To get the smell out of your washing machine and rug you have to do two things

1.) Get the source of the smell into the water and
2.) Keep it in the water for flushing.

To get it into the water I suggest using dish soap. Dish soap is far more design to handle the greasy materials you listed and will dissolve them better than regular laundry detergent.  The draw back to this is that dish soap will form foam quite readily which laundry detergent is formulated not to do. If you have a top loading washer this can cause suds to over flow your washer.  You may obfuscate this issue by monitoring and adding rubbing alcohol as needed to suppress the suds. If front loading washer you do not need to monitor but it will take a long time, as these machines can sense suds and will wait for the suds to die down before restarting. I also suggest using LA's Awesome as it is a very effective degreaser and uses limonene (orange oil) as an emulsifier.

To keep the smell in the water I suggest activated charcoal  which can be bought for use in aquarium filters.  This will absorb the petro-chemicals and carry them away.  I suggest you use activated charcoal after a first attempt with the prior suggest method of soap and degreaser as charcoal can coat your drier affecting the next load or two, but washing a rag can help clean it out.

One of the comments mentions scaling as a possible issue.  LA's awesome has chealating agents and this should be of no issue.  If you choose to etch some scale I would suggest vinegar for this over citric acid.


How is drift due to static electricity eliminated from an analytical balance?

Static charge can affect the measurements made by an analytical balance, as can magnetic materials because the mechanism in modern analytical balances involves a feedback loop between the load that is measured and an electromagnet. Insulating weighing vessels (plastic or glass) tend to accumulate static charge, especially when handled with another insulator (gloves) by the triboelectric effect.

You should first make sure that the body of the balance is grounded properly according to the manual for your model.

As a short-term solution to static charged glass, you might try arranging the sample so that it is father away from the balance. One way might be to place the sample on top of an empty sample bottle on the balance. As you mentioned, a more robust solution for discharging static build-up is to have a commercially available air ionizer nearby. The device will either produce an equal amount of positive and negative ions in the air at high voltage AC or DC electrodes, or with an ionizing radioactive material such as is found in anti-static brushes used for cleaning vinyl audio records.

If you are frequently measuring magnetic materials, many analytical balances can be configured in a bottom-loading arrangement. In this setup a platform is hooked beneath the balance, and the sample is supported sufficiently far away from the force-restoring motor so that it does not influence the measurement.


There are a couple of things wrong here. First off, your final reaction is unbalanced. Once you've fixed the balancing, read the other mistakes:

The ions do not exist in the liquid state! They are solvated/hydrated by the solvent. Since the solvent is water here, we'll say that the ions are in the aqueous (aq) phase instead.
While there is certainly no salt bridge here, there is still an electrolyte - an aqueous solution of sulphuric acid. Hence, you must mention it in your cell notation, between the anode and the cathode.
The $\ce{PbSO4}$ formed at the anode is in solid state. Hence, writing it as $\ce{Pb^2+(s)}$ is incorrect, as it is not dissociated into the ions $\ce{Pb^2+}$ and $\ce{SO4^2-}$.

With all these corrections, your final, correct cell representation should be:

   $$\small{\ce{Pb(s), HSO4^-(aq) | PbSO4(s), H+ | H2SO4 ($\pu{x~M}$) | PbO2(s), HSO4-(aq), H+(aq) | PbSO4(s)}}$$

Some websites (like KhanAcademy) and texts (NCERT 12), cite the lead-acid battery reaction as this instead:

$$
\begin{array}{}
\text{Oxidation}&\ce {Pb(s) + SO4^2-(aq) &-> PbSO4(s) + 2e-}\\
\text{Reduction}&\ce{PbO2 + SO4^2-(aq) + 4H+(aq) + 2e- &-> PbSO4(s) + 2H2O(l)}\\
\text{Total reaction}&\ce{Pb(s) + PbO2(s) +2SO4^2(aq)- + 4H+(aq) &-> 2PbSO4(s) + 2H2O(l)}\\
\end{array}
$$

assuming the bisulphite ion to be further ionized. The cell notation in this case would then be:

   $$\small{\ce{Pb(s), SO4^2-(aq) | PbSO4(s) | H2SO4 ($\pu{x~M}$) | PbO2(s), SO4^2-(aq), H+(aq) | PbSO4(s)}}$$


Can entropy in a closed system remain constant when you heat a liquid?

The quantitative differential classical thermodynamic definition of entropy is $$\mathrm{d}S=\mathrm{d}Q/T$$

Therefore providing heat $Q$ to the acetone increases its entropy as
$$\Delta S=\int_{T_\mathrm{1}}^{T_\mathrm{2}} \frac{C(T)}{T}\cdot \mathrm{d}T$$

where $C$ is the heat capacity of the system.

The statistical thermodynamics definition has an alternative definition to $$S=k_\mathrm{b} \cdot \ln \Omega$$ as $$S=-k_\mathrm{b} \cdot \sum_\mathrm{i} p_\mathrm{i} \cdot \ln p_\mathrm{i}$$ as the sum over all states.

The higher energy states of electrons, of bond vibrations and molecule rotations are not occupied at low temperatures, or are occupied unevenly.
In such a case, even for the same number of states, more evenly distributed probabilities at higher temperatures lead to higher entropy.


Calculating moles of Hydrogen gas in multiple gas container

Using pv = nrt$$1atm\times(0.5L) = (total moles)(.08206\frac{atm\cdot L}{mol\cdot k})(293K) \\ \text{total moles} = .02079mol $$ There are 3 total moles of gas with 2 of those being hydrogen and 1 being oxygen. This means 66% of the total moles is hydrogen and 33% is oxygen. Soooooo 66% times .02079 is the amount of hydrogen moles and 33% times .02079 is the amount of oxygen. TADA!


Can we fill potato chips bags with a gas other than nitrogen?

As Nilay Ghosh said, nitrogen is cheap. Very cheap. Neon is expensive. Argon is cheaper than neon, but considerably more expensive than nitrogen. Helium is also expensive and needs to be used wisely, for important things, e.g., cryogenics. And hydrogen! I can just see the ads: “Buy our chips: they are lighter than air! But avoid open flames and sparks unless you want to be Hindenburged to a crisp (no pun intended)”
Another fill gas to avoid is sulfur hexafluoride. A tennis ball manufacturer once decided to fill tennis balls with sulfur hexafluoride, assuming this would prevent the balls from going flat as a consequence of the high molar mass of sulfur hexafluoride. But the tennis balls exploded on the shelves because air diffused in.
Thankfully, no one has ever tried using nitrous oxide as the fill gas in potato chip bags!


Solubility experiment of different organic compounds

Carboxylic acids easily react with bases forming salts, which are ionic compounds and are usually readily dissolve in polar solvents such as water, which obviously makes 95% of your base solution.

Carbonyl compounds are easily dissolved in strong sulfuric acid as 1) a reaction may take place, introducing -SO3H group to position 3 of aromatic ring of respective carbonyl compound; and 2) a strong acid can form hydrogen bonds with carbonyl oxygen. In both cases affinity of carbonyl compounds towards sulfuric acid is increased.


Compressibility factor of hydrogen at low pressure and constant temperature

In my textbook the opposite meaning is given to $a$ (an attractive parameter) and $b$ (a size parameter), but this may be a matter of differences in labelling, not in meaning. I will for the sake of consistency stick to your convention. For $\ce{He}$ the van der Waals parameters have the following values:

$b = \pu{3.4598 J mol-1 M-1}$
$a = \pu{0.023733 M-1}$

As you can see from the following figure at $RT$ ($\pu{298K}$), while your first approximation is quite accurate for $\ce{He}$ already at $V_\mathrm m \approx \pu{1 L/mol}$, your second approximation fails until higher  $V_\mathrm m$. It is negative, but negliglibly so, and at that point is not particularly useful as an approximation as already $Z \approx 1$.

The approximation fails (at RT) because the condition $b/RT \gt a$ is never observed:
$$\frac{b}{RT} = \pu{0.0014 M-1}$$
whereas
$$a = \pu{0.0237 M-1}$$

You can solve for the temperature at which these two become equal:

$$T_b = \frac{b}{aR}$$

For $\ce{He}$, $T_b = 17.53\ \mathrm K$. Below $T_b$ you may observe $Z<1$.

Sure enough, below that $T$ your second approximation starts to hold, as shown here for $T = 10\ \mathrm K$:

or in terms of $P$:

Note: it is not clear that the virial equation and van der Waals parameters should capture the behavior of the gases over a wide range of $P$ or $V_m$ at such low $T$.


In the resonance structure drawn at the top of your image, the $\ce{p}$ orbital on carbon and the $\ce{p}$ orbital on oxygen share a bonding pair of electrons, a lone pair of electrons reside in a nitrogen $\ce{p}$ orbital, two lone pairs reside in two oxygen $\ce{sp^2}$ orbitals.  So for this case, the most commonly drawn representation, there is one lone pair in a $\ce{p}$ orbital.  In your middle resonance structure, there is a lone pair in an oxygen $\ce{p}$ orbital, and another lone pair in the nitrogen $\ce{p}$ orbital; hence, 2 lone pairs in two different $\ce{p}$ orbitals.  In your bottom resonance structure, there is a lone pair in an oxygen $\ce{p}$ orbital and a shared bonding electron pair in the $\ce{C-N}$ $\ce{pi}$ bond; therefor, one lone pair in a $\ce{p}$ orbital for this case.

This seems to be a poorly worded question.  Since the real molecule is represented by a combination of the 3 resonance structures, the correct answer would be "somewhere between one and two lone pairs in $\ce{p}$ orbitals."  Adding the detail contained in the above paragraph would probably help the grader understand your explanation.


Why does hot oil fry things, but hot water does not?

Oil can get hotter than the b.p. of water, 373 K.

The browning (decomposition) of starches proceeds very slowly (perhaps years for a noticeable change) at 373 K. In addition, the water in the potato will not evaporate away, since the potato is in water (not the water vapor above the liquid, which can be hotter than 373 K).
Since the cooking oil is hotter (e.g. 440 K), starch decomposes (caramelizes) in a few minutes, and dehydration makes the chip crunchy. Bon appétit!


What household compound will neutralize naphthalene-stained washer (for clothes)?

Napthalene is not water soluble, so the temperature doesn't matter as much as the detergent. What makes you think it's napthalene, though? It could be some organic compound in which case you need a good de-greaser (Dawn is the easiest to get, but there are better industrial ones like what cleaning crews may use). If what you're dealing with is an aromatic hydrocarbon (which napthalene is), airing it out will go a long way. I would see how far you can get with a fan blowing it out before you wash the rest out with a good de-greaser.


What is the true caloric impact of Alcohol?

An interesting study, 'Can calories from ethanol contribute to body weight preservation by malnourished rats?', to quote:

  In terms of caloric content, ethanol is a unique drug whose oxidation by alcohol dehydrogenase (ADH) produces 7.1 kcal per gram of ethanol oxidized. However, it is not known to what extent these calories are converted into body weight. There is no controversy about the malnutrition produced by chronic ethanol intake in alcoholics. In addition to the known toxic effects on absorption and synthesis of essential nutrients (1), some important metabolic changes occur. Pirola and Lieber (2) reported that activation of the microsomal ethanol oxidizing system (MEOS) in hepatocytes of alcoholics may consume one third of the caloric content of ethanol and impair ATP synthesis. Lands and Zakhari (3) suggested the existence of a futile cycle in ethanol metabolism which would account for a loss of six ATPs per gram of ethanol oxidized. They proposed that ethanol may be oxidized to acetaldehyde, consuming three ATPs, and acetaldehyde may be again reduced to ethanol, consuming three extra ATPs.
  However, there is some controversy about what happens to ethanol calories when consumption is moderate and not related to the metabolic changes of the liver that characterize chronic ethanol intake. It has been suggested that calories derived from ethanol may function as do calories derived from carbohydrates, producing a thermogenesis value of about 10% (4,5). This thermogenic value may be higher than that of fatty food (3%) and lower than that of proteins (about 20%) (6).

  The contribution of calories provided by ethanol to body weight gain has also been studied but there seems to be no consensus about the data from animal and human studies.

  Moderate ethanol consumption seems to favor an increase in fat storage and could result in weight gain, especially in overweight individuals (7). In an epidemiological study involving 89,538 women and 48,493 men, Colditz et al. (8) reported a strong negative association between alcohol intake and body weight index for women and the absence of this association for men.

  Animal studies are consistent in reporting a decrease in the body weight of rats receiving ethanol solutions as the only source of liquids. Concentrations of ethanol as low as 5% (v/v), which are similar to the ethanol content of a Brazilian beer, or as high as 40% (v/v), solution similar to spirit drinks, are related to decreased body weight gain (9). Similar results have been reported for 20% (v/v) ethanol solution (10).

  Different results have been obtained for malnourished animals. Da-Silva et al. (11), studying rats which had been treated with ethanol for 90 days, reported a significant weight gain by malnourished rats (50% food restriction) drinking a 20% (v/v) ethanol solution when compared to malnourished rats drinking water. A more recent study (12) reported improvement in somatic and motor development and a decrease in the mortality rate of the offspring of malnourished rats drinking low doses of ethanol (5%, v/v). These data suggest that malnourished rats can benefit from ethanol calories.

  In summary, in spite of the large number of studies on the effects of ethanol in well-nourished animals and humans, there is still controversy about how well ethanol-derived calories can be utilized. Fewer studies are available about special physiological conditions such as malnutrition. Over the last few years, scientific research has mainly focused on obesity, an increasing problem in developed countries, which led us to the false belief that malnutrition was no longer a problem worth investigating. However, there are still 800 million malnourished people in the world (13). The decreasing interest of the scientific community in problems related to malnutrition has left many questions without an answer. Ethanol consumption and its consequences on the malnourished organism are among them.

  In view of the importance of malnutrition in Brazil - 22% of the population or 40 million people are malnourished (14), as well as ethanol consumption and alcoholism - 11% of Brazilian population are alcoholics (15), the aim of the present study was to assess the use of ethanol calories in a dose/effect model by evaluating body weight before and after the installation of malnutrition.

My take, ethanol is apparently somewhat unique and there remains some controversy about how well ethanol-derived calories can be utilized. Study classification factors include well-nourished vs. malnourished, low doses (5%) vs high doses (20%), producing significant variations in effects.


What is the difference between melting and dissolving?

Juha invited me to write a summary (see comment on my previous answer ) of the differences between melting and dissolving. I’ll try to outline this roughly in the same order as his. I'm giving this as a new answer since my last answer was quite long as it was.

Differences:

Melting and dissolving are completely different processes on the molecular/atomic level that could not be mistaken for each other if you could observe what was happening at that scale.
If you cool the liquid that arose from melting the solid, to a temperature below it’s melting point, you would see the entire sample solidify. If you cool the solution (dissolved solid + solute) to below the melting point of the solid (solute) in it, you would see no change. (Unless you had a saturated or nearly saturated solution.)
Melting requires only a single substance and energy input while dissolving requires a solvent and a solute that are compatible (“like forces”). (This is actually a pretty huge difference and would potentially affect all of the physical and chemical properties.) Dissolving a solid can be either endo- or exo-thermic.
"Phase change" (wording from the previous summary)  In each case you end up with a liquid. Melting caused a phase change (composition of the substance didn’t change) while dissolving a solid in a liquid is not considered a phase change since a change in composition occurred.

Similarities:

In each case, forces between the particles that comprise the solid are disrupted and that takes energy. (Whether it’s chemical bonds or intermolecular forces depends on the process and on the solid and on your definitions. (See this question.) But melting (rare exception noted in previous comments) is endothermic and dissolving can be either endo- or exo-thermic.
In each case you end up with a liquid. Macroscopically, if you walked into a room and saw the liquid on the table, it would be difficult to say whether this liquid came from a solid that had melted or a solid that had dissolved in a solute and made a solution.
But it would be very easy to determine which you had experimentally in a “dozen” different ways.
Both melting and dissolving require interaction among groups of atoms, molecules, or ions.

There are probably more differences that could be given (how to handle thermodynamic calculations, complexity of the system etc.), and possibly more similarities, but that's enough for me on this topic.


Basically this is due to two reasons.

The disproportionation of chlorine in water looks like this:

$$
\ce{Cl2 + H2O <=> H+ + Cl- + HOCl}
$$

So you need the $\ce{H+}$ to shift the equilibrium to the halogen site and the halogen produced is a gas and will leave the reaction.

The other explanation is that, as $\ce{HOCl}$ is an oxidizer and a $\ce{Cl+}$ transfer-agent, both of these actions happen in an $\ce{S_{N}2}$-type fashion. For the $\ce{Cl+}$ transfer it is called associative activated substitution of a nucleophile, here $\ce{Cl-}$. So it's pretty much like an $\ce{S_{N}2}$ reaction, where
$\ce{H-Cl}$ and $\ce{Cl-OH}$ react.


How can I figure out which coupling constants match which hydrogen?

Coupling constants are mutual. If nuclei $\ce{A}$ and $\ce{B}$ couple and this coupling results in a coupling constant $J_\ce{AB} = \pu{7.4Hz}$ on $\ce{A}$ then $\ce{B}$ will also be split into two signals with a coupling constant $J_\ce{BA} = \pu{7.4Hz}$.

Therefore, you can indeed start at the anomeric hydrogen (which you can identify not only by the coupling pattern but also by its chemical shift) and work your way along the ring, noting whether two hydrogens are in gauche or anti configuration along the way.


What is a hydrogen bond acceptor and donor?

Dissenter's answer which you linked appears to be wrong (as of now). I have left a comment.

The conventional way in which we talk about hydrogen bond acceptors/donors is: in a hydrogen bond $\ce{X-H...Y}$, the hydrogen bond donor is $\ce{X-H}$ and the hydrogen bond acceptor is $\ce{Y}$. (See for example Chem LibreTexts, or this other question on hydrogen bonds.)

Of course, water is capable of acting as both an acceptor and as a donor:

I wouldn't call it a "proton donor" like Wikipedia does, though, as that suggests some kind of acid-base reaction occurring, like

$$\ce{\color{red}{H2O} + \color{blue}{H2O} <=> \color{red}{OH-} + \color{blue}{H3O+}}$$

In this reaction the red water molecule is acting as a proton donor (Brønsted acid) and the blue water molecule is acting as a proton acceptor (Brønsted base). But these terms should only be used in the context of a proton transfer, and not hydrogen bonding.


Apparently, the particles of your solid are either very small, or / and not crystalline.  This may complicate filtration, though this is one of the reasons filter papers used in the chem labs are sold in different grades of paper hardness.
A practical alternative to separate such a slurry from the liquid phase would be to substitute filter and funnel by a centrifuge.  With some luck you still find some with a manual crank (with conic tubes, about $\pu{20 mL}$ each). Based on own experience, this often is faster than setting up and then performing a suction filtration over a glass filter. On small scale, it is simpler to wash the slurry, and clean up the tools, too. Just keep the opposite tubes balanced and don't mind the surprised gaze of your colleagues around.

(credit)


How exactly does a battery work (specifically Pb/H2SO4/H2SO4/PbO2 lead acid battery)?

In order for oxidation-reduction chemical reactions to occur, the electrons must somehow go from the anode electrode where the oxidizing (electron donor) takes place to the cathode electrode where the reducing (electron acceptance) occurs.

Ideally, the electrolyte, in this case sulfuric acid, should allow the movement of positively charged ions but prevent the movement of electrons through it.  This forces the electrons to go through a circuit, external to the cell, to go from anode to cathode thus enabling the redox to occur.  Without electron flow, said redox reaction will not occur.

In the real world, there will be some leakage current through the electrolyte and other parts of battery.  We can eliminate the leakage with a more insulating electrolyte, but this would hurt the charging process and make the battery less rechargeable.  So, what we have is an economical compromise.


Why is nucleus charge squared in Rydbergs formula?

Derivation comes from the formula for electrostatic potential energy.

The electrostatic potential between the point charges of an electron, and a proton would be represented by the formula

$$U_E(r) = k_e \frac{q_eQ_p}{r}$$

However, in the derivation that we are looking at a difference of energy is from an emission or absorption, such that the energy concerned is

$$E_i - E_f$$

The subtraction appears in the Rydberg formula from the distance in the form

$$\left( \frac {1} {n_1^2} - \frac {1} {n_2^2} \right)$$

However, the terms other than the radius are multiplied together and come outside the parentheses.  Even though we are looking at only a single electron, the term for the charge in the nucleus is squared because of this. This can be seen in the Bohr derivation of the Rydberg formula.

$$E = h\nu = E_i - E_f = \frac {m_eq_e^2q_Z^2} {8h^2\epsilon_0} \left( \frac {1} {n_f^2} - \frac {1} {n_i^2} \right)$$

This shows the origin of the squared term for the nuclear charge which correlates with the Rydberg formula for a hydrogen-like element since all terms outside the parentheses other than charge are constant.

$$\frac {1} {\lambda_{vac}} = RZ^2\left( \frac {1} {n_1^2} - \frac {1} {n_2^2} \right)$$


First of all they are not really arbitrary.

But the main point, regarding balancing the equations is that: You can keep track of the total number of electrons per species, which is a well defined value. Like, for one permanganate ($\ce{MnO4-}$) you have 25+4*8+1 electrons, for one $\ce{Mn(II)}$ you have 23 electrons, for one water you have 10 electrons, for one $\ce{H+}$ you have 0 electrons. So for this half-equation, you must have 5 electrons on the left side:

$$\ce{8H+ + MnO4- + 5e- -> Mn(II) + 4H2O}$$

or you can say that, here I assume all H atoms are +, all oxygens are -2, and for consistency, Mn in $\ce{MnO4-}$ is +7. I can dismiss H and O's since all are same, and all change is because of Mn, which goes from +7 to +2 and difference is 5 electrons.

You could do this like that as well: O is -4 everywhere, so H's in $\ce{H2O}$ should be +2 and they were +1 on the left side. They donated 8 electrons. Mn in $\ce{MnO4-}$ should be +15 and it is +2 on the right side, which means it accepted 13 electrons. And the difference is 5 electrons, again.

Bottomline: It is all about consistency. If all elements in each and every species sum up to the charge of this species, whatever you say their formal charges are, it will work -because this just another way of keeping track of total electron count.  So it is better to treat most of them constant like, O is -2, H is +1 almost always.


You have a two part question so I'll answer both of them:

  Question - What makes bonds break apart from heat?

"If you start to heat the system, two things can happen:

The atoms gain enough energy to move into higher, excited states in the potential well. Eventually the additional energy may be high enough to make the total energy positive. The nuclei can escape the potential well - the bond is broken.
The electrons that participated in forming the bond, i.e. those on orbitals overlapping both nuclei, get excited into different, higher energy orbitals. This makes the potential well shallower, and the nuclei can escape it easier - the bond becomes weaker and easier to break. Eventually the potential well can even become so shallow, that no bound state can exist." - Source

  Question - Why do alkali metals ($\ce{Li, Na, K ...}$) break bonds in water ($\ce{H_2O}$)?

"All alkali metals melt as a part of the reaction with water. Water molecules ionise the bare metallic surface of the liquid metal, leaving a positively charged metal surface and negatively charged water ions. The attraction between the charged metal and water ions will rapidly increase the surface area, causing an exponential increase of ionisation. When the repulsive forces within the liquid metal surface exceeds the forces of the surface tension, it vigorously explodes."- Source

I hope that my answers were useful to you.


What is the active ingredient in this "stimulant?"

If you google Schizandrol A ( 1,2,3,10,11,12-hexamethoxy-6,7-dimethyl-5,6,7,8-tetrahydrodibenzo[a,c]cycloocten-6-ol, here's a link to a picture of the structure) you find that it is an herb capable of curing just about anything.  If you poke around a little deeper, you find two repetitive themes 1) it increases focus and 2) synergistically enhances the effect of other chemicals ingested along with it.  You also note this herb has caught the attention of the bodybuilding community.  The lifters use terms like "pump and rush", "being on another planet", etc.  Apparently it can take focus to a whole new level. From what I read it looks like it's always ingested with other stuff so it's hard to separate out what's what.  My guess is that all the ingredients in the bottle come into play and the effect they have together, particularly with the Schizandrol A present, is much greater than any ingredient alone.


How do acids prevent ferrous sulfate from getting oxidised?

No, adding acid to a ferrous sulfate solution will not prevent its being oxidized by bromine water.
Fresh FeSO$_4$ solutions are light blue-green, almost colorless, but pick up O$_2$ readily:
Fe$^3$$^+$ + e$^-$ --> Fe$^2$$^+$       +0.770 V and
O$_2$ + 2 H$_2$O + 4e$^-$ --> 4 OH$^-$    +0.401 V
(The half cell voltages are only indicative of what's going on, because we are not at standard concentrations of 1 M.)
The light-colored solution quickly becomes yellowish and consumes all the oxygen in the vessel, and then sucks in more air and consumes that oxygen, and then things slow down a little, but diffusion still lets a bit more oxygen get in, and the "ferrous" becomes somewhat ferric, and the liquid becomes a muddy color and you wish you could preserve the nice blue-green color somehow.
You could preserve the ferrous if you could somehow complex it and keep it from oxidizing to ferric. But nature prefers to complex the ferric ion: smaller ion, bigger charge. Typical ferric stability constants are larger than ferrous by ten orders of magnitude.
It is instinctive to suppose that metals will oxidize in the presence of O$_2$, and if higher valences are possible, well, then, the metal will go as far as possible. But the oxidation of ferrous to ferric is unfavored (+0.770 V) and Fe$^2$$^+$ needs help from O$_2$. The reduction of water is also unfavored, but is driven forward by the consumption of OH$^-$ ions by precipitation by Fe$^+$$^3$, which also keeps the Fe$^+$$^3$ low.
LeChatelier predicted as much: by increasing the acidity, you reduce the availability of electrons from OH$^-$, and O$_2$ is more able to take electrons from Fe$^+$$^2$.
So you could say that adding acid stabilizes the ferrous ion, but what it really does is inhibit the oxygen molecule. Bromine is totally unaffected.


Does limonene react with calcium hydroxide or acetic acid?

I can't find the results of too many experiments with Limonene.

Actually, there are quite a lot and the feature article Limonene: a versatile chemical of the bioeconomy, Chem. Commun., 2014, 50, 15288-15296 by Rosaria Ciriminna, Monica Lomeli-Rodriguez, Piera Demma Carà, Jose A. Lopez-Sanchez and Mario Pagliaro provides an interesting insight:

d-limonene (1) is available in large quantities as a side product in the production of orange juice
it is widely used as a bio-based solvent to  replace petroleum-based solvents for various purposes, such as

general household cleaners
degreasing of metals
outdoor maintenance cleaners

it is used as an insecticide

In order to to prove comsumer-friendliness of d-limonene and its degradation products, a lot of studies on the reaction of d-limonene with oxygen, leading to endoperoxides, hydroperoxides, and the monoterpene ketone carvone (2), have been performed.

In the ongoing BioProMo project, the bacterial conversion of d-limonene to the carboxylic acid 3 is examined. The latter is a natural preservative.

  I'm curious of the results of combining Limonene and Calcium Hydroxide (Lime) in one experiment and another experiment combining Limonene and White Vinegar.

To my information, these combinations aren't used in industrial or academic research, but in another context.

As outlined above, d-limonene is widely used as a natural replacement for other organic solvents, which aren't easily available outside science and/or industry. As a consequence, d-limonene as a solvent has also found its way into recipes for the clandestine extraction of controlled substances from plant material.

Some of these recipes seem to promote the combination of d-limonene, calcium hydroxide, and white vinegar (as a source of acetic acid) to extract N,N-dimethyltryptamine (DMT) as its acetate from the bark of Mimosa tenuiflora.

While this seems technically possible and reactions between d-limonene, the base, and the acetic acid are not to be expected, the posession of the extraction product is apparently illegal in a variety of countries.


Is lithium is less reactive than sodium?

You can't just take the electrode potential as if that's going to correspond to the rate in which something is going to react.  For example, why do you think that aluminum is so stable in atmospheric conditions versus iron?  Aluminum forms a stable oxide layer that serves to protect it against further reaction, while iron oxide is weak, adheres poorly and is and porous.  There are numerous factors which affect a rate of reaction, and they vary in different situations.

Calculating rates of reaction is a very complicated thing, and it's usually easier / more accurate to just determine the Arrhenius parameters experimentally and use those.


they are the same chemical yes - (the 2 just refers to the position of the side branch.)
I Would not expose the glue to air unnecessarily as any moisture will make it harden . I can not say for you using it on your nails for insurance reasons ( but scientifically they are the same )


If the conditions of the reaction (depending on temperature, pressure and any other gases present) are such that you are producing $\ce{CO2}$, then $\ce{CuO}$ should readily oxidize any $\ce{CO}$ that gets produced all the way to $\ce{CO2}$.

The main reaction should be as you said:

$$\ce{2CuO + C -> 2Cu + CO2}$$

and if you have any formation of $\ce{CO}$, it will be oxidized as:

$$\ce{CuO + CO -> Cu + CO2}$$


Why can mercury oxide be decomposed into mercury and oxygen upon heating?

In simple terms, heating is indeed a physical method. You might have heard of heat being, in fact, molecules/ions moving faster. There’s nothing in there per se that would make it a chemical method.

But. The atomic particles that are moving quicker can bump into each other with more energy (because they are faster — think of a car crash which is worse at high speeds); this can be understood as activation energy, i.e. energy that allows reactions to proceed.

So by adding enough heat energy, you are inducing a chemical change which leads to the (thermal) decomposition of mercury oxide into its elements.


Why is the principal quantum number irrational?

So, first, as I said in my comment, you confused the orbital angular momentum quantum number $l$ with the principal quantum number $n$. Both can take only integer values (positive and non-negative respectively), but for totally unrelated reasons.

Secondly, the orbital angular momentum itself is related to its quantum number $l$ by the following equation,
$$
    \hat{L}^2 \Psi = l (l + 1) \hbar^2 \Psi \, ,
$$
where $\hbar^2{l(l+1)}$ is the square of the magnitude of an electron orbital angular momentum. Thus, while the orbital angular momentum quantum number $l$ is always a non-negative integer, the magnitude of an electron orbital angular momentum $\sqrt{l(l+1)} \hbar$ can take irrational values.

For instance, for a d-electron, $l=2$, and thus, the magnitude of its orbital angular momentum is $\sqrt{2(2+1)} \hbar = \sqrt{6} \hbar = \sqrt{6} h / 2 \pi$.


Can diastereomers form a racemic mixture?

The short answer is no.

A racemic mixture is defined as an equimolar mixture of enantiomers. Since each enantiomer rotates plane polarized light equally, but in opposite directions, the overall optical activity is zero.

Describing diastereomers as optical isomers is incorrect. Diastereomers are stereoisomers (that is, compounds that have the same connectivity of atoms but differ by the orientation of atoms in space) but are not enantiomers. In many cases, diastereomers are not even chiral (have no optical activity). If there's a mixture of diastereomers that are chiral, this is not described as racemic. Most likely that mixture would be optically active, but even if it was optically inactive, that would just be coincidental and not considered racemic.

The situation where a mixture is made up of two enantiomers in equal ratios is so common (and important) that it gets the special name of racemic.


The following reaction schemes are adapted from [1, pp. 76–77].
Amorphous borax $\ce{Na2B4O7}$ forms tetrahydroxoborate in concentrated sodium hydroxide solution:

$$\ce{Na2B4O7 + 7 H2O + 2 NaOH → 4 Na[B(OH)4]}$$

Upon fusion sodium metaborate is formed:

$$\ce{Na2B4O7 + 2 NaOH →[\pu{700-750 °C}] 4 NaBO2 + H2O}$$

In order to form peroxoborate, a peroxide should be introduced in the first place, e.g. by using hydrogen peroxide (see also Preparation of Sodium Peroxoborate):

$$\ce{Na2B4O7 + 2 NaOH + 4 H2O2 + 11 H2O → 2 Na2[B2(O2)2(OH)4] * 6H2O}$$

References

R. A. Lidin, V. A. Molochko, and L. L. Andreeva, Reactivity of Inorganic Substances, 3rd ed.; Khimia: Moscow, 2000. (in Russian)


Why do shape and size have the greatest effect on the optical properties of nanoparticles?

As Surface Plasmon Resonance phenomenon is mainly seen in metal nanoparticles, I don't know whether this answer will help you? The size dependent optical phenomenon of semiconductor nanoparticles (e.g., $\ce{CdS}$) can be explained using the effective mass approximation model.

where r is radius of the particle (the rest details of the above equation can be find in the link).

so the energy gap is inversely proportional to the radius of the particle.

When the radius of the particle  decreases, the band gap increases so it will emit high energy wavelength (i.e., wavelength in blue region) and when the particle size is large the energy gap will be less so it will emit light in red region (lesser energy).


Structure of Niobium and Tantalum carbonyls

From the abstract of J. Chem. Theory Comput., 2011,
7, 2112-2125 (DOI) is seems that niobium forms a dinuclear $\ce{Nb2(CO)12}$ complex, in which two $\ce{Nb(CO)6}$ units are connected by a 340 pm $\ce{Nb-Nb}$ bond.


Reagent choice in the formation of Wittig reagent

I am fairly sure that for both, there are practicality issues at hand.
This is almost certainly so for the choice of phosphine. Even though PPh3 is absolutely terrible from an atom economy point of view, and even though PPh3O can be a real pain to remove after the reaction is done, at least it's a nice white solid which doesn't smell. On the other hand, the simpler alternative PMe3 is a volatile, pyrophoric, and extremely stinky liquid. PPh3 is also substantially cheaper than other alternative phosphines. (There are similar considerations for the Mitsunobu reaction which I have some experience with.)
As for the base, $\ce{BuLi}$ is just one of the most accessible strong bases to use. I don't think you can buy dimsyl sodium from the average chemical supplier. My suspicion is that the main reason why dimsyl anion is somewhat associated with the sulfur ylids is because E. J. Corey and his postdoc M. Chaykovsky popularised them together (e.g., Ref.1). Also note that $\ce{NaH/DMSO}$ mixtures are associated with significant explosion hazards (Ref.2).
Others with more experience will be able to give fuller answers.
References:

E. J. Corey, Michael Chaykovsky, “Methylsulfinyl Carbanion ($\ce{CH3-SO-CH2-}$). Formation and Applications to Organic Synthesis,” J. Am. Chem. Soc. 1965, 87(6), 1345–1353 (https://doi.org/10.1021/ja01084a033).
Qiang Yang, Min Sheng, James J. Henkelis, Siyu Tu, Eric Wiensch, Honglu Zhang, Yiqun Zhang, Craig Tucker, David E. Ejeh, “Explosion Hazards of Sodium Hydride in Dimethyl Sulfoxide, N,N-Dimethylformamide, and N,N-Dimethylacetamide,” Org. Process Res. Dev. 2019, 23(10), 2210–2217 (https://doi.org/10.1021/acs.oprd.9b00276).


What is the ionic equation for reaction of potassium permanganate with potassium iodide?

How stable is $\ce{IO3-}$ in the presence of $\ce{I-}$ under acidic conditions?

It would end up being in an equilibrium with very few $\ce{IO3-}$ anions and lots of iodine, due to the following equilibrium:

$$\ce{2 IO3- + 12 H3O+ + 10 I- <=> 6 I2 + 18 H2O}$$


In chemical resonance, the same principle holds as in electrostatics:

"In general, the boundary structures of a conjugated system with the lowest energy are those
...
- in which the distance between charges of opposite sign is as small as possible and the distance between charges of the same sign is as large as possible
... ."

Wikipedia: Resonance (chemistry) - Major and minor contributors:
"Energetically unfavourable and therefore less favorable structures are minor contributors. With rules listed in rough order of diminishing importance, major contributors are generally structures that
...
3. carry a minimum of formally charged atoms, with the separation for unlike and like charges minimized and maximized, respectively
... ."


The former equation is just a simplified version of the latter.

The important factors are the concentrations (activities) of the positive and negative species.

However, since these values are small ($10^{-7}\,\mathrm{mol\cdot l^{-1}}$) compared to the concentration of water ($55,56\,\mathrm{mol\cdot l^{-1}}$) this distinction would not affect the value of the dissociation constant either.


What is the significance of an organic compound's M peak?

"M" stands for molecular ion.  It is often represented as $\ce{M, M^{+} or M^{+~.}}$. In the mass spec experiment, an electron is knocked out of some of the sample molecules.  The peak at M/q (q=1) represents the molecular weight of the entire molecule, less an electron, before it fragments to smaller ions.  From precise measurement of the molecular ion (e.g. 82.34567890) you can determine the exact molecular formula of your sample compound.  This can help greatly with sample identification.


I have no real knowledge of it, and I've found almost no literature on its handling and storage. From Scientific American 1878:"Unlike lead, however, it acquires only a very slight tarnish on exposure to moist air..." and "In aerated water it tarnishes slightly." My very very old copy of Cotton and Wilkinson (Adv. Inorg. Chem) states that it is a moderately reactive metal and will react with sulfur. And of course we know it readily alloys (I'd say "amalgamates" except that term is specific for mercury alloys) with a large variety of metals. The (III) oxide is white, so unless your scum is white, it ain't that. The (I, sub) oxide is brown-black so that's a possibility. As are mixed oxide /hydroxides. Since the original investigation says it does tarnish in air, I'd take that at face value. There is the obvious possibility that your sample is either contaminated or was sold adulterated. If it's been in contact with any other metal (especially Aluminum) then it is almost certainly contaminated. There's no practical way you'll be able to recover the metal - the expense will far exceed the price to simply replace it. The best way to check for any large contamination is to determine its melting point, it should be very very close to its stated MP, shouldn't even be off by 1°.


What are the bond angles in cyclohexane?

The crystal structure shows a strangely distorted molecule with only $C_\mathrm{i}$ symmetry. Values have quite a wide range, $\angle(\ce{CCH})\approx 100-120^\circ$. See below for all values and xyz coordinates to read in with a molecular viewer. Source:
 R. Kahn, R. Fourme, D. André and M. Renaud, Acta Cryst. 1973, B29, 131-138.

A DF-BP86/def2-TZVPP calculation (gas phase, 0 K) gives a highl $D_\mathrm{3d}$ symmetric structure with
\begin{array}{ll}
\angle(\ce{CCH_{ax}}) & =109.1^\circ,\\
\angle(\ce{CCH_{eq}}) & =110.3^\circ,\\
\angle(\ce{CCC}) &=111.5^\circ,\\
\angle(\ce{H_{eq}CH_{ax}}) &=106.4^\circ.\\
\end{array}
 Also Bond length differ very slightly (and impossibly to determine at room temperature and pressure), with
\begin{array}{ll}
\mathbf{d}(\ce{CC}) &=1.537~\mathrm{\mathring{A}},\\
\mathbf{d}(\ce{CH_{ax}}) &=1.105~\mathrm{\mathring{A}},\\
\mathbf{d}(\ce{CH_{eq}}) &=1.103~\mathrm{\mathring{A}}.\\
\end{array}
The tiny differences can be explained with Bent's rule.
In principle the assignment Goeff already gave you is absolutely valid, especially if you take the dynamic nature of the molecules into account that they have at standard temperature and pressure.
Also see xyz coordinates for visualisation below.

Appendix

Crystal structure

\begin{array}{lr}
R(1-2)          &1.528 \\
R(1-3)          &1.519 \\
R(1-4)          &0.884 \\
R(1-5)          &1.146 \\
R(2-6)          &1.055 \\
R(2-7)          &0.929 \\
R(2-10)         &1.521 \\
R(3-8)          &1.100 \\
R(3-9)          &0.944 \\
R(3-11)         &1.521 \\
R(10-12)        &1.519 \\
R(10-13)        &1.100 \\
R(10-14)        &0.944 \\
R(11-12)        &1.528 \\
R(11-15)        &1.055 \\
R(11-16)        &0.929 \\
R(12-17)        &0.884 \\
R(12-18)        &1.146 \\
A(2-1-3)        &110.4 \\
A(2-1-4)        &106.7 \\
A(2-5)        &105.8          \\
A(1-2-6)        &106.6          \\
A(1-2-7)        &118.2          \\
A(1-2-10)       &111.3          \\
A(3-1-4)        &103.0          \\
A(3-1-5)        &114.0          \\
A(1-3-8)        &110.6          \\
A(1-3-9)        &116.9          \\
A(1-3-11)       &112.4          \\
A(4-1-5)        &116.7          \\
A(6-2-7)        &109.5          \\
A(6-2-10)       &110.6          \\
A(7-2-10)       &100.5          \\
A(2-10-12)      &112.4          \\
A(2-10-13)      &109.1          \\
A(2-10-14)      &100.4          \\
A(8-3-9)        &106.9          \\
A(8-3-11)       &109.1          \\
A(9-3-11)       &100.4          \\
A(3-11-12)      &111.3          \\
A(3-11-15)      &110.6          \\
A(3-11-16)      &100.5          \\
A(12-10-13)     &110.6          \\
A(12-10-14)     &116.9          \\
A(10-12-11)     &110.4          \\
A(10-12-17)     &103.0          \\
A(10-12-18)     &114.0          \\
A(13-10-14)     &106.9          \\
A(12-11-15)     &106.6          \\
A(12-11-16)     &118.2          \\
A(11-12-17)     &106.7          \\
A(11-12-18)     &105.8          \\
A(15-11-16)     &109.5          \\
A(17-12-18)     &116.7          \\
\end{array}

C        2.329589000      3.011988000     -0.044238000
C        3.630888000      2.632672000      0.661249000
C        1.367908000      1.836044000     -0.049671000
H        2.524936000      3.113740000     -0.900292000
H        1.933922000      3.936128000      0.506802000
H        3.388552000      2.469096000      1.674854000
H        4.348450000      3.220000000      0.605369000
H        1.030575000      1.607424000      0.972471000
H        0.592407000      1.921052000     -0.581309000
C        4.247092000      1.383956000      0.049671000
C        1.984112000      0.587328000     -0.661249000
C        3.285411000      0.208012000      0.044238000
H        4.584425000      1.612576000     -0.972471000
H        5.022593000      1.298948000      0.581309000
H        2.226448000      0.750904000     -1.674854000
H        1.266550000     -0.000000000     -0.605369000
H        3.090064000      0.106260000      0.900292000
H        3.681078000     -0.716128000     -0.506802000

Computed (DF-BP86/def2-TZVPP)

C       -1.270112000     -0.733300000      0.229506000
C        0.000000000     -1.466599000     -0.229506000
C       -1.270112000      0.733300000     -0.229506000
H       -1.328386000     -0.766944000      1.332131000
H       -2.167276000     -1.251277000     -0.145050000
H        0.000000000     -1.533888000     -1.332131000
H        0.000000000     -2.502555000      0.145050000
H       -1.328386000      0.766944000     -1.332131000
H       -2.167276000      1.251277000      0.145050000
C        1.270112000     -0.733300000      0.229506000
C        0.000000000      1.466599000      0.229506000
C        1.270112000      0.733300000     -0.229506000
H        1.328386000     -0.766944000      1.332131000
H        2.167276000     -1.251277000     -0.145050000
H        0.000000000      1.533888000      1.332131000
H        0.000000000      2.502555000     -0.145050000
H        1.328386000      0.766944000     -1.332131000
H        2.167276000      1.251277000      0.145050000


I am talking about dissolving in water since an acid needs some sort of solvent to dissociate.

No, it doesn't. You neglect autoprotolysis where the water-free acid itself is the solvent.

\begin{align}
\ce{2H2SO4 &<=> H3SO4+ + HSO4-}\\
\ce{3HF &<=> H2F+ + HF2-}
\end{align}


Why does the Gibbs Free Energy change determine spontaneity of a reaction?

By the second law of thermodynamics, your main goal is to ensure that $$\Delta S_{\mathrm{universe}} \geq 0$$

$$\Delta S_{\mathrm{universe}} = \Delta S_{\mathrm{sys}} + \Delta S_{\mathrm{external}}$$

$\Delta S_{\mathrm{sys}}$ is measurable on your system. $\Delta S_{\mathrm{external}}$ is a bit more challenging, since you're not going to count the microstates of the external system, so you compute indirectly by heat transfer to the surrounding. Thus,

$$\Delta S_{\mathrm{external}} = -\frac{\Delta H}{T}$$

Combine:

$$T\Delta S_{\mathrm{universe}} = T\Delta S_{\mathrm{sys}} - \Delta H$$

or

$$\Delta G = -T\Delta S_{\mathrm{universe}} = \Delta H -T\Delta S$$

And with the sign flip, you're looking for $\Delta G \leq 0$.

Importantly, I should point out that this is not a closed system in the sense that we assume constant-pressure and not constant-volume.

If you're looking at a process that is constant-volume, you need to replace $\Delta H$ (which has pressure-volume work compensation) with $\Delta U$ (which does not) and replace $\Delta G$ (Gibbs free energy, constant pressure) with $\Delta F$ (Helmholtz free energy, constant volume).


Steps in mechanism of pyruvate dehydrogenase

The ylide of thiamine pyrophosphate does not ‘form’; rather it is there. There are two major mesomeric structures for the five-membered ring: One including a double bond from nitrogen to the carbene carbon, the other one with a single bond (lone pair on nitrogen, sextet on the carbene carbon). As with mesomeric forms of benzene, the truth is somewhere between these two extremes.

In the carbene elimination step, you seem to assume just by the existence of the arrows that the deprotonation of the hydroxide must come first. That is not the case. In fact, I would assume the first arrow to be the recreation of the keto unit, followed by the removal of the thiamine. Only then would the proton be lost, since we are now dealing with a protonated ketone which is a very strong acid.


Why do tetrahedral complexes have approximately 4/9 the field split of octahedral complexes?

The ratio is derived in The angular overlap model. How to use it and why J. Chem. Educ., vol. 51, page 633-640.

First, relative expressions (Shaffer Angular Overlap Factors) are derived for overlap integrals between metal and ligand orbitals as a function of angles.

For ligand sigma orbitals, the expressions are a functions of 2 angles, specifying the direction of the ligands in the coordinate system of the metal.

For pi-x and pi-y ligand orbits, there is a third angle for orientation about the ligand-metal axis.

Then, assuming the ligand-metal distances are the same in both the tetrahedral and octahedral cases, and considering the coordinate geometry of the ligands in the two cases, the relationship:

$$\Delta_{T} = -4/9 \Delta_{O}$$

is obtained.

For another discussion of this ratio in the framework of the angular overlap model, as well as extension to other geometries such as square planar, see A New Look at Structure and Bonding in Transition Metal Complexes, Advances in Inorganic Chemistry and Radiochemistry, Volume 21, pages 113-146.


Why can't Pd/C and H2 reduce both the alkene and carbonyl portions of α,β-unsaturated carbonyls?

You hit it right on the nose. The real key piece of information is that given enough time, all the unsaturated bonds will be reduced. This tells you that though the reduction is thermodynamically favorable, it is the difference in the energy barriers ($\ce{\Delta \mathrm{G^{‡}}}$) that prevents the carbonyl reduction from occurring at the same rate as the alkene reduction. This means to produce the alcohol faster, we must manipulate the kinetics of the reaction.

To understand why the carbonyl-reduction transition state is higher in energy, we should consider the differences between the carbonyl and alkene bonds. The bonds are much more polarized, and as such the carbonyl $\pi$ bond is considerably stronger than alkene ($93\ \mathrm{kcal\ mol^{-1}}$ vs. $63\ \mathrm{kcal\ mol^{-1}}$)$^{\mathrm{[1]}}$. This means that at low temperatures, only a small fraction of the molecules have enough energy for the carbonyls to associate onto the $\ce{Pd/C}$ surface. Catalytic hydrogenation of carbonyl compounds (aldehydes, ketones, and especially esters) requires high temperatures and pressures to increase the presence of both the substrate and hydrogen on the catalyst.

$^{\mathrm{[1]}}$ Fox, M. A.; Whitesell, J. K. Organic chemistry; Jones and Bartlett: Sudbury, MA, 1997.


The first step is to set the discriminant equal to zero, and thereby solve for the critical value of $r_5$. When I did this, I got $$r_5a_0=\frac{r_2c_1}{1+2\sqrt{\frac{r_2r_3}{r_1r_4}}}$$.  The next step is to substitute this into the equation for m, and find the two  corresponding equal values of m.


You are aware that, from the open system version of the first law of thermodynamics, the shaft work per unit mass must be equal to the change in enthalpy per unit mass, correct?  So you need to determine the change in enthalpy per unit mass between the gas entering the turbine and the gas exiting the turbine.  Do we agree so far?

We know that enthalpy is a function of state.  So we need to figure out a convenient path between state 1 at 100 bar, 600 K, and state 2 at 20 bar ,445 K such that the equation provided can easily be applied to determine the change in enthalpy.  Consider the following 3 part path:

Hold the temperature at 600 K and lower the pressure to say 1 bar, where the gas behaves as an ideal gas.  Use the equation provided with dT set to 0 to evaluate $\Delta h$ for this step
Hold the pressure constant at 1 atm, and use the ideal gas heat capacity to calculate $\Delta h$ for cooling the gas down from 600K to 445K.
Hold the temperature constant at 445K and raise the pressure from 1 bar to 20 bars.  Use the equation provided with dT set to 0 to evaluate $\Delta h$  for this step.

The sum of the enthalpy changes for these three steps adds up to the enthalpy change for the overall change.


Are monosubstituted cyclopropanes achiral or chiral?

Here is a 3-D conformer from PubChem

As you can clearly see, a plane of symmetry can be sent along the black line perpendicular to the plane of the screen. Hence, the molecule is achiral. If you take a mirror image, you can ultimately super-impose it again on the parent form

Here is an illustrative 3D image(courtesy of andselisk) which clearly shows the plane of symmetry in pink and ring plane in blue.


Theoretical plates in Plate theory of chromatography

I shall take the example of the mixture ethanol-water, first without chromatography. If I start distilling 250 mL of a mixture 10% ethanol and 90% water, the solution boils at 96°C, and the vapor contains 55% water + 45%  ethanol. This vapor is condensed by passing through the cold condenser. So the first fraction of maybe 10 mL condensed vapor contains 45% ethanol. Suppose 250 mL of this first mixture (45% ethanol) is heated in another distilling apparatus. At maybe 85°C, the liquid boils and the first vapor will contain 70% ethanol and 25% water. If this vapor is condensed, the liquid so obtained contains 70% ethanol. These distillation - condensation operations can be repeated, with the following results.
1st distillation. Original liquid : 10% ethanol. Distilled liquid : 45% ethanol
2nd distillation. Original liquid : 45% ethanol. Distilled liquid : 70% ethanol
3rd distillation. Original liquid : 70% ethanol. Distilled liquid : 83% ethanol
4th distillation. Original liquid : 83% ethanol. Distilled Liquid : 90% ethanol.
etc.
nth distillation. Original liquid : 96% ethanol. Distilled liquid : 96% ethanol.
Now suppose all these distillations are carried out in the same apparatus. Instead of having the vapor go through a cold tube and be recovered in a flask out of the condenser, imagine that the vapor is sent into a long vertical tubes situated just above the first container half filled with the mixture 10% ethanol. Suppose this vertical tube contains a lot of successive floors or plates, punched with one or two small holes. In the beginning the vapor made of 45% ethanol will pass through the holes of the first floor or plate, and will be condensed just above this level and may partly go down to the bottom of the setup. But a fraction of this liquid will now boil again and produce a vapor made of 70% ethanol. This vapor passes through the second floor and be condensed just above the second floor. On this second floor the liquid (containing 70% ethanol) will later on be evaporated and produce a vapor containing 83% ethanol. And so on. After n floors the vapor and the liquid will contain 96% ethanol. As a result, the separation ethanol-water improves if the number of floors or plates increases. And you may imagine that the height of the room between two floors is critical. If it is too small, it will be immediately filled. If it is too big, some secondary condensation will happen on the wall of each room.
Now ethanol is not the best example for my purpose because the successive distillations cannot get to pure ethanol. Suppose for the rest of my description that after n distillations, a pure amount of a solvant X can be obtained. Suppose that to get a better that 99% separation you need 8 floors, which are all 5 mm high.
Suppose now that instead of having used a distillation apparatus, you are using of chromatographic column. In order to get a better than 99% separation of pure compounds you need a column measuring 60 cm. The plate volume is : 60 cm / 8 plates = 7.5 cm. Of course, if you use a longer column, the separation might be 99.9% or better.
Of course this explanation is an over-simplification of the reality. But it may help you understand what is a plate volume.
Did you follow me ?


What type of chemical reaction is it when sodium bicarbonate is added to citric acid?

The first step is usually called an acid-base reaction or a proton transfer. Any acid of similar strength to citric or acetic (vinegar) acids will react with bicarbonate this way, and it is very fast in water solution at room temperature. The products are carbonic acid and the sodium salt of the acid used--for example, sodium citrate.

Carbonic acid is energetically unstable relative to carbon dioxide and water, to which it is rapidly converted via what most chemists would call an elimination reaction. I used to think that this occurs instantly, but carbonic acid is a species that actually exists in solution and can be observed and studied.

While there are many named organic reactions, I  agree with Ivan that there is no one universally agreed upon way to name them, so others may wish to weigh in with their opinions about this reaction.


Calorific values are usually quoted under standard conditions making the distinctions a matter of preference and convenience

The apparently different units used in quoting heat of combustion are not as different as they appear. We need to know heat of combustion for many reasons (eg how to bill domestic customers for the correct amount of heating or cooking their gas supply delivers). But the units used are all equivalent once we define a standard set of conditions when measuring them.

For example, a mole of pure methane at standard temperature and pressure (STP) occupies 22.4 L and will weigh about 16g and will burn in air releasing about 890kJ of energy. This means that, if we stick to the same conditions (STP in this case) we can interconvert the energy/mol, energy/mass and energy/volume with a simple calculation.

Which one we choose to use is a matter of convenience. Domestic gas meters might choose the energy/volume as they measure volume flow at a standard pressure and temperature. Liquefied gas ships may choose to measure by mass as the methane is mostly liquefied and the mass may be easier to measure than the volume, temperature or pressure (but, knowing the mass, we can work out the volume at STP to work out how much domestic gas can be produced). If a chemist is working on some reaction scheme, then they will probably prefer molar metrics as they make balancing the equations easier.

But the essential point is that a given amount of substance always has the same heat of combustion. How you measure the amount is a matter of convenience.


Explanation for why nickel turns green in hydrochloric acid

The question is really badly worded. For starters, let’s look at solutions of nickel(II):

Figure 1: Nickel(II) solutions. From left to right: $\ce{[Ni(NH3)6]^2+}$, $\ce{[Ni(en)3]^2+}$, $\ce{[NiCl4]^2-}$, $\ce{[Ni(H2O)6]^2+}$. Image taken from Wikipedia, where a full list of authors is available.

You can ignore the left two but the rightmost is a standard nickel(II) solution (could be $\ce{NiSO4}$) while the second from the right is a chlorido-complex—basically $\ce{NiCl2}$ but dissolved in hydrochloric acid. I wouldn’t call the chloride solution ‘yellow’, more like ‘yellowish-green’ but your mileage may vary. Next, let’s look at solids.

Figure 2: salts of nickel(II) chloride. Left or top: hexahydrate. Right or bottom: anhydrous. Images taken from Wikipedia (hexahydrate, anhydrous), where a full list of authors is available.

The picture on the right is dry, i.e. completely water-free nickel(II) chloride (anhydrous). On the left, the salt contains tightly bound water molecules which can be removed by further drying although the salt itself would be dry in everyday terms; this is the hexahydrate of nickel(II) chloride.

It might seem easy that the salt on the left gives a greenish solution while the salt on the right gives a yellowish one. Here’s the catch: they should both give a greenish solution, because there is no excess of chloride as was required in the solution in figure 1.

But what about the actual question? Well, dissolving nickel in hydrochloric acid means that there should in fact be excess chloride, meaning that yellow (or yellowish green) would be the correct answer.

tl;dr: That question is a perfect opportunity to complain if it was marked as ‘incorrect’.


What are bond lengths and angles for picric acid?

I would suggest to take a crystal structure available for free from Crystallography Open Database, COD#5000146. The entry webpage already has a JSmol script loaded, so you should see the molecular structure loaded on a black area. Selecting RMB (right mouse button) -> Measurements -> Show Measurements allows you to measure linear and angular parameters with mouse clicks.

Alternatively, you can download the CIF-file (5000146.cif) and load it with any crystallographic viewer if you prefer to work offline. For beginners I'd recommend Mercury, there is a free version available for Windows, macOS and Linux:

N. B. The asymmetric unit in this structure contains two molecules of picric acids – with and without H-atoms. Obviously, you need the one with the protons assigned.


What is the product obtained when methyl vinyl ketone is reacted with 1,2-ethanedithiol and the product is further hydrogenated with Raney Nickel?

The answer could be 3 or 4.
1,2-ethanedithiol will react with the carbonyl group to form a dithiane (similar to the ethylene glycol reaction). Reaction of this with Raney Nickel/H2 desulfurises it, so the whole process is effectively a reduction of the carbonyl group to CH2. It depends on the reaction conditions whether the double bond survives. Raney Nickel can reduce alkenes.
You can read more about it here 1 and here 2


What are the fundamental principles that are found throughout chemistry?

This is an inherently subjective question. It's very difficult to come up with an answer that isn't just a fundamental principle of physics. My personal opinion is the following:

  Chemistry is the study of interacting electrons.

Now, obviously the electrons are only near enough to interact because the positive charges of nuclei don't let the electrons fly away. Fundamentally, however, all of chemistry can be described in terms of electron densities, which is the basis of density functional theory.

This is not just a theoretical construct either. In every field, people are talking about different types of bonds and different electronegativities. If you think about organic chemistry, reaction mechanisms are described in terms of electron-pushing, which is some highly abstract way of talking about the dynamics one might find if you could solve the time-dependent Schrodinger equation for arbitrary lengths of time.

Even things that seem to have nothing to do with electrons, such as nuclear magnetic resonance, are often useful because they key in information about the where electrons are localized via chemical shifts.

Also, this definition manages to distinguish chemists from physicists because very often physicists don't care about solving for the energy of something to the sixth decimal place. Whereas, as a computational chemist, I sometimes worry that I might actually need to get that electronic energy out to the ninth decimal place...

So, my opinion is that chemistry is the study of interacting electrons, and chemists care very deeply about the details of those interactions.


Is a compound aromatic if it also has anti-aromatic rings?

The molecule actually optimizes to a planar, delocalized $\pi$ system

Using B3LYP/6-31G(d) optimizations, we see that the molecule optimizes to become planar. (The initial configuration was intentionally bent.)

Moreover, if I plot the HOMO, it's highly delocalized:

Similarly, the LUMO is also delocalized across the entire molecule:

Consequently, I don't see any indication that the "side" benzene rings are localized from the cycloocta-1,5-diyne ring. The bond lengths are in fairly good agreement with the crystal structure:

Here's the crystal structure from J. Am. Chem. Soc., 1975, 97 (3), 658–659. Note there is a profound triple bond character, but the bridging bonds (1.44 Å) are somewhere between a typical aromatic bond length (1.40 Å) and a full single bond (~1.54 Å).


Can HI/P reduction be used to reduce nitroalkanes and nitroarenes to alkyl amines and aryl amines?

I've never seen the $\ce{HI/P}$ method being used for the reduction of aliphatic or aromatic nitro compounds.

  I learnt from my textbook that heating the nitroalkanes and nitroarenes with $\ce{H2/Pt}$ is sufficient for the above conversion.

I've reduced a lot of nitroarenes to their corresponding amines by catalytic hydrogenation under standard conditions in ethanol (or methanol) over $\ce{Pd/C}$. In the cases I examined, heating wasn't necessary.


Basic Understanding of Relative Atomic Mass

It's because there's two hydrogen atoms in water, not one. In order to show you, let's assume we're making HO instead of H2O. This would clearly only require half the amount of hydrogen, but still the same amount of oxygen. Thus we have .5 grams of hydrogen combining with 8 grams of oxygen in a 1:1 relationship. Thus the oxygen weighs 16 times more than the hydrogen. Do you understand?

In H2O they combine in a 2:1 relationship. We have 1 gram of hydrogen combining with 8 grams of oxygen, but because there's 2 hydrogen for every oxygen, we actually need TWICE as many oxygen atoms to have the same amount of atoms. In other words, there will be the same number of atoms in 1g of hydrogen as there is in 16g of oxygen.


Standard Enthalpy of Formation of H2(g) vs H(g)

The value may be used whenever atomic hydrogen is present as a (probably intermediate) species. E.g. take the reaction

$$\ce{2H <=> H2}$$

which is the basic reaction in atomic hydrogen welding.


Solute does not mean adsorbate always.An Adsorbate is a substance that is adsorbed  on a surface of another substance. In a dilute solution, negative adsorption occurs as solvent is adsorbed by the adsorbent and not solute. In positive adsorption, solute is adsorbed.


Determine the concentration of the original NaOH solution

I am going to need a unit for the 1.06. The question needs you to figure out the amount in moles of Nickel Hydroxide. Then multiply this by two as two moles of NaOH are required to form one mole of the Nickel precipitate. Then this is the number of moles that was surplus in the reaction with HCl (NaOH is added in excess - the key to the question). Find the number of moles of HCl using the volume and concentration. Add this number to the number of moles that were used to form the Nickel precipitate and then find the concentration using the moles and the volume.


The value $\pu{1.3653 mol L^{−1} atm^{−1}}$ is the solubility constant (or Henry's law solubility constant), not the solubility. The solubility is defined as the maximum possible concentration (the saturation concentration) of a solute under given solution conditions (e.g. temperature and pressure), whereas the solubility constant $H^{cp}$ defines how solute partitions between the gas (concentration in gas here given as the partial pressure in atmospheres — although this is not stated in the OP it can be inferred from the expected answer) and solution phases (concentration in $\pu{mol/L}):$

$$H^{cp} = \frac{c}{p}= \pu{1.3653 mol L^{−1} atm^{−1}}$$

Assuming atmospheric pressure of 1 atm then $\pu{10 ppm}$ translates into a partial pressure of $\pu{1e-5 atm}$ (if the atmospheric pressure is $p_T$, then the partial pressure of $\ce{SO2}$ is $p = p_T \cdot 10^{-5}$) and a solution concentration $\pu{1.365e-5 M}$  $\;(c=H^{cp} p)$. Proceeding from the definition of $\mathrm{p}K_\mathrm{a}$ the degree of dissociation $(\alpha=0.99886)$ can be computed and from this the $\mathrm{pH} = 4.865,$ which agrees with the expected result.

You should check solubility limits when evaluating this sort of problem, but since the solubility is not given you can ignore this and assume that the limit has not been reached.


How to neutralise path weeder contamination

Amitrole is slightly toxic to various species of freshwater fish$^1$. However, it's quite water soluble and should not be difficult to remove from most surfaces with simple warm-water washing.

Ammonium Thiocyanate is particularly water soluble and should also readily be removed via warm-water washes. Thiocyanate ion has lower solubility with some other cations that would be present, but in general, thiocyanates are quite water soluble. Thiocyanate ion has been shown to be toxic to fish at the tens of mg per liter level, however, $^2$.

Simazine has a very low water solubility, on the order of 5 mg/L (4-5 orders of magnitude less soluble than the other compounds) but has been shown to be lethal to fish at levels on the order of its limit of solubility$^3$. I strongly suspect that residual simazine is thus to blame for the persistent toxicity to your fish.

Simazine is about 100 times more soluble in isopropyl alcohol (rubbing alcohol) than it is in water. I suggest obtaining some 99% isopropyl alcohol and scrubbing/soaking all of the decorative material that you value, as well a the tank itself, thoroughly rinsing the alcohol off with warm water, then drying completely. There is no reason to believe that vinegar would be useful in this capacity.

Anything relatively cheap should probably simply be replaced, particularly difficult to clean items like driftwood, high-surface area gravel or sand, and of course all of the filter media. My biggest concern of the items you mentioned is the crushed coral. If it were me, after the cleaning I would return everything to the tank except the crushed coral and once the water is ready, test it out with a couple fish. If things seem OK for a while, then I would set up the tank with everything including the crushed coral.

As a former aquarist, I understand how much work is involved with my suggestions. I strongly believe, however, that you have identified the culprit in the fish die-offs and that the procedure above has a good chance of resolving the issue. Please don't hesitate to ask for any clarifications in the comments below.

US EPA. 1984 (March). Amitrole: Pesticide Registration Standard and Guidance Document. Office of Pesticides and Toxic Substances, US EPA, Washington, DC.
Speyer, M. R. and Raymond, P. (1988), The acute toxicity of thiocyanate and cyanate to rainbow trout as modified by water temperature and pH. Environmental Toxicology and Chemistry, 7: 565–571. doi:10.1002/etc.5620070705
WSSA Herbicide Handbook Committee. Herbicide Handbook of the Weed Science Society of America, 6th Ed. WSSA, Champaign, IL. 1989


The Kirk-Othmer Encyclopedia of Chemical Technology reports that: "About 60% is made by the [Bayer] hypochlorite–ketazine process, 25% by the peroxide–ketazine route, and the remainder by the Raschig and urea processes." Here is a link to the original article, if you can access it (e.g. at a university library): Hydrazine and Its Derivatives, by Eugene F. Rothgery. The data cited dates back to 2004, so it may have changed somewhat, but probably not too much.

Consider this also: the above makes hydrazine hydrate. Propellant-grade hydrazine must be anhydrous, so there is an extra step in which water is removed from the hydrate, typically by azeotropic distillation with aniline.


Carbocation rearrangement involving three membered rings

Unfortunately you have opened the can of worms that is cyclopropylmethyl carbocations.

It is known that the parent unsubstituted cyclopropylmethyl carbocation is non-classical, and leads to mixtures of cyclopropylmethyl, cyclobutyl, and homoallyl derivatives. However, the presence of substitution can alter the electronic structure quite significantly. There is too much on the topic to discuss here; the interested reader may find reviews for themselves in George Olah's books Onium Ions and Carbonium Ions.

In this case, I would say that it is virtually impossible to predict which major product will be formed. Perhaps that's not the answer you wanted, but I don't want to propagate fake knowledge and claim that we can predict something that we can't.

Looking up this reaction on Reaxys, I find that it has not been done before. However, the reaction of the corresponding alcohol with HBr has been reported before. In different people's hands and under different conditions, it gives different major products; both the cyclopropylmethyl bromide and the homoallyl bromide have been isolated. [Apparently not the cyclobutyl bromide, though; why? Again I don't have a good answer.]


Why is HCl the acid of choice for swimming pools?

There are several reasons not to consider these acids to be interchangeable for use in swimming pools. A total answer covers aspects of chemistry, biology, safety, availability, and cost. As hydrochloric, nitric, and sulfuric acids are all strong acids and can provide the desired acidification, they are very different when you consider their respective counter ions. Do you want chloride, nitrate, or sulfate in your pool water?

Safety First:

Hydrochloric, nitric, and sulfuric acid are all strong acids that are capable of causing chemical burns and present inhalation hazard from their fumes.
Additional safety information can be obtained from PubChem.:
Hydrochloric acid
Sulfuric acid
Nitric acid

Chemistry:

Nitric and sulfuric acids are more corrosive to copper and stainless steel. Copper (I) chloride is significantly less soluble in water than copper salts of nitric and sulfuric acids. Swimming pool heaters are likely to contain copper pipe.
As an added benefit for salt pools, which use an electrochemical chlorine generator, the Cl- ions are converted to chlorine for disinfecting the pool water.

Biology:

As pointed out in another response, nitrates can promote algae growth.

Availability and Cost:

Hydrochloric acid is commonly available as muriatic acid (31.5% HCl, ~10 molar) from home centers, building and construction suppliers, and hardware stores, because it is used in many construction-related activities, with cost ranging from \$5 - 9 (USD) per US gallon. A gallon will contain approximately 37 mole equivalents of H+.
Sulfuric acid is available from auto parts retailers as battery acid (37% H2SO4) and costs approximately \$10 (USD) per US quart ($40 per gallon). A gallon of battery acid will contain approximately 36 mole equivalents of H+ (adjusting for normality). Sulfuric acid costs 4-8 times more than hydrochloric acid for equivalent pH reducing capacity.
Nitric acid is less readily available to the consumer market. One on-line supplier offered reagent grade (70%) for \$125 (USD) per liter.

Added Note:

Cyanuric acid is a weak acid and serves as a stabilizer. Quoting from PubChem:

  Cyanuric acid is used in swimming pools to lower the rate of photochemical reduction of chlorine, hypochlorous acid and hypochlorite ion.


It's called a force sensing resistor. It's basically a film of small conductive particles mixed into a non-conductive polymer matrix on top of a conductive substrate. The more force on the film, the more conductive particles can touch the layer underneath, reducing the resistance. Sensitronics owns the original patent for the technology and their site shows how they're constructed, though you'll probably have to dig up the patent to find out exactly what's in the force sensing layer.


Isopropyl alcohol to clean germs on countertops

A 70% “rubbing” alcohol solution will kill those pesky bacterium usually in ten seconds or less. It will also kill a host of viruses, but not all of them.

For your common household disinfecting, a 70% solution will do the trick. Especially for e. Coli, salmonella, staph, and a few others that would be risk bacterium when preparing poultry, etc.

A final bit of information: 70% will be a more effective carrier than the 90-99% equivalent. Why? Because the water mix (30%) will act as a carrier or transport mechanism to get the alcohol into the bacterium cells in order to terminate them! The 90% mix simply is not very effective at penetrating the cell membranes.

See this for additional information:  https://www.cdc.gov/infectioncontrol/guidelines/disinfection/disinfection-methods/chemical.html


Selective alkylation of carboxylic acid versus phenol

Using the haloform reaction to convert the methyl ketone 1 to the corresponding carboxylic acid 2 is a neat idea. According to an old review by R. C. Fuson and B. A. Bull, published in Chem. Rev., 1934, 15, 275-309 (DOI), this seems to work for 4-hydroxyacetophenone. Your starting material 1 is a phenyl homolog of that.

In 2, you have two acidic protons ($\ce{-COOH}$ and $\ce{-OH}$). Based on our knowledge and by comparison with 4-hydroxybiphenyl ($\mathrm pK_\mathrm a = 9.55$), benzoic acid ($\mathrm pK_\mathrm a = 4.2$) and 2-phenylbenzoic acid ($\mathrm pK_\mathrm a = 3.46$) (all data are taken from the Rubber Book), you know that

the carboxylic acid is more acidic than the phenol
the difference is acidity is quite large

Potassium carbonate ($\ce{K2CO3}$) in a polar aprotic solvent, such as N,N-dimethylformamide (DMF) should be sufficient to deprotonate both centres.

Back to your question on the order of events:

I would add 1 equivalent $\ce{K2CO3}$ of to a solution of 2, stir at room temperature for a while and then add the propargyl bromide, assuming that the large differences in acidities will make sure that the carboxylic acid is deprotonated, while the phenol is not.

Monitoring the reaction by TLC will show whether the esterification is quantitative.

If this is the case, I would dare to run a one-pot reaction, add another equivalent of $\ce{K2CO3}$, stir again and then add the bromohexanol to form the ether.

It might be a good idea to try this first with very small amounts, just enough to check via TLC.


Why don't Schools follow latest IUPAC for naming groups?

The reason for this is that in 1988 (actually more than a few years back), IUPAC changed their naming system for the groups. Originally the transition metals didn't have a group number and weren't counted as a group. However now it is changed so that the transition metals do have a group number. So number instead of carbon being in group 4, it is in group 14. However despite this being changed quite a few years ago, you will still see in several websites and schools like your and mine still following the old IUPAC conventions and say that carbon is in group 4 rather than 14.

The reason that they do this I suppose is that it avoids confusion in students who are just picking up chemistry as they can see there is a clear relationship between valence electrons and group number. However besides this, schools do follow IUPAC conventions.

References:

Pure Appl. Chem. 1988, 60, 431–436, DOI: 10.1351/pac198860030431

(thanks @Loong)


Can we prevent metal corrosion by creating a vacuum around it?

The word corrosion has a range of meanings, but it is generally taken to indicate the chemical reaction of a material with its environment. For example iron corrodes by reacting with oxygen in the air (or water if it's underwater).

If you put the metal in a vacuum then it doesn't have anything around it to react with, so in this case yes putting the metal in a vacuum will stop it corroding.


Reagents used to prepare organic ICP standards, from scratch

Running an ICPOES or ICPMS on organic solvents is not easy, you may have problems such as coke / carbon deposits in the torch.

Also I would like to point out that sodium dicyclopentadienide is very sensitive to oxygen, it decomposes with great ease on contact with air. Ask almost anyone who has made ferrocene in the lab about this question.

I would like to suggest that you consider making an emulsion in water of the organic solution using a detergent such as triton-x100. I have heard of this method for running organic samples in a normal ICPMS but I have never tried using it.

In the ideal world what I would do would be to use a wet combustion (maybe use a acid digestion bomb such as a Parr 45 ml bomb) to convert the organic sample into a aqueous sample.

Another option would be to use neutron activation anaylisis, but these days it is very hard to find a strong neutron source suitable for this type of work. You would have to seal up samples in quartz tubes. I will warn you that even with a nuclear reactor and a good gamma spectrometer this type of work is far from trival.


This question may explain about what happens with the pennies at a microscopic level: What is the product of a copper and vinegar reaction?

Also, in his paper, Stephen DeMeo presents the electrochemical perspective:

  The mechanism that has been proposed involves the formation of a free radical and an electrochemical corrosion process (1). [...], suffice it to say that the mechanism does involve formation of a strong oxidizer, peroxoacetic acid ($\ce{CH3COOOH}$), from dissolved oxygen, as well as the production of a copper oxide and a copper ion.

He also explains why the reaction just happens in the presence of oxygen:

  Since copper is below hydrogen in the electrochemical series, copper is reluctant to displace hydrogen ions from electrolytes.  In terms of standard electrode potential, copper has a potential of 0.34 volts ($\ce{Cu -> Cu^{2+} + 2e-}$), while hydrogen has a greater potential of 0.00 volts ($\ce{2 H+ + 2e- -> H2}$). The electromotive force of the cell would be negative and therefore the reaction would be nonspontaneous under standard conditions. When oxygen is involved in the reaction between copper and acetic acid, the electrochemical series and the standard electrode potentials cannot be used as a guide to determine if the reaction is spontaneous.  A generalized relationship between copper metal and acids is nicely described by Massey: "Although copper metal will dissolve readily in an oxidizing acid such as nitric acid it will not dissolve in a nonoxidizing acid unless either an oxidizing agent or a suitable complexing agent is present also" (6).

Reference: DeMeo, S. Does Copper Metal React with Acetic Acid? J. Chem. Educ. 1997, 74, 844-846. DOI: 10.1021/ed074p844


RO- as leaving group in saponification process

$\ce{OH-}$ of course can, and will, be a leaving group here. The reaction is in equilibrium, $\ce{OH-}$ adds and leaves very many times, but that is a dead end as it does not drive the reaction forward. $\ce{RO-}$ can also be the leaving group - the difference between the basicity of $\ce{OH-}$ and $\ce{RO-}$ is not so great that it does not happen. Even if it happens one time in ten thousand, when it does the cascade it sets off makes the reaction irreversible so the reaction proceeds, if slowly, and the salt of the carboxylate is produced.


Bromination of hexene in presence of UV light or heat

Both products are formed although 3-bromocyclohexene is the major product.

Formation of major product: 3-bromocyclohexene

Under UV light, $\ce{Br2}$ undergoes homolytic splitting to generate $\ce{Br*}$ radicals:

$$\ce{Br2 ->[hv] 2Br*}$$

The formation of 3-bromocyclohexene is an example of substitution of alkanes, which require the free-radical mechanism:

In the first step of the upper mechanism, which is also the rate-determining step, a stable allyl radical is generated, which is stabilized by resonance:

As a result, the activation energy of the first step is significantly lowered.

Formation of minor product: 1,2-dibromocyclohexane

Individual bromine radicals are not electrophilic enough to attack the double bond in the cyclohexene, so the formation of 1,2-dibromocyclohexane requires the ions mechanism, typical for addition reactions (the lower mechanism in the following diagram).

The first step in this mechanism is the rate-determining step. In this step, bromine is ionized, which requires a moderate amount of activation energy, albeit still much higher than the rate-determining step of the upper mechanism.

Conclusion

Therefore, the upper mechanism occurs at a much faster rate than the lower mechanism, which makes the major product 3-bromocyclohexene and the minor product 1,2-dibromocyclohexane.

PS: Many people think that addition reaction is very fast. It is only true in water, where the bromine ion is stabilized by solvation in water.

Disclaimer: The mechanism I used for the addition reaction probably contradicts with your book. However, it does not matter. The point is that an ion is formed which makes the activation energy high.


Very interesting question and not very obvious. As can be seen from the picture, LAB's reduce only tertiary amides. It is stated in the publication(1) about LAB reductions:

  LAB reagents do not
  reduce primary and secondary amides even in refluxing THF.

In the same publication it is provided reason for different outcomes of reductions. They say that:

  We found that the outcome of these reductions depends on
  the sterics of the amide as well as of the LAB reagent

Which means that if LAB and/or amide is sterically hindered it can't interact with nitrogen like in compound 5

In lower part of the picture is described reaction where more hindered LAB leads to amine product 9

(1)Pasumansky, L., Goralski, C. T., & Singaram, B. (2006). Lithium aminoborohydrides: powerful, selective, air-stable reducing agents. Organic process research & development, 10(5), 959-970.


Angular Momentum of an orbital and orbit

Bohr Model came before the quantum theory and it was a success for explaining the non- continuous emission spectrum of excited hydrogen atoms. However, Bohr model was not sufficient for explaining many other phenomena related to the interaction of matter with light, such as the non-continuous  emission spectrum of multi-electronic atoms,  Zeeman Effect,... Therefore, the quantum theory is considered as a more complete theory.
The quantification of the angular orbital momentum should be linked to the second quantum number $l$ not $n$.

Up till now, we use Bohr model in chemistry textbooks in spite of its limitation. This is because it's simple and doesn't require complex mathematical tools as quantum theory.


Extraction of monosodium glutamate (MSG) from kelp

Monosodium glutamate was originally discovered by Kikunae Ikeda using a simple aqueous (hot water, namely his soup) extraction of kelp that was later refined and patented.  That particular species was Laminaria japonica. This along with a more detailed description of the original process and the history of MSG processing is given here.

The sulfuric acid treatment and precipitation process described therein certainly cannot be simply and safely done at home. The main purpose of that is to convert other compounds (e.g. glutamine) into glutamate to give higher yields of pure MSG for commercial usage. As stated in this Wikipedia article:

  "MSG is freely soluble in water",

so, as Ikeda enjoyed the water extract, I would suggest trying that first.  If you need a more concentrated extract, simply let the water dry.  I wouldn't use more than mild heat, maybe an oven on very low setting with a fan, to prevent degradation of your product.

An interesting side note, stated in both sources above, is that Ikeda's discovery of MSG was also recognized as the discovery of a new flavor at that time (in addition to sweet, salty, sour and bitter).


Hybridisation of cobalt in tris(oxalato)cobaltate(III)

As explained in The Chemistry of Iron, Cobalt and Nickel: Comprehensive Inorganic Chemistry, at pages 1104 and 1105:

almost all cobalt(III) complexes are low spin ... only with fluoride ions as ligands are high spin complexes ... found

In table 10, the book specifically lists $\ce{[Co(C2O4)3]}^{3-}$ as low spin and cites to J. Chem. Soc. (A) (1966) 798.
Therefore, according to the historical valence bond theory of transition metal complexes, it would be considered $\mathrm{d^2 sp^3}$ for the following reason:
There are (in the context of the theory) two unoccupied $\ce{3d}$ atomic orbitals in $\ce{Co^{3+}}$.
Six pairs of electrons from the oxalate ions are added to the two $\mathrm{3d}$ orbitals and the $\mathrm{4s}$ and three $\mathrm{4p}$ orbitals.


Finding direction of the reaction using Affinity of the reaction

Firstly, since $\overline{\mu_\alpha} = \overline{g_\alpha}$, the two approaches that you propose are basically the same: $\overline{\mu_\ce{CO2}}-\overline{\mu_\ce{CO}} -\frac{1}{2}\overline{\mu_\ce{O2}}$ is the same as the change in $G$ per mol of forward reactions.

Let's suppose that we can calculate the true, absolute value of the molar Gibbs energy at some reference physical and chemical state, labelled $^*$. The molar Gibbs energy at any arbitrary state is then

$$ \overline{g} = \overline{g}^* + \Delta \overline{g}^\text{chem} + \Delta \overline{g}^\text{phys} $$

Where the to $\Delta g$ terms account for the fact that the state of interest may have a different chemical and/or physical state than the reference state. If we substitute this back into your expression, we find

$$\begin{aligned} A &=\overline{\mu}_\ce{CO2}-\overline{\mu}_\ce{CO} -\frac{1}{2}\overline{\mu}_\ce{O2}
\\
&=\overline{g}_\ce{CO2}-\overline{g}_\ce{CO} -\frac{1}{2}\overline{g}_\ce{O2}
\\
&=(\overline{g}^*_\ce{CO2}+\Delta\overline{g}_\ce{CO2}^\text{chem}+\Delta\overline{g}_\ce{CO2}^\text{phys})
-(\overline{g}^*_\ce{CO}+\Delta\overline{g}_\ce{CO}^\text{chem}+\Delta\overline{g}_\ce{CO}^\text{phys})
-\frac{1}{2}(\overline{g}^*_\ce{O2}+\Delta\overline{g}_\ce{O2}^\text{chem}+\Delta\overline{g}_\ce{O2}^\text{phys})
\end{aligned}
$$

If we collect the $g^*$ components, their sum is

$$ \sum \overline{g}^* = \overline{g}_\ce{CO2}^*-\overline{g}_\ce{CO}^* -\frac{1}{2}\overline{g}_\ce{O2}^* $$

But then, by definition of the reference chemical state,

$\overline{g}^*_\ce{CO2} = \overline{g}^*_\ce{C} + \overline{g}^*_\ce{O2}$ (rearranging the $\ce{CO2}$ into $\ce{C}$ and $\ce{O2}$ does not change $g^*$ because $g^*$ depends on the reference chemical state, not the actual chemical state)
$\overline{g}^*_\ce{CO} = \overline{g}^*_\ce{C} + \frac{1}{2} \overline{g}^*_\ce{O2}$ (rearranging the $\ce{CO}$ into $\ce{C}$ and $\ce{O2}$ does not change $g^*$ because $g^*$ depends on the reference chemical state, not the actual chemical state)

Substituting this back:

$$\begin{aligned} \sum \overline{g}^* &= \overline{g}_\ce{CO2}^*-\overline{g}_\ce{CO}^* -\frac{1}{2}\overline{g}_\ce{O2}^*
\\ &= (\overline{g}^*_\ce{C} + \overline{g}^*_\ce{O2}) - (\overline{g}^*_\ce{C} + \frac{1}{2} \overline{g}^*_\ce{O2}) - \frac{1}{2}\overline{g}_\ce{O2}^*
\\ &= 0
\end{aligned}$$

(The term cancels to zero because the reaction is balanced)

Thus $A$ has no dependence at all on the absolute value of $g$. We can cancel the sum of the $g^*$ components and write

$$ A = (\Delta\overline{g}_\ce{CO2}^\text{chem}+\Delta\overline{g}_\ce{CO2}^\text{phys})
-(\Delta\overline{g}_\ce{CO}^\text{chem}+\Delta\overline{g}_\ce{CO}^\text{phys})
-\frac{1}{2}(\Delta\overline{g}_\ce{O2}^\text{chem}+\Delta\overline{g}_\ce{O2}^\text{phys}) $$

In practice, we typically define state $^*$ by:

Physical State: $T = 25^\circ\,\text{C}$, $P = 1 \, \text{atm}$
Chemical State: pure elements in their most stable form at $T = 25^\circ\,\text{C}$, $P = 1 \, \text{atm}$

and then re-label/re-name the chemical component of $\overline{g}$:

$\Delta \overline{g}^\text{chem} \rightarrow \overline{g}^\circ_f$ the "Gibbs energy of formation" (change in Gibbs energy associated with forming the substance from pure elements at the reference physical state)

To actually evaluate this, you'd

Look up $\sum \overline{g}^\circ_f$ for this particular reaction (which might be labelled as $\Delta G$) in a table, if one is available. If one isn't, you could look up the individual species' $\overline{g}^\circ_f$ (if you have tables which present them), or calculate them from the species' $\overline{h}^\circ_f$ and $\overline{s}^\circ$ (again, from tables).
For each species, evaluate $\Delta \overline{g}^\text{phys}$ (the difference in $\overline{g}$ associated with changing the species' physical state from $T = 25^\circ\,\text{C}$, $P = 1 \, \text{atm}$ to the $T$ and $P$ of interest) using some property model (equation or table) for each substance in turn

My Notation: For some extensive quantity $X$

$x$ represents $X$ per mass
$\overline{x}$ represents $X$ per amount


What does peroxide value exactly measure?

The test measures the amount of hydroperoxide by a redox reaction with iodide:

$\ce{2ROOH + 2H+ +2I- -> 2ROH + I2 + H2O}$

In the course of the reaction, the hydroperoxide is reduced, while iodide is oxidized to iodine. In principle, this is a variation of the iodine clock reaction of hydrogen peroxide.


How to determine the temperature dependence of equilibrium constants?

I agree that the second question seems a bit leading, but personally I would just calculate all quantities of interest as you laid on in your plan. The van't Hoff equation requires the reaction enthalpy change, so I assume using a tabulated value is fair game. If so, you will know whether the forward reaction is endo- or exothermic. If you calculate the equilibrium constant at the unknown temperature as you planned, you can compare it to the equilibrium constant given in the problem (for 727 C). Comparing these two equilibrium constants will tell you which way the reaction proceeded, and with the sign of the enthalpy, you should be able to deduce whether heat was added or lost... i.e. whether the new temperature is higher or lower than the initial state, 727 C (without having to go through the entire calculation).


What does the salt actually do during electrolysis?

A salt like $\ce{Na2SO4}$ is essential in electrolysis. It provides ions $\ce{Na+}$ and $\ce{SO4^{2-}}$ which are attracted by the electrodes in the solution and migrate to them. When they arrive near the electrodes, they are not discharged. But they neutralize the charges of the ions that are produced out of water being destroyed at these electrodes. Let's see what's going on at each electrode.
At the cathode (negative pole), water is decomposed according to : $\ce{2 H2O + 2 e- -> H2 + 2 OH-}$. This is only possible if there are some positive ions available near the cathode to compensate the negative charges appearing in the solution with these newly created $\ce{OH-}$ ions. Without positive ions like $\ce{Na+}$ near the cathode, these new $\ce{OH-}$ ions would repel further electrons coming from the outer circuit and prevent them from arriving to the electrode and from reacting with water. This would stop the electrolysis.
At the anode (positive pole), water is decomposed according to : $\ce{4 H2O -> O2 + 4 H+ + 4 e-}$. This is only possible if there are negative ions arriving near the anode to compensate the positive charges appearing in solution because of these newly created $\ce{H+}$ ions. Without negative ions like $\ce{SO4^{2-}}$ in the solution near the anode, these new $\ce{H+}$ ions would make a positively charged solution. This would prevent further electrons to be ejected from water. This would stop the electrolysis.
At the end, the composition of the solution is modified around the electrodes. Near the cathode, the solution contains more and more $\ce{NaOH}$. Near the anode, the solution contains more and more $\ce{H2SO_{4}}$ Of ourse if you stir and mix anodic and cathodic solutions, the acid and the base neutralize each other, which regenerate the solution of the original salt.


During formation and breaking of bonds, can the terms potential energy and bond energy be used interchangeably?

The answer is a qualified yes. Specifically, the "bond dissociation energy" (often shortened to "bond energy") is a positive number that is the amount of energy one must add to the system to completely break a mole of equivalent bonds. A completely broken bond by definition has 0 potential energy, so bond formation results in a negative potential energy. Thus, the potential energy of the bond is equal in magnitude but opposite in sign to the bond dissociation energy. If we are only talking about the magnitude (which is often the case), they're interchangeable.

In addition to the chemical bonds, a system can have other sources of potential energy such as gravitational potential energy or electrostatic potential energy, so in that sense potential energy is not synonymous with bond energy, but generally it's pretty clear from context when we mean only the potential energy of bonding.


Definition of the B3LYP functional in common QC programs

As a former contributor to Orca who had access to Turbomole sources before that, I can say with confidence that the internal definitions of B3LYP and B3LYP_gaussian are consistent among them and Gaussian (where available). So no calculations should go wrong.

While I cannot speak for Gaussian, for Orca I am certain that the ScalLDAC parameter is badly printed here (and not relevant for many of the newer functionals).

As to what you identified as a typo in Molpro -- It may be that B88 was shoe-horned into the more mainstream definition of GGAs, which can be loosely worded as "a multiplicative correction to the LDA energy" (the energy density, actually). Becke, on the other hand, described the B88 functional originally as an additive correction ($\Delta E$). NWChem makes the distinction explicit in the keyword nonlocal, see the manual section on exchange-correlation potentials. So for Molpro, the exchange accounting should read: $0.2+0.72+0.08=1$, whereas the others use: $0.2+0.8=1$ and keep the $0.72$ implicit.


The species you are proposing would be a methanediol dianion (CH2O22–) with two sodium cations “attached” to it in some way, depending on the nature of the solvent.

While methanediol can indeed readily form an anion, with a pKa of 13.3, I find no mention of the dianion or any measure (or prediction) of the second pKa of methanediol.