent with our $\ce{CH3, CH2}$ and a mono-substituted benzene. We also have a peak at ~δ170, which is a $\ce{-C(O)-{}}$

There are only a few ways you could put these groups together. Rather than using ChemDraw to predict the spectrum and see which is identical to your question, you should rationalise why each possibility is or isn't the correct answer.

Trying to rationalise a bunch of peaks around δ7.4–7.5 is the of minimal interest, and remember that a real spectrum will almost always look different to a simulation. The peaks you should focus on willjustify the chemical shifts for the $\ce{-CH3}$ group and the $\ce{-CH2}$ group in the proton spectrum, and the $\ce{-CH3, -CH2}$ and $\ce{-C(O)-{}}$ in the carbon spectrum. There is only one possible correct answer.


I feel very ashamed to answer my own question, but having found a possible answer, I don’t see why I shouldn’t share it with the community.

I would like to start by bringing in this table of orbital energies (from the appendix I linked to in the question) as a reference point for further explanations:

$$\begin{array}{ccc}
\hline
\text{Element} & \text{Energy of outer s orbital / eV} & \text{Energy of outer p orbital / eV} \\
\hline
\text{Carbon} & -19.43 & -10.66 \\
\text{Silicon} & -14.89  & -7.78 \\
\text{Germanium} & \mathbf{-16.05} & -7.54 \\
\text{Tin} & -14.56 & -7.01 \\
\text{Lead} & \mathbf{-15.12} & -6.81 \\
\hline
\end{array}$$

As you can see, the energies of the p orbitals get higher and higher as you go down the group, while the energies of the s orbitals don't.

The reason why, in general, orbitals rise in energy when going down a group is that even though the nuclear charge is bigger (which has a stronger effect than the rise in the quantum number $n$), the increased screening effect of added electrons prevails over this rise.

Germanium has such a low 4s electron energy because of the so-called d-block contraction. Since it has 10 added protons due to the existence of the d-block, and the d electrons don’t shield this increased nuclear charge well, the s electrons are stabilised. The p electrons don’t feel this effect as much because they are shielded more than the s electrons (less penetration), thus experiencing the rise in the nuclear charge less, while being shielded additionally by the stabilised s electrons (and presumably by the d electrons as well, to a lesser extent).

The lowered energy of the 6s orbital of lead is the result of the equivalent f-block contraction, amplified by the relativistic contraction of the s shell. The p electrons don’t have a noticeable change for the same reason they didn’t in germanium.

Now, tin has d electrons too, so it will experience the d-block contraction – but to a far more limited extent, since the change in nuclear charge is not as big in percentage. To illustrate this concept, it is useful to bring in this formula:

$$E \approx -13.6 \left(\frac{Z}{n}\right)^2 \mathrm{~eV}$$

where $E$ is the energy of the electron (in the hydrogen-like atom!), $Z$ the nuclear charge and $n$ the principal quantum number. Since it refers to hydrogen-like atoms, this formula obviously doesn’t bring in the screening effect. To see why the d-block contraction doesn’t have such a strong influence on the tin atom, we can use this formula to calculate how the non-screened, non-relativistic energy grows from Si to Sn:

$$\begin{array}{cc}
\hline
\text{Element} & \text{Non-screened energy of outer orbital / eV} \\
\hline
\ce{Si} & -296.178 \\
\ce{Ge} & -870.4 \\
\ce{Sn} & -1360 \\
\hline
\end{array}$$

As you can see, the percentual change from Si to Ge is much higher than the change from Ge to Sn – thus, the increase in nuclear charge from Ge to Sn doesn’t have an enormous effect on the electron energy, while the screening effect between Ge and Sn grows steadily: the orbital energy is lowered in the many-electron atom. Hence, the Sn s shell is stabilised, but to a lesser extent.

Now, why does Sn form divalent compounds so easily, when Ge and Si have lower s energies and don’t? There’s a passage in Greenwood’s Chemistry of the Elements (2nd ed), pp 226-7 that explains it:

  The term “inert-pair effect is somewhat misleading since it implies that the energy required to involve $n\mathrm{s^2}$ electrons in bonding increases in the sequence $\ce{Al} < \ce{Ga} < \ce{In} < \ce{Tl}$. […] This is not so […]. The explanation lies rather in the decrease in bond energy with increase in size from Al to Tl so that the energy required to involve the s electrons in bonding is not compensated by the energy released in forming the 2 additional bonds.

Indeed, that is what happens: the $\ce{Sn-X}$ bonds are longer than the $\ce{Si-X}$ bonds, and thus have smaller energy, as permeakra pointed out in his/her comment. However, I believe this is not due to the interference of the d-subshell, but simply to the larger Sn nuclear charge. This increased nuclear charge amplifies the repulsion to the nuclei of the other species (say, oxygen), thus lengthening the bond. Since the s electrons have a very low energy compared to the p ones, they do not participate in this bond – they cannot overlap.

In short, tin forms divalent compounds because:

Its s subshell is more stable than the p subshell, because of screening differences;
Its bonds are longer, due to the increased nuclear charge: thus, the s electrons don’t bond.

The combined effect is smaller in germanium compounds (Ge is more tetravalent), because even though the s shell is more stabilised, the bonds are shorter; it is stronger in lead compounds (Pb is more divalent) because the 6s shell is more stabilised than the Sn 5s shell and the bonds are even longer.


The probability of tunnelling at an energy $E$ is given by $p(E)\approx e^{-bA\sqrt{m}}$ where $A$ is proportional to the area of the potential energy barrier above  energy $E$, i.e. the top part of the potential barrier, $m$ the mass and $b$ some constants, $\pi, \hbar$. etc. Thus for a given mass and energy if the barrier is narrow, so $A$ is small, tunnelling is more likely than if the barrier is wide. At a given energy for the same barrier if the mass is large tunnelling is small. Thus we tend to see tunnelling only with H and D  and not Cl atoms for example. Tunnelling is also important in electron transfer reactions.

As there is not just a single energy in a reaction but a distribution of energies, according to the Boltzmann distribution, it is necessary to modify the expression above to average over the energy but the basic result is the same, which is that the reaction rate constant is reduced by the factor $p(E)$.

[If the potential energy barrier is $V(x)$ then $\displaystyle p(E)=\exp\left(-\frac{2\sqrt{m}}{\hbar}\int_{x_1}^{x_2}\sqrt{V(x)-E}\right)$ where $x_{1,2}$ are the points either side of  the barrier with energy $E$.]


Is there an intramolecular hydrogen bond in guaiacol?

According to a quick calculation on the DF-BP86/def2-SVP level of theory, there is an intramolecular hydrogen bond with a strength of about $\ce{-26.6 kJ/mol}$ according to a quantum theory of atoms in molecule (QTAIM) analysis.

For an explanation of the evaluation of the bond strength see: E. Espinosa, E. Molins, C. Lecomte. Chem. Phys. Lett. 1998, 285, 170-173.
Analysis done with MultiWFN 3.3.8. Calculations with Gaussian 09 D.01.


How to calculate the quantum density of states?

If you really do want to calculate the density of states then you can try direct counting to get the total number of states N and then calculate the density of states $\rho =[N(E+dE)-N(E)]/dE$. This, of course, will only work if you have a few vibrational normal modes as all the combinations of quanta have to be included, e.g. in water with 3 modes, quanta (in the order of increasing energy) are: $(0,0,0), (0,0,1),(0,0,2),(0,1,0), (1,0,0),(0,0,3),(0,1,1)$, etc.  But for a molecule with just a few more atoms, it is quite probable you will get $10^6$ vibrational states /wavenumber at an excess energy of, say,  $5000 \pu{~cm^{-1}}$.

In the field of unimolecular reactions (RRKM models etc.) there has been much effort in calculating total and density of states. There is the Beyer-Swinehart algorithm; Beyer, T., & Swinehart, D. F. 1973, Commun. Assoc. Comput. Machinery, 16, 379 and Stein S & Rabinovitch B. J. Chem. Phys. v58, 2438, 1973. Whitten and Rabinovitch have also modified the classical sum of states to give good agreement with the direct counting (quantum) method. (J. Chem. Phys. 41, 1883, 1964, and 48, 1427 1968). Their modification $^{[1]}$ gives the total sum of states as:

$$ N(E)=\frac{(E+\alpha E_{zpe})^s}{s!\prod_{i=1}^s h\nu_i}$$

where s is the number of (harmonic) oscillators, $E_{zpe}$ is the zero point energy for the i th oscillator,

$$\alpha = 1-\beta w(E')$$

$$\beta = \frac{s-1}{s}\frac{<\nu^2>}{<\nu>^2}$$

and w depends on the range of $E'=E/E_{zpe}$. When $0.1 \lt E' \lt 1; w(E')=(5E'+2.73\sqrt(E')+3.51)^{-1}$, and when $1\lt E' \lt 8; w(E')=\exp(-2.4191(E')^{1/4})$.
The equation in N can be differentiated to find the density of states.

If, however, you want to calculate a rate constant using the Fermi Golden rule it is better not to start with the density of states separated out, but to calculate the Franck-Condon weighted density of states.

The rate constant has the Golden Rule form
$$ k= \frac{2\pi}{\hbar}|<av''|H|bv'>|^2\delta(E_{av'}-E_{bv''})$$

and then this is modified depending on what actual situation is to be analysed, e.g. thermally averaged over the Boltzmann distribution if the molecules are in solution.

In this case the best place to start is probably with the papers

(a)  Heller, Freed & Gelbart, J. Chem. Phys. 56, 2309 (1972),

(b)  Bixon & Jortner, J. Chem. Phys. 48, 715-726, 1968 and

(c) S. H. Lin; J. Chem. Phys. 58, 5760 (1973).

Each deal with non-adiabatic interactions causing radiationless transitions in molecules, i.e. from one vibronic level into a continuum of states, such as singlet to triplet intersystem crossing. There is a summary of different methods in Eyring, Lin & Lin 'Basic Chemical Kinetics' ch 7.

$^{[1]}$ equations given in Steinfeld, Francisco & Hase, 'Chemical Kinetics & Dynamics'. 2nd ed. p333.


The best approach would be to crystallize $\ce{ZnSO4.7H2O}$ and then weigh the required amount and prepare a fresh solution.

How:

Slowly evaporate the solution until it is about one-fifth of its original volume. Caution: Do not boil the solution as it may spit.

Allow the concentrated solution to cool until crystals form.

Filter off the crystals and put the filter paper and crystals on a watch glass and dab dry with another piece of filter paper.

Allow it to dry thoroughly (if you're keen on very accurate concentrations)

I assume by $x\%$ solution you mean $x\%$ weight($\pu{g}$)/volume($\pu{mL}$)
You may proceed to prepare the solution by dissolving $4$ grams in 100 $\pu{mL}$ water. In case you need a different volume, you may use unitary method.
Edit:
Crystallization is certinly more tedious than simple evaporation and volume restoration, but is bound to be very accurate. There might be loss of salt in trace amounts during evaporation, adheration to vessel walls et cetera. But if much precision is not your requirement, you may follow the latter, by all means.


The quantity you refer to has units of kJ/mol. Thus, for 1/2 mole of $\ce {\rm Cl}_{2}$, the quantity is halved to 122 kJ/mol. Your teacher might be referring to the quantity on a per-bond basis, but the number you report agrees with references online for a per-mole basis.


First point of difference rule 1,6-dimethylcyclohex-1-ene

First of all, the ‘lowest sum rule’ mentioned in the picture in the question does not exist in IUPAC nomenclature.

As already explained in a related answer, the most important simplified criteria for the numbering in such cases are:

lower locants for the principal characteristic group that is expressed as suffix
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

  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:

  (…)

  (c) principal characteristic groups and free valences (suffixes);

  (…)

  (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;

  (g) lowest locants for the substituent cited first as a prefix in the name;

  (…)

The compound that is given in the question doesn’t have any principal characteristic group that could be expressed as a suffix. Therefore, Rule (c) is not relevant in this case.

Thus, a low locant is assgined first to the double bond according to Rule (e). Therefore, the name of the parent structure without any further substituents is cyclohex-1-ene. (Without further substituents, however, the locant ‘1’ would be omitted.)

Finally, low locants are assigned to the remaining substituents as a set according to Rule (f).

  P-14.3.5 Lowest set of locants

  The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference; for example, the locant set ‘2,3,5,8’ is lower than ‘3,4,6,8’ and ‘2,4,5,7’.

Thus, the correct name is 1,6-dimethylcyclohex-1-ene rather than 2,3-dimethylcyclohex-1-ene since the locant set ‘1,6’ is lower than ‘2,3’.


In this paper [1] from 1963, various properties of heavy-oxygen water are measured including the melting point of both $\ce{H2^{18}O}$ and $\ce{D2^{18}O}$.

The melting temperatures were measured at:
$$\ce{H2^{18}O}:\ T_\mathrm{mp}=\pu{0.28\pm0.02^\circ C}$$
$$\ce{D2^{18}O}:\ T_\mathrm{mp}=\pu{4.02\pm0.02^\circ C}$$

So, as some commenters mentioned, the melting (or freezing) point of the heavy-oxygen water is slightly larger when it is protium on the heavy-oxygen, and quite a bit larger when deuterium is present. This basically means that the liquid has become more structured on average, which is to say that the average lifetime of a hydrogen bond is slightly longer mostly due to an increase in the average strength of a hydrogen bond.

I would be very surprised if there has been a measurement of $\ce{T2^{18}O}$ because tritium is very rare and not very easy to isolate. Additionally, heavy-oxygen is also quite uncommon, so purifying the $\ce{T2^{18}O}$ to make any reliable measurements would likely be very difficult. I would be very willing to guess, however, that the melting temperature is $\ce{T2^{18}O}:\ T_\mathrm{mp}\approx\pu{4.7^\circ C}$ by just noting that the isotopic substitiution effects seem to be roughly additive.

Extra Credit: Trying to Interpret the Data:

As a note which is really not what you're asking but I think is really interesting, is why there is such a sharp increase in freezing temperature for deuterium (and hence why heavy-oxygen has these effects only on a smaller scale). Indeed, it is true that the $\ce{O-D}$ vibrational frequency in $\ce{D2O}$ is quite a bit smaller than that mode in $\ce{H2O}$, but it really isn't obvious that lowering this frequency should make $\ce{D2O}$ freeze at a higher temperature. For instance, all else being equal, a larger frequency in $\ce{H2O}$ would predict a shorter hydrogen-bond distance because the vibrationally averaged bond length will be longer (because the long bond lengths contribute more than the short ones because the potential is Morse-like). But, what I just said about the vibrationally averaged bond length did not allow the $\ce{O-O}$ distance to relax in $\ce{D2O}$, so if the $\ce{O-O}$ distance shrunk in $\ce{D2O}$, then we could say it's just the lower zero-point energy because then the shorter $\ce{O-O}$ distance would explain the stronger hydrogen-bonding we seem to be observing in $\ce{D2O}$. Well, very interestingly, the $\ce{O-O}$ distance in liquid $\ce{D2O}$ has been both theoretically and experimentally observed to increase slightly or stay exactly the same. See ref. [2] and experimental references therein. So both geometric parameters point towards water actually having a higher melting point even though this obviously isn't true.

What's the deal then? Well most of the answer seems to be that ordinarily we think of nuclei as being classical, i.e. the atom can be located precisely, but this type of reasoning is quite bad for hydrogen. The important point then is still that deuterium is heavier than hydrogen, though not because of the explicit contribution which the frequency lowering would have on the geometric parameters if everything else were the same. Rather, a hydrogen atom is much more spread out than a deuterium atom due to the fact that lighter atoms will essentially behave more quantum mechanically. In the sense of being more wavelike.

The fact that this delocalization effect for hydrogen is most important was demonstrated in ref. [3], where the authors find that the hydrogen-bond length cannot be directly related to the vibrational frequency in a one-to-one sort of way. This is due to a large sensitivity to the $\ce{OHO}$ angle. This then explains what we were missing up above because even though the average bond length in $\ce{H2O}$ is larger than the in $\ce{D2O}$, this sensitivity to the hydrogen-bonding angle greatly decreases the hydrogen-bonding strength in liquid $\ce{H2O}$ compared to $\ce{D2O}$ which does not delocalize as much and hence does not suffer from this angle problem.

As a final fun point, if one does a simulation of liquid $\ce{H2O}$ using classical nuclei, and then the same simulation is done using path integral methods to include nuclear quantum effects, the change in freezing point is a decrease of about $\pu{4^\circ C}$, which can only be due to this delocalization effect I describe.

I basically included all this to say that interpreting what this heavy-oxygen change really means can actually be pretty complicated. Whatever it is, there must be some structural change, and one would always expect isotopic substitution for a heavier isotope to make liquids more structured both from the frequency effects and the nuclear quantum delocalization.

Sorry if that was overkill :)

[1] Steckel, F., & Szapiro, S. (1963). Physical properties of heavy oxygen water. Part 1.—Density and thermal expansion. Transactions of the Faraday Society, 59, 331-343.

[2] Chen, B., Ivanov, I., Klein, M. L., & Parrinello, M. (2003). Hydrogen bonding in water. Physical Review Letters, 91(21), 215503.

[3] Rey, R., Møller, K. B., & Hynes, J. T. (2002). Hydrogen bond dynamics in water and ultrafast infrared spectroscopy. The Journal of Physical Chemistry A, 106(50), 11993-11996.


Bonds are usually used to describe how electrons are shared. "Partially ionic bonds" are polar covalent bonds. Which is not the same as saying that a compound has different types of bonds in the whole compound. I think that your teacher gave you a bad question and a bad answer. If the question had asked whether all the bonds in sodium nitrate are ionic, then it would be a different story.

However, it is true that the bonds between the nitrogen and oxygen are covalent. Nevertheless, stating that sodium nitrate is ionic shouldn't be wrong.


Neutralization of Sodium Azide using Ammonium Cerium Nitrate

I think that you are doing fine!
The azide anion can be used for quite some amazing chemistry, but it's rather toxic and harmful for the environment.
Admittedly, I'm too lazy right now to look up the redox potentials or to write the reaction equation, but
CAN should be able to oxidize $\ce{N3-}$ all the way up to $\ce{NO3-}$, which is safe to be disposed.

Update
Judging from memory can result in bad advices!
A closer look at the SDS of both compounds shows that both have

H410 Very toxic to aquatic life with long lasting effects. and

P273 Avoid release to the environment.

Consequently, the combined solutions (with excess of CAN) should not be poured down the drain, but collected (aqueous inorganic waste).


tl;dr version: You can only define an acid or a base in context of a reaction. Individual compounds are not acids by themselves. They are acids in reactions in which they lose protons (or gain electron pairs).

The Lewis definition is the broadest acid-base definition. Each of the others are specific subsets.
The Arrhenius definition is the simplest:

Acid = something that creates $\ce{H+}$ ions in water.
Base = something that creates $\ce{OH-}$ ions in water.

Thus any compound that can ionize in or react with water, like $\ce{HCl}$ or $\ce{HCO2H}$ is an acid and any compound that contains hydroxide ions or reacts with water to produce them, like $\ce{NaNH2}$, is a base.
However, the Arrhenius definition breaks down in non-aqueous media.
The Brønsted-Lowry definition is broader and more inclusive.

Acid = proton donor - All Arrhenius acids are Brønsted-Lowry acids
Base = proton acceptor - All Arrhenius bases are Brønsted-Lowry bases

The Brønsted-Lowry definition allows us to extend our discussion of acids and bases to, say, solutions in methanol $\ce{CH3OH}$. Note that the Brønsted-Lowry definition requires a proton transfer reaction to be occurring to properly identify acids. For example, water is amphiprotic - it can react as either an acid or a base. Many other compounds are similar. There is no absolute "this is an acid always". Instead, you have "this compound is an acid in this reaction". For example, water is an acid in the first reaction and a base in the second.
$$\ce{H2O + H- -> H2  + OH-}$$
$$\ce{H2O + H2SO4 -> H3O+ + HSO4-}$$
The Lewis definition is broader and more inclusive still. It allows extension of acids and bases into aprotic media and situations:

Acid = electron pair acceptor - all Brønsted-Lowry acids are Lewis acids
Base = electron pair donor - all Brønsted-Lowry bases are Lewis bases

Again, the Lewis definition requires a reaction to occur in order to classify compounds.
However, in all cases, there are compounds that behave as acids much more frequently than they act as bases, and vice versa. For example, $\ce{HNO3}$ is almost always an acid, but not exclusively. The first step of the mechanism of aromatic nitration is considered to be a proton transfer step between sufluric acid and nitric acid in which nitric acid is the base.
$$\ce{HOSO3H + HONO2 <=> HOSO3- + O2NOH2+}$$


Rotational Barriers - Amide and Hydroxamic Acid

You can check out molecular orbitals of N-methylacetamide here. Unticking the spin-box will stop the molecule spinning. The important orbitals are $\ce{HOMO} -1$ and $\ce{HOMO} - 2$. (The actual HOMO is not part of the $\unicode[Times]{x3C0}$-system.)

The same site also features a low-level Simple Hückel Molecular Orbit Calculator. I used that to create the four images below of the $\unicode[Times]{x3C0}$-system of N-hydroxyformamide. From left to right the atoms are $\ce{O-N-C=O}$

The lowest $\unicode[Times]{x3C0}$-orbital is always bonding across the entire $\unicode[Times]{x3C0}$-system. In N-methylacetamide, the second $\unicode[Times]{x3C0}$-orbital has a nodal plain almost directly on the carbon, only slightly moved towards the carbonyl oxygen (theory tells me that, I can’t actually see that good in the model). The third $\unicode[Times]{x3C0}$-orbital is empty and thus does not participate in bonding. This means that each bond ($\ce{N-C}$ and $\ce{C=O}$ can be regarded as something between a single and a double bond — in accordance with the Lewis resonance structures. There is considerable overlap to be lost on rotating the system around $\ce{N-C}$.

In the hydroxamic acid, the second and third $\unicode[Times]{x3C0}$-orbitals (the occupied ones) only have very little electron density in the $\ce{N-C}$ bond. In fact, the lowest MO is largest for the $\ce{O-N}$ bond, the second is (weakly) antibonding concerning $\ce{O-N}$ and has the largest contributions on $\ce{C=O}$. The third is again antibonding along $\ce{O-N}$. Altogether this means that most of the electron density is between $\ce{O-N}$ and between $\ce{C=O}$ — the $\ce{N-C}$ bond only carries very little density and is thus much better described by a single bond. Much less resonance is lost when rotating, so the rotational barrier is lower.


Can the plane of symmetry for a compound be the plane of the paper?

In short: The presence of the $\sigma_h$ plane you see depends on the conformation of the methyl groups.

If your example were either benzene or 1,4-dichlorobenzene, then you were be right, the molecule would contain a plane of symmetry identical to the plane defined by the benzene ring.  This $\sigma_h$ plane then could be described as in the following example by the yellow disk:

(credit to symmetry@otterbein, example symmetry $C_{6h}$, hexakis(Me2N)benzene).

Your example however is about a methyl-substituted benzene. While the carbon atoms of the methyl groups are in the plane of the symmetry, you no longer can apply the same operation (reflection on this mirror plane) on the hydrogens of these methyl groups to yield corresponding, symmetry equivalent hydrogens on the other side of the mirror.  This is only possible if both the methyl groups share a special conformation: for each exactly one hydrogen must reside in this $\sigma_h$ mirror plane; then, the other two hydrogens of the methyl group in question are symmetry equivalent.

In all other conformations, the three hydrogens of one methyl group were no longer symmetry equivalent in respect to this $\sigma_h$ plane.  And it would require only one methyl group to be «off» to erase the $\sigma_h$ plane for the whole molecule.

I would suggest to venture out the interactive tutorials around symmetry@otterbein.  All you need is an internet browser with javascript enabled.  Later, if interested in applying the concepts of symmetry into crystallography (including the transition of Schoenflies notation used in IR-spectroscopy by the  Hermann-Maguin about point groups), you may find its sibling crystals@otterbein a valuable complement.


Like you said Gibbs free energy is the enthalpy minus TdS so basically it's the energy available (free energy) after you take the thermal energy out of the system. The TdS term is the energy due to random thermal fluctuations so sometimes it's more useful to remove that from the enthalpy and deal with the free energy especially when trying to determine spontaneity of reaction.


Stepped surfaces generally refer to surfaces (of a particular index) which have 'steps' like a flight of stairs. A cursory search on google images can yield some useful examples, such as this one:

As you can see from the example, a stepped surface will be a surface which includes steps which go from one layer to the next.


Is there a way of finding solubility/solvent from basic information?

This is a broad question as solubility equally depends e.g., on the temperature of the solvent(s), presence of other (already dissolved) compounds. A database like Reaxys used by research institutes may indicate the publications with these (measured) values, or contain these values itself. Based on known data, programs may be trained to anticipate solubility of unknowns based on a structure drawn. E.g., for X in water background of a relevant program, Marvin, in function of the pH value, or an implementation of ESOL.


This is more of a maths question, but I can give a quick answer.

As you said,

$$\frac{\mathrm{d}}{\mathrm{d}Q}\ln Q = \frac{1}{Q}$$

You could write this as

$$\frac{\mathrm{d}(\ln Q)}{\mathrm{d}Q} = \frac{1}{Q}$$

So, by the chain rule,

$$\begin{align}
\frac{\partial (\ln Q)}{\partial V} &= \frac{\partial Q}{\partial V}\cdot\frac{\mathrm{d}(\ln Q)}{\mathrm{d}Q} \\
&= \frac{1}{Q}\frac{\partial Q}{\partial V}
\end{align}$$

If you are studying physical chemistry, it is a good idea to remember this trick. It appears everywhere.


why do electron pairs not take axial positions in the VSEPR structure of ClF3?

Let's take a look at the electronic configuration of $\ce{ClF3}$.

For $\ce{ClF3}$ actually three isomers are possible.

It is highly unfavorable for the lone pairs to be 90° apart which rules one of the three. In order to rationalize the obervation that the isomer with both lone pairs in equatorial positions is the observed form, it is necessary to count the number of 90° bond pair - lone pair interactions. The observed isomer  has four such angles and the isomer with the lone pairs in axial positions has six which is (presumably) less favorable. The molecule might be called T-shaped based on the atomic positions only.


The structure you've drawn on the left does have a resonance structure, but not the one you've drawn on the right.

Drawing resonance structures out by hand gets quicker over time.  You find that you don't need to actually put in a "C" for carbon, or label the hydrogen atoms.  Huckel's rule generally applies to planar, cyclic molecules with continuous conjugation, although it has also be applied to transition states where an S or P orbital on a shifting C or H moiety completes a loop of continuous conjugation (for example, the 1,5 hydrogen shift in cyclopentadiene).


What does ethylene glycol decompose to when exposed to air for an extended period of time?

This image highlights the oxidation process of ethylene glycol fairly well

( Taken from Ethylene glycol: properties, synthesis, and applications )

Since it has little tendency to evaporate, unless it is kept in direct sunlight, it would probably oxidize to a mixture of the above liquid compounds (glyoxal, glyoxalic acid) and not directly to water. If it is kept in sunlight, the reaction would be photochemically catalysed and a it may degrade to carbon dioxide faster, but according to theoretical principles you would not get much water.
Ethylene glycol is relatively hydrophilic and also has a low tendency to separate from water, so the only major degradation process available to it in soil would be biodegradation by either aerobic or anaerobic process, which is NOT the same as atmospheric oxidation1
Not sure about reaction energetics regarding the same

Most of the danger regarding glycol spills (and other chemicals in the environment) comes from absorption by other organisms; since glycol is comparatively less volatile and more hydrophilic than, say ethanol, it has a greater tendency to be absorbed by organisms.

1 Taken from William R. Roy, in Handbook of Antistatics (Second Edition), 2016


Computation of pH when an acid and base are mixed in solution

You are headed in the right direction.  Pretend that the $\ce{H+}$ ions and the $\ce{OH-}$ ions neutralize each other 1 for 1.  That will leave you with only one of those ions left.

The water ionization won't affect your problem as that ionization is repressed by the excess ion present in the solution.

Now you should be able to solve the problem.


I think the first bit of gas produced probably dissolved in the water, so you didn't see it for a little while.

You can skip your three initial "points" with zero volume, and you're not going to have $\ln(0)$, which is what's hanging you up. Note that you did this in your report, and why!

Do a linear fit with the remaining points. Your y-intercept will be $\ln \mathrm{[A]}_0$ and your slope will be $-k$. It's not a terribly good fit.


Alphabetization rule in case of consecutively placed substituents

Rule P-45.2.2 in the 2013 IUPAC Recommendations reads:

  The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes (other than 'hydro/dehydro') to the parent structure.

The term "lowest set of locants" is defined in P-14.3.5; see Loong's answer here. As such the correct set of locants is 1,2,3, not 1,2,5 as was proposed in another answer since "1,2,3" is lower than "1,2,5". (The other answer has since been deleted.) As far as I can tell you have correctly identified this. However, the ambiguity between these two possible numbering schemes is not resolved by P-45.2.2:

In order to decide one first needs to generate the corresponding names for both numbering schemes. Regardless of locant numbering, the three substituents are listed in alphanumerical order (rule P-14.5). Scheme A would give the name "2-bromo-3-chloro-1-iodocyclopentane", whereas Scheme B would give the name "2-bromo-1-chloro-3-iodocyclopentane", since "bromo", "chloro", and "iodo" come in that order alphabetically.

Turning to the next rule P-45.2.3:

  The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes to the parent structure (other than 'hydro/dehydro' prefixes) in their order of citation in the name.

Since Scheme A gives a name with the locants cited in the order 2,3,1, whereas Scheme B has the locants cited in the order 2,1,3, Scheme B is used to generate the preferred IUPAC name 2-bromo-1-chloro-3-iodocyclopentane.

As for what is missing in your train of thought: From what I understand of your question, you seem to be thinking that the prefixes must be ordered in increasing locant number i.e. the name generated is "1-X-2-Y-3-Zcyclopentane", and that the locant set must then be chosen such that X, Y, and Z come in alphabetical order. As you already noticed with this example, that is not quite a foolproof rule. The correct way to do it is the other way round: first, order the prefixes by alphabetical order, and then choose the locant set in order to minimise the locants.


The use of aluminum in the construction of your ashtray as you've described will not result in the ingestion/inhalation of any form of aluminum in any significant way.

All aluminum that has been exposed to air is covered in a chemically and physically stable oxide. This aluminum oxide has an even lower vapor pressure than aluminum metal, such that the tip of the cigarette will actually be more than $\pu{2000^oC}$ below it's boiling point. By physically stable, I mean that aluminum oxide adheres strongly to the bulk aluminum and will not flake off as in the case of many other metal oxides.

Although your question was about potential safety issues, I will make one comment regarding your discussion of the design. You suggested that heat transfer into the aluminum might play a role in extinguishing the cigarette, which is not the case. So long as there is sufficient oxygen available for the combustion, the cigarette will continue to burn.


Mass of salt required to make a 1.2 g/ml solution

To answer in detail Jan, I didn't make stranged calculus, now I answered more than a year ago I can't tell you if my calculus are the same, but here is the logic.

For any solution (aqueous) we have,
$$m_{solution}=\rho_{solution}\times V \tag1$$

We also have,
$$\rho_{solution}=d\times \rho_{water} \tag2$$

where d is the relative density of the solution and then, merging both $(1)$ and $(2)$,
$$m_{solution}=d \times \rho_{water}\times V \tag3$$

The mass of solute in the solution is a percentage of the masse of the whole solution then, because,

$$n_{solute}=\frac{m_{solute}}{M_{solute}} \tag4$$

We have,

$$n_{solute}=\frac{p\times d \times \rho_{water}\times V}{100\times M_{solute}}\tag5$$

Then find my previous result with $C\times V=n_{solute}$.

Now I think everyone is able to goes from molar to mass concentration then I can stop here. If not ask me.


What type of chemical reaction is sulfuric acid with calcium carbonate?

While double replacement is correct, you could also classify it as a neutralization reaction (what I think would be the most common answer if one had to pick a single description).  I agree with the comment explaining that you can't consider it a decomposition reaction as written.  You would need to explicitly show that as part of the reaction for that to be an acceptable description.


Is there a substance that is hard (ideally abrasive) in solid form but would dissolve in water?

A usual cleaning product that does not pose any health hazard and that is easily washed away is sodium hydrogenocarbonate (aka baking soda). I use it myself, and it works a charm in the kitchen!


"Silverpoint" nibs have a long history in art and significantly predate modern drawing tools such as the lead pencil (made of graphite and clay but possibly called lead as lead metal was one possible material in metal pens). Silverpoint writing and drawing doesn't smudge or fade and many examples have survived almost unchanged over hundreds of years.

There is a useful site with a good history of the original artistic uses of silverpoint pens here. According to this site a wide variety of metals will work. Historically which was used depended on status:

  Wealthy and royal persons who needed to keep records (business inventories and so forth) employed these scribes, and the stylus material served as notice to all of the status of the employer - gold and silver for the upper classes, lead for somewhat lesser beings.

It seems tht many metals will work. The dictionary of Art is quoted as saying:

  Metalpoint. Drawing instrument (the forerunner of the pencil) made from a small, pointed metal tip, usually of lead, silver, copper or gold, encased in a wooden holder.

The current preferred metal for artistic use seems to be fairly pure silver.

  Best results are obtained from a thicker point that is "pure" silver, or ".999 fine." This is much softer than sterling, so it will deposit more metal on the drawing surface

The silverpoint.com web site sells ready made points of both silver and gold.

It isn't obvious what metal or alloy is used in the pens mentioned in the Wired story, but it isn't silver. It apparently contains at least some lead as the pencils come with warnings not to let children chew them!


What is the structure of trifluoromethanesulfonic acid in the gas phase?

Has the structure of trifluoromethanesulfonic acid or a structurally strongly related acid been studied in the gas phase? (Please provide a reference)

Yes. See Gas Phase Structure of Methyl Trifluoromethanesulfonate, CH3OSO2CF3, and Conformational Properties of Covalent Sulfonates Inorg. Chem., 1999, 38 (13), pp 3051–3055 and references cited therein, especially Electron Diffraction Investigation of the Molecular Structure of Trifluoromethanesulfonic acid (triflic acid) Z. Naturforsch. Teil A 1981, 36, 917

  Does the experiment suggest equivalent or non-equivalent oxygens?

Non-equivalent.

The 1981 paper finds

S=O 141.7 ( 1) pm
S-O 155.7 ( 2) pm

  How well can that idea be extended? e.g. should one consider the structure of formic acid to be C2v symmetric rather than Cs?

From the following gas phase study, I don't think that gas phase formic acid monomer is C2v.
http://actachemscand.dk/pdf/acta_vol_23_p2848-2864.pdf

  Do (solid-state) crystal structures or other experiments exist that confirm a different structure for solid TfOH?

Quoting from Trifluoromethyl Chlorosulfonate, CF3OSO2Cl: Gas Phase and Crystal
Structure, Conformation and Vibrational Analysis Studied by
Experimental and Theoretical Methods Inorg. Chem. 2003, 42, 7297-7303

"The structure of triflic acid, HOSO2CF3 (trifluoromethansulfuric acid), was also studied by GED, and an anti orientation
was assumed in the analysis of the diffraction pattern. [reference 6: the 1981 reference above] In
contrast, an X-ray investigation of 2HOSO2CF3 [dot] H2O [ie. the hemihydrate] resulted
in gauche orientation of the O-H bond,[reference 7] in accordance with
theoretical calculations. [references 8 and 9]"


IUPAC nomenclature: putting the number before or in the middle of the name

The name “2-pentanone” and “pentan-2-one” describe the same compound. The name “2-pentanone” is in accordance with obsolete IUPAC Recommendations. The relevant rule of the IUPAC Nomenclature was changed in the 1993 IUPAC Recommendations:

R-0.1.2 Position of locants
Locants (numerals and/or letters) are placed immediately before the part of the name to which they relate, except in the case of traditional contracted forms (see R-2.5).

The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.3.2 Position of locants
Locants (numerals and/or letters) are placed immediately before that part of the name to which they relate, except in the case of the traditional contracted names when locants are placed at the front of their names.

Examples:

hex-2-ene (not 2-hexene)
cyclohex-2-en-1-ol (not 2-cyclohexen-1-ol)
pentan-2-one (not 2-pentanone)
(propan-2-yl)benzene

Examples for traditional contracted names with locants at the front of the name:

2-naphthyl (not naphth-2-yl), the preferred name is naphthalen-2-yl
2-pyridyl (not pyrid-2-yl), the preferred name is pyridin-2-yl


A redox reaction is any reaction in which oxidation and reduction are occurring. The word redox is a portmanteau of REDuction-OXidation. So, if we can correctly identify reduction and oxidation, we can identify redox reactants. In practice, this can be a little challenging, so I will focus first on the second part of your questions (definitions) and then go back to the reaction.

  Ionization is any process that produces ions.

The conversion of a hydrogen atom to a hydrogen ion by loss of an electron is an ionization.
$$\ce{H -> H+ + e-}$$
So to is the reaction of hydrochloric acid and water, which produces hydronium ions and chloride ions.
$$\ce{HCl + H2O -> Cl- + H3O+}$$

  Oxidation is a loss of electrons.

The ionization of hydrogen atoms to hydrogen ions is an oxidation.
$$\ce{H -> H+ + e-}$$

So to is the conversion of carbon monoxide to carbon dioxide:

$$\ce{2CO + O2 -> 2CO2}$$

Note that this type of reaction is what gave us the word oxidation - compounds are reacting with oxygen and atoms end up with more oxygen atoms bound to them.

  Reduction is a gain of electrons.

The ionization of chlorine atoms to chloride ions is a reduction.

$$\ce{Cl + e- -> Cl-}$$

So too is the conversions of acetylene to ethane:

$$\ce{C2H2 + 2H2 -> C2H6}$$

Not all oxidations and reductions are ionizations. There is a simple mnemonic in English for remembering oxidation and reduction: OIL RIG (Oxidation Is Loss Reduction Is Gain).

  Oxidation and reduction happen together.

In principle we can write an equation that shows only oxidation:

$$\ce{H -> H+ + e-}$$

but oxidation cannot happen on its own. That electron that was lost needs to be picked up by something else. This reaction that shows only oxidation is called a "half-reaction" and needs to be coupled with a reduction half-reaction to show a valid chemical reaction:

$$\ce{H -> H+ + e-}$$
$$\ce{Cl + e- -> Cl-}$$
$$\ce{H + Cl ->HCl}$$

In the other examples I gave, both processes are occurring. In the oxidation of carbon monoxide, the oxygen molecules are being reduced.

The surest way to identify oxidation and reduction (and thus redox) is to compare the oxidation states of the reactants and the products. Determining oxidation states/numbers is more complex that I want to put here, so start with the Wikipedia article.

For the carbon monoxide example:

                  reactants    products
C                 +2           +4
O (bonded to C)   -2           -2
O2                0            -2

The carbon atoms are being oxidized (oxidation number increased) and the oxygen atoms in $\ce{O2}$ are being reduced (oxidation number decreases). The oxygen atom already attached to carbon did not change its oxidation number.


Egg-Detection in Pastries: An Analysis in Heat-Alterated Molecule Identification

Egg white consists of mostly Ovalbumin. It is a protein, and so will be hard to distinguish from other proteins in a baked product. Egg white changes structure to a solid polymer on cooking. You could dissolve it in a mixture of detergent and hair perm, returning the Ovalbumin to its uncooked state, and then run it through a column to identify it.

There might also be biological assays you could run using enzymes/antibodies which would only interact with Ovalbumin.

I don't think there are any chemical means to identify a particular cooked protein easily.

Ask on Biology exchange?


First off, as suggested in the comment above, the contribution of 3d orbitals in hypervalency is probably minimal.

It's clear from sampling the inorganic literature that "beyond the octet," the size of the atom matters, as well as the shape of the ligands, size of the surrounding coordination sphere, etc. To some degree, it's a matter of fitting X atoms around the central atom.

If you have a 5th or 6th row element (2nd or 3rd row transition metal), lanthanide, actinide, etc., higher coordination is relatively common. For example, for lanthanide ions, 8-coordinate is the most common coordination number. Still, 8-coordinate iron complexes have been found, e.g. "Combined Mössbauer Spectral and Density Functional Study of an Eight-Coordinate Iron(II) Complex" Inorg. Chem., 2015, 54 (17), pp 8415–8422 or "Square-Antiprismatic Eight-Coordinate Complexes of Divalent First-Row Transition Metal Cations: A Density Functional Theory Exploration of the Electronic–Structural Landscape" Inorg. Chem., 2015, 54 (4), pp 1375–1383.

I'm not aware of a general overview of hypervalency and "what coordination numbers can you get", but there's a nice overview on modern descriptions using multi-center bonding.


It depends on the geometry/construction of the bridge. In a typical glass-tube bridge, some small amount of the ions from the cell may eventually make their way all the way across the salt bridge, but over reasonable amounts of time, it's mostly the ions from the salt bridge that move into the half-cells (and the ions from the half-cells moving into the ends of the bridge). For a small filter paper strip being used as a bridge, the time it takes should be shorter, as there is much less solution in the salt bridge and the distance is likely shorter. Sometimes, glass salt bridges are filled with gels to limit diffusion further (at the expense of increasing the resistance of the bridge).

The goal of using a salt bridge and separating the half-cells is to prevent the solutions of each from directly mixing. If significant quantities of the ions from the half-cells are crossing the bridge, then they can react directly, bypassing the electrode connections. e.g. if we have a simple galvanic cell of $\ce{Zn∣Zn^2+∥Cu^2+∣Cu}$ and $\ce{Cu^2+}$ makes it across the separator to the zinc size, it can be directly reduced onto the zinc electrode, instead of by the electrons flowing through the electrode connections. Having a single beaker with both $\ce{Cu^2+}$ and $\ce{Zn^2+}$ would thus make a poor galvanic cell as most of the reagents would be wasted reacting directly, rather than doing work in the outside circuit.

With a simple porous separator, there are no other ions than the ones in the half-cells, so naturally, these are the ions that flow across, with the goal simply to prevent bulk mixing of the two solutions.


How much ascorbate to neutralize one liter of chlorinated tap water?

This is just a very short explanatory note on the article suggested by Greg E..

The reaction between ascorbic acid and hypochlorous acid is as follows:
$$ \ce{C5H5O5CH2OH + HOCl → C5H3O5CH2OH + HCl + H2O}$$
so for 1 mole of hypochlorous acid you need 1 mole of ascorbic acid. This can be recalculated into a mass ratio by the use of their molecular masses:

$$M(\ce{C5H5O5CH2OH})=176\;\text{g/mole}$$
$$M(\ce{HOCl})=52.5\;\text{g/mole}$$
From this we can see that if the molar ratio is $1:1$, the mass ratio is $176:52.5$ which is $3.3:1$ somewhat higher than the article claims.

I am not exactly sure of the reason, it could be that there is some side reaction of ascorbic acid that also removes some additional hypochlorous acid although I wouldn't know which. Perhaps the article is talking about the amount of ascorbic acid you need to get a concentration which you don't taste anymore?