You're dealing with a mixture of similar hydrated salts instead of a single salt (as in the example of $\ce{Ca(NO3)2}$). The notation $\ce{K2O-nSiO2-xH2O}$ means that there are different crystals with varying proportions of silicon dioxide and water (hydration). i.e. $n$ could be $1$, or $2$, $3$, etc. and $x$ likewise, for any given molecular "speck". A hydrated salt can be de-hydrated with heat (thus changing relative weights of elements), so be careful where you keep the fertilizer, because it will change the math.

This means, to measure a reasonable concentration, you need to consider the reported bulk average quantity of each component. Based on the percent by weight you reported, you can approximate the material as being (by mass -- that is wt%)

$28\%~\ce{K2O}$
$57.5\%~\ce{SiO2}$
(Thus) $14.5\%~\text{water}$

Say you want to know the $\pu{ppm}$ of $\ce{K+}$ for a solution of $\pu{1 g}$ of this fertilizer in a gallon ($\pu{3.78 L}$) of $\ce{H2O}$. Using the good approximation $\pu{1 mg/L} = \pu{1 ppm}$, We have

$$\begin{multline}
\pu{1 g}
\times \frac{\pu{0.28 g}~\ce{K2O}}{\pu{1 g fertilizer}}
\times \frac{\pu{1 mol}~\ce{K2O}}{\pu{94.2 g}~\ce{K2O}}\\
\times \frac{\pu{2 mol}~\ce{K+}}{\pu{1 mol}~\ce{K2O}}
\times \frac{\pu{39.0983 g}~\ce{K+}}{\pu{1 mol}~\ce{K+}}
\times \frac{\pu{1000 mg}}{\pu{1 g}}
\times \frac{\pu{1 gal}}{\pu{3.78 L}}\\
 = \pu{61.5 ppm}~\ce{K+}
\end{multline}$$

Here I guessed a concentration that may be relevant for you, but you can see the way to move forward. I would note, though, that since purity is reported pretty low ($77\%$), you should multiply the above result by $0.77$. Hard to say what the impurities are though, if they are mostly other potassium salts, the concentration could be higher. At any rate it's safe to say the above example is $\approx \pu{45-65 ppm}~\ce{K+}$.

You can also think of the compound like this, based on component masses:

$$\text{Molar mass} = 0.28 \times M(\ce{K2O}) + 0.575 \times M(\ce{SiO2}) + 0.145 \times M(\ce{H2O}) = \pu{63.532 g/mol},$$ and the average molar formula is $\ce{K2O - 3.2SiO2 - 2.7H2O}$.

Finally, I would recall that dissolution of $\ce{K2O}$ in general produces a basic solution, so if cucumbers are fond of acid (I don't know), you may consider adding a little acid ($\ce{HNO3}$, perhaps) to the solution to avoid perturbing the greens. But, you're the farmer!


I think you are mixing things up in your question and it would be helpful to specify what the question in your textbook was.
NaCl is a solid made up from Na$^+$ and Cl$^-$ ions, not from uncharged atoms as stated in your question,  arranged in a periodic crystal lattice. So the redox equations for oxidation and reduction in your question are not relevant for the formation of NaCl from its ions e.g. from a supersaturated solution, where the formation of the solid releases lattice energy. Although the redox equations are relevant, when looking at the formation of NaCl$_{(s)}$ from its elements Na and Cl, which you rightly pointed out yield the favorable noble gas configuration for both. (See Born-Harber cycle for more information on the formation of salts from its elements and the related energies)
As stated in Nilay Ghosh's comment, a question (and answer) comparing the stability of Na vs Na$^+$ can be found here.
Also: Sodium ions do in fact react with other (uncharged) atoms, it's just a question of a strong enough reducing agent under the right environment. Commercially, sodium was produced in the Deville process at 1100 °C with carbon as the reducing agent:
$\ce{Na2CO3 + 2 C -> 2 Na + 3 CO}$


Will sodium sulfite effect the oxidization of toluene to benzoic acid with potassium permanganate?

How do you know your permanganate contains sodium sulphite?

It seems highly unlikely because $\ce{KMnO4}$ is a very strong oxidiser and sulphite is very easily oxidised to sulphate:

$$\ce{SO3^{2-}\to SO4^{2-} +2e-}$$

Also, it seems improbable to find a sodium salt in a potassium salt.

Even if the $\ce{KMnO4}$ contained some "surviving" $\ce{Na2SO3}$, during the oxidation of the toluene, it would quickly be oxidised to sulphate. At worst your benzoic acid yield would suffer slightly.


Can the change in the entropy of the surroundings always be obtained by dividing heat transferred by the temperature at which the transfer occurs?

I disagree with the explanation given in the text.  The entropy change of the surroundings is given by the equation presented in your title only for an ideal constant temperature reservoir.  Such an ideal reservoir is characterized by (a) an infinite capacity to absorb heat without its temperature changing and (b) infinite thermal conductivity, so that temperature is uniform throughout the reservoir, including at the interface with the system (where all the heat transfer occurs). So all the heat transfer between the system and the reservoir occurs at the reservoir temperature $T_{Res}$. In this way, all the irreversibility of the process occurs within the system, irrespective of whether the process is reversible, and the entropy change of the system always satisfies $ΔS≥Q/T_{Res}$.

None of the entropy generation for the process occurs within the ideal reservoir, and the entropy change of the ideal reservoir is thus always $-Q_{sys}/T_{Res}$, irrespective of whether the process is reversible. I should also mention that, although often not stated, the Clausius inequality is supposed to be expressed in terms of the temperature at the interface with the surroundings $T_{Res}$ where all the heat transfer is occurring.

Typically, all the surroundings reservoirs you will encounter in most thermodynamics texts will be ideal constant temperature reservoirs.

The description in your post is inappropriate because, if the reservoir has a constant volume (i.e., constant mass), its temperature can change between the beginning and end of the process.  Its entropy change will not be given then by the equation presented.  In addition, the description omits the requirement that the reservoir most have infinite thermal conductivity (so that the temperature in the reservoir is spatially uniform).


Is soap the natural form of Sodium Lauryl Sulfate?

Conventional soaps are the sodium salts of long chain carboxylic acids (aka fatty acids). These are formed by the hydrolysis of triglycerides, a process known as saponification. This process breaks the triglyceride at its ester linkages and yields three fatty acid molecules and glycerol molecule.

As the aforementioned wikipedia article indicates, sodium lauryl sulfate is formed by refunctionalising the fatty acid lauric acid. In this case, the $\ce{R-OOH}$ (carboxylic acid) head group has been replaced with an $\ce{R-OH}$ (alcohol) group, which is in turn replaced by $\ce{R-OSO3^{-}}$ (sulfate). Here I am using $\ce{R}$ to refer to the long alkyl chain of the molecule.

I'm not sure whether sodium lauryl sufate occurs naturally, however the other molecules you mention are different in the sense that they have carboxylate ($\ce{R-COO^{-}}$) head groups whereas sodium lauryl sulfate has the sulfate head group ($\ce{R-OSO3^{-}}$). Both classes of molecule will act as surfactants as they have polar head groups which are hydrophilic, and nonpolar alkyl tails which have an affinity for fats and oils.

Lauric acid derivatives spotter's guide

(These were drawn using the nifty ChemFig package for (La)TeX.)

The last two (sodium laurate and sodium lauryl sulfate) are sodium salts. Both carboxylic acids ($\ce{R-COOH}$) and hydrogen alkyl sulfates ($\ce{R-OSO3H}$) are fairly willing to ionise to give the corresponding carboxylate ($\ce{R-COO^{-}}$) or sulfate ($\ce{R-OSO3^{-}}$). Sodium ions happily associate with these anions, balancing the negative charge.


Is reduction potential equivalent to the ease of discharge?

Easy is a big word.

Lower potential means it requires less energy, but that doesn't say anything about how quick the reaction will proceed.

A low potential reaction may still have slow kinetics and therefore not be "easy".

If you are comparing ions as to which will produce more product as a result of an electrochemical reaction, you need to find something with low potential and fast kinetics.


Which of the two structures is more stable?

The explanation can[1] be given by analysing their respective resonance[2] structures which is as follows:

(Credits: ChemWriter)

Now, inductive effects would work.

In case of first (c's resonance), positive charge would be stabilised by $\ce{+I}$ effect of methyl group; whereas in the second (d's resonance) one, $\ce{+I}$ effect would become dead upto the positive charge.

Hence, less stable would be (d).

Notes:

[1]: Firstly, I'm not sure if hyperconjugation will stabilize carbocation in resonance structures or not, because it will separate the charges even more. Secondly, in the given structures, it is very hard to differentiate the stability, as contribution of inductive and hyperconjugation are resulting in comparable stabilities. Hence, I tried it explaining by other canonical form.

[2]: resonance will happen due to conjugation of $\ce{\pi}$ bonds in both of the molecules


Reaction of styrene oxide with hydrogen iodide

Your answer is the correct one. Acid-catalysed hydrolysis of epoxides proceeds exactly as shown. In the specific case of styrene oxide, nucleophiles attach themselves to the benzylic carbon1,2, which means that A should be the major product. I'm attaching an image from (1) which gives the products of reaction of methanol with styrene oxide:

If you want to, you can try to drop a note to the authors of your book about this error.
References:

Mohseni, Sharareh, et al. “Theoretical and Experimental Studies on the Regioselectivity of Epoxide Ring Opening by Nucleophiles in Nitromethane without Any Catalyst: Nucleophilic-Chain Attack Mechanism.” Progress in Reaction Kinetics and Mechanism, vol. 39, no. 1, 2014, pp. 89–102. doi:10.3184/97809059274714X13874723178403.
Morrison, Robert Thornton, and Robert Neilson Boyd. Organic Chemistry. 5. ed, Allyn and Bacon, 1987, pp. 780-785.


The former equation assumes
$$[\ce{H2A}]\simeq [\ce{A^2-}]$$
due reaction
$$\ce{ 2 HA- <=> H2A + A^2-}$$
The is possible with 2 simplifying conditions:

The concentration of oxonium resp. hydroxide ions originated from water dissociation is much lower than concentration of the basic resp. acidic ampholyte form.
$$[\ce{H2A}] \gg \sqrt{K_\mathrm{w}}$$
$$[\ce{A^2-}] \gg \sqrt{K_\mathrm{w}}$$
That assures the effect of ampholyte reaction with water dissociation products
$$\ce{ A^2- + H3O+ -> HA- + H2O}$$
$$\ce{ H2A + OH- -> HA- + H2O}$$
is negligible for the ampholyte  form concentration ratios and you can ignore water dissociation.
This condition is critical for low ampholyte concentration near neutral $\mathrm{pH}$.

The concentration of oxonium resp. hydroxide ions is much lower than concentration of acidic resp. basic ampholyte forms.
$$[\ce{H2A}] \gg [\ce{H+}]$$
$$[\ce{A^2-}] \gg [\ce{OH-}]$$
That assures the effect of production of oxonium resp. hydroxide ions to reach the final $\mathrm{pH}$
$$\ce{ HA- + H2O -> A^2- + H3O+}$$
resp.
$$\ce{ HA- + H2O -> H2A + OH-}$$
does not compete with creation of acidic/basic ampholyte forms.
$$\ce{ 2 HA- + -> A^2- + H2A }$$
This condition is not valid for very low /very high target $\mathrm{pH}$, as respective ions are produced in expense of the deviation from the same ratio of ampholyte form concentration.


Did Mendeleev predict the existence of neutrinos?

While it's an interesting thought, I don't think it's very useful to suggest Mendeleev predicted neutrinos. It's far more sensible for it to be a curious coincidence. His suggestions of ghost-like properties similar to neutrinos (very hard to detect, tiny mass, little interaction with matter, etc) are more easily explained as an attempt to justify (properly) why it was so hard to detect: if it had a much larger mass, someone should've weighed it by then; if it interacted more, someone should've managed to capture it physically or chemically, and so on. It's not uncommon for theoretical predictions to temporarily "take refuge" just beyond the limit of observability (for example, proton decay), with varying degrees of justification. In this case, Mendeleev's prediction was wrong in its original terms, and happened to almost match something else entirely because neutrinos really are hard to detect! Also, the connection with radioactivity could very likely have been made merely due to good timing with its discovery. Since no one had a good explanation, why not forward your own!

And as a sidenote, neutrinos definitely have mass, otherwise they could not suffer the observed flavour oscillation.


Spontaneous combustion of sulfur in a steel barrel

Iron sulfide FeS is pyrophoric: flammable, and while it can start just on exposure to air, the presence of UV light will help.  As is usual, one of the key components is the surface to volume ratio.  A massive lump is unlikely to catch fire.  A crusty scale of FeS, such as forms in tanks used to store sour oil (H2S content) is more likely to combust when air is let into the tank.

Deactivation of Pyrophoric Iron Sulfides

From the descriptions in the abstract above, I suspect that I had one of the other iron sulfides.  Whatever I had did not burn fast.


How to to test the concentration of melatonin in a strong acid?

Is there any relatively simple method to determine the concentration?

That depends on your equipment. Having the following at hand would be helpful:

a UV/VIS spectrometer
mortar and pestle
UV-grade ethanol (or methanol)
pure melatonin

How to proceed:

Prepare a stock solution of the pure melatonin in ethanol.
Prepare some dilutions for the expected concentration range.
Measure the UV absorption of all solutions at $\lambda = \pu{277 nm}$.
Draw your calibration curve (absorption vs concentration).
Powderize the tablet using a mortar an a pestle.
Dissolve the powder in a defined volume of ethanol.
Measure the UV absorption of the sample.
Determine the concentration in the sample using your calibration curve.

You have just applied the Lambert-Beer law. Others have done the same and reported the method in Pharmacophore 2014, 5, 252-257 (free PDF).


Why do I see two peaks for methyl salicylate in my GC–MS spectrum?

The short answer: Do not rely on the SIM trace alone.  Amplify the representation of the trace about the TIC (total ion current) until you see a reasonable peak  intensity at the same retention time as in your SIM traces (perhaps a time spectrum, but not a mass spectrum).  Then integrate with the peak-shaped cursor (currently shown in the magenta coloured trace of one SIM) in the trace of the TIC and equally consult the then displayed EI scan at this time interval (in part already seen at the lower part of your photo) to assess the mass spectrum.  Compare this to a trustworthy reference spectrum; e.g. an authentic sample; your spectrometer's library; NIST, SDBS, etc.
The longer answer:
Your experiment coupled a separation technique, either GC, with an analytical, MS.  Often, the mass resolution of these simpler, less-boxy bench-top MS hyphenated to the chromatography (i.e., not a tandem MS-MS detector) is low compared to, e.g., a sector field mass spectrometer.  Thus, observing the SIM traces alone is more like asking «tell me if whatever ion passes the detector with a $m/z = 119.2$», or even worse «tell me if any ion passes the detector in the range of $m/z = 119 - 121$.  You will see this approach often used (sometimes misused) in HPLC-MS if one is not sure about the setup's sensitivity for $\ce{M+}$ vs. an adduct like $\ce{[M + H]+}$.  Or because the software allows you to check for an $\ce{NH+_4}$-adduct after the chromatography run is recorded because your eluent included an ammonium buffer.
Only if you couple the information by the SIM with the integration in the TIC to yield a complete mass spectrum you may collect additional evidence that the peak in the SIM may belong to your target compound.  Otherwise, it could be (but need not to be) caused by an ion which accidentally shares the same ratio of $m/z$ within the setup's resolution in time (think elution of chromatography, scan rate of the MS), and mass (the MS) to your fragment in question; but shows up regardless if it belongs to the defragmentation pattern of your target compound, or not.  Without additional inspection of the spectra it equally may be an impurity, e.g. the sample just before yours was too concentrated and some is eluated (and subsequently characterized by the MS) now altogether with your sample.
Without the separate inspection of these integrated spectra around $t_R = \pu{4.1 min}$ and $\pu{5.1 min}$, the present form of your question does not allow to answer if one of these peaks (and if so, which one) safely may be attributed to methyl salicylate.  A mixture of isomers of compounds, e.g. (cis)/(trans) dienes may elute at different times and -- depending of the $m/z$-ratio you set for the SIM -- equally yield two peaks in the SIM but distinctly different mass spectra.


Actually in pH the p is for potenz or power of hydrogen. The concentration of proton $(\rm H^+\equiv H_3O^+)$ in water is often $10^{-3},10^{-5},10^{-14}$ which is very tedious to write, so pH is defined as $3,5,14$ for these solutions. Or:
$$\rm [H^+]=10^{-pH}\iff \log_{10}[H^+]=-pH\iff pH=-\log_{10}[H^+]$$

Note that pH scale and proton concentration works in reverse direction, one increases so the other decreases.

The pH Scale at normal room temperature is from 0 to 14 (debatable) and 7 is neutral (no excess proton or no deficiency of proton) and 0-7 is acidic (excess proton) wheras 7-14 is basic (deficiency of proton). This is defined because purest form of water at this temperature has pH 7 and pH more than this (and hence concentration of proton less than this) is caused by bases which take up proton and and pH less than this (and hence concentration of proton more than this) is caused by acids which release proton.


Production of hydrogen through the reaction of aluminium metal and sodium hydroxide

Oxide layer would not hold against the strong alkali, and sodium aluminate does not form protective layer either. I suppose cans and foil may be covered by some coating, to make them suitable for food grade applications. Try burning it off, or rather use other sorts of aluminium (cut-up wire, maybe). Normally it would react quite vigorously.


Cyanide is a pretty good ligand for coordination compounds. The electron pair on carbon (which, incidentally, also carries the Lewis structure’s formal charge) is located in the HOMO — much like as in $\ce{CO}$, whose molecular orbitals can be found in this answer by Martin (replace oxygen with nitrogen to arrive at $\ce{CN-}$) — making it a good Lewis base and a good σ donor.

It is remarkable for being a strong-field ligand, i.e. one that often creates low-spin complexes even for 3d-metals (which generally prefer high-spin if there are no competing reasons for low-spin). While one could draw analogies to $\ce{CO}$ or even $\ce{NO+}$ whose π-antibonding orbitals are strongly Lewis acidic and participate in metal to ligand backbonding, it turns out that for cyanide the main reason is a high covalency of the metal–ligand-σ-bond due to the relevant orbitals being very similar in energy. Thus, cyanide is unique in that it both stabilises higher oxidation states by being an anion and low-spin complexes due to its binding modalities.

One might think, in analogy to $\ce{K4[Fe(CN)6]}$, that cyanide preferentially binds to haemoglobin’s iron(II) centre and thereby inhibits oxygen transport, but that is in fact not the case as this centre requires a less-oxidised iron(II). Oxygen (due to the resulting redox-mechanism) and carbon monoxide are much better ligands for haemoglobin.

What cyanide does strongly inhibit is cytochrome c oxidase, the last enzyme in the respiratory cycle which collects electrons from four cytochrome c’s and transfers them to an oxygen molecule to create two water molecules:

$$\ce{4 Fe^2+-cytochrome~$c$ + 4 H+ + O2 -> 4 Fe^3+-cytochrome~$c$ + 2 H2O}$$

This is done by a plethora of metal centres consisting of two haems, to cytochromes and two copper centres. I can’t tell you exactly which one, but there will be at least one which needs accessable iron(III) at some point in the catalytic cycle to which $\ce{CN-}$ can bind well and strongly inhibiting the catalysis of the enzyme.

Since the inhibited process is directly responsible for cellular respiration by transferring electrons to molecular oxygen (and being the only enzyme dealing with molecular oxygen in the respiratory chain), inhibition of this protein will cause the breakdown of the entire respiratory chain and lead to cellular suffocation.

Considering that an amount of $200~\mathrm{mg}$ can be toxic according to Wikipedia, and also considering that a human being contains some 10 trillion cells also according to Wikipedia, that’s still some 400 million cyanide ions per cell demonstrating the efficiency of binding and inhibition.

Note that popular antidotes also directly address the fact that cyanide is a strong ligand:

One antidote, hydroxycobalamin, presents the cyanide ion with a $\ce{Co^{III}}$ centre and a weakly bound, easily displaced hydroxido ligand. Hydroxide is displaced by cyanide creating cyanidocobalamin, a variant of vitamin B12 that the body has no problems dealing with. It is ejected via the kidneys. Injected into the circulatory system, it scavenges cyanides before they can enter cells.
Another antidote, also injected into the circulatory system, is 4-dimethylaminophenol (not to be abbreviated as DMAP to prevent confusion with 4-dimethylaminopyridine). It is an oxidising agent designed to oxidise haemoglobin’s iron(II) to methaemoglobin (containing iron(III)) thereby creating a much better acceptor for cyanide. Since there is a lot of iron(II) (a quick estimate with $8~\mathrm{mmol/l}$ gives $40~\mathrm{mmol}$ haemoglobin in human blood) in the blood for oxygen transport, small amounts of this can be sacrificed to capture cyanide before that reaches the respiratory cycle.


How effective is a level A hazmat suit for immersion in deadly substances?

Note that I'm used to standard lab safety measures including gas masks and masks mith independent air supply but I never used a Hazmat level A suit in research. You might want to take the following with a grain of salt!

    Sodium cyanide

($\ce{NaCN}$) is a solid with a melting point around 560 °C. While it is a solid, jumping into a pool of it shouldn't do any harm. Level A suits are like "your personal bubble" with independant air support. They are however not that heat resistant: Do not jump into a pool of molten sodium cyanide!

    High concentration (100%) Hydrogen Peroxide

3% solutions of hydrogen peroxide ($\ce{H2O2}$) in water are used for wound disinfection, solutions up to 18% are used to bleach hair. (In German, the term wasserstoffblond was used for bleach blonde).

Much higher concentrations or waterfree (100%) ($\ce{H2O2}$) is another league.

Hydrogen peroxide can decompose into water and oxygen:

$$\ce{2 H2O2 ->[\textrm{cat}] H2O + O2}$$

The reaction releases some heat, may happen spontaneously and is catalyzed by some metals. Add some burnable material that can be vapourized and you have a rocket propellant. Think Messserschmitt Me 163.

If the hazmat suit is tight, and the material withstands "bleaching" by the hydrogen peroxide there's still the question whether hooks, valves, visor frames or any other metal parts, that might be part of the whole gear will catalyze the decomposition.

    80% hydrochloric acid

That does not exist in water. Water won't take up that much hydrogen chloride gas under normal conditions. Concentrated hydrochlorid acid (~ 38%) should be fine.

    80% nitric acid

That's close to fuming nitric acid and again a pretty strong oxidant. I'd say that you're safe, but don't take my word for it ;)


Which allotrope of phosphorus is kinetically the most stable?

I think the key to this question lies in the understanding of the difference between thermodynamic and kinetic stability. In this post, I find Thomij's answer the most rigorous and enlightening.

From this you should learn, that the thermodynamically most stable allotrope corresponds to the global minimum on the potential energy surface, while a kinetically stable allotrope, i.e. hindered through an activation barrier, corresponds to local minimum. From this it can easily be seen, that the thermodynamically most stable allotrope is also the kinetically most stable allotrope. This is due to the fact, that the barriers for interconversion of the different allotropes always have to be higher for the most stable one.

In the following two dimensional case you see that the thermodynamically most stable, i.e. the global minimum, is surrounded by two local minima. It must always be true, that $E_\mathrm{a}(1\to2)<E_\mathrm{a}(2\to1)$ if $1$ is a local and $2$ is a global minimum. The same applies to the other case. If this was not true, then $2$ cannot be the global minimum and hence not the thermodynamically most stable modification.

Having established that, it is therefore true, that black phosphorus is the thermodynamically and kinetically most stable form, see Hollemann-Wiberg, p. 680.

However, looking at a different reaction, the case might not be so simple any more. This is even true for different temperatures, as the following graphic will show (quoted from the linked book).


I think you've misunderstood, Pauli's exclusion principle doesn't mean that the four numbers are different it means that no electron can have the same respective values for $n$, $l$, $m_s$, $m_l$ as another electron (in some case some of the numbers themselves can be the same). i.e. electrons in the same suborbital must have different spins.

$n$ indicates the orbital shell
$l$ indicates the type of suborbital, s, p etc
$m_l$ indicates the orientation of the suborbital, px, py etc
$m_s$ indicates the electron spin

Pauli's exclusion principle means that no electron around an atom can occupy identical orbitals, i.e two electron with up spin in a 2s1, as all their electron numbers would be the same as their respective equivalents ($n_1 =n_2 = 2$, $l_1 = l_2 = 0$, $m_{l,1} = m_{l,2} = 0$,  $m_{s,1} = m_{s,2} = -1/2$).


This is a nice question, as it confronts a very replicable and common experience with a well established yet seemingly contradictory fact. As you expected, the smell of metal has nothing to do with the metal actually getting into your nose, as most metals have far too low of a vapor pressure at ordinary temperatures to allow direct detection. The characteristic smell of metal, in fact, is caused by organic substances!

There has been the focus on the specific case of the smell of iron (free-access article!). There are at least two ways in which iron produces a metallic smell. Firstly, acidic substances are capable of corroding iron and steel, releasing phosphorus and carbon atoms present in the metal or alloy. These can react to form volatile organophosphorus compounds such as methylphosphine ($\ce{H3CPH2}$ which have a garlic/metallic odor at small concentrations. From the article:

  The “garlic” metallic odor (see Supporting Information) of the gas product from the acidic dissolution of cast iron is dominated by these organophosphines. We measured an extremely low odor threshold for two key odorants, methylphosphine and dimethylphosphine (6 and 3 ng P/m³, respectively, garlic-metallic odor), which belong therefore to the most potent odorants known. Phosphine ($\ce{PH3}$) is not important for this odor because we found it has a much higher odor detection threshold (>10⁶ ng/m³). A “calcium carbide” (or “burned lime”/“cement”) attribute of the general “garlic” odor is probably caused by unsaturated hydrocarbons (alkynes, alkadienes) that are linked to a high carbon content of iron (Table 1, see Supporting Information).

Also, it turns out that $\ce{Fe^{2+}}$ ions (but not $\ce{Fe^{3+}}$) are capable of oxidizing substances present in oils produced by the skin, namely lipid peroxides. A small amount of $\ce{Fe^{2+}}$ ions are produced when iron comes into contact with acids in sweat. These then decompose the oils releasing a mixture of ketones and aldehydes with carbon chains between 6 and 10 atoms long. In particular, most of the smell of metal comes from the unsaturated ketone 1-octen-3-one, which has a fungal/metallic odour even in concentrations as low as $1\ \mu g\ m^{-3}$ . In short:

  Sweaty skin corrodes iron metal to form reactive $\ce{Fe^{2+}}$ ions that are oxidized within seconds to $\ce{Fe^{3+}}$ ions while simultaneously reducing and decomposing existing skin lipid peroxides to odorous carbonyl hydrocarbons that are perceived as a metallic odor.

In the supporting information for the article (also free-access), the authors describe experiments performed with other metals, including copper:

  Comparison of iron metal with other metals (copper, brass, zinc, etc.): When solid copper metal or brass (copper-zinc alloy) was contacted with the skin instead of iron, a similar metallic odor and GC-peak pattern of carbonyl hydrocarbons was produced and up to one μmole/dm² of monovalent cuprous ion [$\ce{Cu+}$] was detected as a corrosion product (Supporting Figs. S3 to S6). Zinc, a metal that forms $\ce{Zn^{2+}}$ but no stable $\ce{Zn+}$, was hesitant to form metallic odor, except on very strong rubbing of the metal versus skin (that could produce metastable monovalent $\ce{Zn+}$). The use of common color-tests to demonstrate directly on human palm skin the presence of low-valence ions (ferrous and cuprous) from the corrosion of iron, copper and brass alloys is shown in Supporting Figure S6. Alumina powder rubbed on skin did not produce significant odorants. These results provide additional evidence that it is not metal evaporation, but skin lipid peroxide reduction and decomposition by low valence metal ions that produces the odorants.

The last paragraphs of the article summarize the findings:

  In conclusion: 1) The typical “musty” metallic odor of iron metal touching skin (epidermis) is caused by volatile carbonyl compounds (aldehydes, ketones) produced through
  the reaction of skin peroxides with ferrous ions ($\ce{Fe^{2+}}$) that are formed in the sweat-mediated corrosion of iron. $\ce{Fe^{2+}}$ ion containing metal surfaces, rust, drinking water, blood etc., but also copper and brass, give rise to a similar odor on contact with the skin. The human ability to detect this odor is probably a result of the evolutionarily developed but largely dormant ability to smell blood (“blood scent”).

  2) The “garlic-carbide” metallic odor of phosphorus- and carbon-rich cast iron and steel under attack by acid, is dominated by volatile organophosphines. Corroding cast iron
  is an environmental source of C–P compounds that may lead to confusion in the verification and monitoring of the Chemical Weapons Convention (see also ref. [15])

As an aside, this may be why sometimes people recommend getting strong smells off your hands by rubbing them against a metal object. While it probably doesn't work for some metals and for some smelly compounds, it's possible that the metal catalyzes the decomposition of the malodorous substances into less strongly smelling ones.

You can read a little more in this press article on the study.


Are the tars in cannabis considerably more dangerous than those in tobacco?

This is not a complete answer (merely a step in a particular direction, namely that of the falsehood of the claim), but I contacted the author of the book (Introduction to Medicinal Chemistry) Graham Patrick and he wrote the following:

  Dear Jori,

  Many thanks for your e-mail and observations. Unfortunately, that
  statement was made in a very early edition of the textbook published
  over ten years ago and so I doubt if I will be able to track down the
  original reference on which it was based. As you say there is
  controversy and the statement is too sweeping. Work on the 6th edition
  is starting this year and so it will be possible to modify that
  particular passage. Many thanks for your input.

  Best wishes Graham


Will UVC light/ozone affects color on fabrics?

I'm planning to put a low power UVC light bulb into my wardrobe to kill bacteria and prevent bad smell on clothes but I'm wondering will it affect color on the clothes?

If the UV light is strong enough to break bonds to kill bacteria it is certainly strong enough to bleach the organic dyes and degrade the clothing.

  I've done some research but haven't got into a conclusion if UVC light will affect color faster than sunlight or not (because there's no UVC light reaches the earth surface).

This depends on the brightness of the bulb and the sun where you live

  ...but I've read that ozone will cause cracking on some plastic do fabrics also get affected by that?

Yes fabrics are also degraded by ozone.

Keep in mind that though the UV and ozone may cause degradation of your clothes, they may or may not do so faster than mold or normal use would too many factors to say with certainty which is best.


I have worked on a computational project where I had to study the sites of deprotonation in this molecule:

When deciding between mechanisms I had to consider both thermodynamic and kinetic. For example, I had this case: deprotonations of two different hydrogens(bond with N Blu atoms in the picture)were similarly thermodynamically favored (similar free Gibbs energy of reaction) but one H was towards the inside of the ring and the other pointed towards the outside, therefore, because kinetically favored, the mechanism proposed was the one with the external H removed.

Your questions:

  What is the exact distinction between these two terms?

A Thermodynamically favored base means that the Gibbs energy of the rection where an H+ binds to the protonation site(eg.lone pair of the N) is very negative.
However, if this protonation site is difficult to access, then this reaction is very slow or even impossible and a less thermodynamically favored protonation reaction may occur instead(if the molecule has more than one protonation sites or other bases are present).

  Is thermodynamic basicity the one mentioned in common basicity defintions? (arrhenius, lewis, bronsted-lowry)

yes, and the kinetic effect is evident only in limited numbers of cases.

  Why are they measured differently?

To make it easy and avoid competitive reaction considerations, suppose you have only one type of protonation site. If you want to know how easy is for the proton to access that site, you have to determine the speed of the reaction(dynamic experiment). If you want to know only about thermodynamic(energies involved) then you let the reaction go to equilibrium and just measure the pKa. If the deprotonation site is very inaccessible then this thermodynamic experiment may take very long because you have to wait it reaches the equilibrium. In this case, if it takes more than your experiment time scales than you will measure an apparent pKa that is lower than expected because your reaction has not reached the equilibrium yet. This apparent lower pKa is due to the kinetic effect.

  Are they denoted differently, do they have different units and is there some general mathematical formula for each that would help me get deeper understanding and distinction between these two terms?

I don't know if there are other ways to express this other than kinetic and thermodynamic acid/base.
Thermodynamic: you wait for the reaction to reach the equilibrium and using the $\Delta G$ of reaction determine the pKa.
$ΔG = -RT\ln K_\mathrm{a}$
 $pKa = -log(K_\mathrm{a})$
 In case of a difficult to access protonation site you may not wait for enough and measure a lower pKa but this is an apparent pKa because the definition of it is purely thermodynamic.

  Also, how are these two mesaures of basicity related to gas phase proton affinity and intrinsic basicity and pKb?

Not expert of these techniques, I worked only theoretically but as usual, the solvent affects the value of the pka. The most stable conformation in the gas phase may not be the most stable in solution anymore and there's the solvation energy contribution to consider when you determine the Gibbs energy of the protonation reaction. Therefore, the measurements in solution may not be comparable with the gas phase.

Finally, I can give you some example of very evident kinetic effect:
 In the molecule below the lone pair of the nitrogen is difficult to access for an H3O+ molecule and the pKa measured(not the thermodynamic pKa) will be different depending on how much you wait during the titration experiment. I don't know the timescales for this molecule.

or you can have a base trapped in a fullerene molecule:

This is the most extreme case with an infinite equilibrium time. The pKa of the molecule trapped is still the same but you will experimentally measure the pKa of your solvent and conclude it is not basic.


Why copper wires submerged in salt (or fresh) water oxidize much faster while they conduct electrical current?

Electricity in a wire submerged in water effectively turns it into an electrochemical cell, and then it is no surprise that the anode gets oxidized pretty quickly. A modest potential of a few volts would suffice to oxidize any metal, even gold. This works for AC as well (a wire oxidizes during the positive half-wave and then does nothing during the other half-wave).


Mixing solvents, or how does paint thinner work?

I had dissolved some polycaprolactone in that universal thinner above, and it seemed to dissolve quicker than in either toluene or xylene alone.

Some of these universal thinners do contain a lot more than just toluene and the xylenes.

If you want to find out about the composition of these mixtures, a material safety data sheet (MSDS, in German: Sicherheitsdatenblatt), provided by the manufacturer gives some information:

As an example, take this MSDS by a swiss manufacturer, which lists the following ingredients:

\begin{array}{lr}
\textrm{compound} & \textrm{amount [%]}\\
\hline
\textrm{toluene} & 25-50\\
\textrm{xylene} & 10-25\\
\textrm{acetone} & 10-25\\
\textrm{propan-2-ol} & 2.5-10\\
\textrm{4-methyl-pentan-2-one} & 2,5-10\\
\textrm{2-methoxy-1-methylethylacetate} & 2,5-10\\
\end{array}

Particularly the latter two might be much better solvents for the polycaprolactone than toluene or xylene.


Protein Data Bank: Asymmetric Anit/Biological Assembly

Most of the structures in the Protein Databank are determined with X-ray crystallography. Like inorganic compounds and small organic molecules, under the right conditions proteins can form crystals - ordered, repeating patterns of protein molecules, filling a 3D space. That is, you can take a basic building block and rotate and translate this building block to completely fill the three-dimensional space. (2D example).

So what's being modeled in an X-ray crystallography experiment is that basic building block, the one which undergoes symmetry operations (rotation and translation) in order to fill out 3D space. This basic building block is called the asymmetric unit, because it is without the crystallographic symmetries which result in filling space.

But even in the asymmetric unit there can exist noncrystallographic symmetry. These are symmetries (rotations and translations) which relate portions of the asymmetric unit, but aren't lined up quite right to fill space. Because they don't contribute to the overall symmetry of the crystal, normally they're not exactly symmetrical. Usually there are slight differences, based on how the noncrystallographic symmetry fits into the crystallographic one.

For a 2D example, think of a triangle which is tiled on a square grid. The square grid is the crystallographic symmetry. The triangle has noncrystallographic symmetry, because the three-fold rotational axis of the triangle does not contribute to tiling the plane. This means the corners of the triangle are not exactly equivalent, as each corner is in a different position with respect to the neighbors in the next square.

So when an X-ray cryptographer goes to solve the structure of a protein crystal, the primary unit of interest is the asymmetric unit - even if there's noncrystallographic symmetry in the asymmetric unit. So that's what the PDB records as its primary entry - the contents of the asymmetric unit, including the slight differences between any repeated units.

In your case (1iep), that's what happened. Because of the specific way the protein crystallized, there just happens to be two copies of the protein in the asymmetric unit, almost - but not quite - identical to each other. This representation of the asymmetric unit - including both proteins - is what you've downloaded.

Because this might not be what you're interested in, the PDB also provides one or more "Biological Assembly" files. These are processed files which attempt to model what the experimentalists believe* is the actual biologically relevant unit. In your case, the relevant biological unit is a monomer, so they provide two biological assembly files - one for each (distinct) monomer in the asymmetric unit. For multimeric proteins, it sometimes happens that one of the symmetry axes for the multimer lines up with the crystallographic axis. In this case, you'd only get part of the protein in the asymmetric unit download, but the biological assembly file would contain a reconstructed representation of the multimer.

If you want further, extensive detail, I would second the recommendation made here to look at the textbook Crystallography Made Crystal Clear, by Gale Rhodes.

(*The biological unit provided by the PDB is a guess. Sometimes you'll see proteins which should be multimeric represented as monomers, or monomeric proteins represented as multimers. It depends in part on how accurately the depositors annotated what the relevant biological unit was.)


Butane has pretty high  boiling point (around $\ce{0 °C}$). This means, that just cooling air to $\ce{-30 °C}$ should condense a lot (if not all) of butane from it.

However, as I guess, most of butane used in foam production remains in the foam, otherwise the foam would collapse.  So, it is very possible that there is not much to recycle in air itself. The gas, of course, will be released eventually, but only when foam itself is destroyed.


Calculating Redox Potential; Coupled Reactions

$2~Ag + 2~H^+ ~(+2~Cl^{-}) \rightarrow 2~Ag^{+} + H_2 ~ ( + 2~Cl^{-})$

As two half-reactions, it is seen that the silver is oxidized:

$2~Ag \rightarrow 2~Ag^{+} +2~e^- ~~ E^{\circ}= 0.80\,V$

And the hydrogen is reduced:

$2~H^+ + 2~e^- \rightarrow H_2 ~~ E^{\circ}= 0.0\,V$

To calculate the reduction potential you have to solve the following equation:

$\Delta E = E_{acceptor} - E_{donor} = E_{2H^{+}/H_2}- E_{Ag/Ag^{+}}$

$\Delta E =  0\,V - (0.80\,V) = -0.80\,V$

The reactions doesn't run voluntarily if $\Delta E < 0$


Differential equation for the rate of unimolecular reaction

To start with, the equation you wrote for the rate of change of $[A]$ is not correct. It should instead be

$$-\frac{d[A]}{dt}=k_{1}[A][M]-k_{2}[A^*][M].$$

Note that $[A]$ is consumed in the first reaction step and produced in the second. Now, rearrange equation $(1)$ to get

$$k_{1}[A][M]=[A^*](k_{2}[M]+k_{3}).$$

Plug in the above equation into the actual expression for $-d[A]/dt$ to get

$$-\frac{d[A]}{dt}=[A^*](k_{2}[M]+k_{3})-k_{2}[A^*][M].$$

This, of course, can be simplified to yield the desired equation $(2)$:

$$-\frac{d[A]}{dt}=k_{3}[A^*].$$


Can sugars dissolve in liquid ammonia?

Yes, according to Ref.1, liquid ammonia is used to extract sugars in sugar-beet chips:

5.88 kilograms of sugar-beet chips having a moisture content of 5.4 percent and a saccharose content of 68 percent (calculated on the dried substance) are treated in an autoclave of 50-litre capacity with 18 kilograms of liquid ammonia. [...] The extracted pulp, after driving off the ammonia, contains only 1.2 percent of saccharose, which constitutes 0.4 percent of the total sugar.

Accordingly, 99.6% of sugar in sugar-beet chips are isolated in one extraction using liquid ammonia as the solvent. However, it was noted that the saccharose is not present as such in the solution, but is in the form of a compound of saccharose with ammonia. These ammonia-saccharose compounds can be obtained in a well crystallized form, which are composed of $\pu{1 mol}$ of saccharose and $\pu{2 mol}$ of ammonia and even those of other compositions Ref.1). I assume this is analogous to hydration by water molecules in aqueous solutions.
The high solubility of sugar in liquid ammonia seemingly due to H-bonding capabilities similar to water. This is evident by sucrose solubility in nonaqueous solvents such as morpholine (with a $\ce{N-H}$ bond) and N-methylmorpholine (without a $\ce{N-H}$ bond), which are $39.8$ and $\pu{0.37 g}$ in $\pu{100 g}$ of solution, respectively at $\pu{85 ^\circ C}$ (Ref.2). This difference in solubility is analogous to sucrose solubility in water and methanol, which are $210$ and $\pu{1 g}$ in $\pu{100 mL}$ of solution, respectively at $\pu{25 ^\circ C}$ (Wiki, and Ref.3). Further, sucrose is insoluble in diethyl ether.
References:

Georg Hingst, Werner Emte, "Process for recovering sugar from natural products containing it," US Patent 1958, 2,829,985 (Patent Application, 1955, US2829985A) (PDF).
Oleg K. Kononenko, Karl M. Herstein, "Nonaqueous solvents for sucrose," Ind. Eng. Chem. Chem. Eng. Data Series 1956, 1(1), 87–92 (DOI: https://doi.org/10.1021/i460001a017) (PDF).
Z. Bubník, P. Kadlec, "Chapter 5: Sucrose solubility," In Sucrose: Properties and Applications; M. Mathlouthi, P. Reiser, Eds.;  Springer: Boston, MA, 1995, pp. 101-125 (ISBN: 978-1-4613-6150-3).


Why don't covalent compounds dissociate in water?

Many 'covalent' molecules do dissociate in water, HCl (as pointed out in the comments), phenol, acetic acid, for example, whereas some 'ionic' compounds do not to any appreciable extent, e.g. silver chloride, lead sulfate. The general reason is that the 'system', that is molecule plus solvent, will move into its lowest energy state. This means that if breaking the weakest bond and solvating the ions produced, is of lowest energy then this will occur. In the case of HCL, the polar water molecules can solvate making $\ce{H3O^+_{aq} and Cl^-_{aq}}$ where both ions are surrounded by a few water molecules. Water has a very high dielectric constant (relative permittivity) which means that electric fields are 'damped', i.e they are attenuated to short range so that oppositely charged ions can easily separate and hence exist independently in solution. With carboxylic acids the reasoning is similar, the weakest (O-H) bond dissociates, and the ions are solvated. However, dissociation may not be 100% indicating that the energy gain on dissociating is not that great. In insoluble salts, the lattice binding energy and also bond energy is too great compared to the energy gain expected upon solvation and they remain as molecules in a solid. Organic molecules with strong bonds e.g. cyclohexane, benzene, with only C-C and C-H, hardy dissociate in water. The bonds are too strong compared to energy gain on solvation after dissociation. (If dissociation produced radicals, there will be no effective solvation in a polar solvent so would remain at high energy and hence unlikely to occur)


Beer's law with a polychromatic light source

There are several ways to answer the question, so here are my three. The first answer involves using a simple integer wavelengths approximation, i.e., noticing that Figs. 1 and 2 have 1 nm point spacing, and then using a spreadsheet to perform a summation. The answer explained in this part also serves to define the terms used in the second answer, which involves analytic calculation of the relevant overlap integrals. Then the third answer, not discussed in detail, uses a Mueller calculus-based method. Before proceeding to the explanations, the numerical answer to the question is $A_{maximum} = 0.46951302113867$ at $C_{max} = 0.01 M$ and Fig. 4 shows the absorbance as a function of concentration:

The absorbance is 93.9% of the ideal value (i.e., 0.5) at $C_{max}$, despite the source being significantly polychromatic, and the Beer’s law plot is convex, but perhaps less than might have been expected intuitively.

All three of the answers and explanations that follow are based on the model shown in Fig. 4 below:

As per the question’s assumed conditions, the light intensity incident at the monochromator is unity, i.e., 1, with fictive units of Watts. Note that units may be ignored since they will divide out when transmittance is computed later. The 50:50 beam splitter sends half of the light it receives to the sample (upper) beam path and the other half of the light it receives goes to the reference (lower) beam path. Let $I_{incident}(\lambda )$ denote the light intensity incident upon the absorber and $I_{sample}(\lambda )$ denote the attenuated light intensity that gets through the absorber. Then $I_{sample}(\lambda )$ is the incident light intensity on the upper photodetector and transimpedance amplifier (PD-TIA).

From Fig. 4, it is clear that $I_{reference}(\lambda )$ = $I_{incident}(\lambda )$, assuming the mirror is ideal. This is simply the unattenuated incident light intensity on the lower PD-TIA. The photodiodes were stated to have flat responsivities, i.e., spectrally constant responsivities, so their photocurrent responses are independent of wavelength. Hence, the responsivities, in this idealized scenario, cancel out when transmittance is computed later. Likewise for the constant transimpedances, which are assumed equal. Hence, it suffices to simply ‘add up’ the light intensities, for each color of light, incident at each PD-TIA. Now to the answers and explanations.

Answer 1: As per Figs. 1 and 2, assume a wavelength spacing of 1 nm. There is no light outside the region 405 to 445 nm and wavelengths are effectively “discretized”, i.e., are treated as discrete integer values. Therefore, the number of discrete wavelengths, where light intensity is non-zero, is just 39, i.e., from 406 to 444 nm, in 1 nm increments. At 405 nm, $I_{incident}(\lambda )$ is 0 and it increases linearly to 0.5 (at 425 nm). It then decreases linearly to 0 (at 445 nm). The absorbance, $A_{absorber}(\lambda)$, increases linearly from 0.1 (at 405 nm) to 0.9 (at 445 nm). Hence, the transmittance through the absorber, $T_{absorber}(\lambda)$, decreases non-linearly from about 0.7943 (at 405 nm) to about 0.1259 (at 445 nm). As always, $T = 10^{-A}$, so there is no difficulty computing $T_{absorber}(\lambda )$ at any of the relevant wavelengths.

Accordingly, $I_{sample}(\lambda )$ = $I_{incident}(\lambda )$ x $T_{absorber}(\lambda )$ and $I_{reference}(\lambda )$ = $I_{incident}(\lambda )$ x $T_{reference}(\lambda )$ = $I_{incident}(\lambda )$, since $T_{reference}(\lambda )$ = 1. All of the necessary calculations are shown in the spreadsheet screenshot in Fig. 5 below:

In Fig. 5, column A contains the wavelengths and column B is a unitless “x” variable that is relative to 405 nm, i.e., at 405 nm, x = 0 and so forth. Column C contains the incident light intensities, column D has the associated absorbances and column E has the corresponding transmittances through the absorber. Column F has the unity transmittances for the reference beam path, column G has the $I_{sample}(\lambda )$ values and column H has the $I_{reference}(\lambda )$ values. Adding up the 41 values in columns G and H yields total intensities: 3.39164 and 10.000, respectively. Then T is simply 0.339164 and $A_{maximum} = 0.469590$. This is close to the true value, as computed in the second answer, despite the rather coarse “discretization” of the wavelengths.

Answer 2: From the first answer, it is clear how the relevant overlap integrals should be calculated. First, x is the independent (unitless) wavelength variable and it is continuous from 0 to 40. As x increases from 0 to 20, $I_{incident}(x)$ increases from 0 to 0.5, so $I_{incident}(x)$ = x/40. As x increases from 20 to 40, $I_{incident}(x)$ decreases from 0.5 to 0, so $I_{incident}(x)$ = 1- (x/40). For the absorbances, $A_{absorber}(x)$ = 0.1 + 0.02x for x = 0 to 40. Hence, $T_{absorber}(x)$ = $10^{0.1 + 0.02x}$ and $I_{sample}(x)$ = $I_{incident}(x)$ x $10^{0.1 + 0.02x}$.

For the sample beam path, the total intensity, $I_{sample}^{total}$, is the following overlap integral: $$I_{sample}^{total} = \int\limits_0^{40} {I_{sample}(x)\ \mathrm dx} $$

But this is simply the sum of two definite integrals: $$\int\limits_0^{20} {(x/40)\ 10^{0.1 + 0.02x}\ \mathrm dx} + \int\limits_{20}^{40} {[1-(x/40)] 10^{0.1 + 0.02x}\ \mathrm dx}$$

Using the fact that $10^{-x} = e^{-xln(10)}$, and defining $a \equiv -0.02 ln(10)$, then results in
 $$ \frac{e^{5a}}{40} \int\limits_{0}^{20} {xe^{ax}  \mathrm dx} + e^{5a} \int\limits_{20}^{40} {e^{ax}  \mathrm dx} - \frac{e^{5a}}{40} \int\limits_{20}^{40} {xe^{ax}  \mathrm dx}$$

The integrals are easily evaluated, resulting in $$ \frac{e^{5a}}{40a^2} {[(20a - 1) e^{20a} + 1]} + \frac{ e^{5a}}{a}{[e^{40a} – e^{20a}]} - \frac{e^{5a}}{40a^2} {[(40a - 1) e^{40a}] - [(20a - 1) e^{20a}]} $$

With $a \equiv -0.02 ln(10)$, $I_{sample}^{total}$ = 3.39224318670537.

For the reference beam path, the integrals are easier because $T_{reference}(x)$ = 1. Hence $$ I_{reference}^{total} = \int\limits_0^{20} {(x/40) \ \mathrm dx} + \int\limits_{20}^{40} {[1-(x/40)] \ \mathrm dx} = 5 + 5 = 10$$ Alternatively, this is simply the area of the triangle: 0.5 x 40 x 0.5 = 10. Therefore, $T =  I_{sample}^{total}\ / \ I_{reference}^{total} = 3.39224318670537\ /\ 10 = 0.339224318670537$. Then $A_{maximum} = 0.46951302113867$.

The analytical method worked, as expected, but the system was so heavily idealized that it was essentially a “spherical cow” and they don’t milk. Any realistic complications, e.g., even simply using a tungsten bulb light source, make the analytic method harder and it does not take much to make the equations all but analytically intractable. Then numerical methods may be employed, but the difficulty then mostly lies in getting the right model equations: correctly evaluating equations is not necessarily the same as evaluating correct equations.

What about convolution? Convolution, per se, is unnecessary in this heavily idealized problem, for the reasons discussed below. First, note that the spectral profile of $I_{incident}(\lambda )$ is the same as the monochromator’s transmittance profile (shown in Fig. 1), except for having a maximum value of 0.5 rather than unity. The spectral profile is symmetric. Second, the absorber’s transmittance profile is as shown in Fig. 2 and it is also symmetric. Third, the convolution of two real functions involves reversing one of the functions and then computing the overlap integrals as a function of a shift parameter, i.e., a relative displacement of the two functions. Since the two functions to be convolved are both symmetric, nothing needs to be reversed, so the convolution reduces to computing the overlap integrals as a function of a shift parameter, $\tau $. Zero displacement, i.e., $\tau = 0$, can be conveniently defined as when the two functions are perfectly aligned.

The next step is to compute the convolution function either analytically, if possible or desired, or numerically, if necessary or convenient. Then the convolution function would be evaluated at $\tau = 25$ (nm), since this corresponds to the 25 nm shift between the two spectral profiles and $\tau $ is integer valued, i.e., data values are available every nm, as shown in Fig. 5. With finer wavelength spacing, the shift would scale up accordingly. If the convolution is computed analytically, then $\tau $ is a real number.

For the data in Fig. 5, the numerically computed convolution is shown in Fig. 6:

From Fig. 6, the value of the convolution at 425 nm is 3.39164, as in Fig. 5. If the two spectral profiles do not overlap, then the value of the convolution is 10, as expected. But note that all that was needed was the single overlap integral at the known $\tau = 25$ (nm), and that value is 3.39164, as shown above. For the reference beam path, the overlap integral was 10, as shown in the trivially simple calculation above.

This problem is so heavily idealized that there is no necessary reason to compute the convolution function. In a more realistic situation, the computation would be harder and even numerical evaluation of the convolution function might be fruitless.

Answer 3: Figure 4 appears to just be a schematic block diagram, but it is actually a screenshot of a Mueller optical calculus-based computer simulation program. Although discussion of the fundamental principles of the several optical calculuses is beyond the scope of this answer, both Mueller calculus and Jones calculus were invented in the early 1940’s and there is a considerable body of information about them in the literature, e.g., [1-4].

In the simulation depicted in Fig. 4, stacks of Stokes vectors (one per discrete color) are passed from one optical component to the next optical component(s), e.g., from the lamp block to the monochromator block. Each component has a resident stack of matrices, with components as appropriate. When an optical component receives an incoming stack of optical vectors, each vector is multiplied by the corresponding matrix and the resulting stack of processed optical vectors is sent out to the next optical component (or components) “downstream”. All the stacks of optical vectors are received, processed and then sent on in a single computer simulation step.

The plot shown in Fig. 3 was obtained by using 1,001 concentrations: 0 to 0.01 M, in increments of 10 µM. The wavelength range was from 405 to 445 nm, in 1 pm increments. Thus, with every one of the 1,001 simulation steps, stacks of 40,001 Stokes vectors (used in Mueller calculus) were passed, starting with the lamp block and ending with the sample and reference PD-TIA blocks. Both of those blocks, for every step, computed 40,001 intensities, multiplied them by responsivities, summed the resulting photocurrents and then multiplied by the transimpedances.

In the present case, the responsivities were simply constant default values (0.1 A/W) and the transimpedances were $1 k\Omega$. For every step of the simulation, T was computed as the quotient of the real number outputs of the two PD-TIA blocks and $A_{maximum} = -log(T) = 0.46951302122$. Compared with the exact answer, this differs in the ninth decimal place. As well, there was no need to solve any equations or use a spreadsheet: no wheel had to be re-invented.

Another of the advantages of using optical calculus-based spectrometric simulations is that they can easily accommodate non-idealities, such as realistic lamp spectral profiles and photodetector responsivities, imperfect beam splitters and mirrors, polarized light, optical activity, circular dichroism, stray light, light source modulation and so on. In fact, any linear optical effect can be modeled with an optical calculus. In addition, noise effects are easily included, e.g., detector/preamplifier noises and light source noises, including low frequency noises.

Summary: Answer 2 provided an exact solution to the posed question, but it was a bit tedious to obtain and would not be useful if the system was altered in any significant way, e.g., if the photodiode responsivities were those of a real red-sensitive photodiode. Answer 1 was adequately accurate, despite its simplicity, but not particularly amenable to generalizations. Answer 3 gave an accurate answer and made it simple to get the plot shown in Fig. 3. There is no significant difficulty in generalizing to more realistic scenarios.

Lastly, it is clear that using the decisively polychromatic source in Fig. 1 caused a relatively small reduction in $A_{maximum}$ and the Beer’s law plot is convex, but not badly so. The deleterious effect on spectral resolution was not investigated here, but could easily be done, perhaps as another question.

References:

W.A. Shurcliff, Polarized Light, Harvard U. Press, Cambridge, MA ©1962 & 1966.
D.S. Kliger, J.W. Lewis, C.E. Randall, Polarized Light in Optics and Spectroscopy, Academic Press, Boston ©1990.
Mueller calculus article in wikipedia and references therein.
Jones calculus article in wikipedia and references therein.


Do absorbances of different compounds at one wavelength sum up?

[…] I can see no way to use this equation, since i'm only given a wavelength and a whole bunch of coefficients.

We have all the data that we need.

You have already mentioned the Lambert-Beer law.

For the solution of a single compound S in a cuvette of width d, the absorbance $E_\lambda$ at a particular wavelength $\lambda$ is proportional to the concentration $c$ of S.

\[E_\lambda = \log_{10}{\left(\frac{I_0}{I} \right)} = \epsilon_\lambda\cdot c \cdot d \]

A higher concentration of S will result in higher absorbance.

Now imagine that the light-absorbing molecules S are linked together by units that do not absorb the light at the wavelength observed.

Let's compare two equimolar solutions of
$$\ce{-linker-S-linker-S-linker}$$
and
$$\ce{-linker-S-linker-S-linker-S-linker-S-linker-S-linker}$$

The second will show a higher absorbance, since the concentration of S is higher!

Now let's have a look at equimolar solutions of different compounds S1 and S2 and their absorbance.

We have outlined above that the absorbance $E_\lambda$ is proportional to the concentration of the light-absorbing species.

But apparently, different compound to not all absorb the same at a particular wavelength!

Exactly this is expressed with the molar absorption coefficient $\epsilon_\lambda$. The higher $\epsilon_\lambda$, the stronger the attenuation of incoming light. Note that $\epsilon_\lambda$ is

always given for a particular wavelength
a compound-specific constant

Let's tie the pieces together!

When light-absorbing units are linked together by 'blind' linkers, the effective concentration of the absorbing units counts.
Different compounds can absorb at the same wavelength, but due to their specific $\epsilon_\lambda$, they will do this to different amounts.

Consequently, we can sum up the combined absorbances in the following way:

$$E_\lambda =  \left(\epsilon_A \cdot c_A + \epsilon_C \cdot c_C + \epsilon_G \cdot c_G + \epsilon_T \cdot c_T \right)\cdot d$$


Is the titration of sodium carbonate with hydrochloric acid temperature dependent?

I think you might have a different understanding of what constitutes a 'change in the observation/reading' from the textbook, resulting in your confusion.

I am assuming that the textbook is only considering the titration graph per se, pH vs volume of acid added.

You are referring to the time to pH equilibrium during each drop of acid added. For the purposes of all titrations I have encountered, this was never a factor that was discussed. As @Ivan Neretin commented, the reaction is assumed to occur instantly.

As the concentration of the species remains the same, the titration graph will look the same. In other words: The temperature has no influence on the equivalence point.


How do scientists check which kind of unit cell is present in a crystal?

X-ray diffraction. Usually computer will tell you automatically nowadays but in the old days you had to check whether the values you got for the miller planes were possible with the symmetry of different unit cells. They discarded the ones that didnt match until the would find the one unit cell which matched the miller planes from XRD


I believe you can compare many of the models with a search engine.
http://www1.asminternational.org/asmenterprise/apd/AdvancedSearchAPD.aspx


How to calculate the enthalpy change of water?

Yes, but your answer should be just in joules not $\mathrm{J/g}$, and your units are off.

$$(2.6~\mathrm{\color{\red}{g}})\left(4.184~\mathrm{\frac{J}{\color{\red}{g}\ K}}\right)(5~\mathrm{K})= 50~\mathrm{J}$$


Temperature involves kinetic energy; however, bonds and intermolecular forces (IMF's) involve potential energy.  The potential energy of the system increases when water vaporizes, but kinetic energy does not change.

In a liquid, for each IMF broken, another is made, resulting in a net change of zero.

A phase change requires a time interval because it takes time to add energy to the system.


The solutions manual is correct, they are enantiomers.  The compound on the left is (R)-2-chloro-(S)-3-butanol and the compound on the right is (S)-2-chloro-(R)-3-butanol.  Try to use the "R\S" nomenclature system on the 2 chiral carbons, that should help figuring out whether they are identical [(R,R and R,R), (S,S and S,S), (R,S and R,S), (S,R and S,R)] enantiomers [(R,R and S,S), (R,S and S,R), (S,R and R,S)] or diastereomers [(R,R and R,S), (R,R and S,R), (S,S and S,R), (S,S and R,S)].


Why there is a huge difference between the basicity of urea and guanidine?

Urea and guanidine are structurally similar except that the carbonyl oxygen of urea has been replaced by an imine $\ce{NH}$ in guanidine. When it come to protonation, carbonyl oxygen of urea is always protonated over its 2 $\ce{NH2}$ groups (Ref.1 and 2). In guanidine, it is the imine $\ce{NH}$ protonated first. This preference in guanidine is not because a lone pair of $\mathrm{sp^2}$ nitrogen is more basic than that of $\mathrm{sp^3}$ nitrogen (actually it is other way around), but the lone pair of $\mathrm{sp^2}$ nitrogen is the most available pair for protonation over those of two $\mathrm{sp^2}$ nitrogens (lone pairs of two $\mathrm{sp^2}$ nitrogen atoms are contributing to resonance stabilization at the time). However, when protonated, guanidinium ion was highly stabilized by resonance as depicted in the diagram:

Thus, basically, it  comes down to protonation ability of oxygen versus nitrogen. Oxygen is more electronegative than that of nitrogen, thus like to retain its lone pare better than nitrogen (cf. basicity of water and ammonia). Therefore, in that sense alone, guanidine is a better base than urea.

References:

Nanping Wen, Murray H. Brooker, “Urea protonation: Raman and theoretical study,” J. Phys. Chem. 1993, 97(33), 8608-8616 (https://doi.org/10.1021/j100135a013).
B. Valentine, T. E. St. Amour, D. Fiat, “A $\ce{^{17}O}$ NMR study of the protonation of urea,” Organic Magnetic Resonance 1984, 22(11), 697-700 (https://doi.org/10.1002/mrc.1270221105).


Why does the Leidenfrost effect (seemingly) not apply to the case of molten NaCl when it is poured into water?

In the video, TheBackyardScientist offered an explanation for the huge explosive effect when 50 grams of molten NaCl is poured into water. He said that some water penetrated the blob of molten NaCl and blew it apart from the inside. A Leidenfrost effect on the outside of the blob is visible, but not the major effect.

The borax blob did not explode, so it must not have been penetrated by a significant amount of water. Borax is a tetrahydrate; when it melts and then solidifies, it goes thru a glassy stage which might be expected to be somewhat cohesive, perhaps resembling Silly Putty. In addition, it loses a lot of water on melting: an initial amount of 50 grams melts to only 26 grams. (Boric acid solidifies similarly.)

NaCl does not exhibit a glassy stage on cooling (note the crystalline surface on the patty from solidified NaCl). Perhaps 26 grams of NaCl would also not explode but merely show the expected Leidenfrost effect. As a 50-gram blob of NaCl cools, solid (strong) regions form over weak molten interior; the weak part of the blob splits open, allowing water into an interior hot enough to boil the water quickly and blast the blob skins apart. The hardened skins of the blob serve as a partial container - like a pipe bomb.

50 grams of washing soda (sodium carbonate decahydrate) will melt down to 18 grams; again, perhaps this is too little to give an explosion.

The Leidenfrost effect should be expected for small blobs of anything. The explosive effect might be expected only for larger quantities, but including even materials cooling thru a glassy stage, if the amount were large enough.


Ring expansion of two fused rings to a larger ring

I think the possible mechanism for the intermidiate conversions can be as following.Here thr $\ce{R}$ is actually $\ce{-OCHO}$.


Mechanism for formation of amide by reaction of amine with ester

Surely the negatively charged oxygen would protonate before the other
  oxygen would.

Most of the time, but not all of the time.

  The amine group is more basic than the "alcohol group" that must leave
  to form the amide so why wouldn't that protonate first and then just
  leave (i.e no reaction)?

Most of the time, but not all of the time.

Your reasoning is correct, but remember, all of these steps are equilibria.  Although what you say will happen most of the time, some of the time the reaction will take the forward route.  Presumably the reaction is run with a large excess of ammonia to "force" the reaction in the forward direction.


Is micelle formation in a soap solution a chemical reaction?

Chemical 'reactions' are normally defined by changes in 'bonding'.  Related to soap micelles, for the micelle to exist and have integrity enough to be characterized from a situation where they did not previously exist, some 'bonding' must come into being.  That bonding is normally described as 'London Forces' or 'Vander Waals Bonds' or 'Hydrogen bonding' or 'electrostatic' interactions.  Since equivalent concentrations produce equivalent results, the micelle formation is also reproducible.  In the purest sense this does imply that soap micelle formation is a chemical reaction.

The question is more philosophical than anything.


What happens when oxidizing Ni(OH)2 with a strong base?

This is where things get difficult; wouldn't the spare electron be used by the sodium ions to make
  $\ce{Na (s)}$ ?: $\ce{Na+ + e- -> Na (s)}$

This reaction is, you might be knowing, the reverse of one of "the best" known favourable reaction, as the data suggests:
$$\ce{Na->Na+ +e-}\tag{$E^\circ=2.71\;V$}\\
2.71\;V\equiv-261.475\;\text{kJ/mol}\equiv1.0938\times10^{50}(\text{rate})$$
Just imagine the reverse reaction.

  But then,
  $\ce{Na(s) + H2O -> NaOH }$

This then would never actually then happen. Intersetingly see this


Find the pH of an amphiprotic salt dissolved in water at room temperature

$$\begin{align}
\mathrm pH &=\mathrm
pK_\mathrm{a1}+\log \frac {[\ce{HA-}]} {[\ce{H2A}]} \\
\mathrm pH &=\mathrm pK_\mathrm{a2}+\log \frac {[\ce{A^2-}]} {[\ce{HA-}]} \\
\mathrm pH &=\frac{\mathrm pK_\mathrm{a1} + pK_\mathrm{a2} +\log \frac {[\ce{A^2-}]} {[\ce{H2A}]}} {2}\\
\end{align}$$

The last equation is the sum of the equation sides, divided by 2.

We can consider for expected acidic $\mathrm pH$, that we can neglect water dissociation and suppose $\ce{H+}$ is produced and consumed by these 2 reactions:
$$\begin{align}
\ce{HA- + H+ &<=> H2A} \\
\ce{HA-  &<=> H+ + A^2-} \\
\end{align}$$

Therefore:

$$\begin{align}
[\ce{H}]&={[\ce{A^2-}]}-{[\ce{H2A}]} \\
\mathrm pH &=\frac{\mathrm pK_\mathrm{a1} + \mathrm pK_\mathrm{a2} +\log \frac {[\ce{H2A}] + [\ce{H+}] } {[\ce{H2A}]}} {2}\\
\mathrm pH &=\frac{\mathrm pK_\mathrm{a1} + \mathrm pK_\mathrm{a2} +\log \left( 1 + \frac { [\ce{H+}] } {[\ce{H2A}]}\right)}  {2}\\
K_\mathrm{1a}&=\ce{[H+][HA-] /[H2A]} \\
\ce{[H2A]}&=\ce{[H+][HA-]} /K_\mathrm{1a}\\
\mathrm pH &=\frac{\mathrm pK_\mathrm{a1} + \mathrm pK_\mathrm{a2} +\log \left( 1 + \frac { K_\mathrm{1a}} {[\ce{HA-}]}\right)}  {2}\\
\end{align}$$

Therefore, if $\mathrm{p}K_\mathrm{a1}$ is high enough, i.e. if $\ce{H2A}$ is rather a weak acid,
the simplified formula $\mathrm{p}H=(\mathrm{p}K_\mathrm{a1} + \mathrm{p}K_\mathrm{a2})/2$ is right enough.

If $\mathrm{p}K_\mathrm{a1}=\ce{[HA-]}$, the $\mathrm{p}H$ correction is cca $+0.15 = \log (2)/2$

And then we can use this equation to  iterative corrections, with $\ce{[MHA]}$ being concentration of the used acidic  salt. The fraction is the classical formula to calculate ion fraction.

$$\ce{[HA-]}=\ce{[MHA]} \cdot \small
\frac {K_\mathrm{a1}.\ce{[H+]}}
{[\ce{H}]^2+K_\mathrm{a1} \cdot \ce{[H+]} + K_\mathrm{a1} \cdot K_\mathrm{a2}} $$

$$\frac {K_\mathrm{a1}}{\ce{[HA-]}}=
\frac {[\ce{H}]^2+K_\mathrm{a1} \cdot \ce{[H+]} + K_\mathrm{a1} \cdot K_\mathrm{a2}}
{\ce{[MHA]} \cdot \small
\ce{[H+]}}$$

For particular calculation, the first shoot is taking $\mathrm pH$ from the simplified  formula:

$$\begin{align}
\mathrm pH &=\frac{\mathrm pK_\mathrm{a1} + \mathrm pK_\mathrm{a2}}{2}=(1+4)/2=2.5 \\
\end{align}$$

I have realized, the direction of iterations must be reversed, as the formulas lead to divergence. ( It is known mathemathical problem to choose the right iteration with the small derivative $\mathrm{d}f(x)/\mathrm{d}x$.

The simplified equation at the bottom of the post expects  not too acidic or alkaline $\mathrm pH$. It supposes the equilibrium reactions affecting
 $[\ce{H}]$ do not significantly affect ratio of concentrations  $$\frac {[\ce{A^2-}]}{[\ce{H2A}]}$$

As the $\mathrm pH$ is near the both $\mathrm pK_\mathrm a$, it is quite a fair assumption for most cases.

$$\begin{align}
\ce{2 HA- &<=> H2A + A^2-} \\
\frac {[\ce{H2A}]} {[\ce{HA-}]} &=\frac {[\ce{A^2-}]} {[\ce{HA-}]}\\
 [\ce{H2A}] &= [\ce{A^2-}] \\
\mathrm pH &=\frac{\mathrm pK_\mathrm{a1} + \mathrm pK_\mathrm{a2}}{2}  \\
\end{align}$$


What is the difference, if there is, between this two distillation methods?

There is no fundamental difference between the two methods.
In both cases steam is formed and it is passed through the plant material.
The intention is to extract the non volatile water insoluble essential oils from the plant.
The method is called steam distillation.

On the other hand, normally it is not done in this way.
Since generally only the essential oil is of interest, the integrity of the plant does not matter, it is placed in water and the whole is heated and the water is distilled.
The students do this in the lab with cinnamon and orange peels, for instance.
It is much easier because you only have to heat one vessel.
In these two step setups, you must isolate the part containing the plant very well so that the steam does not condense, once the whole system is hot enough.
Preferably, you should also heat the part containing the plant, especially in your second example.


How to determine whether two hydrogen are homotopic, diastereotopic, emantiotopic?

You determine by labeling. Take the two hydrogens in question and label them one at a time to get two molecules. What's the relationship between these two molecules? If the molecules are the same, then the hydrogens were homotopic. If they are enantiomers, the hydrogens are enantiotopic. Otherwise, if they are diastereomeric molecules, the hydrogens are diastereomeric.


Are there any sulfides, sulfates, carbonates that cannot be dissolved by aqua regia?

Chemical Book reports that barium sulfate (barite, or baryte) is

slightly soluble in concentrated sulfuric acid and soluble in carbonate alkali metal solution in which it is converted to barium carbonate; it is insoluble in other kinds of acid or base.

It sounds like the barite is your culprit.  If you try grinding it with an eye towards physically separating the denser gold from it, the Mohs hardness of the mineral is 3-3.5, so there appears to be no issue with the physical grinding process.


Since you did not tell us how much of this final solution you would like to make, let's imagine that you want to make a generic $V_f\ \ce{mL}$ of it.

Take $\text{A}$ as an example:

$$
n_A \ \text{in final solution} =  25000\ \ce{ng/mL} \times V_f\ \ce{mL}
$$

$$
V_{A}\ \text{stock to take} = \frac{n_A}{c_{A\ stock}} = \frac{25000\ \ce{ng/mL} \times V_f\ \ce{mL}}{1\ \ce{mg/mL} \times 10^6\ \ce{ng/mg}} \frac{1000\ce{\ \mu L}}{\ce{mL}}= \frac{25000}{10^3}\ V_f\ \ce{\mu L}
$$

So let's suppose that you want to make $10\ \text{mL}$ of such a solution containing $\text{A}$ at $25,000\ \text{ng/mL}$, then you need to measure out $25000 \times 10\ /\ 1000\ \ce{\mu L} = 250\ \ce{\mu L}$ of the $\text{A}$ stock and bring it up to the final volume with urine.

You will want to carry out these calculations for all your compounds, then add all the corresponding quantities to a calibrated vessel, e.g. a volumetric flask, then bring this mixture up to the final volume with your solvent, i.e. urine.


The carbon atom in $CO_2$ has zero lone pairs of nonbonding electrons.

[By the way, the carbon has $sp$ hybridization, as it requires the bonding orbitals to have the same lobe sign on each side to make two identical $\sigma$ bonds. The $2p_z$ orbital alone is not sufficient to make two identical bonds, because the molecule would then lose the reflection symmetry it ought to have.]

Let's construct the Lewis structure ourselves:

Carbon has four valence electrons. We accept that.
Each oxygen has six valence electrons. We also accept that.

Therefore, the total number of valence electrons is $4+6+6=16$. Carbon is the larger atom, so this is a more usual structure with the largest atom in the center. The skeletal structure so far is:

$\text{O}-\text{C}-\text{O}$

We place three lone pairs of electrons on each oxygen next, to account for $2+2$ bonding and $2\times3+2\times3=12$ (currently) nonbonding electrons, for a total of $16$ as we needed.

Lastly, we use one lone pair from each oxygen to make a $\pi$ bond. This only redistributes electrons, without adding any extra or losing any. That results in:

$:\ddot{\text{O}}=\text{C}=\ddot{\text{O}}:$

And are the formal charges (FCs) minimized?

Carbon owns 4 valence electrons and came in with 4 valence electrons, so $FC = 0$.
Oxygen owns 6 valence electrons and came in with 6 valence electrons, so $FC = 0$ as well.

This must be the actual Lewis structure, and it also is in the correct image you have placed in your question.


Can molecules diffuse upstream when emptying a cup of tea?

Particles/solutes are always diffusing in all directions. The question to ask is, "are they diffusing fast enough to overcome being swept along by the flow?"

Short Answer: No. Don't try it, though.

Long Answer:

One governing equation that can be used to answer this question is the following:

$$
\overline v\!\left(\vec r\right)\cdot \overline\nabla C_i\!\left(\vec r\right) = D_i\nabla^2C_i\!\left(\vec r\right)\tag{1}
$$

This is the (simplifed) steady-state mass transport balance for species $i$ at a given point $\vec r$ in space. The left-hand side represents transport due to convection; the right-hand side encapsulates transport due to diffusion. $\overline v\!\left(\vec r\right)$ is the fluid flow velocity, $C_i\!\left(\vec r\right)$ is the concentration, and $D_i$ is the diffusivity. For concision, the explicit dependence on $\vec r$ will be omitted below.

Given that we're most interested in the transport in the direction of the flow of the tea, if we approximate the stream as a cylinder of uniformly flowing liquid with velocity $v_z$ and assume that concentration variations perpendicular to the flow are unimportant, Eq. $\left(1\right)$ becomes:

$$
v_z{dC_i\over dz} = D_i{d^2C_i\over dz^2} \tag{2}
$$

if we choose the $z$-axis as lying along the stream of tea.

The most useful way to use Eq. $\left(2\right)$ is to non-dimensionalize it. That is, to re-write it so that all of the variables and coefficients have no units. For this problem, $z$ and $C_i$ are the variables to be non-dimensionalized, whereas $v_z$ and $D_i$ are parameters to be grouped into non-dimensional groups.  So, let's define:

$$
\eta = {z\over L}; \Theta = {C_i\over C_i^*} \tag{3}
$$

$L$ is some length scale representative of the system of interest. Here, the best choice is probably the height you're holding the cup over the surface of the toxic waste. $C_i^*$ is a representative concentration of a chemical species of interest, such as the nastiest player in that concentrated toxic waste. To substitute these definitions into Eq. $\left(2\right)$, I'll rewrite them as $z=\eta L$ and $C_i = \Theta C_i^*$:

$$
{v_zC_i^*\over L}{d\Theta\over d\eta} =
  {D_iC_i^*\over L^2}{d^2\Theta\over d\eta^2} \tag{4}
$$

The terms with derivatives in Eq. $\left(4\right)$ are now dimensionless. To make the coefficients non-dimensionless, divide both sides by $D_iC_i^*/L^2$:

$$
{v_zL\over D_i}{d\Theta\over d\eta} = {d^2\Theta\over d\eta^2} \tag{5}
$$

The new grouping of parameters in Eq. $\left(5\right)$ has a special name: the Péclet number, abbreviated $\mathrm{Pe}$.

$$
\mathrm{Pe}{d\Theta\over d\eta} = {d^2\Theta\over d\eta^2} \tag{6}
$$

The magnitude of $\mathrm{Pe}$ indicates the relative importance of convection relative to diffusion in a problem like this one. If the Péclet number is large $\left(\mathrm{Pe}\gg 1\right)$, it means that convection dominates the mass transfer behavior; if it's small $\left(\mathrm{Pe}\ll 1\right)$, then diffusion is the key phenomenon. If it's in the vicinity of unity $\left(\mathrm{Pe} \sim 1\right.$, or approximately $\left.0.3\leq \mathrm{Pe}\leq 3\right)$, then both diffusion and convection must be considered together.

So, can pouring tea into toxic waste poison your tea?

Well, setting aside the questionable wisdom of bringing a drink into a toxic waste area in the first place, what are some representative numbers for $D_i$, $v_z$, and $L$?  The upper limit for the diffusivity of a solute in water is about $2\times 10^{-6}\,\mathrm{cm^2\over s}$ (see, e.g., here). Since we want to know if diffusion can outrace convection, we should give it the best chance possible and selecting this maximum value will do that. (That is to say, maximizing $D_i$, which is in the denominator of the definition of $\mathrm{Pe}$, will tend to minimize $\mathrm{Pe}$.)

Similarly, selecting the minimum practical values of $L$ and $v_z$ will also tend to minimize $\mathrm{Pe}$, giving us a worst-case scenario for whether or not you face a gruesome death from toxic waste poisoning.  Let's say, optimistically(???), that one inch is the minimum height you can hope to pour your tea from without it splashing back into the cup. If we (rather loosely) approximate the  velocity of the tea stream at the toxic waste surface as that of a freely falling particle, the velocity will be $v_z=\sqrt{2gL}$.

So, our estimated Péclet number is (Google Calculator result):

$$
\mathrm{Pe} = {v_zL\over D_i} = {\sqrt{2gL^3}\over D_i} = 8.96\times 10^7
$$

This is a huge Péclet number. There's no way any of that toxic waste will diffuse up into your tea. (Please don't test this, though. Splashing, breathing toxic fumes, falling in, etc. are very real possibilities.)

As a further note, observe that in combining terms to form Eq. $\left(5\right)$, the reference concentration $C_i^*$ cancelled out completely. So, in this case, the answer is the same regardless of how concentrated that toxic waste is.


Often the key with stopping the conduction of heat is to provide as many phase transitions as possible. Most thermal insulators achieve this by having voids of Air which is achieved in a variety of ways, bubble wrap, fibre glass etc.

For a "non comercial" I assume they mean stuff you have lying around the lab?

To attempt this I'd layer blue roll / cotton wool / glass wool with tin foil. The more layers the better.


Are there any fluids which are both viscous and volatile?

Volatility and viscosity are 2 properties going each against the other.

A volatile liquid has rather small molecules a/o  rather weak intermolecular bonds.

A viscous liquid has rather big molecules a/o  rather strong ( often multiple ) intermolecular bonds.

A partial solution to the task is to have composite liquid, a volatile liquid mixed with either viscous liquid ( e.g  glycerine),
either with dissolved thickener ( like polyvinylalcohol or vaseline/wax )


It is a habit to assign greek letters (most often lowercase) to angles. In case of "normal" angles, it's mostly $\alpha, \beta, \gamma$, whereas for dihedrals some higher letters, as you mentioned, $\phi, \psi, \omega$. As noted in comments, whatever you use, clearly state what you mean.

The difference between dihedral and torsional angle is more conceptual, then formal. In torsion you discuss rotation around a bond, dihedral is general angle of two planes, as mentioned in the $\ce{CH_4}$ example. The formal expression for both of them is the same, indeed.

To be noted, there are some well known widely used dihedral angles, where you should stick to the common notation. The most prominent example is the protein backbone arrangement, see Ramachandran plot. In this case you should not invent your own notation.


Aflatoxin versus Benzene, permissible levels

It is usually assumed that there is no finite threshold below which carcinogens will not exert an effect. Therefore, there may be no safe exposure level below which there is no health risk.

Accordingly, a target value that is based solely on possible health risks (considering exposure over a lifetime, possibly even demanding an adequate margin of safety) would be zero.

Nevertheless, permissible limit values for carcinogens should not be set as low as possible since the exposure should not be reduced as much as possible at all costs. Instead, the potential exposure and limit values should be as low as reasonably achievable taking into account economic factors, other benefits and disadvantages, and the ability to detect and remove contaminants.

Hence, there may be similar limit values for various substances with different carcinogenic potentials. On the other hand, there may be different limit values for one substance depending on the exposure pathway (e.g. for different foodstuffs).


Normality = Equivalents of solute/liters of solution

Equivalents are determined by the number of electrons exchanged (or protons in water solutions, or yet other ions in different solvents). For example, if HCl is used to neutralize an alkali, then one mole of HCl will neutralize one mole of NaOH, but it takes two moles of HCl to neutralize one mole of Ca(OH)2 or three moles of Al(OH)3, because they have, respectively, two and three active OH- groups.

There are some pretty clear explanations in more detail at Molarity, Molality and Normality and at Normality.

And then there's Mel Brooks Abby Normal...


Why is predicting products of chemical reactions difficult?

Because you need would need to solve the Dirac equation (relativistic Schrodinger equation) for all the particles in the system to get a completely accurate result.

Even solving the time independent non-relativistic Schrodinger equation for $\ce{H2+}$ in isolation, where there is only one electron and two protons involves approximating the protons as fixed relative to each other.

Now imagine trying solving a system of $10^{23}$ molecules each having multiple nuclei, and dozens or even hundreds of electrons, and including time dependence!


determine density of oxygen gas using modified ideal gas equation

What an awkward, ugly equation that is! Mithoron is correct about the traditional, elegant and understandable form.

Yes, (P+1.01325)*10^5 converts gauge pressure in atmospheres to absolute pressure in Pascals.

z is compressibility - look up the Nelson-Obert generalized compressibility charts.

As for the 8314, the units here should be Pa*m^3/kmol-K. Dividing it by the molecular weight in kg/kmol will give you the specific gas constant for oxygen in Pa-m^3/kg-K. That's already been done in this equation.

Look up compressibility factor on wikipedia then find the Nelson Obert charts online, also critical data for oxygen, and use them to determine whether your oxygen is ideal in the P, T range you are dealing with (z factor between 1.01 and 0.99).  The specific volume of the gas depends on z as well as on P and T, but you can neglect the z correction if it's within a percent or two of the ideal value of 1.

Responding to your subsequent comment:

P, T and z determine the gas's specific volume in m^3/kg, which along with mass flow rate (kg/s) and pipe cross-sectional area (m^2) determine flow velocity (m/s), which along with pipe roughness will determine friction factor, which along with gas viscosity, pipe length and the presence of fittings like elbows and open valves, will determine pressure drop.


In rovibrational spectra, when exactly is the Q branch forbidden?

I would like tell you what I know about the answer to this question.

So, it's known that the selection rules for a fine vibrational spectrum (rotational-vibrational spectrum) for diatomic molecules make $\Delta J = 0$ or $\pm 1$ transitions allowed between the rotational levels of two vibrational states (usually $V''=0$ to $V'=1$, where the symbols have their usual meaning). But one must be careful in realizing that the $\Delta J = 0$ transition is forbidden for all parallel vibrations in the oscillator and only allowed for perpendicular vibrations.

Now, let's see what are parallel and perpendicular vibrations!

Any vibrational mode in infrared active only when the change in the dipole moment vector is sinusoidal w.r.t. time and thus can interact with the electric field of the electromagnetic radiation. If this change in the dipole moment vector is parallel to the major axis of the molecule, then the vibrational is labelled as a parallel mode of vibration. If the change in the dipole moment is perpendicular w.r.t. the major molecular axis, it is a perpendicular mode of vibration.

Let's look at the case of diatomic molecule, as the question is focused on that.

The major molecular axis for it is the bond axis (containing both the atoms). Since a hetero-nuclear diatomic molecule has only one vibrational mode, viz., stretching, it can only occur parallel to the bond axis. Therefore, $\Delta J = 0$ transition is forbidden in the case of parallel vibrations. Thus, on observing the fine infrared spectrum of hetero-nuclear diatomic molecules, we only observe $P$ and $R$ branches. $Q$ branches are not observed.

References:

Fundamentals of Molecular Spectroscopy - Banwell and McCash
Introduction to Molecular Spectroscopy - Gordon M. Barrow
Atomic and Molecular Spectroscopy - Rita Kakkar


First molecule:
The ester group is correct - you have a carbonyl (C=O) carbon bonded to an oxygen and both sides are bonded to an R group (any carbon/hydrogen group).

Phenol is incorrect - for a phenol group, the hydroxy/OH must be directly bonded to the ring, like so:

So instead, what you've labelled comprises of two functional groups: the ring is an arene group, and the OH is a hydroxy/alchohol group.

You haven't labelled the two branch alkyl group on the left of the molecule - this is called an isopropyl group (it has three carbons, so it's basically a propyl group with something attached to the middle carbon). This group might not always be considered a functional group though (maybe ask your lecturer?). Or it can just be called an alkane group.

Second molecule:
The amide group is pretty much correct, except it should only include at most the atoms bonded directly to the nitrogen and carbonyl carbon.

The alkene should only include the double bond (while the whole chain could be considered an alkene, the alkene functional group only contains the double bond).

One that you haven't included is the ether group, R-O-R.

To learn these I would suggest just using a table of common functional groups until you start to remember them. I think the Wikipedia page is good because you can look at your molecule and see what unusual atoms you have (e.g. nitrogen, oxygen) and see if any of the groups under that heading look like they're in your molecule, except it might be missing a few e.g. phenol.


What would this unsaturated cyclic compound be called?

Chemspider tells me that the name should be

  1,1'-Bi(cyclohexylidene)

http://www.chemspider.com/Chemical-Structure.121798.html?rid=e40cd595-d5e5-450e-bdb3-3e558a9da92d

While PubChem mentions:

  Cyclohexylidenecyclohexane

https://pubchem.ncbi.nlm.nih.gov/compound/Cyclohexylidenecyclohexane

  My guess was maybe it's "cyclohexylene cyclohexane"

There's a flaw to this thinking, to name this, I'd suggest you choose one of the cyclohexane rings to act as the word-root. Now you've to just add the name of the substituent.

Recall that (source: http://www.ebi.ac.uk/chebi/searchId.do?chebiId=65236) also click this link for an example (this website particularly has propylidene group as the example):

  An alkylidiene group formed from propane by removal of two hydrogen atoms from the same carbon atom, the free valencies of which are part of a double bond.

So, the substituent in this case would be called "cyclohexyl idene " (without the space). And hence, the name should be cyclohexylidenecyclohexane.


Can a reverse osmosis system remove radioactive elements in contaminated water?

I don't know about cesium in particular, but according to the EPA's web site (see the section called "How will radionuclides be removed from my drinking water?"), both isotopes should be covered under the beta particle section.
I cannot quantify the efficiency part of your question, but presume it's acceptable to the EPA (which may or may not instill confidence to you in terms of whatever methodology they're using).

The following treatment method(s) have proven to be effective in removing radionuclides at levels below their MCLs:
Beta particle and Photon Radiation: ion exchange and reverse osmosis;
(Gross) Alpha Emitters: reverse osmosis;
Radium 226 and Radium 228 (Combined): ion exchange, reverse osmosis, lime softening;
Uranium: Ion exchange, reverse osmosis, lime softening,  coagulation/filtration.


How is the temperature in which vapor pressure is equal to atmospheric pressure the boiling point of the liquid?

Vapour pressure is not merely calculation, it is easily measureable.
Evaporation is due molecules leaving liquid not by overcoming atmospheric pressure, but by overcoming cohesive intermolecular forces. The saturated vapour pressure is a measure of the rate of this massive leaving of molecules.
Boiling is just a special evaporation case. Above boiling point, vapour is always unsaturated and liquid evaporate at its surface regardless of gas composition. Additionally, random processes create miniature bubbles ( or these may be bubbles of formerly dissolved gasses ), that have the same or higher pressure then external pressure and expands themselves.
The result is more or less massive and turbulent evaporation with creation of vapour bubbles.
Liquids can be frequently overheated above their boiling point, if there are absent "bubble seeds". Once available, e.g. by strirring or repulsion of dissolved gas, creation of vapour bubbles quickly cools liquid down to its boiling point.


How easy is it to make animal horns transparent?

I found this description of how the horn is prepared in a post by Tammy L. Austin, at the University of Notre Dame's website (https://www3.nd.edu/~rbarger/www7/hornbook.html):

The horn of oxen and sheep were used to make the laminating structure. The horn was left in cold water for several weeks, which separated the usable part from the bone. It was then heated, first in boiling water then by fire, and pressed by plates and machines to make it smooth and transparent.


Question regarding the colour of Benzoic acid and Benzyl alchohol

none of the components you mentioned (Benozic acid, Benzyl alcohol, sodium benzoate) has a yellow colour.
The colour in organic compouns is closely tied to ther delocalized sytems of pi-electrons and depends on the gap between the highest occupied molecular orbital (HOMO) and the lowest unoccupied molecular orbital (LUMO).
When the energy of this gap corresponds to a wavelength that is visible, the compound absorbs this wavelength and the colour of the compound is the complementary colour.
A good example are condensated aromats. For benzene or naphthalene the gap between HOMO and LUMO is too big therefore, the corresponding wavelength is in the area of UV light.
For anthracene, the absorbed wavelenght is violett and therefore the colour of the compound is yellow.
Back to your question. Because the reaction happends under strong basic conditions, high temperature and with many aromatic compounds, I think sidereactions will occure where multiple aromats condensate together and therefore the solution turns yellow to orange.


Why isn't a polar protic solvent used in the Finkelstein reaction?

The key point is driving the equilibrium where you want it. You could do it by relative reaction rates. The Finkelstein reaction, however, uses a better method: precipitation of one of the products. Acetone is specifically chosen because sodium chloride, bromide and fluoride are insoluble in it while sodium iodide is soluble. So the equation becomes:

$$\ce{R-X + Na+ ({acet.}) + I- ({acet.}) -> R-I + NaX v}\tag{1}$$

Nothing beats the equilibrium-driving force of precipitation.


Intuitive Explanation for why more negative charge on ligand does not imply more splitting effect?

Crystal Field Theory assumes that all of the interactions occuring are purely electrostatic and only concern d-orbitals.  This is clearly not true when dealing with some ligands which create a covalent bond.
Most limitations of CFT all stem from the inability to explain non-electrostatic interactions.  For this reason, it works well when the electrostatic aspect is the dominant factor but it completely fails for some cases such as neutral CO ligands.  While you could easily argue electrostatic forces still exist in this situation, clearly they are not dominant and do not explain any observed behavior while still having an effect.


Redox reactions: identifying what is oxidized vs reduced

Yes sulfur of $\ce{SO4^2-}$ has undergone reduction as the oxidation number reduces from +6 to +4 and Oxygen of $\ce{H2O2}$ has undergone oxidation as its oxidation number increases from -1 to 0.

$$\ce{\overset{\color{red}{+6}}{S}O4^{2-}\+H2\overset{\color{red}{-1}}{O2}\+2H+->\overset{\color{red}{+4}}{S}O2\+\overset{\color{red}{0}}{O2}\+2H2O}$$

NOTE: Oxidation sate of oxygen in peroxide is -1. This because in the neutral molecule hydrogen peroxide the oxidation state of $\ce{H}$ is +1 and as the sum of oxidation numbers of the two hydrogen atoms and the two oxygen atoms must be zero so each oxygen in $\ce{H2O2}$ has -1 oxidation state.