Is there any way to quantitatively find when a reaction has completed?

Option 1: Spectrometry

Easiest method, in most cases accurate enough. If you know that there are no side-reactions, this is a safe bet. Simply measure some characteristic wavelength for one reactant (preferably $\ce{Fe^{3+}}$) and one product (I would go for the $\ce{Cu^{+}}$ there) and then make measurements of your reaction mixture on the fly.

Option 2: Titration

Not that recommendable if you need any reliable results, but if you are low on budget and/or want a didactic effect. Take a little sample from a big flask and do a $\ce{Fe^{2+/3+}}$-redox-titration.

Option 3: Electrolysis

Don't know how to implement that, but basically you would take the sample and measure the amount of electricity needed to convert all $\ce{Fe^{2+}}$ in that sample back to $\ce{Fe^{3+}}$. Probably more accurate than spectrometry as the measurements of currents can be done quite precisely

You could also do some weird stuff like quantitative chlorine-NMR, but I can not think of any benefit that you would have with that.


How to determine the proton concentration in the following solution in mol/L?

You need more than a hint. You need a hard shove...

The overall notion here is that water dissociates

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

and that the equilibrium is given by

$$K_w = 1.00\times10^{-14} = \ce{[H+][OH-]}$$

where $\ce{[H+]}$ is the molar concentration of $\ce{H+}$ ions and $\ce{[OH-]}$ is the molar concentration of $\ce{OH-}$ ions.

NOTE: The pH of a solution is defined as the negative log of the $\ce{H+}$ ion concentration, and pOH of a solution is defined as the negative log of the $\ce{OH-}$ ion concentration. So

$$14 = \text{pH + pOH}$$

Sodium hydroxide is a strong base, so the assumption is that it completely dissociates.

$$\ce{NaOH(solid) <=> Na+(aq) + OH-(aq)}$$

The molecular mass of sodium hydroxide is 40.00 grams/mole.

$$\ce{[OH-]} = \dfrac{0.60\text{ g}}{40.00\text{ g/mole}\times0.80\text{ liter}} = 0.01875\text{ mole/liter}$$

Rearranging the $K_w$ equation

$$\ce{[H+]} = \dfrac{K_w}{\ce{[OH-]}} = \dfrac{1.00\times10^{-14}}{1.875\times10^{-2}} = 5.3\times10^{-13}\text{ mole/liter}$$


Calculation of equilibrium constant given the reaction temperature

By definition, the standard equilibrium constant is:

$K = e^{-\Delta_rG/RT}$

Where $\Delta_rG$ is the difference between the Gibbs Energies of the products and reactants.


What is the isomer distribution in monosubstituted fluorobullvalene?

Extensive NMR studies of substituted bullvalenes were done in the 1960's and 70's, by Oth et al. Much of their original work was published in German. Thankfully, some of these topics were revisited (in English!) during the 1990's by Luz et al., and these included some rather nice low temperature 19F and 13C NMR studies in both solution and solid state (and make it much quicker and easier for me to read).

Let us assign the following isomers 1–4, as is done in the question:

To answer your question:

Oth et al.[1] studied fluorobullvalene at −25 °C, and found no evidence by 19F of isomer 1. The equilibrium populations of isomers 4:3:2 was 78%:7%:15%. However, they could not distinguish between isomers 3 and 2, so it could be 78:15:7. In any case, predominantly isomer 4. They do state, however, that isomer 1 must exist ["Das C-Isomer muss also durchlaufen werden."], as the Cope rearrangement mechanism for isomer 4 to go to isomer 3 must go through isomer 1.

Luz et al.[2] studied the solution state Cope rearrangement of a number of monosubstituted bullvalenes.
At −30 °C, they identify the ratio of isomers 4:3:2 to be 87%:3.4%:9.6%. Again no evidence of isomer 1 by 19F at this temperature. However, at −55 °C, they observe a peak in the 19F spectrum that they ascribe to isomer 1, with an intensity of 0.2% relative to isomer 4. This peak broadens beyond detection at −40 °C. The ratios of isomers at −55 °C, then, was found to be 1.0:0.018:0.065:0.002 for isomers 4:3:2:1.

And just out of interest:

Of course, the other monsubstituted bullvalenes all have different preferred isomer distributions in solution. In the solid state, all monosubstituted bullvalenes studied by Luz crystallize as well-ordered single isomer crystals. Fluorobullvalene[3] crystallizes exclusively as isomer 4. Cyanobullvalene, bullvalenecarboxylic acid[4], and (ethylthio)bullvalene[5] crystallize exclusively as isomer 3. Bromo- and iodobullvalene[6] crystallize entirely as isomer 2.

Useful references for you:

Oth, J. F. M.; Merényi, R.; Röttele, H.; Schröder, G. Fluor-, chlor- und jodbullvalen. Tetrahedron Lett. 1968, 9 (36), 3941–3946. DOI: 10.1016/S0040-4039(00)72372-0.
Poupko, R.; Zimmermann, H.; Müller, K.; Luz, Z. Dynamic NMR Investigation of the Cope Rearrangement in Solutions of Monosubstituted Bullvalenes. J. Am. Chem. Soc. 1996, 118 (34), 7995–8005. DOI: 10.1021/ja954004t.
Müller, K.; Zimmermann, H.; Krieger, C.; Poupko, R.; Luz, Z. Reaction Pathways in Solid-State Processes. 1. Carbon-13 NMR and X-ray Crystallography of Fluorobullvalene. J. Am. Chem. Soc. 1996, 118 (34), 8006–8014. DOI: 10.1021/ja954005l.
Müller, K.; Zimmermann, H.; Krieger, C.; Poupko, R.; Luz, Z. Reaction Pathways in Solid-State Processes. 2. Carbon-13 NMR and X-ray Crystallography of Cyanobullvalene and Bullvalenecarboxylic acid. J. Am. Chem. Soc. 1996, 118 (34), 8015–8023. DOI: 10.1021/ja954006d.
Luger, P.; Roth, K. X-Ray, n.m.r., and theoretical studies of the structures of (ethylthio)bullvalene. J. Chem. Soc., Perkin Trans. 2 (1972-1999) 1989, 649–655. DOI: 10.1039/P29890000649.
Luz, Z.; Olivier, L.; Poupko, R.; Müller, K.; Krieger, C.; Zimmermann, H. Bond Shift Rearrangement of Chloro-, Bromo-, and Iodobullvalene in the Solid State and in Solution. A Carbon-13 and Proton NMR Study. J. Am. Chem. Soc. 1998, 120 (22), 5526–5538. DOI: 10.1021/ja9728029.


If you write the ionic equation, it seems that chlorine gas will produce. However, we see their half equation and standard potential.

$$\ce{ Cr2O7^2- + 14H+  + 6e-  ->  2Cr^3+  + 7H2O}$$
Standard potential: $E^\circ = \pu{+1.33V}$

$$\ce{ 6Cl-  ->  3Cl2 + 6e-}$$
Standard potential: $E^\circ = \pu{-1.36V}$

Thus, the standard potential cell for this ionic you say is equal to  $$\pu{+1.33V} + (\pu{-1.36V}) = \pu{-0.03V}$$

Because the standard potential cell is negative, so, it is call not energetically feasible or this reaction will not happen.

P.S.: The explanation only work under standard conditions.


The NMR spectrum seems to show that the tautomeric state of creatinine is that of the keto-form (bottom structure).

If the ketone is the only functional group, the keto form is more stable than the enol form. One the other hand, if the carbon-carbon bond is part of a conjugated system, the enol form might be more stable (e.g. acetylacetone or phenol).

You also suggest either the imino or the amino tautomer. There is a Raman study that addresses this question and also raises the dependence on pH.


You are correct. The system does positive work. Textbooks introduce the negative to show that the system loses energy as it does expansion work. As the gas expands, it does positive work but it loses this energy in the process. Work, like heat, is basically a transfer of energy.


Aluminium carbonate reacts easily with water/water vapour to form aluminium hydroxide as you can see on the Wikipedia page.

$$\ce{Al2(CO3)3 + 3 H2O → 2 Al(OH)3 + 3 CO2}$$

So, if you want to store aluminium carbonate, you have to make sure it is free from moisture.

Why it is unstable:

Smaller and more charged metal cations (i.e. metal cations with high charge density) tend to distort the electron cloud of the carbonate ions (polarisation).

Polarisation eventually leads to abstraction of oxygen from the carbonate ion, making one of the $\ce{C-O}$ bond in the carbonate weak, which makes it very prone to attack by water and heat (similar to the decomposition of copper (II) carbonate when compared to alkali metal carbonates).

The reaction itself doesn't release a lot of energy; hence why it is rather unstable and not reactive in chemistry terms.


The DLR (German Aerospace Center) has developed a couple of DIY instructions for school education. Among these is a PDF in German for the construction of a wind tunnel.

For the smoke, they suggest to use either

incense (the sticks or cones that you usually find in asia shops or esoteric shops) or a
fog machine, like the ones used for entertainment (dance clubs, etc.)

In the latter case, a mixture of 1,2-propanediol and water is heated and pressed through a nozzle at elevated temperature. Given that properly insulated machines, build for safe use at home, are available from DIY electronics shops for roughly 30 € up, I wouldn't start to build one myself.


Why is it necessary to correct the pressure of a gas that is collected over water?

Generally, gases which are insoluble in water are collected by putting them into an inverted vessel via a tube, so that the gas rises to the top and 'displaces' the water downward. Take a look at the following picture—

Now, the problem is that water has a certain vapour pressure at any temperature. In other words, whenever water is exposed to a gaseous phase, it will start to evaporate until the partial pressure of the water vapour in the gas phase reaches a certain value (this value depends on the temperature). This value is known as the 'vapou pressure of water'.

In this case, the gas you collected by displacing the water will be contaminated by water vapour. Therefore, the pressure of the gas which you measure will be the sum of the partial pressure of the product gas, and partial pressure of water vapour.
$$\begin{align*}
Measured\;pressure &= Partial\;pressure\;of\;product\;+\;partial\;pressure\;of\;water\;vapour \\
&=Actual\;pressure + Vapour\;Pressure
\end{align*}$$ This means that you must correct the pressure, by subtracting the vapour pressure of water from the measured value to get the actual pressure of the produced gas.

Note: There are standard tables which list the value of vapour pressure of water at different temperatures. These tables are generally used for the correction.


pH of sodium metasilicate anhydrous vs pentahydrate when dissolved in water

As explained in the comment by permeakra, due to the structural differences between anhydrous and hydrated Na2SiO3, their solutions will also be intrinsically different, at least for a while.
Whether anhydrous salt will turn to hydrate or vice versa depends on your ambient humidity and temperature. Anyway, I'd not recommend leaving either of them in contact with open air, for fear they might react with CO2. Silicic acid is so weak that it can be displaced even by carbonic acid.
Well, theoretically, the information about the SiO2:Na2O ratio is already present in the chemical formula Na2SiO3 (with or without water). However, the industrial-grade product may not quite match that formula. In fact, it may be a wild mixture of different silicates, where each individual compound cannot and need not be characterized. For many large-scale applications it suffices to know just the said ratio. Not sure about your application, though.


Is storing diluted (~1M) sulfuric acid, HCl, NaOH, etc. in plastic PET bottles a bad idea?

Here you can find thorough and clear instructions about using plastic bottles for acids and bases:

http://www.corning.com/lifesciences/emea/en/technical_resources/product_guid/bottles/material.aspx

A common rule is to not store strong acids and bases (like sulfuric and hydrochloric acids, and NaOH) in plastic bottles if you are not sure about the type of plastics.

Polystyrene bottles are OK for strong alkali.

Here is a PDF document on this topic:

http://csmedia2.corning.com/lifesciences/media/pdf/PSUG_Storage_Bottles_CLS_GL_025.pdf


Why do some dentists claim mercury "leaks out" of amalgam fillings?

You are labouring under a misconception about the exact meaning of the word "compound".

Normally when we describe, for example, ethanol as a compound of hydrogen, oxygen and carbon we don't mean that the component elements can "leak" out of it without a chemical reaction taking place.

Mercury amalgams are casually described as "compounds" but not in the same sense as ethanol. Amalgams are mercury alloys with varying compositions but they don't, usually, consist of strongly bonded molecules like ethanol with strong chemical bonds preventing the molecules falling apart. They are closer to mixtures than what we normally think of as compounds. Having said that they are not just mixtures as they can have very different bulk properties than their components.

The issue with mercury amalgams used in fillings is that they are mostly, but not completely, inert. Some mercury will evaporate from them and some chemical and physical processes can increase the amount that evaporates. But good estimates suggest only micrograms/day which is less than typical workplace exposure limits so probably not a source of legitimate worry. The dentists who still use mercury fillings are likely exposed to more than this and this is a better reason to limit their use. The subject is controversial with a lot of disagreement among studies. This is well summarised by the Wikipedia article on the controversy.


Find the number of copper spheres in a box

Here is your formulation :

$m_{\ce{Cu}} + m_{\ce{Pb}} = m_{\text{tot}}$

$\rho_{\ce{Cu}}V_{\ce{Cu}} + \rho_{\ce{Pb}}V_{\ce{Pb}} = m_{\text{tot}}$

$\rho_{\ce{Cu}}V_{\ce{Cu}} + \rho_{\ce{Pb}}(V_{\text{tot}}-V_{\ce{Cu}}) = m_{\text{tot}}$

$\rho_{\ce{Cu}}V_{\ce{Cu}} + \rho_{\ce{Pb}}(420-V_{\ce{Cu}}) = 5400$

$V_{\ce{Cu}}(\rho_{\ce{Cu}}-\rho_{\ce{Pb}}) + 420\rho_{\ce{Pb}}= 5400$

Put values for $\rho_{\mathrm{Cu}}$ and $\rho_{\mathrm{Pb}}$ in the equation, and rest is simple arithmetic.


In general, we can derive the expression for the density of any substance as a function of temperature using the definition of thermal expansion.

$$\frac{\mathrm dV}{\mathrm dT}=\alpha V\tag1$$

By differentiating $\rho=m/V$, we can show that $\mathrm d\rho/\rho = -\mathrm dV/V$. So using this and equation $\text{(1)}$,

$$\frac{\mathrm d\rho}{\mathrm dT}=-\alpha V\tag2$$

Integrating this above expression and approximating the expression,

$$\rho = \rho^\circ\cdot\left(1-\alpha\Delta T\right)\tag3$$

where $\rho^\circ$ is the initial density at the initial temperature, $\alpha$ is the volumetric thermal expansion coefficient and $\Delta T$ is the temperature.
For water, $\alpha$ is typically $207\times10^{-6}\ \mathrm{K^{-1}}$ at $20\ \mathrm{^\circ C}$. Using this, we can observe how temperature affects the density of water.


Water has two lone pairs. H$^+$ is a naked nucleus with a positive charge. It is extremely reactive and would bind to any neutral molecule. Water has two lone pairs of electrons and H$^+$ binds one of them to form H$_3$O$^+$.

In absence of water it would bind other things. For example, it would bind methane to form unstable CH$_5^+$ that breaks down to CH$_3^+$ + H$_2$.


Does acetic acid give a positive result with the iodoform test?

The iodoform reaction proceeds by the mechanism shown below. (The final proton transfer need not occur between the two partners, any other hydroxide could abstract the acid’s proton and any water molecule can protonate iodoform.)

This can fail for carboxylic acids or esters in one of two ways:

The first step. The $\mathrm{p}K_\mathrm{a}$ value of the proton being abstracted is $26.5$ for acetone and $29.5$ for ethyl acetate according to the Bordwell $\mathrm{p}K_\mathrm{a}$ table. This can be attributed to the second oxygen of an ester donating electron density into the carbonyl group and thus lowering the tendency to form the enol(ate). But we can still argue that enough ester will be deprotonated for the iodination to occur. (Note that this fails entirely for a free acid, since the molecule is already charged. A second deprotonation to produce the endioldiate is very unfavourable.)

The final fragmentation step. In esters, $\ce{R}$ also contains an oxygen atom, so there are a total of three pathways by which the tetrahedral intermediate can fragment (one being redissociation of the hydroxide). In the ketone’s case, only one is productive, namely the formation of iodoform which precipitates and drives the equilibrium towards its formation. The ester can fragment under displacement of the alcoholate (productive, gives an acid) which is much more favourable than displacing a $\ce{CI3-}$ group due to the $\mathrm{p}K_\mathrm{a}$ value of both’s conjugate acids.
The resulting acid is merely deprotonated. The attack of a nucleophilic (hydroxide) species to the resulting anion is strongly disfavoured.

It is also to be noted that an ester can always be hydrolysed to the corresponding acid under basic aquaeous conditions. Thus, at any intermediate an acetate anion could generate inhibiting any further reactions.


What do the subscripts 1/2 and 3/2 for the p-orbitals refer to?

Essentially, this derives from Dirac’s relativistic quantum mechanics.

In nonrelativistic quantum mechanics—the Schrödinger equation that you are probably highly familiar with—p orbitals are triply degenerate while d orbitals are 5-way degenerate. However, nonrelativistic quantum mechanics does not include the electronic property known as spin unless it is introduced in a separate step.

Dirac’s relativistic quantum mechanics solves this problem by introducing $c$ as a finite constant. Almost magically, his equations directly result in a quantum number for the spin, $\vec s$. However, the result is that the azimuthal quantum number no longer behaves like a true quantum number (i.e. it no longer describes stationary states) but thankfully the sum $\vec j = \vec l + \vec s$ does. Therefore, instead of considering just one $\vec l$ as a descriptor to decide between s, p and d orbitals, we are stuck with $\left |\vec l - \vec s\right |$ and $\left | \vec l + \vec s\right |$. For $\vec l = 1$ (a p orbital) this gives us the values of $\vec j = \frac12$ and $\vec j = \frac32$. There are two p orbitals of the $\mathrm p_\frac32$ type and one of the $\mathrm p_\frac12$ type. These are no longer degenerate, i.e. $\mathrm p_\frac12$ has a lower energy than the $\mathrm p_\frac32$ ones.

The orbital of the $\mathrm p_\frac12$ type happens to be spherical in nature. For a more thorough background, I recommend Nicolau’s extensive answer which I mostly paraphrased in the above paragraphs and which contains links to further materials.


Why can curcumin cross the blood-brain barrier, but not congo red?

Preface: both molecules could act as bidentate ligands but the β-diketo structure found in curcumin is typically a much better bidentate ligands — as the 2-oxophenol moieties would be. So I would expect curcumin to be a much better ligand than congo red.

The key difference between the two molecules is comparing the extent of their lipophilicity and their hydrophilicity. Curcumin has a rather low water solubility of $3.2~\mathrm{mg/l}$, in line with the fact that it has rather little features that can enhance solubility. Basically, it boils down to the two keto-groups and the phenol moiety.

Congo red has two sulphonic acid groups and two aromatic amino groups. This means that there is a much greater relative charge present, and indeed its solubility in water is reported to be $1.16 \times 10^{2}~\mathrm{g/l}$ — notice the different unit. So congo red is already much more hydrophilic than curcuminis.

In terms of lipophilicity, the sulphonic acid groups present in congo red reduce its lipophilicity while the amino groups themselves don’t have too strong an effect. On curcumin’s side, the phenol groups can comfortably hydrogen bond to the neighbouring methoxy residues and the β-diketo fragment can tautomerise to a keto-enol form and form another intramolecular hydrogen bond. Thus, curcumin is a lot more capable of coping with low-hydrogen bond solvents and is thus expected to have a high lipophilicity.

This is the edge that curcumin has over congo red to cross the blood-brain barrier.


A drop of water in a tin of sugar: Which one's the solvent, the sugar or the water?

You could imagine stirring the sugar enough for the water molecules to be uniformly distributed throughout - it would then be homogeneous.

However, even then, to refer to the mixture as a solution of water in sugar is unhelpful, not least because referring to a slightly damp solid as a solution will only confuse.  A definition needs to be useful, as well as being a set of criteria to be met.

Consider adding more water, and stirring till uniformly distributed.  At some point, the mixture will become a thick syrupy liquid.  Add yet more, and it will eventually resemble an ordinary solution.  At what arbitrary point do you say one is a solution of the other?

As a device to get students thinking about the interactions that take place when a solid dissolves in a liquid, or a liquid in another liquid, it's an excellent question, but not one that has either a right or a wrong answer, other than "neither", or indeed "both".

What is definitely wrong is to insist that there's only one right answer to the question.


I can speak based on my experience studying chemical engineering as an undergraduate at an ABET accredited university in the United States and working in industry.

A typical curriculum would include a focus on fundamentals which can then be applied in specific industries or in different cases rather than an in depth study of specific manufacturing processes or specific chemical reactions. This process is intended to develop a core of knowledge to approach specific processes or reactions and also as a preparation for learning more

So for example an undergraduate might start off taking calculus, and general chemistry and proceed to additional chemistry like organic and possibly phsyical chemistry. Starting in chemical engineering a student will encounter fundamental fields of chemical engineering which are fundamental tools such as mass and energy balance, thermodynamics, fluid dynamics, heat transfer, mass transfer, and chemical kinetics. These are all tools which can be used as a chemical engineer to evaluate processes. Also a review and study of types of what's called unit operations would be included.

Unit Operations are like the building blocks of many processes. The include operations like heat exchangers, distillation or absorbion columns, cooling towers, reactors, tanks, and more.

An understanding of chemical engineering fields such as kinetics or thermodynamics or heat transfer combined with an understanding of unit operations can be combined to learn more about a specific process.

I am sure there are some folks out there with more insight so I'll be very interested to hear from them as well!


Hybrid functional calculations using different approach

In hybrid functional such as B3LYP, PBE0, HSE etc, some part of HF
  exchange and/or correlation is added to some part of DFT X/C part in
  order to get better results.

First, there is no correlation in HF, only exchange, so you can not add part of the HF correlation into DFT, as you said. In principle, of course, if you want to add not only exchange from the wave function theory (WFT) methods but also some correlation, you could use other that the HF sources, but the conventional hybrid DFT corresponds to mixing in just (a part of) the HF exchange.

Now, you very question of

  why don't we just use the full HF solution that we already have in our
  hand?

is addressed in almost any textbook on DFT. I suggest that you read, for instance, Section 6.6 Hybrid Functionals in Koch & Holthausen, specifically, pages 78-80. The authors provide a relatively simple argument for for the well-known failure of the exact exchange/density functional
correlation combination approach.

In short, the source of the problem is that the separation of an exchange-correlation entity (would it be energy, or hole, or something else) into exchange and correlation contributions is artificial. Only the whole exchange-correlation entity has a physical meaning, while the above mentioned individual contribution are introduced just for convenience. Usually, it means that exchange and correlation contributions has to be chosen consistently so that their sum, an exchange-correlation entity, has some desired properties.

For instance, the exchange-correlation hole in molecular systems is usually relatively localized, so that the exchange hole and the correlation hole must both be either localized or delocalized (in a very special way so that combined they result in a localized hole). In LDA- and GGA-DFT both the exchange and the correlation hole are localized, but when we mix HF exchange and LDA/GGA-DFT correlation in the most straightforward way, i.e.,
$$
E_{\mathrm{XC}} = E_{\mathrm{X}}^{\mathrm{HF}} + E_{\mathrm{C}}^{\mathrm{DFT}} \, ,
$$
then, I quote:

  We combine the exact, delocalized exchange hole with a localized model
  hole for correlation. Because the cancellation between the two
  individual holes cannot take place, the resulting total hole has the wrong characteristics.

1) Koch, W., & Holthausen, M. C. (2001). A Chemist's Guide to Density Functional Theory, Second Edition, Wiley-VCH Verlag GmbH.


Let's say you go to Stonehenge.
You see a big rock there.
You can observe the state of the rock.
But how much work was done to bring the rock there?
How far away did the rock come from? Was it a downhill or uphill journey?  Was the stone dragged, moved with wheels, or pulled on a sled over snow?

So you can know the exact state of the rock, but not how much work was done to bring it to that state.

Therefore work is not a state function. The others given in the exercise, i.e. internal energy, pressure, volume, temperature are all state functions.


Yellowish tinge during titration of oxalic acid with potassium permanganate

Without knowing the concentrations of the involved substances and the values of some other parameters, we can only guess what has happened during your experiment.

The intended analytical reaction is the reduction of purple permanganate to colourless $\ce{Mn^2+}$:

$$\ce{MnO4- + 8H+ + 5e- <=> Mn^2+ + 4H2O}$$

If you add the permanganate solution too quickly, the reduction may be incomplete because not enough reducing agent (here: oxalic acid) or acid $(\ce{H+})$ is available in the reaction zone:

$$\ce{MnO4- + 4H+ + 3e- <=> MnO2 + 2H2O}$$

The observed colour may be caused by a cloud of finely dispersed particles of dark brown manganese(IV) oxide $(\ce{MnO2})$, which dissolves when the solution is well mixed again so that the intended reaction continues.

The same reaction may occur when you don’t add enough sulfuric acid, since $\ce{H+}$ is consumed during the reaction (8 mol $\ce{H+}$ per 1 mol $\ce{MnO4-}$).


Why is an atom with excess neutrons unstable?

Pulling out Theoretical Nuclear Physics by Blatt and Weisskopf (published in 1952, so a bit dated in areas but still a good introduction), one finds (in Chapter VI.1 The Systematics of Stable Nuclei) that the stability of isobaric nuclei (all the same $A$) is determined by:

(1) The symmetry effect: the charge independence and the exchange character of nuclear forces, together with the Pauli exclusion principle, strongly depresses the energy of nuclei with equal or nearly equal numbers of protons and neutrons.

(2) The charge effect: the effect of the Coulomb repulsion of the protons favors nuclei with fewer protons than neutrons. The neutron-proton mass difference also enters the charge effect, but only as a small correction. The charge effect increases in importance with increasing nuclear charge.

(3) In a few special cases a third effect is important: the spin dependence of nuclear forces favors parallel spin over anti-parallel sin of a pair of (extra) nucleons.

So, yes, you picked up on the Coulomb bit (number 2 above), but it is the symmetry effects of the strong force that are more important, particularly for light nuclei. Until one gets to some nuclear physics, the characteristics of the weak and strong forces just aren't familiar to us humans.


Something I don't understand about batteries

We manually put Cu ions on the electrolyte, it then reacts with the excess electrons on the electrode and makes Cu atoms.


Why ground-state configuration is consistent ("additive") from one element to the next?

Well, this configurations arise in what is usually called the orbital approximation and the consistent pattern is due to the very nature of this approximation.

In the orbital approximation the overall electronic wave function describing the atom is represented as the product of the wave functions for the individual electrons known as atomic orbitals. This can be sketched as follows.
$$
\newcommand{\op}[1]{\hat{#1}}
\newcommand{\el}[1][]{_{\mathrm{e}}}
\newcommand{\nuc}[1][]{_{\mathrm{n}}}
\newcommand{\elel}[1][]{_{\mathrm{ee}}}
\newcommand{\elnuc}[1][]{_{\mathrm{en}}}
$$

Electronic Schrödinger Equation

The electronic wave function $\psi_{\mathrm{e}}$ is an eigenfunction of the electronic Hamiltonian $\op{H}\el$
$$
\op{H}\el \psi\el = E\el \psi \el \, ,
$$
$$
\op{H}\el
=
\op{T}\el + \op{V}\elnuc + \op{V}\elel
=
- \sum\limits_{i=1}^{n} \frac{1}{2} \nabla_{i}^{2}
- \sum\limits_{i=1}^{n} \frac{Z}{r_{i}}
+ \sum\limits_{i=1}^{n} \sum\limits_{j > i}^{n} \frac{1}{r_{ij}} \, ,
$$
which indeed is different for different atoms. Thus, the electronic wave function can also be expected to be different. However, the exact analytical solutions of the electronic Schrödinger equation are known only for one and two-electron systems, such as $\ce{H}$, $\ce{H-}$, $\ce{He}$, $\ce{Li+}$, etc., while for all other chemical systems only approximate solutions can be found.

Separation of variables

The electronic Schrödinger equation is partial differential equation and one of the most frequently used analytical methods to solve such equations is technique known as the separation of variables aimed to broke one compound partial differential equations of many variables into a set of simpler partial differential equations of fewer variables. The main disadvantage of the separation of variables is that it is by no means a universal approach: for some equations a solution of the form mentioned above make the separation of variables possible, for others it is of no use.

The most common way of separation of variables for partial differential equation in a function $f(x_1, x_2, \dotsc, x_n)$ of $n$ variables $x_{i}$ is writing a solution as a product of $n$ functions $f_{i}(x_{i})$ each of which is a function of one variable only
$$
    f(x_1, x_2, \dotsc, x_n) = f_1(x_1) f_2(x_2) \dotsm f_n(x_n) \, ,
$$
and then substituting solution of this form into the equation.
Using algebraic manipulation one could try to break the resulting equation into a set of $n$ independent ordinary differential equations for each function $f_{i}(x_{i})$ and if succeed then one could solve these independent equations and by plugging all $f_{i}(x_{i})$ back into the product obtain the solution of the starting equation.

Orbital approximation

So, it is tempting to try to separate the electronic coordinates by representing $\psi\el$ as a product of one-electron wave functions, however, it is fairly well known that this attempt will be successful only if the potential energy can be expressed as the sum of the separate potentials for each particle. This clearly is not the case for the electronic Hamiltonian: the potential energy of the Coulomb repulsions between the electrons $\op{V}\elel$ prevents the separation of electronic coordinates
\begin{equation*}
    \psi\el(\vec{r}_{1}, \vec{r}_{2}, \dotsc, \vec{r}_{n}) \neq \psi_{1}(\vec{r}_{1}) \psi_{2}(\vec{r}_{2}) \dotsb \psi_{n}(\vec{r}_{n})\, .
\end{equation*}
So, exact separation in this form is not possible, but what about approximate one? Can we somehow approximate $\op{V}\elel$ with some model potential which will allow the separation of electronic coordinates? To some extent, the answer is, yes, we can.

In the orbital approximation it is indeed assumed that the many-electron wave function can be written as a product of one-electron wave functions called orbitals. As it was already said, it will work only if the exact $\op{V}\elel$ potential is replaced by the model potential $\op{V}_{\mathrm{MF}} = \sum\nolimits_{i=1}^{n} \op{v}_{\mathrm{MF}}(r_{i})$ which allows the separation. The approximate Hamiltonian is
$$
\op{H}_{\mathrm{MF}}
=
\op{T}\el + \op{V}\elnuc + \op{V}_{\mathrm{MF}}
=
- \sum\limits_{i=1}^{n} \frac{1}{2} \nabla_{i}^{2}
- \sum\limits_{i=1}^{n} \frac{Z}{r_{i}}
+ \sum\limits_{i=1}^{n} \op{v}_{\mathrm{MF}}(r_{i}) \, .
$$
Physically it means that electrons do not instantaneously interact with each other, but rather each and every electron interacts with the average, or mean, electric field created by all other electrons. Hence, another name of the approximation, the mean field approximation.

Even the mean-field Hamiltonian still looks different for different atoms, however, it is now a sum of one-electron Hamiltonian operators
$$
\op{H}_{\mathrm{MF}}
=
\sum\limits_{i=1}^{n}\op{h}(r_{i}) \, ,
$$
where
$$
    \op{h}(r_{i})
    =
    - \frac{1}{2} \nabla_{i}^2
    - \frac{Z}{ r_{i} }
    + \op{v}_{\mathrm{MF}}(r_{i}) \, .
$$
And in contrast to an eigenfunction $\psi\el$ of the true electronic Hamiltonian $\op{H}\el$ an eigenfunction $\psi_{\mathrm{MF}}$ of the mean-field Hamiltonian $\op{H}_{\mathrm{MF}}$ can be written as products of one-electron wave functions (orbitals)
$$
    \psi_{\mathrm{MF}} = \psi_{1}(\vec{r}_{1}) \psi_{2}(\vec{r}_{2}) \dotsb \psi_{n}(\vec{r}_{n}) \, ,
$$
where $\psi_{i}(\vec{r}_{i})$ are solutions of the corresponding one-electron Schrödinger equations
$$
    \op{h}_{i} \psi_{i}(\vec{r}_{i}) = \varepsilon_{i} \psi_{i}(\vec{r}_{i}) \, .
$$
Now this resulting equations that defines orbitals are not that different: there will be just a small difference in one-electron Hamiltonians due to slightly  different nuclear charges as you move along a group. Of course, there will be more and more occupied orbitals as you proceed through the group, but equations themselves and consequently orbitals will be quite simillar.


Your premise is incorrect. $\unicode[Times]{x3c3}_\mathrm{p-d}$ are very much possible. Consider any hexahalidometal complex such as $\ce{[FeF6]^3-}$. Since fluoride is essentially unhybridised as iron would be, too, we can separate the atomic orbitals into irreducible representations of the $O_\mathrm{h}$ point group. We find that iron’s d-orbitals transform as $\mathrm{t_{2g} + e_g}$ and the 18 fluoride p-orbitals as $\mathrm{a_{1g} + e_g + 2t_{1u} + t_{1g} + t_{2g} + t_{2u}}$. If we separate out those 12 p-orbitals that are perpendicular to the iron-fluoride axis, the ones that remain are $\mathrm{a_{1g} + e_g + t_{1u}}$. You may immediately realise that $\mathrm{e_g}$ is an irreducible representation of both and thus that the fluorides’ p-orbitals can and will mix with the iron’s $\mathrm{d}_{x^2 - y^2}$ and $\mathrm{d}_{z^2}$.

Professor Klüfers included the calculation results of said ion in his internet scriptum of his coordination chemistry course. The ones of relevance are:

You may be slightly surprised at the shape since you cannot recognise the typical $\mathrm{d}_{z^2}$ shape. But as Brandhorst, Grunenberg and Tamm explain, these two orbital shapes are mathematically equivalent descriptions for the $\mathrm{e_g}$ orbitals.[1]

The molecular orbitals shown are both of antibonding type (formally $\mathrm{e_g^*}$). The corresponding bonding orbitals are much lower in energy and centred on the fluoride ligands. Both of the depicted ones are occupied only by a single electron in α spin.

References

[1] K. Brandhorst, J. Grunenberg, M. Tamm J. Chem. Educ. 2008, 85, 1692. DOI: 10.1021/ed085p1692.


What exactly is the "oxygen reduction reaction"?

According to your cited paper the reaction they care most about is $$\ce{O2 + 4H+ + 4e- -> 2H2O}$$ as you are suggesting. This should also be the total reaction and the most interesting one for applications, e.g. in fuel cells.

There are various suggestions for the mechanistics of the Oxygen Reduction Reaction. A quite recent review was published by Nagappan Ramaswamy and Sanjeev Mukerjee. A very early general scheme can be found at electronicstructure.wikidot.com, where they cited Halina S. Wroblowa et.al..


Electrochemistry conditions for voltage to increase in a galvanic cell

Yes, your voltage will change in accordance with the Nernst equation

$$\Delta E_\text{cell} = - \frac{RT}{zF}\ln Q_{\text{rxn}}$$

So, in your copper-zinc battery, where

$$\ce{Cu^2+(aq) + Zn(s) -> Cu(s) + Zn^2+(aq)}$$

with a standard $E^\circ_\text{cell} \approx \pu{1.10 V},$ setting the $\left[\ce{Cu^2+}\right] = \pu{4 M}$ should raise your cell potential by

$$\Delta E_\text{cell} = -\frac{RT}{2F}\ln\frac{1}{4} = \pu{+0.0178 V}$$

for a new cell potential (at $\pu{25 °C})$ of

$$E_{\text{cell}} = \pu{1.10 V} + \pu{0.0178 V} \approx \pu{1.12 V}$$


Why is it possible to separate NaOH from certain compounds using acetone?

Ask yourself what kind of reaction happens?
What glassware have you used to remove impurities?

By controlling pH and polarities you are able to separate compounds because of different affinities between them and solvents. You change molecular solvatation of your analyte and impurities. Then by controlling boiling point, using a rotavap, you may be able to separate two phases that are miscible. Or by making impurities insoluble you can remove by mechanical operation such as centrifugation. Such post-treatments are very common in chemistry and they scale up very well.

So, there is no chemical reaction (except acid/base reaction if any) there are physical reactions (solvatation, phase extraction, phase change: evaporation, condensation, crystallisation and sedimentation).


Why does this region fit NMR spectroscopy technique?

The insert in the Ld ortholog occurs in a beta sheet and suggests the presence of additional strands in that sheet compared to the human variant. The structure of the human protein has been determined by crystallography. It would therefore appear that it should be possible to determine the structure of the Ld protein the same way (via crystallography), provided a crystallizable construct is identified.

Based on the available reference provided, the insertion feature from LdRab5a appears to make the protein unstable in vitro and difficult to crystallize, leaving NMR as the only good alternative for a structure determination. However, it seems the insertion feature also rendered the protein too unstable to be able to complete long NMR experiments, so the authors chose to remove the insert in order to stabilize the protein and perform NMR.


How can a phase be thermodynamically stable at room temperature if it only appears at high temperature?

$\alpha$-alumina is thermodynamically stable at room temperature, and also at high temperature (around $1,200-1,400 \space ^{\circ}C$) - however, alpha alumina is not what is typically formed when alumina is made from the various aluminum ores (ref). Typically, one of the several other crystal structures will form, with different particle sizes and amounts of water (hydration). These and all other phases that are found along the way to $\alpha$-alumina are called transition aluminas.

This link explains what happens upon heating. In general, first the crystal is dehydrated, which causes it to form a different, dehydrated phase. Then, as the temperature continues to rise, the alumina crystal structure rearranges into more compact structures, until eventually, near the sintering point, the atoms have sufficient kinetic energy that they can "find" the most thermodynamically stable configuration - $\alpha$-alumina.

In other words, the transition alumina phases are kinetically trapped in structures that are not the most thermodynamically stable, but heating allows them to get out of that trap and into the most favorable state.


What is the difference between ΔfH and H-H° in NIST–JANAF tables?

As a side note, "JANAF" stands for Joint Army–Navy–Air Force; these tables were designed to provide quick access to the thermodynamic data for rocket scientists.
Here, $Δ_\mathrm{f}H^\circ$ is the standard enthalpy of formation. The difference $H-H^\circ(T_\mathrm{r})$ is required to calculate the standard enthalphy of reaction $Δ_\mathrm{r}H^\circ$ if in the given temperature range a reaction takes place.
There is a thorough explanation in Silbey's Physical Chemistry [1, p. 62]:

The JANAF tables give standard enthalpies of formation at a series of temperatures, and so these values may be used directly to calculate enthalpies of reaction. [...] Some thermodynamic tables give values of $\bar{H}^\circ_T - \bar{H}^\circ_{298}$ to assist in the calculation of $Δ_\mathrm{r}H^\circ$ for a chemical reaction or phase transition:
$$\bar{H}^\circ_T - \bar{H}^\circ_{298} = \int_\pu{298 K}^T \bar{C}^\circ_P \mathrm{d}T$$
Depending on the table, $Δ_\mathrm{r}H^\circ$ for phase transitions in the intervening temperature range may be added to the right-hand side of the equation.

NIST-JANAF tables are given for a reference temperature $T_\mathrm{r}$ and a standard state pressure $p^\circ$. Unless stated otherwise, $T_\mathrm{r} = \pu{298.15 K}$ and $p^\circ = \pu{0.1 MPa}$ [2].
References

Silbey, R. J.; Alberty, R. A.; Bawendi, M. G. Physical Chemistry, 4th ed.; Wiley: Hoboken, NJ, 2005. ISBN 978-0-471-21504-2.
NIST-JANAF Thermochemical Tables, 4th ed.; Chase, M. W. J., Ed.; American Institute of Physics: Washington, DC : New York, 1998. ISBN 978-1-56396-831-0.


What is the transparent film that covers water-soluble dishwasher tablets?

It is made using Polyvinyl Alcohol (Mowiflex material).


Your assessment is correct. The rules concerning the prefixes ‘benzyl’ and ‘benzylidene’ have been changed. In the 1993 IUPAC Recommandations, these prefixes could be substituted, but only on the ring. According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), these prefixes cannot be substituted for preferred IUPAC names, although for general nomenclature restricted substitution is permitted (see P-29.6.2).

Except for the use of enclosing marks to set off parts of the name dealing with specific structural features, the suggested name is correct. The preferred IUPAC name is 3-{(E)-[(4-methylphenyl)methylidene]amino}benzoic acid.

Note that enclosing marks are used around compound (see Rule P-29.1.2) and complex (see Rule P-29.1.3) prefixes. When multiple types of enclosing marks are required, the nesting order is {[({[( )]})]}, etc.

Concerning the use of ‘amino’ in this case instead of ‘aza’, Rule P-62.2.3 explains

  (…)

  Preferred IUPAC names for prefixes corresponding to $\ce{-NHR}$, $\ce{-NRR'}$, or $\ce{-NR2}$ are formed by prefixing the names of the groups $\ce{R}$ and $\ce{R'}$ to the prefix ‘amino’, for example ‘methylamino’ for $\ce{-NH-CH3}$. Prefixes such as azanyl and azanylidene may be used in general nomenclature.


Using Beer-Lambert law to calculate light intensities

In my opinion, Beer–Lambert law is no use. I would rather write
$$\frac{\mathrm dI}{I\,\mathrm dl} = \pu{0.8 m^-1},\tag{1}$$
which leads to the integrated formula
$$\ln\left(\frac{I_0}{I}\right) = 0.8 l.\tag{2}$$
And, if $\displaystyle\frac{I}{I_0} = 0.5,$ the corresponding depth is
$$l = \frac{\ln 2}{0.8} = \pu{0.846 m}.\tag{3}$$


Why does ozone have higher entropy than oxygen?

You are mixing up two different quantities (entropy of reaction, and molar entropies of the two gases involved in the reaction) and either your text book is in error, or you have misread it somehow. For that reaction, as written, the formation of ozone from oxygen,
$\Delta G$ is positive and $\Delta S$ is negative (the opposite of what you wrote in your question). It is unfavourable on both enthalpic and entropic grounds. Ozone is thermodynamically unstable with respect to oxygen under standard conditions.

Your experience, regarding the sign of $\Delta S$ and its relation to the change in the number of molecules in a gas-phase reaction, is by-and-large a good guide, because the translational contribution to the entropy typically dominates over the rotational and vibrational contributions in a gas. This reaction is consistent with this rule of thumb.

The relevant thermodynamic quantities are the standard Gibbs free energy, enthalpy, and entropy, of formation of ozone, since the oxygen is in its standard state. Taking the figures from this copy of CRC we have
$$
\Delta_{\text{f}} G^\circ (\text{O}_3) = 163.2 \, \text{kJ}\,\text{mol}^{-1}
\quad
\Delta_{\text{f}} H^\circ (\text{O}_3) = 142.7 \, \text{kJ}\,\text{mol}^{-1}
$$
and for the standard entropies
$$
S^\circ (\text{O}_3) = 238.9 \, \text{J}\,\text{K}^{-1}\text{mol}^{-1}
\quad
S^\circ (\text{O}_2) = 205.2 \, \text{J}\,\text{K}^{-1}\text{mol}^{-1}
$$
So indeed the standard molar entropy of ozone is a little higher than that of oxygen (as the comments say, if you want to explain this precisely you need to take into account internal degrees of freedom, as well as properly calculating the translational contributions), but this does not lead to a positive entropy of reaction. To calculate this, i.e. the entropy of formation of ozone, we need
$$
\Delta_{\text{f}} S^\circ (\text{O}_3)
= S^\circ (\text{O}_3) - \tfrac{3}{2} S^\circ (\text{O}_2)
= -68.9 \, \text{J}\,\text{K}^{-1}\text{mol}^{-1}
$$
remembering that all these standard quantities of formation are per mole of ozone. Equivalently you can calculate this number from $[\Delta_{\text{f}} H^\circ (\text{O}_3)-\Delta_{\text{f}} G^\circ (\text{O}_3)]/T$ giving the same answer.

So the reaction has a negative $\Delta S$ and a positive $\Delta G$.

[EDIT following OP comment]

This means that, in the oxygen-ozone equilibrium, both forward and backward reactions are happening, at the same rates,
but the rate constants are such that the equilibrium concentration (or partial pressure) of O$_3$ is much lower than that of O$_2$.

This is not how things happen in the stratosphere, nor in most methods for preparing ozone. Generally, these involve a multi-step process, in which
the first step is the dissociation of O$_2$ into two oxygen atoms,
by absorbing a photon of UV light, or through an electrical discharge (for example). Wikipedia provides a reasonable summary of the Ozone-oxygen cycle, or Chapman cycle, with links to further resources, while various production methods are described on the Ozone page. The above equilibrium thermodynamics relations cannot be applied to processes like this, because of the injection of energy.

Instead, in the case of the Chapman cycle, a steady-state kinetic scheme can be set up for creation and destruction of O$_3$, and O atoms, where some of the rate constants depend on the intensity of UV light. The net result is a steady (small) population of O$_3$, and O atoms, accompanied by the conversion of energy from UV light into heat which is dissipated in the atmosphere. In real life, the situation is more complicated still, because of the roles of other species and reactions: books such as RP Wayne's Chemistry of Atmospheres give a lot more detail.

[2nd EDIT following further OP comments]

For the equilibrium between oxygen and ozone, the standard Gibbs free energy change for the reaction $\Delta_r G^\circ$ may be related to the equilibrium constant $K$ (see e.g. here)
$$
\Delta_r G^\circ = -RT \ln K
$$
where $K$ is expressed (for ideal gases) in terms of the partial pressures
relative to the standard pressure $p^\circ$ (1 bar)
$$
K = \frac{(p_{\text{O}_3}/p^\circ)^2}{(p_{\text{O}_2}/p^\circ)^3}
$$
For large positive $\Delta_r G^\circ$, $K$ is extremely small,
so the equilibrium partial pressure of ozone is extremely small,
if we are given that the partial pressure of oxygen has a reasonable value.
Indeed, if we take (for the reaction as written, producing two moles of ozone)
$$
\Delta_r G^\circ=2\times\Delta_{\text{f}} G^\circ (\text{O}_3) = 326.4 \, \text{kJ}\,\text{mol}^{-1}
$$
then $K=\exp(-\Delta_r G^\circ/RT)\approx 6\times10^{-58}$;
if the partial pressure of oxygen is $\approx 0.21$ bar
(as in the atmosphere around us),
then the equilibrium partial pressure of ozone is $p_{\text{O}_3}\approx 2.4\times 10^{-30}$ bar.

It is wrong to conclude that the reaction producing ozone "cannot happen"
because the free energy change is positive.
$\Delta_r G^\circ$ is simply the free energy change associated with
the complete conversion of reactants into products.
In reality, an equilibrium is always established somewhere in between.
In this case, it is very far over to one side, but not 100%.
If $\Delta_r G^\circ$ were not so large (but still positive) the
equilibrium would not be so far in favour of reactants.
Thermodynamically, you can think of the position of equilibrium
as being determined by both $\Delta_r G^\circ$ and the entropy of mixing of the reactants with the products:
this will always result in an equilibrium position that is neither 100% reactants
nor 100% products.

It is also possible to give a dynamical interpretation of chemical equilibria
in terms of forward and reverse reaction rates,
but I emphasize that this argument only applies, at least in this simplified form, to one-step chemical reactions. The oxygen-ozone equilibrium certainly involves multiple steps
(nobody is pretending that three molecules of oxygen miraculously collide
together at once). To see why this more complicated mechanism
does not affect the argument above about the equilibrium constant, see
How is it that the equilibrium constant does not depend on the mechanism? and you might also find Transition state and free energy useful. So consider the dynamical equilibrium
$$
\text{A} + \text{B} \leftrightharpoons \text{C} + \text{D}
$$
and assume that the forward and reverse reactions are simple one-step processes
obeying the rate equations
$$
\text{Rate}(\rightarrow) = k_\rightarrow [\text{A}][\text{B}],
\qquad\text{and}\qquad
\text{Rate}(\leftarrow) = k_\leftarrow [\text{C}][\text{D}].
$$
At equilibrium both forward and reverse reactions are happening at the same rate because otherwise the concentrations would be changing.
We can write
$$
\text{Rate}(\rightarrow) = \text{Rate}(\leftarrow)
\qquad\Rightarrow\qquad
K = \frac{[\text{C}][\text{D}]}{[\text{A}][\text{B}]}
= \frac{k_\rightarrow}{k_\leftarrow}
$$
So, if $\Delta_r G^\circ$ is large and positive, $K\ll 1$,
and $k_\leftarrow \gg k_\rightarrow$.
The rate constant for the reverse reaction is much larger than that for the
forward reaction.
However, both reactions are still happening, and the rates of these reactions
are equal.

As I said, you can't apply this argument directly to a multi-step reaction scheme, but the two links I gave above show how the same ideas apply in such a case.
Also, most of the above material is covered in standard physical chemistry texts.


How to calculate the mass of phosgene in the equilibrium mixture of carbon dioxide and chlorine?

We have an equilibrium mixture for the reaction

$$\ce{CO (g) + Cl2 (g) <=> COCl2 (g)}$$

We should recognize that

$$K_\mathrm{eq} = \frac{[\text{products}]}{[\text{reactants}]} = \frac{[\ce{COCl2}]}{[\ce{CO}][\ce{Cl2}]}$$

Note that the reactants and products all have stoichiometric coefficients of 1 so we can write our $K_\mathrm{eq}$ equation out in this way.

Because we are given concentrations of reactants in the equilibrium mixture, we can go ahead and plug in the values that we know and solve for $[\ce{COCl2}]$.

$$\begin{align}
2.9 \times 10^{10} &= \frac{[\ce{COCl2}]}{(\pu{1.8 \times 10^-6 M})(\pu{6.3 \times 10^-7 M})} \\
[\ce{COCl2}] &= \pu{0.033 M}
\end{align}$$

We are asked how many grams of $[\ce{COCl2}]$ is present in the reaction mixture.  Concentration and mass are related via the following equation:

$$[\ce{COCl2}] = \frac{n(\ce{COCl2})}{V} = \frac{m(\ce{COCl2})}{M(\ce{COCl2}) \cdot V}$$

or

$$m(\ce{COCl2}) = [\ce{COCl2}] \cdot M(\ce{COCl2}) \cdot V$$

where $n$ is the amount of $\ce{COCl2}$ (in g), $M$ is the molar mass (in g/mol), and $V$ is the volume (in L).

Let us plug in all of our known values and solve.

$$\begin{align}
m(\ce{COCl2}) &= (\pu{0.033 mol L-1})(\pu{98.906 g mol-1})(\pu{5.69 L}) \\
&= \pu{18.6 g}
\end{align}$$


Why do we consider all gasses to be in one phase?

Just because two fluids are completely miscible does not guarantee that they are uniformly mixed at all times.

For example, if you have a glass half full with sugar water solution, you can gently pour pure water to form a layer on top.  This is the basis for sucrose density gradient centrifugal separation and layered drinks.

So the same can occur for gases, a non-equilibrium state can exist for considerable time where the gases aren't mixed even though they are completely miscible.


Do Acids/Bases undergo dilution/bio-degradation when exposed to air?

What you heard about NaOH is right: it surely got completely carboxylated in 3 months, maybe even in 3 days. Not that the resulting compounds are completely neutral to the environment, but yes, they lack that corrosive ability.

As to the more general question of what happens to inorganic substances once they are out in the open: dilution is quite possible, biodegradation is not so likely. Some of them, like hydrochloric acid (and most other acids), would pretty soon end up reacting with some random compounds out there to yield relatively harmless products which would become normal part of the environment. Others (think of heavy metals) may have long-lasting adverse effects.


Why did we introduce the concept of oxidation number in chemistry?

Oxidation number is a deliberate approximation to make keeping track of electrons in molecules easy
You are quite right to say that oxidation number isn't "true". It is clearly a simplified idea and not a perfect representation of "reality" whatever that is.
But there is a point to it. That point is to enable chemists to keep track of the electron count in molecules. Electron distributions may be infinitely variable in bonds but electrons are discrete and keeping count of them is important. In transition metal complexes, for example, it is important to make sure all the bonding electrons are correctly accounted for. Many transition metals can have variable oxidation states with very different properties and the (crude) approximation of where those electrons are helps us classify those differences even when the metal-ligand bonds are covalent and the electron distributions are not clear-cut or ionic.
So there is no intent to correctly represent the distribution of charge in bonds. But pretending the oxidation number is an integer helps keep track of things and is often very useful at explaining properties.


If x and y are 2 extensive properties of a system then is x/y always intensive?

Extensive properties are those that are directly proportional to the mass of the system, under the restriction that all intensive properties are held constant.
The restriction that "all intensive properties are held constant" means we are not changing the inherent nature of the system; rather, we are only changing its size.
For instance, if we keep all the intensive properties constant, and double the mass of the system, all the extensive properties will double.
Thus extensive properties all have the form $X=nZ$, where n is the number of particles in the system, and Z is some intensive property.  [We can use n in place of mass because the restriction that all intensive properties are held constant means that the composition doesn't change.]
Consequently, if we have two extensive properties, X1 and X2, their ratio is:
$$\frac{X1}{X2}=\frac{nZ1}{nZ2}=\frac{Z1}{Z2}$$
Since the n's cancel out, the ratio of two extensive properties becomes the ratio of two intensive properties.  And since intensive properties are independent of n, their ratio is also independent of n.
Based on the above definition of extensivity, volume is only extensive in 3-dimensional systems, because it's only in 3-D systems that $V\propto n$.  Area is only extensive in 2-D systems, because it's only in 2-D systems that $A\propto n$.
I have never before considered resistance in the context of extensivity, but I believe the reason you are running into trouble is that resistance depends both on the size of the system and its geometry.  Given that, I would conclude that resistance is neither intensive nor extensive.
You could make resistance extensive by constraining the geometry, specifically by constraining A to be constant, in which case resistance would be a 1-D extensive property that depends upon L: $R \propto L \propto n$.
However, I don't believe you could turn resistance into an extensive property by constraining L to be constant, since then you would have: $R \propto A^{-1} \propto n^{-2}$.


There are! First off, you're correct that Hartree-Fock and DFT produce effective one-particle states $\phi_i$ which are the molecular orbitals (MOs) in the case of HF or the Kohn-Sham (KS) orbitals in the case of DFT (I'll focus on Hartree-Fock since I am more familiar with it than DFT). The MOs are defined via expansion coefficients ($C_{\mu i}$) in the basis set $\{\chi_\mu\}$ such that $\phi_i = \sum_\mu C_{\mu i} \chi_{\mu}$. The more physically meaningful quantity one works with is the one-particle reduced density matrix (1PRDM) defined as $\Gamma_{pq} = \sum_{\mu \nu} C^*_{\mu p} P_{\mu \nu} C_{\nu q}$ where $P_{\mu \nu} = \sum_i C_{\mu i}^*C_{\nu i}$. Diagonalization of the 1PRDM produces what are known as natural atomic orbitals (NAOs) $\phi^{\rm{NAO}}_i = \sum_p a_{p i} \phi_p$ where the coefficients $a_{p i}$ are defined by the eigenvalue relation $\Gamma_{p q} a_{q i} = \rho_i a_{p i}$ where $\rho_i$ is the occupation number corresponding to th $i^{\rm{th}}$ NAO. Since $\Gamma$ is Hermitian, the NAOs are orthonormal and form a perfectly suitable basis to perform any quantum chemistry calculation, however, their value lies in the fact that NAOs actually show you where the electron density is so it's a very useful visualization tool. Once you have NAOs, all of the analysis you want is available. There are things called natural bond orbitals (NBOs) which rotate NAOs in such a way that they maximize the electron density between 2 atoms, thus simulating the notion of a chemical bond. Here, you find the orbitals describing your $\sigma$- and $\pi$-bonds. The point I'm trying to make is that what's important for this kind of analysis is having $\Gamma_{pq}$.
So the answer to your question depends on whether or not you calculate $\Gamma_{pq}$ for a variety of post-HF methods, such as configuration interaction (CI), coupled-cluster (CC), and others. If we denote the Hartree-Fock determinant as $|\Phi\rangle$, the 1PRDM from before is simply $\Gamma_{pq} = \langle \Phi | a^+_p a_q | \Phi \rangle$. The wavefunction originating from post-HF calculations is $|\Psi_\mu\rangle = (1 + C_\mu)|\Phi\rangle$ in the case of CI and $|\Psi_\mu\rangle = R_\mu e^T |\Phi\rangle$ in the case of equation-of-motion (EOM) CC. The analogous expression of the 1PRDM for the $\mu^{\rm{th}}$ excited state ($\mu = 0$ is the ground state) is
$$\Gamma^{\mu}_{pq} = \langle \Psi_\mu | a^+_p a_q | \Psi_\mu \rangle$$
You can calculate this object directly for the various types of wavefunction that come out of correlated electronic structure calculations. It's not light work for sure, but if you know your way around many-body algebra (e.g. Slater's Rules, diagrammatics, Wick's Theorem), you can calculate this object and diagonalize it to find correlated bonding orbitals just as you would in Hartree-Fock. Also note that in the case of CC, the bra state $\langle \Psi_\mu | \neq [|\Psi_\mu\rangle]^+$ and instead we use the biorthogonal left-CC parameterization $\langle \Psi_\mu | = \langle \Phi | L_\mu e^{-T}$ where $L_\mu$ is a linear de-excitation operator complementary to $R_\mu$. These and other technicalities are of course well-documented in the CC literature. The key observation here is that we never explicitly build a correlated $|\Psi_\mu\rangle$ defined via Slater determinants directly. In the case of CC wavefunctions, you'd blow up your computer since $e^T$ is an object as large as the full CI wavefunction for any level of truncation. So the way things are done is you obtain $|\Psi_\mu\rangle$ indirectly though a set of CI coefficients or CC cluster amplitudes and calculate the 1PRDM, again, completely defined through this sequence of finite coefficients. It may still be a single-particle orbital, but it has many-body information embedded within it.
That being said, these calculations are available in most electronic structure programs.


What elastic polymeric material look like at molecular level?

Broadly, the elastic polymeric materials are structural with the mixture of networks of molecular chains and liquid. There may be a distinction between the terms "elastic polymeric materials" and "elastic polymers". Key features contain repetition of some parts so open to bioengineering such as bioprinting. Each specific elastic polymeric material differs at molecular level. For example some materials can have long repetitive carbon chains while some materials can have short repetitive carbon chains where each structure is attached together by some forces. There are different kinds of forces where their degree is relative to the molecule:

Brainstorming on the question "What look like?" What does it mean?

what is meant by "look like"?
intermolecular forces? Repulsive forces? Attraction forces?
EM fields between each charged particles? How are quantum phenomena in materials such as doping considered?
What does the electron current look like over the material?
What is the distribution of static charge?

Perhaps relevant

The book Edited by Xiang Yang Liu and Jing-Liang Li "Soft Fibrillar Materials Fabrication and Applications".

  "As one of the most important classes of soft materials, supramolecular materials are of a mixture of networks of molecular
  chains/fibrils and a liquid. These self-assembled fibrous/ molecular
  architectures exhibit various functionalities, ie. superhydrophobicity
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  structures.

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  fibers and functional fibers, with respect both to the fundamentals
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  this interdisciplinary field."


In ion chromatography there are multiple retention mechanisms, the ion-exchange is the primary one. However, let us keep it simple.

You should be able to explain why monovalents retain less than divalents? Use Coulomb's law! The force of attraction between the ion and the ion-exchange on stationary phase is proportional to the product of their charges and inversely proportional to the hydrated radius. Think about it why fluoride comes out before chloride, almost universally on all IC columns.

Another complication, this explanation breaks down with large anions like iodide, thiocyanate and perchlorate. They retain extremely long on most IC columns. They are called polarisable ions. These ions are sticking because of some other reasons.

Coming to divalents and trivalents: The retention of divalents and trivalents are very complicated. It depends on the chemistry of the stationary phase and its cross-linking and the very nature of the charged site on the ion-exchanger. However you can do a ball-park at an elementary level. Hard to predict in general.

Is phosphate ion always 3(-) or it can have 2(-) or 1(-) charge? Is the ionization pH dependent? The pKa (third proton) is >12. Is the mobile phase pH above 12?


Q is known as the ionic product which is defined as product of concentration of ions of the electrolyte, each raised to the power of their coefficients in the balanced chemical equation in solution at any concentration. $\pu{K_{sp}}$ is the same but it is applicable only for a saturated solution.

To find the solubility for the salt with the given $\pu{K_{sp}}$ value, For $\ce{AgCl}$, we have the equilbrium,
$$ \ce{ AgCl(s) <=> Ag+(aq) + Cl-(aq)}$$
If initally we have $‘a ‘$ moles of AgCl, at equilbrium we will have $’a-s ‘$ and we have $\ce{[Ag+]=[Cl-]=s}$. Hence we get the relation as,
$$ \pu{K_{sp}} = s^2 $$
This is the maximum concentration of the ions present in solution at equilbrium.(because $\pu{K_{sp}}$ = Q).
When Q just crosses $\pu{K_{sp}}$, salt is precipitated out.

Since they are products of ionic concentration, we can’t directly subract Q and $\pu{K_{sp}}$ to find the amount of salt precipitated.

I’m just working backwards to get $\ce{K_{sp}}$.

Let mass of $\ce{AgCl}$ precipitated be $’m ‘$ g.
No. of moles precipitated $n = \frac{m}{M}$.  Let the volume be $’V’$ litres,
$\ce{[AgCl]}$ precipitated $x = \frac{n}{V}$

$$ \ce{Ag+ + Cl- -> AgCl} $$

We know initial concentrations of the ions. Let $\ce{[Ag+]}$ and $\ce{[Cl-]}$ be $’p‘$ and $’q‘$ respectively.
Then solubility product is,
$$ \ce{K_{sp}} = (p-x)(q-x) $$

Hope that makes sense.


Confirming the presence of 4-bromoacetanilide during bromination of acetanilide

I advise against fishing in the mixture in this case. Try to isolate the product, purify it and determine its properties.

Have a look at the melting points of the starting material and the possible monobromination products:

acetanilide 114 °C
2-bromoacetanilide 96.5–100.5 °C
3-bromoacetanilide 81–85 °C
4-bromoacetanilide 165–169 °C

This promising! You can of course use further spectroscopical methods (NMR), but I'd start with the melting point.

You mentioned possible complications for the workup, but I disagree.

The reaction is typically performed in glacial acetic acid with a slight excess of bromine to make sure that there's (almost) no starting material left.

For workup, you pour the whole mixture into a beaker with water that has been placed in a bowl with ice. If you want/need to destroy an excess of bromine, add some bisulfite solution until the aqueous phase is colourless.

There should be some crystals. Filter them off, and recrystallize the material from ethanol.


Reaction of iron(III) oxide with nitric acid

Yes, reacting iron(III) oxide with nitric acid would yield iron(III) nitrate.

$$\ce{Fe2O3 + 6HNO3 → 2Fe(NO3)3 + 3H2O}$$

  Iron(III) oxide react with nitric acid to produce iron(III) nitrate
  and water. Nitric acid - diluted solution. This reaction takes place
  slowly.(source)


How can alkali metals form covalent bonds?

For the most part, they don't.
However, lithium sometimes can. This is due to their valence shells being relatively close to the nucleus, so they have some electronegativity and can form somewhat covalent bonds.

For more info see this.


Why are reactions with higher activation energy more sensitive to temperature?

The first thing to note is that the rate constant with a large activation barrier will be far smaller than one with a small barrier at a given temperature. If the barrier is very small (e.g. 5 kJ/mol) at, say, room temperature, almost all collisions between the reactive molecules have sufficient energy to react and the barrier is crossed and this leads to a fast reaction. This means that increasing the temperature cannot lead to much of an increase, because the reaction is going almost as fast as it can at the lower temperature.

Now consider the case with a large barrier. In this case the rate constant is far smaller than when the barrier is small, and very few collisions have enough energy  to cross the barrier. Now increase the temperature and the number of filled energy levels the molecule has increased greatly, exponentially so by the Boltzmann distribution.  Now very many more collisions have sufficient energy to react and so the rate constant increases greatly, although it may still be small in value.

If you plot the Arrhenius equation (given in your text) vs temperature the slope of this gives the rate of change with temperature. With larger activation energies the slope gets larger. This is the effect also described in your text.

The figure shows rate constants calculated with the activation energies $E_a$ shown (in kJ/mol). You can see how rapidly the rate constant increases with temperature when the activation energy is large, i.e. the slope is greater when the activation energy is larger, than when the activation energy is small. (The rate constant $\displaystyle k=A_0e^{-E_a/(RT))}$ are with $A_0 =10,\; (E_a=10);\; A_0= 10^3,\;(E_a=20)$ and $A_0=10^6,\;(E_a=40)$ with $A_0$ chosen only to get them on the same scale).

The gradient of the rate constant is $\displaystyle \frac{dk}{dT}=E_a\frac{A_0e^{-E_a/(RT)}}{RT^2}$ which shows that the slope depends on the activation energy.


Hydrolysis of Aspartame with a warmed alkali

This looks like it's from an exam so there's a very high possibility that the person setting the question and answer intended the ester group on the right to be something else. I am largely of the opinion that the answer should be A.

However, there is one way I can imagine that you can try justify "B" being the answer (although it is a long shot):

  Under alkali conditions, we can expect the acid group from aspartic
  acid to be deprotonated to carboxylate which can act as an
  intramolecular nucleophile.

  We take note that intramolecular reactions are generally faster than
  intermolecular reactions. The carboxylate can attack either the ester
  group or the amide group. An attack on the amide results in a
  5-membered ring transition state, and an attack on the ester results
  in a 9-membered ring transition state. Since 5-member rings are more
  stable, the carboxylate attacks and breaks bond 1, leaving bond 2
  alone. Aqueous alkali then destroys the acid anhydride that is formed.

That said, this is highly unlikely given that esters are less stable to base-catalyzed hydrolysis than amides. So if Bond 1 Breaks, Bond 2 must have also broken.


Formation of green precipitation whilst adding potassium fluoride to aqueous solution of copper(II) sulfate

$\ce{CuSO4}$ exists as $\ce{[Cu(H2O)4]SO4}$ in solution. It is blue in colour due to the presence of $\ce{[Cu(H2O)4]^{2+}}$ ions.

Now if aqueous $\ce{KF}$ is added, the solution turns green due to formation of complexes $\ce{[CuF4]^2-}$ and $\ce{[CuF6]^4-}$(Initially copper(II) fluoride($\ce{CuF2}$) is formed but it formed complexe in present of water). The complete reaction is given by:

$$\ce{[Cu(H2O)4]SO4 + 4KF -> K2[CuF4] + K2SO4 + 4H2O}$$


Normal dioxygen ($\ce{O2}$) exists as a ground state, triplet biradical. This is an example of a molecule that, thermodynamically should be quite reactive, yet is kinetically unreactive - once again, a case of kinetics vs. thermodynamics.

Because spin must be conserved in a reaction, if ground state, triplet $\ce{O2}$ were to react with something, it would have to produce a triplet product. For most compounds, the triplet state is considerably higher in energy than the singlet state.  Hence reaction with ground state, triplet $\ce{O2}$ would produce a higher energy, triplet product. Since ground state, triplet dioxygen doesn't have any excess energy, the activation energy required to produce a high energy triplet product just isn't there - hence, no reaction.

On the other hand, singlet $\ce{O2}$ is a spin-paired excited state with plenty of excess energy (the first singlet lies some 95 kJ/mol above the ground state triplet).  Basically, singlet $\ce{O2}$ behaves like any other electrophilic (due to the electronegativity of oxygen), reactive (due to its excess energy), compound with a double bond. It readily undergoes the Diels-Alder, ene and even 2+2 cycloaddition reactions - again, just like any other reactive olefin.


How to predict a COSY NMR spectra for an AMX spin system?

An AMX spin system represents three nuclei coupled to each other, with their signals separated by a large chemical shifts compared to the coupling between them (such that $\Delta \delta >> 3J$). This system is said to be first-order, and the A, M and X signals are each a doublet of doublets, and the couplings can be extracted by inspection.

The simple COSY spectrum of this will show the three core signals along the diagonal, with cross peaks appearing for all signals ie each signal will have two cross peaks. Below is a (very basic) schematic for an AMX COSY.

The appearance of the cross peaks will depend on the type of COSY experiment you are running, such as a COSY-45 or COSY-dqf etc, but this should give you the general picture for what you are asking.


How can I know the colour change of this reaction?

Potassium dichromate (VI) is the species that turns from orange and to green when reduced. Most of the organic substances I came across for testing purposes are colourless, so I would assume that compound B would be as well.

It is just like you are expected to know the colour change of potassium manganate (VII) from purple to colourless when it is reduced.

Those two oxidants were used frequently in my labs, and I'm pretty sure you'll be using them often as well.

I hope this answers your question! :)


The effect of a voltmeter on the electron flow of a voltaic cell

An ideal voltmeter has an infinite resistance, so does not let any current flow in the system. However nothing is ideal in real life, it consumes a very small amount of current to move the magnet and needle arrangement or electronics. An ideal ammeter has zero resistance. Assume that you have a water tank on your ceiling which supplies water to your kitchen tap. If I add a pressure gauge before your closed kitchen tap, it will read a positive pressure, much greater than the atmospheric pressure. Note that there is no water flow. Forget about the textbook picture of measuring galvanic cell voltages using a voltmeter (as shown in chemistry textbooks), as nobody every measures cell potential this way. A reliable approach to measure cell potential is called null-point detection (you can search more about it). As the name indicates, it draws zero current at the point of balance condition. It is one of the most precise and accurate techniques in classical potentiometry.

Just think a little deeper, the two electrodes in a galvanic cell do not "know" each other and they are oblivious to each other's presence. So current flow is not a requirement to measure potential difference. Each single electrode is in contact with its ion in the solution. If you have a zinc and copper galvanic cell, each has its own electrode potential e.g. a Zn rod dips in Zn2+ soln and Cu rod dips in Cu2+ solution. Electrode potential is the thermodynamic tendency of ions (in solution) to reduce themselves without any current flow, in this case Zn2+ --> Zn and Cu2+ --> Cu.  The voltmeter simply reads the difference of potential between zinc and copper, however it does not know the individual electrode potential of zinc and copper.


Does plant material oxidize from 02 gas (air or solved in water) and/or oxygen in H20?

Okay first, it's the oxygen in the air that causes oxidation, also dissolved oxygen in water, but if you let your tobacco get wet enough for the latter to be significant you will be getting more rot than curing ...
Anyway, I don't think a sealed jar would be a good idea. Ammonia build-up, wet rot, etc would tend to be an issue.
You could build a proper curing/fermenting kiln to cure the leaf, there's DIY instructions here: https://www.seedman.com/wkiln.htm for one type.  But if you're using the finished product for dry snuff rather than smoking or chewing, you might not be too bothered about the taste/smell so careful fermenting is probably not that important to you. Then easiest is to just hang them somewhere like the garage and spray them with a plant mister when they get too dry like this guy: https://dengarden.com/gardening/Tobacco-Growing-and-Curing-at-Home or if it gets too cold where you live to use the garage, in a closet, wooden box or linen chest.


What are the products formed when CH3CH2OH reacts with NaH?

Consider firstly the ease of reaction: the hydrogen in $\ce{NaH}$ is basic $\ce{H-}$ and the hydrogen in ethanol is acidic $\ce{H+}$. The easiest reaction then, is a simple acid-base neutralization.

The reaction products you propose requires that a $\ce{C-O}$ bond is broken, or a situation wherein which the hydride attacks a carbon center which then ejects a hydroxide ion $\ce{OH-}$. In reality, however, you would have to eject $\ce{O^{2-}}$ which is much harder to remove.


How does heat transfer between molecules happen in deep?

A1: There is no displacement of the wall, forces are in equilibrium with zero net force. There are initiated stronger vibrations on warmer wall side by energetic hot air molecules. And weaker vibrations on colder wall side by energy leaching cold water molecules .
A2a: If the wall initial temperature was the cold water temperature, then yes. Otherwise no, as the mean kinetic energy of wall molecules can generally goes both ways, depending on initial temperatures of air, wall and water, weighted by air and water thermal capacity a conductivity.
A2b: Yes
A3: No, as there is no displacement. Hot air with in average more energetic molecules causes stronger vibrations of wall molecules, which cause faster motion of water molecules.
Imagine the whole scenario in more familiar macro world. 2 boxes of ideally elastic balls in gravityless ISS station, separated by a wall of balls interconnected by elastic strings holding all of them at their place.
Now, balls in one of boxes move faster. What will happen during the time ?

The mean Kinetic energy of molecules is proportional to absolute temperature. Their mean speed is then proportional to the square root of absolute temperature and momentum to the absolute temperature.
Pressure on the walls is proportional to momentum change rate, what is proportional to molecule momentum and the rate of collisions. For gases in constant volume, both is proportional to square root of absolute temperature, so pressure is proportional to absolute temperature of ( ideal ) gas.
But pressure is not essential for molecular mechanism of heat transfer.
Molecules of matter at given temperature do not have the same speed and kinetic energy. During their collisions, it fluctuates all the time. What is constant is their mean value.
For monoatomic ideal gas, the mean kinetic energy of 1 mole of the gas is $E_\mathrm{k}=3/2RT$, where R is the universal gas constant about 8.314 J/K/mol. The speed distribution follows the Maxwell-Boltzmann distribution.
It means that for given kinetic energy and speed of a molecule, we cannot say what is the temperature of matter the molecule belongs.
When 2 different regions of matter have different temperature, they exchange heat, what means molecular kinetic energy.
Molecules from adjacent regions mutually exchange kinetic energy by random way by collisions. As by random collisions, they can both gain or lose the energy.
Molecules from both colder and warmer regions mutually both transfer and accept the energy. If temperatures are equal, their mean kinetic energy (*) is equal and the rates of energy flow are equal in both directions. If one region is warmer, the rate of passing energy is greater than the rate of accepting energy.

(*) It is more complicated, as more exactly, it means the mean kinetic energy per each way the molecule can move, called degree of freedom. But the above is still valid for monoatomic gases.


Python package for modelling chemical reactions

This is not really my area of expertise, but a quick search for "python chemical reactions" revealed three hits I've never seen before that may be of interest, the first one being closest to what you want.

chempy

from chempy import ReactionSystem  # The rate constants below are arbitrary
rsys = ReactionSystem.from_string("""2 Fe+2 + H2O2 -> 2 Fe+3 + 2 OH-; 42
    2 Fe+3 + H2O2 -> 2 Fe+2 + O2 + 2 H+; 17
    H+ + OH- -> H2O; 1e10
    H2O -> H+ + OH-; 1e-4
    Fe+3 + 2 H2O -> FeOOH(s) + 3 H+; 1
    FeOOH(s) + 3 H+ -> Fe+3 + 2 H2O; 2.5""")  # "[H2O]" = 1.0 (actually 55.4 at RT)
from chempy.kinetics.ode import get_odesys
odesys, extra = get_odesys(rsys)
from collections import defaultdict
import numpy as np
tout = sorted(np.concatenate((np.linspace(0, 23), np.logspace(-8, 1))))
c0 = defaultdict(float, {'Fe+2': 0.05, 'H2O2': 0.1, 'H2O': 1.0, 'H+': 1e-7, 'OH-': 1e-7})
result = odesys.integrate(tout, c0, atol=1e-12, rtol=1e-14)
import matplotlib.pyplot as plt
_ = plt.subplot(1, 2, 1)
_ = result.plot(names=[k for k in rsys.substances if k != 'H2O'])
_ = plt.legend(loc='best', prop={'size': 9}); _ = plt.xlabel('Time'); _ = plt.ylabel('Concentration')
_ = plt.subplot(1, 2, 2)
_ = result.plot(names=[k for k in rsys.substances if k != 'H2O'], xscale='log', yscale='log')
_ = plt.legend(loc='best', prop={'size': 9}); _ = plt.xlabel('Time'); _ = plt.ylabel('Concentration')
_ = plt.tight_layout()
plt.show()  # doctest: +SKIP

OEChem

Cantera

There is RDKit and Pybel as well, but those are more traditional cheminformatics tools.


What is the solvent and what is the solute?

In a binary homogeneous mixture, the component present in the larger amount is generally called the "solvent". The component present in the smaller amount is generally the "solute".

In your first example, however, it is more complicated as coffee is already a complex solution (but mainly water). Milk is not a homogeneous mixture. It is a fat/protein emulsion (also mainly water).

The other three are solutions: salt water is a solution of a solid dissolved in a liquid (which is the most familiar type), air freshener (one dispersed) is a solution of one gas in others (but air itself is a mixture), and brass is a solid solution.


I don't really like the picture but let's try to do this with that.
So, you probably know about the NaCl lattice? In there sodium is surrounded by six chlorides, which forms an octahedron. Chloride has six sodium around it to form another octahedron. And you have probably heard that also other compounds crystallize in this lattice, let's say sodium bromide (so increasing the anion in size) or potassium chloride. It does not change the crystal structure.

Now, what about more complicated units. Let's deform that spherical anion to an elongated one. What about calcium carbide? The C-C unit with its triple bond is unlikely cleaved during the crystallization, so we can count the $C_2^2-$ unit as one part. This forms an elongated cell but still, it is a NaCl-type lattice.

If we make it flat instead we end up with something like carbonate for example. As you may no the carbonate unit is flat, it's basically a triangle with the carbon in the center. If we see this unit as one part then we can just form another NaCl-lattice. And as the jump from sodium chloride to potassium chloride was possible, we can assume that if calcium carbonate crystallizes in a deformed NaCl-lattice, then magnesium carbonate will do the same.

And the final step is to combine them now, instead of separating the two phases, 1/2 of the calcium-ions are just exchanged by magnesium to form the dolomite-structure.

So you could say that carbonate forms a cubic packing where 50% of the octahedral sites are occupied by calcium and the 50% by magnesium for example. All carbonate-units are parallel to each other therefore creating one direction in which they do not require that much space causing the deformation of the cell.

The metals are coordinated by (and I had to load this in my solid state software as well to see it) six oxygen on the carbonates, or if you take the center carbon of each carbonate unit also by six carbonates, which forms an octahedron again. On the other hand, each of these center carbons is coordinated by three oxygen in a triangle or if you count the counter ions by six metals in the shape of an octahedron.

In your structure you can start off by looking at the dark atoms, which are the carbons, and connect them to the red ones, so can see it will always form triangles. For all the others it's hard to see in this picture. If you want I can also highlight you the different polyhedra in my software so you can see it better. Because this is just a unit cell so it is cut at some positions creating the feeling that some of the connections are wrong. You'd have to watch several unit cells aligned to see the full structure of it. But if you imagined the yellow and the blue ones to be the same you'd end up with the Calcium carbonate or magnesium carbonate structure.

EDIT: I also added the crystal structure now:

EDIT II: Because there were still some problems I tried to further explain the structure a bit. I don't know how others see this structure but I like this layer explanation and it's also the one you can find in many books. So please just look through the new attachment and tell me if this helps you more.


Looking at the structures of those two acids might help. In both cases, the acidic hydrogens are part of hydroxyl groups. Because of polarity and the inductive effect, the O-H bond is weaker than C-H bonds which are very strong. As a result, it's nearly impossible for Carbon to lose a hydrogen without some kind of elimination reaction, whereas the O-H bond can autoprotolyze with other O-H bonds (even water). The weakness of the O-H bond is the basis for the acidity of those hydrogens. C-H bonds are essentially non-acidic, having pKa values around 45-50.


Varying potential diffrence between two graphite strips in NaCl solution

What type of voltmeter are you using? to verify the answer to my question, you need to determine 1) the precision and 2) the detection limit of the instrument .

The precision is a measure of the error in a typical measurement. The smaller the standard deviation in repeat measurements, the higher the precision. To get a guess at the precision, prepare a system that should generate a known potential difference in the same range as your current measurements (like 100 mV). If the variation in that value is similar to what you are seeing, then you are recording the noise.

The detection limit is the lowest value your instrument can reliably measure. Generally, you cannot reliably measure values smaller than the twice the precision of your instrument. If you determine a standard deviation of 50 mV when determining precision, then 100 mV is your the detection limit. If this is true, you are recording the noise.

The precision and detection limit of an instrument depend on quality (price), but also on purpose. A voltmeter designed to be used by an electrician will have high precision in the range 1-200 V, but low precision below 1 V (because that range is not as important). A sensitive voltmeter used for electrochemical measurements may have high precision between 1 uV and 1 V, and behave miserably outside that range.


Your interpretation is correct.

The molar heat of fusion is the amount of heat necessary to melt a mole of ice at its melting (or freezing) point. That means, to get 1 mole of water from 1 mole of ice at its melting point (0 degrees Celsius) it would require an input of $6.01\ \mathrm{kJ}$.

The conceptual reasoning is that it requires so much energy to break the bonds and allow more motion (but not completely).

Conversely, $6.01\ \mathrm{kJ}$ would be released upon the freezing of water at that point (1 mole of water anyway).


How can one explain niobium’s weird electronic configuration?

There is an explanation to this that can be generalized, which dips a little into quantum chemistry, which is known as the idea of pairing energy. I'm sure you can look up the specifics, but basically in comparing the possible configurations of $\ce{Nb}$, we see the choice of either pairing electrons at a lower energy, or of separating them at higher energy, as seen below:

d:   ↿ ↿ ↿ _ _        ↿ ↿ ↿ ↿ _        ↿ ↿ ↿ ↿ ↿     ^
                 OR               OR               |
s:   ⥮               ↿                _            Energy gap (E)

The top row is for the d-orbitals, which are higher in energy, and the bottom row is for the s-orbital, which is lower in energy. There is a quantifiable energy gap between the two as denoted on the side (unique for every element). As you may know, electrons like to get in the configuration that is lowest in energy. At first glance, that might suggest putting as many electrons in the s-orbital (lower energy) as possible, and then filling the rest in the d-orbital. This is known as the Aufbau principle and is widely taught in chemistry classes. It's not wrong, and works most of the time, but the story doesn't end there. There is a cost to pairing the electrons in the lower orbital, two costs actually, which I will define now:

Repulsion energy: Pretty simple, the idea that e- repel, and having two of them in the same orbital will cost some energy. Normally counted as 1 C for every pair of electrons.

Exchange energy: This is a little tricky, and probably the main reason this isn't taught until later in your chemistry education. Basically (due to quantum chemistry which I won't bore you with), there is a beneficial energy associated with having pairs of like energy, like spin electrons. Basically, for every pair of electrons at the same energy level (or same orbital shell in this case) and same spin (so, if you had 2 e- in the same orbital, no dice, since they have to be opposite spin), you accrue 1 K exchange energy, which is a stabilizing energy. (This is very simplified, but really "stabilizing energy" is nothing more than negative energy. I hope your thermodynamics is in good shape!) The thing with exchange (or K) energy is that you get one for every pair, so in the case:

↿ ↿ ↿

from say a p-subshell, you would get 3 K, for each pair, while from this example:

⥮ ↿ ↿ ↿ ↿

from a $\ce{d^6}$, you would get 10 K (for each unique pair, and none for the opposite spin e-)

This K is quantifiable as well (and like the repulsion energy is unique for each atom).

Thus, the combination of these two energies when compared to the band gap determines the state of the electron configuration. Using the example we started with:

d:   ↿ ↿ ↿ _ _         ↿ ↿ ↿ ↿ _         ↿ ↿ ↿ ↿ ↿    ^
s:   ⥮           OR   ↿           OR   _            |
PE:  3K + 1C          6K + 0C          10K + 0C     Energy gap (E)

You can see from the example that shoving 1 e- up from the s to the d-subshell results in a loss of 1C (losing positive or "destabilizing" repulsive energy) and gaining 3K (gaining negative or "stabilizing" exchange energy). Therefore, if the sum of these two is greater than the energy gap (i.e. 3K - 1C > E) then the electron will indeed be found in the d shell in $\ce{Nb}$'s ground state. Which is indeed the case for $\ce{Nb}$.

Next, lets look at perhaps exciting the second s e- up to the d-subshell. We gain 4 additional K but don't lose any C, and we must again overcome the energy gap for this electron to be found in the d-subshell.

It turns out that for $\ce{Nb}$: 4K + 0C < E  (remember that C is considered a negative value, which we're not losing any of), so $\ce{Nb}$ is ultimately found in the $\ce{5s^1 4d^4}$ configuration.


How does the inductive effect, explain the reactivity of tertiary alcohols with halogen acids?

Secondary and tertiary alcohols react with strong acid (and bases) such $\ce{HX}$ following SN1 mechanism.

$$\ce{ROH <=> HX + ROH_2^+ + X^-} \tag{1}$$
$$\ce{ROH_2^+ <=> R^+ +H_2O} \tag{2} $$
$$\ce{R+ +X- \rightarrow  RX} \tag{3} $$

In these reactions, the rate determining step is the formation of the carbocation (2). So the successive attack of the nucleophilic chlorides is dependent on that.

Here is how the inductive effect comes into play. The inductive effect of $\ce{R}$ groups of the alcohols stabilize the carbocation (figure show the stability of carbocations from the more stable to the less) and then increase the chances of a nucleophilic attack!


Is carbon dioxide soluble in petroleum ether, and why?

Carbon dioxide is, in fact, more soluble in nonpolar solvents than in water.

From the Wikipedia page on carbon dioxide: Solubility in water
1.45 g/L at 25 °C (77 °F), 100 kPa
(Alternative source: CRC handbook. Solubility is about twice as much at 0°C.)

From IUPAC, Solubility Data Series: Carbon Dioxide in Non-Aqueous Solvents at Pressures Less Than 200 kPa, chart on page 3: mole fraction of CO2 in hexane at 101.3 kPa, 300 K = 0.012, which corresponds to 4 g/L.
(Solubility is only slightly (10-20%) higher at 0°C.)

So the forum posts I read were wrong.

As to why, I suppose it's sufficient to argue that CO2 is a nonpolar molecule, and like dissolves like.
(It's true because it's Latin: Similia similibus solvuntur! )


I think you may be missing a key point to catalysis; that the catalyst is not net consumed or altered by the reaction. If you were to put the catalyst in the chemical equation as in your second reaction as a reactant, you would technically need to also put it on the right hand side as a product. However, this is not the way a catalyzed reaction is typically indicated. A common way to indicate that a catalyst is used would be as follows:

$$\ce{48KNO3 + 5C12H22O11 ->[Fe2O3] 55H2O + 36CO2 + 24K2CO3 + 24N2}$$


The above is the mechanism for Fries Rearrangement for a brominated ester( doesn't matter really, for any ester the mechanism essentially remains to be the same).As per what I infer from your question, I think you mean to ask why even in the presence of $\ce{O^+-AlCl3}$ which is a strongly deactivating group, the major product is not meta-substituted.
As you can clearly see, in the next step itself a stable acylium is formed and the formal charge on $\ce{O}$ is no longer positive.So, as phenoxide ion is an activating group and o,p-director, two possible products are expected which are ortho and para substituted.However, at higher temperatures, the major product is ortho substituted which is driven by chelation.
NOTE:
As far as I know, both intramolecular and intermolecular reaction mechanisms are able to explain the major products and it is still not confirmed which one is the actual mechanism.
Image source:
https://www.synarchive.com/named-reactions/fries-rearrangement


"HBr is donating a proton to LiOH": Why doesn't it produce LiOH2?

I think it would produce something like "LiOH2,"

You are missing the charge. It should be something like $\ce{LiOH2+}$, which is an unconventional way of writing $\ce{Li+(aq) + H2O(l)}$.

So where is my misunderstanding?

You have to get used to interpreting the (aq) designation after ionic compounds. To help you out, here is the equation written with the ions shown as individual species (more typing, but closer to the actual situation):
$$\ce{Li+(aq) + OH-(aq) + H+(aq) + Br-(aq) -> Li+(aq) + H2O(l) + Br-(aq)}$$


Evaporating the water of a solution of only carbonate ions and water gives?

I would love to add some quantum chemical bone to the stew we are trying to cook here.

Let us assume that it is possible to have a solution of carbonate ions, like already discussed by Nicolau in the comments. Then Rauru is also right that you will have to consider several equilibria:
\begin{aligned}\ce{
(1) && 2H2O &<=> H3+O + {}^{-}OH \\
(2) && CO3^{2-} + H2O &<=> HCO3- + {}^{-}OH\\
(3) && HCO3- + H2O &<=> H2CO3 + {}^{-}OH
}\end{aligned}

Now $(3)$ seems a little far fetched, so we will ignore it.

When you do now remove water from the equilibrium through (gentle) evaporation, you will have a mixture of only $\ce{CO3^{2-} + HCO3- + {}^{-}OH}$. Further assuming this is possible you will always have some collisions within this ensemble. Statistical it is therefore always possible to also remove $\ce{HCO3- + {}^{-}OH}$ via $(2)$ from the system. Also single entities of carbonate ions will repel each other as states by G M. So basically it all boils down to on question: Is carbonate stable in vacuum? What happens to carbonate if you put it into vacuum?

Now we have some quite good approximations to these particular conditions (gas phase and 0K), but I do not want to go into detail. I did a calculation on the carbonate with Gaussian09 and as a cheap shot BP86/cc-pVTZ level of theory. The only thing I want to show is the highest occupied molecular orbital.

What can clearly see is that this is no bonding orbital. There are lone pair contributions from all Oxygen atoms (due to imposed symmetry constraints). In addition to the high negative charge, electrons will not be stabilized by the field of the nuclei. Also, Orbitals only vanish at infinity, giving electrons a non-zero probability to actually leave the compound. Furthermore, there are two electrons in this orbital, so the spins of these are paired, which also costs energy.

My suspicion therefore is that the carbonate at least loses one electron. To where? To anywhere. (Also mentioned by Nicolau in the comments.) I do not have any actual proof for this, as it quite hard to find people actually doing experiments and/ or calculations on that topic. I remember that one question of a Professor of mine in the first semester was: "What happens to sulphate if you shoot it into space?" The answer was given without any further explanation and I never bothered to ask: "It looses an electron."

Note:

Another point that should always be considered when dealing with gas phase reactions is, that when ions have enough kinetic energy they might react in the strangest ways. One of them might be
$$\ce{2CO3^{2-} -> 2CO2- + O2^{2-}}.$$

Also collision with any particle may cause electrons to be transferred.


Acid catalysed dehydration of 5-methylidenecyclopent-2-en-1-ol

I'd say your mechanism is acceptable. However, the final product, pentafulvene (PIN: 5-Methylidenecyclopenta-1,3-diene) is said to be not stable at room temperature. Generally, fulvenes are thermally unstable, sensitive to oxygen and photosensitive (Ref.1). They are also prone to acid- and cation-catalysed polymerisations. For example, there have been numerous reports of pentafulvenes undergoing dimerization via Diels–Alder cycloadditions at room temperature similar to that of cyclopentadiene (e.g., Ref.2):

Based on these facts, I'd like to suggest an alternative path for the reaction giving much stable cyclopentaenone derivative:

References:

Ellen Swan, Kirsten Platts, Anton Blencowe, "An overview of the cycloaddition chemistry of fulvenes and emerging applications," Beilstein J. Org. Chem. 2019, 15, 2113–2132 (doi:10.3762/bjoc.15.209).
Beat Uebersax, Markus Neuenschwander, Hans‐Peter Kellerhals, "Fulven‐Dimere: Synthese, Strukturbeweis and thermisches Verhalten," Helvetica Chimica Acta 1982, 65(1), 74-88 (https://doi.org/10.1002/hlca.19820650109).